Percentage elongation is the increase in gauge length compared to the original length.

##### Percentage Elongation Formula

Percentage elongation can be expressed as;

\(Percentage\space\ elongation\space\ = \frac{Increase\space\ in \space\ length \times\ 100}{Original\space\ length}\)Symbolically it can be expressed as;

\(Percentage\space\ elongation=\frac{\delta l\times\ 100}{l}\)##### Percentage Elongation Example

Find percentage elongation of a steel rod, when it is subjected to an axial pulling force of 50 KN.whose diameter is 30 mm and it is 3 meters long. Take E= 200 X 10^9 Newton per meter square.

##### Solution

Given data is;

Diameter of rod = D = 30 mm

Length of rod = l = 3 m = 3 x 1000 = 3000 mm

Axial Pulling Force = P = 50 KN = 50 X 1000 = 5 X 10^4 KN

Young’s Modulus = E = 200 X 10^9 N/meter square = 200 X 10^3 N/millimeter square

Required;

Percentage Elongation = ?

\(Percentage\space\ elongation=\frac{\delta l\times\ 100}{l}\)Original length is known. Now we have to find change in length because of pulling force on steel rod;

\(Cross\space\ Area =A= \frac{\pi \times\ D^{2}}{4}\)put steel diameter = 30 mm

Hence, Cross sectional Area = A = 706.858 millimeter square.

Find value of stress in steel rod;

Stress = Force/Area

\(Stress = \frac{5\times\ 10^{4}}{706.858}\)Stress = 70.736 N/millimeter square = 70.736 MPa

Now find strain from young’s modulus formula

Young’s Modulus = Stress/ Strain

Strain = Stress/young’s Modulus

\(e\space\ = \frac{f}{E}\space\ =\frac{70.736}{200\times 10^{3}}\)Strain = e = 0.000354

As we know that;

Strain = Change in length/original length

Hence;

Change in length = Strain X Original length

Change in length = 0.000354 X 3000 = 1.062 mm

Now every thing is known, put values in percentage elongation formula and find the percentage elongation as a result of pulling force;

\(Percentage\space\ elongation=\frac{\delta l\times\ 100}{l}\)Percentage Elongation = (1.062 x 100)/ 3000

Percentage Elongation = 0.0345%