# ANNEXES ANNEXES Annex A Checkings

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```					                                   ANNEXES

Annex A: Checkings of pillars, beams and crosses

   Pillars
   Beams
   Crosses

Annex B: Plans of portics and joints

Annex C: Checkings of Joints

Annex D: Checking and verification efforts in the shoe

Annex E: Plans of the building
ANNEX A: Checkings of pillars, beams and crosses

Legend

ᵝ: Buckling coefficient
LK: Buckling length (m)
Cm: Moment coefficient
Nt: Tensile
Nc: Resistance to compression
MY: Flexion strength Y
MZ: Flexion strength Z
VY: Shear strength Y
VZ: Shear strength Z
MYVZ: Resistance to bending moment Y and Z combined shear
MZVY: Resistance to bending moment Z and Y combined shear
NMYMZ: Resistance to bending and combined axial
NMYMZVYVZ: Resistance to bending, shear and combined axial
Mt: Torsional strength
MtVZ: Shear resistance and torque combined Z
MtVY: Shear resistance and torque combined Y
 : Limiting slenderness
x: Distance from the origin of the bar
ƞ: Utilization rate (%)

Pillars........................................................................................................... 2
Beams.......................................................................................................... 37
Crosses ....................................................................................................... 72
PILLARS

PILLAR 1

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       83 . 469 t
Must be satisfied:                                  0 . 110  1
N r , Rd       756 . 881 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (3).
Nt,Ed: Axial traction applicant bad calculation = 83.469 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

3
N c , Ed       127.298 t
                               0 . 168  1
N c , Rd       756.881 t

N c , Ed       127.298 t
                               0 . 770  1
N b , Rd       165.426 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·Q1 +1.5·V1 (4) +1.05·N1.
Nc,Ed: Axial compression applicant bad calculation = 127.298 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.22 · 270.0 cm2 · 2803.26 kp/cm2 = 165.426 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.
1
                                    0.91 (  y) and 0.22(  z ) ≤ 1
                ( )
2            2

Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 0.69 (  y) and 2.70(  z).
α: Reduction coefficient for buckling.= 0.21(αy) and 0.34(αz).

4
A· f y
 : Reduced slenderness                                       = 0.55 (  y ) and 1.95 (  z ).
N cr

Ncr: Elastic critical elastic buckling. = 198.511 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 2508.902 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z             2
= 198.511 t
L kz
c) Elastic critical axial buckling torque.
1            ·E ·I w 
2

N cr ,T    2 · G ·I t       2     =∞
io              L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20 cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. = 12 m
Lkz: effective length for buckling with the axis Z. = 12 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 26.14 cm
io  (i y  i z  y 0  z 0 )                   ( 25 ,17  7 . 08  0  0 )           26 . 14 cm
2       2             2    2   0 .5            2           2          0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                1 0 7 . 313 t ·m
         Ed
                              0 . 596  1
M   c , Rd
1 8 0 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1(2)

Where:

5
MEd: Bending moment calculation applicant awful = 107.313 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  1 . 042 t ·m
          Ed
                     0 . 027  1
M    c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1 (1)
+1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 1.042 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed             17 . 198 t
                                    0 . 114  1
V c , Rd          150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 17.198 t
The shear resistance of calculating VcRd is given by:
f yd
VCRd = A v                   150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93.00 cm2

6
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 34.84
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         0 . 171 t
                                 0 . 001  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 0.171 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v               301.520 t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270.0 cm2
d: Web depth. = 540 mm
tw: Web thickness. = 15.5 mm

7
Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 17.198 ≤ 75.259
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(2).

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 0.171 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(1)+1.05·N1.

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                            
 M y , Ed              M z , Ed 
                                                 1  0.030 ≤ 1
 M N , Rd , y 
                       M N , Rd , z 
              

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k yy ·                            k yz ·                 1  0.343 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k zy ·                            k zz ·                 1  0.846 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(1)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 127.298 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.

8
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 2
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M      pl , Rd
 179 . 781 tm  M   pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                  N , Rd , z
 M   pl , Rd , z
·1          =38.993 tm

     1 a  

 2
  5·n  5  1

N c , Ed
n                       0 . 168  1
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993 tm)
A  2·b·t f
a                              0 . 33  0 . 5
A
A: Area of the gross section. = 270.0 cm2
b: width of the wing. = 30 cm
tf: thickness of the wing = 30 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 067 tm
                                           0 . 019  1
M T , Rd           3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(2).
MT,Ed: torque calculation applicant awful.= 0.067 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  3.599 t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

9
PILLAR 2

Tensile (Eurocode 3 EN 1993-1-1):

This colum is not working in Tensile strength.

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       20.504 t
                               0 . 027  1
N c , Rd       756.881 t

N c , Ed       20.504 t
                               0 . 124  1
N b , Rd       165.426 t

The calculation effort lousy applicant occurs for the combination of actions G
+0.9·V1 (1) +1.5·N1.
Nc,Ed: Axial compression applicant bad calculation = 20.504 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.0 cm2 · 2803.26 kp/cm2 = 165.426 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.22 · 270.0 cm2 · 2803.26 kp/cm2 = 165.426 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00

10
χ: Reduction coefficient for buckling.
1
                                            0.91 (  y) and 0.22(  z ) ≤ 1
                        ( )
2                2

Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 0.69 (  y) and 2.70(  z).
α: Reduction coefficient for buckling.= 0.21(αy) and 0.34(αz).

A· f y
 : Reduced slenderness                                              = 0.55 (  y ) and 1.95 (  z ).
N cr

Ncr: Elastic critical elastic buckling. = 198.511 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 2508.902 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z                2
= 198.511 t
L kz
c) Elastic critical axial buckling torque.
1              ·E ·I w 
2

N cr ,T       2
· G ·I t       2     =∞
io                L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20 cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. = 12 m
Lkz: effective length for buckling with the axis Z. = 12 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 26.14 cm
io  (i y  i z  y 0  z 0 )                        ( 25 . 17  7 . 08  0  0 )            26 . 14 cm
2       2               2        2   0 .5               2          2          0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

11
Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                1 5 9 . 852 t ·m
         Ed
                       0 . 888  1
M   c , Rd
180 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of G+1.5·V1 (3)
+1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 159.852 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                10 . 460 t ·m
         Ed
                    0 . 268  1
M   c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35G +
1.5·V1(4).

Where:

MEd: Bending moment calculation applicant awful = 10.460 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::

12
V Ed        19 . 511 t
                                 0 . 130  1
V c , Rd       150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 19.511 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v               150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 34.84
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         1 . 682 t
                                 0 . 006  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 1.682 t

13
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                  301.520 t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270 cm2
d: Web depth. = 540 mm
tw: Web thickness. = 15.5 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 19.511 ≤ 75.259
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(3)+1.05·N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 1.682 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                         
 M y , Ed                M z , Ed       
                                                1  0.788≤ 1
 M N , Rd , y
                
         M N , Rd , z
                


14
N c , Ed                            M                         M z , Ed
 
y ,Ed
 k yy ·                             k yz ·                 1  0.936 ≤ 1
 y · A· f yd                  LT ·W pl , y · f yd            W pl , z · f yd

N c , Ed                            M                        M z , Ed
 
y ,Ed
 k zy ·                            k zz ·                 1  0.577 ≤ 1
 y · A· f yd                  LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(3)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 17.646 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M     pl , Rd
 180 . 110 tm  M    pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                 N , Rd , z
 M pl , Rd , z ·1          =38.993 tm

     1 a  
 2
  5·n  5  1

N c , Ed
n                      0 . 023  1
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993 tm)
A  2·b·t f
a                              0 . 33  0 . 5
A
A: Area of the gross section. = 270.0 cm2
b: width of the wing. = 30.0 cm
tf: thickness of the wing = 30.0 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 011 tm
                                          0 . 003  1
M T , Rd           3 . 599 tm

15
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(2)+1.05·N1
MT,Ed: torque calculation applicant awful.= 0.011 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd           ·W T · f yd  3.599 t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

PILLAR 3

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed        0 . 285 t
Must be satisfied:                                            0 . 001  1
N r , Rd        756 . 881 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (4).
Nt,Ed: Axial traction applicant bad calculation = 0.285 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       20.959 t
                               0 . 028  1
N c , Rd       756.881 t

N c , Ed       20.959 t
                               0 . 078  1
N b , Rd       269.171 t

16
The calculation effort lousy applicant occurs for the combination of actions 1.35·G
+0.9·V1 (2) +1. 5·N1.
Nc,Ed: Axial compression applicant bad calculation = 20.959 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.36 · 270.0 cm2 · 2803.26 kp/cm2 = 269.171 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.
1
                                  0.95(  y) and 0.36(  z ) ≤ 1
              ( )
2            2

Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 0.61 (  y) and 1.79(  z).
α: Reduction coefficient for buckling.= 0.21(y) and 0.34(z)

A· f y
 : Reduced slenderness                                = 0.41(  y ) and 1.46 (  z ).
N cr

Ncr: Elastic critical elastic buckling. = 236.24 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.

17
 ·E ·I y
2

N cr , y              2
 2869.85 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z             2
= 236.24 t
L kz
c) Elastic critical axial buckling torque.
1              ·E ·I w 
2

N cr ,T       2
· G ·I t       2     =∞
io                L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. = (11·1,02) m
Lkz: effective length for buckling with the axis Z. = 11 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 21.21 cm
io  (i y  i z  y 0  z 0 )                   ( 25 . 17  7 . 08  0  0 )           26 . 14 cm
2       2             2    2   0 .5             2        2             0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                 61.058 t ·m
         Ed
                              0 . 339  1
M   c , Rd
1 8 0 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35 G
+1.5·V1(2) +1.05N1

Where:

MEd: Bending moment calculation applicant awful = 61.058 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:

18
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  1 . 926 t ·m
          Ed
                     0 . 049  1
M    c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35G
+1.5·V1 (1).

Where:

MEd: Bending moment calculation applicant awful = 1.926 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed              8 . 007 t
                                    0 . 053  1
V c , Rd          150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 8.007 t
The shear resistance of calculating VcRd is given by:
f yd
VCRd = A v                   150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

19
Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 34.84
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         0 . 363 t
                                 0 . 001  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 0.393 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v               301.520 t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270 cm2
d: Web depth. = 540 mm
tw: Web thickness. = 15.5 mm

20
Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                8.007 ≤ 75.259
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 0.393 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(1).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                            
 M y , Ed              M z , Ed 
                                                 1  0.116 ≤ 1
 M N , Rd , y 
                       M N , Rd , z 
              

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k yy ·                            k yz ·                 1  0.367 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k zy ·                            k zz ·                 1  0.241 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(2)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 16.661 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.

21
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M      pl , Rd
 180 . 110 tm  M   pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                  N , Rd , z
 M   pl , Rd , z
·1          =38.993 tm

     1 a  

 2
  5·n  0 . 11  1

N c , Ed
n                       0 . 022
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993tm)
A  2·b·t f
a                              0 . 33  0 . 5
A
A: Area of the gross section. = 270 cm2
b: width of the wing. = 30 cm
tf: thickness of the wing = 30 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 013 tm
                                           0 . 009  1
M T , Rd           3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).
MT,Ed: torque calculation applicant awful.= 0.033 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  3.599 t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

22
PILLAR 4

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       53 . 811 t
Must be satisfied:                                           0 . 071  1
N r , Rd       756 . 881 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (2).
Nt,Ed: Axial traction applicant bad calculation = 53.811 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       93.711 t
                               0 . 124  1
N c , Rd       756.881 t

N c , Ed       93.711 t
                               0 . 566  1
N b , Rd       165.426 t

The calculation effort lousy applicant occurs for the combination of actions 1.35·G
+1.5·V1 (4) +1. 05·N1.
Nc,Ed: Axial compression applicant bad calculation = 93.711 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2

23
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.22 · 270.0 cm2 · 2803.26 kp/cm2 = 165.426 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.
1
                                           0.91(  y) and 0.22(  z ) ≤ 1
                       ( )
2                2

Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 0.69 (  y) and 2.70(  z).
α: Reduction coefficient for buckling.= 0.21(y) and 0.34(z)

A· f y
 : Reduced slenderness                                         = 0.55(  y ) and 1.95(  z ).
N cr

Ncr: Elastic critical elastic buckling. = 198.511 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 2508.902 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z               2
= 198.511 t
L kz
c) Elastic critical axial buckling torque.
1            ·E ·I w 
2

N cr ,T    2 · G ·I t       2     =∞
io              L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20cm4
Iw: constant warping of the section = 10970000.00 cm6

24
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. = 12 m
Lkz: effective length for buckling with the axis Z. = 12 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 21.21 cm
io  (i y  i z  y 0  z 0 )            ( 25 . 17  7 . 08  0  0 )           26 . 14 cm
2     2        2    2   0 .5             2        2             0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                136.965 t ·m
         Ed
                       0 . 760  1
M   c , Rd
1 8 0 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1(1).

Where:

MEd: Bending moment calculation applicant awful = 136.965 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                1 . 926 t ·m
         Ed
                    0 . 012  1
M   c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35G
+1.5·V1 (4)+1.05N1.

Where:

MEd: Bending moment calculation applicant awful = 1.926 t·m

25
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed         23 . 268 t
                                  0 . 155  1
V c , Rd       150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 23.268 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                          = 34.84
tw

72
λmax: Maximum slenderness =                            · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                          = 0.92
fy

26
fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed          0 . 076 t
                                 0 . 001  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 0.393 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                301.520 t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270 cm2
d: Web depth. = 540 mm
tw: Web thickness. = 15.5 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED             23.268 ≤ 75.259
2

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(1).

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED             0.076 ≤ 150.760
2

27
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05N1.

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                             
 M y , Ed                     M z , Ed          
                                                         1  0.563 ≤ 1
 M N , Rd , y
                     
         M N , Rd , z
                   


N c , Ed                               M                          M z , Ed
 
y ,Ed
 k yy ·                                 k yz ·                 1  0.837 ≤ 1
 y · A· f yd                    LT ·W pl , y · f yd              W pl , z · f yd

N c , Ed                               M                          M z , Ed
 
y ,Ed
 k zy ·                                 k zz ·                 1  0.609 ≤ 1
 y · A· f yd                    LT ·W pl , y · f yd              W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(1)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 33.802 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M     pl , Rd
 180 . 110 tm  M    pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                    N , Rd , z
 M pl , Rd , z ·1          =38.993 tm

     1 a  
 2
  5·n  0 . 225  1

N c , Ed
n                          0 . 045
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993tm)
A  2·b·t f
a                                  0 . 33  0 . 5
A

28
A: Area of the gross section. = 270 cm2
b: width of the wing. = 30 cm
tf: thickness of the wing = 30 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed       0 . 049 tm
                                0 . 014  1
M T , Rd       3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(1).
MT,Ed: torque calculation applicant awful.= 0.049 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd            ·W T · f yd  3.599 t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

PILLAR 5

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed        8 . 681 t
Must be satisfied:                                            0 . 011  1
N r , Rd        756 . 881 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (3)+1.05N1.
Nt,Ed: Axial traction applicant bad calculation =8.681 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

29
Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       16.181 t
                               0 . 021  1
N c , Rd       756.881 t

N c , Ed       16.181 t
                               0 . 084  1
N b , Rd       192.643 t

The calculation effort lousy applicant occurs for the combination of actions 1.35·G
+1.5·V1 (1).
Nc,Ed: Axial compression applicant bad calculation = 16.181 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.0 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.25 · 270.0 cm2 · 2803.26 kp/cm2 = 192.643 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.
1
                                    0.92(  y) and 0.25(  z ) ≤ 1
                ( )
2            2

Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 0.66 (  y) and 2.37(  z).

30
α: Reduction coefficient for buckling.= 0.21(y) and 0.34(z)

A· f y
 : Reduced slenderness                                       = 0.50(  y ) and 1.79 (  z ).
N cr

Ncr: Elastic critical elastic buckling. = 236.245 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 2985.800 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z             2
= 236.245 t
L kz
c) Elastic critical axial buckling torque.
1            ·E ·I w 
2

N cr ,T    2 · G ·I t       2     =∞
io              L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. = 11 m
Lkz: effective length for buckling with the axis Z. = 11 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 21.21 cm
io  (i y  i z  y 0  z 0 )                   ( 25 . 17  7 . 08  0  0 )           26 . 14 cm
2       2             2    2   0 .5             2          2           0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                11.297 t ·m
         Ed
                              0 . 063  1
M   c , Rd
1 8 0 . 110 t ·m

31
The calculation effort lousy applicant occurs for the combination of 1.35 G
+1.5·V1(1).

Where:

MEd: Bending moment calculation applicant awful = 11.297 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  9 . 276 t ·m
          Ed
                     0 . 238  1
M    c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1 (4).

Where:

MEd: Bending moment calculation applicant awful = 9.276 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed              1 . 974 t
                                    0 . 013  1
V c , Rd          150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 1.974 t
The shear resistance of calculating VcRd is given by:

32
f yd
VCRd = A v               150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 34.84
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         4 . 917 t
                                 0 . 016  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 4.917 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v               301.520 t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2

33
A: Gross sectional area.= 270 cm2
d: Web depth. = 540 mm
tw: Web thickness. = 15.5 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED              1.974 ≤ 75.259
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(1).

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 3.792 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05N1.

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                         
 M y , Ed                  M z , Ed       
                                                  1  0.238 ≤ 1
 M N , Rd , y
                  
         M N , Rd , z
                


N c , Ed                        M                          M z , Ed
 
y ,Ed
 k yy ·                             k yz ·                 1  0.238 ≤ 1
 y · A· f yd                LT ·W pl , y · f yd           W pl , z · f yd

N c , Ed                        M                          M z , Ed
 
y ,Ed
 k zy ·                             k zz ·                 1  0.238 ≤ 1
 y · A· f yd                LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
G+1.5·V1(4).

34
Where:
Nc,Ed: Axil applicant compression calculation. = 16.661 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M     pl , Rd
 180 . 110 tm  M   pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                 N , Rd , z
 M   pl , Rd , z
·1          =38.993 tm

     1 a  
 2
  5·n  0 . 005  1

N c , Ed
n                      0 . 001
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993tm)
A  2·b·t f
a                             0 . 33  0 . 5
A
A: Area of the gross section. = 270 cm2
b: width of the wing. = 30 cm
tf: thickness of the wing = 30 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 010 tm
                                          0 . 003  1
M T , Rd           3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(2)+1.05N1.
MT,Ed: torque calculation applicant awful.= 0.010 t m
The resistant torque calculation MT,Rd is given by:

35
1
M T , Rd        ·W T · f yd  3.599 t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

36
BEAMS

BEAM 1

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed        0 . 568 t
Must be satisfied:                                  0 . 001  1
N t , Rd       756 . 881 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (1).
Nt,Ed: Axial traction applicant bad calculation = 0.568 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 270.00 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

37
N c , Ed           6.008 t
                                   0 . 008  1
N c , Rd        756.881 t

The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1 (3) +1.05·N1.
Nc,Ed: Axial compression applicant bad calculation = 6.008 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.00 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 013
N Cr

A· f y
 : Reduced slenderness                            = 1.28
N cr

Ncr: Elastic critical elastic buckling. = 460.942 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 5825.658 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z            2
= 469.942 t
L kz
c) Elastic critical axial buckling torque.

38
1            ·E ·I w 
2

N cr ,T    2 · G ·I t       2     =∞
io              L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20 cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. =7.875 m
Lkz: effective length for buckling with the axis Z. = 7.875 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 26.55 cm
io  (i y  i z  y 0  z 0 )           ( 25 ,17  7 . 08  0  0 )           26 . 14 cm
2      2     2    2    0 .5            2        2             0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                31 . 856 t ·m
         Ed
                     0 . 177  1
M   c , Rd
180 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (2) +1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 31.856 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3

39
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                 36 . 397 t ·m
          Ed
                     0 . 933  1
M    c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (3) +1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 36.397 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed              8 . 051 t
                                    0 . 053  1
V c , Rd          150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 8.051 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                   150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93.00 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

40
d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 34.84
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed        14 . 250 t
                                 0 . 047  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 14.250 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v              301.520t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270.00 cm2
d: Web depth. = 540 mm
tw: Web thickness. =15.5 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.

41
V cRd
V ED                8.051 ≤ 75.259
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 3.400 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(3)+1.05·N1.

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                            
 M y , Ed                     M z , Ed         
                                                        1  0.935 ≤ 1
 M N , Rd , y
                     
         M N , Rd , z
                  


N c , Ed                              M                          M z , Ed
 
y ,Ed
 k yy ·                                k yz ·                 1  0.699 ≤ 1
 y · A· f yd                   LT ·W pl , y · f yd              W pl , z · f yd

N c , Ed                              M                          M z , Ed
 
y ,Ed
 k zy ·                                k zz ·                 1  0.976 ≤ 1
 y · A· f yd                   LT ·W pl , y · f yd              W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(3)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 6.008 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M    pl , Rd
 180 . 110 tm  M    pl , Rd , y
 180 . 110 tm
1  0 . 5·a

42
    na 
2

na M                 N , Rd , z
 M pl , Rd , z ·1          =38.993 tm

     1 a  
 2
  5·n  0 . 004  1

N c , Ed
n                   0 . 008
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993 tm)
A  2·b·t f
a                          0 . 33  0 . 5
A
A: Area of the gross section. = 270.00 cm2
b: width of the wing. = 30.00 cm
tf: thickness of the wing = 30.00 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 022 tm
                                           0 . 006  1
M T , Rd           3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(3).
MT,Ed: torque calculation applicant awful.= 0.022 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  3.599t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

Z shear resistance and torque combined (EN 1993-1-1 Eurocode 3: 2005)

V Ed              8 . 051 t
                                            0 . 053  1
V pl ,T , Rd        150 . 506 t

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

VEd: Applicant for calculating shear bad.= 8.051 t
MT,Ed: Torque calculation applicant awful.= 0.001 t·m

43
The shear resistance reduced calculation Vpl, T, Rd is given by:

 T , Ed
V pl ,T , Rd     1                     ·V pl , Rd  150 . 506 t
f yd
1 . 25 ·
3

Where:
Vpl, Rd: Shear resistance of calculation = 150.517 t
M T , Ed
τT, Ed: Shear stresses due to torsion =                                      0 . 31 kN        2
Wt                     cm

WT: Module torsional strength = 222.40 cm3

BEAM 2

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed        3 . 918 t
Must be satisfied:                                                  0 . 007  1
N t , Rd       554 . 485 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (4).
Nt,Ed: Axial traction applicant bad calculation = 3.918 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 197.80 cm2 · 2803.26 kp/cm2 = 554.485 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00
Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       17.029 t
                                       0 . 031  1
N c , Rd       554.485 t

The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1 (2) +1.05·N1.
Nc,Ed: Axial compression applicant bad calculation = 17.029 t

44
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 197.80 cm2 · 2803.26 kp/cm2 = 554.485 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 022
N Cr

A· f y
 : Reduced slenderness                             = 0.84
N cr

Ncr: Elastic critical elastic buckling. = 791.188 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 4217.771 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z            2
= 791.188 t
L kz
c) Elastic critical axial buckling torque.
1              ·E ·I w 
2

N cr ,T       2
· G ·I t       2     =∞
io                L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 57680.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 10820.00 cm4
It: Moment of inertia of uniform torsion = 355.70 cm4
Iw: constant warping of the section = 3817000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2

45
Lky: effective length for buckling with the axis Y. =5.375 m
Lkz: effective length for buckling with the axis Z. = 5.375 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 18.61 cm
io  (i y  i z  y 0  z 0 )           (17 , 08  7 . 40  0  0 )           18 . 61 cm
2     2       2     2   0 .5            2        2             0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 17.08 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.40 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                10 . 415 t ·m
         Ed
                    0 . 214  1
M   c , Rd
90 . 601 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (1) +1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 10.415 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 90.601 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 3232.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                21 . 666 t ·m
         Ed
                    0 . 700  1
M   c , Rd
30 . 948 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1 (2).

Where:

MEd: Bending moment calculation applicant awful = 21.666 t·m

46
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 30.948 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1104.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed        7 . 457 t
                                0 . 085  1
V c , Rd       87 . 397 t

Where:
VEd: Applicant for calculating shear badly. = 7.457 t
The shear resistance of calculating VcRd is given by:
f yd
Vc,Rd = A v              87.397 t
3

Where:
Av: Cross-cutting area = h · tw = 54.00 cm2
h: Song of the section. = 400 mm
tw: Web thickness. = 13.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  26.07 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                          = 26.07
tw

72
λmax: Maximum slenderness =                            · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

47
fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed          7 . 259 t
                                 0 . 030  1
V c , Rd       243 . 223 t

Where:
VEd: Applicant for calculating shear badly. = 7.259 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               243.223t
3

Where:
Av: Cross-cutting area = A - d · tw = 150.28 cm2
A: Gross sectional area.= 197.80 cm2
d: Web depth. = 352 mm
tw: Web thickness. =12.5 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED               6.486 ≤ 43.699
2

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(1)

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED               7.259 ≤ 121.611
2

48
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(2).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                             
 M y , Ed                     M z , Ed          
                                                         1  0.690 ≤ 1
 M N , Rd , y
                     
         M N , Rd , z
                   


N c , Ed                               M                          M z , Ed
 
y ,Ed
 k yy ·                                 k yz ·                 1  0.525 ≤ 1
 y · A· f yd                    LT ·W pl , y · f yd              W pl , z · f yd

N c , Ed                               M                          M z , Ed
 
y ,Ed
 k zy ·                                 k zz ·                 1  0.737 ≤ 1
 y · A· f yd                    LT ·W pl , y · f yd              W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(2)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 17.029 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M     pl , Rd
 90 . 601 tm  M   pl , Rd , y
 90 . 601 tm
1  0 . 5·a

    na 
2

na M                    N , Rd , z
 M pl , Rd , z ·1          =30.948 tm

     1 a  
 2
  5·n  0 . 155  1

N c , Ed
n                          0 . 031
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 554.485 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (90.601 tm and 30.948 tm)
A  2·b·t f
a                                  0 . 27  0 . 5
A

49
A: Area of the gross section. = 197.80 cm2
b: width of the wing. = 30.00 cm
tf: thickness of the wing = 24.00 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed          0 . 013 tm
                                          0 . 005  1
M T , Rd          2 . 399 tm

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(2)+1.05N1.
MT,Ed: torque calculation applicant awful.= 0.013 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd               ·W T · f yd  2.399t m
3
Where:
WT: Torsional modulus. = 148.21 cm3

Z shear resistance and torque combined (EN 1993-1-1 Eurocode 3: 2005)

V Ed             0 . 587 t
                                          0 . 007  1
V pl ,T , Rd        87 . 296 t

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

VEd: Applicant for calculating shear bad.= 0.587 t
MT,Ed: Torque calculation applicant awful.= 0.007 t·m
The shear resistance reduced calculation Vpl, T, Rd is given by:

 T , Ed
V pl ,T , Rd        1                     ·V pl , Rd  87 . 296 t
f yd
1 . 25 ·
3

Where:
Vpl, Rd: Shear resistance of calculation = 87.397 t
M T , Ed
τT, Ed: Shear stresses due to torsion =                                  4 . 67 kN        2
Wt                     cm

WT: Module torsional strength = 148.21 cm3

50
BEAM 3

Tensile (Eurocode 3 EN 1993-1-1):

The check was not necessary, since there is no axial traction

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed           9.976 t
                                   0 . 013  1
N c , Rd        756.881 t

The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1(1) +1.05·N1.
Nc,Ed: Axial compression applicant bad calculation = 9.976 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.00 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 010
N Cr

A· f y
 : Reduced slenderness                          = 0.87
N cr

Ncr: Elastic critical elastic buckling. = 989.351 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):

51
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 12503.995 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z             2
= 989.351 t
L kz
c) Elastic critical axial buckling torque.
1              ·E ·I w 
2

N cr ,T       2
· G ·I t       2     =∞
io                L kt   
Where:
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20 cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. =5.375 m
Lkz: effective length for buckling with the axis Z. = 5.375 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 26.55 cm
io  (i y  i z  y 0  z 0 )                  ( 25 ,17  7 . 08  0  0 )           26 . 14 cm
2       2            2    2   0 .5            2        2             0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                28 . 054 t ·m
         Ed
                           0 . 156  1
M   c , Rd
180 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (1).

Where:

MEd: Bending moment calculation applicant awful = 28.054 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m

52
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                 30 . 370 t ·m
          Ed
                     0 . 779  1
M    c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1(4).

Where:

MEd: Bending moment calculation applicant awful = 30.370 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed             10 . 952 t
                                    0 . 073  1
V c , Rd          150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 10.952 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                   150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93.00 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

53
Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 34.84
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         4 . 383 t
                                 0 . 015  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 4.383 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v              301.520t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270.00 cm2
d: Web depth. = 540 mm
tw: Web thickness. =15.5 mm

54
Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                9.630 ≤ 75.259
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(1).

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 4.383 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                            
 M y , Ed              M z , Ed 
                                                 1  0.778 ≤ 1
 M N , Rd , y 
                       M N , Rd , z 
              

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k yy ·                            k yz ·                 1  0.549 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k zy ·                            k zz ·                 1  0.788 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(4).
Where:
Nc,Ed: Axil applicant compression calculation. = 3.999 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.

55
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 2
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M      pl , Rd
 180 . 110 tm  M   pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                  N , Rd , z
 M   pl , Rd , z
·1          =38.993 tm

     1 a  

 2
  5·n  0 . 025  1

N c , Ed
n                       0 . 005
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993 tm)
A  2·b·t f
a                              0 . 33  0 . 5
A
A: Area of the gross section. = 270.00 cm2
b: width of the wing. = 30.00 cm
tf: thickness of the wing = 30.00 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 108 tm
                                           0 . 030  1
M T , Rd           3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).
MT,Ed: torque calculation applicant awful.= 0.108 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  3.599t m
3

Where:
WT: Torsional modulus. = 222.40 cm3

56
Z shear resistance and torque combined (EN 1993-1-1 Eurocode 3: 2005)

V Ed              0 . 834 t
                                           0 . 006  1
V pl ,T , Rd        148 . 715 t

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

VEd: Applicant for calculating shear bad.= 0.834 t
MT,Ed: Torque calculation applicant awful.= 0.107 t·m
The shear resistance reduced calculation Vpl, T, Rd is given by:

 T , Ed
V pl ,T , Rd        1                     ·V pl , Rd  148 . 715 t
f yd
1 . 25 ·
3

Where:
Vpl, Rd: Shear resistance of calculation = 150.517 t
M T , Ed
τT, Ed: Shear stresses due to torsion =                                         48 . 16 kN        2
Wt                      cm

WT: Module torsional strength = 222.40 cm3

BEAM 4

200x200x18mm

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       17 . 367 t
Must be satisfied:                                                     0 . 049  1
N t , Rd       354 . 040 t

The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1(3)+1.05N1.
Nt,Ed: Axial traction applicant bad calculation = 17.367 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 126.30 cm2 · 2803.26 kp/cm2 = 354.040 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2

57
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       17.652 t
                               0 . 050  1
N c , Rd       354.040 t

N c , Ed       17.652 t
                               0 . 119  1
N b , Rd       147.853 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1(1).
Nc,Ed: Axial compression applicant bad calculation = 17.652 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 126.30 cm2 · 2803.26 kp/cm2 = 354.040 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.42·126.30 cm2 · 2803.26 kp/cm2 = 147.853 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.
1
                                    0.42 ≤ 1
                ( )
2            2

58
Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 1.52
α: Reduction coefficient for buckling.= 0.49

A· f y
 : Reduced slenderness                               = 1.23
N cr

Ncr: Elastic critical elastic buckling. = 232.293 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 232.293 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z           2
= 232.293 t
L kz
Where:
Iy: Moment of inertia of the gross section with the axis Y = 6818.49 cm4
Iz: moment of inertia of the gross section with the axis Z = 6818.49 cm4
E: Modulus of elasticity = 2140673 kp/cm2
Lky: effective length for buckling with the axis Y. = 7.875 m
Lkz: effective length for buckling with the axis Z. = 7.875 m

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                16 . 139 t ·m
         Ed
                       0 . 642  1
M   c , Rd
25 . 153 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (4) +1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 16.139 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 25.153 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1

59
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 897.26 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  2 . 732 t ·m
          Ed
                     0 . 109  1
M    c , Rd
25 . 153 t ·m

The calculation effort lousy applicant occurs for the combination of G + 1.5·V1(4).

Where:

MEd: Bending moment calculation applicant awful =2.732 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 25.153 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 897.26 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed             8 . 206 t
                                   0 . 086  1
V c , Rd          95 . 554 t

Where:
VEd: Applicant for calculating shear badly. = 8.206 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v                  95.554 t
3

Where:
Av: Cross-cutting area = h · tw = 59.04 cm2
d: Song of the section. = 164 mm
tw: Web thickness. = 18 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

60
d        72
          ·  9.11 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                         = 9.11
tw

72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         0 . 486 t
                                 0 . 004  1
V c , Rd       108 . 851 t

Where:
VEd: Applicant for calculating shear badly. = 0.486 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               108.851 t
3

Where:
Av: Cross-cutting area = A - d · tw = 67.26 cm2
A: Gross sectional area.= 126.30 cm2
d: Web depth. = 164 mm
tw: Web thickness. = 18 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.

61
V cRd
V ED                7.233 ≤ 47.777
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+0.9·V1(3)+1.5·N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                0.486 ≤ 54.426
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                            
 M y , Ed              M z , Ed 
                                                 1  0.499 ≤ 1
 M N , Rd , y 
                       M N , Rd , z 
              

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k yy ·                            k yz ·                 1  0.755 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k zy ·                            k zz ·                 1  0.532≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(4)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 7.991 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively (16.139 t·m and 2.420 t·m).
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.

62
1 n
M    N , Rd , y
 M      pl , Rd
 25 . 153 tm  M   pl , Rd , y
 25 . 153 tm
1  0 . 5·a
1 n
M    N , Rd , z
 M      pl , Rd , z
=25.153 tm
1  0 . 5·a
  1 . 661
1 . 66
                                  1 . 661  6
1  1 . 13 ·n
2

N c , Ed
n                      0 . 023
N   pl , Rd

Npl,Rd: compressive strength of the gross section.=354.040 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively (25.153 t·m both).
A  2·b·t f
a                              0 . 43  0 . 5
A
A: Area of the gross section. = 126.30 cm2
b: width of the wing. = 20.00 cm
h: ridge section = 200.00 cm
tf: thickness of the wing = 18 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed             0 . 098 tm
                                             0 . 005  1
M T , Rd           19 . 259 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(3)+1.05·N1.
MT,Ed: torque calculation applicant awful.= 0.098 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                   ·W T · f yd  19.259t m
3
Where:
WT: Torsional modulus. = 1189.96 cm3

Z shear resistance and torque combined (EN 1993-1-1 Eurocode 3: 2005)

V Ed                6 . 814 t
                                             0 . 071  1
V pl ,T , Rd          95 . 383 t

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

63
VEd: Applicant for calculating shear bad.= 6.814 t
MT,Ed: Torque calculation applicant awful.= 0.034 t·m
The shear resistance reduced calculation Vpl, T, Rd is given by:

 T , Ed
V pl ,T , Rd      1                      ·V pl , Rd  95 . 383 t
f yd
1 . 25 ·
3

Where:
Vpl, Rd: Shear resistance of calculation = 95.554 t
M T , Ed
τT, Ed: Shear stresses due to torsion =                                        2 . 89 kN        2
Wt                     cm

WT: Module torsional strength = 1192.46 cm3

BEAM 5

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed        3 . 368 t
Must be satisfied:                                                    0 . 004  1
N t , Rd       756 . 881 t

The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1 (1)+1.05N1.
Nt,Ed: Axial traction applicant bad calculation = 3.368 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 270.00 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       19.106 t
                                        0 . 025  1
N c , Rd       756.881 t

64
The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (3).
Nc,Ed: Axial compression applicant bad calculation = 19.106 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 270.00 cm2 · 2803.26 kp/cm2 = 756.881 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 2
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 019
N Cr

A· f y
 : Reduced slenderness                             = 0.87
N cr

Ncr: Elastic critical elastic buckling. = 989.351 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
c) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 12503.995 t
L ky

d) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z            2
= 989.351 t
L kz
e) Elastic critical axial buckling torque.
1            ·E ·I w 
2

N cr ,T    2 · G ·I t       2     =∞
io              L kt   
Where:

65
Iy: Moment of inertia of the gross section with the axis Y = 171000.00 cm4
Iz: moment of inertia of the gross section with the axis Z = 13530.00 cm4
It: Moment of inertia of uniform torsion = 667.20 cm4
Iw: constant warping of the section = 10970000.00 cm6
E: Modulus of elasticity = 2140673 kp/cm2
G: shear modulus = 825688 kp/cm2
Lky: effective length for buckling with the axis Y. =5.375 m
Lkz: effective length for buckling with the axis Z. = 5.375 m
Lkt: effective length for torsional buckling. = 0 m
i0: polar turning radius of the gross section with respect to the torsion center
= 26.55 cm
io  (i y  i z  y 0  z 0 )           ( 25 ,17  7 . 08  0  0 )           26 . 14 cm
2     2       2    2    0 .5            2        2             0 .5

iy: radius of gyration of the gross section with respect to the principal
axes of inertia Y = 25.17 cm
iz: radius of gyration of the gross section with respect to the principal
axes of inertia Z = 7.08 cm
y0, z0: Coordinates of the center of torque in the direction of the
principal axes Y and Z, respectively, relative to the center of gravity of
the section = 0 cm.

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                35 . 579 t ·m
         Ed
                     0 . 198  1
M   c , Rd
180 . 110 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (1) +1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 35.579 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 180.110 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 6425.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

66
M                 29 . 491 t ·m
            Ed
                     0 . 756  1
M   c , Rd
38 . 993 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1 (2).

Where:

MEd: Bending moment calculation applicant awful = 29.491 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 38.993 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 1391.00 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed             13 . 751 t
                                      0 . 091  1
V c , Rd            150 . 517 t

Where:
VEd: Applicant for calculating shear badly. = 13.751 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                    150.517 t
3

Where:
Av: Cross-cutting area = h · tw = 93.00 cm2
h: Song of the section. = 600 mm
tw: Web thickness. = 15.5 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  34.84 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                           = 34.84
tw

67
72
λmax: Maximum slenderness =                           · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                         = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed          9 . 539 t
                                 0 . 032  1
V c , Rd       301 . 520 t

Where:
VEd: Applicant for calculating shear badly. = 9.539 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               301.520t
3

Where:
Av: Cross-cutting area = A - d · tw = 186.30 cm2
A: Gross sectional area.= 270.00 cm2
d: Web depth. = 540 mm
tw: Web thickness. =15.5 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED             12.456 ≤ 75.259
2

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(1)+1.05·N1.

68
Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                9.539 ≤ 150.760
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(2).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                             
 M y , Ed                 M z , Ed          
                                                     1  0.762 ≤ 1
 M N , Rd , y 
                          M N , Rd , z
                   


N c , Ed                           M                          M z , Ed
 
y ,Ed
 k yy ·                             k yz ·                 1  0.603 ≤ 1
 y · A· f yd                  LT ·W pl , y · f yd            W pl , z · f yd

N c , Ed                           M                          M z , Ed
 
y ,Ed
 k zy ·                             k zz ·                 1  0.802 ≤ 1
 y · A· f yd                  LT ·W pl , y · f yd            W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
G+1.5·V1(2).
Where:
Nc,Ed: Axil applicant compression calculation. = 3.871 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively.
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M     pl , Rd
 180 . 110 tm  M    pl , Rd , y
 180 . 110 tm
1  0 . 5·a

    na 
2

na M                N , Rd , z
 M pl , Rd , z ·1          =38.993 tm

     1 a  

 2
  5·n  0 . 025  1

69
N c , Ed
n                   0 . 005
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 756.881 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (180.110 tm and 38.993 tm)
A  2·b·t f
a                          0 . 33  0 . 5
A
A: Area of the gross section. = 270.00 cm2
b: width of the wing. = 30.00 cm
tf: thickness of the wing = 30.00 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed          0 . 050 tm
                                           0 . 014  1
M T , Rd          3 . 599 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(3).
MT,Ed: torque calculation applicant awful.= 0.050 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  3.599t m
3
Where:
WT: Torsional modulus. = 222.40 cm3

Z shear resistance and torque combined (EN 1993-1-1 Eurocode 3: 2005)

V Ed              5 . 960 t
                                            0 . 040  1
V pl ,T , Rd        150 . 480 t

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

VEd: Applicant for calculating shear bad.= 5.960 t
MT,Ed: Torque calculation applicant awful.= 0.001 t·m
The shear resistance reduced calculation Vpl, T, Rd is given by:

 T , Ed
V pl ,T , Rd        1                      ·V pl , Rd  150 . 480 t
f yd
1 . 25 ·
3

70
Where:
Vpl, Rd: Shear resistance of calculation = 150.517 t
M T , Ed
τT, Ed: Shear stresses due to torsion =               1 . 01 kN        2
Wt                     cm

WT: Module torsional strength = 222.40 cm3

71
CROSSES

CROSS 1

D=180mm t=18mm

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       24 . 445 t
Must be satisfied:                                  0 . 095  1
N t , Rd       256 . 804 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (1).
Nt,Ed: Axial traction applicant bad calculation = 24.445 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 91.61 cm2 · 2803.26 kp/cm2 = 256.804 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2

72
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       32.279 t
                               0 . 126  1
N c , Rd       256.804 t

N c , Ed       32.279 t
                              0 . 365  1
N b , Rd       88.337 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1 (3) +1.05·N1.
Nc,Ed: Axial compression applicant bad calculation = 32.279 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 91.61 cm2 · 2803.26 kp/cm2 = 256.804 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.34· 91.61 cm2 · 2803.26 kp/cm2 = 88.337 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.
1
                                   0.34 ≤ 1
               ( )
2            2

Where:

73

  0 . 5· 1   ·  0 . 2   (  )
2
= 1.80
α: Reduction coefficient for buckling.= 0.49

A· f y
 : Reduced slenderness                               = 1.41
N cr

Ncr: Elastic critical elastic buckling. = 128.394 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 128.394 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z            2
= 128.394 t
L kz
Where:
Iy: Moment of inertia of the gross section with the axis Y = 3042.33 cm4
Iz: moment of inertia of the gross section with the axis Z = 3042.33 cm4
E: Modulus of elasticity = 2140673 kp/cm2
Lky: effective length for buckling with the axis Y. = 7.075 m
Lkz: effective length for buckling with the axis Z. = 7.075 m

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                 0 . 288 t ·m
         Ed
                        0 . 022  1
M   c , Rd
1 3 . 297 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35G
+1.5·V1(1)

Where:

MEd: Bending moment calculation applicant awful = 0.288 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 13.297 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 474.34 cm3

74
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  9 . 471 t ·m
          Ed
                     0 . 712  1
M    c , Rd
13 . 297 t ·m

The calculation effort lousy applicant occurs for the combination of G+1.5·V1(4).

Where:

MEd: Bending moment calculation applicant awful = 9.471 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 13.297 t·m

Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 474.34 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed              0 . 191 t
                                    0 . 002  1
V c , Rd          94 . 389 t

Where:
VEd: Applicant for calculating shear badly. = 0.191 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v                   94.389 t
3

Where:
A
Av: Cross-cutting area = 2·                      = 58.32 cm2

A: Area of the section. = 91.61 cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed             3 . 581 t
                                    0 . 038  1
V c , Rd          94 . 389 t

Where:

75
VEd: Applicant for calculating shear badly. = 3.581 t
The shear resistance of calculating VcRd is given by:
f yd
VCRd = A v                94.389 t
3

Where:
A
Av: Cross-cutting area = 2·                       = 58.32 cm2

A: Area of the section. = 91.61 cm2

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED              0.191 ≤ 47.194
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(3)+1.05N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED              3.581 ≤ 47.194
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                         
 M y , Ed              M z , Ed       
                                              1  0.503 ≤ 1
 M N , Rd , y 
                       M N , Rd , z
                


76
N c , Ed                          M                            M z , Ed
 
y ,Ed
 k yy ·                              k yz ·                 1  0.474 ≤ 1
 y · A· f yd                  LT ·W pl , y · f yd             W pl , z · f yd

N c , Ed                          M                            M z , Ed
 
y ,Ed
 k zy ·                              k zz ·                 1  0.758≤ 1
 y · A· f yd                  LT ·W pl , y · f yd             W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(4)+1.05N1.
Where:
Nc,Ed: Axil applicant compression calculation. = 3.283 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively (0.222 t·m and 9.426 t·m).
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.

 1 . 04 ·M                     (1  n          )  13 . 297 tm  M                  13 . 297 .tm
1 .7
M   N , Rd , y                     pl , Rd , y                                         pl , Rd , y

 1 . 04 ·M                     (1  n          )  13 . 297 tm  M                  13 . 297 .tm
1 .7
M   N , Rd , z                    pl , Rd , z                                          pl , Rd , z

 2
  5·n  2  1

N c , Ed
n                      0 . 013
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 256.804 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (13.297 tm both)

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           0 . 671 tm
                                        0 . 061  1
M T , Rd         19 . 942 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).
MT,Ed: torque calculation applicant awful.= 0.671 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  19.942 t m
3

77
Where:
WT: Torsional modulus. = 676.07 cm3

CROSS 2

D=350mm t=30mm

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       46 . 150 t
Must be satisfied:                                          0 . 055  1
N t , Rd       845 . 444 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1(1).
Nt,Ed: Axial traction applicant bad calculation = 46.150 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 301.59 cm2 · 2803.26 kp/cm2 = 845.444 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       65.228 t
                               0 . 077  1
N c , Rd       845.444 t

The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1 (3)+1.05N1.
Nc,Ed: Axial compression applicant bad calculation = 65.228 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 301.59 cm2 · 2803.26 kp/cm2 = 845.444 t
Where:

78
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 034
N Cr

A: Area of the gross section of Class 1, 2 and 3 = 301.59 cm2.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00

A· f y
 : Reduced slenderness                            = 0.67
N cr

Ncr: Elastic critical elastic buckling. = 1903.546 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 1903.546 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z           2
= 1903.546 t
L kz
Where:
Iy: Moment of inertia of the gross section with the axis Y = 38943.18 cm4
Iz: moment of inertia of the gross section with the axis Z = 38943.18 cm4
E: Modulus of elasticity = 2140673 kp/cm2
Lky: effective length for buckling with the axis Y. = 6.574 m
Lkz: effective length for buckling with the axis Z. = 6.574 m

79
Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  3 . 374 t ·m
          Ed
                     0 . 039  1
M    c , Rd
86 . 369 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G +
1.5·V1 (1).

Where:

MEd: Bending moment calculation applicant awful = 3.374 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 86.369 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 3081.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                 31 . 101 t ·m
          Ed
                     0 . 360  1
M    c , Rd
86 . 369 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1(2).

Where:

MEd: Bending moment calculation applicant awful = 31.101 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 86.369 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 3081.00 cm3
Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed              0 . 972 t
                                    0 . 003  1
V c , Rd          310 . 745 t

80
Where:
VEd: Applicant for calculating shear badly. = 0.972 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               310.745 t
3

Where:
A
Av: Cross-cutting area = 2·                   = 192.00 cm2

A: Area of the section. = 301.59 cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed          6 . 899 t
                                 0 . 022  1
V c , Rd       310 . 745 t

Where:
VEd: Applicant for calculating shear badly. = 2.050 t
The shear resistance of calculating V cRd is given by:
f yd
VCRd = A v                262.191 t
3

Where:
A
Av: Cross-cutting area = 2·                   = 192.00 cm2

A: Area of the section. = 301.59 cm2

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED             0.915 ≤ 155.373
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(3)+1.05·N1.

81
Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 6.899 ≤ 155.373
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(2).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                              
 M y , Ed                   M z , Ed           
                                                        1  0.130 ≤ 1
 M N , Rd , y 
                            M N , Rd , z
                    


N c , Ed                           M                             M z , Ed
 
y ,Ed
 k yy ·                               k yz ·                 1  0.233 ≤ 1
 y · A· f yd                   LT ·W pl , y · f yd              W pl , z · f yd

N t , Ed                          M                             M z , Ed
 
y ,Ed
 k zy ·                               k zz ·                 1  0.375≤ 1
 y · A· f yd                   LT ·W pl , y · f yd              W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(2).
Where:
Nc,Ed: Axil applicant traction calculation. = 9.175 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively (0.479 t·m and 31.089 t·m).
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
(1  n )
M    N , Rd , y
M      pl , Rd , y
 86 . 369 tm  M    pl , Rd , y
 86 . 369 .tm
1  0 . 5·a w

(1  n )
M    N , Rd , z
 M     pl , Rd , z
 86 . 369 tm  M    pl , Rd , z
 86 . 369 .tm
1  0 . 5·a f

  2

Where:

82
N t , Ed
n=                  0 . 011
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 845.444 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (86.369 tm both)

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed        7 . 303 tm
                                     0 . 101  1
M T , Rd        72 . 032 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(2).
MT,Ed: torque calculation applicant awful.= 7.303 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd              ·W T · f yd  72.032 t m
3
Where:
WT: Torsional modulus. = 4450.65 cm3

CROSS 3

D=180mm t=18mm

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       12 . 029 t
Must be satisfied:                                                0 . 047  1
N t , Rd       256 . 804 t

The calculation effort lousy applicant occurs for the combination of actions
1.35G+1.5·V1(1).
Nt,Ed: Axial traction applicant bad calculation = 12.029 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 91.61 cm2 · 2803.26 kp/cm2 = 256.804 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.

83
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       30.510 t
                               0 . 119  1
N c , Rd       256.804 t

N c , Ed       30.510 t
                              0 . 316  1
N b , Rd       96.600 t

The calculation effort lousy applicant occurs for the combination of actions G+
+1.5·V1 (3) +1.05·N1.
Nc,Ed: Axial compression applicant bad calculation = 32.279 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 91.61 cm2 · 2803.26 kp/cm2 = 256.804 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

The resistance of buckling Nb,Rd compressed in a bar is given by:
Nb,Rd = χ ·A · fyd = 0.38· 91.61 cm2 · 2803.26 kp/cm2 = 86.600 t
Where:
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00
χ: Reduction coefficient for buckling.

84
1
                                            0.38 ≤ 1
                       ( )
2                2

Where:

  0 . 5· 1   ·  0 . 2   (  )
2
= 1.66
α: Reduction coefficient for buckling.= 0.49

A· f y
 : Reduced slenderness                                         = 1.33
N cr

Ncr: Elastic critical elastic buckling. = 145.054 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 145.054 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z               2
= 145.054 t
L kz
Where:
Iy: Moment of inertia of the gross section with the axis Y = 3042.33 cm4
Iz: moment of inertia of the gross section with the axis Z = 3042.33 cm4
E: Modulus of elasticity = 2140673 kp/cm2
Lky: effective length for buckling with the axis Y. = 6.657 m
Lkz: effective length for buckling with the axis Z. = 6.657m

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                 0 . 370 t ·m
         Ed
                        0 . 028  1
M   c , Rd
1 3 . 297 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35G
+1.5·V1(3)

Where:

MEd: Bending moment calculation applicant awful = 0.370 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 13.297 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1

85
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 474.34 cm3

Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                  5 . 301 t ·m
          Ed
                     0 . 399  1
M    c , Rd
13 . 297 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G+
1.5·V1(2)+1.05·N1.

Where:

MEd: Bending moment calculation applicant awful = 5.301 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 13.297 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 474.34 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed             0 . 210 t
                                   0 . 002  1
V c , Rd          94 . 389 t

Where:
VEd: Applicant for calculating shear badly. = 0.210t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v                  94.389 t
3

Where:
A
Av: Cross-cutting area = 2·                     = 58.32 cm2

A: Area of the section. = 91.61 cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)
Must be satisfied::

86
V Ed         3 . 052 t
                                0 . 032  1
V c , Rd       94 . 389 t

Where:
VEd: Applicant for calculating shear badly. = 3.052 t
The shear resistance of calculating VcRd is given by:
f yd
VCRd = A v                94.389 t
3

Where:
A
Av: Cross-cutting area = 2·                  = 58.32 cm2

A: Area of the section. = 91.61 cm2

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED             0.199 ≤ 47.194
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(1)+1.05N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED             0.155 ≤ 47.194
2

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(3).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:

87
                              
 M y , Ed                   M z , Ed 
                                                        1  0.157 ≤ 1
 M N , Rd , y 
                            M N , Rd , z 
              

N t , Ed                          M                             M z , Ed
 
y ,Ed
 k yy ·                               k yz ·                 1  0.416 ≤ 1
 y · A· f yd                   LT ·W pl , y · f yd              W pl , z · f yd

N c , Ed                           M                             M z , Ed
 
y ,Ed
 k zy ·                               k zz ·                 1  0.416≤ 1
 y · A· f yd                   LT ·W pl , y · f yd              W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(2).
Where:
Nt,Ed: Axil applicant traccion calculation. = 1.246 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively (0.203 t·m and 5.261 t·m).
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.

 1 . 04 ·M                     (1  n          )  13 . 297 tm  M                  13 . 297 .tm
1 .7
M    N , Rd , y                     pl , Rd , y                                         pl , Rd , y

 1 . 04 ·M                     (1  n          )  13 . 297 tm  M                  13 . 297 .tm
1 .7
M    N , Rd , z                    pl , Rd , z                                          pl , Rd , z

 2
1 . 66
                               26
1  1 . 13 n
2

N c , Ed
n                    0 . 005
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 256.804 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (13.297 tm both)

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed            0 . 369 tm
                                         0 . 034  1
M T , Rd          10 . 942 tm

Applicants efforts produced dismal spreadsheet for the combination of 1.35·G
+1.5·V1(3)+1.05N1.
MT,Ed: torque calculation applicant awful.= 0.369 t m

88
The resistant torque calculation MT,Rd is given by:
1
M T , Rd           ·W T · f yd  10.942 t m
3
Where:
WT: Torsional modulus. = 676.07 cm3

CROSS 4

200x200x18mm

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       22 . 148 t
Must be satisfied:                                           0 . 063  1
N t , Rd       354 . 040 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1(1).
Nt,Ed: Axial traction applicant bad calculation = 22.148 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 126.30 cm2 · 2803.26 kp/cm2 = 354.040 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       20.757 t
                               0 . 059  1
N c , Rd       354.040 t

The calculation effort lousy applicant occurs for the combination of actions 1.35G
+1.5·V1(3)+1.05N1.
Nc,Ed: Axial compression applicant bad calculation = 20.757 t
The compressive strength calculation Nc,Rd is given by:

89
Nc,Rd = A · fyd = 126.30 cm2 · 2803.26 kp/cm2 = 354.040 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 033
N Cr

A· f y
 : Reduced slenderness                             = 0.75
N cr

Ncr: Elastic critical elastic buckling. = 633.857 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 633.857 t
L ky

b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z            2
= 633.857 t
L kz
Where:
Iy: Moment of inertia of the gross section with the axis Y = 6818.49 cm4
Iz: moment of inertia of the gross section with the axis Z = 6818.49 cm4
E: Modulus of elasticity = 2140673 kp/cm2
Lky: effective length for buckling with the axis Y. = 4.767 m
Lkz: effective length for buckling with the axis Z. = 4.767 m

90
Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                18 . 818 t ·m
         Ed
                    0 . 470  1
M   c , Rd
25 . 153 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G
+1.5·V1 (4) +1.05·N1

Where:

MEd: Bending moment calculation applicant awful = 11.818 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 25.153 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 897.26 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                4 . 207 t ·m
         Ed
                    0 . 167  1
M   c , Rd
25 . 153 t ·m

The calculation effort lousy applicant occurs for the combination of G + 1.5·V1(4).

Where:

MEd: Bending moment calculation applicant awful =2.732 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 25.153 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 897.26 cm3

Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::

91
V Ed         4 . 243 t
                                 0 . 044  1
V c , Rd       95 . 554 t

Where:
VEd: Applicant for calculating shear badly. = 4.243 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               95.554 t
3

Where:
Av: Cross-cutting area = h · tw = 59.04 cm2
d: Song of the section. = 164 mm
tw: Web thickness. = 18 mm

Shear buckling of the soul (EN 1993-1-1 Eurocode 3)
Although no transverse stiffeners are arranged, it is not necessary to check the
buckling resistance of the soul, since it satisfies:

d        72
          ·  11.11 < 55.46
tw  
Where:

d
λw: Slenderness of the soul.=                           = 11.11
tw

72
λmax: Maximum slenderness =                             · =55.46


ƞ: coefficient to consider the additional resistance due to plastic regime
hardening of the material.= 1.20

f ref
ɛ: Reduction factor =                          = 0.92
fy

fref: Reference Yield.= 2395.51 kp/cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:
V Ed         1 . 166 t
                                  0 . 011  1
V c , Rd       108 . 851 t

Where:
VEd: Applicant for calculating shear badly. = 1.166 t

92
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               108.851 t
3

Where:
Av: Cross-cutting area = A - d · tw = 67.26 cm2
A: Gross sectional area.= 126.30 cm2
d: Web depth. = 164 mm
tw: Web thickness. = 18 mm

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                4.243 ≤ 47.777
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                1.166 ≤ 54.426
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(4).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                       
 M y , Ed              M z , Ed 
                                            1  0.326 ≤ 1
 M N , Rd , y 
                       M N , Rd , z 
              

93
N c , Ed                             M                          M z , Ed
 
y ,Ed
 k yy ·                               k yz ·                 1  0.621 ≤ 1
 y · A· f yd                  LT ·W pl , y · f yd              W pl , z · f yd

N c , Ed                             M                          M z , Ed
 
y ,Ed
 k zy ·                               k zz ·                 1  0.491≤ 1
 y · A· f yd                  LT ·W pl , y · f yd              W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(4)+1.05·N1
Where:
Nc,Ed: Axil applicant compression calculation. = 17.648 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively (11.818 t·m and 3.714 t·m).
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.
1 n
M   N , Rd , y
 M      pl , Rd
 25 . 153 tm  M      pl , Rd , y
 25 . 153 tm
1  0 . 5·a
1 n
M   N , Rd , z
 M      pl , Rd , z
=25.153 tm
1  0 . 5·a
  1 . 665
1 . 66
                                 1 . 665  6
1  1 . 13 ·n
2

N c , Ed
n                         0 . 050
N   pl , Rd

Npl,Rd: compressive strength of the gross section.=354.040 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively (25.153 t·m both).
A  2·b·t f
a                                 0 . 43  0 . 5
A
A: Area of the gross section. = 126.30 cm2
b: width of the wing. = 20.00 cm
h: ridge section = 200.00 cm
tf: thickness of the wing = 18 mm

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:

94
M T , Ed        0 . 098 tm
                                  0 . 069  1
M T , Rd       19 . 259 tm

Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(4)+1.05·N1.
MT,Ed: torque calculation applicant awful.= 1.334 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd           ·W T · f yd  19.259t m
3
Where:
WT: Torsional modulus. = 1189.96 cm3
CROSS 5

D=350mm t=30mm

Tensile (Eurocode 3 EN 1993-1-1):

N t , Ed       15 . 033 t
Must be satisfied:                                            0 . 018  1
N t , Rd       845 . 444 t

The calculation effort lousy applicant occurs for the combination of actions
G+1.5·V1(4).
Nt,Ed: Axial traction applicant bad calculation = 15.033 t
The tensile strength Nt,Rd is:
Nt,Rd= A·fyd = 301.59 cm2 · 2803.26 kp/cm2 = 845.444 t
Where:
A: Gross area of the cross section of the bar.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Compressive strength (EN 1993-1-1 Eurocode 3)
Must be satisfied:

N c , Ed       35.103 t
                                 0 . 042  1
N c , Rd       845.444 t

95
The calculation effort lousy applicant occurs for the combination of actions
1.35·G+1.5·V1(2)+1.05N1.
Nc,Ed: Axial compression applicant bad calculation = 35.103 t
The compressive strength calculation Nc,Rd is given by:
Nc,Rd = A · fyd = 301.59 cm2 · 2803.26 kp/cm2 = 845.444 t
Where:
Class: Class section, as the deformation capacity and development of the
plastic resistance of compressed flat elements of a section = 1
A: Area of the gross section to the sections of Class 1, 2 and 3.
fyd: Resistance of the steel.
fyd = fy / ɣMo =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣMo: Partial safety of the material = 1.00

Buckling resistance (EN 1993-1-1 Eurocode 3)

If the slenderness ratio is λ≤0.2 or NC,Ed / Ncr ≤ 0.04 you can ignore the effect of
buckling, and see only the cross section resistance.

Where:
N C , Ed
 0 . 020
N Cr

A: Area of the gross section of Class 1, 2 and 3 = 301.59 cm2.
fyd: Resistance of the steel.
fyd = fy / ɣM1 =2803.26 kp/cm2
fy: Elastic limit = 2803.26 kp/cm2
ɣM1: Partial safety of the material = 1.00

A· f y
 : Reduced slenderness                             = 0.69
N cr

Ncr: Elastic critical elastic buckling. = 1798.263 t
The elastic critical elastic buckling Ncr is the smaller of the values obtained
in a), b) and c):
a) Axil elastic critical for buckling about the axis Y.
 ·E ·I y
2

N cr , y     2
 1798.263 t
L ky
b) Elastic critical buckling elastic bending about the Z axis
 · E ·I z
2

N cr , z           2
= 1798.26 t
L kz

96
Where:
Iy: Moment of inertia of the gross section with the axis Y = 38943.18 cm4
Iz: moment of inertia of the gross section with the axis Z = 38943.18 cm4
E: Modulus of elasticity = 2140673 kp/cm2
Lky: effective length for buckling with the axis Y. = 6.764 m
Lkz: effective length for buckling with the axis Z. = 6.764 m

Flexural axis Y (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                1 . 725 t ·m
         Ed
                    0 . 020  1
M   c , Rd
86 . 369 t ·m

The calculation effort lousy applicant occurs for the combination of 1.35·G +
1.5·V1 (4)+1.05N1.

Where:

MEd: Bending moment calculation applicant awful = 1.7254 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,y · fyd = 86.369 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1
Wpl,y : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 3081.00 cm3
Flexural axis Z (EN 1993-1-1 Eurocode 3)

Must be satisfied:

M                21 . 903 t ·m
         Ed
                    0 . 254  1
M   c , Rd
86 . 369 t ·m

The calculation effort lousy applicant occurs for the combination of G +1.5·V1(1).

Where:

MEd: Bending moment calculation applicant awful = 21.903 t·m
The bending moment resistance calculation Mc,Rd is given by:
Mc,Rd= W pl,z · fyd = 86.369 t·m
Where:
Class: Class of the section as the deformation capacity and development
of the plastic resistance of flat elements of a simple bending section. = 1

97
Wpl,z : Plastic modulus for the fiber with higher voltage, for sections 1 and 2
class.= 3081.00 cm3
Shear strength Z (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed          0 . 924 t
                                 0 . 003  1
V c , Rd       310 . 745 t

Where:
VEd: Applicant for calculating shear badly. = 0.924 t
The shear resistance of calculating V cRd is given by:
f yd
Vc,Rd = A v               310.745 t
3

Where:
A
Av: Cross-cutting area = 2·                   = 192.00 cm2

A: Area of the section. = 301.59 cm2

Shear strength Y (EN 1993-1-1 Eurocode 3)
Must be satisfied::
V Ed          7 . 845 t
                                 0 . 025  1
V c , Rd       310 . 745 t

Where:
VEd: Applicant for calculating shear badly. = 7.845 t
The shear resistance of calculating VcRd is given by:
f yd
VCRd = A v                262.191 t
3

Where:
A
Av: Cross-cutting area = 2·                   = 192.00 cm2

A: Area of the section. = 301.59 cm2

Resistance to bending moment Y and Z combined shear (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.

98
V cRd
V ED                 0.805 ≤ 155.373
2
Applicants efforts produced dismal spreadsheet for the combination of
1.35·G+1.5·V1(2)+1.05·N1.

Resistance to bending moment (Z) and shear (Y) combined (EN 1993-1-1
Eurocode 3)
It is not necessary to reduce the design moment resistance, as the applicant for
calculating shear bad VED does not exceed 50% of the design shear strength
VcRd.
V cRd
V ED                 7.845 ≤ 155.373
2
Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(1).

Resistance to bending and combined axial (EN 1993-1-1 Eurocode 3)
Must be satisfied:
                           
 M y , Ed               M z , Ed 
                                                 1  0.062 ≤ 1
 M N , Rd , y 
                        M N , Rd , z 
              

N c , Ed                       M                          M z , Ed
 
y ,Ed
 k yy ·                            k yz ·                 1  0.172 ≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

N t , Ed                      M                          M z , Ed
 
y ,Ed
 k zy ·                            k zz ·                 1  0.272≤ 1
 y · A· f yd               LT ·W pl , y · f yd           W pl , z · f yd

Applicants efforts produced dismal spreadsheet for the combination of actions
1.35·G+1.5·V1(1)+1.05N1.
Where:
Nc,Ed: Axil applicant traction calculation. = 17.308 t m
My, Ed, Mz, Ed: Applicants for calculating bending moments dismal, according to the
axes Y and Z, respectively (0.155 t·m and 21.530 t·m).
Class: Class of the section as the deformation capacity and development of the
plastic resistance of flat elements, for axial and bending simple. = 1
MN, Rd,y and MN, Rd, z: small plastic bending moment resistance calculation, about
Y and Z axes, respectively.

99
(1  n )
M   N , Rd , y
M           pl , Rd , y
 86 . 369 tm  M   pl , Rd , y
 86 . 369 .tm
1  0 . 5·a w

(1  n )
M   N , Rd , z
 M        pl , Rd , z
 86 . 369 tm  M   pl , Rd , z
 86 . 369 .tm
1  0 . 5·a f

  2
Where:
N t , Ed
n=                        0 . 020
N   pl , Rd

Npl,Rd: compressive strength of the gross section.= 845.444 t
Mpl,Rd,y , Mpl,Rd,z: Flexural strength of the gross section plastic conditions
with respect to Y and Z axes, respectively. (86.369 tm both)

Torsional strength (EN 1993-1-1 Eurocode 3: 2005)
Must be satisfied:
M T , Ed           7 . 783 tm
                                                 0 . 108  1
M T , Rd           72 . 032 tm

Applicants efforts produced dismal spreadsheet for the combination of
G+1.5·V1(2).
MT,Ed: torque calculation applicant awful.= 7.783 t m
The resistant torque calculation MT,Rd is given by:
1
M T , Rd                 ·W T · f yd  72.032 t m
3
Where:
WT: Torsional modulus. = 4450.65 cm3

100
ANNEX B: Plans of portics and joints
ANNEX C: Checkings of Joints

In this file, we will check all the joints of the portics, the tall and the short one.
We have used the Eurocode 3 to determinate the number of bolts, the lenght of
the weldings, the dimensions of the joint... All the joints are showed like a
croquis in the Annex B.

J1

8 ,9  f yo  t 0  b1  b 2 
2
 1 
                                                 kn  
0 ,5
N i , Rd                                                                     
sin(  i )  2  b o                                  M 5 
b0            150
                               7 ,5
2  t0         2  10
b1  h1  b 2  h 2             120  120  120  120
                                                                                  0 ,8
4b0                              4·150
161 ,83 kN

42 , 07  10 mm
2        2
 0 , Ed
3
275  10 kN                                                           k n  1( tension )
f yo                      mm
2

n                                                        0 ,1398             0 , 4·n         0 , 4  0 ,1398
M5                   1                                       k  1,3           1,3                   1, 23  1( comp )
 n
                                 0 ,8

8 , 9  275 N                   10
2

mm
2
 120  120             1 
 1  57 º  N 1 , Rd                                                          2  150   7 ,5  1   1   639 ,37 kN
0 ,5

sin( 57 º )                                            
8 ,9  275 N                    10
2

mm
2
 120  120              1 
 1  53 º  N 2 , Rd                                                          2  150   7 , 5  1   1   671 , 42 kN
0 ,5

sin( 53 º )                                             

 0 ,55 ·h o  e  0 , 25 ·ho   0 ,55 ·150  37  0 , 25 ·150   82 ,5  37  37 ,5

b1 / 2         120
             0 ,8  0 ,35
b0            150
b1 / 2                                                           210000 N             2
120                            E                               mm
             12  1, 25 ·                  1, 25                              34 ,54
t1 / 2          10                             f y1                  275 N        2
mm

b1 / 2                         b0         120                        150
 0 ,1  0 , 01                          0 ,1  0 , 01          0 ,8  0 , 25
b0                            t0         150                        10

b1 / 2          120
             12  35
t1 / 2          10

b0          150
15                          15  35
t0          10

b1  b 2                                120  120
0 ,6                        1,3  0 , 6                        1,3  0 , 6  1  1,3
2 b1                                   2·120

g  20  t 1  t 2  10  10  20

b1  h1  b 2  h 2                120  120  120  120
                                                                            0 ,8
4b0                               4·150

g                                  20
 1,5·1                         1,5·1  0 ,8   0 ,13  0 ,3
b0                                  150
g                                   20
 0 ,5·1                        0 ,5·1  0 ,8   0 ,13  0 ,1
b0                                  150

275 N    2·10
10 f yo ·t 0           10         mm
b eff        ·           ·b i      ·               ·120  80 mm
b 0 f y 1 ·t i        150 275 N
2·10
mm
t0                     10

J2
8 , 9  f yo  t 0  b1  b 2 
2
 1 
                                                                 kn  
0 ,5
N i , Rd                                                                                     
sin(  i )  2  b o                                                 M 5 
b0              150
                                       7 ,5
2  t0            2  10
b1  h1  b 2  h 2                       120  120  100  100
                                                                                                   0 , 73
4b0                                            4·150
105 , 04 kN

42 , 07  10 mm
2          2
 0 , Ed
275  10
3    kN                                                           k n  1( tension )
f yo                                               mm
2

n1                                                                         0 , 0911              0 , 4·n          0 , 4  0 , 0911
M5                                  1                                          k  1, 3           1, 3                    1, 254  1( comp )
 n
                                   0 ,8
105 , 04 kN

34 , 07  10 mm
2        2
 0 , Ed
275  10
3    kN                                                           k n  1( tension )
f yo                                              mm
2

n2                                                                          0 ,1398                0 , 4·n          0 , 4  0 ,1398
M5                                   1                                         k n  1, 3           1, 3                   1, 23  1( comp )

                                     0 ,8

8 , 9  275 N                       10
2

mm
2
 120  100              1 
 1   2  55 º  N 1 , Rd                                                                          2  150   7 , 5  1   1   600 , 05 kN
0 ,5

sin( 55 º )                                              

 0 ,55 ·h o  e  0 , 25 ·ho   0 ,55 ·150  37 ,5  0 , 25 ·150   82 ,5  37 ,5  37 ,5

b1         120
                0 ,8  0 , 35
b0         150
b2         100
                0 , 66  0 , 35
b0         150

b1                                                                              210000 N                   2
120                                        E                                              mm
                 12  1, 25 ·                             1, 25                                         34 ,54
t1          10                                       f y1                          275 N               2
mm

b2                                                                               210000 N                  2
100                                        E                                              mm
                10  1, 25 ·                             1, 25                                         34 ,54
t2          10                                        f y1                         275 N               2
mm

b1                              b0             120                                 150
 0 ,1  0 , 01                                     0 ,1  0 , 01                   0 ,8  0 , 25
b0                                  t0         150                                  10
b2                              b0             100                                 150
 0 ,1  0 , 01                                       0 ,1  0 , 01                 0 , 66  0 , 25
b0                                  t0         150                                  10
b1          120
              20  35
t1          10
b2          100
              10  35
t2           10

b0        150
15                           15  35
t0          10

b1  b 2                           120  100
0 ,6                         1,3  0 , 6                 1,3  0 , 6  0 ,916  1,3
2 b1                            2·120

g  25  t 1  t 2  10  10  20

g                                 25
 1,5·1                      1,5·1  0 , 73   0 ,166  0 , 4
b0                                150
g                                  25
 0 ,5·1                     0 ,5·1  0 , 73   0 ,166  0 ,135
b0                                 150

275 N    2· 10
10 f yo ·t 0          10         mm
b eff       ·           ·b1      ·                ·120  80 mm
b 0 f y 1 ·t 1       150 275 N        ·10
2
mm
t0                    10

275 N    2· 10
10 f yo ·t 0           10         mm
b eff       ·           ·b 2      ·                ·100  66 , 67 mm
b 0 f y 1 ·t 2        150 275 N        ·10
2
mm
t0                     10

J3
8 ,9  f yo  t 0  b 1  b 2 
2
 1 
N i , Rd                                                              kn 
0 ,5
                                                  
sin(  i )  2  b o                                       M 5 
b0               150
                                    7 ,5
2 t0             2  10
b1  h1  b 2  h 2                     100  100  100  100
                                                                                            0 , 6667
4b 0                                     4·150
54 , 77 kN

34 , 07  10 mm
2          2
    0 , Ed

275  10
3     kN                                                        k n  1( tension )
f yo                                     mm
2

n1                                                                    0 , 0585               0 , 4·n         0 , 4  0 , 0911
                                   1                                       k n  1,3           1,3                    1, 27  1( comp )
M 5                                                                  
                                   0 ,8

8 ,9  275 N                     10
2

mm
2
 100  100             1 
 1   2  55 º  N 1 , Rd                                                                    2  150   7 ,5  1   1   545 ,503 kN
0 ,5

sin( 55 º )                                             

 0 ,55 ·h o  e  0 , 25 ·ho   0 ,55 ·150  37 ,5  0 , 25 ·150   82 ,5  37 ,5  37 ,5

b1 / 2          100
                  0 , 66  0 , 35
b0             150

b1 / 2                                                                              210000 N                     2
100                                      E                                                  mm
                  10  1, 25 ·                            1, 25                                          34 ,54
t1 / 2           10                                      f y1                          275 N                 2
mm

b1 / 2                                 b0            100                                   150
 0 ,1  0 , 01                                     0 ,1  0 , 01                       0 , 66  0 , 25
b0                                     t0           150                                   10

b1 / 2          100
                  10  35
t1 / 2           10

b0               150
15                                   15  35
t0              10

b1  b 2                                            100  100
0 ,6                              1,3  0 , 6                                       1,3  0 , 6  1  1,3
2 b1                                                  2·100

g  37  t 1  t 2  10  10  20

g                                            37
 1,5·1                                 1,5·1  0 , 66   0 , 246  0 , 495
b0                                           150
g                                             37
 0 ,5·1                                0 ,5·1  0 , 66   0 , 246  0 ,165
b0                                           150
275 N    2· 10
10 f yo ·t 0           10         mm
b eff       ·           ·b i      ·                ·100  66 , 67 mm
b 0 f y 1 ·t i        150 275 N        ·10
2
mm
t0                     10

J4

Strength to endure: 35,329t = 346,46kN

Kind of Bolts: 8.8(fy=800N/mm2 and M20(As=245mm2 ; d=20mm; d0=22mm)

0 , 6· f y · A s       0 , 6·800 ·245
F v , Rd                                            94 ,1kN
1, 25                  1, 25

346 ,36 kN
Number of bolts =                                      3, 68 bolts  4 bolts
94 ,1kN / bolt

e 1  1, 2·d 0  26 , 4 mm
e 2  1, 5·d 0  33 mm
p 1  2 , 2·d 0  48 , 4 mm
p 2  3 , 0·d 0  66 mm

Also, we have to check the welding that is between the hollow profile and the
small plate in each side.

fu
3
f v ,Wd 
w  M 2

fu tensile strength of steel plates welded

γM2 = 1.25 partial safety factor of the welded joint

βW correlation coefficient depending on the type of steel parts soldiers
The throat thickness of fillet weld must be such that:

a  0 , 7  e min
a  0 , 7  10 mm
a  7 mm

We have taken a= 5mm. The length of the weld will be:

F Ed
Lw 
2  a  f v ,Wd

346 , 46  10 N
3

Lw 
2  5 mm  222 , 7 N           2
mm
L w  155 ,57 mm

J5

Strength to endure: 50,195t = 492,24kN

Kind of Bolts: 8.8(fy=800N/mm2 and M24(As=353mm2 ; d=24mm; d0=26mm)

0 , 6· f y · A s       0 , 6·800 ·353
F v , Rd                                            135 , 6 kN
1, 25                  1, 25

492 , 24 kN
Number of bolts =                                        3, 63 bolts  4 bolts
135 , 6 kN / bolt

e 1  1, 2·d 0  31 , 2 mm
e 2  1, 5·d 0  39 mm
p 1  2 , 2·d 0  57 , 2 mm
p 2  3 , 0·d 0  78 mm
Also, we have to check the welding that is between the hollow profile and the
small plate in each side.

fu
3
f v ,Wd 
w  M 2

fu tensile strength of steel plates welded

γM2 = 1.25 partial safety factor of the welded joint

βW correlation coefficient depending on the type of steel parts soldiers

The throat thickness of fillet weld must be such that:

a  0 , 7  e min
a  0 , 7  10 mm
a  7 mm

We have taken a= 5mm. The length of the weld will be:

F Ed
Lw 
2  a  f v ,Wd

492 , 24  10 N
3

Lw 
2  5 mm  222 , 7 N            2
mm
L w  221 , 03 mm

J6
8 ,9  f yo  t 0  b1  b 2 
2
 1 
                                                        kn  
0 ,5
N i , Rd                                                                            
sin(  i )  2  b o                                         M 5 
b0              200
                                   8 . 33
2  t0          2  12
b1  h1  b 2  h 2                 140  140  140  140
                                                                                          0 ,7
4b0                                   4·200
311 , 74 kN

50 , 07  10 mm
2         2
 0 , Ed
3
275  10 kN                                                                 k n  1( tension )
f yo                        mm
2

n1                                                                0 , 2264             0 , 4·n         0 , 4  0 , 2264
M5                       1                                            k  1,3           1,3                    1,171  1( comp )
 n
                                 0 ,7

8 ,9  275 N                     10
2

mm
2
 140  140              1 
 1   2  53 º  N 1, Rd                                                                  2  200   8 ,33  1   1   619 ,148 kN
0 ,5

sin( 53 º )                                              

 0 ,55 ·h o  e  0 , 25 ·ho   0 ,55 ·200  30  0 , 25 ·200   110  30  50

b1 / 2          140
               0 , 7  0 ,35
b0             200

b1 / 2                                                                    210000 N                   2
140                               E                                          mm
               14  1, 25 ·                  1, 25                                          34 ,54
t1 / 2          10                               f y1                       275 N                2
mm

b1 / 2                              b0         140                           200
 0 ,1  0 , 01                              0 ,1  0 , 01                  0 , 7  0 , 26
b0                                 t0         200                            12

b1 / 2          140
               14  35
t1 / 2          10

b0         200
15                             16 , 667  35
t0             12

b1  b 2                                    140  140
0 ,6                           1,3  0 , 6                               1,3  0 , 6  1  1,3
2 b1                                       2·140

g  22  t 1  t 2  10  10  20

g                                       22
 1,5·1                           1,5·1  0 , 7   0 ,15  0 , 45
b0                                       200
g                                       22
 0 ,5·1                           0 ,5·1  0 , 7   0 ,15  0 ,15
b0                                       200
275 N    2· 12
10 f yo ·t 0          10          mm
b eff       ·           ·b i      ·                ·140  100 ,8 mm
b 0 f y 1 ·t i        200 275 N        ·10
2
mm
t0                    12

J7

8 , 9  f yo  t 0  b1  b 2 
2
 1 
                                                   kn  
0 ,5
N i , Rd                                                                        
sin(  i )  2  b o                                   M 5 
b0              200
                                   8 . 33
2  t0              2  12
b1  h1  b 2  h 2                 140  140  120  120
                                                                                      0 , 65
4b0                              4·200

304 ,95 kN

50 , 07  10 mm
2     2
 0 , Ed
3
275  10 kN                                                            k n  1( tension )
f yo                          mm
2

n1                                                            0 , 2215             0 , 4·n         0 , 4  0 , 2215
M5                       1                                       k  1,3           1,3                    1,163  1( comp )
 n
                                0 , 65
190 , 27 kN

42 , 07  10 mm
2     2
 0 , Ed
3
275  10 kN                                                           k n  1( tension )
f yo                          mm
2

n1                                                            0 ,164               0 , 4·n          0 , 4  0 ,164
M5                       1                                     k n  1,3           1, 3                  1,198  1( comp )

                                  0 , 65

8 ,9  275 N                 10
2

mm
2
 140  120              1 
 1   2  53 º  N 1 , Rd                                                             2  200   8 ,33  1   1   353 , 798 kN
0 ,5

sin( 53 º )                                              

 0 ,55 ·h o  e  0 , 25 ·ho   0 ,55 ·200  30  0 , 25 ·200   110  30  50

b1           140
               0 , 7  0 , 35
b0           200
b2           120
               0 , 6  0 , 35
b0           200
b1                                                             210000 N              2
140                            E                                 mm
              14  1, 25 ·                1, 25                                34 ,54
t1          10                            f y1                  275 N            2
mm

b2                                                             210000 N              2
120                            E                                  mm
              12  1, 25 ·                1, 25                                34 ,54
t2           10                            f y1                     275 N        2
mm

b1                            b0        140                         200
 0 ,1  0 , 01                           0 ,1  0 , 01          0 , 7  0 , 26
b0                            t0         200                        12
b2                            b0        120                         200
 0 ,1  0 , 01                           0 ,1  0 , 01          0 , 6  0 , 26
b0                             t0        200                        12

b1          140
              14  35
t1          10
b2          120
              12  35
t2           10

b0        200
15                           16 , 667  35
t0          12

b1  b 2                               140  120
0 ,6                         1,3  0 , 6                          1,3  0 , 6  0 ,92  1,3
2 b1                                  2·140

g  35  t 1  t 2  10  10  20

g                                  35
 1,5·1                       1,5·1  0 , 65   0 ,175  0 ,975
b0                                  200
g                                  35
 0 ,5·1                       0 ,5·1  0 , 65   0 ,175  0 ,175
b0                                  200

275 N    2· 12
10 f yo ·t 0         10          mm
b eff       ·           ·b1      ·                ·140  100 ,8 mm
b 0 f y 1 ·t 1       200 275 N        ·10
2
mm
t0                   12

275 N    2· 10
10 f yo ·t 0          10          mm
b eff       ·           ·b 2      ·                ·120  86 , 4 mm
b 0 f y 1 ·t 2        200 275 N        ·10
2
mm
t0                    12

J8
8 ,9  f yo  t 0  b1  b 2 
2
 1 
                                                          kn  
0 ,5
N i , Rd                                                                              
sin(  i )  2  b o                                           M 5 
b0               200
                                    8 . 33
2  t0             2  12
b1  h1  b 2  h 2                   120  120  120  120
                                                                                            0 ,6
4b0                                    4·200
178 , 05 kN

42 , 07  10 mm
2         2
 0 , Ed
3
275  10 kN                                                                 k n  1( tension )
f yo                        mm
2

n1                                                                  0 ,154               0 , 4·n         0 , 4  0 ,154
M5                        1                                           k n  1,3           1,3                  1,197  1( comp )

                                  0 ,6

8 , 9  275 N                     10
2

mm
2
 120  120                    1 
 1   2  53 º  N 1 , Rd                                                                   2  200   8 , 33       1     530 , 698 kN
0 ,5

sin( 53 º )                                                   1 

 0 ,55 ·h o  e  0 , 25 ·ho   0 ,55 ·200  30  0 , 25 ·200   110  30  50

b1 / 2          120
                0 , 6  0 ,35
b0             200

b1 / 2                                                                     210000 N                    2
120                                E                                           mm
                12  1, 25 ·                   1, 25                                          34 ,54
t1 / 2           10                                f y1                       275 N                2
mm

b1 / 2                               b0          120                           200
 0 ,1  0 , 01                                0 ,1  0 , 01                  0 , 6  0 , 26
b0                                  t0          200                            12

b1 / 2          120
                12  35
t1 / 2           10

b0          200
15                              16 , 667  35
t0              12
b1  b 2                                120  120
0 ,6                   1,3  0 , 6                            1,3  0 , 6  1  1,3
2 b1                                    2·120

g  48  t 1  t 2  10  10  20

g                                  48
 1,5·1                        1,5·1  0 , 6   0 , 24  0 , 6
b0                              200
g                                  48
 0 ,5·1                       0 ,5·1  0 , 6   0 , 24  0 , 20
b0                                  200

275 N    2· 12
10 f yo ·t 0          10          mm
b eff       ·           ·b i      ·                ·120  86 , 4 mm
b 0 f y 1 ·t i        200 275 N        ·10
2
mm
t0                    12

J9

Strength to endure: 41,165t = 403,69kN

Kind of Bolts: 8.8(fy=800N/mm2 and M20(As=245mm2 ; d=20mm; d0=22mm)

0 , 6· f y · A s       0 , 6·800 ·245
F v , Rd                                              94 ,1kN
1, 25                     1, 25

403 , 69 kN
Number of bolts =                                         4 , 29 bolts  6 bolts
94 ,1kN / bolt

e1  1, 2·d 0  26 , 4 mm
e 2  1,5·d 0  33 mm
p 1  2 , 2·d 0  48 , 4 mm
p 2  3 , 0·d 0  66 mm

Also, we have to check the welding that is between the hollow profile and the
small plate in each side.

fu
3
f v ,Wd 
w  M 2
fu tensile strength of steel plates welded

γM2 = 1.25 partial safety factor of the welded joint

βW correlation coefficient depending on the type of steel parts soldiers

The throat thickness of fillet weld must be such that:

a  0 , 7  e min
a  0 , 7  12 mm
a  8 , 4 mm

We have taken a= 5mm. The length of the weld will be:

F Ed
Lw 
2  a  f v ,Wd

403 , 69  10 N
3

Lw 
2  5 mm  222 , 7 N           2
mm
L w  181 , 27 mm

J10

Strength to endure: 86,218t = 845,51kN

Kind of Bolts: 8.8(fy=800N/mm2 and M30(As=561mm2 ; d=30mm; d0=33mm)

0 , 6· f y · A s       0 , 6·800 ·561
F v , Rd                                            215 , 4 kN
1, 25                  1, 25

845 ,51 kN
Number of bolts =                                        3,92 bolts  4 bolts
215 , 4 kN / bolt
e 1  1, 2·d 0  39 , 6 mm
e 2  1, 5·d 0  49 . 5 mm
p 1  2 , 2·d 0  72 . 6 mm
p 2  3 , 0·d 0  99 mm

Also, we have to check the welding that is between the hollow profile and the
small plate in each side.

fu
3
f v ,Wd 
w  M 2

fu tensile strength of steel plates welded

γM2 = 1.25 partial safety factor of the welded joint

βW correlation coefficient depending on the type of steel parts soldiers

The throat thickness of fillet weld must be such that:

a  0 , 7  e min
a  0 , 7  12 mm
a  8 , 4 mm

We have taken a= 5mm. The length of the weld will be:

F Ed
Lw 
2  a  f v ,Wd

845 , 51  10 N
3

Lw 
2  5 mm  222 , 7 N            2
mm
L w  379 , 66 mm

J11
This type of joint only servant to join the two beams because these are very
long. We put the screws are metric M20 minimal as they usually work in
compression and only have the function of uniting the two parts.

Strength to endure: 22,807t = 223,66kN
Kind of Bolts: 8.8(fy=800N/mm2 and M20(As=245mm2 ; d=20mm; d0=22mm)

0 , 6· f y · A s       0 , 6·800 ·245
F v , Rd                                            94 ,1kN
1, 25                  1, 25

223 . 66 kN
Number of bolts =                                      2 ,37 bolts  4 bolts ( 2 and 2 )
94 ,1kN / bolt

e1  1, 2·d 0  26 , 4 mm
e 2  1,5·d 0  33 mm
p 1  2 , 2·d 0  48 , 4 mm
p 2  3 , 0·d 0  66 mm

Also, we have to check the welding that is between the hollow profile and the
small plate in each side.

fu
3
f v ,Wd 
w  M 2

fu tensile strength of steel plates welded

γM2 = 1.25 partial safety factor of the welded joint

βW correlation coefficient depending on the type of steel parts soldiers
The throat thickness of fillet weld must be such that:

a  0 , 7  e min
a  0 , 7  12 mm
a  8 , 4 mm

We have taken a= 5mm. The length of the weld will be:

F Ed
Lw 
2  a  f v ,Wd

223 , 66  10 N
3

Lw 
2  5 mm  222 , 7 N            2
mm
L w  100 , 43 mm

J12

Strength to endure: 55,868t = 547,88kN

Kind of Bolts: 8.8(fy=800N/mm2 and M20(As=245mm2 ; d=20mm; d0=22mm)

0 , 6· f y · A s       0 , 6·800 ·245
F v , Rd                                            94 ,1kN
1, 25                  1, 25

547 . 88 kN
Number of bolts =                                      5 ,82 bolts  8 bolts ( 4 and 4 )
94 ,1kN / bolt

e1  1, 2·d 0  26 , 4 mm
e 2  1,5·d 0  33 mm
p 1  2 , 2·d 0  48 , 4 mm
p 2  3 , 0·d 0  66 mm

Also, we have to calculate the welding part of each side. The resistance per unit
area of the weld angle is equal to fV,Wd, whose value is derived from:

fu
3
f v ,Wd 
w  M 2
fu tensile strength of steel plates welded

γM2 = 1.25 partial safety factor of the welded joint

βW correlation coefficient depending on the type of steel parts soldiers

The throat thickness of fillet weld must be such that:

a  0 , 7  e min
a  0 , 7  12 mm
a  8 , 4 mm

We have taken a= 5mm. The length of the weld will be:

F Ed
Lw 
2  a  f v ,Wd

547 ,88  10 N
3

Lw 
2  5 mm  222 , 7 N          2
mm
L w  246 , 01 mm
ANNEX D: Checking and verification efforts in the shoe

Having obtained feedback on the knots that form in encastaments land values
are taken for the bending moment greater (M *) in the direction plan of the
portico, the value of axillary major (N *) and shear value greater (T *). It first
determinesm the eccentricity is subjected to the plate, using the equation:

M
M *               M       22 , 22 tm
e                                             0 , 251 m
N *       N        N       88 , 218 t


It makes the study of plate previously classified according to the eccentricity
from in the following cases:

1. e<a/6
2. a/6<e<0.375a
3. e>0.375a

a: is the largest of the lengths of the plate.=1000mm=1mm

1m
 0 ,1666  0 , 251 m  0 . 375 ·1m
6

Considering the type of structure under study is necessary to follow the
procedure pertaining to the case 2.

The following is the calculation of checking one of the motherboards,
specifically the type of the pillars of the portico. To make it checks the
combination that has transmitted more than normal effort in the shoe of a pillar
located approximately in the center of the ship.

N Ed  88 , 218 t  865 ,1kN
V Ed  0 , 231 t  2 , 27 kN
M   Ed
 22 , 22 tm  217 ,97 kNm

Value calculation of internal forces
The conditions of equilibrium to determine the value calculation of internal
forces.
N Ed  T Ed  C Ed  0
N Ed  T Ed ·a t  C Ed ·a c  0

at: is the distance between the central axis of the bolt plate. In this case it 440 mm.

ac: is the distance between the central axis of the plate and half the wing pillar. In this
case is 300mm.

M         N Ed ·a c
T Ed        Ed
 56,16kN
ac  at

C Ed  T Ed  N Ed  808 ,95 kN

Effective Area

To calculate it, first find the value of a 1 and b1. These are the minimum values shown
below:

a1  5 a  5 m                               b1  5 b  4 m
a 1  a  h  1,12 m                         b1  b  h  0 , 92 m

The parameters used in these formulas are shown in the figure below:

a=1m b=0,8m a1=1,12m b1=0,92 m

Now you can calculate the concentration factor (k j) to be less than 5, the bearing
resistance of the surface settlement (fjd) and the distance c which will calculate the
effective area of the plate subjected to compression.
a 1 ·b1
kj                          1 . 135  5
a b

2                   2
f   j
       ·k j · f ck         1 . 135  15 N                  2    11 . 35 N        2
3                   3                                mm                     mm

275
f yd                            1 . 05
c  tf                         30 mm                          85 mm
3  f jd                            11 . 35
3
1 . 05

The effective area is the one white. In case checking will take another way because of
having signs in both directions. In this case, therefore, c is the distance measured from
the outer face of the signs.

For the calculation, however, need to know the effective area subjected to
compression. Therefore, it is considered the part marked with a red box.

A eff , c  94000 mm
2

Strength Concrete

C Ed
Internal efforts:  Ed                                            8,6 N         2
Aeff , c                      mm

Local resistance to compression:  C , Rd  k j  f cd  16 , 214 N                                    2
mm

   Ed
  C , Rd  OK

Strength of the motherboard (compressed area)

c
Internal efforts: M                      Ed
   Ed
·c·         31067 , 5 Nmm
2                          mm
2
275         120
Resistance: M        Rd
 f yd  W                          209 . 523 ,81 Nmm
1, 05        18                             mm

M   Ed
 M   Rd
 OK
ANNEX E: Plans of the building

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