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Can someone provide the derivation for the following equation:

$$\int_l Q_Z(x) dx = M_y (x) \tag1$$

or equivalently :

$$\frac{dM_y}{dx}=Q_Z\tag2.$$

I am interested in the $1D$ isotropic material version.

enter image description here

Appendix

When drawing the shear and moment diagram for this case, we would get something like this: enter image description here

Since the $Q_z$ is positive, the slope of $M_y$ has to be positive, because of equation $(2).$

But if $M_y$ has to have a positive slope and has to be zero at the free end, than it has to look like in the picture, i.e. it has to have a negative value.

Can that be a priori recognized from the equation $(1)$ or is it just a convention that when a beam curves downwards that it will be negative, and when it curves upwards, it will be positive, like explained here?

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  • $\begingroup$ Draw a free body diagram. $\endgroup$ Mar 12 at 21:44

1 Answer 1

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We just integrate and evaluate $x$ from $0$ to $l$.

$$\int_l Q_Z(x) dx = Q_z(x) \ x\biggr\rvert _o^l = Q_z(x)l= M_y(x)$$

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  • $\begingroup$ Thanks for the answer. I just added an appendix to my question, can you please confirm if my reasoning is correct? $\endgroup$
    – User198
    Mar 13 at 15:16
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    $\begingroup$ @User198, yes it is a convention. bending moment is positive when it is acting anticlockwise to the right of section. $\endgroup$
    – kamran
    Mar 13 at 17:47

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