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I am working on this question on my statics assignment:
Approximately at point B there are two forces, 20kN and 50kN.So far, I have only calculated global equilibrium and got the following values:
$D_y=(\frac{80}{7})kN$
$D_x=(\frac{20}{\frac{\sqrt{13}}{2}})kN$
$A_y\approx89.094 kN$
My question here is, when I'm drawing the shear force diagram do I sum up the 20kN force and the 50kN force into one resultant and graph it? Or, do I draw them separately. I need help. enter image description here

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To answer your question, you just need to break the inclined force into its component X and Y forces:

$$\begin{alignat}{4} f_x &= \dfrac{20}{\sqrt{1^2+1.5^2}}&&=11.09\text{ kN} \\ f_y &= -1.5f_x&&=-16.64\text{ kN} \\ \end{alignat}$$

and then add the vertical component to the 50 kN force.

That being said, your reactions (other than $D_x$) are incorrect:

$$\begin{align} \sum M_A &= -66.64\times2 - 6\times4\times\left(2+\dfrac{4}{2}\right) + 200 + 9D_y - 2\times2\times\left(9 + \dfrac{2}{2}\right) - 200 = 0 \\ \therefore D_y &= 29.92\text{ kN}\\ \sum F_y &= A_y + 29.92 - 66.64 - 6\times4 - 2\times2 = 0 \\ \therefore A_y &= 64.72\text{ kN} \end{align}$$

You can then draw the diagram as normal, remembering to add the inclined force's vertical component to the vertical force at B:

enter image description here


Diagram obtained with Ftool, a free 2D frame analysis program.

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  • $\begingroup$ I noticed a few minutes ago that I took the moment about the wrong point. Thanks for your help. By the way, what software did you use to draw the shear force diagram? $\endgroup$ – AugieJavax98 May 20 '18 at 1:29
  • $\begingroup$ @AugieJavax98, I used Ftool, a free 2D frame analysis program. $\endgroup$ – Wasabi May 20 '18 at 1:36

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