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Reaction forces and moment of simple beam, triangular distributed load

작성자 Uploader : pmsix 작성일 Upload Date: 2023-07-14변경일 Update Date: 2023-07-14조회수 View : 161

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As shown in the figure, the reaction force of a simple beam of length L(m) acting on a triangularly distributed load with a load at point B of P(kN) can be calculated as follows. No Image
Calculate the reaction force RA by using that the sum of moments at point B is zero.

∑MB = RA*L - (1/2)PL*L/3 = 0


Using that the sum of the vertical force is 0,

∑PV = RA + RB - (1/2)PL = 0,


Meanwhile, the moment at a point away from point A by x (m) can be obtained as follows.
(Clockwise + )

Mx = RA*x - (P/L)*x*(x/2)*(x/3) = (1/6)*(P*L*x - (P/L)*x^3)


The location xm (m) where the moment becomes maximum can be calculated by differentiating Mx.

Mx′ = (1/6)*(PL - 3(P/L)*x^2) = 0

Thus, (PL/6)(1-3(x/L)^2) = 0, x/L = (1/3)^(1/2)


By substituting xm into Equation 3, the maximum moment Mmax can be obtained.

Mmax = (1/6)*(P*L*L*(1/3)^(1/2) - (P/L)*(L*(1/3)^(1/2))^3)
          = (P*L^2/6)(1/3)^(1/2)(2/3)
          = (P*L^2/9)(1/3)^(1/2)


The Mmax can also be calculated from the moment equation No.3.


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