A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,790 kip·in; distance to extreme fiber (c) = 12.7000 in; moment of inertia (I) = 1,250 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) = 1,790 kip·in, distance to extreme fiber (c) = 12.7000 in, moment of inertia (I) = 1,250 in⁴.

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

5. Step 5 — Evaluate:

   $$\sigma = 18.1864\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 18.1864 ksi to

   $$\sigma = M c / I$$

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

Why the other options are there

• 36.3728 — kept a factor of two that cancels in the correct rearrangement.
• 9.0932 — dropped that same factor in the other direction.
• 20.0050 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Principal Stresses

A shaft under combined bending and torsion is checked using principal stresses. Given normal stress x (sigma_x) = 5.0000 ksi; normal stress y (sigma_y) = -14.0000 ksi; shear stress (tau_xy) = 9.6000 ksi, determine the max principal stress (sigma_1) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Principal stresses (plane stress).
• Everything except sigma_1 is given, so isolate sigma_1 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.
• Principal stresses on a stressed element are the maximum and minimum normal stresses found from Mohr's circle relations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

2. Step 2 — Rearrange symbolically for sigma_1:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

3. Step 3 — List the givens: normal stress x (sigma_x) = 5.0000 ksi, normal stress y (sigma_y) = -14.0000 ksi, shear stress (tau_xy) = 9.6000 ksi.

4. Step 4 — Substitute the given values:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + 9.6000^2}$$

5. Step 5 — Evaluate:

   $$\sigma_{1} = 9.0059\ \text{ksi}$$

6. Step 6 — Check: returning sigma_1 = 9.0059 ksi to

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

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

Why the other options are there

• 18.0118 — kept a factor of two that cancels in the correct rearrangement.
• 4.5030 — dropped that same factor in the other direction.
• 9.9065 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Principal Stresses

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 14.3000 in; moment of inertia (I) = 580.0 in⁴; bending stress (sigma) = 21.3400 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) = 14.3000 in, moment of inertia (I) = 580.0 in⁴, bending stress (sigma) = 21.3400 ksi.

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

5. Step 5 — Evaluate:

   $$M = 865.5\ \text{kip·in}$$

6. Step 6 — Check: returning M = 865.5 kip·in to

   $$\sigma = M c / I$$

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

Why the other options are there

• 1,731 — kept a factor of two that cancels in the correct rearrangement.
• 432.8 — dropped that same factor in the other direction.
• 952.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Principal Stresses

A plate with combined normal and shear stress is analyzed for principal stresses. Given normal stress x (sigma_x) = -18.0000 ksi; normal stress y (sigma_y) = -13.5000 ksi; shear stress (tau_xy) = 12.8000 ksi, determine the max principal stress (sigma_1) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Principal stresses (plane stress).
• Everything except sigma_1 is given, so isolate sigma_1 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.
• Principal stresses on a stressed element are the maximum and minimum normal stresses found from Mohr's circle relations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

2. Step 2 — Rearrange symbolically for sigma_1:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

3. Step 3 — List the givens: normal stress x (sigma_x) = -18.0000 ksi, normal stress y (sigma_y) = -13.5000 ksi, shear stress (tau_xy) = 12.8000 ksi.

4. Step 4 — Substitute the given values:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + 12.8000^2}$$

5. Step 5 — Evaluate:

   $$\sigma_{1} = -2.7538\ \text{ksi}$$

6. Step 6 — Check: returning sigma_1 = -2.7538 ksi to

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

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

Why the other options are there

• -5.5075 — kept a factor of two that cancels in the correct rearrangement.
• -1.3769 — dropped that same factor in the other direction.
• -3.0291 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Principal Stresses

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,860 kip·in; moment of inertia (I) = 190.0 in⁴; bending stress (sigma) = 48.7600 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) = 2,860 kip·in, moment of inertia (I) = 190.0 in⁴, bending stress (sigma) = 48.7600 ksi.

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

5. Step 5 — Evaluate:

   $$c = 3.2393\ \text{in}$$

6. Step 6 — Check: returning c = 3.2393 in to

   $$\sigma = M c / I$$

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

Why the other options are there

• 6.4786 — kept a factor of two that cancels in the correct rearrangement.
• 1.6197 — dropped that same factor in the other direction.
• 3.5632 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Principal Stresses

A pressure vessel wall's principal stresses are computed from biaxial loading. Given normal stress x (sigma_x) = -14.5000 ksi; normal stress y (sigma_y) = 8.0000 ksi; shear stress (tau_xy) = 7.8000 ksi, determine the max principal stress (sigma_1) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Principal stresses (plane stress).
• Everything except sigma_1 is given, so isolate sigma_1 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.
• Principal stresses on a stressed element are the maximum and minimum normal stresses found from Mohr's circle relations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

2. Step 2 — Rearrange symbolically for sigma_1:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

3. Step 3 — List the givens: normal stress x (sigma_x) = -14.5000 ksi, normal stress y (sigma_y) = 8.0000 ksi, shear stress (tau_xy) = 7.8000 ksi.

4. Step 4 — Substitute the given values:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + 7.8000^2}$$

5. Step 5 — Evaluate:

   $$\sigma_{1} = 10.4395\ \text{ksi}$$

6. Step 6 — Check: returning sigma_1 = 10.4395 ksi to

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

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

Why the other options are there

• 20.8790 — kept a factor of two that cancels in the correct rearrangement.
• 5.2198 — dropped that same factor in the other direction.
• 11.4835 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Principal Stresses

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,590 kip·in; distance to extreme fiber (c) = 11.2000 in; bending stress (sigma) = 6.9800 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) = 2,590 kip·in, distance to extreme fiber (c) = 11.2000 in, bending stress (sigma) = 6.9800 ksi.

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

5. Step 5 — Evaluate:

   $$I = 4156\ \text{in⁴}$$

6. Step 6 — Check: returning I = 4,156 in⁴ to

   $$\sigma = M c / I$$

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

Why the other options are there

• 8,312 — kept a factor of two that cancels in the correct rearrangement.
• 2,078 — dropped that same factor in the other direction.
• 4,571 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Principal Stresses

A shaft under combined bending and torsion is checked using principal stresses. Given normal stress x (sigma_x) = 8.0000 ksi; normal stress y (sigma_y) = 0.0000 ksi; shear stress (tau_xy) = 14.6000 ksi, determine the max principal stress (sigma_1) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Principal stresses (plane stress).
• Everything except sigma_1 is given, so isolate sigma_1 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.
• Principal stresses on a stressed element are the maximum and minimum normal stresses found from Mohr's circle relations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

2. Step 2 — Rearrange symbolically for sigma_1:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

3. Step 3 — List the givens: normal stress x (sigma_x) = 8.0000 ksi, normal stress y (sigma_y) = 0.0000 ksi, shear stress (tau_xy) = 14.6000 ksi.

4. Step 4 — Substitute the given values:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + 14.6000^2}$$

5. Step 5 — Evaluate:

   $$\sigma_{1} = 19.1380\ \text{ksi}$$

6. Step 6 — Check: returning sigma_1 = 19.1380 ksi to

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

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

Why the other options are there

• 38.2761 — kept a factor of two that cancels in the correct rearrangement.
• 9.5690 — dropped that same factor in the other direction.
• 21.0518 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Principal Stresses

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,800 kip·in; distance to extreme fiber (c) = 5.0000 in; moment of inertia (I) = 920.0 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) = 3,800 kip·in, distance to extreme fiber (c) = 5.0000 in, moment of inertia (I) = 920.0 in⁴.

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

5. Step 5 — Evaluate:

   $$\sigma = 20.6522\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 20.6522 ksi to

   $$\sigma = M c / I$$

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

Why the other options are there

• 41.3043 — kept a factor of two that cancels in the correct rearrangement.
• 10.3261 — dropped that same factor in the other direction.
• 22.7174 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Principal Stresses

A plate with combined normal and shear stress is analyzed for principal stresses. Given normal stress x (sigma_x) = -19.0000 ksi; normal stress y (sigma_y) = -10.5000 ksi; shear stress (tau_xy) = 12.0000 ksi, determine the max principal stress (sigma_1) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Principal stresses (plane stress).
• Everything except sigma_1 is given, so isolate sigma_1 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.
• Principal stresses on a stressed element are the maximum and minimum normal stresses found from Mohr's circle relations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

2. Step 2 — Rearrange symbolically for sigma_1:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

3. Step 3 — List the givens: normal stress x (sigma_x) = -19.0000 ksi, normal stress y (sigma_y) = -10.5000 ksi, shear stress (tau_xy) = 12.0000 ksi.

4. Step 4 — Substitute the given values:

   $$\sigma_{1} = \dfrac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + 12.0000^2}$$

5. Step 5 — Evaluate:

   $$\sigma_{1} = -2.0196\ \text{ksi}$$

6. Step 6 — Check: returning sigma_1 = -2.0196 ksi to

   $$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

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

Why the other options are there

• -4.0392 — kept a factor of two that cancels in the correct rearrangement.
• -1.0098 — dropped that same factor in the other direction.
• -2.2216 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Principal Stresses