If stress at x = ℓ/3 from bottom is ( W/A +2 /γ * W/A) then find γ: (Refer to the figure showing a vertical bar subjected to load W)

The correct answer is 6.

To find the value of $\gamma$, we analyze the stress in the vertical bar shown in the figure.
The figure shows a vertical bar suspended from a fixed support at the top. The bar has a cross-sectional area $A$ and is subjected to:
- An external load $W$ attached to the bottom.
- The bar's self-weight, labeled as $w$, where we take $w=W$ to express the stress in terms of the load $W$.

At any section located at a distance $x$ from the bottom of the bar, the total axial force $F$ is the sum of the external load $W$ and the weight of the portion of the bar of length $x$ below that section:

$F=W+W��\frac{x}{\ell }$

The stress $\sigma$ at this section is the axial force divided by the cross-sectional area $A$:

$\sigma =\frac{F}{A}=\frac{W}{A}+\frac{W}{A}·\frac{x}{\ell }$

Now, we calculate the stress at a distance $x=\frac{\ell }{3}$ from the bottom:

$\sigma =\frac{W}{A}+\frac{W}{A}·\frac{\ell /3}{\ell }=\frac{W}{A}+\frac{1}{3}·\frac{W}{A}$

We are given that the stress at $x=\frac{\ell }{3}$ is:

$\sigma =\frac{W}{A}+\frac{2}{\gamma }·\frac{W}{A}$

By comparing the two expressions, we find:

$\frac{2}{\gamma }=\frac{1}{3}$

Solving for $\gamma$:

$\gamma =2·3=6$