**Show that Strain energy line due to bending of the beam**

Solution –

Let, due to bending the filament of length dl at a distance z from the neutral layer is extended by x work done necessary in Stretching the filament is stored up in the filament as a potential energy.

Work done in stretching the filament by dx,

= force along the filament x dx

= Stress x cross section of filament x dx

Total work done in stretching the filament by x,

Summing up all such filament, we get total potential energy stored up in a beam of length dl due to bending

If G be the bending moment,

Thus potential energy stored up in the whole beam of length l,

NB – For light cantilever –

**Prove that a shear is equivalent to a compressional and an equal extensional at right angles to each other**

Consider a cube ABCD having each side equal to L. When a tangential force F is applied so that face ABCD is sheared to the position A’B’CD through an angle θ.

Let AA’ = BB’ = l

The diagonal DB is elongated DB’ and the diagonal AC is shortened to A’C.

Now DB = DM = L**√**2

Since angle of shear (θ) very small therefore ΔAA’N and ΔBB’M are isosceles right angle triangles

Extensional strain along the diagonal DB

Compressional strain along the diagonal AC,

Therefore, a shear θ is equivalent to an extensional strains (θ/2) and a compressional strain (θ/2); each.

**Calculate the strain energy or work done in twisting a rod of circular cross -section**

The work done in straining the shaft with in the elastic limit is called strain energy. consider a shaft of a certain diameter and a particular length subjected to a gradually applied torque. Let θ be the angle of twist. Energy is stored in the shaft due to this angular distortion. This is called torsional energy.

Let a wire of length l and radius a is twisted at an angle θ.

If c be the torsional couple per unit twist of the rod, then couple required to produce a twist Φ in the rod.

= cΦ

Work done in twisting the rod further by dΦ,

dw = cΦdΦ

Total work done in twisting the rod by an angle θ

### Explain why a hollow cylinder is stronger than a solid cylinder of same length, mass and material.

Consider two cylinders of same length, mass and material ; one of them is solid another is hollow.

Let the radius of solid cylinder is r.

The inner and outer radius of hollow cylinder is r1 and r2 respectively.

Torsional rigidity for solid cylinder,

Torsional rigidity for hollow cylinder,

i.e. torsional rigidity (torsional couple per unit twist) of hollow cylinder is greater than solid cylinder.

#### A horizontal wire of length 2l and cross -section is tightly fixed to rigid supports at both ends. A weight w is suspended from the mid point of the wire. Show that the depression d of the midpoint of the wire is given by,

#### where Y is Young’s modulus of the material of the rod.

soln –

If T be the tension along the stretched wire, then

**Longitudinal stress**

**Longitudinal strain**

…………………….. proved.

**Stress vs Strain curve**

Figure shows a stress vs strain curve for a metal wire which is gradually being loaded.

i) Initial part of of the graph- the wire is perfectly elastic.

ii) After point A – stress is not proportional to strain and curve AB is obtained. This OB of the graph is called elastic region and point B is called elastic point.

iii) After point B – strain is increased rapidly than stress. If load is removed at any point C, the wire doesn’t come back to its original length. Thus the material is said to have acquired a permanent set. The fact that stress strain curve is not retraced on reversing the action is called elastic hysteresis.

(iv) If the load is increased beyond the point C, there is a large increase in strain of the wire and for developing constraints.Ultimately breaks at point D- called fracture point D. This region is called Plastic region.

#### 1. Find the depression of a uniformly loaded beam supported horizontally on two knife edges –

Let us take a uniformly loaded beam of weight per unit length is ρ , supported horizontally on two knife edges k1 and k2.

So, the expressions for depression,

The maximum depression ** **at the midpoint i.e. x = l/2

#### 2. Depression of a uniformly loaded beam clamped horizontally at both ends –

Let us consider a uniformly loaded beam of length l, weight per unit length ρ is clamped at the both ends.

So, the expressions for depression,

#### 3. Depression due to uniform beam clamped horizontally at both ends and a concentrated load w is applied at the midpoint, weight of the beam being neglected

So, the expressions for depression,

The maximum depression ** **at the midpoint i.e. x = l/2

*** IF NECESSARY , I WILL PROVIDE THE DETAILED CALCULATION OF 1, 2, 3, 4, DEPENDING ON YOUR QUERIES AND RESPONSES.

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