Given
a = 3.05 m
w1 = 5.95 kN/m
w2 = 4.6 kN/m
R1 = (w1*a/2)a/3
= (1/6)a^2*w1
= (1/6)3.05^2*5.95
= 9.22 KN - m
R2 = (w2*a)(a + a/2)
= (3/2)a^2*w2
= (3/2)3.05^2*4.6
= 64.18
R = 64.18-9.22
= 54.96
A couple moment acting at point A is 54.96 kN-m
counter-clockwise.
Thanks For the left figure below, replace the distributed loads by an equivalent resultant force and...
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