Elastic Strain Energy

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

2 formulas

10 exam-style examples

~49 min

All Mechanics of Materials lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

If the strain remains within the elastic limit, the work done during deflection (extension) of a member will be transformed into
potential energy and can be recovered.
If the final load is P and the corresponding elongation of a tension member is δ, then the total energy U stored is equal to the
The strain energy per unit volume is
(Use these values if the speciﬁc alloy and temper are not listed on Table 2 below)

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

Elongation of a steel hanger rod

A 25 mm diameter steel rod, 6.0 m long, carries 120 kN in tension. With E = 200 GPa, find the stress, the strain and the elongation.

Given

$d = 25 mm$
$L = 6.0 m$
$P = 120 kN$
$E = 200 GPa$

Find

σ, ε and δ

Start with the thinking

Area first — every axial answer depends on it.
Keep units consistent: N and mm give MPa directly.

Figure 1 — schematic for Elongation of a steel hanger rod

Step-by-step solution

Area — A = πd²/4 = π(25)²/4 = 491 mm²
Stress
$\sigma = P/A = 120,000/491 = 244 MPa$
Strain
$\varepsilon = \sigma/E = 244/200,000 = 0.00122$
Elongation — δ = PL/(AE) = εL = 0.00122(6,000)
Result
$\delta = 7.33 mm$

Answer:

\sigma = 244 MPa, \varepsilon = 0.00122, \delta = 7.33 mm

Why the other options are there

δ = 0.733 mm (metres and millimetres mixed)
σ = 61 MPa (diameter used as area)

Reference: FE Reference Handbook — Mechanics of Materials — Uniaxial loading

Example 2

Axial stress and elongation of a steel rod — Elastic Strain Energy

A steel rod of area 5.50 in² and length 170 in. carries 29 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 29 kip$
$A = 5.50 in^{2}$
$L = 170 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 29/5.50 = 5.27 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 29(170)/(5.50 \times 29,000)$
Evaluate
$\delta = 0.0309 in$
Strain check
$\varepsilon = \delta/L = 0.000182 = \sigma/E = 0.000182 ✓$

Answer:

σ ≈ 5.27 ksi; δ ≈ 0.031 in.

Why the other options are there

5,273 ksi (psi/ksi confusion)
896.4 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 3

Axial stress and elongation of a steel rod — Elastic Strain Energy (2)

A steel rod of area 2.00 in² and length 286 in. carries 100 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 100 kip$
$A = 2.00 in^{2}$
$L = 286 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 100/2.00 = 50.00 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 100(286)/(2.00 \times 29,000)$
Evaluate
$\delta = 0.4931 in$
Strain check
$\varepsilon = \delta/L = 0.001724 = \sigma/E = 0.001724 ✓$

Answer:

σ ≈ 50.00 ksi; δ ≈ 0.493 in.

Why the other options are there

50,000 ksi (psi/ksi confusion)
14,300 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 4

Axial stress and elongation of a steel rod — Elastic Strain Energy (3)

A steel rod of area 8.00 in² and length 182 in. carries 53 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 53 kip$
$A = 8.00 in^{2}$
$L = 182 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 53/8.00 = 6.63 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 53(182)/(8.00 \times 29,000)$
Evaluate
$\delta = 0.0416 in$
Strain check
$\varepsilon = \delta/L = 0.000228 = \sigma/E = 0.000228 ✓$

Answer:

σ ≈ 6.63 ksi; δ ≈ 0.042 in.

Why the other options are there

6,625 ksi (psi/ksi confusion)
1,206 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 5

Axial stress and elongation of a steel rod — Elastic Strain Energy (4)

A steel rod of area 3.50 in² and length 67 in. carries 67 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 67 kip$
$A = 3.50 in^{2}$
$L = 67 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 67/3.50 = 19.14 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 67(67)/(3.50 \times 29,000)$
Evaluate
$\delta = 0.0442 in$
Strain check
$\varepsilon = \delta/L = 0.000660 = \sigma/E = 0.000660 ✓$

Answer:

σ ≈ 19.14 ksi; δ ≈ 0.044 in.

Why the other options are there

19,143 ksi (psi/ksi confusion)
1,283 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 6

Axial stress and elongation of a steel rod — Elastic Strain Energy (5)

A steel rod of area 7.00 in² and length 80 in. carries 104 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 104 kip$
$A = 7.00 in^{2}$
$L = 80 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 104/7.00 = 14.86 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 104(80)/(7.00 \times 29,000)$
Evaluate
$\delta = 0.0410 in$
Strain check
$\varepsilon = \delta/L = 0.000512 = \sigma/E = 0.000512 ✓$

Answer:

σ ≈ 14.86 ksi; δ ≈ 0.041 in.

Why the other options are there

14,857 ksi (psi/ksi confusion)
1,189 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 7

Axial stress and elongation of a steel rod — Elastic Strain Energy (6)

A steel rod of area 7.50 in² and length 235 in. carries 104 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 104 kip$
$A = 7.50 in^{2}$
$L = 235 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 104/7.50 = 13.87 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 104(235)/(7.50 \times 29,000)$
Evaluate
$\delta = 0.1124 in$
Strain check
$\varepsilon = \delta/L = 0.000478 = \sigma/E = 0.000478 ✓$

Answer:

σ ≈ 13.87 ksi; δ ≈ 0.112 in.

Why the other options are there

13,867 ksi (psi/ksi confusion)
3,259 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 8

Axial stress and elongation of a steel rod — Elastic Strain Energy (7)

A steel rod of area 5.25 in² and length 67 in. carries 97 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 97 kip$
$A = 5.25 in^{2}$
$L = 67 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 97/5.25 = 18.48 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 97(67)/(5.25 \times 29,000)$
Evaluate
$\delta = 0.0427 in$
Strain check
$\varepsilon = \delta/L = 0.000637 = \sigma/E = 0.000637 ✓$

Answer:

σ ≈ 18.48 ksi; δ ≈ 0.043 in.

Why the other options are there

18,476 ksi (psi/ksi confusion)
1,238 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 9

Axial stress and elongation of a steel rod — Elastic Strain Energy (8)

A steel rod of area 7.25 in² and length 297 in. carries 103 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 103 kip$
$A = 7.25 in^{2}$
$L = 297 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 103/7.25 = 14.21 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 103(297)/(7.25 \times 29,000)$
Evaluate
$\delta = 0.1455 in$
Strain check
$\varepsilon = \delta/L = 0.000490 = \sigma/E = 0.000490 ✓$

Answer:

σ ≈ 14.21 ksi; δ ≈ 0.145 in.

Why the other options are there

14,207 ksi (psi/ksi confusion)
4,219 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 10

Axial stress and elongation of a steel rod — Elastic Strain Energy (9)

A steel rod of area 5.50 in² and length 176 in. carries 78 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

$P = 78 kip$
$A = 5.50 in^{2}$
$L = 176 in$
$E = 29,000 ksi$

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$\sigma = P/A$
Substituting
$\sigma = 78/5.50 = 14.18 ksi$
Elongation
$\delta = PL/(AE)$
Substituting
$\delta = 78(176)/(5.50 \times 29,000)$
Evaluate
$\delta = 0.0861 in$
Strain check
$\varepsilon = \delta/L = 0.000489 = \sigma/E = 0.000489 ✓$

Answer:

σ ≈ 14.18 ksi; δ ≈ 0.086 in.

Why the other options are there

14,182 ksi (psi/ksi confusion)
2,496 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy