Mohr's Circle – Stress, 2D — FE Civil Lecture

Mohr's Circle – Stress, 2D

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

5 formulas

10 exam-style examples

~55 min

All Mechanics of Materials lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

To construct a Mohr's circle, the following sign conventions are used.
1. Tensile normal stress components are plotted on the horizontal axis and are considered positive. Compressive normal stress
2. For constructing Mohr's circle only, shearing stresses are plotted above the normal stress axis when the pair of shearing
stresses, acting on opposite and parallel faces of an element, forms a clockwise couple. Shearing stresses are plotted below
the normal axis when the shear stresses form a counterclockwise couple.
The circle drawn with the center on the normal stress (horizontal) axis with center, C, and radius, R, where
The two nonzero principal stresses are then:

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

Maximum transverse shear stress

The same 200 × 400 mm beam carries a shear force of 90 kN. What is the maximum transverse shear stress?

Given

$V = 90 kN$
Rectangular section 200 × 400 mm

Find

τ_max

Start with the thinking

For a rectangle the parabolic distribution peaks at 1.5 V/A.
Average shear stress is not the answer the exam wants.

Step-by-step solution

Area
$A = 200(400) = 80,000 mm^{2}$
Average shear
$V/A = 90,000/80,000 = 1.13 MPa$
Rectangular peak
$\tau_max = 1.5 V/A$
Substitute
$\tau_max = 1.5(1.13)$
Result
$\tau_max = 1.69 MPa at the neutral axis$

Answer:

\tau_max = 1.69 MPa

Why the other options are there

1.13 MPa (average reported)
2.25 MPa (factor 2 used instead of 1.5)

Reference: FE Reference Handbook — Mechanics of Materials — Transverse shear

Example 2

Principal stresses from a plane stress state

At a point, σx = 80 MPa, σy = −20 MPa and τxy = 30 MPa. Find the principal stresses and the maximum in-plane shear.

Given

σx = 80 MPa
σy = −20 MPa
τxy = 30 MPa

Find

σ₁, σ₂, τ_max

Start with the thinking

Average stress is the circle centre; the radius is the maximum shear.
Sign of τxy affects the angle, not the magnitudes.

Figure 2 — schematic for Principal stresses from a plane stress state

Step-by-step solution

Centre — σ_avg = (σx + σy)/2 = (80 − 20)/2 = 30.0 MPa
Half-difference — (σx − σy)/2 = 50.0 MPa
Radius — R = √(50² + 30²) = √3,400 = 58.3 MPa
Principal stresses
$\sigma_{1} = 30.0 + 58.3 = 88.3 MPa and \sigma_{2} = 30.0 - 58.3 = -28.3 MPa$
Maximum shear
$\tau_max = R = 58.3 MPa$

Answer:

\sigma_{1} = 88.3 MPa, \sigma_{2} = -28.3 MPa, \tau_max = 58.3 MPa

Why the other options are there

σ₁ = 110 MPa (σy sign dropped)
τ_max = 30 MPa (applied shear reported)

Reference: FE Reference Handbook — Mechanics of Materials — Mohr's circle

Example 3

Flexural stress — solve for bending stress — Mohr's Circle – Stress, 2D

A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,870 kip·in; distance to extreme fiber (c) = 5.4000 in; moment of inertia (I) = 120.0 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 4,870 kip·in
$distance to extreme fiber (c) = 5.4000 in$
$moment of inertia (I) = 120.0 in^{4}$

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) = 4,870 kip·in, distance to extreme fiber (c) = 5.4000 in, moment of inertia (I) = 120.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 219.2\ \text{ksi}$
Step 6 — Check: returning sigma = 219.2 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 219.2\ \text{ksi}

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 4

Flexural stress — solve for moment — Mohr's Circle – Stress, 2D (2)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 3.4000 in; moment of inertia (I) = 440.0 in⁴; bending stress (sigma) = 14.5300 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 3.4000 in$
$moment of inertia (I) = 440.0 in^{4}$
$bending stress (sigma) = 14.5300 ksi$

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) = 3.4000 in, moment of inertia (I) = 440.0 in⁴, bending stress (sigma) = 14.5300 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 1880\ \text{kip·in}$
Step 6 — Check: returning M = 1,880 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 1880\ \text{kip·in}

Why the other options are there

3,761 — kept a factor of two that cancels in the correct rearrangement.
940.2 — dropped that same factor in the other direction.
2,068 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 5

Flexural stress — solve for distance to extreme fiber — Mohr's Circle – Stress, 2D (3)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,290 kip·in; moment of inertia (I) = 1,070 in⁴; bending stress (sigma) = 43.3700 ksi, determine the distance to extreme fiber (c) in in.

Given

moment (M) = 2,290 kip·in
$moment of inertia (I) = 1,070 in^{4}$
$bending stress (sigma) = 43.3700 ksi$

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) = 2,290 kip·in, moment of inertia (I) = 1,070 in⁴, bending stress (sigma) = 43.3700 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 20.2646\ \text{in}$
Step 6 — Check: returning c = 20.2646 in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 20.2646\ \text{in}

Why the other options are there

40.5292 — kept a factor of two that cancels in the correct rearrangement.
10.1323 — dropped that same factor in the other direction.
22.2910 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 6

Flexural stress — solve for moment of inertia — Mohr's Circle – Stress, 2D (4)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,910 kip·in; distance to extreme fiber (c) = 8.8000 in; bending stress (sigma) = 46.1100 ksi, determine the moment of inertia (I) in in⁴.

Given

moment (M) = 3,910 kip·in
$distance to extreme fiber (c) = 8.8000 in$
$bending stress (sigma) = 46.1100 ksi$

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) = 8.8000 in, bending stress (sigma) = 46.1100 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$I = 746.2\ \text{in⁴}$
Step 6 — Check: returning I = 746.2 in⁴ to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 746.2\ \text{in⁴}

Why the other options are there

1,492 — kept a factor of two that cancels in the correct rearrangement.
373.1 — dropped that same factor in the other direction.
820.8 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 7

Flexural stress — solve for bending stress (case 2) — Mohr's Circle – Stress, 2D (5)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,590 kip·in; distance to extreme fiber (c) = 11.7000 in; moment of inertia (I) = 1,080 in⁴, determine the bending stress (sigma) in ksi.

Given

moment (M) = 1,590 kip·in
$distance to extreme fiber (c) = 11.7000 in$
$moment of inertia (I) = 1,080 in^{4}$

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) = 1,590 kip·in, distance to extreme fiber (c) = 11.7000 in, moment of inertia (I) = 1,080 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma = 17.2250\ \text{ksi}$
Step 6 — Check: returning sigma = 17.2250 ksi to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma = 17.2250\ \text{ksi}

Why the other options are there

34.4500 — kept a factor of two that cancels in the correct rearrangement.
8.6125 — dropped that same factor in the other direction.
18.9475 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 8

Flexural stress — solve for moment (case 2) — Mohr's Circle – Stress, 2D (6)

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 3.1000 in; moment of inertia (I) = 350.0 in⁴; bending stress (sigma) = 48.6900 ksi, determine the moment (M) in kip·in.

Given

$distance to extreme fiber (c) = 3.1000 in$
$moment of inertia (I) = 350.0 in^{4}$
$bending stress (sigma) = 48.6900 ksi$

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) = 3.1000 in, moment of inertia (I) = 350.0 in⁴, bending stress (sigma) = 48.6900 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 5497\ \text{kip·in}$
Step 6 — Check: returning M = 5,497 kip·in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 5497\ \text{kip·in}

Why the other options are there

10,995 — kept a factor of two that cancels in the correct rearrangement.
2,749 — dropped that same factor in the other direction.
6,047 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 9

Flexural stress — solve for distance to extreme fiber (case 2) — Mohr's Circle – Stress, 2D (7)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,440 kip·in; moment of inertia (I) = 1,610 in⁴; bending stress (sigma) = 27.7500 ksi, determine the distance to extreme fiber (c) in in.

Given

moment (M) = 3,440 kip·in
$moment of inertia (I) = 1,610 in^{4}$
$bending stress (sigma) = 27.7500 ksi$

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) = 3,440 kip·in, moment of inertia (I) = 1,610 in⁴, bending stress (sigma) = 27.7500 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 12.9876\ \text{in}$
Step 6 — Check: returning c = 12.9876 in to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 12.9876\ \text{in}

Why the other options are there

25.9753 — kept a factor of two that cancels in the correct rearrangement.
6.4938 — dropped that same factor in the other direction.
14.2864 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 10

Flexural stress — solve for moment of inertia (case 2) — Mohr's Circle – Stress, 2D (8)

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,760 kip·in; distance to extreme fiber (c) = 6.2000 in; bending stress (sigma) = 47.4600 ksi, determine the moment of inertia (I) in in⁴.

Given

moment (M) = 1,760 kip·in
$distance to extreme fiber (c) = 6.2000 in$
$bending stress (sigma) = 47.4600 ksi$

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) = 1,760 kip·in, distance to extreme fiber (c) = 6.2000 in, bending stress (sigma) = 47.4600 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$I = 229.9\ \text{in⁴}$
Step 6 — Check: returning I = 229.9 in⁴ to
$\sigma = M c / I$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 229.9\ \text{in⁴}

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D