Mechanical Properties and their tests

Tensile test: a standard specimen (gauge length $L_0$, area $A_0$) is gripped in a universal testing machine and pulled axially at a slow, constant rate until fracture. Load $P$ and extension $\Delta L$ are recorded. Then

$$\sigma = \frac{P}{A_0}, \qquad e = \frac{\Delta L}{L_0}$$

give the engineering stress and engineering strain. The test gives modulus, yield strength, tensile strength and ductility.

Stress-strain diagram of mild steel

 stress
   |            D (UTS)
   |          ,-'-.
   |    B C ,'     `.
   |   /`--'         `. E
   |  /                `x  (fracture)
   | / A
   |/
   +------------------------> strain
   O

Salient points

• O-A: proportional limit. Stress is proportional to strain (Hooke's law), $\sigma = E e$. The slope is Young's modulus (about 200 GPa for steel).
• A-B: elastic limit. The specimen returns to its original length on unloading.
• B: upper yield point; C: lower yield point. Dislocations break away from interstitial (carbon, nitrogen) atmospheres, so stress drops and the specimen extends at nearly constant load (Luders band). Permanent deformation starts here.
• C-D: strain hardening. Strain increases with stress as dislocation density rises, with uniform elongation along the gauge length.
• D: ultimate tensile strength (UTS). The maximum engineering stress, $P_{max}/A_0$. A neck starts to form here.
• D-E: necking. Deformation concentrates in the neck, area falls rapidly, so engineering stress falls although true stress still rises.
• E: fracture (cup-and-cone type for ductile steel).

Properties obtained

Yield strength, UTS, percentage elongation $=\dfrac{L_f - L_0}{L_0}\times100$, percentage reduction in area $=\dfrac{A_0 - A_f}{A_0}\times 100$, resilience and toughness (areas under the curve).

A ductile material undergoes large plastic deformation before fracture; a brittle material fractures with little or no plastic deformation.

 stress
   |    ductile           brittle
   |     ,--.              /
   |   ,'    `.           /
   |  /        x         /
   | /                  x
   |/                  /
   +------------------------> strain

Offset yield strength

Aluminium, copper and high-strength steels show no sharp yield point. The yield strength is then taken by the 0.2% offset method:

1. Mark a strain of 0.002 on the strain axis.
2. Draw a line from this point parallel to the straight elastic portion.
3. The stress at which this line cuts the curve is the 0.2% proof stress (yield strength).

True stress $\sigma_T$ is the load divided by the instantaneous cross-sectional area $A_i$; true strain $\varepsilon_T$ is the sum of incremental strains based on the instantaneous length.

$$\sigma_T = \frac{P}{A_i}, \qquad \varepsilon_T = \int_{L_0}^{L_i}\frac{dL}{L} = \ln\frac{L_i}{L_0}$$

Derivation

Strain is $e = \dfrac{L_i - L_0}{L_0}$, so $\dfrac{L_i}{L_0} = 1 + e$. Therefore

$$\varepsilon_T = \ln(1 + e)$$

Plastic deformation conserves volume: $A_0 L_0 = A_i L_i$, so $A_i = \dfrac{A_0}{1+e}$. Then

$$\sigma_T = \frac{P}{A_i} = \frac{P}{A_0}(1+e) = \sigma\,(1 + e)$$

Validity

Both equations hold only up to the start of necking (uniform deformation), because after necking the strain is no longer uniform and volume constancy cannot be applied along the gauge length. Beyond necking, $\sigma_T$ is found from measured neck area and $\varepsilon_T = \ln(A_0/A_i)$.

Flow curve

In the plastic region $\sigma_T = K\varepsilon_T^{\,n}$, where $K$ is the strength coefficient and $n$ the strain-hardening exponent (0.1-0.5 for metals). Necking begins at $\varepsilon_T = n$.

Why true stress keeps rising

After the UTS the engineering stress falls because the load falls while $A_0$ is used. But the neck area shrinks even faster, so load per actual area ($\sigma_T$) keeps increasing due to strain hardening until fracture.

Given

$d_0 = 12.5$ mm, $L_0 = 50$ mm, $d_f = 8.5$ mm, $L_f = 62.5$ mm.

Original area:

$$A_0 = \frac{\pi}{4}(12.5)^2 = 122.72\ \text{mm}^2$$

Final (neck) area:

$$A_f = \frac{\pi}{4}(8.5)^2 = 56.75\ \text{mm}^2$$

(a) Young's modulus

$$\sigma = \frac{18000}{122.72} = 146.7\ \text{MPa}, \qquad e = \frac{0.0367}{50} = 7.34\times10^{-4}$$ $$E = \frac{\sigma}{e} = \frac{146.7}{7.34\times10^{-4}} = 1.998\times10^{5}\ \text{MPa} \approx 200\ \text{GPa}$$

(b) Yield strength

$$\sigma_y = \frac{36000}{122.72} = 293.4\ \text{MPa}$$

(c) Ultimate tensile strength

$$\sigma_{u} = \frac{52000}{122.72} = 423.7\ \text{MPa}$$

(d) Fracture stress

$$\sigma_f = \frac{41000}{122.72} = 334.1\ \text{MPa (engineering)}$$ $$\sigma_{Tf} = \frac{41000}{56.75} = 722.5\ \text{MPa (true)}$$

(e) Ductility

$$\%\,EL = \frac{62.5 - 50}{50}\times100 = 25\%$$ $$\%\,RA = \frac{122.72 - 56.75}{122.72}\times 100 = 53.8\%$$

(f) Modulus of resilience

$$U_r = \frac{\sigma_y^2}{2E} = \frac{(293.4)^2}{2 \times 1.998\times10^{5}} = 0.215\ \text{MJ/m}^3$$

(1 MPa = 1 MJ/m$^3$ per unit strain, so the result is in MJ/m$^3$.)

Answer: E = 200 GPa, σy = 293.4 MPa, σu = 423.7 MPa, σf = 334.1 MPa (true 722.5 MPa), elongation 25%, reduction in area 53.8%, Ur = 0.215 MJ/m³.

(a) n and K

Taking logs of the power law $\sigma_T = K\varepsilon_T^{\,n}$ for the two points:

$$n = \frac{\ln(\sigma_2/\sigma_1)}{\ln(\varepsilon_2/\varepsilon_1)} = \frac{\ln(520/450)}{\ln(0.12/0.05)} = \frac{0.1449}{0.8755} = 0.165$$ $$K = \frac{\sigma_1}{\varepsilon_1^{\,n}} = \frac{450}{(0.05)^{0.165}} = 738\ \text{MPa}$$

(b) Necking

Necking begins when the strain-hardening rate equals the stress, $d\sigma_T/d\varepsilon_T = \sigma_T$, which gives $\varepsilon_T = n$.

$$\varepsilon_{T,neck} = n = 0.165$$ $$\sigma_{T,neck} = K n^{n} = 738 \times (0.165)^{0.165} = 548\ \text{MPa}$$

Engineering UTS: $\sigma_T = \sigma(1+e)$ with $1 + e = e^{\varepsilon_T} = e^{0.165} = 1.180$, so

$$\sigma_{UTS} = \frac{548}{1.180} = 465\ \text{MPa}$$

(c) Engineering to true

$$\sigma_T = \sigma(1+e) = 400 \times 1.10 = 440\ \text{MPa}$$ $$\varepsilon_T = \ln(1+e) = \ln 1.10 = 0.0953$$

This is valid because $e = 0.10$ is less than the engineering strain at necking ($1.180 - 1 = 0.18$).

Answer: n = 0.165, K = 738 MPa; necking at ε_T = 0.165, σ_T = 548 MPa, engineering UTS = 465 MPa; (c) σ_T = 440 MPa, ε_T = 0.0953.

Properties from the tensile test

• Young's modulus $E$: slope of the elastic line, a measure of stiffness.
• Yield strength $\sigma_y$: stress at which plastic deformation begins (0.2% offset if no clear yield).
• Ultimate tensile strength: maximum engineering stress, $P_{max}/A_0$.
• Ductility: percentage elongation and percentage reduction in area.
• Resilience: energy absorbed elastically, $U_r = \sigma_y^2/2E$ (area under the elastic part).
• Toughness: total energy absorbed per unit volume up to fracture (total area under the curve).
• Poisson's ratio: ratio of lateral to axial strain (about 0.3 for steel).

Effect of temperature

As temperature rises:

1. Yield strength, UTS and modulus decrease, because thermal energy helps dislocations to move.
2. Ductility (elongation, reduction in area) increases.
3. Toughness generally increases, and the yield point of mild steel becomes less sharp.

 strength / ductility
  |\  strength
  | \_______
  |         \___
  |  ductility  /
  |      _____/
  +-------------------> temperature

At low temperature BCC metals (carbon steel) show a ductile-to-brittle transition: yield strength rises, but fracture becomes brittle. FCC metals (Cu, Al, austenitic steel) stay ductile. At high temperature, time-dependent deformation (creep) also appears under constant load.