Question

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of the uniform distributed load that can be applied to the beam without risk of causing the strut to buckle. take f.s. = 2 against bucklin

Answer 1

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$. This means it is safe

The maximum intensity w = 11.211KN/m