Find CD, EF, AF, AB 15 ft E F B CO с 15 kips H OD +12 ft 12 ft-- 12 ft-12 ft-12 ft- I'll edit this later 2 in compression, 2 in tension, 3 are 36 kips, 1 is 45 kips
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- Determine the nominal strength of a A36 steel column (Fy = 36 ksi), W 14 X 68 with a total length of 50 feet (both pin-end connection). W14x68 A = 20 in.^2 d = 14 in. tw = 0.415 in. bf = 10 in. tf = 0.72 in. T = 10-7/8 in. k = 1.31 in. k1 = 1.0625 in. gage = 5-1/2 in. rt = 2.71 in. d/Af = 1.94 Ix = 722 in.^4 Sx = 103 in.^3 rx = 6.01 in. Iy = 121 in.^4 Sy = 24.2 in.^3 ry = 2.46 in.Plz do not use gpt if u use chatgpt i will downvote you again don't use chatgptA W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 7 m long, pinned at the top fixed at bottom, and has additional support in the weak direction 3 m from the top. Properties of the section are as follows: A = 14,500 mm^2 Ix = 258.6 x 10^6 mm^4 Iy = 84.375 x 10^6 mm^4 1. Calculate the critical slenderness ratio of the member. a. 26.208 b. 45.882 c. 36.706 d. 27.529 2. calculate the nominal axial load capacity of the column a. 3104 kN b. 4851 kN c. 4213 kN d. 5344 kN 3. calculalte the service axial dead load if the service axial live load is twice as that of the dead load. Use LRFD. a. 1354 kN b. 992 kN c. 1093 kN d. 634 kN
- A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 8 m long, pinned at the top fixed at bottom with additional lateral support at mid height in the weak direction. The properties are as follows: Ag = 14,500 sq.mm, Ix= 258.6 x10^6 mm ^4 Iy=84.375 x 10^6 mm^4. 57. Calculate the flexural buckling stress Fcr in MPaEquivalent:G = 6M = 4L = 8Equivalent G = 6M = 4L = 8
- The solid 34 mm diameter steel [E = 200 GPa] shaft supports two belt pulleys. For the loading shown, determine: (a) the shaft deflection at pulley B. (b) the shaft deflection at pulley C. Assume P = 810 N, Q = 480 N, LAB = 260 mm, LBC = 400 mmf. Calculate the maximum safe load "P" in KN Two A36 16mm Thick Steel Plates Are Connected By Four Rivets With Fv=152 MPa As Shown. Given: Rivet Dia 20 mm x=100 mmA W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 7 m long, pinned at the top fixed at bottom, and has additional support in the weak direction 3 m from the top. Properties of the section are as follows: A = 14,500 mm^2 Ix = 258.6 x 10^6 mm^4 Iy = 84.375 x 10^6 mm^4 1. calculate the nominal axial load capacity of the column a. 3104 kN b. 4851 kN c. 4213 kN d. 5344 kN 2. calculalte the service axial dead load if the service axial live load is twice as that of the dead load. Use LRFD. a. 1354 kN b. 992 kN c. 1093 kN d. 634 kN
- Steel Design Two channels having the given properties shown is placed at a distance of 300 mm to back and is properly connected by a pair of lacings as shown. Properties of one channel A = 5595 mm2 d = 305 mm x = 17mm Ix = 67.3 x 106 mm4 Iy = 2.12 x 106 mm4 rx = 19.3 mm Assume K = 1.0 Determine the safe axial load in kN, that the column section could carry. Unsupported height of column is 6m.A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 8 m long, pinned at the top fixed at bottom with additional lateral support at mid height in the weak direction. The properties are as follows: Ag = 14,500 sq.mm, Ix= 258.6 x10^6 mm ^4 Iy=84.375 x 10^6 mm^4. 59. If the columns sustains a service axial dead load of 1 490 KN Calculate the safe service axial live load in KN by ASD.A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 8 m long, pinned at the top fixed at bottom with additional lateral support at mid height in the weak direction. The properties are as follows: Ag = 14,500 sq.mm, Ix= 258.6 x10^6 mm ^4 Iy=84.375 x 10^6 mm^4. 60. If the columns sustains a service axial live load of 990 KN calculate the safe service axial dead load in KN by LRFD