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- Compute 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 b₁ 13.0 y 2.70 Cw 43.8 tw 0.705 Ix 293F1-18 F2-18 F3-27 F4-11H.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 LL
- 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 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 20.7Use 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 20.7Structural Analysis: Solve the F4-5
- AW14×142 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 carry130% v - + 7 Annotate T| Edit Trial expired Unlock Full Version ENGR 263 A A A 4.8 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is a single overhanging beam. Beam AB supports a 500-lb/ft uni- formly distributed floor load on the beam overhang only. Overneia 4 Fr Stee ream (reactions) R. Free-body diagram (negligible weight) Loading conditionWhich set of elements contains Zero Force Member? I s/2 K- F, s/2 F I D F A S. F2 F1 AB - BD - DE - EG- G DC - CE - HG - JI- HJ
- 1) 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 14O 88 130% v - + I Annotate T| Edit Trial expired Unlock Full Version ENGR 263 + A A A T O 4.10 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is an overhanging beam that supports a uniformly distributed roof load of 500 lb/ft. It also carries concentrated loads from a rooftop HVAC unit (4000 lb), an interior hanging display support (2000 lb), and a marquee sign (3000 lb). Marquee overnang Displau SLIPPort II Raof Kiots Ol I W = 500 Ib/Ft A B 4000 b 2000 000 I Steel beam !! I1 3 FE 5 FE - Ft RoOFtop HVAC unit 10 FE 4 Ft Marquee sign< R R2 Steel beam (negligible weight) Free-body diagram Dispiay Support Loading conditionB2-8m RC Beam Sec B-B Sec A-A Sec A-A C1 (0,8x0,8)m Sec A-A- Sec 8-B A Sec A-A- 143.9* Sec A-A Sec A- Sec B-B Sec C-C Sec C-C Soc B-B Sec Sec C-C -3.185m Sac A-A-D 83=4.5m Sec B-B -Sec 8-8 C1 (0,8x0,8)m DBT Figure 2: Plan of Reinforced Concrete Beams Sec A-A -C1 (0.8x0.8)m