Homework 10-1: Compute the net area of each of the given members. Use standard- size bolt holes for all problems. (A = 1.93 in²) 7/8 in bolts 7/8 in bolts L4X4 X14 Bolts are 7/8" 7
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- A) 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 *Calculate the critical net area (mm²) for the tension member shown. The bolts are M22. 40 65 90 115 90 65 PL22X360The bolted connection shown in Figures is connected with 3/4-in-diameter bolts in standard holes. The plate material is A36 steel. By using strength level design (LRFD), find the available tensile strength of the plates. u=0.8 (5)-¾" bolts 4" C12× 20.7 | 4½" ,3" , 4½" |
- 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 VUZMGLFor the two lines of M20 bolts shown in the figure, determine the design tensile strength (kN) of the plate if it is 12mm thick. Use the pitch (s) that gives it a net area DEFG equal to the one along ABC. D 50 3. 50 50 IG ic Pitch PitchDetermine the effective net area of a W 360 x 91 shown. The holes were punched for M20 bolts. 65 mm 0000 P O O DO 65 mm 60 85 85 85 mm mm mm mm
- For the cross section shown, calculate the tensile capacity of the steel plate. Assume that the bolt hole is 1/16 inch larger than bolt diameter. Plate 1/2" x 11" .T 7/8" dia. bolt, typ. ASTM A325NDetermine 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 IFind øR, considering bolt limit states only. (S275 Steel, fy = 2804 kgf/cm2, fu = 4385 kgf/cm2, Bearing type connection 8.8 M20 bolts, ti = 15 mm, t2 = 18 mm) PE >P 50 mm 80 mm 80 mm 50 mm
- Example 3.9 FIGURE 3.20 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. 312" 23/4" S15 × 50 1½" 1/2" b e la C dProblem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts T12 mm te12 mm P 300 75 75 75 Edge distances = 75 mm %3D dnole for tensile and rupture = dp +3 mm %3D dnole for bearing strength for Lec = d + 1.5 mm Fy = 248 MPa Fu = 400 MPa %3D En 330 MPa %3D Use LRFD design method. Determine the design strength due to tensile rupture strength of plates. (kN)The plate shown in the figure 2.19 has a thickness of 12 mm. Diameter of bolts is 16 mm. A 36 steel is used with Fy= 248 MPa and F= 400 MPa. Standard nominal hole diam. of 16 mm bolts = 17 mm. 12 mm 75 mm 75 mm 75 mm 75 mm Figure 2.19 T 300 mm Compute the minimum pitch "S" for which only two and one half bolts need b. subtracted at any one section in calculating the net area. Compute the effective net area of the section if the reduction factor u = 0.75. Compute the allowable tensile strength of this section.