Question

If the roller at A and the pin at B can support a load upto 4kN and 8kN, repsectively, determine the maximum intensity of the distributed load w, measured in kN/m, so that failure of the supports does not occur.

Answer 1

Solution :

Taken moment about point B;

$\sum M_B = 0$

$\left[(w)(4)(2)-N_A \sin 30^\circ (3)(\sin 30^\circ)-N_A \cos 30^\circ(3 \cos 30^\circ +4) \right] =0$

$N_A = 1.2376w$

$\sum F_x = 0$

$N_A \sin 30^\circ - B_x = 0$

$B_x = 1.2376w \times \sin 30^\circ$

$B_x = 0.6188w$

$\sum F_y = 0$

$B_y+N_A \cos^\circ - (w \times 4) = 0$

$B_y+(1.2376w) \cos 30^\circ - (w \times 4)=0$

$B_y = 2.9282w$

Now calculating the resultant force at B,

$F_B= \sqrt{B_x^2+B_y^2}$

$F_B= \sqrt{(0.6188w)^2+(2.9282w)^2}$

     = 2.9929w

For no failure,

$N_A<4 \ kN$

2.9929 w < 4 kN

$w < 3.232 \ kN/m$

And,

$F_B < 8 \ kN$

2.9929 w < 8 kN

w < 2.673 kN/m

For the safe operation, w = 2.673 kN/m