Solution to Problem 318 Torsion | Strength of Materials Review at MATHalino

Problem 318
A solid aluminum shaft 2 in. in diameter is subjected to two torques as shown in Fig. P-318. Determine the maximum shearing stress in each segment and the angle of rotation of the free end. Use G = 4 × 10⁶ psi.

Solution 318

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$\tau_{max} = \dfrac{16T}{\pi D^3}$

For the 2-ft segment:

$\tau_{max2} = \dfrac{16(600)(12)}{\pi (2^3)} = 4583.66 \, \text{psi}$ answer

For the 3-ft segment:

$\tau_{max3} = \dfrac{16(800)(12)}{\pi (2^3)} = 6111.55 \, \text{psi}$ answer

Angle of twist

$\theta = \dfrac{TL}{JG}$

$\theta = \dfrac{1}{JG} \Sigma TL$

$\theta = \dfrac{1}{\frac{1}{31}\pi (2^4)(4 \times 10^6)} [ \, 600(2) + 800(3) \, ] \, (12^2)$

$\theta = 0.0825 \, \text{rad}$

$\theta = 4.73^\circ$ answer

Tags:

Torque

Angle of Twist

Torsional Shearing Stress

Aluminum Shaft

Shearing Stress

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