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- F1-18 F2-18 F3-27 F4-11Compute the compressive strength of a WT 12x81 of A992 steel. The effective length with respect to the x-axis is 25 feet 7inches, the effective length with respect to the y axis is 20feet, and the effective length with respect to the z-axis is 20 feet WT12X81 A 23.9 t₁ 1.22 ly 221 d 12.5 k des 1.72 J 9.22 bf 13.0 y 2.70 Cw 43.8 tw 0.705 293Use A36 steel and compute the nominal strength of the column shown. The member ends are fixed in all directions (x, y, and z). use recommended value of k. A 6.08 tw 0.282 b/t 5.87 ly 3.86 ro 4.93 no 11.5 C12X20.7 d 12.0 tf 0.501 h/t w 36.3 J 0.369 H 0.899 bf 2.94 K des 1.13 1x 129 Cw 112 rts 0.983 10' C12 x 20.7
- Use A36 steel and compute the nominal strength of the column shown. The member ends are fixed in all directions (x, y, and z). use recommended value of k. A 6.08 tw 0.282 b/t 5.87 ly 3.86 ro 4.93 ho 11.5 C12X20.7 d 12.0 tf 0.501 h/t w 36.3 J 0.369 H 0.899 bf 2.94 K des 1.13 1x 129 Cw 112 0.983 10' C12 x 20.7H.W: draw the I.L for RA, RC, RF, RD, MD, VCL, VG and MG. A 2 m B -2 m C G 0.4 m 1.6 m D 2 m- E 2 m LLIntegrate the following: ,S (xª – 5x³ + 6) dx - 1. 6. su(1 – dx 2.
- Structural Analysis: Solve the F4-5AW14×142 is used as a column having a length of 9m long. It is hinged at the upper end and fixed at the lower end but there is a lateral bracing perpendicular to the minor axis of the W section at a point 5.4m above the bottom support. It is assumed to be pinned connected at the bracing point. Using A36 steel F, = 248MP and the NSCP Specifications. E, = 200000MP Properties of the W14X142 A = 26967.69mm² d = 374.65mm by t, = 27.00mm Sy = 1396.18 × 10³mm³ r, = 160.53mm ry = 100.84mm tw = 17.27mm I, = 695.11 × 10ʻmm* %3D = 393.70mm Iy = 274.71 x 10ómm* Sz = 3719.86 × 10³mm³ Assume K, = 0.8 for 9.0m length K, = 1.9 for the 3.6m length K, = 0.80 for the 5.4m length a. Compute the critical slenderness ratio b. Compute the allowable stress F. c. Compute the allowable critical load that the column can carry1) Column AB is made of solid 410, cold worked Steel with the following specification Sy=85 KSI, E-29000 KSI Length: 9.5 in. Diameter: 425 in. Ends: Free & Fixed For the above column calculate: a) Type of column failure (show proof) b) The critical load that would cause failure. c) Allowable force with Factor of Safety of 2.5. Answer Unit Value a 13 14 14
- Calculate the axial load carrying capacity of the welded column section as shown. Column is fixed at both the ends and it is of length 5 m. Use Fe410 grade of steel. N SEMONTICS 310 ¡Y Y - 18 Z 12 K 25 450 25(Q-11.) Compute the clear gap between the wheels if the centre to centre spacing is 280 mm and the radius of contact area is 110 mm.Situation 5: A W4x13 has a length of 3.5 m. Fy 248 MPa. The properties are given below. Properties of W shape A 2,470 mm^2 d-106 mm bf - 103 mm tf-8.76 mm kdes 15.1 mm tw- 7.11 mm rx-43.7 mm ry= 25.4 mm 1. Which of the following most nearly gives the axial capacity (Pa) of the column if one end is fixed and the other is pinned connected? 226 KN 215 KN. O 244 KN O 260 KN