Compute for the maximum allowable distributed load oo that can be applied to the simply supported beam as shown in Figure M-Va. The beam's cross section is detailed in Figure M-Vb. w(N/m) Allowable Stresses: Fiber Stress = +13.5 MPa / -12 MPa Transverse Shear Stress at NA = 900 kPa Transverse Shear Stress at Junction = 800 B 2 m Figure M-Va 2 m Draw the flexural and shear stress distributions of the cross section. ←→→→→→ T 40 40 40 Figure M-Vb 40 120 mm

find the maximum tensile and compressive bending stress make P= 4p x 12= 48KN P= 5P x 12= 60KN W= 9kN/m M= 50 kN.m. note: dimensions are in mm

The light rigid bar ABCD shown is pinned at B and connected to two vertical rods. Assuming that the bar was initially horizontal and the rods stress-free, determine the stress in each rod after the load after the load P = 30 kips is applied.

3. An A36 Steel Cantilever Beam is shown below, also showing its cross section. Determine the following: Given the formula of the Maximum allowable bending stress of steel, fb = 0.66Fy. Where Fy is the Yield Stress of Steel. Where the bending stress for a beam is (fb) = Mc/I, where c is the distance of the centroid to the extreme most fiber of the beam (either top-most or bottom-most distance). Does the Steel fail under the loading given? Use Fy = 450 Mpa. Use the Moment in (a) and the moment of inertia in (c).