Question

The ultimate tensile strength of a circular rod is 400 MPa and the elongation up to maximum load is 35%. If the max. load reached is 220 KN. Then the initial rod diameter is

Answer 1

Given :

The ultimate tensile strength of the  rod  =  400 MPa

The elongation up to maximum load  = 35%

To Find :

If the maximum load reached is 220 KN , then the initial diameter of rod = ?

Solution :

Since we know that :

$Pressure = \frac{force}{area}$

So, $400 MPa = \frac{220 KN}{A}$     ( A is cross sectional area )

Or, $400 = \frac{220}{\pi R^2}$

Or, $R = \sqrt{\frac{220 \times 10^3}{400 \times 3.14 \times 10^6}}$

So, R = 0.012 m

Now increase in length :

$l' = l + \frac{35}{100}l$               ( it is given that length increase by 35% )

Or, $l' = \frac{135l}{100}$

Or, $l' = \frac{27l}{20}$

Now , since volume remains same , so :

$\pi (r')^2 l' = \pi r^2 l$    ( r' is final radius and r is initial radius )

Or, $(r')^2 \frac{27}{20}l = r^2l$

Or, $r' = \sqrt {\frac{20}{27}}r$

Or, $r = \sqrt {\frac{27}{20}}r'$

        =$\sqrt{1.35} R$

        =1.16 R

        =$1.16 \times 0.012$

        = 0.019 m

So initial rod diameter =$2 \times r$

                                     =$2 \times 0.019$

                                     = 0.038 m

                                     = 3.8 cm

Hence , the initial diameter of rod is 3.8  cm .