Uniaxial compression test data for a solid metal bar of length 1 m is shown in the figure.

The bar material has a linear elastic response from O to P followed by a non-linear response. The point P represents the yield point of the material. The rod is pinned at both the ends. The minimum diameter of the bar so that it does not buckle under axial loading before reaching the yield point is _______ mm (round off to one decimal place).

The correct answer is 56.9.

1. Identification of Material Properties from the Graph:
Based on the provided uniaxial compression test graph:
- The material exhibits a linear elastic response from the origin O up to the yield point P.
- At the yield point P, the yield stress is:
${\sigma }_y=100\text{MPa}=100\times {10}^6\text{Pa}$
- The corresponding yield strain is:
${\epsilon }_y=0.002\text{mm/mm}$
(Note: Although the strain axis in the diagram shows the value under point P as 0.02, it precedes 0.005 and is logically 0.002 mm/mm).

2. Calculation of Young's Modulus (E):
Since the material is linear elastic up to point P, Hooke's Law applies:
$E=\frac{{\sigma }_y}{{\epsilon }_y}$
Substituting the values:
$E=\frac{100\times {10}^6\text{Pa}}{0.002}=5\times {10}^{10}\text{Pa}=50\text{GPa}$

3. Formulation for Buckling and Yielding limits:
Let d be the minimum diameter of the solid metal bar in meters.
- The cross-sectional area A of the bar is:
$A=\frac{\pi }{4}{d}^2$
- The moment of inertia I_min for a circular cross-section is:
$I_{\text{min}}=\frac{\pi }{64}{d}^4$
- The bar is pinned at both ends, meaning its effective length L_e equals its physical length L:
$L_e=L=1\text{m}$

To prevent the bar from buckling before it reaches the yield point, the Euler critical buckling load P_b must be greater than or equal to the load at yielding P_y:
$P_b\ge P_y$
Where:
$P_b=\frac{{\pi }^{2}E{I}_{\text{min}}}{L_e^2}$
and
$P_y={\sigma }_{y}A$

4. Finding the Minimum Diameter:
Equating the buckling load to the yielding load at the limiting condition:
$\frac{{\pi }^{2}E\left(\frac{\pi }{64}{d}^4\right)}{L^2}={\sigma }_y\left(\frac{\pi }{4}{d}^2\right)$
Simplifying the expression by canceling common terms:
$\frac{{\pi }^{2}E{d}^2}{16L^2}={\sigma }_y$
Solving for d²:
$d^2=\frac{16L^2{\sigma }_y}{{\pi }^{2}E}$

Substituting the numerical values into the equation:
$d^2=\frac{16\times 1^2\times {10}^8}{{\pi }^2\times 5\times {10}^{10}}$
$d^2=\frac{16}{500{\pi }^2}=\frac{16}{4934.802}\approx 0.003242{\text{m}}^2$
Taking the square root:
$d\approx 0.05694\text{m}=56.94\text{mm}$

Rounding off to one decimal place, the minimum diameter of the bar required is 56.9 mm.