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 PL22X360Find 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 VUZMGL
- The 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" |Determine 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 mmDetermine 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 I
- 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 A325NFind 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 mmExample 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 d
- Problem. 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)Compute the net area of each tension member. Note that for a standard-size bolt hole, hole dia = bolt dia (in mm) + 3mm or hole dia. = bolt dia (in inches) + 1/8in. 1. PL 25 x 300 mm 25 mm Ø bolts 2. 19 mmØ bolts W12 x 40 - 3. An L8 x 4 x % with two lines of %-in dia. bolts in the long leg and one line of - in dia. bolts in the short leg.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.