A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,150 kip·in; distance to extreme fiber (c) = 7.8000 in; moment of inertia (I) = 1,670 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) = 4,150 kip·in, distance to extreme fiber (c) = 7.8000 in, moment of inertia (I) = 1,670 in⁴.

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

5. Step 5 — Evaluate:

   $$\sigma = 19.3832\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 19.3832 ksi to

   $$\sigma = M c / I$$

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

Why the other options are there

• 38.7665 — kept a factor of two that cancels in the correct rearrangement.
• 9.6916 — dropped that same factor in the other direction.
• 21.3216 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

A steel wide-flange beam's stresses in beams are evaluated under a point load. Given bending moment (M) = 930.0 kip-in; distance to extreme fiber (c) = 8.7000 in; moment of inertia (I) = 175.0 in^4, determine the bending stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Bending stress in beams.
• 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.
• Stresses in beams from bending are computed from the flexure formula relating moment, distance to extreme fiber, and moment of inertia.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = \dfrac{M c}{I}$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = \dfrac{M c}{I}$$

3. Step 3 — List the givens: bending moment (M) = 930.0 kip-in, distance to extreme fiber (c) = 8.7000 in, moment of inertia (I) = 175.0 in^4.

4. Step 4 — Substitute the given values:

   $$\sigma = \dfrac{930.0 8.7000}{175.0}$$

5. Step 5 — Evaluate:

   $$\sigma = 46.2343\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 46.2343 ksi to

   $$\sigma = \dfrac{M c}{I}$$

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

Why the other options are there

• 92.4686 — kept a factor of two that cancels in the correct rearrangement.
• 23.1171 — dropped that same factor in the other direction.
• 50.8577 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Stresses in Beams

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 11.6000 in; moment of inertia (I) = 1,700 in⁴; bending stress (sigma) = 46.0500 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) = 11.6000 in, moment of inertia (I) = 1,700 in⁴, bending stress (sigma) = 46.0500 ksi.

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

5. Step 5 — Evaluate:

   $$M = 6749\ \text{kip·in}$$

6. Step 6 — Check: returning M = 6,749 kip·in to

   $$\sigma = M c / I$$

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

Why the other options are there

• 13,497 — kept a factor of two that cancels in the correct rearrangement.
• 3,374 — dropped that same factor in the other direction.
• 7,424 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

A cantilevered beam's stresses in beams are computed at the fixed support. Given distance to extreme fiber (c) = 2.0000 in; moment of inertia (I) = 2,310 in^4; bending stress (sigma) = 35.1000 ksi, determine the bending moment (M) in kip-in.

Start with the thinking

• The governing relation printed in this handbook section is Bending stress in beams.
• 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.
• Stresses in beams from bending are computed from the flexure formula relating moment, distance to extreme fiber, and moment of inertia.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = \dfrac{M c}{I}$$

2. Step 2 — Rearrange symbolically for M:

   $$M = \dfrac{\sigma I}{c}$$

3. Step 3 — List the givens: distance to extreme fiber (c) = 2.0000 in, moment of inertia (I) = 2,310 in^4, bending stress (sigma) = 35.1000 ksi.

4. Step 4 — Substitute the given values:

   $$M = \dfrac{35.1000 2310}{2.0000}$$

5. Step 5 — Evaluate:

   $$M = 40541\ \text{kip-in}$$

6. Step 6 — Check: returning M = 40,541 kip-in to

   $$\sigma = \dfrac{M c}{I}$$

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

Why the other options are there

• 81,081 — kept a factor of two that cancels in the correct rearrangement.
• 20,270 — dropped that same factor in the other direction.
• 44,595 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Stresses in Beams

A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,550 kip·in; moment of inertia (I) = 690.0 in⁴; bending stress (sigma) = 14.6000 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) = 4,550 kip·in, moment of inertia (I) = 690.0 in⁴, bending stress (sigma) = 14.6000 ksi.

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

5. Step 5 — Evaluate:

   $$c = 2.2141\ \text{in}$$

6. Step 6 — Check: returning c = 2.2141 in to

   $$\sigma = M c / I$$

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

Why the other options are there

• 4.4281 — kept a factor of two that cancels in the correct rearrangement.
• 1.1070 — dropped that same factor in the other direction.
• 2.4355 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

A simply supported timber beam's stresses in beams are checked at midspan. Given bending moment (M) = 485.0 kip-in; moment of inertia (I) = 2,015 in^4; bending stress (sigma) = 35.8000 ksi, determine the distance to extreme fiber (c) in in.

Start with the thinking

• The governing relation printed in this handbook section is Bending stress in beams.
• 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.
• Stresses in beams from bending are computed from the flexure formula relating moment, distance to extreme fiber, and moment of inertia.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = \dfrac{M c}{I}$$

2. Step 2 — Rearrange symbolically for c:

   $$c = \dfrac{\sigma I}{M}$$

3. Step 3 — List the givens: bending moment (M) = 485.0 kip-in, moment of inertia (I) = 2,015 in^4, bending stress (sigma) = 35.8000 ksi.

4. Step 4 — Substitute the given values:

   $$c = \dfrac{35.8000 2015}{485.0}$$

5. Step 5 — Evaluate:

   $$c = 148.7\ \text{in}$$

6. Step 6 — Check: returning c = 148.7 in to

   $$\sigma = \dfrac{M c}{I}$$

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

Why the other options are there

• 297.5 — kept a factor of two that cancels in the correct rearrangement.
• 74.3680 — dropped that same factor in the other direction.
• 163.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Stresses in Beams

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,410 kip·in; distance to extreme fiber (c) = 9.6000 in; bending stress (sigma) = 33.8600 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) = 2,410 kip·in, distance to extreme fiber (c) = 9.6000 in, bending stress (sigma) = 33.8600 ksi.

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

5. Step 5 — Evaluate:

   $$I = 683.3\ \text{in⁴}$$

6. Step 6 — Check: returning I = 683.3 in⁴ to

   $$\sigma = M c / I$$

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

Why the other options are there

• 1,367 — kept a factor of two that cancels in the correct rearrangement.
• 341.6 — dropped that same factor in the other direction.
• 751.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

A steel wide-flange beam's stresses in beams are evaluated under a point load. Given bending moment (M) = 1,150 kip-in; distance to extreme fiber (c) = 9.4000 in; bending stress (sigma) = 59.7000 ksi, determine the moment of inertia (I) in in^4.

Start with the thinking

• The governing relation printed in this handbook section is Bending stress in beams.
• 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.
• Stresses in beams from bending are computed from the flexure formula relating moment, distance to extreme fiber, and moment of inertia.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = \dfrac{M c}{I}$$

2. Step 2 — Rearrange symbolically for I:

   $$I = \dfrac{M c}{\sigma}$$

3. Step 3 — List the givens: bending moment (M) = 1,150 kip-in, distance to extreme fiber (c) = 9.4000 in, bending stress (sigma) = 59.7000 ksi.

4. Step 4 — Substitute the given values:

   $$I = \dfrac{1150 9.4000}{59.7000}$$

5. Step 5 — Evaluate:

   I = 181.1\ \text{in^4}

6. Step 6 — Check: returning I = 181.1 in^4 to

   $$\sigma = \dfrac{M c}{I}$$

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

Why the other options are there

• 362.1 — kept a factor of two that cancels in the correct rearrangement.
• 90.5360 — dropped that same factor in the other direction.
• 199.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Stresses in Beams

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,560 kip·in; distance to extreme fiber (c) = 11.3000 in; moment of inertia (I) = 1,700 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) = 1,560 kip·in, distance to extreme fiber (c) = 11.3000 in, moment of inertia (I) = 1,700 in⁴.

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

5. Step 5 — Evaluate:

   $$\sigma = 10.3694\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 10.3694 ksi to

   $$\sigma = M c / I$$

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

Why the other options are there

• 20.7388 — kept a factor of two that cancels in the correct rearrangement.
• 5.1847 — dropped that same factor in the other direction.
• 11.4064 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

A cantilevered beam's stresses in beams are computed at the fixed support. Given bending moment (M) = 1,575 kip-in; distance to extreme fiber (c) = 8.2000 in; moment of inertia (I) = 530.0 in^4, determine the bending stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Bending stress in beams.
• 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.
• Stresses in beams from bending are computed from the flexure formula relating moment, distance to extreme fiber, and moment of inertia.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = \dfrac{M c}{I}$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = \dfrac{M c}{I}$$

3. Step 3 — List the givens: bending moment (M) = 1,575 kip-in, distance to extreme fiber (c) = 8.2000 in, moment of inertia (I) = 530.0 in^4.

4. Step 4 — Substitute the given values:

   $$\sigma = \dfrac{1575 8.2000}{530.0}$$

5. Step 5 — Evaluate:

   $$\sigma = 24.3679\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 24.3679 ksi to

   $$\sigma = \dfrac{M c}{I}$$

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

Why the other options are there

• 48.7358 — kept a factor of two that cancels in the correct rearrangement.
• 12.1840 — dropped that same factor in the other direction.
• 26.8047 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Stresses in Beams