A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 48.0000 kip·in; outer radius (c) = 3.3000 in; polar moment of inertia (J) = 212.0 in⁴, determine the shear stress (tau) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Torsional shear stress.
• Everything except tau is given, so isolate tau symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = T c / J$$

2. Step 2 — Rearrange the relation so that tau stands alone on the left-hand side.

3. Step 3 — List the givens: torque (T) = 48.0000 kip·in, outer radius (c) = 3.3000 in, polar moment of inertia (J) = 212.0 in⁴.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\tau = 0.7472\ \text{ksi}$$

6. Step 6 — Check: returning tau = 0.7472 ksi to

   $$\tau = T c / J$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1.4943 — kept a factor of two that cancels in the correct rearrangement.
• 0.3736 — dropped that same factor in the other direction.
• 0.8219 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

A drill rod's torsional strain is measured during a twist test. Given radial distance (r) = 3.8500 in; angle of twist (phi) = 0.0410 rad; shaft length (L) = 33.0000 in, determine the shear (torsional) strain (gamma).

Start with the thinking

• The governing relation printed in this handbook section is Torsional strain (circular shaft).
• Everything except gamma is given, so isolate gamma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Torsional strain in a circular shaft varies linearly with radial distance from the axis for a given angle of twist.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\gamma = \dfrac{r \phi}{L}$$

2. Step 2 — Rearrange symbolically for gamma:

   $$\gamma = \dfrac{r \phi}{L}$$

3. Step 3 — List the givens: radial distance (r) = 3.8500 in, angle of twist (phi) = 0.0410 rad, shaft length (L) = 33.0000 in.

4. Step 4 — Substitute the given values:

   $$\gamma = \dfrac{3.8500 0.0410}{33.0000}$$

5. Step 5 — Evaluate:

   $$\gamma = 0.0048$$

6. Step 6 — Check: returning gamma = 0.0048 to

   $$\gamma = \dfrac{r \phi}{L}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0096 — kept a factor of two that cancels in the correct rearrangement.
• 0.0024 — dropped that same factor in the other direction.
• 0.0053 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Torsional Strain

A mechanics of materials problem uses Torsional shear stress. Given outer radius (c) = 1.2000 in; polar moment of inertia (J) = 315.0 in⁴; shear stress (tau) = 22.5400 ksi, determine the torque (T) in kip·in.

Start with the thinking

• The governing relation printed in this handbook section is Torsional shear stress.
• Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = T c / J$$

2. Step 2 — Rearrange the relation so that T stands alone on the left-hand side.

3. Step 3 — List the givens: outer radius (c) = 1.2000 in, polar moment of inertia (J) = 315.0 in⁴, shear stress (tau) = 22.5400 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$T = 5917\ \text{kip·in}$$

6. Step 6 — Check: returning T = 5,917 kip·in to

   $$\tau = T c / J$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 11,834 — kept a factor of two that cancels in the correct rearrangement.
• 2,958 — dropped that same factor in the other direction.
• 6,508 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

A machine shaft's torsional strain distribution is checked across its cross section. Given angle of twist (phi) = 0.0320 rad; shaft length (L) = 40.0000 in; shear (torsional) strain (gamma) = 0.0033, determine the radial distance (r) in in.

Start with the thinking

• The governing relation printed in this handbook section is Torsional strain (circular shaft).
• Everything except r is given, so isolate r symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Torsional strain in a circular shaft varies linearly with radial distance from the axis for a given angle of twist.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\gamma = \dfrac{r \phi}{L}$$

2. Step 2 — Rearrange symbolically for r:

   $$r = \dfrac{\gamma L}{\phi}$$

3. Step 3 — List the givens: angle of twist (phi) = 0.0320 rad, shaft length (L) = 40.0000 in, shear (torsional) strain (gamma) = 0.0033.

4. Step 4 — Substitute the given values:

   $$r = \dfrac{0.0033 40.0000}{0.0320}$$

5. Step 5 — Evaluate:

   $$r = 4.1625\ \text{in}$$

6. Step 6 — Check: returning r = 4.1625 in to

   $$\gamma = \dfrac{r \phi}{L}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 8.3250 — kept a factor of two that cancels in the correct rearrangement.
• 2.0813 — dropped that same factor in the other direction.
• 4.5788 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Torsional Strain

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 471.0 kip·in; polar moment of inertia (J) = 35.0000 in⁴; shear stress (tau) = 27.7100 ksi, determine the outer radius (c) in in.

Start with the thinking

• The governing relation printed in this handbook section is Torsional shear stress.
• Everything except c is given, so isolate c symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = T c / J$$

2. Step 2 — Rearrange the relation so that c stands alone on the left-hand side.

3. Step 3 — List the givens: torque (T) = 471.0 kip·in, polar moment of inertia (J) = 35.0000 in⁴, shear stress (tau) = 27.7100 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$c = 2.0591\ \text{in}$$

6. Step 6 — Check: returning c = 2.0591 in to

   $$\tau = T c / J$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 4.1183 — kept a factor of two that cancels in the correct rearrangement.
• 1.0296 — dropped that same factor in the other direction.
• 2.2650 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

A drive shaft's torsional strain at the outer fiber is computed under applied torque. Given radial distance (r) = 1.0000 in; shaft length (L) = 47.0000 in; shear (torsional) strain (gamma) = 0.0081, determine the angle of twist (phi) in rad.

Start with the thinking

• The governing relation printed in this handbook section is Torsional strain (circular shaft).
• Everything except phi is given, so isolate phi symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Torsional strain in a circular shaft varies linearly with radial distance from the axis for a given angle of twist.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\gamma = \dfrac{r \phi}{L}$$

2. Step 2 — Rearrange symbolically for phi:

   $$\phi = \dfrac{\gamma L}{r}$$

3. Step 3 — List the givens: radial distance (r) = 1.0000 in, shaft length (L) = 47.0000 in, shear (torsional) strain (gamma) = 0.0081.

4. Step 4 — Substitute the given values:

   $$\phi = \dfrac{0.0081 47.0000}{1.0000}$$

5. Step 5 — Evaluate:

   $$\phi = 0.3788\ \text{rad}$$

6. Step 6 — Check: returning phi = 0.3788 rad to

   $$\gamma = \dfrac{r \phi}{L}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.7576 — kept a factor of two that cancels in the correct rearrangement.
• 0.1894 — dropped that same factor in the other direction.
• 0.4167 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Torsional Strain

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 235.0 kip·in; outer radius (c) = 3.5000 in; shear stress (tau) = 26.3900 ksi, determine the polar moment of inertia (J) in in⁴.

Start with the thinking

• The governing relation printed in this handbook section is Torsional shear stress.
• Everything except J is given, so isolate J symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = T c / J$$

2. Step 2 — Rearrange the relation so that J stands alone on the left-hand side.

3. Step 3 — List the givens: torque (T) = 235.0 kip·in, outer radius (c) = 3.5000 in, shear stress (tau) = 26.3900 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$J = 31.1671\ \text{in⁴}$$

6. Step 6 — Check: returning J = 31.1671 in⁴ to

   $$\tau = T c / J$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 62.3342 — kept a factor of two that cancels in the correct rearrangement.
• 15.5836 — dropped that same factor in the other direction.
• 34.2838 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

A drill rod's torsional strain is measured during a twist test. Given radial distance (r) = 3.2500 in; angle of twist (phi) = 0.0700 rad; shear (torsional) strain (gamma) = 0.0099, determine the shaft length (L) in in.

Start with the thinking

• The governing relation printed in this handbook section is Torsional strain (circular shaft).
• Everything except L is given, so isolate L symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Torsional strain in a circular shaft varies linearly with radial distance from the axis for a given angle of twist.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\gamma = \dfrac{r \phi}{L}$$

2. Step 2 — Rearrange symbolically for L:

   $$L = \dfrac{r \phi}{\gamma}$$

3. Step 3 — List the givens: radial distance (r) = 3.2500 in, angle of twist (phi) = 0.0700 rad, shear (torsional) strain (gamma) = 0.0099.

4. Step 4 — Substitute the given values:

   $$L = \dfrac{3.2500 0.0700}{0.0099}$$

5. Step 5 — Evaluate:

   $$L = 23.0030\ \text{in}$$

6. Step 6 — Check: returning L = 23.0030 in to

   $$\gamma = \dfrac{r \phi}{L}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 46.0061 — kept a factor of two that cancels in the correct rearrangement.
• 11.5015 — dropped that same factor in the other direction.
• 25.3033 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Torsional Strain

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 372.0 kip·in; outer radius (c) = 3.6000 in; polar moment of inertia (J) = 37.0000 in⁴, determine the shear stress (tau) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Torsional shear stress.
• Everything except tau is given, so isolate tau symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = T c / J$$

2. Step 2 — Rearrange the relation so that tau stands alone on the left-hand side.

3. Step 3 — List the givens: torque (T) = 372.0 kip·in, outer radius (c) = 3.6000 in, polar moment of inertia (J) = 37.0000 in⁴.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\tau = 36.1946\ \text{ksi}$$

6. Step 6 — Check: returning tau = 36.1946 ksi to

   $$\tau = T c / J$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 72.3892 — kept a factor of two that cancels in the correct rearrangement.
• 18.0973 — dropped that same factor in the other direction.
• 39.8141 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

A machine shaft's torsional strain distribution is checked across its cross section. Given radial distance (r) = 1.2000 in; angle of twist (phi) = 0.0770 rad; shaft length (L) = 9.0000 in, determine the shear (torsional) strain (gamma).

Start with the thinking

• The governing relation printed in this handbook section is Torsional strain (circular shaft).
• Everything except gamma is given, so isolate gamma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Torsional strain in a circular shaft varies linearly with radial distance from the axis for a given angle of twist.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\gamma = \dfrac{r \phi}{L}$$

2. Step 2 — Rearrange symbolically for gamma:

   $$\gamma = \dfrac{r \phi}{L}$$

3. Step 3 — List the givens: radial distance (r) = 1.2000 in, angle of twist (phi) = 0.0770 rad, shaft length (L) = 9.0000 in.

4. Step 4 — Substitute the given values:

   $$\gamma = \dfrac{1.2000 0.0770}{9.0000}$$

5. Step 5 — Evaluate:

   $$\gamma = 0.0103$$

6. Step 6 — Check: returning gamma = 0.0103 to

   $$\gamma = \dfrac{r \phi}{L}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0205 — kept a factor of two that cancels in the correct rearrangement.
• 0.0051 — dropped that same factor in the other direction.
• 0.0113 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Torsional Strain