Astandard 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