Help understanding why bending moment is integral of shear

Question

I am aware of the basics of Euler-Bernoulli beam theory but there is one thing that I have never understood satisfactorily. Where $M$ is the bending moment, and $V$ the shear force:

$$\frac{dM}{dx}=V\,\text{ or }M(x)=\int_0^x V(t)\,dt\qquad(*).$$

The explanation I see most widely posits an element of the beam and says that it is equilibrium and so the sum of the moments about a point is zero. For example, a small element with extents $x$ and $x+\Delta x$, and then it calculates the moment about the midpoint. Say from Figure 8.6.2 here.

Where we have $V(x+\Delta x)=V(x)+\Delta V=V+\Delta V$ and similar for $M(x+\Delta x)$, I understand where the terms $-V\cdot \Delta x/2$ and $-(V+\Delta V)\cdot \Delta x/2$ come from but for the life of me I don't understand how we can just include the moments at $x$ and $x+\Delta x$ into this sum: $$-V\cdot \frac{\Delta x}{2}-(V+\Delta V)\cdot \frac{\Delta x}{2}-M+(M+\Delta M)=0.$$

Can someone either explain why this makes sense or perhaps come up with a different explanation of either of (*).

I can kind of feel that we are adding up all the moments in the element... but in my head $M$ is somehow already included via $-V\cdot \Delta x/2$ and $M+\Delta M$ comes into $-(V+\Delta V)\cdot \Delta x/2$

Answer 1

Let $\Delta V = w_x*\Delta x$, and $\sum M_B = 0$

$- V_x*\Delta x -\Delta V*\frac{\Delta_x}{2} - M_x + M_{x + \Delta x} = 0$

$- V_x*\frac{\Delta_x}{2} - V_x*\frac{\Delta_x}{2} - \Delta V*\frac{\Delta_x}{2} - M_x + M_{x + \Delta x} = 0$

$-V_x*\frac{\Delta_x}{2} - (V_x + \Delta V)*\frac{\Delta_x}{2} - M_x + M_{x+\Delta x} = 0$

Note $M_{x+\Delta x} = M_x + \Delta M$

$-V_x*\frac{\Delta_x}{2} - (V_x + \Delta V)*\frac{\Delta_x}{2} - M_x + (M_x+\Delta M) = 0$, which is identical to

Numerical Example:

$V_A = 1$, $\Delta V = 0.1$, $V_B = -1.1$

$M_B = M_A + V_A*(1) + \Delta V * (1/2) = 2.05$, or

$M_B - M_A - V_A*(1) - \Delta V * (1/2) = 0$

Since $\sum F = 0$ and $\sum M = 0$, the beam segment is in equilibrium.

Answer 2

Short answer to address your confusion is that moment $M(x)$ and transverse force $V(x)$ are basically independent quantities if you are looking just at a specific coordinate $x$ (e.g. there will be sections, where $M$ is zero and $V$ is not and also vice versa). In other words, there is no proportionality between these quantities $M(x)\propto V(x)$.

So your intuition that $M$ is somehow already included in the equilibrium using $-V\cdot \Delta x/2$ is not correct. (This would require the above proportionality to be true.)

Answer 3

I still don't understand the conventional explanation given by r13 but their patience got me there in the end.

Definition: The bending moment at $x$ is the sum of all the moments about $x$ to the left of $x$.

There is a moment at $x=0$ called $M_A$. Cut the beam from $z=0$ to $z=x$ into elements of width $\Delta z$. Each element comprises a force of $w(z)\Delta z$ and the distance between the element and $x$ is $x-z$ to give a moment due to the element of:

$$\Delta M=w(z)\Delta z(x-z).$$

Sum these (passing to the integral):

$$ \begin{aligned} M(x)&=M_A+\int_{z=0}^{z=x}w(z)(x-z)\,dz \\ &=M_A+\int_0^x xw(z)\,dz-\int_0^xzw(z)\,dz \\ &=M_A+x\int_0^xw(z)\,dx-\int_0^x zw(z)\,dx \\&=M_A+x\cdot V(x)-\int_0^x zw(z)\,dz \end{aligned}$$ Consider $\int_0^x z\,w(z)\,dz$ with parts using $u=z$ and $dv=w(z)\,dz$ so that $du=dz$ and $v=V(z)$: $$\begin{aligned} \int zw(z)\,dz&=zV(z)-\int V(z)\,dz \\ \implies\int_0^xzw(z)\,dz&=xV(x)-\int_0^xV(z)\,dz \end{aligned}$$

Plugging into the above yields: $$M(x)=M_A+\int_0^x V(z)\,dz\implies \frac{dM}{dx}=V(x),$$ as required.