Chapter 10, Problem 10.5QPP

Using a relevant image such as a 3D architectural rendering of a warehouse landscape in Edmond.

A simple beam has a span of 10 m and supports a total uniformly distributed load of 12 kN/m. Properties of W 480×86: Property Value A = 10800 mm² bf = 180 mm tf = 15 mm Ix = 383.13 × 10⁶ mm⁴ tw = 12 mm d = 480 mm Questions: Calculate the maximum shear in the beam in kN. Calculate the average shear stress in the beam in MPa. Calculate the maximum shear stress in the beam in MPa.   Given the properties of the wide flange: Property Value d = 530 mm   bf = 210 mm   tw = 18 mm   tf = 16 mm   Question: Compute the value of rₜ, defined as the radius of gyration of the compression flange plus 1/3 of the compression web area about the y-axis.

Given an existing two-story steel structure with interior columns spaced as shown in Fig.2. The columns are spaced at 18 ft in the North-South direction and at 30 ft in the East-West direction. An interior lower-story column is to be removed by adding newsteel girder as shown in Fig. 4. The floor dead loads and the roof dead loads are 70 psfand 18 psf respectively. The floor live loads and the roof live loads are 50 psf and20 psf respectively. All existing steel materials are ASTM A36 steel (Fy=36 ksi). Newgirder is ASTM A992 steel (Fy= 50 ksi). All columns are W8x31. Use the LRFD Method.Assumptions:1- The loads given include column and beam self weights.2- Existing beam and new girder are simply supported at both ends.3- New girder top flange is laterally braced at mid span and at girder ends only.4- Columns are continuous from foundation to roof and are prevented from sway atfloor level and at roof level in both directions.5- Columns are pin supported at foundation, at floor level,…