A standard steel pipe (D = 3.81 in.; d = 3.24 in.) supports a concentrated load of P = 930 lb. The span length of the cantilever beam is L = 3.7 ft. Determine the magnitude of: (a) the maximum horizontal shear stress τmax in the pipe. (b) the maximum tension bending stress σmax in the pipe.

a) τmax = 586.78 P. S. I.

b) σmax = 15942.23 P. S. I

Explanation:

D = 3.81 in

d = 3.24 in

P = 930 lb

L = 3.7 ft = 44.4 in

a) The maximum horizontal shear stress can be obtained as follows

τ = V*Q / (t*I)

where

V = P = 930 lb

Q = (2/3) * (R³ - r³) = (1/12) * (D³ - d³) = (1/12) * ((3.81 in) ³ - (3.24 in) ³)

⇒ Q = 1.7745 in³

t = D - d = 3.81 in - 3.24 in = 0.57 in

I = (π/64) * (D⁴-d⁴) = (π/64) * ((3.81 in) ⁴ - (3.24 in) ⁴) = 4.9341 in⁴

then

τ = (930 lb) * (1.7745 in³) / (0.57 in*4.9341 in⁴)

⇒ τmax = 586.78 P. S. I.

b) We can apply the following equation in order to get the maximum tension bending stress in the pipe

σmax = Mmax * y / I

where

Mmax = P*L = 930 lb*44.4 in = 41292 lb-in

y = D/2 = 3.81 in / 2 = 1.905 in

I = 4.9341 in⁴

then

σmax = (41292 lb-in) * (1.905 in) / (4.9341 in⁴) = 15942.23 P. S. I