# Why is the stress different when different methods are used?

A cantilever beam, and a uniform load acting on the beam (along with its intensity), is shown in the first picture. Its free body diagram and 2 diferent methods to solve for the stress at the site shown by dotted lines is shown in the second picture.

I - (Second figure of second picture) If we first calculate the resultant of the uniform load, the magnitude of stress would be the vector sum of R1 and P(x+l) (Resultant of entire load). (The position of the resultant would surely be in the left section).

II - (Third figure of second picture) If we calculate the resultant after separating into sections, the magnitude of stress would be the vector sum of R1 and P(x).

What is the change to be made to obtain the same stress by both methods ?

Edit: The stress referred is shear stress.

The reaction moment is not shown because it is irrelevant for calculating the stresses (shear) in this case.

• Something is wrong. The top diagram shows a cantilever beam but the top figure of the bottom diagram appears to be a simply supported beam. Are you sure these are part of the same problem? Aug 29, 2019 at 9:34
• Actually, I didn't show the the horizontal reaction and the reaction moment because I thought that they were irrelevant for calculating the stresses.
– Zam
Aug 29, 2019 at 9:36
• Who’s talking about a horizontal reaction? I see no horizontal load in these diagrams. I’m saying there is no $R_2$ reaction in the top (cantilever) beam so why is it shown in the bottom diagram? Aug 29, 2019 at 9:47
• @Bob D Sorry. It was my mistake. I corrected (edited) it.
– Zam
Aug 29, 2019 at 10:07
• I found Beer and Johnston to be good, even over 50 years ago when I took statics. Aug 29, 2019 at 13:05

I - (Second figure of second picture) If we first calculate the resultant of the uniform load, the magnitude of stress would be the vector sum of R1 and P(x+l) (Resultant of entire load). (The position of the resultant would surely be in the left section).

The second figure is incorrect. You are showing the entire load $$P(x+L)$$ as acting to the left of the cut section but only part of the load acts to the left of the section. The total load to the left of the section is $$Px$$.

II - (Third figure of second picture) If we calculate the resultant after separating into sections, the magnitude of stress would be the vector sum of R1 and P(x).

The third figure is correct, for the reason I stated above regarding the second figure. But you should show the vertical shear at the section acting downwards, as shown in my diagrams below, in order for the FBD to be complete.

The basic mistake you are making is using the total equivalent load when analyzing shear. After using the total equivalent load in determining reactions, you must revert to the distributed load when determining bending moment and shear force anywhere in the beam.

There are two steps in solving this type of problem.

The first step is to determine the reactions at all supports needed for static equilibrium. For that purpose you can replace the distributed load with a single equivalent force located midpoint of the beam, which is what you did. Then you can determine the moment and upward vertical reactions at the cantilever support at the left. I show the applicable FBD in the upper left diagram below.

The next step is to determine the shear force (and bending moment, if desired) at any desired section, in your case indicated by the dotted line. Now you must remove the equivalent concentrated load and replace it with the distributed load. I show that in the upper right figure.

Next cut the beam at the desired section and analyze the free body diagram for either the left or right sections. I show that in the bottom figure. In this case the simplest approach is to use the FBD for the section at the right. The total load to the right of the section equals $$PL$$ acting downward. For there to be equilibrium the vertical shear acting up must equal $$PL$$.

Is there a reason why the total equivalent load cannot be used while analyzing shear ? Considering that the usage of the total equivalent load (like centre of mass) do not make any difference to the model and is just a simplification of the model, why can't it be used ?

The short answer is because you will get the wrong value for shear, unless you happen to be only interested in the shear at the middle of the beam where the load on each side is the same.

As I showed, the shear at the section of interest equals $$PL$$ based on the amount of the load to the right of the section. You should get the same answer for the left side of the section, but if you use the load shown in your middle diagram, $$P(x+L)$$, you will not get the same answer. That is because $$P(x+L)$$ is the total beam load, not just the load to the left of the section which should be $$Px$$. If you solve for the downward shear force $$V$$ on the section using $$P(x+L)$$ in the middle diagram, you will get $$V=0$$, when it should be $$VL$$. If you use $$Px$$ the sum of the forces will be

$$+R_{1}-Px-V=0$$ $$+P(x+L)-Px-V=0$$ $$V=PL$$

Which is the same as shear on the right section.

A word of advice. If you wish to progress in statics and mechanics of solids, you will get wrong answers if you leave in concentrated loads for distributed loads when analyzing the internal bending moments and shear forces in beams. Use them only for determining reactions at supports. Then revert back to the actual loading configuration.

Hope this helps.

• Thanks. Please also add: Is there a reason why the total equivalent load cannot be used while analyzing shear ? Considering that the usage of the total equivalent load (like centre of mass) do not make any difference to the model and is just a simplification of the model, why can't it be used ?
– Zam
Aug 29, 2019 at 14:58
• @Zam OK I will revise my answer in a bit to respond Aug 29, 2019 at 17:02
• @Zam See my revised answer. I hope this does it for you. Aug 29, 2019 at 20:39