True stress

Materials Science · FE Reference Handbook section

Materials Science

24 formulas

10 exam-style examples

~60 min

All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The elastic modulus (also called modulus, modulus of elasticity, Young's modulus) describes the relationship between
engineering stress and engineering strain during elastic loading. Hooke's Law applies in such a case.
Key mechanical properties obtained from a tensile test curve:
• Ductility (also called percent elongation): Permanent engineering strain after failure
• Ultimate tensile strength (also called tensile strength): Maximum engineering stress
• Yield strength: Engineering stress at which permanent deformation is first observed, calculated by 0.2% offset method.
• Creep: Time-dependent deformation under load. Usually measured by strain rate. For steady-state creep this is:
• Fatigue: Time-dependent failure under cyclic load. Fatigue life is the number of cycles to failure. The endurance limit is
the stress below which fatigue failure is unlikely.
• Fracture toughness: The combination of applied stress and the crack length in a brittle material. It is the stress intensity
when the material will fail.
The critical value of stress intensity at which catastrophic crack propagation occurs, KIc, is a material property.
Representative Values of Fracture Toughness

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

True stress — solve for true stress — True stress

A materials engineer converts engineering stress to true stress for a necked specimen. Given engineering stress (sigEng) = 30.0000 ksi; engineering strain (epsEng) = 0.2150 in/in, determine the true stress (sigT) in ksi.

Given

$engineering stress (sigEng) = 30.0000 ksi$
$engineering strain (epsEng) = 0.2150 in/in$

Find

true stress (sigT), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigT is given, so isolate sigT 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 1 — schematic for True stress — solve for true stress — True stress

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigT:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 3
$List the givens: engineering stress (sigEng) = 30.0000 ksi, engineering strain (epsEng) = 0.2150 in/in$
Step 4 — Substitute the given values:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 5 — Evaluate:
$sigT = 36.4500\ \text{ksi}$
Step 6 — Check: returning sigT = 36.4500 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigT = 36.4500\ \text{ksi}

Why the other options are there

72.9000 — kept a factor of two that cancels in the correct rearrangement.
18.2250 — dropped that same factor in the other direction.
40.0950 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 2

True stress — solve for engineering stress — True stress (2)

A student computes the true stress at a given point on the stress-strain curve. Given engineering strain (epsEng) = 0.1250 in/in; true stress (sigT) = 62.4000 ksi, determine the engineering stress (sigEng) in ksi.

Given

$engineering strain (epsEng) = 0.1250 in/in$
$true stress (sigT) = 62.4000 ksi$

Find

engineering stress (sigEng), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigEng is given, so isolate sigEng 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 2 — schematic for True stress — solve for engineering stress — True stress (2)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigEng:
$sigEng = \dfrac{\sigma_T}{1+\varepsilon_{eng}}$
Step 3
$List the givens: engineering strain (epsEng) = 0.1250 in/in, true stress (sigT) = 62.4000 ksi$
Step 4 — Substitute the given values:
$sigEng = \dfrac{\sigma_T}{1+\varepsilon_{eng}}$
Step 5 — Evaluate:
$sigEng = 55.4667\ \text{ksi}$
Step 6 — Check: returning sigEng = 55.4667 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigEng = 55.4667\ \text{ksi}

Why the other options are there

110.9 — kept a factor of two that cancels in the correct rearrangement.
27.7333 — dropped that same factor in the other direction.
61.0133 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 3

True stress — solve for engineering strain — True stress (3)

The true stress exceeds the engineering stress as the specimen elongates. Given engineering stress (sigEng) = 62.0000 ksi; true stress (sigT) = 107.2 ksi, determine the engineering strain (epsEng) in in/in.

Given

$engineering stress (sigEng) = 62.0000 ksi$
$true stress (sigT) = 107.2 ksi$

Find

engineering strain (epsEng), in in/in

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except epsEng is given, so isolate epsEng 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 3 — schematic for True stress — solve for engineering strain — True stress (3)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for epsEng:
$epsEng = \dfrac{\sigma_T}{\sigma_{eng}} - 1$
Step 3
$List the givens: engineering stress (sigEng) = 62.0000 ksi, true stress (sigT) = 107.2 ksi$
Step 4 — Substitute the given values:
$epsEng = \dfrac{\sigma_T}{\sigma_{eng}} - 1$
Step 5 — Evaluate:
$epsEng = 0.7290\ \text{in/in}$
Step 6 — Check: returning epsEng = 0.7290 in/in to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

epsEng = 0.7290\ \text{in/in}

Why the other options are there

1.4581 — kept a factor of two that cancels in the correct rearrangement.
0.3645 — dropped that same factor in the other direction.
0.8019 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 4

True stress — solve for true stress (case 2) — True stress (4)

A materials engineer converts engineering stress to true stress for a necked specimen. Given engineering stress (sigEng) = 27.5000 ksi; engineering strain (epsEng) = 0.1300 in/in, determine the true stress (sigT) in ksi.

Given

$engineering stress (sigEng) = 27.5000 ksi$
$engineering strain (epsEng) = 0.1300 in/in$

Find

true stress (sigT), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigT is given, so isolate sigT 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 4 — schematic for True stress — solve for true stress (case 2) — True stress (4)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigT:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 3
$List the givens: engineering stress (sigEng) = 27.5000 ksi, engineering strain (epsEng) = 0.1300 in/in$
Step 4 — Substitute the given values:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 5 — Evaluate:
$sigT = 31.0750\ \text{ksi}$
Step 6 — Check: returning sigT = 31.0750 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigT = 31.0750\ \text{ksi}

Why the other options are there

62.1500 — kept a factor of two that cancels in the correct rearrangement.
15.5375 — dropped that same factor in the other direction.
34.1825 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 5

True stress — solve for engineering stress (case 2) — True stress (5)

A student computes the true stress at a given point on the stress-strain curve. Given engineering strain (epsEng) = 0.1500 in/in; true stress (sigT) = 42.5000 ksi, determine the engineering stress (sigEng) in ksi.

Given

$engineering strain (epsEng) = 0.1500 in/in$
$true stress (sigT) = 42.5000 ksi$

Find

engineering stress (sigEng), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigEng is given, so isolate sigEng 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 5 — schematic for True stress — solve for engineering stress (case 2) — True stress (5)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigEng:
$sigEng = \dfrac{\sigma_T}{1+\varepsilon_{eng}}$
Step 3
$List the givens: engineering strain (epsEng) = 0.1500 in/in, true stress (sigT) = 42.5000 ksi$
Step 4 — Substitute the given values:
$sigEng = \dfrac{\sigma_T}{1+\varepsilon_{eng}}$
Step 5 — Evaluate:
$sigEng = 36.9565\ \text{ksi}$
Step 6 — Check: returning sigEng = 36.9565 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigEng = 36.9565\ \text{ksi}

Why the other options are there

73.9130 — kept a factor of two that cancels in the correct rearrangement.
18.4783 — dropped that same factor in the other direction.
40.6522 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 6

True stress — solve for engineering strain (case 2) — True stress (6)

The true stress exceeds the engineering stress as the specimen elongates. Given engineering stress (sigEng) = 52.0000 ksi; true stress (sigT) = 72.7000 ksi, determine the engineering strain (epsEng) in in/in.

Given

$engineering stress (sigEng) = 52.0000 ksi$
$true stress (sigT) = 72.7000 ksi$

Find

engineering strain (epsEng), in in/in

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except epsEng is given, so isolate epsEng 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 6 — schematic for True stress — solve for engineering strain (case 2) — True stress (6)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for epsEng:
$epsEng = \dfrac{\sigma_T}{\sigma_{eng}} - 1$
Step 3
$List the givens: engineering stress (sigEng) = 52.0000 ksi, true stress (sigT) = 72.7000 ksi$
Step 4 — Substitute the given values:
$epsEng = \dfrac{\sigma_T}{\sigma_{eng}} - 1$
Step 5 — Evaluate:
$epsEng = 0.3981\ \text{in/in}$
Step 6 — Check: returning epsEng = 0.3981 in/in to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

epsEng = 0.3981\ \text{in/in}

Why the other options are there

0.7962 — kept a factor of two that cancels in the correct rearrangement.
0.1990 — dropped that same factor in the other direction.
0.4379 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 7

True stress — solve for true stress (case 3) — True stress (7)

A materials engineer converts engineering stress to true stress for a necked specimen. Given engineering stress (sigEng) = 69.0000 ksi; engineering strain (epsEng) = 0.0600 in/in, determine the true stress (sigT) in ksi.

Given

$engineering stress (sigEng) = 69.0000 ksi$
$engineering strain (epsEng) = 0.0600 in/in$

Find

true stress (sigT), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigT is given, so isolate sigT 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 7 — schematic for True stress — solve for true stress (case 3) — True stress (7)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigT:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 3
$List the givens: engineering stress (sigEng) = 69.0000 ksi, engineering strain (epsEng) = 0.0600 in/in$
Step 4 — Substitute the given values:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 5 — Evaluate:
$sigT = 73.1400\ \text{ksi}$
Step 6 — Check: returning sigT = 73.1400 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigT = 73.1400\ \text{ksi}

Why the other options are there

146.3 — kept a factor of two that cancels in the correct rearrangement.
36.5700 — dropped that same factor in the other direction.
80.4540 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 8

True stress — solve for engineering stress (case 3) — True stress (8)

A student computes the true stress at a given point on the stress-strain curve. Given engineering strain (epsEng) = 0.1800 in/in; true stress (sigT) = 111.0 ksi, determine the engineering stress (sigEng) in ksi.

Given

$engineering strain (epsEng) = 0.1800 in/in$
$true stress (sigT) = 111.0 ksi$

Find

engineering stress (sigEng), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigEng is given, so isolate sigEng 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 8 — schematic for True stress — solve for engineering stress (case 3) — True stress (8)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigEng:
$sigEng = \dfrac{\sigma_T}{1+\varepsilon_{eng}}$
Step 3
$List the givens: engineering strain (epsEng) = 0.1800 in/in, true stress (sigT) = 111.0 ksi$
Step 4 — Substitute the given values:
$sigEng = \dfrac{\sigma_T}{1+\varepsilon_{eng}}$
Step 5 — Evaluate:
$sigEng = 94.0678\ \text{ksi}$
Step 6 — Check: returning sigEng = 94.0678 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigEng = 94.0678\ \text{ksi}

Why the other options are there

188.1 — kept a factor of two that cancels in the correct rearrangement.
47.0339 — dropped that same factor in the other direction.
103.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 9

True stress — solve for engineering strain (case 3) — True stress (9)

The true stress exceeds the engineering stress as the specimen elongates. Given engineering stress (sigEng) = 53.0000 ksi; true stress (sigT) = 30.5000 ksi, determine the engineering strain (epsEng) in in/in.

Given

$engineering stress (sigEng) = 53.0000 ksi$
$true stress (sigT) = 30.5000 ksi$

Find

engineering strain (epsEng), in in/in

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except epsEng is given, so isolate epsEng 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 9 — schematic for True stress — solve for engineering strain (case 3) — True stress (9)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for epsEng:
$epsEng = \dfrac{\sigma_T}{\sigma_{eng}} - 1$
Step 3
$List the givens: engineering stress (sigEng) = 53.0000 ksi, true stress (sigT) = 30.5000 ksi$
Step 4 — Substitute the given values:
$epsEng = \dfrac{\sigma_T}{\sigma_{eng}} - 1$
Step 5 — Evaluate:
$epsEng = -0.4245\ \text{in/in}$
Step 6 — Check: returning epsEng = -0.4245 in/in to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

epsEng = -0.4245\ \text{in/in}

Why the other options are there

-0.8491 — kept a factor of two that cancels in the correct rearrangement.
-0.2123 — dropped that same factor in the other direction.
-0.4670 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

Example 10

True stress — solve for true stress (case 4) — True stress (10)

A materials engineer converts engineering stress to true stress for a necked specimen. Given engineering stress (sigEng) = 27.5000 ksi; engineering strain (epsEng) = 0.2800 in/in, determine the true stress (sigT) in ksi.

Given

$engineering stress (sigEng) = 27.5000 ksi$
$engineering strain (epsEng) = 0.2800 in/in$

Find

true stress (sigT), in ksi

Start with the thinking

The governing relation printed in this handbook section is True stress.
Everything except sigT is given, so isolate sigT 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.
True stress accounts for the instantaneous cross-sectional area, converted from engineering stress and strain.

Figure 10 — schematic for True stress — solve for true stress (case 4) — True stress (10)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
Step 2 — Rearrange symbolically for sigT:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 3
$List the givens: engineering stress (sigEng) = 27.5000 ksi, engineering strain (epsEng) = 0.2800 in/in$
Step 4 — Substitute the given values:
$sigT = \sigma_{eng}(1+\varepsilon_{eng})$
Step 5 — Evaluate:
$sigT = 35.2000\ \text{ksi}$
Step 6 — Check: returning sigT = 35.2000 ksi to
$\sigma_T = \sigma_{eng}(1+\varepsilon_{eng})$
reproduces the given quantities, and both sides carry the same units.

Answer:

sigT = 35.2000\ \text{ksi}

Why the other options are there

70.4000 — kept a factor of two that cancels in the correct rearrangement.
17.6000 — dropped that same factor in the other direction.
38.7200 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Stress

True stress

Handbook notes for this section

Core formulas for this FE topic

vT=AvT = AvT=A

σT = true stressNeeds technical review

F=appliedforceF = applied forceF=appliedforce

A=actualcross−sectionalareaA = actual cross-sectional areaA=actualcross−sectionalarea

σ=Eε\sigma = E\varepsilonσ=Eε

whereE=elasticmoduluswhere E = elastic moduluswhereE=elasticmodulus

df=n−RTdf = n - RTdf=n−RT

A=pre−exponentialconstantA = pre-exponential constantA=pre−exponentialconstant

n=stresssensitivityn = stress sensitivityn=stresssensitivity

Q=activationenergyforcreepQ = activation energy for creepQ=activationenergyforcreep

R=idealgaslawconstantR = ideal gas law constantR=idealgaslawconstant

T=absolutetemperatureT = absolute temperatureT=absolutetemperature

R=vmin⁡R = v \minR=vmin

whereR=stressratiowhere R = stress ratiowhereR=stressratio

ForR=−1andhighcyclefatigue,basedontheBasquinequation:For R = -1 and high cycle fatigue, based on the Basquin equation:ForR=−1andhighcyclefatigue,basedontheBasquinequation:

N=dnN =d nN=dn

N=cyclestofailureN = cycles to failureN=cyclestofailure

σr = completely (fully) reversed stressNeeds technical review

AandB=materialconstantsA and B = material constantsAandB=materialconstants

KIC=YvraKIC = Yv raKIC=Yvra

KIC=fracturetoughnessaKIC = fracture toughness aKIC=fracturetoughnessa

σ=appliedengineeringstress\sigma = applied engineering stressσ=appliedengineeringstress

a=cracklengtha = crack lengtha=cracklength

Y=geometricalfactorEXTERIORCRACK(Y=1.1)INTERIORCRACK(Y=1)Y = geometrical factor EXTERIOR CRACK (Y = 1.1) INTERIOR CRACK (Y = 1)Y=geometricalfactorEXTERIORCRACK(Y=1.1)INTERIORCRACK(Y=1)

Worked exam-style examples