Select the lightest available W8 section to support the axial loads Pp = 75 k and 125 k if KL = 13 ft and F, = 50 ksi.

A W 16 x50 is used as a beans to carry a uniform we and dead load including own weight of 37 kN/m and an axial tension load of T acting through the centroid, of the member. It has a simple span of 5.7 m. The compression flange of the member is laterally supported against local buckling.Use A 36 steel, Fy = 248 MPa.Properties of W 16x 50A = 9483.85 mm^2d= 412.75 mmbf= 179.65 mmtf= 15 95 mmtw= 9.65 mmSx= 1324 × 10^3 mm^3Sy = 172 x 10^3 mm^3 Questions:Determine the Safe axial load T that the beam could carry Hint Answer: 432.69 KN

Problem 1.A compression member is 30 feet long. It is made up of ASTM A500 Grade B (Fy= 46 ksi and Fu = 58 ksi) rectangular HSS 12x8x3/16. Assume k-value equals to 1.0 and  Q= 1.0 . If the member carries an axial load composed of 26 kips dead load and 77 kips live load, is the member adequate? Use E=29000 ksi. a. Solve for the gross area.b. Compute for the design strength against squash load.c. Solve for the moment of inertia and radius of gyration for both axes. (X-X and Y-Y)d. Compute for the design strength against global buckling.e. Solve for kL/rfor both axes.(X-X and Y-Y)f. Solve for Fcr.g. Solve for design strength against local buckling.Check first if Local Buckling is applicable.h. Compute for the ultimate load to be carried by the member. Use LRFDLoad Combination.i. Does this member have enough strength? Why or why not?

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I need help finding the compressive resistance of the column. A column is braced at mid height to prevent movement in the weak axis only. The column is 10m in length and pinned at both ends. The column has been specified as a W360 x 91 G40.21 345W steel. Given data: d=353mm , b=254mm, t=16.4mm, w=9.5mm, d-2t = 320mm, rx= 152mm, ry= 62.3mm, A = 11500mm^2, Fy= 345MPa

The service load bending moments on a beam are 128 kip-ft for dead load and 107 kip-ft for live load. The beam is 12 in. wide, f'c is 3000 psi and fy is 60 ksi. Determine Mu. Mt. p. R. the depth of the beam and the As of tensile reinforcing required. How many no. 8 bars or No. 10 bars would have the required As?

Of the types of loads imposed on the Member Load (Beam) are: Concentrated Force, Uniform Force Staad pro True O False The section L75x75x5 SD is double angle iron Staad pro O True False

A bracket of length 1.5m is rigidly attached to the simply-supported beam at its end. The beam is subjected to the combined effect of M,, M, and compressive force P caused by the factored loads as shown. There are lateral supports only at the ends. The member is made of HE300A section and steel grade is S275 (Fy-275MPA, Fu= 430 MPa). The loads acting are determined from LRFD combinations. Determine whether the member can safely carry the applied loads. Multiply first-order moment diagrams by a factor of 1.2 to account for second-order effects. No need to check shear limit state. You do not need to additionally consider the beam self-weight. E Use the following simple expression to calculate the value of L:: L, = T*rts 0.7*Fy y P = 400kN 1.5 m F = 150kN 2 т 4 т X: Lateral support НЕЗ00А

steel structure (AISC ) masnual

Select a WT section for the compression member shown in Figure. The load is the total service load, with a live-to-dead load ratio of 2.5:1. Use Fy 5 50 ksi. a. Use LRFD. b. Use ASD 21' 175 k 1