Question 6.2.2: The inclined beam in Figure 1 is subjected to the vertical f...

Goal Calculate the loads acting on the beam at A and B.
Given Length and orientation of beam, the types of supports at A and B, and information to describe the distributed load.
Assume The system is planar, upward force is positive, and that the weight of the beam is negligible.
Draw A free-body diagram of the beam is shown in Figure 2. We have established two coordinate systems, one aligned with the horizontal and vertical (xy), which we will use for calculating F, the other along the slope of the inclined beam (x* y*), which makes calculating the forces at A and B less complex.
Formulate Equations and Solve First we find the total equivalent force and its location. Next we apply equilibrium conditions to find the loads acting at A and B. We use two approaches. In the first we use (6.13) and (6.15) to find the equivalent total force and its location. In the second approach we use the information in Appendix C for a standard line load distribution.

Total  forcein  y direction =F_{y}=\int_{span}^{}{\omega  dx}                (6.13)
X_{C}=\frac{x\left(\int_{span}^{}{\omega dx} \right) }{F_{y}} \underbrace{=}_{substituting   in from  (6.13)  for   F_{y}} =\frac{x\left(\int_{span}^{}{\omega dx} \right)}{\int_{span}^{}{\omega dx} }        (6.15)

Approach 1: We begin by determining \omega \left(x\right) . Using a linear distribution with unknown constants k_{0} and k_{1} to represent the load distribution:

\omega \left(x\right) =k_{0}+ k_{1}x

and applying the boundary conditions \omega \left(0\right)= = 0 and \omega \left(L\cos \theta \right) = – \omega _{0} to determine the constants results in

\omega \left(0\right) =k_{0}+ k_{1}\left(0\right) =0                                    k_{0}=0
\omega \left(L\cos \theta\right) =k_{0}+ k_{1}\left(L\cos \theta\right) = – \omega _{0}                                k_{1}=- \omega _{0} /\left(L\cos \theta\right)

Therefore , \omega \left(x\right)=\frac{- \omega _{0}}{L\cos \theta}x
Using (6.13) to find the total force,

F=\int_{0}^{L\cos \theta}{\omega \left(x\right)dx}=\int_{0}^{L\cos \theta}{\frac{- \omega _{0}}{L\cos \theta}xdx } = \left[\frac{- \omega _{0}}{L\cos \theta}\frac{x^{2} }{2} \right] ^{L\cos \theta}_{0}=\frac{- \omega _{0}}{2}\frac{L^{2}\cos^{2} \theta}{L\cos \theta} =\frac{- \omega _{0}L\cos \theta}{2}                       (1)

Using (6.15) to find the centroid,

X_{C}=\frac{\int_{0}^{L\cos \theta}{\omega \left(x\right)xdx} }{F} = \frac{\int_{0}^{L\cos \theta}{\frac{- \omega _{0}}{L\cos \theta}x^{2}dx} }{F} =\frac{\left[\frac{- \omega _{0}}{L\cos \theta}\frac{x^{3} }{3} \right]^{L\cos \theta}_{0} }{F} =\frac{\frac{- \omega _{0}L^{2}\cos^{2} \theta}{3} }{\frac{- \omega _{0}L\cos \theta}{2} } =\frac{2}{3} L\cos \theta

Approach 2: We use Appendix C to find the total equivalent force and its location. Because \omega _{0} is given in force units per horizontal length, we must use the horizontal length (L cos \theta) as the base of our triangle.

F=\frac{bh}{2} =\frac{\left(L\cos \theta\right)\left(- \omega _{0}\right) }{2}=\frac{- \omega _{0}L\cos \theta }{2}

The centroid is located at 2/3 the distance from the vertex:

X_{C}=\frac{2}{3} L\cos \theta

These values, the same as we calculated using the integral approach, are shown in Figure 3, which is also a free-body diagram of the beam.
To find the loads at supports A and B, we refer to the free-body diagram and use a coordinate system that has its x^{*} axis along the beam to apply equilibrium equations parallel and perpendicular to the beam.

\sum{F_{x^{*}}} =0=F\sin \theta -B_{x^{*}}

We substitute for F from (1), noting from Figure 3 that F sin is in the positive x^{*} direction. There are two sign conventions in play here, one that says negative loads act downward, and a second that we are using to sum forces in the x^{*} direction.

B_{x^{*}}=F\sin \theta=\left\lgroup\frac{\omega _{0}L\cos \theta}{2} \right\rgroup \sin \theta=\frac{\omega _{0}L\cos \theta \sin \theta}{2}                              (2)

Using B as the moment center gives

\sum{M_{z@B} } =0\left(\curvearrowleft+ \right)=-A_{y^{*}}L+ F\cos \theta \left(L/3\right)
A_{y^{*}}=\frac{F\cos \theta}{3}= \frac{\omega _{0}L\cos \theta}{2}\frac{\cos \theta }{3} =\frac{\omega _{0}L\cos^{2} \theta}{6}                          (3)
\sum{F_{y^{*}}} =0=A_{y^{*}}+B_{y^{*}} – F\cos \theta
B_{y^{*}}=F\cos \theta – A_{y^{*}}

Substituting for F from (1) and for A_{y^{*}} from (3) gives

B_{y^{*}}=\frac{\omega _{0}L\cos \theta}{2}\cos \theta – \frac{\omega _{0}L\cos \theta}{6}=\frac{\omega _{0}L\cos^{2} \theta}{3}                         (4)

Check The solution can be checked by summing the moments about A, and plugging in the results. In addition, note that B_{y^{*}} is twice as big as A_{y^{*}}; this is as expected since F is located two times closer to B than to A.