Question

A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa. If a 610N load( about that exerted by the mass of an average human) is applied to the other end, what will be the elongation of the brass rod in mm?

Answer 1

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$  

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm