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```					Design of Steel Structures (3rd Civil)   New Cairo Academy                   2008-2009

3. DESIGN OF TENSION MEMBERS

3.1 INTRODUCTION

Tension members are encounters in most steel structures. They occur as principal
structural members in bridges and roof trusses, in truss structures such as transmission and
microwave towers and wind bracing systems in multi-story buildings. These members can
be defines as “ Members which carry only axial tensile forces”. Some examples of these
members are shown in figure (3.1) and their cross-sections are shown in figure (3.2).

Figure (3.1) Samples of Steel Structures Containing Tension Members

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                   1
Design of Steel Structures (3rd Civil)   New Cairo Academy                           2008-2009

Figure (3.2) Samples of Cross-Sections of Tension Members

3.2 TENSION MEMBER RESISTANCE:
The maximum resistance of a tension member “Tr” is calculated using the following
equation:

Tr = Anet . Ft                                                         (3-1)

where Anet is the effective net area of the cross section and Ft is the allowable stress of the steel
in tension which is defined by the Egyptian Code of Practice, ECP 2001, as:

Ft = 0.58 Fy                                                           (3-2)

3.2.1 Allowable Tensile Stresses for Steel

The allowable tensile strength for steel is calculated using equation (3-2) in which, Fy is the
nominal yield stress of steel. Table (3-1) presents the nominal yield stress and the nominal
ultimate stress for different steel grades used in structural design.

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                            2
Design of Steel Structures (3rd Civil)   New Cairo Academy                        2008-2009

Table (3-1) Nominal Yield Stresses and Ultimate Stresses for Different Steel Grades
Nominal Values of Yield Stress Fy and Ultimate Strength Fu
Thickness “t”
Grade               t ≤ 40 mm                     40 mm ≤ t ≤ 100mm
of Steel        Fy              Fu                Fy               Fu
t/cm2           t/cm2             t/cm2            t/cm2
St 37         2.40            3.60              2.15             3.40
St 44         2.80            4.40              2.55             4.10
St 52         3.60            5.20              3.35             4.90

According to table (3-1) and equation (3-2) , Ft can be taken from table (3-2):

Table (3-2) Allowable Tensile Stresses for Different Steel Grades
t ≤ 40 mm           40 mm ≤ t ≤ 100mm
St 37                1.40                    1.30
St 44                1.60                    1.50
St 52                2.10                     2.0

The above values for the allowable stress, Ft, may be increased by 20% if secondary stresses

3.2.2 Effective Net Area “Anet”

Whenever tension member is to be fastened by means of bolts or rivets, holes must be
provided at the connection. As a result, the member cross-sectional area at the connection
is reduced and the strength of the member may also be reduced depending on the size and
location of the holes. The effective area in resisting the tensile stress of the cross-section is
called the effective net area Anet.

The tension member resistance must be calculated based on the effective net area of the
cross-section Anet which can be calculated as follows:

i-      For Lined up Holes:

Whenever there is only one hole or multiple holes lined up ransverse to the loading
direction, there is only one potential line may exist. The controlling failure line is that
which passes through the bolt hole/ or holes (line AB in Figure).

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                        3
Design of Steel Structures (3rd Civil)   New Cairo Academy                    2008-2009

Anet = Agross – n (d+ 2 mm) t                            (3-3)

Where n = number of bolts in the cross section perpendicular to the direction of
t = the thickness of plate element at which the hole lies.
d = bolt dimeter
It is noted that the hole diameter exceeds the bolt diameter by 2 mm.

ii-     For Staggered Holes:
Whenever there is more thaone hole and the holes are not lined up transverse to the loading
direction, more than one potential line may exist. The controlling failure line is that which
gives the minimum area. In the opposite figure, there are two lines of staggered holes. The
failure line may be through one hole (section A-B), or it might be along a diagonal path
(section A-C).

Anet (Path A-B) = Agross – (d+2mm) . t                          (3-4)

Anet (Path A-C) = Agross – 2. (d+2mm) . t +( s2/4g)xt           (3-5)

The net area Anet is the minimum of the two values.
3.3 MEMBERS USED IN TRUSSES

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                    4
Design of Steel Structures (3rd Civil)              New Cairo Academy                            2008-2009

Unsymmetric Sections (Single Angles)
Used in Vertical and diagonal members of truss
Minimum radius of gyration is iv = 0.2 a

Symmetric Sections (Star-Shaped)
Used in Vertical members of truss at intersection with long. Bracing, and long tension members

Minimum radius of gyration is iu = 0.385 a

Symmetric Sections (two angles back-to-back)
Used in chord members of truss and web members of high values of force.

Minimum radius of gyration is ix = 0.3 a
Maximum radius of gyration is iy = 0.45 a

3.4 APPLICATION ON TRUSS MEMBERS WITH BOLTED CONNECTIONS

Truss members are considered symmetric sections when the center of gravity of the
members coincides with the center of gravity of the connecting gusset plate.

3.4.1 Symmetric Sections:

Sections composed of two angles back-to-back and two
angles star shaped (where there is no eccentricity at the
location of the C.G. of the member and the C.G. of the
gusset plate) are considered symmetric sections.

Net Area = Gross Area – Area of Bolt’s holes
Anet = Agross - AHoles

Where the diameter of the bolt hole = diameter of the bolt + 2 mm for drilled bolts.

3.4.2 Unsymmetric Sections (single angle)

In single angle truss members where there is eccentricity between the C.G. of the member
Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                5
Design of Steel Structures (3rd Civil)     New Cairo Academy                  2008-2009

and the C.G. of the gusset plate at the location of the connection (the system shown in
figure), the net area of the member is calculated as follows:

3.A1
Anet= A1 + A2 (             )                                        (3-6)
3 A1 + A2

Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg.

3.4.3 Unsymmetric sections (double angles).

The two angles are considered unsymmetric when they lie on one side of the gusset plate
as shown in figure. To allow for the eccentricity of the connection (the system shown in
figure), the net area of the member is calculated as follows:

5.A1
Anet = A1 + A2 (         )                                         (3-7)
5 A1 + A2
Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg.

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                    6
Design of Steel Structures (3rd Civil)   New Cairo Academy               2008-2009

3.5 APPLICATION ON TRUSS MEMBERS WITH WELDED CONNECTIONS

3.5.1 Symmetric Sections:

Anet = Agross

3.5.2 Unsymmetric sections ( e.g. single angle ):

3.A1
Anet = A1 + A2 (            )
3 A1 + A2
Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg
(there are no bolt holes to be deducted from areas A1 and A2).

3.5.3Unsymmetric Sections (Double Angles):

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                              7
Design of Steel Structures (3rd Civil)   New Cairo Academy                   2008-2009

3.A1
Anet = A1 + A2 (            )
3 A1 + A2
Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg
(there are no bolt holes to be deducted from areas A1 and A2).

3.6

STIFFNESS REQUIREMENTS FOR TENSION MEMBERS:

To avoid sag, vibration and slack of tension member the ECP, 2001defined:

3.6.1 The Maximum Slenderness Ratio

The maximum slenderness ratio of a tension member (excluding wires) :

1. Tension member in                                                 λ= leff / r
Buildings                                                                   300
Bridges:
Railway Bridges                                                     160
Vertical hungers                                                    300
Bracing system                                                      200

3.6.2 The length/depth ratio (Recommended Values)

Another stiffness requirements for tension members (excluding wires) is:
Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                  8
Design of Steel Structures (3rd Civil)        New Cairo Academy                             2008-2009

Tension member in                                       l/d
Buildings                                               60
Railway Bridges                                         30

3.7 DESIGN PROCEDURES:

Step1: Estimate the cross section:

Force
Bolted connections – Symmetric Sections Arequired =
0.85 Ft

Force
Bolted connections – Unsymmetric Sections Arequired =
0.85 x 0.85 x Ft
Force
Welded connections – Symmetric Sections Arequired =
Ft
Force
Welded connections – Unsymmetric Sections Arequired =
0.85 x Ft
Then choose a suitable cross-section

Step2. Check of the Chosen Member:

Force
1-Check of the actual Stress fact                                    f act =         ≤ 0.58 Fy
Anet
l
2-Check of the length top depth ratio             ≤ 60 (for diagonals and chord members only)
d

3-Check member stiffness                  λ max ≤ 300
l b −in             l
λ max is the maximum of λin =                    and λ out = b −out in case of two angles back-to back;
rx                  ry
lb − max
and λ max =            in case of two angles star-shaped
ru −1L

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                                   9
Design of Steel Structures (3rd Civil)   New Cairo Academy                  2008-2009

lb − max
and λ max =            in case of single angle sections.
rv −1L

4-minimum angle leg “a” shall be ≥ 3 d + t

Where d is the bolt diameter and t is the angle leg thickness.

Example 1:

Design the lower chord tension member “A” shown in the Figure. The force in the member
is 30t (Case II) and the bolts used in the connections are 16 non-pretensioned bolts.
Data: Force = 30 t (Case II ) Length = 3.00 m Lx= 3.0 m and Ly = 9.0 m

Estimation of the cross section:
Force
Areq =
0.85 Ft
Areq = 30 / ( 0.85 x 1.4 x 1.20 ) = 21.0 cm2
Choose 2 angles back-to-back for the lower chord. A of 1L = 21.0/2= 10.50 cm2
Knowing that for 2 angles back-to-back with equal legs:
rx=0.3 a and ry = 0.45 a
λx = Lx / (rx =0.3a)=300/0.3 a= 300 a req=3.33 cm
λy = Ly / (ry=0.45 a)= 900/(0.45x300) = 6.67 cm

For construction : a-t > 3d = 4.8 cm.

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                 10
Design of Steel Structures (3rd Civil)   New Cairo Academy                             2008-2009

From the tables choose 2Ls 80x80x8

Check:

1. Strength:
Anet = Agross - AHoles
Anet = 2 ( 12.3 - (1.6+0.2) * 0.8 ) = 21.72 cm2.
Fact = 30 / 21.72 = 1.38 t/cm2 < 1.4x1.2 ( case II ) =1.68 t/cm2
Safe

2. Stiffness:
l/d = 300 / 8 = 37.5                                                     < 60 OK
lx/rx = 300 / 2.42 = 124                                                < 300 OK
Iy = 2 ( Iy (1L) + A(1L) • { ey + 0.5 • tg. pl. }2 ) = 2 ( 72.3 + 12.3 ( 2.26+0.5x1.0 )2 ) = 332 cm4
A = 2 x 12.3 = 24.6 cm2
Iy        332
ry =            =          =3.67
A2 L       24.6
Ly/ry = 900 / 3.67 = 245                                                         < 300 OK

3. Construction requirement:
a = 8.0 cm > 3 d + t = 3x1.6+0.9 = 5.7 cm OK

Example 2:
Design the same member of Example 1 with welded connections.

Estimation of the cross section:

Areq = 30 / (1.4 x 1.2) = 17.86 cm2
Choose 2 angles back-to-back for the lower chord. A of 1L = 17.86/2 = 8.92 cm2

For 2 angles back-to-back with equal legs:
rx=0.3 a and ry=0.45 a
rx = Lx / (rx=0.3 a)=300/(0.3x300) = 3.33 cm
ry = Ly / (ry=0.45 a) = 900/(0.45x300) = 6.67 cm

From the tables choose 2Ls 70x70x7
Check:

1. Strength:
Anet = Agross = 2 x 9.4= 18.8 cm2.
Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                                11
Design of Steel Structures (3rd Civil)      New Cairo Academy                           2008-2009

fca = 30 / 18.8 = 1.595 t/cm2<           1.4x1.2 ( case II ) =1.68 t/cm2. Safe

2. Stiffness:
l/d = 300/7 = 42.85 < 60           O.K. Safe

l x 300
=         = 142.8 < 300 O.K.
rx     2 .1
t gusset 2                       1
I y = 2[ I y (1L ) + A1L (e y +         ) ] = 2[42.4 + 9.4(1.97 + ) 2 ] = 199.5 cm4
2                             2
2
A = 2x9.4 = 18.8 cm
Iy         199.5
ry        =             = 3.26
A           18.8
ly     900
=          = 276 < 300 O.K.
ry 3.26

Example 3:

Design the same tension member of Example 1 using unequal angles. Use                        16 non-
pretensioned bolts in all the truss connections and 10 mm gusset plates.

Data: Force = + 30 t (Case II), Member length = 3.0 m, Lin = 3.0 m, Lout=9.0 m

Estimation of the cross section:
Areq = 30 / ( 0.85 x 1.4 x 1.2 ) = 21.0 cm2.
Choose 2 angles back-to-back for the lower chord. Area of 1L = 21.0/2= 10.50 cm2

Choose 2 Ls back-to-back 65x100x7
(choose unequal angle and use the longer leg to add
stiffness for the member around the Y-Y axis as
shown in the Figure.

Check:
1. Strength:
Anet = 2 ( 11.2 - 1.8 x 0.7 ) = 19.88 cm2.
•       Fact = 30 / 19.88 = 1.51 t/cm2 < 1.4x1.20 t/cm2.                              Safe

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                             12
Design of Steel Structures (3rd Civil)    New Cairo Academy                 2008-2009

2. Stiffness:
•       l/d = 300 / 6.5 = 46.2                                       < 60 OK

•        lx/ry = 300 / 1.84 = 163                                   < 300 OK

Iy = 2 [ 113 + 11.2 { (1.0/2 + 3.23 }2] = 537.7 cm4
537.7
ry =             = 4.899
2 x11.2
ly/ry = 900 / 4.899 = 184                                          < 300 OK

3. Construction requirement:
a = 6.5 cm > 3d+t = 3x1.6+0.7 = 5.5 cm OK

Example 4
Design a diagonal member in a truss if the tensile force in the member is 6 t (Case I) and
the member length is 3.6 m. The bolts used in all the truss connections are M16 non-
pretensioned bolts grade 4.6, with 10 mm thick gusset plates.
Data: Force = + 6t (Case I)           Length = 3.6 m.

Estimation of the member cross section:

Assume the section is single angle (unsymmetric section)

Areq = 6.0 / ( 0.85 x 0.85 x 1.4 ) = 5.93 cm2

For 1 L: rv=0.2 a
L / (rv=0.2 a)=300                  a = 360/(0.2x300) = 6.0 cm

for construction : a > 3d+ t a-t > 4.8 cm.

To satisfy the above 3 requirements: choose 1L 60x60x6

Check:

1. Strength:

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                 13
Design of Steel Structures (3rd Civil)   New Cairo Academy                     2008-2009

A1 = 6.0x0.6 - (1.6+0.2)x0.6 = 2.52 cm2

A2 = (6.0-0.6)x0.6 = 3.24 cm2

Anet =A1 + A2{ 3A1 / (3A1 + A2 )} = 2.52 + 3.24 {3 x 2.52 / ( 3 x 2.52 + 3.24 )}
= 4.79 cm2

•        Fact = 6 / 4.79 = 1.25 t/cm2       < 0.58 fy = 1.40 t/cm2           Safe

2. Stiffness

•        l/d = 360 / 6.0 = 60                                                OK

•        l/rv = 360 / 1.17 = 308> 300                                     unsafe

Use L 65x65x7 and recheck stiffness only:

l/rv = 360 / 1.26 = 286         < 300                                        OK

Example 5
Design the shown tension member if the length of the member is 6.0 m and it carries a
force of 6 t. The bolts used in the connections are M16 non-pretensioned bolts with 10 mm
gusset plates.

Estimation of the member cross section:

Areq = 6/ ( 0.85 x 1.4 ) = 5.04 cm2
(Assuming that the cross section will be symmetric about the guest plate: Star-shaped
cross-section)
Prof. Ahmed Abdel-Salam Ismail El-Serwi                                            14
Design of Steel Structures (3rd Civil)   New Cairo Academy                   2008-2009

A of one angle = 5.04/2 = 2.5 cm2
For star-shaped angle: ru=0.385a
a = L / 0.385 a < 300
a > 600/0.385x300 = 5.2 cm

L/d < 60 , d> 600/60 = 10 cm (d=2.a + tG)

Construction requirements:
a-t > 4.8 cm.

From table choose 2 Ls star-shaped 55x55x5.

Check:

1.Strength:
Anet = 2 ( 5.32 - (1.6+0.2)x0.5) = 8.84 cm2
•      Fact = 6/8.84 = 0.67 t/cm2< 0.58 fy t/cm2 safe.

2. Stiffness:
l/d = 600 / ( 2x5.5+1.0 ) = 50       > 60        OK
l/ru = 600 /2.09 = 287         < 300 OK

Example 6:

Design the lower chord tension member “A” shown in the Figure using circular hollow
section (tube). The force in the member is 30t (Case II) and the member is connected to the
truss members by using weld.
Data: Force = 30 t (Case II ) Length = 3.00 m Lx= 3.0 m and Ly = 9.0 m
Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                  15
Design of Steel Structures (3rd Civil)   New Cairo Academy                         2008-2009

Estimation of the cross section:
Force                                                               D=133 mm
Areq =
Ft
Areq = 30 / (1.4 x 1.20 ) = 17.86 cm2
Choose CHS 133x5
t=5 mm
A = 20.1 cm2
rx= ry = 4.53 cm

Check:

2. Strength:
Agross= 20.1 cm2
fact = 30 / 20.1 = 1.49 t/cm2 < 1.4x1.2 ( case II ) =1.68 t/cm2
Safe

2. Stiffness:
l/d = 300 / 13.3 = 22.55                                           < 60 OK
lx/rx = 300 / 4.53 = 66.23                                        < 300 OK

Prof. Ahmed Abdel-Salam Ismail El-Serwi                                                        16

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