Strength of Materials Formulas with Worked Examples

On this page

1. Normal stress
2. Strain and Hooke’s law
3. Axial deformation
4. Direct shear stress
5. Thermal stress
6. Bending stress
7. Torsional shear stress
8. Angle of twist
9. Beam deflection
10. Euler buckling load
Using these formulas in design

Strength of materials answers one question: will this part carry its load without breaking or bending too much? These ten formulas cover most of the hand calculations in a first design check, before any finite element analysis. Every example uses SI units: newtons, metres and pascals.

Units to keep straight

1 Pa = 1 N/m². 1 MPa = 10⁶ Pa = 1 N/mm². Working in newtons and millimetres gives stress directly in MPa, which is how material strengths are usually listed. Mild structural steel: E ≈ 200 GPa, G ≈ 80 GPa, yield strength about 250 MPa.

A simply supported beam with a central load F. Each support pushes up with F/2, the beam sags by δ at mid-span, and bending puts the top fibres in compression and the bottom fibres in tension, with zero stress at the neutral axis.

1. Normal stress

\sigma = \frac{F}{A}

Symbol	Meaning	SI unit
$\sigma$	Normal (tensile or compressive) stress	Pa
$F$	Axial force	N
$A$	Cross-sectional area at right angles to the force	m²

Assumptions: the force acts along the axis through the centroid of the section, the section is uniform, and the point considered is away from holes, shoulders and the load point, where stress concentrates.

Worked example

A steel rod of 20 mm diameter carries a tensile load of 30 kN. Its yield strength is 250 MPa. Find the stress and the factor of safety.

A = \frac{\pi d^2}{4} = \frac{\pi (0.020)^2}{4} = 3.142 \times 10^{-4}\ \text{m}^2

\sigma = \frac{30\,000}{3.142 \times 10^{-4}} = 95.5 \times 10^{6}\ \text{Pa}

Factor of safety = yield strength / working stress = 250 / 95.5 = 2.6.

Answerσ = 95.5 MPa, factor of safety 2.6

2. Strain and Hooke’s law

\varepsilon = \frac{\delta}{L} \qquad \sigma = E\,\varepsilon

Symbol	Meaning	SI unit
$\varepsilon$	Normal strain	none (m/m)
$\delta$	Change in length	m
$L$	Original length	m
$E$	Young’s modulus (modulus of elasticity)	Pa

Assumptions: the material is linear elastic, so the stress stays below the proportional limit, and it is isotropic.

Worked example

Find the strain in the rod from example 1. For steel, E = 200 GPa.

\varepsilon = \frac{\sigma}{E} = \frac{95.5 \times 10^6}{200 \times 10^9} = 4.77 \times 10^{-4}

Answerε = 4.77 × 10⁻⁴ (0.048 %)

3. Axial deformation

\delta = \frac{F L}{A E}

Symbol	Meaning	SI unit
$\delta$	Extension or shortening	m
$F$	Axial force	N
$L$	Length	m
$A$	Cross-sectional area	m²
$E$	Young’s modulus	Pa

Assumptions: uniform bar, constant force along its length, linear elastic material. For a stepped bar, add up $FL/AE$ for each step.

Worked example

The rod from example 1 is 2 m long. How much does it stretch?

\delta = \frac{30\,000 \times 2}{3.142 \times 10^{-4} \times 200 \times 10^{9}} = 9.55 \times 10^{-4}\ \text{m}

Answerδ = 0.95 mm

4. Direct shear stress

\tau = \frac{F}{n A}

Symbol	Meaning	SI unit
$\tau$	Average shear stress	Pa
$F$	Shear force	N
$n$	Number of shear planes (1 = single shear, 2 = double shear)	none
$A$	Area of one shear plane	m²

Assumptions: the stress is spread evenly over the shear plane (it is really an average), and bending of the pin is ignored.

Worked example

A 10 mm diameter pin holds a clevis joint in double shear under a load of 12 kN.

A = \frac{\pi (0.010)^2}{4} = 7.854 \times 10^{-5}\ \text{m}^2

\tau = \frac{12\,000}{2 \times 7.854 \times 10^{-5}} = 76.4 \times 10^{6}\ \text{Pa}

Answerτ = 76.4 MPa

5. Thermal stress

\sigma = E\,\alpha\,\Delta T

Symbol	Meaning	SI unit
$\sigma$	Stress when expansion is fully prevented	Pa
$E$	Young’s modulus	Pa
$\alpha$	Coefficient of linear thermal expansion	1/K
$\Delta T$	Temperature change	K

Assumptions: the ends are completely fixed, the temperature is uniform, and $E$ and $\alpha$ do not change over the range. If the part is free to expand, there is no stress and it simply grows by $\delta = \alpha L \Delta T$ .

Worked example

A steel bar fixed between two rigid walls warms up by 40 K. For steel, α = 12 × 10⁻⁶ /K.

\sigma = 200 \times 10^{9} \times 12 \times 10^{-6} \times 40 = 96 \times 10^{6}\ \text{Pa}

Answerσ = 96 MPa in compression

6. Bending stress

\sigma_b = \frac{M\,y}{I} = \frac{M}{Z}

Symbol	Meaning	SI unit
$\sigma_b$	Bending stress at distance $y$ from the neutral axis	Pa
$M$	Bending moment at the section	N·m
$y$	Distance from the neutral axis (largest at the outer fibre)	m
$I$	Second moment of area about the neutral axis	m⁴
$Z$	Section modulus, $I/y_{max}$	m³

For a rectangle $I = bh^3/12$ and $Z = bh^2/6$ . For a solid circle $I = \pi d^4/64$ .

Assumptions: plane sections stay plane, the material is linear elastic with the same modulus in tension and compression, and the beam is straight and long compared with its depth.

Worked example

A rectangular steel beam 50 mm wide and 100 mm deep carries a bending moment of 5 kN·m. Find the largest bending stress.

I = \frac{0.05 \times 0.1^3}{12} = 4.167 \times 10^{-6}\ \text{m}^4, \qquad y_{max} = 0.05\ \text{m}

\sigma_b = \frac{5000 \times 0.05}{4.167 \times 10^{-6}} = 60 \times 10^{6}\ \text{Pa}

Answerσ = 60 MPa (compression on one face, tension on the other)

7. Torsional shear stress

\tau = \frac{T\,r}{J}, \qquad \text{solid shaft: } \tau_{max} = \frac{16\,T}{\pi d^3}

Symbol	Meaning	SI unit
$\tau$	Shear stress at radius $r$	Pa
$T$	Torque	N·m
$r$	Radius (outer radius for the maximum)	m
$J$	Polar second moment of area, $\pi d^4/32$ for a solid shaft	m⁴

Assumptions: circular shaft (solid or hollow), linear elastic material, and a point away from keyways and shoulders. Non-circular sections need different formulas.

Worked example

A solid shaft of 40 mm diameter transmits a torque of 500 N·m.

\tau_{max} = \frac{16 \times 500}{\pi \times 0.040^3} = 39.8 \times 10^{6}\ \text{Pa}

Answerτ = 39.8 MPa

8. Angle of twist

\theta = \frac{T L}{G J}

Symbol	Meaning	SI unit
$\theta$	Angle of twist	rad
$T$	Torque	N·m
$L$	Length of shaft	m
$G$	Shear modulus (modulus of rigidity)	Pa
$J$	Polar second moment of area	m⁴

Assumptions: the same as for torsional stress, with constant torque and section along the length.

Worked example

The shaft from example 7 is 1 m long and made of steel (G = 80 GPa).

J = \frac{\pi \times 0.040^4}{32} = 2.513 \times 10^{-7}\ \text{m}^4

\theta = \frac{500 \times 1}{80 \times 10^{9} \times 2.513 \times 10^{-7}} = 0.0249\ \text{rad}

Answerθ = 0.0249 rad = 1.43°

9. Beam deflection

Simply supported beam with a central point load:

\delta_{max} = \frac{F L^3}{48\,E I}

Symbol	Meaning	SI unit
$\delta_{max}$	Largest deflection	m
$F$	Point load	N
$L$	Span	m
$E$	Young’s modulus	Pa
$I$	Second moment of area	m⁴

The other standard cases, with $w$ a uniform load in N/m:

Beam and load	Largest deflection	Where
Cantilever, point load at the free end	$FL^3/3EI$	Free end
Cantilever, uniform load	$wL^4/8EI$	Free end
Simply supported, central point load	$FL^3/48EI$	Mid-span
Simply supported, uniform load	$5wL^4/384EI$	Mid-span

Assumptions: small deflections, linear elastic material, constant section, and deflection from shear ignored (fine for slender beams).

Worked example

The beam from example 6 spans 2 m between simple supports and carries 10 kN at mid-span. (Its largest bending moment is FL/4 = 5 kN·m, so the bending stress is the 60 MPa found above.)

\delta_{max} = \frac{10\,000 \times 2^3}{48 \times 200 \times 10^{9} \times 4.167 \times 10^{-6}} = 2.0 \times 10^{-3}\ \text{m}

Answerδ = 2.0 mm (span/1000)

10. Euler buckling load

P_{cr} = \frac{\pi^2 E I}{(K L)^2}

Symbol	Meaning	SI unit
$P_{cr}$	Critical (buckling) load	N
$E$	Young’s modulus	Pa
$I$	Smallest second moment of area of the section	m⁴
$L$	Column length	m
$K$	Effective length factor: 1 pinned–pinned, 2 fixed–free, 0.7 fixed–pinned, 0.5 fixed–fixed	none

Assumptions: a long, slender, perfectly straight column with a central load and linear elastic material. Short, stocky columns fail by crushing before they buckle, so check that the stress at $P_{cr}$ is well below the yield strength.

Worked example

A solid steel bar of 20 mm diameter and 1 m length is pinned at both ends.

I = \frac{\pi \times 0.020^4}{64} = 7.854 \times 10^{-9}\ \text{m}^4

P_{cr} = \frac{\pi^2 \times 200 \times 10^{9} \times 7.854 \times 10^{-9}}{(1 \times 1)^2} = 15\,500\ \text{N}

The stress at that load is 15 500 / 3.142 × 10⁻⁴ = 49 MPa, far below yield, so the bar buckles long before it crushes and Euler’s formula applies.

AnswerP_cr ≈ 15.5 kN

Using these formulas in design

Hand calculations come first because they are quick and show which dimension matters. Bending stress falls with the square of the beam depth and deflection with its cube, so making a beam deeper is far more effective than making it wider. Torsional stress falls with the cube of the shaft diameter.

When the shape is complicated, with holes, fillets or welded joints, we check the hand calculation with a finite element analysis of the CAD model. The two should agree away from the local details; if they don’t, one of them has an error.

Key points

Work in N and mm to get stresses straight in MPa.
Stress tells you whether the part breaks or yields; deflection and buckling tell you whether it still works.
Every formula here assumes linear elastic material and simple geometry; holes and sharp corners raise local stress.
Compare the result with the material’s yield strength and keep a sensible factor of safety.

Strength of materials formulas with units and examples

1. Normal stress

2. Strain and Hooke’s law

3. Axial deformation

4. Direct shear stress

5. Thermal stress

6. Bending stress

7. Torsional shear stress

8. Angle of twist

9. Beam deflection

10. Euler buckling load

Using these formulas in design

See it in CAD

Knurled Shaft

Foldable Round Table

Read next

Thermodynamics formulas with units and examples

Fluid mechanics formulas with units and examples

Manufacturing calculations: machining, sheet metal and tool life

Need this designed or explained for your team?