Applications of elasticity - class-XI

Modulus of elasticity = $\dfrac { stress }{ strain } $     and, stress = $\dfrac { Force }{ Area } $

In 1676, Robert Hooke presented his law of elasticity, now called Hooke's law.

Maximum load supported by the cable is directly proportional to the breaking stress.

Since,

Breaking stress = F / A

where F is the force and A is the cross-sectional area.

As we see that the breaking stress is independent of the length of the cable. So, if the cable is cut in two equal parts, the maximum load that can be supported by either parts of the cable remain the same as before.

Young's modulus measures modulus of elasticity of a material. 

Young's modulus is an intrinsic property of the material itself and does not depend on the dimensions (radius or length) of the wire.

A small differential element $dx$ at distance $x$ from the bottom of chord

The wire breaks under its own weight when the stress at the top equals the breaking stress. Stress = (m * g) / A = (rho * V * g) / A = rho * L * g. Therefore, L = breaking_stress / (rho * g) = 10^6 / (3e3 * 10) = 10^6 / 3e4 = 33.33 m.

Length of cantilever, $L= 50cm=0.5m$
Deflection at the end $w(L)= 15mm$
Find: Deflection $w(x)$ at 30cm from rigid end.
Depression at a point in a cantilever beam with load at one end is given by,$ w(x)=\dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI } $
We have,
$\dfrac { w(x) }{ w(L) } =\dfrac { \dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI }  }{ \dfrac { PL^{ 3 } }{ 3EI }  } $

$\dfrac { w(0.3m) }{ w(0.5m) } =\dfrac { \dfrac { P{ (0.3) }^{ 2 }(3(0.5)-0.3) }{ 6EI }  }{ \dfrac { P(0.5)^{ 3 } }{ 3EI }  } $

Depression in Solid cylinder $\delta _{1}= 8mm$
Its radius $r _{1}= 3mm$
Outer radius of Hollow cylinder $R _{2}= 4mm$
Inner radius of Hollow cylinder $r _{2}= 2mm$
Let depression in this cylinder be $\delta _{2}$.
Depression $\delta =\dfrac { W{ l }^{ 3 } }{ 12\pi{r }^{ 4 }Y } $
From the above equation we know that $\delta$ is proportional to $\dfrac{1}{r^{4}}$
Hence, we have
$\dfrac{\delta _{1}}{\delta _{2}}= \dfrac{{R _{2}}^{4}-{r _{2}}^{4}}{r _{1}^{4}}$
Substituting the values in above equation, we get
$\delta _{2}=2.7mm$

Bending moment $C=\dfrac{Y{I} _{G}}{R}$
Therefore, $C$ is proportional to $\dfrac{Y}{R}$

$\displaystyle \frac{\Delta l}{L}=\frac{F}{AY}:or:F=\frac{AY\Delta l}{L}$ (using standard equation)

Bending Moment $C=\dfrac { Y{ I } _{ G } }{ R } $
where $Y= 2\times { 10 }^{ 11 }N/{m}^{2}$
$R= 10cm= 0.1m$
${ I } _{ G }=\dfrac{\pi{R}^{4}}{4} = \dfrac{\pi{0.1}^{4}}{4} =7.85\times{10}^{-5} m^{4}$
Put these in the expression for $C$,
$C=1.57N/m$

The change in length is calculated by integrating the strain over the length of the wire, where the radius r(x) varies linearly from r1 at the top to r2 at the bottom. Using the formula delta L = integral of (F dx) / (pi * r(x)^2 * Y), the result for the given values yields 10^-3 m.

The relationship between Young's modulus (Y), Bulk modulus (B), and Shear modulus (G) is given by the standard elastic constant formula 9/Y = 1/B + 3/G. This is a fundamental identity in the theory of elasticity.

Young's modulus is defined as stress divided by strain. For a given force and cross-section, the wire that stretches the least (A) has the highest Young's modulus, as Y = (F * L) / (A * delta L).

The depression (deflection) of a beam supported at its ends is inversely proportional to its Young's modulus (Y). A stiffer material (higher Y) results in less depression for a given load.

$\displaystyle \delta=\displaystyle\frac{Wl^3}{3YI}$, where $W=$load$,\ l=$length of beam$,\ I=$moment of inertia$=\dfrac{b{d}^{3}}{12}$ for rectangular beam, and for square beam$,\ b=d.$ Thus, ${I} _{1}=\dfrac{{b}^{4}}{12}$

Now, for circular cross section, $\displaystyle I _2=\left[\dfrac{\pi r^4}{4}\right]$

$\therefore \delta _1=\dfrac{Wl^3\times 12}{3Yb^4}=\dfrac{4Wl^3}{Yb^4}$

and $\delta _2=\dfrac{Wl^3}{3Y(\pi r^$/4)}=\displaystyle\frac{4Wl^3}{3Y(\pi r^4)}$

Thus, $\dfrac{\delta_1}{\delta_2}=\dfrac{3\pi r^4}{b^4}=\dfrac{3\pi r^4}{(\pi r^2)^2}=\dfrac{3}{\pi}$
$(\because b^2=\pi r^2$ as they have same cross sectional area)

Young's modulus: $ Y=\dfrac { \dfrac { F }{ a }  }{ \dfrac { \Delta l }{ l }  } =\dfrac { Fl }{ a(\Delta l) } $

Critical buckling stress of a column formula is given by 

When a wire is stretched to 3 times its length, its volume remains constant, so the cross-sectional area becomes 1/3 of the original. Since R = rho * (L/A), the new resistance becomes rho * (3L) / (A/3) = 9 * (rho * L/A) = 9R.