This notebook refers to, simplifies and reproduces part of the lecture notes given at MIT Open Course Ware with permission from the author, Prof. David Roylance.
last updated: November 12, 2014

1. TENSILE RESPONSE OF MATERIALS

Part 1: Introduction To Elasticity

Tensile stress is a stress experienced by a material when a force is applied to the material.

• Tensile Stress:$$\sigma = \frac{P}{A_{0}}$$
  $\sigma$ = tensile stress, unit: $Nm^{-2}$
  $P$ = applied load, unit: $N$
  $A_{0}$ = original cross sectional area, unit: $m^2$

The ultimate tensile stress (UTS) is the maximum stress applicable to a material before the material broke/fractured.

• Ultimate Tensile Stress (UTS):$$\sigma_{f} = \frac{P_{f}}{A_{0}}$$
  $\sigma_{f}$ = ultimate tensile stress (UTS), unit: $Nm^{-2}$
  $P_{f}$ = load at fracture, unit: $N$
  $A_{0}$ = original cross sectional area, unit: $m^2$

Tensile stress caused by the material own weight and as a function of its length, $y$.

• Weight density:$$\gamma = \rho{g}$$
  $\gamma$ = weight density, unit: $Nm^{-3}$
  $\rho$ = material density, unit: $kgm^{-3}$
  $g$ = gravity, value: $9.8{\space}ms^{-2}$

• Weight supported by the cross-section at $y$: $$W(y) = \gamma{V}$$
  $V$ = volume of material below $y$, unit: $m^{3}$

• Tensile stress as a function of $y$:
  $$\sigma(y) = \frac{W(y)}{A} = \gamma{y}$$

Stiffness is a measure of the pulling force that is needed to induce a given deformation in a material.

• Hooke's Law:
  $$P =k{\delta}$$
  $P$ = applied load, unit: $N$
  $k$ = proportionality constant named $stiffness$
  $\delta$ = deformation length, unit: $m$

• $k$ is also influenced by geometry of the material e.g. wire, spring etc. Therefore, additional definition was made so that $k$ only relate to material properties: $$\epsilon =\frac{\delta}{L_{0}}$$
  where $\epsilon$ is the dimensionless measure of stretching called strain which is deformation displacement per unit length.

• From the Hooke's Law: $$\frac{P}{A_{0}} =E\frac{\delta}{L_{0}}$$ therefore
  $$\sigma =E\epsilon$$$$ $$ wherein $E$ is know as Young's Modulus or the modulus of elasticity. Hereon, $k$ and $\delta$ can be represented as: $$ $$ $$k =\frac{AE}{L}$$ and $$\delta =\frac{PL}{AE} = \left(\frac{P}{A}\right)\left(\frac{L}{E}\right)=\sigma\left(\frac{L}{E}\right)$$$$ $$It is easy to see that tensile stress is independent of material properties but not the strain, $\epsilon$ (e.g. polymer elongated longer than steel at the same amount load).

Exercises:

1. Determine the stress and total deformation of an aluminum wire, 30 m long and 5 mm in diameter, subjected to an axial load of 250 N.

2. Two rods, one of nylon and one of steel, are rigidly connected as shown. Determine the stresses and axial deformations when an axial load of F = 1 kN is applied.

3. A steel cable 10 mm in diameter and 1 km long bears a load in addition to its own weight of W = 150 N. Find the total elongation of the cable.

4. Show that the effective stiffnesses of two springs connected in (a) series and (b) parallel is

(a) series: $\hspace{2cm}$ $\frac{1}{k_{eff}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}$
(b) parallel: $\hspace{1.85cm}$ $k_{eff} = k_{1} + k_{2}$

For springs connected in series, both springs experience the same load, $P$:

$P = k_{eff}\delta_{eff}$

wherein $\delta_{1}$ = $P/k_{1}$ and $\delta_{2}$ = $P/k_{2}$ and $\delta_{eff} = \delta_{1} +\delta_{2}$ so

$\delta_{eff} = \frac{P}{k_{1}} + \frac{P}{k_{2}}$

or

$\frac{P}{k_{eff}} = \frac{P}{k_{1}} + \frac{P}{k_{2}}$

and this gives (by dividing with $P$):

$\frac{1}{k_{eff}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}$

For springs connected in parallel, both springs experience different loads, $P_{1}$ and $P_{2}$ but the same deformation:

$P_{1} + P_{2} = P$
$\delta_{1} = \delta_{2} = \delta_{eff}$

and

$P_{1} = k_{1}\delta_{1}$
$P_{2} = k_{2}\delta_{2}$

so

$k_{1}\delta_{1} + k_{2}\delta_{2} = k_{eff}\delta_{eff}$

and $\delta_{eff}$ is equal to $\delta_{1}$ as well as $\delta_{2}$ so by dividng it with any $\delta$ will give:

$k_{eff} = k_{1} + k_{2}$

7. A tapered column of modulus E and mass density ρ varies linearly from a radius of $r_{1}$ to $r_{2}$ in a length L. Find the total deformation caused by an axial load P.

From the Hooke's Law:
$$\frac{P}{A_{0}} =E\frac{\delta}{L_{0}}$$ $$ $$ so
$$\delta =\frac{PL}{E}\frac{1}{A}$$
and $A = \pi{r^{2}}$ and $L = y$ at $r$ so $$ $$ $$d\delta =\frac{Pydy}{E}\left(\frac{1}{\pi{r^{2}}}\right)dr$$ $$ $$ if the $r$ varies from $r_2$ to $r_1$ (and $y$ varies from 0 to $L$) then $$ $$ $$\int_{0}^{L}\int_{r_{2}}^{r_{1}}d\delta = \frac{P}{\pi{E}}\int_{0}^{L}\int_{r_{2}}^{r_{1}}\frac{y}{r^{2}}dydr$$

Therefore: $$\delta = \frac{PL^{2}}{2\pi{E}}\left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right]$$ $$ $$ the area under the $f(x) = \frac{1}{r^{2}}$ is always positive (as shown figuratively in plot below).

8. A tapered column of modulus $E$ and mass density $ρ$ varies linearly from a radius of $r_{1}$ to $r_{2}$ in a length $L$, and is hanging from its broad end. Find the total deformation due to the weight of the bar.

From the Exercise 7:
$$\delta =\frac{PL}{E}\frac{1}{A}$$
and $P = \rho{g}Ay$ at any $r$ so $$ $$ $$d\delta =\frac{(\rho{g}Ay)dy}{E}\left(\frac{1}{A}\right) = \frac{(\rho{g})ydy}{E}$$ $$ $$ Thus $\delta$ is independent of $A$ and so is with $r$ $$ $$ $$\int_{0}^{L}d\delta = \int_{0}^{L}\frac{(\rho{g})ydy}{E}$$

Therefore: $$\delta = \frac{\rho{g}L^{2}}{2E}$$

9. A rod of circular cross section hangs under the influence of its own weight, and also has an axial load $P$ suspended from its free end. Determine the shape of the bar, i.e. the function $r(y)$ such that the axial stress is constant along the bar’s length.