3. Determine the LRFD design tensile strength and the ASD allowable tensile strength for an A36 (Fy= = 36 ksi and Fu = 58 ksi) L6 x 6 x 3/8 inches that is connected at its ends with one line of four 7/8-in-diameter bolts in standard holes 3 inches on center in one leg of the angle. Note: L6 x 6 x 3/8 (Ag = 4.38 in²; y = x=1.62 in)
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- Material 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|YISA 100 x 75 x 10 mm (cross-sectional area = 1650 mm2) Serves as a tensile member. This angle is welded to a gusset plate along A and B appropriately as shown. Assuming the yield strength of the steel to be 260 N/mm². ISA (100 × 75 × 10 mm) B Gusset base With reference to the information given, Determine Area of connected leg Area of outstanding leg i Net effective area of the memberThe butt connection shows 8-22 mm diameter bolts spaced as shown below. P- 50 100 50 50 100 50 16 mm +HHHH 40 80 40 12 mm Steel strength and stresses are: Yield strength, Fy = 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Based on the gross area of the plate. Based on the net area of the plate. Based on block shear strength. Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions:
- Determine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. X = 0.7 Y = 3.8 Z=1/2 Round your answer to 2 decimal places. Your Answer: Incorrect The answer is 77.15 ± 1%. 1/2" X" bolts 1½" Gusset plate -L3 X 3 X Z Tu IA 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 LLBBA 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 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 LLBBA) The 3/4 in x 15 in plate has 4 rows (or horizontal lines) of bolts. Each row has 3 bolts. The bolt diameter is 7/8 in. Determine the gross area (Ag) and the net area (An) 3/4 15 O O O O O O Q O O 3" 3 27 3 784 Bolts Pet *Determine the allowable (ASD) tensile strength (kN) of a WT205X74.5 having two lines of M22 bolts as shown. There are four balts in each line, 75 mm on center. The yield strength of the steel is 345 MPa and the ultimate tensile strength is 483 MPa. Neglect the block shear strength.
- 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 VUZMGLThe 200 mm x 150 mm x 12/5 mm angle has one line of 20 mm diameter bolts in each leg. The bolts are 75 mm on center in each line and are staggered 37.5mm with respect to each other. Fy 248 MPa, Fu-400 MPa. Standard nominal hole diameter of 20 mm bolt is 21mm. Area of the angular section= 4355 mm2. Reduction coefficient U-0.75 CIVIL ENGINEERING STEEL DESIGN Compute The following: a) The effective net area b.)The design strength for LRFD. c) The allowable strength for ASD. 200 75 (1125 67.5 150 1125 1125The diagonal at the left to the connection is a double angle 90 mm x 90 mm x 8 mm, with area of 2700 mm?, bolted to the 8 mm thick gusset plate. Bolt diameter = 16 mm Bolt hole diameter = 18 mm Bolt bearing capacity, Fp = 480 MPa Bolt shear strength, Fv = 68 MPa Steel plate strength and stresses are as follows: Yield strength, Fy 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 0.60 Fy Allowable tensile stress on the net area = %3D 0.50 Fu Allowable shear stress on the net area = 0.30 Fu Bolt bearing capacity, Fp = 1.2 Fy Calculate the allowable tensile load, P(kN) under the following conditions: %3D