Answer:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
The principal stress in the pipe at point A, which is located on the outer surface of the pipe is 132.773 psi
It is given that l = 12 in
The force F = 60 lb
The radiuses are as follows:
outer radius, R = 3.90 in/2
R = 1.95 in
Inner radius, r = 3.65 in/2
r = 1.825 in
The angle between the force applied and the distanced from the axis is 30°
So we get the torque:
T = l×Fsin 30°
T = 12 × 60 × (0.5)
T = 360 lb-in
Now, the angle of rotation or angular displacement ωt is given as
ωt = π(R⁴ - r⁴)/(2R)
ωt = π((1.95 in)⁴ - (1.825 in)⁴)/(2×1.95 in)
ωt = 2.7114 in³
Then the principal stress in the pipe at point A is:
principal stress = T/ωt
principal stress = (360 lb-in)/(2.7114 in³)
principal stress = 132.773 lb/in²
principal stress = 132.773 psi
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