$\theta_{st} = \theta_{br}$
$\left( \dfrac{TL}{JG} \right)_{st} = \left( \dfrac{TL}{JG} \right)_{br}$
$\dfrac{T_{st} L}{\frac{1}{32}\pi (2^4)(12 \times 10^6)} = \dfrac{T_{br} L}{\frac{1}{32}\pi (3^4 - 2^4)(6 \times 10^6)}$
$\dfrac{T_{st}}{192 \times 10^6} = \dfrac{T_{br}}{390 \times 10^6}$ → Equation (1)
Applied Torque = Resisting Torque
$T = T_{st} + T_{br}$ → Equation (2)
Equation (1) with Tst in terms of Tbr and Equation (2)
$T = \dfrac{192 \times 10^6}{390 \times 10^6} T_{br} + T_{br}$
$T_{br} = 0.6701T$
Equation (1) with Tbr in terms of Tst 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 (Tbr = 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 (Tst = 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
Comments
Anonymous (guest)
April 15, 2023 - 3:35am
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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?
John123 (guest)
March 16, 2024 - 11:33pm
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I think it was cancelled out
I think it was cancelled out since the length of the steel shaft and bronze are equal