# Why do we use simple bending equation when bending moment is caused due to shear force?

Simple bending equation is derived assuming pure bending(when a couple is applied onto a beam with no shear stress).We tend to use the above derived equation when moment is caused due to shear stress?How far is this correct?

• Look up "St Venant's Principle" on the web, or in your textbook. – alephzero Oct 7 '18 at 9:46
• What do you mean by "simple bending equation"? Do you mean the one that calculates the stress due to bending moment? – Wasabi Oct 7 '18 at 12:29
• M/I = sigma/y = E/R – Shwetha Oct 7 '18 at 13:26
• Theory of simple bending in strength of materials – Shwetha Oct 7 '18 at 13:27
• @alephzero While not wrong, your comment isn't likely to be helpful to anyone who doesn't already understand complex beam theory. How do remoteness/proximity and decay rates come into play, and what are they being applied to? – Phil Sweet Oct 7 '18 at 16:23

We only can ignore the shear when the effects it has on shape and displacement are small enough to not be a concern. In general, we can't ignore them. If we ignore shear, then we are saying that the beam ends (and all cross sections) are constrained to remain perpendicular to the beam's deflected surfaces. Real shear stresses don't let that happen. But for a long beam relative to the thickness, the shear strain can't develop very quickly along the length and the perpendicular assumption is mostly true. But for short, thick beams, shear strain has a large impact on the deformed shape.

One of the first comprehensive approaches to handle this was Timoshenko beam theory.

The equation you're talking about is the following:

$$\sigma = \dfrac{My}{I}$$

The reason this equation only considers bending moment even though "moment is caused [by] shear stress", is because, well, that fact is irrelevant. There is no relationship between a given shear force and the resultant bending moment, since different beams (with different spans, support conditions, etc) might have the same shear force at a given point but wildly different bending moments.

Bending moment causes stresses according to the equation above. Shear stress causes other stresses which follow a different behavior. The effects are completely independent.

Also, while bending moment generates longitudinal stresses (parallel to the beam's central axis), shear stress is transversal (perpendicular to the central axis... or at an angle to it).

We also choose to separate these effects because it makes life easier. For example, look at this beam:

Shear diagram

Bending moment

We can separate this beam into two parts: between the supports and the loads, and the midspan between the loads.

Looking at the midspan, we have bending moment but no shear force. So whatever stresses are felt by the beam in that region are entirely due to the bending moment. So just use the equation above.

Now, looking at the area between the supports and the loads, we have both bending moment and shear force. So the beam's internal stress state in this region is a combination of both forces. But still, you can just calculate the bending moment component using the equation above and others for the shear force component.

To be clear, you can't just add those results since they operate in different orientations, but you could combine them to obtain the principal stresses, we just don't bother doing so.

Basically, it is easier to work with each force component in isolation than simultaneously. And given that there's no benefit to doing it all at once, we choose the easier route. (Obviously, in cases where the components actually influence each other, we take that into consideration, such as cases of flexo-compression or -tension).

• In the area between the supports and the loads,couple is varying.we derived the equation for constant couple.Shouldnt this bring a lot of difference? Could you elaborate on this? – Shwetha Oct 7 '18 at 15:03
• @Shwetha, the couple is varying along the beam's length, but this equation only describes a single point along the beam. If you're looking at point $x=0.5m$, the couple is constant (duh!) and this equation is a perfect representation of the stresses at that point. If you're looking at $x=0.6m$, the couple is different and this equation is a perfect representation of the different stresses at that point. – Wasabi Oct 7 '18 at 15:18
• @Shwetha: Wait a day or so to see if someone writes a better answer, but if not, please click the checkmark next to my answer to set this question as "Answered". – Wasabi Oct 7 '18 at 15:58
• Sure will do that – Shwetha Oct 7 '18 at 16:32