Question

A cylindrical specimen is subjected to plastic deformation in tension up to a uniform elongation of 10\%. The final cross-sectional area of the gage section is found to be 20 mm$^2$. The initial cross-sectional area of the gage section is ............ mm$^2$ (rounded off to an integer).

Hint:

For plastic deformation problems, apply volume constancy: $A_0 L_0 = A_f L_f$. Uniform elongation directly relates the change in length.

Answer 1

Step 1: Apply volume constancy principle.
During plastic deformation of metals: \[ A_0 L_0 = A_f L_f \] where $A_0$ = initial area, $L_0$ = initial length, $A_f$ = final area, $L_f$ = final length. Step 2: Relating length with elongation.
Uniform elongation = 10%. \[ L_f = L_0 (1 + 0.10) = 1.1 L_0 \] Step 3: Substituting values.
\[ A_0 L_0 = A_f L_f \] \[ A_0 = A_f \frac{L_f}{L_0} = 20 \times 1.1 \] \[ A_0 = 22 \, mm^2 \] Final Answer: \[ \boxed{22 \, mm^2} \]