Question

A beam of size 150 mm wide, 250 mm deep carries a uniformly distributed load of w kN/m over entire span of 4 m. A concentrated load 1 kN is acting at a distance of 1.2 m from the left support. If the bending stress at a section 1.8 m from the left support is not to exceed
3.25 N/mm2 find the load w.​

Answer 1

Given : B = 150 mm ; D = 250 mm

           UDL = w kN/m

           span ($l$) = 4 m

P @ 1.2 m from left support = 1 KN

Bending stress (σ) @ 1.8 m from left support = 3.25 N/mm²

first let us find the reactions at support Ra and Rb using the equations of equilibrium.

Ra + Rb = (4w + 1) KN ------(1)

taking moment about left support,

⇒ $\frac{w4^2}{2} + (1*1.2) - (Rb*4) = 0$

⇒ Rb = ( 2w + 1.2) KN.

from (1), Ra  = (4w + 1) - (2w + 1.2)

⇒ Ra  = (2w - 0.2) KN.

Now let us find the value of the bending moment (M) at 1.8 m left support in terms of w .

Taking moment @ 1.8 m from left support.

(Ra x 1.8) - (1 x 1.2)  - (w 1.8² / 2) =  [email protected]

⇒ (2w - 0.2)x1.8  - 1.2 - 1.62w  = [email protected]

⇒ [email protected] = (1.98w - 0.84) KN.m = (1.98w - 0.84)x10⁶ N.mm

from bending equation we have:

$\frac{M}{I} = \frac{stress}{y} = \frac{E}{R}$

$I = \frac{BD^3}{12} = \frac{150 * 250^3}{12}$

⇒ I = 195.312 x 10⁶ mm⁴

y = D/2 = 125 mm ; stress(σ) = 3.25 N/mm²

⇒ $\frac{(1.98w - 0.84)*10^6}{195.312*10^6} = \frac{3.25}{125}$

∴ w =  2.988 ≅ 3KN/m

HOPE THIS HELPS YOU !!    : )