I'm working on the statics problem below:
enter image description here I'm tasked to draw the shear and bending moment diagram for the frame. I'm trying to work out the reactions and I'm taking the moment about pin E.Here's my working so far:
$$\Sigma M_E=0$$ $$(10)(5)-100-(2*10)(\frac{1}{2}*10+5)+(10)(15)-100-(\frac{30}{\sqrt{2}})+(0.5*5*4)(\frac{1}{3}*5)+A_y(20)=0$$ My question here is, for the triangular UDL, is the distance one-third or two thirds? My lecturer did a similar problem and he said that the distance is two thirds but I don't quite get why it is two thirds. Someone help.


It is 2/3 and also if we assume the moment direction positive clockwise, to be consistent with the rest of your positive signs, then that vertical loading's moment is negative, the correct sign and factor is (-0.5*5*4) (2/3*5).

The reason the distance is 2/3 is because C.G. of a the vertical 5m load triangle is at 1/3 from its base, which here is 2/3s from E.

If the loading triangle was reversed increasing up your assumption would be correct.


As for the title of your question, any statics textbook should show you the centroid of a right triangle. This wikipedia article gives the centroids of common shapes.

When replacing a distributed load with an equivalent point load, that point load will act through the centroid of the distributed load. In this case, the equivalent point load for the triangular distributed load will act (1/3)*5 up from Point A.

When summing the moments about a point it is important to indicate your sign convention (clockwise positive or counterclockwise positive). Either convention is acceptable so long as you are consistent throughout. Looking at the work you showed, there appear to be some inconsistencies. For example P1 and M1 (see sketch below for notation) should both produce clockwise rotation about Point E, however your work shows them producing moments with opposite signs.

I won’t write out the whole equation for summing moments about Point E because I don’t want to do the homework for you. You’re on the right track though!

Free Body Diagram

  • $\begingroup$ nice diagram... $\endgroup$ – kamran May 23 '18 at 0:34
  • $\begingroup$ Thanks for that. Yes I am aware of the triangle's centroid. I was just unsure about the distance. Your diagram is very helpful. I can now see why it is 2/3. Thanks again. $\endgroup$ – AugieJavax98 May 23 '18 at 6:28

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