A uniform light slender beam AB of section modulus EI is pinned by a frictionless joint A to the ground and supported by a light inextensible cable CB to hang a weight W as shown. If the maximum value of W to avoid buckling of the beam AB is obtained as βπ2EI,where π is the ratio of circumference to diameter of a circle, then the value of β is

The correct option is 0.0924 m^-2.

1. System Representation and Joint Equilibrium
From the given system diagram, the horizontal uniform beam AB of length $L=2.5\text{ m}$ is pinned at end A and supported at end B by a cable BC making an angle of 30° with the horizontal. A weight $W$ hangs vertically from joint B.
Let $T$ be the tension in the cable CB. By analyzing the static equilibrium of forces at joint B:

• Vertical force equilibrium:
$T\sin (30°)=W$
Since $\sin (30°)=0.5$, we have:
$T=2W$

• Horizontal force equilibrium:
The horizontal component of the cable tension acts as a compressive force $P$ on the beam AB:
$P=T\cos (30°)$
Substituting $T=2W$:
$P=2W\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}W$

2. Buckling Criterion of the Beam
Since the beam AB is pinned at both ends, the critical Euler buckling load $P_{\text{cr}}$ is given by:
$P_{\text{cr}}=\frac{{\pi }^{2}EI}{L^2}$
To avoid buckling of the beam, the compressive force must satisfy:
$P\le P_{\text{cr}}$
Substituting the relation for $P$:
$\sqrt{3}W\le \frac{{\pi }^{2}EI}{L^2}$
Therefore, the maximum weight $W_{\text{max}}$ to avoid buckling is:
$W_{\text{max}}=\frac{1}{\sqrt{3}{L}^2}{\pi }^{2}EI$

3. Calculating the Coefficient $\beta$
Comparing this expression with the given form $W_{\text{max}}=\beta {\pi }^{2}EI$, we find:
$\beta =\frac{1}{\sqrt{3}{L}^2}$
Substituting $L=2.5\text{ m}$:
$\beta =\frac{1}{\sqrt{3}\times {\left(2.5\right)}^2}$
$\beta =\frac{1}{1.73205\times 6.25}=\frac{1}{10.8253}\approx 0.092376{\text{ m}}^{-2}$
Rounding to four decimal places gives 0.0924 m^-2.