5 points with a we ght o 44.0 kN with the wei ht un m ly distributed. The slab i

Question

5 points with a we ght o 44.0 kN with the wei ht un m ly distributed. The slab is held up b one support 20% of the way c oss the bridge, and a se A parti lar bride e with a len th co s sts of a harizontal sla the picture, a truck with a weight of mg- 24.0 kN rd support 80% o tew ay cross the t dge As sho snin s driving from left to right across the bridge. 0.4L 0.2J 0.2L in above. (a) Onee the truck is completely off the bridge, how much force will the left support exert on the horizontal slab? We're looking for an answer in kilonewtons (kN), which is what values are given in above. b) when the truck is at the position show forces where its fui eght can be taken to act at point 40% of the w ey across the b dge, calculate the force that each support exerts on the horizontal seb. We're looking for two answers-the value of each of the support The force exerted by the left support is kN. The force exerted by the right support is kN. (c) As the truck continues to drive to the right across the bridge, what will happen to the magnitudes of the support forces? Select all the correct statements, assuming the truck is entirey on the bridge. The left support force will increase. The left support force will decrease. The right support force will increase. The right support force will decrease. Both support forces will increase. Both support forces will decrease. Neither support force will change. (d) A: the instant the truck's weight s directly above the right support, what is the magnitude of the force the right support exerts on the horlzontal slab? t the instant the truck

Explanation / Answer

(A) without the truck, Figure is syymetric.

F1 = 44/2 = 22 kN

(B) balancing moment about the left end,

(0.2L FL) - (0.4L x 24) + (0.8L)(FR) - (0.5L x 44) = 0

0.2 FL + 0.8 FR = 31.6 ,..... (i)

balancing forcem,

FL + FR = 68

solving eqwuation,

FL = 38 kN and FR = 30 kN

(c) left support will decrease.

right support force will increase

(d) (0.2L FL) - (0.8L x 24) + (0.8L)(FR) - (0.5L x 44) = 0

0.2 FL + 0.8 FR = 41.2 ,..... (i)

FL + FR = 68

FL = 22 kN and FR =46 kN

Ans: FR = 46 kN

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