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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 (f) Section is adequate. 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 member are 50 kips dead load + 30 kips live load. 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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