Cylindrical Pressure Vessel

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

12 formulas

10 exam-style examples

~60 min

All Mechanics of Materials lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

For internal pressure only, the stresses at the inside wall are:
For external pressure only, the stresses at the outside wall are:
For vessels with end caps, the axial stress is:
When the thickness of the cylinder wall is about one-tenth or less of inside radius, the cylinder can be considered as thin-walled.
In which case, the internal pressure is resisted by the hoop stress and the axial stress.

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 — Cylindrical Pressure Vessel

A mechanics of materials problem uses Flexural stress. Given moment (M) = 630.0 kip·in; distance to extreme fiber (c) = 5.5000 in; moment of inertia (I) = 930.0 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 630.0 kip·in
$distance to extreme fiber (c) = 5.5000 in$
$moment of inertia (I) = 930.0 in^{4}$

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) = 630.0 kip·in, distance to extreme fiber (c) = 5.5000 in, moment of inertia (I) = 930.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 3.7258\ \text{ksi}$
Step 6 — Check: returning sigma = 3.7258 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 3.7258\ \text{ksi}

Why the other options are there

7.4516 — kept a factor of two that cancels in the correct rearrangement.
1.8629 — dropped that same factor in the other direction.
4.0984 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 2

Flexural stress — solve for moment — Cylindrical Pressure Vessel (2)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 3.2000 in; moment of inertia (I) = 230.0 in⁴; bending stress (sigma) = 59.6300 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 3.2000 in$
$moment of inertia (I) = 230.0 in^{4}$
$bending stress (sigma) = 59.6300 ksi$

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) = 3.2000 in, moment of inertia (I) = 230.0 in⁴, bending stress (sigma) = 59.6300 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 4286\ \text{kip·in}$
Step 6 — Check: returning M = 4,286 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 4286\ \text{kip·in}

Why the other options are there

8,572 — kept a factor of two that cancels in the correct rearrangement.
2,143 — dropped that same factor in the other direction.
4,714 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 3

Flexural stress — solve for distance to extreme fiber — Cylindrical Pressure Vessel (3)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,440 kip·in; moment of inertia (I) = 1,070 in⁴; bending stress (sigma) = 23.1600 ksi, determine the distance to extreme fiber (c) in in.

Given

moment (M) = 2,440 kip·in
$moment of inertia (I) = 1,070 in^{4}$
$bending stress (sigma) = 23.1600 ksi$

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) = 2,440 kip·in, moment of inertia (I) = 1,070 in⁴, bending stress (sigma) = 23.1600 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 10.1562\ \text{in}$
Step 6 — Check: returning c = 10.1562 in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 10.1562\ \text{in}

Why the other options are there

20.3125 — kept a factor of two that cancels in the correct rearrangement.
5.0781 — dropped that same factor in the other direction.
11.1719 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 4

Flexural stress — solve for moment of inertia — Cylindrical Pressure Vessel (4)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 720.0 kip·in; distance to extreme fiber (c) = 10.5000 in; bending stress (sigma) = 18.0800 ksi, determine the moment of inertia (I) in in⁴.

Given

moment (M) = 720.0 kip·in
$distance to extreme fiber (c) = 10.5000 in$
$bending stress (sigma) = 18.0800 ksi$

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) = 720.0 kip·in, distance to extreme fiber (c) = 10.5000 in, bending stress (sigma) = 18.0800 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$I = 418.1\ \text{in⁴}$
Step 6 — Check: returning I = 418.1 in⁴ to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 418.1\ \text{in⁴}

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 5

Flexural stress — solve for bending stress (case 2) — Cylindrical Pressure Vessel (5)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,900 kip·in; distance to extreme fiber (c) = 14.6000 in; moment of inertia (I) = 1,180 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 4,900 kip·in
$distance to extreme fiber (c) = 14.6000 in$
$moment of inertia (I) = 1,180 in^{4}$

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) = 4,900 kip·in, distance to extreme fiber (c) = 14.6000 in, moment of inertia (I) = 1,180 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 60.6271\ \text{ksi}$
Step 6 — Check: returning sigma = 60.6271 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 60.6271\ \text{ksi}

Why the other options are there

121.3 — kept a factor of two that cancels in the correct rearrangement.
30.3136 — dropped that same factor in the other direction.
66.6898 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 6

Flexural stress — solve for moment (case 2) — Cylindrical Pressure Vessel (6)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 8.0000 in; moment of inertia (I) = 1,810 in⁴; bending stress (sigma) = 22.5800 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 8.0000 in$
$moment of inertia (I) = 1,810 in^{4}$
$bending stress (sigma) = 22.5800 ksi$

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) = 8.0000 in, moment of inertia (I) = 1,810 in⁴, bending stress (sigma) = 22.5800 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 5109\ \text{kip·in}$
Step 6 — Check: returning M = 5,109 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 5109\ \text{kip·in}

Why the other options are there

10,217 — kept a factor of two that cancels in the correct rearrangement.
2,554 — dropped that same factor in the other direction.
5,620 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 7

Flexural stress — solve for distance to extreme fiber (case 2) — Cylindrical Pressure Vessel (7)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,460 kip·in; moment of inertia (I) = 1,710 in⁴; bending stress (sigma) = 9.3500 ksi, determine the distance to extreme fiber (c) in in.

Given

moment (M) = 4,460 kip·in
$moment of inertia (I) = 1,710 in^{4}$
$bending stress (sigma) = 9.3500 ksi$

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) = 4,460 kip·in, moment of inertia (I) = 1,710 in⁴, bending stress (sigma) = 9.3500 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 3.5849\ \text{in}$
Step 6 — Check: returning c = 3.5849 in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 3.5849\ \text{in}

Why the other options are there

7.1697 — kept a factor of two that cancels in the correct rearrangement.
1.7924 — dropped that same factor in the other direction.
3.9434 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 8

Flexural stress — solve for moment of inertia (case 2) — Cylindrical Pressure Vessel (8)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,630 kip·in; distance to extreme fiber (c) = 5.8000 in; bending stress (sigma) = 21.8000 ksi, determine the moment of inertia (I) in in⁴.

Given

moment (M) = 3,630 kip·in
$distance to extreme fiber (c) = 5.8000 in$
$bending stress (sigma) = 21.8000 ksi$

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) = 3,630 kip·in, distance to extreme fiber (c) = 5.8000 in, bending stress (sigma) = 21.8000 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$I = 965.8\ \text{in⁴}$
Step 6 — Check: returning I = 965.8 in⁴ to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 965.8\ \text{in⁴}

Why the other options are there

1,932 — kept a factor of two that cancels in the correct rearrangement.
482.9 — dropped that same factor in the other direction.
1,062 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 9

Flexural stress — solve for bending stress (case 3) — Cylindrical Pressure Vessel (9)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,910 kip·in; distance to extreme fiber (c) = 8.9000 in; moment of inertia (I) = 500.0 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 4,910 kip·in
$distance to extreme fiber (c) = 8.9000 in$
$moment of inertia (I) = 500.0 in^{4}$

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) = 4,910 kip·in, distance to extreme fiber (c) = 8.9000 in, moment of inertia (I) = 500.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 87.3980\ \text{ksi}$
Step 6 — Check: returning sigma = 87.3980 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 87.3980\ \text{ksi}

Why the other options are there

174.8 — kept a factor of two that cancels in the correct rearrangement.
43.6990 — dropped that same factor in the other direction.
96.1378 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 10

Flexural stress — solve for moment (case 3) — Cylindrical Pressure Vessel (10)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 12.7000 in; moment of inertia (I) = 1,590 in⁴; bending stress (sigma) = 28.6300 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 12.7000 in$
$moment of inertia (I) = 1,590 in^{4}$
$bending stress (sigma) = 28.6300 ksi$

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) = 12.7000 in, moment of inertia (I) = 1,590 in⁴, bending stress (sigma) = 28.6300 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 3584\ \text{kip·in}$
Step 6 — Check: returning M = 3,584 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 3584\ \text{kip·in}

Why the other options are there

7,169 — kept a factor of two that cancels in the correct rearrangement.
1,792 — dropped that same factor in the other direction.
3,943 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel