Question

4. The soil below creates a trapezoidal distributed loading in response to the three point loads. Ignore the weight of the concrete slab in this problem. What do the intensities wi and w2 need to be in order to support the three column loadings? In other words, find wı and w2 so that the sum of the forces on the concrete slab equals zero and the sum of all the moments equals zero. (20 pts.) 80 KN 60 KN 50 KN 1 m 1 m -2.5 m -3.5 m Concrete Slab Soil W2 wi

Answer 1

60 KN 80 KN SO KN 1 m 2.5 m -1 m 3.5 m A B w w

zoso om The sum of the forces on the concrete slab equal to (1+2.5+3.5+1 E Fg = 0 Take itve; vave _x8 * (W, -W2) +8 W2 -60-80-50 =0 ( W,- W2 +W2 ) 8–60–80–50 = 0 (W+W2.18 -190 = 0 190X2 2 2 WitW2 8 W, t W2 = 47.5 0 Now take moment about point A equal to reso EMA = 0 (6ox 1)+(80x3.5) + (50x7) -(W2 x8x) - Ž (W, -W₂ ) x8x8 690-32 W2 10.6667 (W, W2)=0 32 W2 +10.6667 (W, -W2) = 690 في حا from B W, + W2 = 47.5 W2 47.5W, from @ 32 W2 + 10.6667 CW, W2) = 690 32(47.5-W) + 10.6667 (W, -W2 ) = 690

32(47.5-W1) +10.6667(W, -67.5-W,)) = 690 1520 – 32 W, +10.6667W, – 506: 668 + 10.6667W, =690 -10.660 W, = -323.332 323.332 10. 666 W, = 30.314 kN/m Wi from o Wit W2 = 47.5 30.314 + W2 67.5 W2 - 17.186 kN/m