A4.3- Design a square short column for P₁ = 1800 kN, Mfx = 250 kN.m, Mfy: f your final design p has to be between 0.01 and 0.02. = 50 kN.m. In Given: f'c = 25 MPa fy = 400 MPa Longitudinal bars: Use 25M Ties: 10M@ 300 Clear cover to the ties: 40mm 25 P₁ = 0.05 20 20 15 P. MPa 2½8 10 5 P₁ = 0.04 P₁ = 0.03 P₁ = 0.02 P₁ = 0.01 0 0 1 2 fs/fy = 0.0x 3 M, Agh . MPa f/fy= 0.5 Y = 0.80 b fy = 25 MPa = 400 MPa jfs/fy = 1.0- h 5 6 7

Use an elastic analysis and compute the extra load in the weld (in kips per inch of length) caused by the eccentricity.

A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.

If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.