Question

Consider 2 steel rods, A and B, B has three times the area and twice the length of A, so young modulus of B will be what factor times young modulus of A?

Answer 1

Answer:

  $\frac{Y}{Y_o}$ = 2/3

Explanation:

The yuong modulus of a rod is defined as the relationship between the tensile strength and the strain

         Y =  $\frac{ \frac{F}{A} }{\frac{\Delta L}{L_o} }$

let's use the subscript "o" for rod A

         I = $\frac{ \frac{F}{A_o} }{ \frac{\Delta L}{L_o} }$

tells us that rod B has

         A = 3 A₀

         L = 2 L₀

we substitute

         Y = $\frac{ \frac{F}{A} }{ \frac{\Delta L}{L} }$

          Y = $\frac{ \frac{F}{3A_o} }{ \frac{\Delta L}{ 2L_o} }$

         y = ⅔  $\frac{ \frac{F}{A_o}}{ \frac{\Delta }{L_o} }$

substituting the value of Y₀

          Y = ⅔  Y₀

           $\frac{Y}{Y_o}$ = 2/3