Stress measures how hard the material is being worked. Strain measures how much it has moved as a result. The two are separate ideas, and keeping them separate matters: stress is a statement about force, strain is a statement about geometry, and only a material model links them. This chapter builds the geometric half.
A bar of original length \(L_0\) is stretched to a new length \(L\), as in Figure 8.5.1. The change in length is the deformation \(\delta = L - L_0\), and dividing it by the original length gives the engineering strain, also called the linear strain,
\[ \boxed{\varepsilon := \frac{L - L_0}{L_0} = \frac{\delta}{L_0}} \tag{8.5.1}\]
which is dimensionless. Since the numbers involved are usually small, strain is often quoted in percent or in microstrain, where \(1000~\mu\varepsilon\) means \(\varepsilon = 0.001\). Steel yields at roughly \(0.2\%\) strain, so the deformations we work with are genuinely tiny, and that smallness is what makes the linear theory of this book usable.
Definition 8.5.1 describes the bar as a whole, and it is adequate only when the stretching is uniform. If the cross section or the load varies along the bar, the strain varies with it, and we need a local definition. Let \(u(x)\) denote the displacement of the material point that was originally at \(x\). A small segment between \(x\) and \(x + \Delta x\) has its two ends displaced by \(u(x)\) and \(u(x + \Delta x)\), so its elongation is the difference between the two, and its strain is that difference divided by its original length \(\Delta x\). Taking the segment to zero length gives
\[ \varepsilon := \frac{\text{elongation}}{\text{original length}} = \frac{u(x+\Delta x) - u(x)}{\Delta x} \;\longrightarrow\; \boxed{\varepsilon = \frac{du}{dx}} \tag{8.5.2}\]
so the strain is the gradient of the displacement field. 8.5.1 is the special case in which that gradient happens to be constant. This is the form we generalise in Chapter 8.7, where \(u\) becomes a vector field and the single derivative becomes a matrix of partial derivatives.
Engineering strains measured against different reference lengths cannot be added.
Suppose we stretch the bar in two stages, first from \(L_0\) to \(L_1\) and then from \(L_1\) to \(L_2\), as in Figure 8.5.3. If each stage is measured against the length the bar had when that stage started, the two strains are \(\delta_1/L_0\) and \(\delta_2/L_1\), and their sum is not the total strain \((L_2 - L_0)/L_0\) measured against the original length. The reference length changed underneath us. For the small deformations of linear elasticity the discrepancy is negligible, but in metal forming, in polymer testing and anywhere else that strains reach tens of percent it is not.
The repair is to measure every increment against the current length. A change \(\Delta L\) at the current length \(L\) contributes an increment of strain
\[ \Delta \varepsilon = \frac{\Delta L}{L} \]
and in the limit \(\Delta L \to 0\) this becomes \(d\varepsilon = dL/L\). Integrating from the original to the final length accumulates the increments,
\[ \int_0^{\varepsilon_L} d\varepsilon = \int_{L_0}^{L} \frac{1}{L}\, dL \quad\Rightarrow\quad \boxed{\varepsilon_L = \ln\!\left(\frac{L}{L_0}\right)} \tag{8.5.3}\]
which is the logarithmic strain, also called true strain or Hencky strain. It adds correctly, because \(\ln(L_2/L_0) = \ln(L_2/L_1) + \ln(L_1/L_0)\) by the functional equation of the logarithm.
The two definitions must agree when the deformation is small, and a Taylor expansion of 8.5.3 about \(\delta = 0\) shows how.
L_0, delta = sp.symbols('L_0 delta', real=True)
eps_log = sp.log((L_0 + delta)/L_0)
series = sp.series(eps_log, delta, 0, 4).removeO()
ltx(r"\varepsilon_L = \ln\left(\frac{L_0+\delta}{L_0}\right) =", series, r"+ \cdots")\[ \varepsilon_L = \ln\left(\frac{L_0+\delta}{L_0}\right) =\dfrac{\delta}{L_{0}} - \dfrac{\delta^{2}}{2 L_{0}^{2}} + \dfrac{\delta^{3}}{3 L_{0}^{3}}+ \cdots \]
Written in terms of the engineering strain \(\varepsilon = \delta/L_0\), this is
\[ \varepsilon_L = \varepsilon \underbrace{- \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - \cdots}_{O(\varepsilon^n),\; n \ge 2} \]
so the two measures differ only at second order. At \(\varepsilon = 0.002\), which is where steel yields, the difference is two parts in a thousand of the strain itself and no instrument in a teaching laboratory would resolve it. At \(\varepsilon = 0.5\) the difference is \(19\%\). The plot below shows where the linear approximation starts to fail.
L0 = 1
approximations = [sp.series(eps_log, delta, 0, n).removeO().subs(L_0, L0)
for n in (2, 3, 4)]
fig, ax = plt.subplots(figsize=(6.5, 4.5))
fplot(eps_log.subs(L_0, L0), (delta, -0.5, 1), ax=ax, color='r', linewidth=2,
label=r"$\varepsilon_L = \ln(L/L_0)$")
for expr, order in zip(approximations, (1, 2, 3)):
fplot(expr, (delta, -0.5, 1), ax=ax, linewidth=1,
label=f"Taylor, order {order}")
ax.axhline(0, color='k', linewidth=0.8)
ax.set_xlabel(r"$\delta$")
ax.set_ylabel(r"$\varepsilon$")
ax.grid(True)
ax.legend(loc='upper left')
plt.show()
The first-order approximation is the engineering strain itself, and it tracks the logarithmic strain closely over roughly \(\lvert\delta\rvert < 0.1\), which for \(L_0 = 1\) means strains up to about \(10\%\). The approximation is worse in compression than in tension, which is a consequence of the logarithm running to \(-\infty\) as the bar is squashed to zero length while the engineering strain merely reaches \(-1\).
A rod is compressed from \(140~\text{mm}\) to \(120~\text{mm}\). Determine both strain measures.
L_0_val, L_val = 140, 120
delta_val = L_val - L_0_val
eps_eng = sp.Rational(delta_val, L_0_val)
eps_log_val = sp.log(sp.Rational(L_val, L_0_val))
ltx(r"\delta &=", delta_val, r"~\text{mm}",
r"\\ \varepsilon &=", eps_eng.evalf(),
r"\\ \varepsilon_L &=", eps_log_val.evalf(), aligned=True, precision=4)\[ \begin{aligned}\delta &=-20~\text{mm}\\ \varepsilon &=-0.1429\\ \varepsilon_L &=-0.1542\end{aligned} \]
The deformation is \(-14\%\) of the original length, well outside the range where the two measures agree, and they differ by about \(8\%\) of their own value. Which one to report depends on what the number is for: a stress analysis in the elastic range uses engineering strain because that is what the elastic constants are defined against, while a forming simulation or a tensile test taken to fracture uses logarithmic strain because only that one accumulates correctly.
Materials deform when heated, without any load being applied. Over the temperature ranges met in ordinary engineering the deformation is proportional to the temperature change, with a proportionality constant \(\alpha\) called the coefficient of thermal expansion, measured in \(1/^\circ\text{C}\). The resulting deformation and length are
\[ \delta = L_0 \alpha \Delta T = L_0\, \varepsilon_{\text{temp}}, \qquad L = L_0 + L_0 \alpha \Delta T \tag{8.5.4}\]
Thermal strain and mechanical strain add, since both are geometric statements about the same bar,
\[ \varepsilon_{\text{tot}} = \varepsilon_{\text{elast}} + \varepsilon_{\text{temp}} = \frac{\sigma}{E} + \alpha \Delta T \tag{8.5.5}\]
where the elastic part has been written using Hooke’s law, which we take up in Chapter 8.6.
8.5.5 is the source of a class of problems that catches students out. If a bar is free to expand, heating it produces strain but no stress, because \(\sigma = 0\) and all of \(\varepsilon_{\text{tot}}\) comes from the thermal term. If the bar is clamped at both ends so that \(\varepsilon_{\text{tot}} = 0\), heating it produces no strain but a compressive stress \(\sigma = -E\alpha\Delta T\), which for steel amounts to roughly \(2.5~\text{MPa}\) per degree. This is why bridges have expansion joints and why pipework has bends in it.
Normal strain measures the stretching of a line. Shear strain measures the change in angle between two lines that were originally perpendicular. Take the square element in Figure 8.5.8, hold its lower edge fixed and apply a shear stress along its upper edge. The square becomes a parallelogram, and the angle by which the vertical edges have tilted is the shear angle \(\gamma\).
The area is preserved by the deformation, which requires \(a^2 = a\,h\cos\gamma\) where \(h = \lvert AB'\rvert\) is the length of the deformed edge. For small angles \(\cos\gamma \approx 1\), so \(h \to a\): the edges neither stretch nor shorten, they only rotate. The upper edge has then translated by \(\delta = \lvert BB'\rvert\) over a height \(a\), and the shear strain is defined as that ratio,
\[ \boxed{\gamma \approx \tan\gamma = \frac{\lvert BB'\rvert}{a} = \frac{\delta}{a}} \tag{8.5.6}\]
which is the exact analogue of \(\varepsilon = \delta/L_0\) with the height playing the role of the reference length. Shear strain is dimensionless and, like normal strain, is small in every application this book treats.
The material relation for shear has the same form as Hooke’s law,
\[ \boxed{\tau = G\gamma} \tag{8.5.7}\]
where \(G\) is the shear modulus. It is not an independent material property. For an isotropic material it is fixed by the Young’s modulus and the Poisson’s ratio through
\[ \boxed{G = \frac{E}{2(1+\nu)}} \tag{8.5.8}\]
which we derive in Chapter 8.8. For steel, with \(E = 210\,000~\text{MPa}\) and \(\nu = 0.3\), this gives \(G \approx 81\,000~\text{MPa}\), so a material resists shearing roughly two and a half times less stiffly than it resists stretching. That single number explains why thin-walled sections are so much better in bending than in torsion, and it will return in Chapter 8.14 when we compare the Euler-Bernoulli and Timoshenko beam models.