Answer:
A) θ = 4.9 x 10^(-3) rad
B) τ_max = 1.173 ksi
C) τ_a = 4.786 ksi
Explanation:
We are given;
diameter; d = 2.5 inches = 0.2083 ft
Length; L = 60 inches = 5 ft
Torque; T = 300 lb.ft
Shear modulus; G = 11.5 x 10^(6) psi = 11.5 x 144 x 10^(6) lb/ft² = 1.656 x 10^(9) lb/ft²
A) Now, formula to determine angle of twist is given as;
T/I_p = Gθ/L
Where I_p is polar moment of inertia
θ is angle of twist.
Now I_p = πd⁴/32 = π(0.2083)⁴/32 = 1.85 x 10^(-4) ft⁴
Thus, making θ the subject, we have;
TL/GI_p = θ
θ = (300 x 5)/(1.656 x 10^(9) x 1.85 x 10^(-4))
θ = 4.9 x 10^(-3) rad
B) Maximum shear stress is given by the formula ;
τ_max = (Gθ/L)(d/2)
From earlier, (Gθ/L) = T/I_p
Thus, (Gθ/L) = 300/1.85 x 10^(-4) = 1621621.6216
Thus,
τ_max = 1621621.6216 x (0.2083/2)
τ_max = 168891.89 lbf/ft²
Converting to ksi = 168891.89/144000 ksi = 1.173 ksi
C) Shear stress at radial distance is given as;
τ_a = (Gθ/L)•r_a
r_a is given as 5.1 inches = 0.425m
τ_a = 1621621.6216 x 0.425 = 689189.189 lbf/ft²
Converting to ksi = 689189.189/144000 ksi = 4.786 ksi
