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)
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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)
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- 3. Determine Rn based on bolt limit states. Slip critical connection is used for design and Grade 10.9 M22 Bolts with Class A surface. Check minimum edge distance and spacing with TSSDC 2016. All members are $355. There is no filler between plates and Du=1.0. Fy= 355 MPa; Fu=510 MPa; E=200000 MPa. All units are in mm. $R.=? 100 100 O 60 Geset Plate 75 Plate Gusset Plate $R.=? *20*201 PlateExample 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 dA) 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 *
- 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 Net Area for the following Plate Section consider all possible Cases. The Plate thickness is 1-in. and the bolts' diameter is 7/8-in. s = 6.0 " g = 2.0 "The 1 × 8 plate is shown in the figure. The holes are for 3/4-in Ø bolts. Compute the net area of each of the given members. (Ans. 6.44 in^2)
- A bolted connection shown is made up of 2L150 x 100 x 9.5 mm. The angle is made of A-36 steel and the bolts have diameter of 22 mm. Properties of 2L150 x 100 x 9.5 mm: A = 4660 mm2 t = 9.5 mm Fy = 248 MPa Fu = 400 MPa Properties of 20 mm A-325 bolts: Fu = 400 MPa 50 50 50 50 50 50,50 힝힝이어 Gusset Plate 9.5mm 100mm Gusset Plate 100mm 31.25 62.5 T 56.25 a. Nominal dimension of hole diameter, dn = 25 mm b. Determine the ultimate shear capacity (kN) of the connection. c. Determine the ultimate tearing capacity (kN) of the connection if the shear lag factor, U = 0.75.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 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.
- A miscellaneous channel MC 300 x 67 is bolted as shown on the figure 2.20 by a 0 mm diam. bolts. A 36 steel is used with Fy= 248 MPa and F₂ = 400 MPa. Standard ominal hole diameter for 20 mm bolt = 21 mm. " Prof. of MC 300 x 67 A = 8250mm² d = 300 mm tw= 17.8 mm ty= 17.5 mm by=100.3 mm 17.8 m 231 23 231 62 5 87.5 187.5 50 50 50 50 50 91910 91919 O e Figure 2.20 Compute the net area of the channel section. Compute the effective net area of the channel section. Compute the allowable tensile strength of the channel section. CIVIL ENGINEERING-STEEL DESIGN18. Determine the net width for a 12×1/2 in. plate with 3/4 in. bolts placed in three lines as shown in Figure P4.18. P4.18 O O O 3 in 4 in.3 in. 2 in. 4 in. 4 in. 2 in. 12 in.Hhello, Thanks a lot for your help, please I have the following points: 1-Is it allowed to use Fult=65 ksi? 2-Why we consider Wt section? this procedure as I understand is applied when the connection is to flange. 3-The net area estimated is only for one bolt, i believe An=11.80-(4*7/8*0.305)=10.733 inch^2. 4-x bar which I consider is from case 2 only since case 7 will not work, x bar is only Tw/2 value?. U value=1-(0.1525/6.50)=0.977. Could you please advise? Thanks.