# Solution to Problem 322 Torsion

**Problem 322**

A solid steel shaft is loaded as shown in Fig. P-322. Using G = 83 GPa, determine the required diameter of the shaft if the shearing stress is limited to 60 MPa and the angle of rotation at the free end is not to exceed 4 deg.

# Solution to Problem 321 Torsion

**Problem 321**

A torque T is applied, as shown in Fig. P-321, to a solid shaft with built-in ends. Prove that the resisting torques at the walls are T_{1} = Tb/L and T_{2} = Ta/L. How would these values be changed if the shaft were hollow?

# Solution to Problem 320 Torsion

**Problem 320**

In Prob. 319, determine the ratio of lengths b/a so that each material will be stressed to its permissible limit. What torque T is required?

# Solution to Problem 319 Torsion

**Problem 319**

The compound shaft shown in Fig. P-319 is attached to rigid supports. For the bronze segment AB, the diameter is 75 mm, τ ≤ 60 MPa, and G = 35 GPa. For the steel segment BC, the diameter is 50 mm, τ ≤ 80 MPa, and G = 83 GPa. If a = 2 m and b = 1.5 m, compute the maximum torque T that can be applied.

# Derivation of the Double Angle Formulas

The Double Angle Formulas can be derived from Sum of Two Angles listed below:

$\sin (A + B) = \sin A \, \cos B + \cos A \, \sin B$ → Equation (1)

$\cos (A + B) = \cos A \, \cos B - \sin A \, \sin B$ → Equation (2)

$\tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \, \tan B}$ → Equation (3)

# Solution to Problem 318 Torsion

**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 × 10^{6} psi.

# Solution to Problem 317 Torsion

**Problem 317**

A hollow bronze shaft of 3 in. outer diameter and 2 in. inner diameter is slipped over a solid steel shaft 2 in. in diameter and of the same length as the hollow shaft. The two shafts are then fastened rigidly together at their ends. For bronze, G = 6 × 10^{6} psi, and for steel, G = 12 × 10^{6} psi. What torque can be applied to the composite shaft without exceeding a shearing stress of 8000 psi in the bronze or 12 ksi in the steel?

# Solution to Problem 316 Torsion

**Problem 316**

A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in Fig. P-316. Determine the maximum permissible value of T subject to the following conditions: τ_{st} ≤ 83 MPa, τ_{al} ≤ 55 MPa, and the angle of rotation of the free end is limited to 6°. For steel, G = 83 GPa and for aluminum, G = 28 GPa.

# Solution to Problem 315 Torsion

**Problem 315**

A 5-m steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from the left end where 20 kW are removed. At the right end, 30 kW are removed and another 20 kW leaves the shaft at 1.5 m from the right end. (a) Find the uniform shaft diameter so that the shearing stress will not exceed 60 MPa. (b) If a uniform shaft diameter of 100 mm is specified, determine the angle by which one end of the shaft lags behind the other end. Use G = 83 GPa.