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The book I'm referring to for studying beams, said that its quite cumbersome to construct shear force and bending moment diagrams, by writing the SF and BM equations for each segment of the beam and then plotting the equations. To solve this problem we can form relationships between shear force, Bending moment and loads and then use them to construct SFD and BMD.

So, the book starts off by developing a relationship between shear force and load. In doing so, it says that consider a beam with some arbitrary distributed load. At any distance x, we take an element between two sections. The distributed load on this element can be assumed uniform with intensity say, $q$.

enter image description here

Then it says that shear force and bending moments will be developed at each section of this element, and in general they can vary along the length. The book then assumes the direction of shear force and BM to be what it considers "positive".

The author has taken shear force as positive if it tries to rotate the element clockwise, and BM positive when it tries to compress the upper part of the element.

applying the equilibrium condition to the element yields,

$$V_1 - V_2-qdx=0$$ $$V_2-V_1=-qdx$$ $$\frac{V_2-V_1}{dx}=-q$$

Thus $V_2-V_1$ will be differential

$$\frac{dV}{dx}=-q$$

The book came at this result by taking the SF and BM as positive (as is shown in the figure).

If I take the direction of SF and BM opposite to what the books takes (i.e. shear force tends to produce an antickw rotation of the element) I get

$$\frac{dV}{dx}=q$$

The relations are different depending on what sign of SF and BM I take. However I have seen the book using the first relation (with a -ve q) even in places where the element is acted upon by -ve shear force (i.e. when the shear force tends to rotate the element antickw)

Why the equation $$\frac{dV}{dx}=-q$$ applies even when the shear force tends to produce an antickw rotation, even though it was derived for clockwise rotation shear force.

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The reason the convention picks the shear positive when on a differential element it tries to rotate clockwise is that this rotation will produce a positive conventional moment, bending the beam into a smile.

This convention makes the calculation easy and because conventionally a horizontal beam is analyzed from left to right and the positive shear direction is vertically up, this convention works.

Why the equation $\frac{dV}{dx}=−q$ applies even when the shear force tends to produce an antickw rotation, even though it was derived for clockwise rotation shear force.

Because if you have a section like this it means either q is applied to head up or $V2<V1$ then we have $\ -(-q)=q\ $ and this will cause an anticlockwise spin. But still, if we stay with convention it will work.

you can Pick up any direction as positive as long as you stay consistent throughout the length of the beam.

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  • $\begingroup$ I saw a case where q can act downwards and we can have -ve SF (rotates element ACW). I'm sharing the image - drive.google.com/file/d/1H29y49voSlo4u07lxALjd35Oz6-BOYvW/… $\endgroup$
    – Harshit Rajput
    Commented Jan 8, 2022 at 18:04
  • $\begingroup$ In this image, if I consider an element dx, in the portion where the shear force is -ve, then the free body diagram (haven't shown moments) of the element will be as shown (SF gives an ACW rotation). Applying equilibrium equations, we get $V_2-V_1=qdx$ which shows that $V_2-V_1$ will be positive, but that is not what the SFD says. SFD says $V_2-V_1$ will be negative. There is a contradiction, so there's definitely some loophole in my understanding. $\endgroup$
    – Harshit Rajput
    Commented Jan 8, 2022 at 18:06
  • $\begingroup$ @HarshitRajput, you have picked a region where both v1 and v2 are negative. but v2 is greater in absolute value, so v2-v1 is still negative. The right half-length of the beam the shear is all negative because it has already canceled the support reaction at left, wl/2. $\endgroup$
    – kamran
    Commented Jan 8, 2022 at 18:23
  • $\begingroup$ Oh Got it, the equilibrium equation that I have written which has V1 and V2, those are magnitudes. so V2-V1=qdx represents that the absolute value of V2 is higher than V1. Right on this? $\endgroup$
    – Harshit Rajput
    Commented Jan 8, 2022 at 18:35
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    $\begingroup$ yes, The absolute value is higher but the algebraic value is less. In engineering, a big part is keeping track of signs. $\endgroup$
    – kamran
    Commented Jan 8, 2022 at 18:39
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"The author has taken shear force as positive if it tries to rotate the element clockwise,..."

This sign convention indicates the positive direction of force is up, and the couple produces a CW rotation about the center of the small element. "q" is assumed to have the same direction of $V_2$ (pointing down) as shown in the first equilibrium equation, and it produces no rotation about the center of the small element.

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