A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the inside diameter is d = 101 mm.

(a) Compute the value of Q for the pipe.

(b) If the allowable shear stress for the pipe shape is 83 MPa, determine the maximum load P that can be applied to the cantilever beam.

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2 * (A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π * (56 mm) ² = 3136 π mm²

y₁ = 4*R / (3*π) = 4*56 / (3*π) mm = 224 / (3*π) mm

A₂ = π*r² = π * (50.5 mm) ² = 2550.25 π mm²

y₂ = 4*r / (3*π) = 4*50.5 / (3*π) mm = 202 / (3*π) mm

then

Q = 2 * (3136 π mm²*224 / (3*π) mm-2550.25 π mm²*202 / (3*π) mm)

⇒ Q = 62437.833 mm³

b) If τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I) ⇒ V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64) * (D⁴-d⁴) = (π/64) * ((112 mm) ⁴ - (101 mm) ⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3) * (R³-r³)

⇒ Q = (4/3) * ((56 mm) ³ - (50.5 mm) ³) = 62437.833 mm³

then we have

V = (83 N/mm²) * (11 mm) * (2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N