Exercise set

Mechanical Stress Analysis Exercises

Worked mechanical engineering exercises for stress analysis covering axial stress, bending, torsion, von Mises stress, stress concentration, deflection, buckling, thermal stress, fatigue, pressure loads, and strain-gauge validation.

These exercises practise first-pass mechanical stress analysis for components, shafts, brackets, beams, columns, pressure parts, and validation tests. The goal is not only to calculate stress. The goal is to connect loads, geometry, material properties, failure modes, factors of safety, stiffness, fatigue, and measured evidence.

Assume linear elastic behaviour, small deformation, and simplified geometry unless an exercise states otherwise. Real design work should also check manufacturing tolerances, surface condition, residual stress, weld details, contact pressure, corrosion, temperature, load spectra, inspection requirements, and applicable codes or design standards.

How to Use These Exercises

For each calculation, define:

  1. the load path and boundary conditions;
  2. whether the stress is nominal, local, principal, shear, or equivalent;
  3. the material data source and the material condition;
  4. the governing failure mode: yield, fracture, fatigue, buckling, deflection, leakage, or loss of function;
  5. the validation evidence needed before the result is used for release.

The common failure in stress analysis is not arithmetic. It is solving the wrong load case, using the wrong failure mode, or interpreting a simplified stress as if it represented the real component.

Use the exercises as release gates: accept or reject a static margin, change a stress-concentration detail, require a deflection limit, escalate a buckling review, revise a fatigue assumption, question a pressure-boundary screen, or hold model validation until measured strain, boundary conditions, and uncertainty match the prediction being used.

Exercise 1: Axial Stress, Strain, and Elongation

A steel tie rod carries an axial tensile load:

F=18\ \text{kN}

The rod cross-sectional area is:

A=240\ \text{mm}^2

The free length is:

L=0.60\ \text{m}

Use:

E=200\ \text{GPa}

Find axial stress, strain, and elongation.

Solution

Axial stress is:

\displaystyle \sigma=\frac{F}{A}

Convert area:

A=240\ \text{mm}^2=240\times10^{-6}\ \text{m}^2

Then:

\displaystyle \sigma=\frac{18{,}000}{240\times10^{-6}}=75{,}000{,}000\ \text{Pa}=75\ \text{MPa}

Elastic strain is:

\displaystyle \epsilon=\frac{\sigma}{E}=\frac{75\times10^6}{200\times10^9}=3.75\times10^{-4}

This is:

\epsilon=375\ \mu\epsilon

Elongation is:

\Delta L=\epsilon L=(3.75\times10^{-4})(0.60)=2.25\times10^{-4}\ \text{m}
\Delta L=0.225\ \text{mm}

Engineering Comment

The nominal stress is simple because the load path is simple. A real tie rod design would still check threads, shoulders, clevis pins, bearing pressure, preload, corrosion allowance, fatigue, and whether the measured elongation is consistent with the intended load.

Exercise 2: Cantilever Bracket Bending with Stress Concentration

A cantilever bracket carries a vertical load:

P=650\ \text{N}

at a distance:

L=140\ \text{mm}

from the fixed section. The rectangular section at the wall has width:

b=35\ \text{mm}

and height:

h=12\ \text{mm}

Assume bending about the strong axis and a stress concentration factor:

K_t=1.7

The material yield strength is:

\sigma_y=355\ \text{MPa}

Estimate nominal bending stress, local elastic stress, and static yield factor.

Solution

The bending moment at the fixed section is:

M=PL=(650)(140)=91{,}000\ \text{N mm}

The second moment of area is:

\displaystyle I=\frac{bh^3}{12}=\frac{35(12^3)}{12}=5040\ \text{mm}^4

The outer-fiber distance is:

\displaystyle c=\frac{h}{2}=6\ \text{mm}

Nominal bending stress:

\displaystyle \sigma_{nom}=\frac{Mc}{I}=\frac{91{,}000(6)}{5040}=108\ \text{MPa}

Local elastic stress estimate:

\sigma_{local}=K_t\sigma_{nom}=1.7(108)=184\ \text{MPa}

Static yield factor:

\displaystyle N_y=\frac{\sigma_y}{\sigma_{local}}=\frac{355}{184}=1.93

Engineering Comment

The static yield screen passes, but it does not prove fatigue life. A bracket root is often controlled by weld toe geometry, fillet radius, load reversals, bolt flexibility, surface finish, and whether the assumed fixed boundary is realistic.

Exercise 3: Combined Bending and Torsion in a Shaft

A solid circular shaft has diameter:

d=30\ \text{mm}

It carries bending moment:

M=220\ \text{N m}

and torque:

T=160\ \text{N m}

For a ductile steel shaft, estimate bending stress, torsional shear stress, von Mises equivalent stress, and yield factor for:

\sigma_y=250\ \text{MPa}

Use:

\displaystyle \sigma_b=\frac{32M}{\pi d^3}

and:

\displaystyle \tau=\frac{16T}{\pi d^3}

with M and T in \text{N mm}.

Solution

Convert moments:

M=220{,}000\ \text{N mm}
T=160{,}000\ \text{N mm}

Bending stress:

\displaystyle \sigma_b=\frac{32(220{,}000)}{\pi(30^3)}=83.0\ \text{MPa}

Torsional shear stress:

\displaystyle \tau=\frac{16(160{,}000)}{\pi(30^3)}=30.2\ \text{MPa}

For uniaxial normal stress plus shear:

\sigma_e=\sqrt{\sigma_b^2+3\tau^2}
\sigma_e=\sqrt{83.0^2+3(30.2^2)}=98.1\ \text{MPa}

Yield factor:

\displaystyle N_y=\frac{250}{98.1}=2.55

Engineering Comment

The calculation is a smooth-shaft static check. A real rotating shaft often needs a fatigue check at shoulders, keyways, grooves, threads, spline roots, press fits, corrosion pits, and bearing transitions. Stress concentration can dominate the design even when the nominal von Mises stress looks low.

Exercise 4: Principal Stresses and Von Mises Stress

A finite element result at a point in a plate gives plane-stress components:

\sigma_x=95\ \text{MPa}
\sigma_y=35\ \text{MPa}
\tau_{xy}=42\ \text{MPa}

Calculate principal stresses and von Mises equivalent stress.

Solution

The average normal stress is:

\displaystyle \sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}=\frac{95+35}{2}=65\ \text{MPa}

The Mohr-circle radius is:

\displaystyle R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}
R=\sqrt{30^2+42^2}=51.6\ \text{MPa}

Principal stresses:

\sigma_1=65+51.6=116.6\ \text{MPa}
\sigma_2=65-51.6=13.4\ \text{MPa}

Plane-stress von Mises stress:

\sigma_e=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}
\sigma_e=\sqrt{95^2-(95)(35)+35^2+3(42^2)}
\sigma_e=110.5\ \text{MPa}

Engineering Comment

Von Mises stress is useful for ductile yielding, but it is not a universal failure measure. Brittle materials, adhesive joints, composites, castings, cracks, contact stresses, and welded details may require principal stress, fracture, or specialized criteria.

Exercise 5: Beam Deflection and Bending Stress

A simply supported steel beam carries a central point load:

P=1.2\ \text{kN}

The span is:

L=1.4\ \text{m}

The rectangular section has width:

b=40\ \text{mm}

and vertical height:

h=60\ \text{mm}

Use:

E=200\ \text{GPa}

Estimate maximum deflection and maximum bending stress.

Solution

The second moment of area is:

\displaystyle I=\frac{bh^3}{12}

In SI units:

\displaystyle I=\frac{(0.040)(0.060^3)}{12}=7.20\times10^{-7}\ \text{m}^4

Maximum deflection for a simply supported beam with central point load is:

\displaystyle \delta_{max}=\frac{PL^3}{48EI}
\displaystyle \delta_{max}=\frac{(1200)(1.4^3)}{48(200\times10^9)(7.20\times10^{-7})}
\delta_{max}=4.76\times10^{-4}\ \text{m}=0.48\ \text{mm}

Maximum bending moment:

\displaystyle M_{max}=\frac{PL}{4}=\frac{(1200)(1.4)}{4}=420\ \text{N m}

Outer-fiber distance:

c=0.030\ \text{m}

Bending stress:

\displaystyle \sigma_{max}=\frac{M_{max}c}{I}=\frac{(420)(0.030)}{7.20\times10^{-7}}=17.5\ \text{MPa}

Engineering Comment

The stress is low for steel, but stiffness may still control if the beam supports alignment-sensitive equipment, seals, optics, gears, or sensors. A part can be strong enough and still fail functionally because deflection is excessive.

Exercise 6: Euler Buckling Factor for a Slender Column

A pinned-pinned column carries compressive load:

P=24\ \text{kN}

The unsupported length is:

L=1.8\ \text{m}

The effective length factor is:

K=1.0

The material modulus is:

E=200\ \text{GPa}

The least second moment of area is:

I=1.1\times10^{-7}\ \text{m}^4

Estimate Euler critical load and buckling factor.

Solution

Euler critical load is:

\displaystyle P_{cr}=\frac{\pi^2EI}{(KL)^2}
\displaystyle P_{cr}=\frac{\pi^2(200\times10^9)(1.1\times10^{-7})}{(1.0\times1.8)^2}
P_{cr}=67{,}000\ \text{N}=67\ \text{kN}

Buckling factor:

\displaystyle N_{buckling}=\frac{P_{cr}}{P}=\frac{67}{24}=2.8

Engineering Comment

The ideal factor is not the whole stability assessment. Real columns have initial crookedness, eccentric loading, imperfect end fixity, residual stress, local buckling, material nonlinearity, and connection flexibility. Compression members should be checked with the design method appropriate to their slenderness and code context.

Exercise 7: Fully Constrained Thermal Stress

An aluminium bar is fully restrained from expanding. Its temperature rises by:

\Delta T=45^\circ\text{C}

Use:

E=70\ \text{GPa}

and:

\alpha=23\times10^{-6}\ /\text{K}

Estimate the thermal stress magnitude. If yield strength is 240\ \text{MPa}, estimate the yield factor.

Solution

For fully restrained uniform expansion:

\sigma=E\alpha\Delta T
\sigma=(70\times10^9)(23\times10^{-6})(45)=72.5\times10^6\ \text{Pa}
\sigma=72.5\ \text{MPa}

The stress is compressive if the bar is heated and prevented from expanding.

Yield factor:

\displaystyle N_y=\frac{240}{72.5}=3.31

Engineering Comment

The simplified result assumes complete restraint and uniform temperature. Real thermal stress depends on temperature gradients, joint stiffness, sliding supports, creep, relaxation, contact pressure, different materials, and thermal cycling. Lower stress may still be unacceptable if it changes alignment, preload, sealing, or fatigue life.

Exercise 8: Goodman Fatigue Screen

A rotating component has local alternating stress:

S_a=80\ \text{MPa}

and local mean stress:

S_m=55\ \text{MPa}

The corrected endurance strength for the design condition is:

S_e=180\ \text{MPa}

The ultimate tensile strength is:

\sigma_{UTS}=520\ \text{MPa}

Use the Goodman relation to estimate utilization and fatigue safety factor:

\displaystyle \frac{S_a}{S_e}+\frac{S_m}{\sigma_{UTS}}\leq \frac{1}{N}

Solution

Goodman utilization is:

\displaystyle u=\frac{S_a}{S_e}+\frac{S_m}{\sigma_{UTS}}
\displaystyle u=\frac{80}{180}+\frac{55}{520}=0.444+0.106=0.550

The implied fatigue safety factor is:

\displaystyle N=\frac{1}{u}=\frac{1}{0.550}=1.82

Engineering Comment

The screen passes for a required factor below 1.82, but only if the stresses and fatigue properties are credible. Fatigue design should check stress concentration, notch sensitivity, surface finish, size, temperature, corrosion, residual stress, spectrum loading, required reliability, and inspection strategy.

Exercise 9: Thin-Walled Pressure Vessel Stress

A thin cylindrical pressure vessel has internal pressure:

p=1.2\ \text{MPa}

inner diameter:

D=0.45\ \text{m}

and wall thickness:

t=6\ \text{mm}

Estimate hoop stress and longitudinal stress. Compare hoop stress with an allowable membrane stress of 120\ \text{MPa}.

Solution

For a thin-walled cylinder:

\displaystyle \sigma_h=\frac{pD}{2t}
\displaystyle \sigma_l=\frac{pD}{4t}

Convert thickness:

t=0.006\ \text{m}

Hoop stress:

\displaystyle \sigma_h=\frac{(1.2\times10^6)(0.45)}{2(0.006)}=45\ \text{MPa}

Longitudinal stress:

\displaystyle \sigma_l=\frac{(1.2\times10^6)(0.45)}{4(0.006)}=22.5\ \text{MPa}

Hoop-stress utilization:

\displaystyle u=\frac{45}{120}=0.375

Engineering Comment

The membrane stress screen is not a pressure-vessel design code. A real vessel review must include weld efficiency, corrosion allowance, nozzle reinforcement, external loads, cyclic pressure, thermal stress, hydrotest pressure, buckling under vacuum or external pressure, inspection, and code-required material factors.

Exercise 10: Strain-Gauge Validation of a Stress Prediction

A strain gauge is bonded near a bracket root in the direction of principal tensile strain. During proof loading, the measured strain is:

\epsilon_{meas}=510\ \mu\epsilon

Use:

E=200\ \text{GPa}

Assume uniaxial elastic stress at the gauge location. The pre-test model predicted local stress:

\sigma_{pred}=96\ \text{MPa}

Estimate measured stress and percentage difference from the model.

Solution

Convert strain:

510\ \mu\epsilon=510\times10^{-6}

Measured stress:

\sigma_{meas}=E\epsilon_{meas}
\sigma_{meas}=(200\times10^9)(510\times10^{-6})=102\times10^6\ \text{Pa}
\sigma_{meas}=102\ \text{MPa}

Relative difference from prediction:

\displaystyle \delta_{model}=\frac{\sigma_{meas}-\sigma_{pred}}{\sigma_{pred}}
\displaystyle \delta_{model}=\frac{102-96}{96}=0.0625=6.25\%

Engineering Comment

A 6.25 percent difference may be excellent or unacceptable depending on the validation plan. The engineer should check gauge placement, alignment, temperature compensation, adhesive condition, bridge calibration, load-cell accuracy, boundary conditions, and whether the measured strain direction matches the model result being compared.

Review Checklist

When reviewing a mechanical stress-analysis result, ask:

  • Is the load case explicit, including preload, thermal load, pressure, shock, fatigue spectrum, and abnormal operation where relevant?
  • Is the reported stress nominal, local, principal, shear, membrane, bending, equivalent, or measured strain-derived stress?
  • Does the material evidence match product form, heat treatment, surface condition, weld state, temperature, corrosion exposure, and inspection basis?
  • Is the governing failure mode yield, fracture, fatigue, buckling, deflection, leakage, wear, or loss of alignment?
  • Are stress concentrations, boundary-condition stiffness, contact pressure, residual stress, and load path eccentricity represented?
  • Is the margin larger than modeling uncertainty, measurement uncertainty, manufacturing variation, and service scatter?
  • Does validation compare the same physical quantity at the same location, direction, load level, and boundary condition?

The final answer should not say only that a part “passes.” It should state what was checked, what assumptions control the result, and what could invalidate the conclusion in the real assembly.

REF

See also