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2 years ago

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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

1 answer:

8090 [49]2 years ago

7 0

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

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Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

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Answer:

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$F=G\frac{Mm}{R^2}$

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m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

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$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

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$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

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Replacing the values that we know in the equation we have:

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2 years ago

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2 years ago