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Very rough sketch but I'm having trouble with a Shear and Bending Moment problem, and I'm pretty new to the topic.

This is the original diagram:

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I already simplified the left side (not sure if it's right). However, I'm having a hard time picturing on how to simplify these other loads below:

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How do you simplify a trapezoid and a right triangle on top of each other?

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  • $\begingroup$ I'd have thought it would be more reliable to draw the SF diagram for each individual load and then add them up. $\endgroup$ Feb 8, 2023 at 6:38

4 Answers 4

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Not sure why your simplifications still leaves the them as load distributions though when you can use weighted averages via integration to reduce each distribution it to a point force acting at some point within the original distribution. Does your problem require you to leave to as a distribution (it might, depending on what you are doing).

Find the equation for each line segment of the distribution. You know the peak, you know it falls off to zero and you know the x-axis span. Chop up each line as necessary to work with the overlap.

But you don't even need to do that. You can treat each load distribution independently and convert each to a point force. That's what I would do.

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  • $\begingroup$ the way I understand it if the aim is the shear and bending diagram, then replacing the distributed load with concentrated loads will produce erroneous results. $\endgroup$
    – NMech
    Dec 14, 2021 at 8:54
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The right hand load slopes up from zero at G, to point F reaching loading of w/2, then it slopes down from there to the the point at 4+4=8 ft from G, to reach zero load.

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The ? is an internal hinge, which made this system solvable using simple statics. (4 unknown with 3 equilibrium equations available and an additional condition - the internal hinge does not have rotational restraint, at which M = 0).

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For simplicity, the external loads are not shown.

Note: The distributed loads can be replaced by equivalent concentrated loads to further simplify the problem.

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  • $\begingroup$ Why went through the tedious route to simplify the loads? As you are obviously beyond the entry-level on structural analysis, you should have already mastered the concept/method of equivalent load, so I leave it for you to figure out how the equivalent concentrated loads are derived/calculated. $\endgroup$
    – r13
    Dec 14, 2021 at 22:36
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I believe you had a minor error on you left simplified result. you 2*w1 should be reduced by w1 resulting in just w1. I came up with the following diagram.

The key to the right varying load, is that the peak values above and below line up and are equal but opposite so they will sum to 0. Since the bottom force becomes 0 at the midway point, you just need to determine the force in the top tapered load also at the mid point. Since the top goes from max to zero over that portion, it will be 1/2 max at the mid point.

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