Question

A solid steel bar of circular cross section has diameter d 5 2.5 in., L 5 60 in., and shear modulus of elasticity G 5 11.5 3 106 psi. The bar is subjected to torques T 5 300 lb-ft at the ends. Calculate the angle of twist between the ends. What is the maxi- mum shear stress and the shear stress at a distance rA 5 1.0 in. measured from the center of the bar

Answer 1

The angle of twist between the ends of the steel bar is approximately 1.074e-4 radians. The maximum shear stress is approximately 38.101 psi. The shear stress at a distance rA = 1.0 in. from the center of the bar is approximately 15.244 psi.

Step-by-step explanation:

To calculate the angle of twist between the ends of the steel bar, we can use the expression:

θ = (TL)/(2πG)

Using the given values, we have:

θ = (300 lb-ft * 60 in.)/(2π * 11.5 * 10^6 psi)

Performing the calculations, we find that the angle of twist is approximately 1.074e-4 radians.

The maximum shear stress can be calculated using the expression:

τ = (16Td)/(πd^3)

Substituting the given values, we get:

τ = (16 * 300 lb-ft * 12 in.)/(π * (2.5 in.)^3)

Calculating the shear stress, we find that the maximum shear stress is approximately 38.101 psi.

The shear stress at a distance rA = 1.0 in. from the center of the bar can be calculated using the expression:

τ = (16TrA)/(πd^3)

Substituting the given values, we have:

τ = (16 * 300 lb-ft * 12 in.)/(π * (2.5 in.)^3)

Calculating the shear stress, we find that the shear stress at a distance rA = 1.0 in. is approximately 15.244 psi.

Answer 2

A) θ = 4.9 x 10^(-3) rad

B) τ_max = 1.173 ksi

C) τ_a = 4.786 ksi

Step-by-step 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