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Answered Same Day Dec 20, 2021

Solution

Robert answered on Dec 20 2021
115 Votes
Mechanics Quiz
For any given problem with distributed load, first make it into a single concentrated load acting the resultant centroid.
1. This problem can be idealized into three concentrated loads acting as a result of the three distributed loads, two with a
gradient/slope(non-uniform/increasing distributed load) on the sides and one uniform load in the middle. These act at the
espected centroidal X coordinates of the area formed by the X-F/unit length graphs.
A B
D
C
EF w=3kN/m
A
9kN/m
7kN/m
4m 3m 5m
• First one has 7kN/m for 4m, decreasing⇒ total load is area of trapezoid= 12 ×4× (3+7)=20kN acting at the centroid.
For a trapezoid, we see it is at a distance of one third from bigge
longer base. So here it is 43 from A.
• For second rectangle, total load is 3× 3 = 9kN acting at 4 + 32 = 112 from A.
• For this last trapezoid, total load= 12 × 5× (3 + 9) = 30kN acting at 4 + 3 + 103 = 313 .
Thus we get the following diagram after adding the adequate reaction forces. Solving this we have as follows.
A B
20kN
9kN
30kN
4
3
11
2
31
3
ByAy
Ax
• ∑Fx = 0⇒ Ax = 0
• ∑Fy = 0⇒ Ay +By = 59000N
• ∑Mz = 0 about A⇒ By × 12 = (20000)43 + (9000) 112 + (30000) 313
On solving these three we get solutions as follows. Ax = 0, Ay = 26819.44444, By = 32180.55556
2. The problem is as follows. Here again, we see that the trapezoid distributed load can be replaced by a single resultant
load= 12 × 8 × (150 + 250) = 1600 acting at 163 from left edge. Also adding the reaction forces for the joint as given, we
have the FBD as follows. Thus we have as follows.
w1 = 150/ft
w2 = 250/ft
P = 500
A
B5 3 4...
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