A person unexpectedly hangs on the swinging arm of the pecto

A person unexpectedly hangs on the swinging arm of the pectoral fly machine at left, producing a bending moment of 1500 lb-in. at an angle of 10 degrees from side a on the arm. If the cross-section is as shown, determine the bending stress at the points A and B indicated. Take the dimensions to be a = 1:5 in., h = 2 in., and t = 0:25 in.

Solution

Two components of the bending moment are acting on the beam

Perpendicular to b is

Mb=1500 x cos10=1477.21

parallel to b:

Ma=1500 x sin 10 = 260.47

Ia=(1.5x2^3-2x1.5^3)/12=0.718

Ib=(1.5^3x2-1^3x1.5)/12=0.437

Then

sigma b=MaYa/Ia

sigma b=260.47 x 1/0.718=362.77   psi

sigma a=-MaYa/Ia+MbYb/Ib

sigma a= - 260.47 x 0.5/0.718+1477.21 x 0.5/0.437

sigma a=1508.78 psi