# Slope of the Bending moment diagram where a point force acts

Consider a beam that is arbitrarily loaded and at any distance x, we take an element in the beam between two sections, where a point force acts as shown.

Let us assume the directions of shear force and bending moment to be as shown in the figure.

Applying the equation of equilibrium we get,

$$V_1-P-V_2=0$$ $$V_2=V_1-P$$

Thus, $$V_x=+V_1$$ and $$V_{x+dx}=+(V_1-P)$$

Note: Shear force which tries to rotate the element clockwise is taken as +ve and Bending moment which tends to compress the upper part of the beam is taken as positive.

We conclude that the shear force abruptly changes as we go from left to right across the point force.

Taking moment about a point on the right face of the element:

$$-M_1-V_1dx+Pdx/2+M_2=0$$

$$M_2=M_1+V_1dx-Pdx/2$$

Thus, $$M_x=+M_1$$ and $$M_{x+dx}= +(M_1+V_1dx-Pdx/2)$$

The change in the bending moment $$M_2-M_1$$ is differential, thus as we go from left to right across a point force the bending moment essentially remains constant.

## My trouble starts from here:

If I try determining the slope of the BMD at x,

$$\frac{dM}{dx}=\frac{M_{x+dx}-M_x}{dx}=\frac{+(M_1+V_1dx-Pdx/2)-+M_1}{dx}=V_1-P/2$$

However the book I'm referring to says that - "Even though the bending moment M does not change at a concentrated load, its rate of change dM/dx undergoes an abrupt change".

I don't get how the slope undergoes an abrupt change, as from the analysis the slope is turning out to be $$V_1-P/2$$ at x.

Because (to my understanding your problem is more in the interpretation of derivatives rather structural mechanics) I will try to explain it in parallel example. Hopefully it will make sense.

Imagine a car accelerating with $$a [m/s^2]$$, and at some point in time t suddenly decelerating at $$- a [m/s^2]$$.

The acceleration in time t will abruplty change from positive to minus.

IF the velocity in time $$t$$ was $$v_0$$ then the velocity at time $$t+\delta t$$ ill be $$v(t+\delta t) = v_0 - a \delta t$$. So only a miniscule difference.

You can apply the same logic if you equate:

• shear force (V) with acceleration
• moment (M) with velocity
• x direction with time t

the derivative of velocity with respect to time is Acceleration .

$$a = \frac{\delta V}{\delta t}$$

The derivative of bending moment with respect to distance x is the shear force.

$$V = \frac{\delta M}{\delta x}$$

• I looked into what you're saying and yes I feel there is a flaw in my math. I came up with something, let me know if it makes sense to you. I'm determining the derivative without knowing whether the BM function is differentiable at x. If I evaluate the derivative at $x^-$ and $x^+$ I will get $V1$ and $V1-P$ respectively. Since, derivative at $x, x^-,x^+$ is not equal thus the BM function is not differntiable at x. Commented Jan 13, 2022 at 19:35
• @HarshitRajput to be honest a) I never really thought about it that way, b) I wouldn't be able to express it precisely in mathematical terms, because I am not that good at mathematical definitions. When I answered I was focusing on the physical interpretation of the discontinuity of the derivative.
– NMech
Commented Jan 13, 2022 at 23:59