Manufacturing Properties of Materials

Given data

$d_0 = 12\ \text{mm},\ L_0 = 50\ \text{mm},\ L_f = 64\ \text{mm},\ d_f = 8.4\ \text{mm}$

Initial area:

$$A_0 = \frac{\pi}{4}(12)^2 = 113.10\ \text{mm}^2$$

Area at fracture:

$$A_f = \frac{\pi}{4}(8.4)^2 = 55.42\ \text{mm}^2$$

(a) Yield strength

$$\sigma_y = \frac{P_y}{A_0} = \frac{28\,500}{113.10} = 252.0\ \text{MPa}$$

(b) Ultimate tensile strength

$$\sigma_{UTS} = \frac{P_{max}}{A_0} = \frac{42\,400}{113.10} = 374.9\ \text{MPa}$$

(c) Percentage elongation

$$\%EL = \frac{L_f - L_0}{L_0}\times 100 = \frac{64-50}{50}\times 100 = 28\%$$

(d) Percentage reduction in area

$$\%RA = \frac{A_0 - A_f}{A_0}\times 100 = \frac{113.10 - 55.42}{113.10}\times 100 = 51\%$$

(e) True fracture stress and strain

True stress uses the instantaneous (smallest) area:

$$\sigma_{f} = \frac{P_f}{A_f} = \frac{33\,000}{55.42} = 595.5\ \text{MPa}$$

True strain at fracture (volume constant, so $L/L_0 = A_0/A$):

$$\varepsilon_f = \ln\frac{A_0}{A_f} = \ln\frac{113.10}{55.42} = 0.713$$

The true fracture stress (595 MPa) is much higher than the engineering stress at fracture ($33\,000/113.1 = 291.8$ MPa) because of the large reduction of area at the neck.

Answer: $\sigma_y = 252$ MPa, UTS $= 374.9$ MPa, elongation $= 28\%$, reduction in area $= 51\%$, true fracture stress $= 595.5$ MPa, true fracture strain $= 0.713$.

The flow curve is the plastic part of the true stress-true strain curve: $\sigma = K\varepsilon^{n}$, where K is the strength coefficient (MPa) and n the strain-hardening exponent.

Finding n and K

Divide the two readings:

$$\frac{450}{300} = \left(\frac{0.20}{0.05}\right)^n \quad\Rightarrow\quad 1.5 = 4^n$$ $$n = \frac{\ln 1.5}{\ln 4} = \frac{0.4055}{1.3863} = 0.292$$ $$K = \frac{300}{(0.05)^{0.292}} = \frac{300}{0.4164} = 720.5\ \text{MPa}$$

Flow stress at $\varepsilon = 0.40$

$$\sigma = 720.5\,(0.40)^{0.292} = 720.5\times 0.7649 = 551.1\ \text{MPa}$$

Average flow stress from 0 to 0.40

$$\bar{Y}_f = \frac{K\varepsilon_{max}^{\,n}}{1+n} = \frac{551.1}{1.292} = 426.6\ \text{MPa}$$

This average value is used to compute forming forces and work in rolling, drawing and forging, because the metal strain hardens during the process.

Meaning of n

• n is the slope of the flow curve on a log-log plot. It lies between 0 (perfectly plastic) and 1 (perfectly elastic).
• A large n (about 0.2-0.5, e.g. annealed copper, stainless steel) means strong strain hardening; the metal stretches uniformly and is good for stretch forming and deep drawing.
• A small n (about 0.05-0.15, e.g. cold-worked steel) means little hardening and early necking.
• For a tensile test, the true strain at the onset of necking equals n.

Answer: $n = 0.292$, $K = 720.5$ MPa, flow stress at 0.40 strain $= 551$ MPa, average flow stress $= 427$ MPa.

Tribology is the science of interacting surfaces in relative motion: it covers friction, wear and lubrication. In manufacturing it controls forming loads, tool life, surface finish and energy use.

Friction

• Sliding (Coulomb) friction: $F = \mu N$, valid at light loads (cold working with good lubrication, $\mu = 0.05$-$0.2$).
• Sticking (shear) friction: at high pressure in hot working the real contact area equals the apparent area, so the friction stress is a fraction of the shear yield stress, $\tau = m k$ ($m$ = friction factor from 0 to 1). Used in forging and hot rolling.
• Friction raises forming force, causes non-uniform flow and heats the tool.

Wear

Functions of lubricants

1. Separate the surfaces and reduce friction and wear.
2. Carry away heat (coolants) and wash away chips.
3. Improve surface finish and prevent welding.
4. Protect the work and die from corrosion.

Types of lubricants

• Liquids: mineral oils, soluble (emulsion) oils, synthetic fluids, with extreme-pressure (EP) additives.
• Solids: graphite, molybdenum disulphide, soap, glass (for hot extrusion).
• Gases and mists: compressed air, aerosol of oil.
• Regimes: hydrodynamic (full film), boundary (thin molecular film) and mixed.

Viscosity and temperature

Viscosity $\eta$ (shear stress divided by shear rate, Pa.s) decreases as temperature rises, approximately exponentially. A lubricant for hot working must therefore keep an adequate film at the die temperature, which is why solid lubricants or glass are used above about 700 degree C. Viscosity rises with pressure, which helps film formation in rolling and drawing.