$\theta_{st} = \theta_{br}$

$\left( \dfrac{TL}{JG} \right)_{st} = \left( \dfrac{TL}{JG} \right)_{br}$

$\dfrac{T_{st} L}{\frac{1}{32}\pi (2^4)(12 \times 10^6)} = \dfrac{T_{br} L}{\frac{1}{32}\pi (3^4 - 2^4)(6 \times 10^6)}$

$\dfrac{T_{st}}{192 \times 10^6} = \dfrac{T_{br}}{390 \times 10^6}$ → Equation (1)

Applied Torque = Resisting Torque

$T = T_{st} + T_{br}$ → Equation (2)

Equation (1) with T_{st} in terms of T_{br} and Equation (2)

$T = \dfrac{192 \times 10^6}{390 \times 10^6} T_{br} + T_{br}$

$T_{br} = 0.6701T$

Equation (1) with T_{br} in terms of T_{st} and Equation (2)

$T = T_{st} + \dfrac{390 \times 10^6}{192 \times 10^6} T_{br}$

$T_{st} = 0.3299T$

Based on hollow bronze (T_{br} = 0.6701T)

$\tau_{max} = \left[ \dfrac{16TD}{\pi(D^4 - d^4)} \right]_{br}$

$8000 = \dfrac{16(0.6701T)(3)}{\pi (3^4 - 2^4)}$

$T = 50 789.32 \, \text{lb}\cdot\text{in}$

$T = 4232.44 \, \text{lb}\cdot\text{ft}$

Based on steel core (T_{st} = 0.3299T):

$\tau_{max} = \left[ \dfrac{16TD}{\pi D^3} \right]_{st}$

$12\,000 = \dfrac{16(0.3299T)}{\pi (2^3)}$

$T = 57\,137.18 \, \text{lb}\cdot\text{in}$

$T = 4761.43 \, \text{lb}\cdot\text{ft}$

Use T = 4232.44 lb·ft. *answer*

## On the compatilibity equation

On the compatilibity equation, what happened on the "L" for both the bronze and steel, and how did you solve for the values in equation 1?

## I think it was cancelled out

In reply to On the compatilibity equation by Anonymous (not verified)

I think it was cancelled out since the length of the steel shaft and bronze are equal