A beam with an unsymmetrical cross-section has a width of 120 mm, a depth of 200 mm, and a thickness of 10 mm on one side and 20 mm on the other side. The centroid of the cross-section is located at a distance of 50 mm from the larger side. The beam is subjected to a bending moment of 50 kN-m about the y-axis. Determine

the maximum tensile and compressive stresses developed in the beam

1 answer

To determine the maximum tensile and compressive stresses developed in the beam, we need to calculate the moment of inertia of the cross-section with respect to the y-axis, and then use the bending stress formula:

σ = M*y/I

where:
σ = bending stress
M = bending moment (50 kN-m)
y = distance from the centroid to the furthest edge of the beam (100 mm)
I = moment of inertia

The moment of inertia of an unsymmetrical cross-section can be calculated by summing the moments of inertia of the individual rectangles that make up the cross-section. The moment of inertia of a rectangle about its centroidal axis is given by:

I_rect = (1/12)*b*h^3

For the larger rectangle (20 mm x 200 mm):
I_larger = (1/12)*20*200^3 = 10666667 mm^4

For the smaller rectangle (10 mm x 200 mm):
I_smaller = (1/12)*10*200^3 = 2666667 mm^4

The total moment of inertia of the cross-section about the y-axis is:
I_total = I_larger + (200-50)^2*20 + I_smaller = 10666667 + 150^2*20 + 2666667 = 12116667 mm^4

Now, we can calculate the maximum tensile and compressive stresses:

Maximum tensile stress:
σ_max = M*y_max/I_total = 50*100/12116667 = 0.041 kN/mm^2 or 41 MPa

Maximum compressive stress:
σ_min = -M*y_min/I_total = -50*50/12116667 = -0.021 kN/mm^2 or -21 MPa

Therefore, the maximum tensile stress developed in the beam is 41 MPa, and the maximum compressive stress developed in the beam is 21 MPa.