Week 11 — Normal Stress and Strain + Mechanical Properties

Overview

Week 11 introduces the mechanics of deformable bodies: how axial forces produce internal stress and strain, and how a uniaxial tensile test reveals a material’s mechanical properties (Young’s modulus, yield stress, UTS, fracture stress, ductility, Poisson’s ratio). EGD102 considers only axial and shear loading (not bending or torsion). The lecture also covers engineering design using a factor of safety to set the allowable stress below failure. These results underpin Portfolio 10/11 and feature heavily on the final exam.

Key concepts

Term	Symbol	Units	Definition
Axial (normal) force	$P$ or $F$	N	Internal force along the axis of a member (tensile = pulling apart, compressive = pushing together).
Cross-sectional area	$A$	m² or mm²	Area perpendicular to $P$.
Axial (normal) stress	$\sigma$	Pa (= N/m²); engineering use MPa = N/mm²	Intensity of force per unit area within a material.
Original (initial) length	$L_0$	m or mm	Length before loading.
Final length / length at fracture	$L$, $L_f$	m or mm	Length under load / at fracture.
Change in length (elongation)	$\Delta L$, $\delta$	m or mm	$L-L_0$.
Axial (normal) strain	$\epsilon$ (or $ϵ$)	dimensionless (or %)	Change in length per unit original length.
Young’s modulus / modulus of elasticity	$E$	Pa (typically GPa)	Stiffness: slope of the linear-elastic region of $\sigma$–$\epsilon$.
Hooke’s Law	—	—	$\sigma =E\epsilon$ in the linear-elastic region.
Elastic deformation	—	—	Recoverable; load removed → no permanent deformation.
Plastic deformation	—	—	Permanent; occurs after the yield point.
Yield stress	${\sigma }_y$	MPa / GPa	Transition stress from elastic to plastic behaviour.
Ultimate tensile strength	${\sigma }_{UTS}$ or UTS	MPa / GPa	Largest stress on the engineering $\sigma$–$\epsilon$ curve.
Fracture stress	${\sigma }_{failure}$	MPa / GPa	Stress at which the material breaks.
Engineering stress / strain	—	—	Computed using initial cross-section $A_0$ and length $L_0$.
True stress / strain	—	—	Computed using instantaneous cross-section $A$.
Ductility (% elongation)	%EL	%	Plastic tensile strain at fracture, expressed as a percentage.
Reduction in area	%AR	%	Percentage change in cross-section between $A_0$ and necked $A_f$.
Offset yield stress	—	MPa	Yield stress read from the curve by intersecting a line drawn from a chosen strain offset (e.g. 0.2%), parallel to the elastic slope.
Factor of safety	$SF$ / $FOS$	dimensionless	Ratio of failure stress to allowable design stress.
Allowable stress	${\sigma }_{allowable}$	MPa	Design stress in the material once $SF$ is applied; always lower than ${\sigma }_{failure}$.
Longitudinal strain	${\epsilon }_{long}$	—	Strain along the loading axis.
Lateral strain	${\epsilon }_{lat}$	—	Strain transverse to loading (diameter change).
Poisson’s ratio	$\nu$	dimensionless (0 to 0.5)	Negative ratio of lateral to longitudinal strain.

Typical Poisson’s ratios (slide 22): Aluminium 0.33; Brass 0.34; Lead 0.43; Steel 0.30; Cast Iron 0.22–0.3; Titanium 0.34; Tungsten 0.28; Concrete 0.1–0.2; Glass 0.22; Clay 0.41.

Typical factors of safety (slide 19): Structural members in buildings 2.0; Pressurised vessels 3.5–4.0; Automobile 3.0; Aircraft and spacecraft 1.2–3.0. Ductile materials generally use lower $SF$ values; brittle materials use higher.

Core formulas

Axial stress (slide 5): $\sigma [\text{Pa}]=\frac{P[\text{N}]}{A[{\text{m}}^2]},\sigma [\text{MPa}]=\frac{P[\text{N}]}{A[{\text{mm}}^2]}$

(Force is always entered in newtons; 1 MPa = 1 N/mm² = $10^6$ Pa.)

Axial strain (slide 8): $\epsilon =\frac{L-L_0}{L_0}=\frac{\Delta L}{L_0}$

Hooke’s Law / Young’s modulus (slide 12): $E=\frac{\Delta \sigma }{\Delta \epsilon }=\frac{{\sigma }_2-{\sigma }_1}{{\epsilon }_2-{\epsilon }_1}⟺\sigma =E\epsilon$

Engineering vs. true stress and strain (slide 13): ${\sigma }_{eng}=\frac{F}{A_0},{\epsilon }_{eng}=\frac{\delta }{L_0}=\frac{\Delta L}{L_0};{\sigma }_{true}=\frac{F}{A},{\epsilon }_{true}=\frac{\delta }{L_0}$

Ductility — percentage elongation and reduction in area (slide 15): $\%EL=\frac{L_f-L_0}{L_0}\times 100,\%AR=\frac{A_f-A_0}{A_0}\times 100$

Factor of safety / allowable stress (slides 18–19): $SF=FOS=\frac{{\sigma }_{failure}}{{\sigma }_{allowable}}⟺{\sigma }_{allowable}=\frac{{\sigma }_{failure}}{SF}$

Poisson’s ratio (slide 21): ${\epsilon }_{long}=\frac{L-L_0}{L_0}=\frac{\Delta L}{L_0},{\epsilon }_{lat}=\frac{d-d_0}{d_0}=\frac{\Delta d}{d_0}$ $\nu =-\frac{{\epsilon }_{lat}}{{\epsilon }_{long}},0\le \nu \le 0.5$

Worked examples

Example 1 — Average normal stress (slide 6)

Bar carries $P=10$ kN with $A=500$ mm².

$\sigma =\frac{P}{A}=\frac{10000\text{N}}{500{\text{mm}}^2}=20{\text{N/mm}}^2=20\text{MPa}$

The bar experiences an average normal (tensile) stress of 20 MPa.

Example 2 — Two rods supporting a lamp (slide 7)

An 80 kg lamp hangs from two rods that meet at $B$. Rod $AB$ runs up-left at $30°$ above horizontal (diameter 10 mm); rod $BC$ runs up-right along a 3-4-5 triangle (diameter 8 mm), so $\sin {\theta }_{BC}=3/5$, $\cos {\theta }_{BC}=4/5$. Find the average normal stress in each rod.

Step 1 — Weight at $B$. $W=mg=80(9.81)=784.8\text{N}$.

Step 2 — Equilibrium at joint $B$. Let $F_{AB}$ and $F_{BC}$ be the rod tensions (forces in the rods directed away from $B$).

$\sum F_x=0$: $-F_{AB}\cos 30°+F_{BC}(4/5)=0$

$\sum F_y=0$: $F_{AB}\sin 30°+F_{BC}(3/5)-784.8=0$

From $\sum F_x$: $F_{AB}=F_{BC}(4/5)/\cos 30°=0.9238F_{BC}$.

Substitute into $\sum F_y$: $0.9238F_{BC}(0.5)+0.6F_{BC}=784.8$ → $1.0619F_{BC}=784.8$ → $F_{BC}\approx 739\text{N}$.

Then $F_{AB}=0.9238(739)\approx 683\text{N}$.

Step 3 — Cross-sectional areas. $A_{AB}=\frac{\pi }{4}(10{)}^2=78.54{\text{mm}}^2,A_{BC}=\frac{\pi }{4}(8{)}^2=50.27{\text{mm}}^2$

Step 4 — Stresses. ${\sigma }_{AB}=\frac{683}{78.54}\approx 8.69\text{MPa},{\sigma }_{BC}=\frac{739}{50.27}\approx 14.7\text{MPa}$

Example 3 — Final length from a strain (slide 9)

Brass cylinder, $d=10$ mm, $L_0=500$ mm, pulled by 100 kN, strain = 0.5% = 0.005.

$\Delta L=\epsilon \cdot L_0=0.005\times 500=2.5\text{mm}$ $L_f=L_0+\Delta L=500+2.5=502.5\text{mm}$

Example 5 — Cable car (slide 20)

Cable car mass 2500 kg, steel cable $d=12$ mm, $E_{steel}=207$ GPa, cable length 2.4 m, ${\sigma }_y=300$ MPa, $SF=1.2$, person mass 65 kg.

(a) Strain in the cable. Load $W=2500(9.81)=24525\text{N}$. Area $A=\frac{\pi }{4}(12{)}^2=113.10{\text{mm}}^2$.

$\sigma =\frac{24525}{113.10}=216.84\text{MPa}=216.84\times 10^6\text{Pa}$ $\epsilon =\frac{\sigma }{E}=\frac{216.84\times 10^6}{207\times 10^9}\approx 1.047\times 10^{-3}\approx 0.00105$

(The slide reports ${\epsilon }_{long}=0.00104$; rounding-consistent.)

(b) Extension. $\Delta L=\epsilon L_0=0.00105\times 2400\approx 2.51\text{mm}$.

(c) Number of people. Allowable stress with $SF$: ${\sigma }_{allowable}=\frac{{\sigma }_y}{SF}=\frac{300}{1.2}=250\text{MPa}$ Allowable force: $F_{allow}={\sigma }_{allowable}A=250\times 113.10=28275\text{N}$. Available for passengers: $28275-24525=3750\text{N}$. Weight per person $=65(9.81)=637.65\text{N}$. $n=\frac{3750}{637.65}\approx 5.88⟹\boxed{n=5\text{people}}$ (Round down: 6 would exceed the allowable stress.)

(d) Reduction in area (Example 5 cont., slide 23). Using $\nu =0.30$ and ${\epsilon }_{long}=0.00104$: ${\epsilon }_{lat}=-\nu {\epsilon }_{long}=-0.30(0.00104)=-3.12\times 10^{-4}$ $\Delta d={\epsilon }_{lat}{d}_0=-3.12\times 10^{-4}(12)=-3.744\times 10^{-3}\text{mm}$ New diameter $d_f=12-0.003744=11.9963\text{mm}$. $A_f=\frac{\pi }{4}(11.9963{)}^2=113.027{\text{mm}}^2$ $\Delta A=A_f-A_0=113.027-113.097\approx -0.0706{\text{mm}}^2$ The cross-section reduces by about 0.07 mm².

Things to practise

The Tutorial 11 worksheet drills exactly these procedures — work each one through fully, then compare with the formulas above.

1. Exercise 1. Specimen $L_0=300$ mm, $d=12$ mm. Force increased from 2.5 kN to 9 kN, elongation 0.225 mm. Find (a) the increase in stress $\Delta \sigma$, (b) the strain (deformation is linear elastic). Hint: $\Delta P=6.5$ kN; $A=\pi d^2/4$; $\epsilon =\Delta L/L_0$.
2. Exercise 2. Bar $L_0=125$ mm, $A=437.5{\text{mm}}^2$, axial force 40 kN, stretch 0.05 mm. Find $E$. Hint: compute $\sigma$ and $\epsilon$, then $E=\sigma /\epsilon$.
3. Exercise 3. Magnesium-alloy tensile test ($d=12$ mm, gauge 50 mm). Read the modulus of elasticity (slope of linear region) and the 0.2% offset yield strength from the $\sigma$–$\epsilon$ plot.
4. Exercise 4. Two wires $AB$ (0.6 m, 3 mm dia, 45°) and $BC$ (0.9 m, 5 mm dia, 30°) support a 6750 N platform. Using the bilinear $\sigma$–$\epsilon$ curve (knee at $\sigma =406$ MPa, $\epsilon =0.002$; second point $\sigma =560$ MPa, $\epsilon =0.01$), find the elongation of each wire. Hint: solve joint equilibrium for $F_{AB}$, $F_{BC}$; compute stress; pick the correct branch of the bilinear curve; then $\Delta L=\epsilon L_0$.
5. Exercise 5. 50-kg flowerpot from wires $AB$ (at 30°) and $BC$ (at 45°); failure stress 350 MPa, $SF=2.5$. Find the minimum diameter of each wire. Hint: equilibrium → $F$; ${\sigma }_{allow}=350/2.5=140$ MPa; $A_{min}=F/{\sigma }_{allow}$; $d_{min}=\sqrt{4A/\pi }$.
6. Exercise 6. Aluminium rod, $L_0=500$ mm, $d_0=40$ mm, $E=70$ GPa, $\nu =0.33$, axial load 100 kN. Find $\Delta L$ and $\Delta d$. Hint: $\sigma =P/A$, ${\epsilon }_{long}=\sigma /E$, $\Delta L={\epsilon }_{long}{L}_0$; ${\epsilon }_{lat}=-\nu {\epsilon }_{long}$; $\Delta d={\epsilon }_{lat}{d}_0$.

Common pitfalls

• Units of stress. $1\text{MPa}=1{\text{N/mm}}^2=10^6\text{Pa}$. If you mix kN with mm², the stress comes out in kN/mm² = GPa — always convert kN to N first. “Force is ALWAYS in Newtons” (slide 5).
• Engineering vs. true stress. Engineering uses $A_0$ (initial); true uses instantaneous $A$. After necking, the engineering curve falls (slide 13) while the true curve keeps rising.
• Yield vs. UTS vs. fracture stress. Yield is the elastic–plastic transition; UTS is the maximum point on the engineering curve; fracture stress is at breakage and (on the engineering curve) is usually lower than UTS due to necking.
• 0.2% offset method. Use 0.2% strain on the x-axis (i.e. $\epsilon =0.002$ or 0.2% on a percentage axis). Draw a line parallel to the linear elastic slope from that offset; read the intersection.
• Sign of Poisson’s ratio. The definition includes a minus sign, so $\nu >0$ for normal materials in tension (where ${\epsilon }_{long}>0$, ${\epsilon }_{lat}<0$). $\nu$ stays between 0 and 0.5.
• Strain percentages vs. decimals. $0.5\%$ strain = 0.005, not 0.5. Convert before multiplying by $L_0$.
• Trig assignment in joint problems. Identify whether the given angle is measured from horizontal or vertical, and whether the 3-4-5 triangle’s “3” or “4” is the vertical leg, before writing equilibrium equations.
• Rounding people / parts down. Allowable load gives a fractional count of people; always round down so the stress stays within the allowable limit.
• GPa vs. MPa for $E$. Most engineering materials sit in the GPa range — if $E$ comes out at “207 MPa” for steel, you’ve made a unit error by a factor of 1000.

Source citations

• Lecture 11 slides — Lecture 1: Normal stress and strain + Mechanical Properties Normal, EGD102 CTP1 2026 (EGD102-Physics/Lecture11_CTP1.pdf):
  • Structural actions (slide 3); axial loading (4); axial stress with unit conversions (5); Example 1 (6); Example 2 (7); axial strain (8); Example 3 (9); elastic deformation and Hooke’s Law (11); Young’s modulus (12); engineering vs. true stress/strain (13); stress-strain curve regions (14); ductility, %EL, %AR (15); 0.2% offset yield method (16); Example 4 stress-strain diagram task (17); factor of safety (18); rearranged $SF$ form and rule-of-thumb table (19); Example 5 cable car (20); Poisson’s ratio definition (21); Poisson’s ratio table (22); Example 5 continued — reduction in area (23); learning activities (25); textbook acknowledgement: Wolfson, R. 2020, Essential University Physics, Vol. 1, 4th ed. (26).
• Tutorial 11 (EGD102-Physics/Tutorial 11.pdf):
  • Problem-solving approach Model–Visualise–Solve–Assess (slide 2); axial stress and strain recap (3–4); elastic deformation and Young’s modulus recap (5–6); Exercise 1 (7); Exercise 2 (8); stress-strain curve and offset yield recap (9–11); Exercise 3 magnesium alloy (12); Exercise 4 two-wire platform (13); factor of safety recap (14); Exercise 5 flowerpot (15); Poisson’s ratio recap (16–17); Exercise 6 aluminium rod (18); learning activities (19).