Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E: Exercises

Toolbox we will lean on the whole way down:

Level 1 — Recognition

Problem 1.1 (L1)

A round strut has diameter $d = 20 mm$ . Its cross-section is a circle. What is its cross-sectional area $A$ in $m^{2}$ ?

Recall Solution 1.1

WHAT we need: the area the force spreads across. WHY: stress is force per area, so we can compute nothing until $A$ is in $m^{2}$ .

Radius $r = d /2 = 10 mm = 0.010 m$ . Area of a circle: $A = π r^{2} = π (0.010)^{2} = π \times 1 0^{- 4} \approx 3.14 \times 1 0^{- 4} m^{2} .$ Notice $20 mm = 0.020 m$ ; squaring metres, not millimetres, is what keeps SI honest.

Problem 1.2 (L1)

A bar of original length $L = 2.5 m$ stretches by $Δ L = 1.25 mm$ . Write the strain as a pure number and as microstrain ( $1 μ ε = 1 0^{- 6}$ ).

Recall Solution 1.2

WHAT: the fractional stretch. WHY divide by $L$ : the same $1.25 mm$ means a lot on a short bar and nothing on a long one — only the ratio describes the material. $ε = \frac{Δ L}{L} = \frac{1.25 \times 1 0 ^{- 3}}{2.5} = 5 \times 1 0^{- 4} .$ In microstrain: $5 \times 1 0^{- 4} = 500 \times 1 0^{- 6} = 500 μ ε$ .

Problem 1.3 (L1)

A material has $E = 70 GPa$ . Under a stress $σ = 35 MPa$ (still elastic), what is the strain?

Recall Solution 1.3

WHY this formula: $σ = E ε$ is Hooke's Law for materials; rearrange to get strain from stress and stiffness: $ε = σ / E$ . $ε = \frac{35 \times 1 0 ^{6}}{70 \times 1 0 ^{9}} = 5 \times 1 0^{- 4} .$ The GPa must become $70 \times 1 0^{9} Pa$ , or the answer is off by $1000$ .

Level 2 — Application

Problem 2.1 (L2)

An aluminium tie ( $E = 70 GPa$ ), length $L = 1.5 m$ , area $A = 2 \times 1 0^{- 4} m^{2}$ , carries a tensile force $F = 1.4 \times 1 0^{4} N$ . Find the stress, the strain, and the extension $Δ L$ .

Recall Solution 2.1

Step 1 — stress (WHAT: crowding of force). $σ = \frac{F}{A} = \frac{1.4 \times 1 0 ^{4}}{2 \times 1 0 ^{- 4}} = 7 \times 1 0^{7} Pa = 70 MPa .$ Step 2 — strain (WHY: undo stiffness via Hooke's law). $ε = \frac{σ}{E} = \frac{7 \times 1 0 ^{7}}{70 \times 1 0 ^{9}} = 1 \times 1 0^{- 3} .$ Step 3 — extension (WHY: undo the "divide by $L$ " that defined strain). $Δ L = ε L = 1 0^{- 3} \times 1.5 = 1.5 \times 1 0^{- 3} m = 1.5 mm .$

Problem 2.2 (L2)

A steel rod ( $E = 200 GPa$ ) of length $L = 0.8 m$ and area $A = 5 \times 1 0^{- 5} m^{2}$ must not stretch more than $0.2 mm$ . What is the largest force it may carry (assume it stays elastic)?

Recall Solution 2.2

WHAT we solve for: $F$ , given a cap on $Δ L$ . WHY rearrange: the deflection equation $Δ L = F L / (A E)$ contains $F$ ; isolate it. $F = \frac{A E Δ L}{L} = \frac{( 5 \times 1 0 ^{- 5} ) ( 200 \times 1 0 ^{9} ) ( 0.2 \times 1 0 ^{- 3} )}{0.8} .$ Numerator: $5 \times 1 0^{- 5} \times 2 \times 1 0^{11} = 1 \times 1 0^{7}$ ; times $0.2 \times 1 0^{- 3}$ gives $2000$ . Divide by $0.8$ : $F = \frac{2000}{0.8} = 2500 N .$

Problem 2.3 (L2)

A titanium cable ( $E = 110 GPa$ ) must limit its stretch to $Δ L = 0.4 mm$ over $L = 1.2 m$ while carrying $F = 6600 N$ . Find the minimum cross-sectional area.

Recall Solution 2.3

Forecast first: less stretch allowed $\Rightarrow$ we need more material, so expect a small but non-tiny $A$ . WHY: solve $Δ L = F L / (A E)$ for the unknown geometry $A$ .

= \frac{7920}{4.4\times10^{7}} = 1.8\times10^{-4}\ \text{m}^2 = 1.8\ \text{cm}^2.$$ **Verify:** $\Delta L = \dfrac{6600\times1.2}{1.8\times10^{-4}\times1.1\times10^{11}} = \dfrac{7920}{1.98\times10^{7}} = 4\times10^{-4}\ \text{m} = 0.4\ \text{mm}$ ✓.

Level 3 — Analysis

Problem 3.1 (L3)

A strut is heated so it wants to expand, but its ends are rigidly clamped so its length cannot change. The material has $E = 70 GPa$ and, if free, would have expanded by a fractional amount $ε_{free} = 8 \times 1 0^{- 4}$ . What stress does the clamp induce, and is it tension or compression? (This is a Thermal Stress setup.)

Recall Solution 3.1

WHAT is happening: heat tries to lengthen the bar by strain $ε_{free}$ , but the clamps force the net length change to zero. To cancel a would-be expansion, the clamps must push the bar back — i.e. squeeze it. So the mechanical strain is $- ε_{free}$ (a compression), see the figure.

Figure — Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

WHY the sign flips: strain is signed. A stretch is $+$ , a squeeze is $-$ . Since the wall stops the expansion, the material sits as if compressed by the same amount it wanted to grow. $σ = E ε = (70 \times 1 0^{9}) (- 8 \times 1 0^{- 4}) = - 5.6 \times 1 0^{7} Pa = - 56 MPa .$ The minus sign means compression — magnitude $56 MPa$ . If it had been cooled instead, the bar would try to shrink, the clamps would hold it stretched, and the sign would be $+$ (tension).

Problem 3.2 (L3)

Two struts A and B are made of the same aluminium and carry the same force $F$ . Strut B has twice the diameter of A. Compare (i) their stresses and (ii) their strains.

Recall Solution 3.2

WHY area matters: stress is $F / A$ , and area scales with the square of diameter, $A = π (d /2)^{2}$ . Doubling $d$ multiplies $A$ by $4$ .

Look at the two circles below: B's circle has $4 \times$ the area of A's.

(i) Stress: $σ_{B} = F / A_{B} = F / (4 A_{A}) = σ_{A} /4$ . Strut B feels one quarter the stress. (ii) Strain: same material means same $E$ , so $ε = σ / E$ scales the same way: $ε_{B} = ε_{A} /4$ . So the fat strut both feels less crowded and stretches proportionally less. Geometry, not material, did all the work here.

Problem 3.3 (L3)

A steel wire ( $E = 200 GPa$ , yield stress $σ_{y} = 250 MPa$ ) of area $A = 1 \times 1 0^{- 6} m^{2}$ carries $F = 300 N$ . Is the wire still in the elastic region where $σ = E ε$ is valid? If so, find its strain.

Recall Solution 3.3

WHY check yield first: Hooke's Law only holds below yield; past it the material enters plastic behaviour and our formulas lie. $σ = \frac{F}{A} = \frac{300}{1 \times 1 0 ^{- 6}} = 3 \times 1 0^{8} Pa = 300 MPa .$ Compare: $σ = 300 MPa > σ_{y} = 250 MPa$ . The wire has yielded — it is not in the elastic region, so $ε = σ / E$ is not valid and the wire is permanently deformed (or breaking). The honest answer is "you may not use $σ = E ε$ here." See the Stress-Strain Curve for where the straight line ends.

Level 4 — Synthesis

Problem 4.1 (L4)

A launch strut must satisfy three rules at once under $F = 4.0 \times 1 0^{4} N$ over $L = 2.0 m$ of aluminium ( $E = 70 GPa$ , yield $σ_{y} = 240 MPa$ ): (a) stress must stay below yield with a Factor of Safety of $2$ (i.e. $σ \leq σ_{y} /2$ ); (b) extension must not exceed $2.0 mm$ . Find the minimum area that meets both, and say which rule governs.

Recall Solution 4.1

WHY two rules: strength and stiffness are different limits (a ceiling vs a slope), so we compute the area each demands separately, then take the larger — the binding constraint.

Rule (a), strength. Allowed stress $σ_{allow} = σ_{y} /2 = 120 MPa = 1.2 \times 1 0^{8} Pa$ . Need $σ = F / A \leq σ_{allow}$ : $A_{a} = \frac{F}{σ _{allow}} = \frac{4.0 \times 1 0 ^{4}}{1.2 \times 1 0 ^{8}} = 3.33 \times 1 0^{- 4} m^{2} .$

Rule (b), stiffness. Need $Δ L = F L / (A E) \leq 2.0 \times 1 0^{- 3} m$ :

= \frac{8.0\times10^{4}}{1.4\times10^{8}} = 5.71\times10^{-4}\ \text{m}^2.$$ **Decide:** the strut must satisfy *both*, so take the larger: $A_{\min} = \max(A_a, A_b) = 5.71\times10^{-4}\ \text{m}^2$. **Stiffness governs** — the stretch limit forces a fatter strut than strength alone would. See [[Spacecraft Load Paths and Struts]] for why deflection so often wins in pointing-critical structures.

Problem 4.2 (L4)

Along a single load path, force $F = 1.0 \times 1 0^{4} N$ passes through a steel section (part 1: $E_{1} = 200 GPa$ , $L_{1} = 0.5 m$ , $A_{1} = 1 \times 1 0^{- 4} m^{2}$ ) and then an aluminium section (part 2: $E_{2} = 70 GPa$ , $L_{2} = 0.5 m$ , $A_{2} = 1 \times 1 0^{- 4} m^{2}$ ) in series. Find the total extension.

Recall Solution 4.2

WHY add: in a series load path the same $F$ flows through each part, and the total stretch is the sum of each part's stretch — see Spacecraft Load Paths and Struts.

= \frac{5000}{2\times10^{7}} = 2.5\times10^{-4}\ \text{m}.$$ $$\Delta L_2 = \frac{FL_2}{A_2 E_2} = \frac{10^{4}\times0.5}{10^{-4}\times7\times10^{10}} = \frac{5000}{7\times10^{6}} = 7.14\times10^{-4}\ \text{m}.$$ **Total:** $\Delta L = \Delta L_1 + \Delta L_2 = 2.5\times10^{-4} + 7.14\times10^{-4} = 9.64\times10^{-4}\ \text{m} \approx 0.96\ \text{mm}.$ The softer aluminium contributes most of the stretch even though the parts are the same size — lower $E$ means more give.

Level 5 — Mastery

Problem 5.1 (L5)

A composite tension member is built from a CFRP core ( $E_{c} = 150 GPa$ , $A_{c} = 2 \times 1 0^{- 4} m^{2}$ ) and an aluminium sleeve ( $E_{a} = 70 GPa$ , $A_{a} = 2 \times 1 0^{- 4} m^{2}$ ) bonded so they share the same length and therefore the same strain. They together carry $F = 5.0 \times 1 0^{4} N$ . Find (a) the shared strain, (b) the force in each material, and (c) the stress in each.

Recall Solution 5.1

WHY same strain, not same stress: the two are glued side by side and stretch together, so the geometry (strain) is common, and each material develops its own stress via its own $E$ . This is the opposite of the series case in 4.2.

Set up force balance. Each carries $F_{i} = σ_{i} A_{i} = E_{i} ε A_{i}$ with the same $ε$ . They sum to the total: $F = ε (E_{c} A_{c} + E_{a} A_{a}) .$ (a) Shared strain. The stiffness sum: $E_{c} A_{c} = 150 \times 1 0^{9} \times 2 \times 1 0^{- 4} = 3.0 \times 1 0^{7}$ ; $E_{a} A_{a} = 70 \times 1 0^{9} \times 2 \times 1 0^{- 4} = 1.4 \times 1 0^{7}$ ; total $= 4.4 \times 1 0^{7} N$ .

= 1.136\times10^{-3}.$$ **(b) Force in each** ($F_i = E_i A_i \varepsilon$): $$F_c = 3.0\times10^{7}\times1.136\times10^{-3} = 3.41\times10^{4}\ \text{N},$$ $$F_a = 1.4\times10^{7}\times1.136\times10^{-3} = 1.59\times10^{4}\ \text{N}.$$ Check: $F_c + F_a = 5.0\times10^{4}\ \text{N}$ ✓ — the stiffer CFRP grabs the bigger share. **(c) Stress in each** ($\sigma_i = F_i/A_i$): $$\sigma_c = \frac{3.41\times10^{4}}{2\times10^{-4}} = 1.705\times10^{8}\ \text{Pa} \approx 171\ \text{MPa},$$ $$\sigma_a = \frac{1.59\times10^{4}}{2\times10^{-4}} = 7.95\times10^{7}\ \text{Pa} \approx 80\ \text{MPa}.$$ Same strain, different stress — exactly because $E$ differs. The stiff phase carries the heaviest load, a key idea when reading a [[Stress-Strain Curve]] for each phase.

Recall One-line self-test after all five levels

Which is shared in a series load path, and which in a parallel (bonded) one? ::: Series shares the force (extensions add); parallel bonded shares the strain (forces add).