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3 years ago

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5-81 The shaft is made of A-36 steel and is fixed at end D, while end A is allowed to rotate 0.005 rad when the torque is applie

d. Determine the torsional reactions at these supports.

1 answer:

mel-nik [20]3 years ago

3 0

Answer:

The answer is "7.73 kip-ft".

Explanation:

Using this equation, calculate the polar moment of shaft inertia:

$J= \frac{\pi }{32} d^4\\\\$

$= \frac{\pi}{ 32} \times 6^4\\\\= 127.234 \ \ in^4$

Equate the number of moments in x:

$\sum M_x =0$

$\to T_A-T_B +T_C+T_D =0\\\\\to T_A-40 +20+T_D =0$

Calculate the torsional response at A using the superposition method:

$\to \phi_A =(\phi_A)_{T} - (\phi_A)_{T_A}\\\\$

$= \frac{T_B L_{BC}}{JG}+ \frac{T_C L_{CD}}{JG} - \frac{T_A L_{AD}}{JG}\\\\$

$0.05= \frac{40 \times 12 \times 2 \times 12}{127.234 \times 11 \times 10^3}+ \frac{20 \times 12 \times 1.5 \times 12 }{127.234 \times 11 \times 10^3} - \frac{T_A \times 12 \times 5 \times 12}{127.234 \times 11 \times 10^3}$

$0.05= (8.22 \times 10^{-3})+ (3.086 \times 10^{-3})- (5.14 \times 10^{-4} \ T_A)\\\\$

$\to T_A =12.27 kip \ ft \\\\$

The torsional answer in help A is 12.27 kip- ft. It replace the necessary equation values:

$\to T_A-40 +20+T_D =0 \\\\\to 12.27-40+20+T_D=0\\\\T_D=7.73 kip \ ft$

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