Stress and strain – axial load

The stress-strain diagram plots nominal stress $\sigma = P/A_0$ against strain $\varepsilon = \delta/L_0$ for a specimen pulled in a universal testing machine.

 stress
   |              U
   |        Y1   _.-"-._
   |         /\_/        '.
   |        /Y2             \ F
   |       /
   |      / P,E
   |     /
   |    /
   +------------------------- strain

(P = proportional limit, E = elastic limit, Y1/Y2 = upper/lower yield point, U = ultimate point, F = fracture.)

Important points

1. Proportional limit (P): the stress is proportional to strain up to here, so Hooke's law $\sigma = E\varepsilon$ holds. The slope of the straight line is the modulus of elasticity $E$ (about 200 GPa for steel).
2. Elastic limit (E): the specimen returns to its original length on unloading. It lies very close to P.
3. Yield point (Y1, Y2): the strain increases suddenly with almost no increase in stress (plastic flow). There is an upper yield point and a lower yield point; design uses the yield stress $\sigma_y$ (about 250 MPa for mild steel).
4. Strain hardening: after yielding the material regains strength and the stress rises again to the ultimate point.
5. Ultimate stress (U): the maximum nominal stress. Beyond U a local reduction in area (necking) starts.
6. Fracture (F): the specimen breaks. The nominal stress falls because it is calculated on the original area, although the true stress keeps rising.

Mild steel is ductile: the elongation at fracture is 20 to 30 percent and it shows a cup and cone fracture.

Cast iron (brittle material)

For materials with no clear yield point (aluminium, high-strength steel) the 0.2 % offset yield stress is used.

Free-body and axial forces

 |A-----30mm-----B----25mm----C---20mm---D
 |<--- 1.0 m --->|<-- 0.8 m ->|<- 0.6 m ->
 R_A <---        60kN->   <-40kN     20kN->

Take forces to the right as positive. Cut each segment and balance the loads on the free side (right-hand side):

(a) Stresses

$A_{AB} = \frac{\pi}{4}(30)^2 = 706.9\ \text{mm}^2$, $A_{BC} = 490.9\ \text{mm}^2$, $A_{CD} = 314.2\ \text{mm}^2$.

(b) Total change in length

$$\delta = \sum \frac{F L}{A E}$$ $$\begin{aligned} \delta_{AB} &= \frac{40000\times1000}{706.9\times200000} = +0.283\ \text{mm}\\ \delta_{BC} &= \frac{-20000\times800}{490.9\times200000} = -0.163\ \text{mm}\\ \delta_{CD} &= \frac{20000\times600}{314.2\times200000} = +0.191\ \text{mm} \end{aligned}$$ $$\delta = 0.283 - 0.163 + 0.191 = +0.311\ \text{mm (elongation)}$$

(c) Reaction at A

Equilibrium of the whole bar: $-R_A + 60 - 40 + 20 = 0$, so $R_A = 40$ kN acting to the left (equal to the force in AB).

Answer: forces $+40$, $-20$, $+20$ kN; stresses $56.6$, $-40.7$, $63.7$ MPa; total elongation $= 0.311$ mm; $R_A = 40$ kN.

When a bar is prevented from expanding freely, a compressive thermal stress develops.

Data

$L = 2500\ \text{mm}$, $A = \frac{\pi}{4}(25)^2 = 490.9\ \text{mm}^2$, $\Delta T = 70 - 20 = 50\,^\circ\text{C}$.

(a) Rigid walls

Free thermal expansion:

$$\delta_T = \alpha L \Delta T = 12\times10^{-6}\times 2500\times 50 = 1.5\ \text{mm}$$

The walls allow no expansion, so the load must shorten the bar by 1.5 mm. Thermal strain $\alpha\Delta T$ is cancelled by elastic strain:

$$\sigma = E\alpha\Delta T = 200000\times 12\times10^{-6}\times 50 = 120\ \text{MPa (compressive)}$$ $$P = \sigma A = 120\times 490.9 = 58.9\times10^3\ \text{N}$$

(The stress does not depend on the length.)

(b) Walls yield by 0.6 mm

Only the difference between free expansion and the wall movement must be suppressed:

$$\delta_{\text{restrained}} = 1.5 - 0.6 = 0.9\ \text{mm}$$ $$\sigma = E\,\frac{\delta_{\text{restrained}}}{L} = 200000\times\frac{0.9}{2500} = 72\ \text{MPa}$$ $$P = 72\times 490.9 = 35.3\ \text{kN}$$

Answer: (a) $\sigma = 120$ MPa (compression), $P = 58.9$ kN; (b) $\sigma = 72$ MPa, $P = 35.3$ kN.

Section areas

$A_s = \frac{\pi}{4}(30)^2 = 706.9\ \text{mm}^2$, $A_c = \frac{\pi}{4}(50^2 - 30^2) = 1256.6\ \text{mm}^2$.

(a) Load sharing

Equilibrium: $P_s + P_c = 150\times10^3$ N. Compatibility (same shortening, same length): $\varepsilon_s = \varepsilon_c$, so

$$\frac{\sigma_s}{E_s} = \frac{\sigma_c}{E_c} \;\Rightarrow\; \sigma_c = \sigma_s\frac{E_c}{E_s} = 0.5\,\sigma_s$$ $$\sigma_s A_s + \sigma_c A_c = 150000 \;\Rightarrow\; \sigma_s\,(706.9 + 0.5\times1256.6) = 150000$$ $$\sigma_s = 112.3\ \text{MPa}, \qquad \sigma_c = 56.2\ \text{MPa}$$ $$P_s = 112.3\times706.9 = 79.4\ \text{kN}, \qquad P_c = 56.2\times1256.6 = 70.6\ \text{kN}$$

(check: $79.4 + 70.6 = 150$ kN)

$$\delta = \frac{\sigma_s L}{E_s} = \frac{112.3\times600}{200000} = 0.337\ \text{mm (shortening)}$$

(b) Temperature rise of 40 degrees C

Copper tends to expand more than steel, but they are held together. Copper is held back (extra compression) and steel is pulled along (extra tension). The internal forces are equal and opposite, $P_t$:

$$\alpha_c\Delta T L - \frac{P_t L}{A_cE_c} = \alpha_s\Delta T L + \frac{P_t L}{A_sE_s}$$ $$P_t = \frac{(\alpha_c - \alpha_s)\Delta T}{\dfrac{1}{A_cE_c} + \dfrac{1}{A_sE_s}} = \frac{5\times10^{-6}\times 40}{\dfrac{1}{1256.6\times10^5} + \dfrac{1}{706.9\times2\times10^5}} = 13.31\ \text{kN}$$

Thermal stresses: steel $= +13310/706.9 = +18.8$ MPa (tension), copper $= -13310/1256.6 = -10.6$ MPa (compression).

Final stresses (add to the load stresses, compression negative):

Check: $-93.5\times706.9 - 66.8\times1256.6 = -150$ kN.

Answer: (a) $\sigma_s = 112.3$ MPa, $\sigma_c = 56.2$ MPa (both compressive), $P_s = 79.4$ kN, $P_c = 70.6$ kN, shortening $0.337$ mm; (b) final $\sigma_s = -93.5$ MPa, $\sigma_c = -66.8$ MPa.

(a) Definitions and derivation

• Poisson's ratio $\nu$ is the ratio of lateral strain to axial strain in a uniaxial stress state: $\nu = -\varepsilon_{lat}/\varepsilon_{axial}$ (0.25 to 0.33 for metals).
• Bulk modulus $K$ is the ratio of hydrostatic (uniform) pressure to the volumetric strain it produces: $K = p/(-\varepsilon_v)$.

Generalized Hooke's law for a cube under three stresses:

$$\varepsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right]$$

and similarly for $\varepsilon_y,\varepsilon_z$. Under hydrostatic pressure $p$, $\sigma_x=\sigma_y=\sigma_z=-p$:

$$\varepsilon_x=\varepsilon_y=\varepsilon_z = \frac{-p(1-2\nu)}{E}$$ $$\varepsilon_v = \varepsilon_x+\varepsilon_y+\varepsilon_z = -\frac{3p(1-2\nu)}{E} \;\Rightarrow\; K = \frac{p}{-\varepsilon_v} = \frac{E}{3(1-2\nu)}$$ $$E = 3K(1-2\nu)$$

(b) Numerical

Strains:

$$\begin{aligned} \varepsilon_x &= \frac{100 - 0.3(-60+40)}{200000} = \frac{106}{200000} = 5.3\times10^{-4}\\ \varepsilon_y &= \frac{-60 - 0.3(100+40)}{200000} = \frac{-102}{200000} = -5.1\times10^{-4}\\ \varepsilon_z &= \frac{40 - 0.3(100-60)}{200000} = \frac{28}{200000} = 1.4\times10^{-4} \end{aligned}$$

Volumetric strain:

$$\varepsilon_v = (5.3 - 5.1 + 1.4)\times10^{-4} = 1.6\times10^{-4}$$

Check: $\varepsilon_v = \frac{1-2\nu}{E}(\sigma_x+\sigma_y+\sigma_z) = \frac{0.4\times80}{200000} = 1.6\times10^{-4}$.

$$V = 100\times80\times50 = 4\times10^5\ \text{mm}^3, \qquad \Delta V = \varepsilon_v V = 64\ \text{mm}^3\ \text{(increase)}$$

Elastic constants:

$$K = \frac{E}{3(1-2\nu)} = \frac{200}{3\times0.4} = 166.7\ \text{GPa}, \qquad G = \frac{E}{2(1+\nu)} = \frac{200}{2.6} = 76.9\ \text{GPa}$$

Answer: $\varepsilon_x = 5.3\times10^{-4}$, $\varepsilon_y = -5.1\times10^{-4}$, $\varepsilon_z = 1.4\times10^{-4}$; $\Delta V = +64\ \text{mm}^3$; $K = 166.7$ GPa; $G = 76.9$ GPa.

Tapered circular bar

Let $x$ be measured from the end of diameter $d_1$. The diameter at $x$ varies linearly:

$$d_x = d_1 + \frac{d_2 - d_1}{L}x = d_1 + kx, \qquad k = \frac{d_2-d_1}{L}$$

The axial force is the same ($P$) at every section, but the area changes. Elongation of a small length $dx$:

$$d\delta = \frac{P\,dx}{A_xE} = \frac{4P\,dx}{\pi E\,(d_1+kx)^2}$$ $$\delta = \frac{4P}{\pi E}\int_0^L \frac{dx}{(d_1+kx)^2} = \frac{4P}{\pi E}\left[-\frac{1}{k(d_1+kx)}\right]_0^L = \frac{4P}{\pi E k}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)$$

Substituting $k = (d_2-d_1)/L$:

$$\boxed{\delta = \frac{4PL}{\pi E\,d_1d_2}}$$

Check: when $d_1 = d_2 = d$ this gives $\frac{4PL}{\pi E d^2} = \frac{PL}{AE}$, the prismatic bar result.

Bar hanging under its own weight

Axial force at a section $x$ below the support carries the weight of the bar below it. Measure $y$ from the free end: $F_y = w A y$.

$$d\delta = \frac{F_y\,dy}{AE} = \frac{wAy\,dy}{AE} \;\Rightarrow\; \delta = \frac{w}{E}\int_0^L y\,dy = \frac{wL^2}{2E}$$

Since total weight $W = wAL$, $\delta = \dfrac{WL}{2AE}$: the elongation is the same as if half the weight acted at the free end.