Question

if the beam is made from a material having an allowable tensile and compressive stress of (125 MPa tensile) and (150 MPa compressive), determine the maximum allowable internal moment M that can be applied to the beam.

Answer 1

Given data The allowable tensile stress, 25 MPa 125(109) Pa The allowable compressive stress, alew 150 MPa -150(10) Pa Equivalent force moment system: 300 Resolving the moment M along y and axes M M sin 30° -0.5M M,M cos 30 -0.8660M (since it is directed towards the negative z-axis Cross Section of the beam 0.15 m 0.15 m 0.3 m 0.15 m 0.15 m 0.15 m The location of the centroid of the cross-section is given by, VA 7 0.3(0.6)(0.3) 0.375(0.15) (0.15) 0.6(0.3) 0.15(0.15) 0.2893 m (from top face)

The moment of inertia of the cross-section about the principal centroidal y axis, bd 12 0.6)(03-0.15)/(0.15) 1.3078(10)m -(0.6 (0.3 12 The moment of inertia of the cross-section about the principal centroidal : axis, bd 12 (0.3)(0.63) 0.3(0.6) (0.3 - 0.2893) (0.15) (0.15)+ 0.15(0.15) (0.375 - 0.2893) 5.2132(10-3) m 4 The maximum bending occurs at either at top left corner or bottom right corner. Let the top left corner be A and the bottom right corner be B The bending stress at A 0.5M (0.15) (0.8660M) (0.2893) 125(10) -$.2132(103(10) 125(106) 1.3078(10-3 M- 1185906.82 N m 1186 kN·m The bending stress at B, which will be in compression,

(-0.8660M)(-0.3107) 0.5M (-0.15) 5.2132(10 1.3078(10*) 150(10° -51.6135M - 57.3482M --108.9617M M 1376630.5 N.m 1376.630 kN m 1377 kN m From the two values of M, we select the lowest value, since beyond that limit failure occurs. Therefore, the maximum allowable bending moment, M 1186 kN -m.