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

Taken moment about point B;

`$\sum M_B = 0$`

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

`$N_A = 1.2376w$`

`$\sum F_x = 0$`

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

`$B_x = 1.2376w * \sin 30^\circ$`

`$B_x = 0.6188w$`

`$\sum F_y = 0$`

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

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

`$B_y = 2.9282w$`

Now calculating the resultant force at B,

`$F_B= √(B_x^2+B_y^2)$`

`$F_B= √((0.6188w)^2+(2.9282w)^2)$`

= 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