**In this section we are going to discuss three points**

**Elongation in Taper Rod**

#### Let us consider a circular taper rod of length *l* whose diameter is changing uniformly from d to D.

*l*

#### Let us consider a small element of a thickness of dx at x distance from one end. The relationship between dx,d, D,l is given by applying the similar triangle property and we get

Let us call that as **equation (1)**

#### δx is the elongation in the element dx. So by Hook’s law.

**Stress = E * Strain**

Putting values and we reach this expression as follows

##### Now we want to find the elongation in whole tapered rod. So all we have to do is integrate the above expression from 0 to l and we will finally get the expression Elongation in tapered rod.

**Remember above expression for elongation in a taper rod.**

Let us double check the formula that we have arrived with so many calculations and integrations.

**This formula should be valid for uniform diameter rod of d.**

With this Elongation in Taper rod is over and write down that expression so that it can come handy in exams.

**Elongation in an object due to slef weight**

#### When an object remains hanging from one end. It’s length increases due to self-weight. Let’s find out how much length increases and what is expression for it.

#### Let us assume a bar of length *l* having density ρ is Fixed at one end vertically.

*l*

##### In these type of problems we have to start with assuming an element of thickness dy at some distance y as shown in the figure

#### If we make a free body diagram of this hanging bar then we see that only Mg force will be there downwards, where M is mass of the whole bar.

**Similarly, we calculate the force up to small section dy that will be = m*g.**

#### Where m = mass of small element up to y having area of cross-section Ac

**Force up to section y = ρ Ac y g**

#### Stress up to section y = Force/Area = ρ Ac y g/Ac

**σ = ρyg**

##### This is the expression we have to remember

**Let’s see special cases **

**Stress at the bottom At bottom, y = 0 ⇒ σ = 0****Stress at top At top y=l ⇒ σ = ρlg,**

**↑ This maximum stress in the bar**

#### Now we see the elongation in the element dy due to above calculated stress.

### By hook’s law

##### We can easily calculate total elongation with integrating above expression from limits 0 to l

**Remember above expression for elongation in a bar due to self-weight.**

Now in exams suppose we are given to calculate elongation in tapered rod due to self weight?How will you solve it? Please let me know in comments

**Volumetric Strain**

##### Well in exams you won’t see much questions from this part. But sometimes there may be a multiple choice question from this part.

Don’t ignore this topic just make one reading and you won’t forget it.

To have a better understanding of volumetric strain. Let’s assume a cuboid of length l, breadth b, height h.

Volume of that cuboid is given by V = lbh

If due to stresses, its dimensions changes in length or breadth or height or any of them ( all of them). Then it will affect the volume i.e. volume will also change.

**Change in volume dV = δ(lbh)**

#### dV = bh(δl) + lh(δb) + lb(δh)

These are the volumetric strain in x,y,z directions respectively.

**Since the stresses are in all three directions so both lateral and longitudinal strain will be there.**

Due to σx change in length

Similarly there will be a change in length due to σy, σz.

Which will be calculated with the help of **Poisson’s Ratio formula**

Let’s calculate the change in length due to stresses in y & z direction.

**Net change in length = δLx + δLy + δLz**

##### So finally we get expression

So remember these final expressions these are very important.

**Let’s see some special cases, from which directly conceptual question comes to exams**

##### If there is uni-axial loading i.e. Force is only in one direction let’s say x , then stress will be only in one direction σx

**In above expressions put σy = σz = 0, we get **