Consider a long steel bar under a tensile stress due to forces $F$ acting at the edges along the length of the bar (figure). Consider a plane making an angle $\theta$ with the length. What are the tensile and shearing stresses on this plane?
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?
Consider the adjacent diagram.
Let the cross-sectional area of the bar be $A$. Consider the equilibrium of the plane aa'. A force $F$ must be acting on this plane making an angle $\frac{\pi}{2}-\theta$ with the normal $O N$. Resolving $F$ into components, along the plane (FP) and normal to the plane.
$$\begin{aligned} & F_P=F \cos \theta \\ & F_N=F \sin \theta \end{aligned}$$
$$\begin{aligned} &\text { Let the area of the face aa' be } A^{\prime} \text {, then }\\ &\begin{aligned} \frac{A}{A^{\prime}} & =\sin \theta \\ \therefore \quad A^{\prime} & =\frac{A}{\sin \theta} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { The tensile stress } & =\frac{\text { Normal force }}{\text { Area }}=\frac{F \sin \theta}{A^{\prime}} \\ & =\frac{F \sin \theta}{A / \sin \theta}=\frac{F}{A} \sin ^2 \theta \\ \text { Shearing stress } & =\frac{\text { Parallel force }}{\text { Area }} \\ & =\frac{F \cos \theta}{A / \sin \theta}=\frac{F}{A} \sin \theta \cdot \cos \theta \\ & =\frac{F}{2 A}(2 \sin \theta \cdot \cos \theta)=\frac{F}{2 A} \sin 2 \theta \end{aligned}$$
$$\begin{aligned} &\text { (a) For tensile stress to be maximum, } \sin ^2 \theta=1\\ &\begin{array}{lrl} \Rightarrow & \sin \theta & =1 \\ \Rightarrow & \theta & =\frac{\pi}{2} \end{array} \end{aligned}$$
$$\begin{aligned} &\text { (b) For shearing stress to be maximum, }\\ &\begin{aligned} & \Rightarrow \quad \sin 2 \theta =1 \\ & \Rightarrow \quad 2 \theta =\frac{\pi}{2} \\ & \Rightarrow\quad \theta =\frac{\pi}{4} \end{aligned} \end{aligned}$$
(a) A steel wire of mass $\mu$ per unit length with a circular cross-section has a radius of 0.1 cm . The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming, the wire to be uniform and lateral strains $<<$ longitudinal strains, find the extension in the length of the wire. The density of steel is $7860 \mathrm{~kg} \mathrm{~m}^{-3}$. (Young's modulus $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$.
(b) If the yield strength of steel is $2.5 \times 10^8 \mathrm{Nm}^{-2}$, what is the maximum weight that can be hung at the lower end of the wire?
Consider the diagram when a small element of length $d x$ is considered at $x$ from the load $(x=0)$
(a) Let $T(x)$ and $T(x+d x)$ are tensions on the two cross-sections a distance $d x$ apart, then -$-T(x+d x)+T(x)=d m g=\mu d x g$ (where $\mu$ is the mass/length). $(\because d m=\mu d x)$
$$d T=\mu g d x \quad[\because d T=T(x+d x)-T(x)]$$
$$\Rightarrow \quad T(x)=\mu g x+C \quad \text { (on integrating) }$$
$$\begin{aligned} &\begin{aligned} \end{aligned}\\ &\text { At } x=0, T(0)=M g \quad \Rightarrow \quad C=M g\\ &\begin{array}{ll} \therefore & T(x)=\mu g x+M g \end{array} \end{aligned}$$
$$\begin{aligned} &\text { Let the length } d x \text { at } x \text { increase by } d r \text {, then }\\ &\begin{aligned} \text { Young's modulus } Y & =\frac{\text { Stress }}{\text { Strain }} \\ \frac{T(x) / A}{d r / d x} & =Y \end{aligned} \end{aligned}$$
$$\begin{array}{ll} \Rightarrow & \frac{d r}{d x}=\frac{1}{Y A} T(x) \\ \Rightarrow & r=\frac{1}{Y A} \int_0^L(\mu g x+M g) d x \end{array}$$
$$\begin{aligned} & =\frac{1}{Y A}\left[\frac{\mu g x^2}{2}+M g x\right]_0^L \\ & =\frac{1}{Y A}\left[\frac{m g L^2}{2}+M g L\right]\quad \text{(m is the mass of the wire)} \end{aligned}$$
$$\begin{aligned} & A=\pi \times\left(10^{-3}\right)^2 \mathrm{~m}^2 \\ & Y=200 \times 10^9 \mathrm{Nm}^{-2} \\ & m=\pi \times\left(10^{-3}\right)^2 \times 10 \times 7860 \mathrm{~kg} \end{aligned}$$
$$\begin{aligned} \therefore \quad r & =\frac{1}{2 \times 10^{11} \times \pi \times 10^{-6}} \quad\left[\frac{\pi \times 786 \times 10^{-3} \times 10 \times 10}{2}+25 \times 10 \times 10\right] \\ & =\left[196.5 \times 10^{-6}+3.98 \times 10^{-3}\right] \approx 4 \times 10^{-3} \mathrm{~m} \end{aligned}$$
$$\begin{aligned} &\text { (b) Clearly tension will be maximum at } x=L\\ &\therefore \quad T=\mu g L+M g=(m+M) g \quad[\because m=\mu L] \end{aligned}$$
$$ \text { The yield force }=\left(\text { Yield strength } Y \text { ) area }=250 \times 10^6 \times \pi \times\left(10^{-3}\right)^2=250 \times \pi \mathrm{N}\right.$$
At yield point $$\quad$$ T = Yield force
$$\begin{array}{lrl} \Rightarrow & (m+M) g & =250 \times \pi \\ \therefore & m & =\pi \times\left(10^{-3}\right)^2 \times 10 \times 7860<< M \\ \text { Hence, } & M g & \approx 250 \times \pi \\ & M & =\frac{250 \times \pi}{10}=25 \times \pi \approx 75 \mathrm{~kg} . \end{array}$$
A steel rod of length $2 l$, cross-sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young's modulus for steel, find the extension in the length of the rod. (assume the rod is uniform)
Consider an element of width $d r$ at $r$ as shown in the diagram.
Let $T(r)$ and $T(r+d r)$ be the tensions at $r$ and $r+d r$ respectively.
Net centrifugal force on the element $=\omega^2 r d m$ (where $\omega$ is angular velocity of the rod)
$=\omega^2 r \mu d r$ $(\because \mu=$ mass/length $)$
$$\begin{aligned} &\begin{array}{rrr} \Rightarrow & T(r)-T(r+d r) & =\mu \omega^2 r d r \\ \Rightarrow & -d T & =\mu \omega^2 r d r \end{array}\\ &\text { [ } \because \text { Tension and centrifugal forces are opposite] } \end{aligned}$$
$$\begin{array}{rlr} \therefore \quad -\int_\limits{T=0}^T d T & =\int_\limits{r=l}^{r=r} \mu \omega^2 r d r & {[\because T=0 \text { at } r=l]} \\ T(r) & =\frac{\mu \omega^2}{2}\left(l^2-r^2\right) & \end{array}$$
Let the increase in length of the element $d r$ be $\Delta r$
So, Young's modulus $Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T(r) / A}{\frac{\Delta r}{d r}}$
$$\begin{array}{ll} \therefore & \frac{\Delta r}{d r}=\frac{T(r)}{A}=\frac{\mu \omega^2}{2 Y A}\left(l^2-r^2\right) \\ \therefore & \Delta r=\frac{1}{Y A} \frac{\mu \omega^2}{2}\left(l^2-r^2\right) d r \end{array}$$
$$\begin{aligned} \therefore \Delta=\text { change in length in right part } & =\frac{1}{Y A} \frac{\mu \omega^2}{2} \int_0^l\left(l^2-r^2\right) d r \\ & =\left(\frac{1}{Y A}\right) \frac{\mu \omega^2}{2}\left[l^3-\frac{l^3}{3}\right]=\frac{1}{3 Y A} \mu \omega^2 L^2 \end{aligned}$$
$$\therefore \quad \text { Total change in length }=2 \Delta=\frac{2}{3 Y A} \mu \omega^2 l^2$$
An equilateral triangle $A B C$ is formed by two Cu rods $A B$ and $B C$ and one Al rod. It is heated in such a way that temperature of each rod increases by $\Delta T$. Find change in the angle $A B C$. [Coeffecient of linear expansion for Cu is $\alpha_1$, coefficient of linear expansion for Al is $\alpha_2$ ]
Consider the diagram shown
Let $$l_1=A B, l_2=A C, l_3=B C$$
$$\therefore \quad \cos \theta=\frac{l_3^2+l_1^2-l_2^2}{2 l_3 l_1} \quad \text { (assume } \angle A B C=\theta \text { ) }$$
$$\begin{aligned} & \Rightarrow \quad 2 l_3 l_1 \cos \theta=l_3^2+l_1^2-l_2^2 \\ & \text { Differentiating } 2\left(l_3 d l_1+l_1 d l_3\right) \cos \theta-2 l_1 l_3 \sin \theta d \theta \end{aligned}$$
$$=2 l_3 d l_1+2 l_1 d l_1-2 l_2 d l_2$$
$$\begin{aligned} \text{Now,}\quad & d l_1=l_1 \alpha_1 \Delta t \quad \text { (where, } \Delta t=\text { change in temperature) } \\ & d l_2=l_2 \alpha_1 \Delta t \Rightarrow d l_3=l_3 \alpha_2 \Delta t \end{aligned}$$
$$\begin{aligned} &\text { and } &l_1=l_2=l_3=l \end{aligned}$$
$$\left(l^2 \alpha_1 \Delta t+l^2 \alpha_1 \Delta t\right) \cos \theta+l^2 \sin \theta d \theta=l^2 \alpha_1 \Delta t+l^2 \alpha_1 \Delta t-l^2 \alpha_2 \Delta t$$
$$\sin \theta d \theta=2 \alpha_1 \Delta t(1-\cos \theta)-\alpha_2 \Delta t$$
Putting $$\theta=60^\circ\quad$$ (for equilateral triangle)
$$\begin{aligned} d \theta \times \sin 60^{\circ} & =2 \alpha_1 \Delta t\left(1-\cos 60^{\circ}\right)-\alpha_2 \Delta t \\ & =2 \alpha_1 \Delta t \times \frac{1}{2}-\alpha_2 \Delta t=\left(\alpha_1-\alpha_2\right) \Delta t \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad d \theta & =\text { change in the angle } \angle A B C \\ & =\frac{\left(\alpha_1-\alpha_2\right) \Delta T}{\sin 60^{\circ}}=\frac{2\left(\alpha_1-\alpha_2\right) \Delta T}{\sqrt{3}} \quad(\because \Delta t=\Delta T \text { given }) \end{aligned}$$
In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{Y \pi r^4}{4 R} . Y$ is the Young's modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
Consider the diagram according to the question, the bending torque on the trunk of radius $r$ of the tree $=\frac{Y \pi r^4}{4 R}$
where $R$ is the radius of curvature of the bent surface.
When the tree is about to buckle $W d=\frac{Y \pi r^4}{4 R}$
If $R \gg h$, then the centre of gravity is at a height $l \approx \frac{1}{2} h$ from the ground.
$$\begin{aligned} & \text { From } \quad \triangle A B C R^2 \approx(R-d)^2+\left(\frac{1}{2} h\right)^2 \\ & \text { if } d \ll R, \quad R^2 \approx R^2-2 R d+\frac{1}{4} h^2 \\ & \therefore \quad d=\frac{h^2}{8 R} \end{aligned}$$
If $\omega_0$ is the weight/volume
$$\frac{Y \pi r^4}{4 R}=\omega_0\left(\pi r^2 h\right) \frac{h^2}{8 R}\quad$$ $[\because$ Torque is caused by the weight]
$$ \Rightarrow \quad h \approx\left(\frac{2 Y}{\omega_0}\right)^{1 / 3} r^{2 / 3}$$
$$\text { Hence, critical height }=h=\left(\frac{2 Y}{\omega_0}\right)^{1 / 3} r^{2 / 3}$$