Question

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is 3.3 × 106 psi for aluminum and 5.9 × 106 psi for brass. If T = 14 kip·in, a=1.1 in. and b=2.2 in., determine (a) The reaction torque at A (b) Maximum shear stress in BC

Answer 1

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

Answer 2

The reaction Torque at A and the maximum shear stress in BC are respectively; T_ab = 697 lb.in; T_bc = 13302 lb.in and τ_bc = 6.352 Ksi

What is Torque and Maximum Shear Stress?

The torques in cylinders AB and BC are statically indeterminate. Matching the rotation φ_b for each cylinder.

For Cylinder AB; c = ¹/₂d = ¹/₂ * 1.1 = 0.55 in

Polar Moment of Inertia; J = ¹/₂ *πc⁴ = ¹/₂ * π * 0.55⁴ = 0.1437 in⁴

Rotation; φ_b = (T_ab * L)/GJ

We are given;

G = 3.3 * 10⁶

L = 12 in

Thus;

φ_b = (12/(3.3 * 10⁶ * 12))T_ab

φ_b = 2.53 × 10⁻⁵ T_ab

For Cylinder BC; c = ¹/₂d = ¹/₂ * 2.2 = 1.1 in

Polar Moment of Inertia; J = ¹/₂ *πc⁴ = ¹/₂ * π * 1.1⁴ = 2.2998 in⁴

Rotation; φ_b = (T_bc * L)/GJ

We are given;

G = 5.9 * 10⁶

L = 18 in

Thus;

φ_b = (18/(5.9 * 10⁶ * 2.2998))T_bc

φ_b = 1.3266 × 10⁻⁶ T_bc

Matching expressions for φ_b, we have;

2.53 × 10⁻⁵ T_ab = 1.3266 × 10⁻⁶ T_bc

Thus; T_bc = 19.0717T_ab    -----(eq 1)

Equilibrium of connection at B;

T_ab + T_bc - T = 0

We are given T = 14 kip·in = 14 × 10³ lb.in

Thus;

T_ab + T_bc - (14 × 10³) = 0

T_ab + T_bc = 14000     ----(eq 2)

Combining equation 1 and 2, we have;

T_ab = 697 lb.in

T_bc = 13302 lb.in

B) Formula for maximum shear stress is;

τ = Tc/J

Maximum shear stress in BC is;

τ_bc = (13302 * 1.1)/2.2998

τ_bc = 6.352 Ksi

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