CE Board Exam Randomizer

⬅ Back to Subject Topics

Torsion in Circular Shafts

Concept Concept Concept Concept Concept Concept Concept Concept Concept Concept Concept

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:

$$T=\frac{P}{2\pi f}$$

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.
$T=Torque$
$L= Length\ of\ shaft$
$\rho=distance\ from\ the\ center\ to\ the\ point\ of\ interest$
$J=polar\ moment\ of\ inertia\ (I_x+I_y)$
$G=shear\ modulus\ or\ modulus\ of\ rigidity$
$\gamma=shear\ strain$

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.

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 1: – Diagram

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.

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 2: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 2: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 2: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 2: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 2: – Diagram
$$\theta_{A/D}= \theta_{AB}+\theta_{BC}+\theta_{CD}$$

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

Solution

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.

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 3: – Diagram

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 TC=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.

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 4: – Diagram

Problem:

Refer to the image shown:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 5: – Diagram

Problem:

Refer to the image shown:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 6: – Diagram

Problem:

Refer to the image shown:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 7: – Diagram

Problem:

Refer to the image shown:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram

See images:

Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram Torsion (Torque and Shear Stress) | Mechanics of Deformable Bodies – Problem 8: – Diagram
Scroll to zoom

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q139

PSAD - Mechanics of Deformable Bodies / Torsion / Engr. Janclyde Espinosa (Clidez)

A torque wrench is used to tighten the pipe shown.
Given:
S1 = 400mm S2 = 250mm S3 = 100mm
Modulus of Rigidity G = 78MPa
Allowable shear stress = 60MPa

q139

The diameter of the solid pipe is 20mm. How much is the maximum force P (N) that can be applied based on the allowable shear stress?

  1. 145
  2. 160
  3. 120
  4. 100

For a hollow pipe with 50mm outside diameter and a thickness of 6mm, compute the maximum force P (kN) that can be applied such that the angle of twist at A does not exceed 5º.

  1. 42.80
  2. 28.80
  3. 32.70
  4. 51.40

The torque applied to tighten the hollow pipe is 200N-m. If the outside diameter of the pipe is 50mm and the pipe thickness is 6mm, determine the maximum shear stress (MPa) in the pipe.

  1. 12.20
  2. 13.50
  3. 16.80
  4. 10.10

Solution pending in psadquestions/q139.json.

Question Bank: q150

PSAD - Mechanics of Deformable Bodies / Torsion / Engr. Janclyde Espinosa (Clidez)

An aluminum bar of solid circular cross-section is twisted by torques T acting at the ends. The dimensions and shear modulu of elasticity are as follows:
L = 1.4m, d = 35mm, and G = 28GPa

Determine the torsional stiffness of the bar.

  1. 2950N-m
  2. 1810N-m
  3. 2059N-m
  4. 1905N-m

If the angle of twist of the bar is 7º, what is the maximum shear stress?

  1. 43MPa
  2. 48MPa
  3. 36MPa
  4. 28MPa

If the angle of twist of the bar is 5º, what is the maximum shear strain (in radians)?

  1. 0.001090
  2. 0.000244
  3. 0.000644
  4. 0.000997

Part 1.

$L=1.4$ m $=1400$ mm; $d=35$ mm; $G=28{,}000$ MPa
$J = \frac{\pi d^4}{32} = \frac{\pi(35)^4}{32} = 147{,}324 \text{ mm}^4$
Torsional stiffness $= \frac{GJ}{L} = \frac{28{,}000 \times 147{,}324}{1{,}400} = 2{,}946{,}480 \text{ N·mm/rad}$
$\boxed{\approx 2{,}950 \text{ N·m/rad}}$

Part 2.

$\theta = 7^\circ = 7 \times \frac{\pi}{180} = 0.1222$ rad
Torque: $T = k_T \cdot \theta = 2{,}946{,}480 \times 0.1222 = 360{,}060 \text{ N·mm}$
$\tau_\max = \frac{T \cdot c}{J} = \frac{360{,}060 \times 17.5}{147{,}324} = \frac{6{,}301{,}050}{147{,}324}$
$\boxed{\approx 43 \text{ MPa}}$

Part 3.

$\theta = 5^\circ = 0.08727$ rad
$T = 2{,}946{,}480 \times 0.08727 = 257{,}120 \text{ N·mm}$
$\tau = \frac{257{,}120 \times 17.5}{147{,}324} = 30.54 \text{ MPa}$
Max shear strain: $\gamma = \frac{\tau}{G} = \frac{30.54}{28{,}000}$
$\boxed{= 0.001091 \text{ rad}}$

Question Bank: q422

PSAD - Mechanics of Deformable Bodies / Torsion / Engr. Janclyde Espinosa (Clidez)

A torque of 12kN-m is applied on a shaft 3m in length. The diameter of the shaft is 100mm. Modulus of rigidity = 83GPa

Determine the angle of twist in degrees.

  1. 2.531
  2. 0.044
  3. 0.056
  4. 3.209

Determine the maximum torsional shear stress in MPa.

  1. 61.12
  2. 122.23
  3. 72.17
  4. 144.33

Solution pending in psadquestions/q422.json.

Question Bank: q560

PSAD - Mechanics of Deformable Bodies / Torsion / Engr. Deguma

Two shafts A and B transmit the same power. The speed of shaft A is 250 rpm and that of shaft B is 300 rpm. Which of the following is true?

Answer:

  1. shaft A has greater diameter
  2. shaft B has greater diameter
  3. both shafts have equal torque
  4. shafts have equal shearing stress

Solution pending in psadquestions/q560.json.

Question Bank: q598

PSAD - Mechanics of Deformable Bodies / Torsion / Mastermatician

Refer to the figure shown:

q598

The shaft in which the maximum shearing stress occurs is at:

  1. CD
  2. DE
  3. AB
  4. BC

The magnitude of the maximum shearing stress (in MPa).

  1. 85.8
  2. 58.8
  3. 85.5
  4. 88.8

Solution pending in psadquestions/q598.json.

Question Bank: q747

PSAD - Mechanics of Deformable Bodies / Torsion / Mastermatician

Find the minimum outer diameter of a hollow shaft that will not twist by more than 5º when subjected to a stress of 50MPa on a length of 2m. Use G=83GPa.

  1. 28mm
  2. 25mm
  3. 24mm
  4. 20mm

Solution pending in psadquestions/q747.json.

Question Bank: q748

PSAD - Mechanics of Deformable Bodies / Torsion / Mastermatician

The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall at A and jointed together by a pin at B. Because of a slight misalignment of the pre-drilled holes, the torque T0=750N-m was applied to the steel rod before the pin could be inserted into the holes. Determine the torque in each component after T0 was removed. Use G=80GPa for steel and G=28GPa for aluminum.

q748
  1. 251N-m
  2. 260N-m
  3. 274N-m
  4. 246N-m

Solution pending in psadquestions/q748.json.

Question Bank: q749

PSAD - Mechanics of Deformable Bodies / Torsion / Mastermatician

A steel shaft and an aluminum tube are connected to a fixed wall and to a rigid disk as shown. Knowing that the initial stresses are zero, determine the maximum torque T that can be applied to the disk if the allowable stresses are 120MPa in the steel shaft and 70MPa in the aluminum tube. Use G=77GPa for steel and G=27GPa for aluminum.

q749
  1. 6325N-m
  2. 6915N-m
  3. 2450N-m
  4. 3690N-m

Solution pending in psadquestions/q749.json.

Question Bank: q759

PSAD - Mechanics of Deformable Bodies / Torsion / Engr. Janclyde Espinosa (Clidez)

Formula-mode item rendered with fixed values for lecture/PDF export.

A hollow steel shaft shall not twist by more than 5.6º in a length of 8m when subjected to a torque of 14.5kN-m. The steel shaft is made of steel which has a modulus of rigidity of 83GPa. The outer diameter of the shaft is 300mm.

Determine the minimum inner diameter of the shaft (in mm).

  1. 298.64218558192886
  2. 296.64218558192886
  3. 294.64218558192886
  4. 292.64218558192886

Determine the maximum shearing stress (in MPa) developed in the shaft.

  1. 152.10544431130896
  2. 62.12652283396525
  3. 39.32776735137176
  4. 28.926501378521966

Solution pending in psadquestions/q759.json.

Question Bank: q760

PSAD - Mechanics of Deformable Bodies / Torsion / Engr. Janclyde Espinosa (Clidez)

Formula-mode item rendered with fixed values for lecture/PDF export.

A flanged bolt coupling consists of 12 - 20mm∅ steel bolts evenly spaced around a bolt circle 340 mm in diameter and 6 - 12mm∅ aluminum bolts on a concentric bolt circle 240mm in diameter. Gst=83GPa and Gal=28GPa. The allowable stresses are as follows:
σst=250MPa
σal=34MPa

Determine the maximum force that can be carried by the aluminum bolts.

  1. 3845.309407993907
  2. 6732.938472328515
  3. 10681.415022205298
  4. 18702.60686757921

Determine the maximum force that can be carried by the steel bolts.

  1. 44855.58510812998
  2. 78539.81633974482
  3. 16148.010638926793
  4. 16148.010638926793

Determine the maximum torque that can be applied.

  1. 94274016.39434075
  2. 165068941.03315598
  3. 162989848.10683507
  4. 96353109.32066168

Solution pending in psadquestions/q760.json.