A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,640 kip·in; distance to extreme fiber (c) = 6.9000 in; moment of inertia (I) = 630.0 in⁴, determine the bending stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except sigma is given, so isolate sigma 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:

   $$\sigma = M c / I$$

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

3. Step 3 — List the givens: moment (M) = 2,640 kip·in, distance to extreme fiber (c) = 6.9000 in, moment of inertia (I) = 630.0 in⁴.

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

5. Step 5 — Evaluate:

   $$\sigma = 28.9143\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 28.9143 ksi to

   $$\sigma = M c / I$$

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

Why the other options are there

• 57.8286 — kept a factor of two that cancels in the correct rearrangement.
• 14.4571 — dropped that same factor in the other direction.
• 31.8057 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 470.0 kip·in; outer radius (c) = 0.5000 in; polar moment of inertia (J) = 485.0 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) = 470.0 kip·in, outer radius (c) = 0.5000 in, polar moment of inertia (J) = 485.0 in⁴.

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

5. Step 5 — Evaluate:

   $$\tau = 0.4845\ \text{ksi}$$

6. Step 6 — Check: returning tau = 0.4845 ksi to

   $$\tau = T c / J$$

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

Why the other options are there

• 0.9691 — kept a factor of two that cancels in the correct rearrangement.
• 0.2423 — dropped that same factor in the other direction.
• 0.5330 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain

A bolted lap joint's shear stress-strain relation determines slip. Given shear modulus (G) = 7,240 ksi; shear strain (gamma) = 0.0065 rad, determine the shear stress (tau) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Shear stress-strain (shear modulus).
• 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.
• Shear stress-strain behavior in the elastic range follows a linear relation through the shear modulus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = G \gamma$$

2. Step 2 — Rearrange symbolically for tau:

   $$\tau = G \gamma$$

3. Step 3

   $$List the givens: shear modulus (G) = 7,240 ksi, shear strain (gamma) = 0.0065 rad$$

4. Step 4 — Substitute the given values:

   $$\tau = 7240 0.0065$$

5. Step 5 — Evaluate:

   $$\tau = 47.0600\ \text{ksi}$$

6. Step 6 — Check: returning tau = 47.0600 ksi to

   $$\tau = G \gamma$$

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

Why the other options are there

• 94.1200 — kept a factor of two that cancels in the correct rearrangement.
• 23.5300 — dropped that same factor in the other direction.
• 51.7660 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Shear Stress-Strain

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 9.9000 in; moment of inertia (I) = 160.0 in⁴; bending stress (sigma) = 29.0100 ksi, determine the moment (M) in kip·in.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except M is given, so isolate M 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:

   $$\sigma = M c / I$$

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

3. Step 3 — List the givens: distance to extreme fiber (c) = 9.9000 in, moment of inertia (I) = 160.0 in⁴, bending stress (sigma) = 29.0100 ksi.

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

5. Step 5 — Evaluate:

   $$M = 468.8\ \text{kip·in}$$

6. Step 6 — Check: returning M = 468.8 kip·in to

   $$\sigma = M c / I$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain

A mechanics of materials problem uses Torsional shear stress. Given outer radius (c) = 5.7000 in; polar moment of inertia (J) = 380.0 in⁴; shear stress (tau) = 14.8000 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) = 5.7000 in, polar moment of inertia (J) = 380.0 in⁴, shear stress (tau) = 14.8000 ksi.

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

5. Step 5 — Evaluate:

   $$T = 986.7\ \text{kip·in}$$

6. Step 6 — Check: returning T = 986.7 kip·in to

   $$\tau = T c / J$$

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

Why the other options are there

• 1,973 — kept a factor of two that cancels in the correct rearrangement.
• 493.3 — dropped that same factor in the other direction.
• 1,085 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain

A rivet in single shear is analyzed using shear stress-strain behavior. Given shear strain (gamma) = 0.0096 rad; shear stress (tau) = 94.1000 ksi, determine the shear modulus (G) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Shear stress-strain (shear modulus).
• Everything except G is given, so isolate G 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.
• Shear stress-strain behavior in the elastic range follows a linear relation through the shear modulus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = G \gamma$$

2. Step 2 — Rearrange symbolically for G:

   $$G = \dfrac{\tau}{\gamma}$$

3. Step 3

   $$List the givens: shear strain (gamma) = 0.0096 rad, shear stress (tau) = 94.1000 ksi$$

4. Step 4 — Substitute the given values:

   $$G = \dfrac{94.1000}{0.0096}$$

5. Step 5 — Evaluate:

   $$G = 9802\ \text{ksi}$$

6. Step 6 — Check: returning G = 9,802 ksi to

   $$\tau = G \gamma$$

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

Why the other options are there

• 19,604 — kept a factor of two that cancels in the correct rearrangement.
• 4,901 — dropped that same factor in the other direction.
• 10,782 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Shear Stress-Strain

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,950 kip·in; moment of inertia (I) = 380.0 in⁴; bending stress (sigma) = 42.0300 ksi, determine the distance to extreme fiber (c) in in.

Start with the thinking

• The governing relation printed in this handbook section is Flexural 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:

   $$\sigma = M c / I$$

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

3. Step 3 — List the givens: moment (M) = 3,950 kip·in, moment of inertia (I) = 380.0 in⁴, bending stress (sigma) = 42.0300 ksi.

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

5. Step 5 — Evaluate:

   $$c = 4.0434\ \text{in}$$

6. Step 6 — Check: returning c = 4.0434 in to

   $$\sigma = M c / I$$

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

Why the other options are there

• 8.0868 — kept a factor of two that cancels in the correct rearrangement.
• 2.0217 — dropped that same factor in the other direction.
• 4.4477 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 370.0 kip·in; polar moment of inertia (J) = 452.0 in⁴; shear stress (tau) = 13.9100 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) = 370.0 kip·in, polar moment of inertia (J) = 452.0 in⁴, shear stress (tau) = 13.9100 ksi.

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

5. Step 5 — Evaluate:

   $$c = 16.9928\ \text{in}$$

6. Step 6 — Check: returning c = 16.9928 in to

   $$\tau = T c / J$$

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

Why the other options are there

• 33.9855 — kept a factor of two that cancels in the correct rearrangement.
• 8.4964 — dropped that same factor in the other direction.
• 18.6920 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain

A rubber mount's shear stress-strain response is tested for stiffness. Given shear modulus (G) = 5,300 ksi; shear stress (tau) = 80.0000 ksi, determine the shear strain (gamma) in rad.

Start with the thinking

• The governing relation printed in this handbook section is Shear stress-strain (shear modulus).
• 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.
• Shear stress-strain behavior in the elastic range follows a linear relation through the shear modulus.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\tau = G \gamma$$

2. Step 2 — Rearrange symbolically for gamma:

   $$\gamma = \dfrac{\tau}{G}$$

3. Step 3

   $$List the givens: shear modulus (G) = 5,300 ksi, shear stress (tau) = 80.0000 ksi$$

4. Step 4 — Substitute the given values:

   $$\gamma = \dfrac{80.0000}{5300}$$

5. Step 5 — Evaluate:

   $$\gamma = 0.0151\ \text{rad}$$

6. Step 6 — Check: returning gamma = 0.0151 rad to

   $$\tau = G \gamma$$

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

Why the other options are there

• 0.0302 — kept a factor of two that cancels in the correct rearrangement.
• 0.0075 — dropped that same factor in the other direction.
• 0.0166 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Shear Stress-Strain

A mechanics of materials problem uses Flexural stress. Given moment (M) = 390.0 kip·in; distance to extreme fiber (c) = 2.9000 in; bending stress (sigma) = 58.7500 ksi, determine the moment of inertia (I) in in⁴.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except I is given, so isolate I 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:

   $$\sigma = M c / I$$

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

3. Step 3 — List the givens: moment (M) = 390.0 kip·in, distance to extreme fiber (c) = 2.9000 in, bending stress (sigma) = 58.7500 ksi.

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

5. Step 5 — Evaluate:

   $$I = 19.2511\ \text{in⁴}$$

6. Step 6 — Check: returning I = 19.2511 in⁴ to

   $$\sigma = M c / I$$

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

Why the other options are there

• 38.5021 — kept a factor of two that cancels in the correct rearrangement.
• 9.6255 — dropped that same factor in the other direction.
• 21.1762 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shear Stress-Strain