Find the available strength of the S-shape shown in Figure 3.20. The holes are for 3/4-inch-diameter bolts. Use A36 steel. 31/2" 23/4" S15 × 50 1½" 1/2" Ma .b C e d
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- A WT 8×13 of A992 steel is used as a tension member. The connection is with 7/8-in. diameter bolts as shown in Figure. Compute the nominal block shear strength. Note: A992 steel Fy = 50 ksi Fu = 65 ksi Ubs = 1.0 %3D 12"+3">|| 42" 7.85" 0.25" Round your answer to 3 decimal places.Find the available strength of the S-shape shown in the figure. The holes are for 3/4-inch- diameter bolts. Use A36 steel. Use LRFD Use ASD (1 (2) 3½" I S15 × 50 $23/4" CIVIL ENGINEERING- STEEL DESIGN 1½" 12" e b С d Answer VUZMGLA double-angle shape is shown. The steel is A36, and the holes are for 12 mm diameter bolts. Assume that A, = 0.75A.. a. Determine the design tensile strength for LRFD. b. Determine the allowable strength for ASD. Ag = 1560 mm² (Angle) 2L 125 x 75 x 8 LLBB
- A Channel C6x13 is used as a tension member. The holes are for 5/8 inch in diameter of bolts. Using A36 steel with Fy=36 ksi and Fu=58 ksi. The gross area of C6x13 is 3.81 in? and Assume U =0.85 Using LRFD what is the design strength based on Tensile Rupture/Fracture 4 @ 2" a 1 1/2 Ob 3" Oc 1 1/2 " I d e C6 x 13 Round your answer to 3 decimal places.Compute the tensile capacity for the ½ inch thick plate shown. The bolts are ¾ inch diameter high strength bolts. The steel is A36. 24" -P, 1 3"A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 215 x 3 x %6 LLBB Determine the design tensile strength for LRFD. Select one: а. 66 b. 156 c. 78 d. 132
- For the bolted tension member is shown in the figure. Determine the effective area, Ae. Given, x = 1.68 inch, Ag = 5.75 inch?. (4)- S/8" A325N bolts L6 x 6 x 1/2 -Tu 3" 3" 3"A PL % x 7 tension member is connected with three 1-inch-diameter bolts, as show in Figure P3.2-1. The steel is A36. Assume that A, = A, and compute the following. a. The design strength , Pu based on gross area ... b. The design strength, Pu based on effective area PL x 7 3/8" O O 7" Cross Sectional area of PL3/8x7 Note: (Fy = 36ksi, Fu = 58ksi) a) Blank 1 b) Blank 2Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|Y
- A single-angle tension member, an L3 12 x 3 12 x 38, is connected to a gusset plate with 7/8-inch-diameter bolts as shown in the figure. A36 steel is used. Fy = 50 ksi, Fu = 58 ksi, E = 29000 ksi. Ag = 2.50 in^2. Assume that the effective net area is 85% of the computed net area. ... O O O L3/2 x 3/2 x 3/8 Section Determine the following: a) Design strength based on gross area (kips). Blank 1 b) Design strength based on net area (kips). Blank 21. The tension member shown in Figure 3.4-2 is a PL 58 x 10, and the steel is A36. The bolts are 3/4-inch in diameter. a. Draw the different potential failure lines. b. Compute the effective net area. 22" 2½" 2" 4 3" *₁ 3" ✓ 2" 2. A double-channel shape, 2C10 x 20 steel, Fy = 50 ksi and Fu = 70 ksi, is used for a built-up tension member as shown in the figure. The holes are for 12-inch-diameter bolts. PNA a. Determine the design strength for LRFD. b. Determine the allowable strength for ASD. YD Shape A 2" 2" -X Area, Depth, d | 3²²-|-1 3²² | 00 "There is no elevator to success, you have to take the stairs." - Zig Ziglar God bless. Ⓒ O O O Web Thickness, tw in. in. O 4" C10-30 x25 x20 5.87 10.0 10 0.379 % x15.3 4.48 10.0 10 0.240 4 4" Table 1-5 C-Shapes Dimensions Flange Width, b₁ 2 in. in. in.2 in. C15x50 14.7 15.0 15 0.716 1168 3.72 3% 0.650 % x40 11.8 15.0 15 0.520 x33.9 10.0 15.0 15 0.400 % 14 3.52 32 0.650 % %/16 3.40 3% 0.650 % C12-30 8.81 12.0 12 0.510 4 3.17 3% x25 7.34 12.0 12…A double-angle shape is shown in Figure 3.5. The steel is A36, and the holes are for ½/2-inch-diameter bolts. Assume that A₂ = 0.75A. a. Determine the design tensile strength for LRFD. b. Determine the allowable strength for ASD. Section 215 x 3 x 16 LLBB