Strength of Materials: free practice, theory and problems

1. Stress and strain

What is it about?

Strength of materials is about how materials behave under load: how much stress builds up inside the material, and how much it deforms. While statics in MAPE1300 finds the external forces and the support reactions, strength of materials goes further and asks whether the material can actually withstand them. This is the foundation for sizing rods, wires, bolts and shafts so that they neither yield nor fracture.

The simplest and most important case is a straight bar loaded by an axial force: the force is (approximately) evenly distributed over the cross-section, and the relationship between stress and strain is linear until the material starts to yield.

Concepts and formulas

• Normal stress: $\sigma = F/A$, measured in N/mm² = MPa.
• Strain: $\varepsilon = \Delta L/L$ (dimensionless, often given in permille).
• Hooke's law in the elastic range: $\sigma = E\varepsilon$.
• Elongation: $\delta = \dfrac{FL}{AE}$.
• Shear stress $\tau = F/A$ (the force acts parallel to the cut surface) and shear strain $\gamma$, with $\tau = G\gamma$.
• Relation between Young's modulus, shear modulus and Poisson's ratio: $G = \dfrac{E}{2(1+\nu)}$.
• Thermal strain: $\varepsilon_T = \alpha\Delta T$. If the bar can expand freely, the stress is zero. If it is restrained at both ends, $\sigma = E\alpha\Delta T$ appears.
• Yield strength $R_e$, tensile strength $R_m$ and safety factor $n = R_e/\sigma$.

How to solve the problems

1. Find the cross-sectional area $A$ and the axial force $F$ acting at the section.
2. Compute the normal stress $\sigma = F/A$, and compare it with $R_e/n$ if you are checking safety.
3. Find the strain $\varepsilon = \sigma/E$ and the elongation $\delta = \varepsilon L$.
4. For a temperature change: decide first whether the bar can expand freely (no stress) or is restrained (stress $E\alpha\Delta T$).
5. If the bar has several segments (different length, area or material), compute each segment's elongation separately and add them.

Example

A steel bar with a diameter of 12 mm and a length of 800 mm carries 15 kN in tension. $E = 210$ GPa. Find the stress, strain and elongation.

1. Area: $A = \pi(6\text{ mm})^2 \approx 113.1$ mm².
2. Stress: $\sigma = F/A = 15,000/113.1 \approx 132.6$ MPa.
3. Strain: $\varepsilon = \sigma/E = 132.6/210,000 \approx 0.632$ ‰.
4. Elongation: $\delta = \varepsilon L \approx 0.000632\cdot 800 \approx 0.505$ mm.

Answer: $\sigma \approx 132.6$ MPa, $\varepsilon \approx 0.632$ ‰ and $\delta \approx 0.505$ mm.

Common mistakes

• Believing a freely moving bar develops thermal stress when heated. It only develops stress when the expansion is prevented.
• Mixing mm and m, or GPa and MPa, in the same expression.
• Using the radius when the problem gives the diameter (or the reverse) in the area formula $A = \pi d^2/4$.
• Using Hooke's law ($\sigma = E\varepsilon$) for stresses above the yield strength, where it no longer holds.

Concepts in this part

Practise stress and strain in the app →

2. Torsion and bending

What is it about?

Shafts transmit moment by twisting (torsion), and beams transmit transverse load by bending. Both produce stresses that are NOT evenly distributed over the cross-section, unlike the pure axial stress from the previous unit: the stress is zero at a central point (the center in torsion, the neutral axis in bending) and largest farthest away. That is exactly why the shape of the cross-section matters so much for how stiff and strong a shaft or beam becomes.

Concepts and formulas

• Torsion in a circular shaft: the shear stress is linear with distance from the center, $\tau = \dfrac{Tr}{J}$, largest at the surface: $\tau_{max} = \dfrac{T(d/2)}{J} = \dfrac{16T}{\pi d^3}$.
• Polar second moment of area for a solid circular shaft: $J = \dfrac{\pi d^4}{32}$. For a hollow shaft (outer $D$, inner $d$): $J = \dfrac{\pi(D^4-d^4)}{32}$.
• Angle of twist over the length $L$: $\theta = \dfrac{TL}{GJ}$ (radians).
• Bending stress at distance $y$ from the neutral axis: $\sigma = \dfrac{My}{I}$, largest at the outer fiber: $\sigma_{max} = M/W$, with $W = I/y_{max}$.
• The shear stress in a bent beam is largest at the neutral axis (for a rectangle: $\tau_{max} = 1.5V/A$), not at the outer fibers where the bending stress is largest.
• Power and torque: $P = T\omega$, where $\omega$ is the angular velocity in rad/s.

How to solve the problems

1. Torsion: find $J$ for the cross-section, compute $\tau_{max}$ at the surface and, if needed, $\theta$ over the relevant length.
2. Bending: find $I$ for the cross-section and the distance $y_{max}$ to the outer fiber (or use $W$ directly), compute $\sigma_{max} = M/W$.
3. When sizing (finding the required diameter): rearrange the formula, for example $d = \sqrt[3]{16T/(\pi\tau_{allow})}$.
4. Keep track of units: use N and mm consistently, so $\sigma$ and $\tau$ come out directly in MPa.
5. Check that the answer is reasonable: the shear stress is zero at the center of a shaft and zero at the neutral axis in bending, largest at the surface/outer fiber.

Example

A solid steel shaft ($G = 80$ GPa) has a diameter of 50 mm and a length of 1.5 m. It transmits a torque of 800 Nm. Find the maximum shear stress and the angle of twist.

1. Polar second moment of area: $J = \pi\cdot 50^4/32 \approx 613,592$ mm⁴.
2. Maximum shear stress: $\tau_{max} = T(d/2)/J = 800,000\cdot 25/613,592 \approx 32.6$ MPa.
3. Angle of twist: $\theta = TL/(GJ) = 800,000\cdot 1500/(80,000\cdot 613,592) \approx 0.0244$ rad $\approx 1.40^\circ$.

Answer: $\tau_{max} \approx 32.6$ MPa and $\theta \approx 1.40^\circ$.

Common mistakes

• Using $J = \pi d^4/64$ (that is $I$, the second moment of area about a diameter) instead of $J = \pi d^4/32$ in the torsion formula.
• Believing the shear stress in torsion is largest at the center. It is zero there and largest at the surface.
• Believing the bending stress is largest at the neutral axis. It is the other way around: $\sigma = 0$ at the neutral axis, largest at the outer fiber.
• Mixing degrees and radians in $\theta = TL/(GJ)$ – the formula gives radians.
• Forgetting that doubling the diameter gives 16 times the $J$ (and therefore 16 times the torque for the same $\tau_{max}$).

Concepts in this part

Practise torsion and bending in the app →

3. Stress transformation and buckling

What is it about?

At any point in a loaded body, the stress looks different depending on which cut you look at. Stress transformation is about finding the worst directions: the principal stresses (the largest and smallest normal stress, with no shear) and the largest shear stress. This is needed to decide whether a ductile or brittle material yields or fractures, since many materials do not care about the stress in one particular direction, but about a combination of all of them.

Buckling is something completely different: a slender column in compression can bow out sideways and collapse long before the material itself has reached the yield strength. It is a stability failure, not a strength failure, and depends strongly on the column's length and end conditions.

Concepts and formulas

• Plane stress state: $\sigma_x$, $\sigma_y$, $\tau_{xy}$. Center of Mohr's circle: $c = (\sigma_x+\sigma_y)/2$. Radius: $R = \sqrt{((\sigma_x-\sigma_y)/2)^2+\tau_{xy}^2}$.
• Principal stresses: $\sigma_{1,2} = c \pm R$ (the shear stress is zero in these directions). Maximum in-plane shear stress: $\tau_{max} = R$.
• Von Mises stress (plane state): $\sigma_v = \sqrt{\sigma^2+3\tau^2}$ for a uniaxial stress $\sigma$ combined with shear $\tau$, more generally $\sigma_v=\sqrt{\sigma_1^2-\sigma_1\sigma_2+\sigma_2^2}$. Yielding occurs when $\sigma_v = R_e$.
• Tresca's criterion: yielding when $\tau_{max} = R_e/2$, somewhat more conservative than von Mises.
• Radius of gyration: $i = \sqrt{I/A}$. Slenderness ratio: $\lambda = L_e/i$, where $L_e = KL$ is the effective buckling length.
• Effective length factor $K$: pinned–pinned $K=1$, fixed–fixed $K=0.5$, fixed–pinned $K\approx0.7$, fixed–free $K=2$.
• Euler's critical load: $P_{cr} = \dfrac{\pi^2EI}{(KL)^2}$, valid as long as the column is still elastic (high slenderness).

How to solve the problems

1. Stress transformation: compute the center $c$ and the radius $R$ of Mohr's circle, and find $\sigma_{1,2} = c\pm R$ and $\tau_{max} = R$.
2. For a yield check: insert the principal stresses into the von Mises or Tresca expression and compare with $R_e$ (possibly divided by a safety factor).
3. Buckling: find the correct $K$ from the end conditions, compute $L_e = KL$, and insert it into Euler's formula.
4. The radius of gyration and the slenderness ratio are used to decide whether Euler's formula applies at all (high $\lambda$) or whether the column will yield instead.
5. Check the order of magnitude: doubling the buckling length gives a quarter of the critical load, not half of it.

Example

A point has $\sigma_x = 80$ MPa, $\sigma_y = -20$ MPa and $\tau_{xy} = 30$ MPa. Find the principal stresses and the maximum shear stress.

1. Center: $c = (80-20)/2 = 30$ MPa.
2. Radius: $R = \sqrt{((80-(-20))/2)^2+30^2} = \sqrt{50^2+30^2} \approx 58.3$ MPa.
3. Principal stresses: $\sigma_1 = 30+58.3 \approx 88.3$ MPa, $\sigma_2 = 30-58.3 \approx -28.3$ MPa.
4. Maximum in-plane shear stress: $\tau_{max} = R \approx 58.3$ MPa.

Answer: $\sigma_1 \approx 88.3$ MPa, $\sigma_2 \approx -28.3$ MPa, $\tau_{max} \approx 58.3$ MPa.

Common mistakes

• Using $(\sigma_x-\sigma_y)$ instead of $(\sigma_x-\sigma_y)/2$ in the radius formula.
• Believing the highest shear stress in space equals $R$ from the plane circle – in 3D the true maximum shear stress can be larger if $\sigma_2$ and $\sigma_3=0$ have opposite signs.
• Using the wrong $K$ factor, especially confusing pinned–pinned ($K=1$) with fixed–free ($K=2$).
• Believing the critical load is proportional to the length. It is inversely proportional to the length squared.
• Using Euler's formula for a short, thick column that would actually yield before it buckles.

Concepts in this part

Practise stress transformation and buckling in the app →