# Problem No

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```					CE 470 – Mid-term Exam                                     Student Name: ____________________________

Problem No. 1
Total points = 50
Determine the design tensile strength of the C10 x 25 channel section made from A36 steel. The
member end connection is made using bolts that are ¾ in. in diameter made from A325 steel.
The bolt connection layout is shown below.
(1) Determine the design tensile strength:                                                        (45 pts)

(a) Consider only the limit states of gross yielding and net section fracture.
(b) For net section fracture, check all fracture paths (at least 3) you deem appropriate.
(c) Calculate U (shear lag factor) using the equation U=1-x/L≤0.9
(2) Answer the following questions qualitatively, i.e., without any calculations.                  (5 pts)

(a) What will be the block shear failure path for the member?
(b) Can the block shear strength be computed using the standard AISC equations, or will
you need to modify them? If so, How?

2”    3” 2.5”

2”

3”

3”
C10x25
2”

Solution
2
Fy  36 ksi    Fu  58 ksi           db  0.75 in       Ag  7.34    in    t w  0.526 in

Part 1

Gross Yielding    Pny  0.9 Fy  Ag            Pny  237.816 kips

x
x  0.617 in   L  5.5    in        U  1          U  0.888        Less than 0.9
L

1
CE 470 – Mid-term Exam                                      Student Name: ____________________________

2”   3” 2.5”

2”
Fracture Path
3”

3”
C10x25
2”

Net Fracture Case 2

An  Ag  3  db 
1                             2                                   2
           t w           An  5.959 in          Ae  U An   Ae  5.291 in
         8

7

Pnf2  0.75Fu  Ae             Pnf2  322.205kips
5

2”   3” 2.5”

2”
Fracture Path
3”

3”
C10x25
2”

2”   3” 2.5”

2”                                                                                            Af/2
Fracture Path
3”                                                                        T
Aredg
3”
C10x25
2”                                                                                        Af/2

2
CE 470 – Mid-term Exam                                              Student Name: ____________________________

Net Fracture Case 4                  T  8 in    ( AISC)      s  2.5 in   g  3 in
Aredg  T t w                     Aredg  4.208       in
Af  Ag  Aredg                    Af  3.132     in
2
wn  T  3  db    1
1       s
                                   wn  5.896      in
     8      4 g
2                                   2
An  Af  wn  t w                  An  6.233 in            Ae  U An    Ae  5.534 in

7

Pnf3  0.75Fu  Ae                   Pnf3  280.848kips
6

2”       3” 2.5”

2”                                                                                                    Af/2
Fracture Path
3”                                                                               T
Aredg
3”
C10x25
2”                                                                                                   Af/2

Part 2
By examining the results in Part 1, the block shear failure path would be as follows:
2”       3” 2.5”

2”
Fracture Path
3”

3”
C10x25
2”

In this case, no modifications to the AISC equation is required. However, if the results in Part 1
indicate that Path 2 governs, then the AISC equations would have to be modified as follows:

         s
2
Agt  t w  6  2        
         4 g 

An t  Agt   db 
1
                 t w
            8

Ultimately, the equation for the gross area is computed as follows:
       s2 
 g    Where g is the vertical spacing.
Agt  t w 
       4g 


3
CE 470 – Mid-term Exam                                 Student Name: ____________________________

Problem No. 2
Total points=50
(a) Using the tension member design Tables in the AISC manual, select a double angle tension
member (with equal legs) to carry a factored design load of 150 kips.
(b) Is the net section fracture strength noted in Table 3-7 correct? If not, then where does the
approximation appear in this Table?
(c) Assuming ¾ in. diameter A325 bolts in shear/bearing for the connection system, how many
such bolts do you need for the connection?
(d) Assuming Le=2 in., s=3.0 in., sketch a reasonable bolt layout, what is the actual net section
fracture design strength?
(e) What is the block shear strength assuming that the shear fracture strength dominates, i.e., Rn
= 0.75 [0.6FuAnv + FyAgt] ≤ 0.75 [0.6FuAnv + FuAnt]
(f) Is the original selected member still valid? Do not re-design, just state what needs to be
changed to make it valid.

Solution
(a) The most economical section from Table 3-7 is a 2L3½ x 3 x 3/8 Py = 150 k, Pn = 151 k
(b) When Ae < 0.75 Ag, tabulated values of design tension rupture cannot be used. Tables
assume Ae = 0.75 Ag

(c) For bearing strength, assuming Le  Lefull and s  sfull Rn = 78.3 kips/in (Table 7-12)
Rn = 78.3 kips/in
Pu

Rn  78.3t 2 Rn  58.725        kips of 1 bolt         n           n  2.554 bolts
Rn

Pu
Shear Strength 1 bolt:     Rn  31.8 kips (Table 7-10)     n           n  4.717 bolts
Rn
Make      n  5 bolts
Bearing                
Rn  58.725n      Rn  293.625 kips
Shear                
Rn  31.8n        Rn  159       kips

4
CE 470 – Mid-term Exam                                               Student Name: ____________________________

d) Bolt layout shown below. Path shown for block shear in Part (e).

2”        3”          3”             3”        3”        2”                  2”

2
Ag  2.32     in

An  Ag  
1        3
              t         An  1.992 in
2
An  2 An
2
An  3.984 in
8        4                                      2 Angles
x
x  0.823 in           L  15 in            U  1            U  0.945        U  0.9
L
2
Ae  An  U        Ae  3.585 in


Pn  0.75Fu  Ae                 Pn  155.964 kips

(e) Path shown above in (d).

An v  t  14  4.5 
1       3 
                                    An v  3.773 in
              8        4 

Agt  t 2                                         Agt  0.75      in
An t  t  2  0.5      
1 3
            8 4                    An t  0.586 in
                 
Fy  Agt  27 kips             Fu  An t  33.984 kips           Use Equation Below :

Rn  0.75 ( 0.6 Fu  An v  Fy  Agt )           Rn  118.737 kips

2 Angles        2 Rn  237.473 kips

5
CE 470 – Mid-term Exam                                             Student Name: ____________________________

Problem No. 3
(Total points=50)
Design the welded connection (only) for the member shown below. The given loads are service
loads. Assume Fy=50 ksi for the angle tension member, and Fy=36 ksi for the gusset plate. Show
the connection design on a sketch complete with all dimensions.

Note: You will have to consider both (angle and gusset) base metal strengths to compare with the
fillet weld strength.

t=3/8 in.

D = 45 kips
L 5 x 5 x 5/16
L = 136 kips

5                    3
Fya  50 ksi         Fyg  36 ksi        D  45 kips      L  136 kips       t a         in     t g        in
16                   8
Pu  1.2 D  1.6 L       Pu  271.6 kips

5                     3                           1
t min          in    t max        in   amax  t min          amax  0.25    in
16                    8                           16
3
amin           in      Table J2.4 Make          a  0.25 in         Fexx 70 ksi
16
Strength Weld Metal                                                         L
Rn ( Lw)  0.750.6 Fexx 0.707a Lw  5.56762500 w
 Lw
Base Strenght Gusset Rn ( Lw)  0.9 0.6 Fyg  t g Lw  7.29000000000 00000000
Base Strenght Angle                                                                     Lw
Rn ( Lw)  0.9 0.6 Fya t a Lw  8.43750000000 00000000
Strength Weld Metal Governs
271.6
Lw                      Lw  48.782 in
5.56762500
Lwmin  4 a  1.00        in
Lwmax 100 a  25.00 in              Theref ore, in one strip, the eff ective length of the
w eld can't be greater than 25 in.

6
CE 470 – Mid-term Exam         Student Name: ____________________________

in.
t=3/822 in.

D = 45
5 in.
L 5 x 5 x 5/16
L = 136
22 in.

in.
t=3/825 in.
0.5 in.
D = 45
L 5 x 5 x 5/16
0.5 in.                                                                    L = 136
25 in.

t=3/8 in. in.
0.25 in. 25

D = 45
L 5 x 5 x 5/16
L = 136
0.25 in.          25 in.

7
CE 470 – Mid-term Exam                                Student Name: ____________________________

Problem No. 4.
Total points = 50
Design a bolted splice connection for a single angle tension member. The service loads acting on
The splice must be slip-critical for service loads, and have adequate shear/bearing strength for
factored loads. Assume 7/8 in. diameter A325 bolts.
(1) Design and check the splice to be slip-critical at service loads
(2) Design and check the shear/bearing strength of splice at ultimate loads.
(3) Design the thickness and width of the gusset plate
(4) Provide a sketch of the splice detail.
There is no need to check the design strength of the angle at ultimate loads. Assume that the
angle design works.

splice plate

L 4 x 3.5 x 1/2                                   L 4 x 3.5 x 1/2

Solution
7
D  50 kips   L  30 kips    db        in   t  0.5 in
8
Part 1
SL  D  L  80   kips
Rn  10.2 kips   For Slip Critical, One Bolt f rom Table 7-16   Use 8 bolts    n  8


Rn  10.2n   Rn  81.6 kips

8
CE 470 – Mid-term Exam                                                 Student Name: ____________________________

Part 2
Pu  1.2 D  1.6 L         Pu  108         kips
Shear (Table 7-10)                   
Rn  21.6n             Rn  172.8 kips ok

Bearing
Lemin  1.125 Table J3.4                 Make Le  1.25 in

smin  2.6667db          smin  2.333 in
spref  3 db            spref  2.625 in             Use     s  3 in

For Edge:                    
Rn  40.8t          Rn  20.4 kips
For Interior Spacing:                   
Rn  91.3t          Rn  45.65 kips

Rn  45.657  20.4 Rn  339.95 kips ok
Note Assume Gusset Plate has same thickness

Part 3
Make U=0.85            From AISC 16.1-178             U  0.85
Pu                              2
Ag                  Ag  3.333 in
0.9 36
Ag
width              width  6.667 in              Make      b  7          in
0.5
Ag  b  0.5        Ag  3.5       in

An  Ag   db 
1
                0.5
       8
Gross Yielding          Py  Ag  0.9 36     Py  113.4            kips

Net Fracture                     
Pn  0.75U An  58         Pn  110.925 kips
Part 4

splice plate

L 4 x 3.5 x 3” 3”
1.25”
3”
1/2                   3”     3” 3”       3”          1.25”
L 4 x 3.5 x 1/2

9

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