Determine the LRFD design strength and the ASD allowable str block shear. A36 steel and 7/8 in diameter bolts in in -L6×3/1×/
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- 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.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. 132A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. Ae 1 2 Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN AL Section 2L5 x 3 x 516 LLBB
- A 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 LLBBDetermine the LRFD design strength and the ASD allowable strength. Neglect block shear. A36 steel and 7/8-in Ø bolts N VA N SZ 2-MC 12 x 40 00 K All 2 in 2 in 3-in 3/in 2/in4. An UPE 300 is to be used as a tension member and it is bolted to a 10mm thick gusset plate with bearing type 8-M20 8.8 bolts as shown in the figure below. a. Check all spacing and edge distance requirements. b. Compute the allowable tensile strength of the channel UPE 300. Don't consider shear strength of the bolts and bearing strength of the bolt holes. [all dimensions are in mm] Material both for gusset plate and UPE 300: S275 F-275N/mm² Fu=430N/mm² UPE 300 Ag = 5660 mm² tw 9.5 mm x = 28.9 mm 70 160 70 + 10mm thick gusset plate +50+ 80 -80-80-50 o +50+80 -80 80-50 UPE 300
- Plate A, PL 16mm x 375mm, is connected to another plate B with 10- 22 mm o bolts as shown. The steel is A36 steel with Fy= 250 MPa and F,= 400 MPa. Assume that plate B has adequate strength. Use ASD. Plate B 375a 75 m 2 75mm Plate A 75 mm 75mm 150 50150 Compute the nominal strength based on yielding in KN.O Determine the allowable tensile strength of Plate A in KN.O Compute the allowable tensile strength at section a-1-2-2-4-c in KN.03. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mmCivil Engineering Base plate bending capacity for tension in 310UB 40.4 column for base plate size 500X250X25 in grade 250 steel, 4-M30 4.6/S holding bolts with hole diameter 33 mm, D=pitch 300 mm, gauge distance C=100 mm design data. Yield strength of plate is 250 MPa 4-BOLT PIN N
- 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 2L5 x 3 x16 LLBB Determine the design tensile strength for LRFD. Select one: а. 78 b. 66 С. 132 d. 1561. 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…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"