Question

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 applied. Determine the torsional reactions at these supports.

Answer 1

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:

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

Answer 2

The torsional reactions at the given supports are;

T_a = 12.3 Kip.ft and T_d = 7.7 Kip.ft

What are Torsional Reactions from Torque?

From the image of the shaft attached, if we take a free body diagram and torsion about x, we will have;

∑Mₓ = 0;

T_a + T_d + 20 - 40 = 0     -----(1)

Using the method of superposition, we have;

Φ_a = (Φ_a)_T - (Φ_a)_T_a

Formula for rotation Φ is;

Φ = TL/(JG)

Thus;

0.005 = $[\frac{40 * 12 * 2 * 12}{(\pi/2) *3^{4} * 11 * 10^{3}} + \frac{20 * 12 * 1.5 * 12}{(\pi/2) *3^{4} * 11 * 10^{3}}] - \frac{T_{A} * 12 * 5 * 12}{(\pi/2) *3^{4} * 11 * 10^{3}}$

Solving for T_a in this gives;

T_a = 12.3 Kip.ft

Thus, from eq 1;

12.3 + T_d + 20 - 40 = 0

T_d = 20 - 12.3

T_d = 7.7 Kip.ft

Read more about Torsional reactions from torque at; https://brainly.com/question/20691242