Shearing Force and Bending Moment Sign Conventions

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

6 formulas

10 exam-style examples

~57 min

All Mechanics of Materials lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

1. The bending moment is positive if it produces bending of the beam concave upward (compression in top fibers and tension
2. The shearing force is positive if the right portion of the beam tends to shear downward with respect to the left.
POSITIVE BENDING NEGATIVE BENDING
POSITIVE SHEAR NEGATIVE SHEAR
The relationship between the load (w), shear (V), and moment (M) equations are:

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Flexural stress — solve for bending stress — Shearing Force and Bending Moment Sign Conventions

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,880 kip·in; distance to extreme fiber (c) = 3.4000 in; moment of inertia (I) = 1,450 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 1,880 kip·in
$distance to extreme fiber (c) = 3.4000 in$
$moment of inertia (I) = 1,450 in^{4}$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 1,880 kip·in, distance to extreme fiber (c) = 3.4000 in, moment of inertia (I) = 1,450 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 4.4083\ \text{ksi}$
Step 6 — Check: returning sigma = 4.4083 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 4.4083\ \text{ksi}

Why the other options are there

8.8166 — kept a factor of two that cancels in the correct rearrangement.
2.2041 — dropped that same factor in the other direction.
4.8491 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 2

Flexural stress — solve for moment — Shearing Force and Bending Moment Sign Conventions (2)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 14.4000 in; moment of inertia (I) = 1,910 in⁴; bending stress (sigma) = 50.7900 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 14.4000 in$
$moment of inertia (I) = 1,910 in^{4}$
$bending stress (sigma) = 50.7900 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that M stands alone on the left-hand side.
Step 3 — List the givens: distance to extreme fiber (c) = 14.4000 in, moment of inertia (I) = 1,910 in⁴, bending stress (sigma) = 50.7900 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 6737\ \text{kip·in}$
Step 6 — Check: returning M = 6,737 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 6737\ \text{kip·in}

Why the other options are there

13,473 — kept a factor of two that cancels in the correct rearrangement.
3,368 — dropped that same factor in the other direction.
7,410 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 3

Flexural stress — solve for distance to extreme fiber — Shearing Force and Bending Moment Sign Conventions (3)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,910 kip·in; moment of inertia (I) = 1,770 in⁴; bending stress (sigma) = 5.1000 ksi, determine the distance to extreme fiber (c) in in.

Given

moment (M) = 2,910 kip·in
$moment of inertia (I) = 1,770 in^{4}$
$bending stress (sigma) = 5.1000 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that c stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 2,910 kip·in, moment of inertia (I) = 1,770 in⁴, bending stress (sigma) = 5.1000 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 3.1021\ \text{in}$
Step 6 — Check: returning c = 3.1021 in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 3.1021\ \text{in}

Why the other options are there

6.2041 — kept a factor of two that cancels in the correct rearrangement.
1.5510 — dropped that same factor in the other direction.
3.4123 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 4

Flexural stress — solve for moment of inertia — Shearing Force and Bending Moment Sign Conventions (4)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 170.0 kip·in; distance to extreme fiber (c) = 7.1000 in; bending stress (sigma) = 14.5800 ksi, determine the moment of inertia (I) in in⁴.

Given

moment (M) = 170.0 kip·in
$distance to extreme fiber (c) = 7.1000 in$
$bending stress (sigma) = 14.5800 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that I stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 170.0 kip·in, distance to extreme fiber (c) = 7.1000 in, bending stress (sigma) = 14.5800 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$I = 82.7846\ \text{in⁴}$
Step 6 — Check: returning I = 82.7846 in⁴ to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 82.7846\ \text{in⁴}

Why the other options are there

165.6 — kept a factor of two that cancels in the correct rearrangement.
41.3923 — dropped that same factor in the other direction.
91.0631 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 5

Flexural stress — solve for bending stress (case 2) — Shearing Force and Bending Moment Sign Conventions (5)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,730 kip·in; distance to extreme fiber (c) = 11.1000 in; moment of inertia (I) = 760.0 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 1,730 kip·in
$distance to extreme fiber (c) = 11.1000 in$
$moment of inertia (I) = 760.0 in^{4}$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 1,730 kip·in, distance to extreme fiber (c) = 11.1000 in, moment of inertia (I) = 760.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 25.2671\ \text{ksi}$
Step 6 — Check: returning sigma = 25.2671 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 25.2671\ \text{ksi}

Why the other options are there

50.5342 — kept a factor of two that cancels in the correct rearrangement.
12.6336 — dropped that same factor in the other direction.
27.7938 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 6

Flexural stress — solve for moment (case 2) — Shearing Force and Bending Moment Sign Conventions (6)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 8.7000 in; moment of inertia (I) = 1,280 in⁴; bending stress (sigma) = 24.0500 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 8.7000 in$
$moment of inertia (I) = 1,280 in^{4}$
$bending stress (sigma) = 24.0500 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that M stands alone on the left-hand side.
Step 3 — List the givens: distance to extreme fiber (c) = 8.7000 in, moment of inertia (I) = 1,280 in⁴, bending stress (sigma) = 24.0500 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 3538\ \text{kip·in}$
Step 6 — Check: returning M = 3,538 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 3538\ \text{kip·in}

Why the other options are there

7,077 — kept a factor of two that cancels in the correct rearrangement.
1,769 — dropped that same factor in the other direction.
3,892 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 7

Flexural stress — solve for distance to extreme fiber (case 2) — Shearing Force and Bending Moment Sign Conventions (7)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,430 kip·in; moment of inertia (I) = 560.0 in⁴; bending stress (sigma) = 55.9200 ksi, determine the distance to extreme fiber (c) in in.

Given

moment (M) = 2,430 kip·in
$moment of inertia (I) = 560.0 in^{4}$
$bending stress (sigma) = 55.9200 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that c stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 2,430 kip·in, moment of inertia (I) = 560.0 in⁴, bending stress (sigma) = 55.9200 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 12.8869\ \text{in}$
Step 6 — Check: returning c = 12.8869 in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 12.8869\ \text{in}

Why the other options are there

25.7738 — kept a factor of two that cancels in the correct rearrangement.
6.4435 — dropped that same factor in the other direction.
14.1756 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 8

Flexural stress — solve for moment of inertia (case 2) — Shearing Force and Bending Moment Sign Conventions (8)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,910 kip·in; distance to extreme fiber (c) = 14.9000 in; bending stress (sigma) = 27.8500 ksi, determine the moment of inertia (I) in in⁴.

Given

moment (M) = 3,910 kip·in
$distance to extreme fiber (c) = 14.9000 in$
$bending stress (sigma) = 27.8500 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that I stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 3,910 kip·in, distance to extreme fiber (c) = 14.9000 in, bending stress (sigma) = 27.8500 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$I = 2092\ \text{in⁴}$
Step 6 — Check: returning I = 2,092 in⁴ to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 2092\ \text{in⁴}

Why the other options are there

4,184 — kept a factor of two that cancels in the correct rearrangement.
1,046 — dropped that same factor in the other direction.
2,301 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 9

Flexural stress — solve for bending stress (case 3) — Shearing Force and Bending Moment Sign Conventions (9)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 340.0 kip·in; distance to extreme fiber (c) = 13.7000 in; moment of inertia (I) = 220.0 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 340.0 kip·in
$distance to extreme fiber (c) = 13.7000 in$
$moment of inertia (I) = 220.0 in^{4}$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.
Step 3 — List the givens: moment (M) = 340.0 kip·in, distance to extreme fiber (c) = 13.7000 in, moment of inertia (I) = 220.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 21.1727\ \text{ksi}$
Step 6 — Check: returning sigma = 21.1727 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 21.1727\ \text{ksi}

Why the other options are there

42.3455 — kept a factor of two that cancels in the correct rearrangement.
10.5864 — dropped that same factor in the other direction.
23.2900 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 10

Flexural stress — solve for moment (case 3) — Shearing Force and Bending Moment Sign Conventions (10)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 7.3000 in; moment of inertia (I) = 420.0 in⁴; bending stress (sigma) = 27.7800 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 7.3000 in$
$moment of inertia (I) = 420.0 in^{4}$
$bending stress (sigma) = 27.7800 ksi$

Find

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

Step 1 — State the governing relation:
$\sigma = M c / I$
Step 2 — Rearrange the relation so that M stands alone on the left-hand side.
Step 3 — List the givens: distance to extreme fiber (c) = 7.3000 in, moment of inertia (I) = 420.0 in⁴, bending stress (sigma) = 27.7800 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 1598\ \text{kip·in}$
Step 6 — Check: returning M = 1,598 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 1598\ \text{kip·in}

Why the other options are there

3,197 — kept a factor of two that cancels in the correct rearrangement.
799.2 — dropped that same factor in the other direction.
1,758 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions