If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the outer surface of the pipe. Express your answers, separated by a comma, to three significant figures.

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

ill do it for 100, also this isnt 80 its 40

i'm on my way to history not a mistory.