Q.A brass wire $1.8\ \text{m}$ long at $27\ ^\circ\text{C}$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39\ ^\circ\text{C}$, what is the tension developed in the wire, if its diameter is $2.0\ \text{mm}$? Coefficient of linear expansion of brass $= 2.0 \times 10^{-5}\ \text{K}^{-1}$; Young's modulus of brass $= 0.91 \times 10^{11}\ \text{Pa}$.

The shear modulus (modulus of rigidity) of an ideal liquid is zero.

Shear modulus G is defined as G = shear stress / shear strain. A liquid (unlike a solid) has no fixed shape; when a tangential (shear) force is applied to a liquid layer, it does not develop a restoring elastic shear stress — it simply flows/deforms continuously. Since a liquid offers essentially zero resistance to shape-changing (shear) deformation whil …

The Force–Displacement graph for F = kx is a straight line passing through the origin, with slope equal to k.

Given F = kx, F is directly proportional to x (k is a constant, e.g., a spring/elastic constant, as in Hooke's Law).

To draw the graph:

• Put displacement x on the horizontal axis and force F on the vertical axis.
• Since F = kx is a linear equation in x with zero intercept, the graph is a straight line that starts at the origin (0,0) and rises with a constant slope equal to k, i.e., for every unit increase in x, F increases by k. …

Within the elastic limit, the stress produced in a body is directly proportional to the strain it undergoes.

When a deforming force is applied to an elastic body, it develops an internal restoring force per unit area called stress, and undergoes a fractional change in dimension called strain. Hooke's law states that as long as the deformation stays within the elastic limit of the material (i.e., the body can still fully recover its original shape when the force is removed), the stress developed is directly proportional to the strain produced:

$\text{Stress} \propto \text{Strain} \quad \Rightarrow \quad \text{Stress} = k \times \text{Strain}$

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Bending a wire repeatedly at the same point strains it beyond its elastic limit again and again, causing progressive internal damage (elastic fatigue) until it fractures, even though one such bend alone would not break it.

When a wire is bent, the material on the outer side of the bend is stretched (tensile strain) while the material on the inner side is compressed. If the bend is sharp, the strain at that point can exceed the elastic limit of the material, so the material does not fully return to its original state when straightened — some permanent (plastic) deformation and internal microscopic damage remains.

When this bending is repeated again and again at the same point, each cycle adds a little more damage: microscopic cracks form and grow within the material, and the local heating produced by the repeated internal friction/deformation further reduces the wire's elastic strength at that spot. This cumulative loss of strength through repeated stress cycles is called elastic fatigue.

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Bending a wire back and forth repeatedly causes elastic fatigue — the metal at the bend loses its elasticity and develops cracks, so it eventually breaks even without a large single force.

When a wire is bent, the layer on the outside of the bend is stretched (put under tensile stress) and the layer on the inside is compressed. If the bend is sharp enough, the local stress at the fold can exceed the material's elastic limit even though the wire as a whole isn't under much load.

Each time we bend the wire back and forth, the material at that point undergoes this stress reversal again. Every cycle produces tiny, permanent internal deformations (dislocations) that the material cannot fully recover from — this cumulative loss of elastic strength with repeated stress cycles is called elastic fatigue. Over many cycles, microscopic cracks form at the bend and grow with each further flex, until the cross-section remaining is too weak to bear even the wire's own handling force, and it snaps.

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