CE Board Exam Randomizer

Torsion in Circular Shafts

Power Transmission

Shafts are often used to deliver mechanical power. If a shaft carries a torque $T$ while rotating with angular speed $\omega$, the transmitted power is given by $P = T\omega$. Since the angular speed can also be written as $\omega = 2\pi f$, where $f$ is the rotational frequency, the torque may be rewritten as:

In SI, power $P$ is commonly expressed in watts $(1.0\ \text{W} = 1.0\ \text{N}\cdot\text{m/s})$ while frequency $f$ is given in hertz $(1.0\ \text{Hz} = 1.0\ \text{rev/s})$, so the formula yields torque in $\text{N}\cdot\text{m}$. In U.S. customary units, if $P$ is stated in $\text{lb}\cdot\text{in./s}$ and $f$ in hertz, the same relation gives torque in $\text{lb}\cdot\text{in.}$ Because power is frequently reported in horsepower $(1.0\ \text{hp} = 550\ \text{lb}\cdot\text{ft/s} = 396 \times 10^3\ \text{lb}\cdot\text{in./min})$, an equivalent and more convenient form of the equation is also often used.

Statically Indeterminate Torsion Problems

The solution process for statically indeterminate torsion cases is much like the method used for indeterminate axially loaded members:

1. Sketch the needed free-body diagrams and write the equilibrium equations.
2. Set up compatibility equations based on the restrictions on angle of twist.
3. Use the torque-twist relationships to rewrite the compatibility equations in terms of torque.
4. Solve the equilibrium and compatibility equations to obtain the unknown torques.

Problem: Solid Steel Shaft | Minimum Diameter and Torsional Shear Stress

What is the minimum diameter (mm) of a solid steel shaft that will not twist through more than 3˚ in a 6m length when subjected to a torque of 12 kNm? What maximum shearing stress (MPa) is developed? Use G = 83 GPa.

See images:

Problem: Relative Angle of Twist | Multiple Torques Applied

An aluminum shaft with a constant diameter of 50mm is loaded by torques applied to gears to it as shown. Using G=28GPa, determine the relative angle of twist in degrees of gear D relative to gear A.

See images:

$$\theta_{A/D}= \theta_{AB}+\theta_{BC}+\theta_{CD}$$

Since the shaft has the same diameter and material throughout, the polar moment of inertia $J$ and the shear modulus $G$ are constant. Thus, $J$ and $G$, as well as the conversion factor for moment to transform it into $N\cdot mm$, may be factored out of the summation when computing the total angle of twist.

$$ \frac{\theta}{A} = \frac{1000^2}{\left(\frac{\pi}{32}(50^4)\right)(28(1000))} \left[ 800(2) - 300(3) + 600(2) \right] \left(\frac{180^\circ}{\pi \text{ rad}}\right) $$

$$ \theta_{A/D} = 6.34^\circ \text{(counterclockwise relative to gear A)} $$

Torque Diagram

To construct the torque diagram, we move from the left end of the shaft toward the right and keep track of the algebraic sum of torques acting on the shaft. A positive torque causes the diagram to step upward, while a negative torque causes it to step downward.

Step-by-Step Construction

1. At point D, an applied torque of $800\ \text{Nm}$ acts on the shaft.
2. Between D and C (length $2\ \text{m}$), the internal torque remains constant:
   $$ T = +800\ \text{Nm} $$
3. At point C, a torque of $1100\ \text{Nm}$ is applied in the opposite sense. The new internal torque becomes
   $$ T = 800 - 1100 = -300\ \text{Nm} $$
4. Between C and B (length $3\ \text{m}$), the torque diagram stays constant at
   $$ T = -300\ \text{Nm} $$
5. At point B, a torque of $900\ \text{Nm}$ acts in the positive direction:
   $$ T = -300 + 900 = 600\ \text{Nm} $$
6. Between B and A (length $2\ \text{m}$), the torque remains constant:
   $$ T = 600\ \text{Nm} $$
7. At point A, the applied torque of $600\ \text{Nm}$ brings the diagram back to zero:
   $$ 600 - 600 = 0 $$

---

Torque Diagram

Problem: Minimum Diameter given the Power and Frequency

A solid steel shaft in a rolling mill transmits 20kW of power at 2Hz. Determine the smallest safe diameter of the shaft if the shear stress is not to exceed 40MPa and the angle of twist is limited to 6º in a length of 3m. Use G=83GPa.

See images:

Problem: Statically Indeterminate Member | Compound Shaft in Torsion

The compound shaft composed of steel, aluminum, and bronze segments, carries the two torques as shown. If T_C=250lb-ft, determine the maximum shear stress developed in each material. The moduli of rigidity for steel, aluminum, and bronze are 12x10⁶psi, 4x10⁶psi, and 6x10⁶psi, respectively.

See images:

Problem 5:

Refer to the image shown:

See images:

Problem 6:

Refer to the image shown:

See images:

Problem 7:

Refer to the image shown:

See images:

Problem 8:

Refer to the image shown:

See images: