# Fully restrained beam moment diagram

1. Jun 25, 2014

### soronemus

1. The problem statement, all variables and given/known data
Find the shear and bending moment diagram for a fully restrained beam with a length of 5m, a uniform distributed load of 1kN/m from x=0 to 2m, a point load of 1kN at 2m, and another uniform distributed load of .9kN/m from x=2, to x=5.

2. Relevant equations

3. The attempt at a solution
I am running into trouble finding the reaction points at A and B (the two fixed ends). The beam is statically indeterminate so I cannot sum the forces in the y direction, and sum the moments about A. I am a mechanical engineering graduate who may have briefly gone over this in my strengths class but that was a long time ago. I was reading up on deflection formulas and if I can get all the information I need if I can write the moment equation, but that is also giving me trouble (I cannot figure out how to break it up correctly to get the free body diagram for each section and obtain the moment equation to integrate/derivate). Any help would be appreciated.

2. Jun 25, 2014

### SteamKing

Staff Emeritus
The following attachment may provide a refresher to help you tackle this problem:

Set up the problem with the normal equations of statics as if the beam were determinate. Because you have several different loads applied to the same beam, writing the slope and deflection equations might get a little complicated. Since linearity applies, you can always treat each load separately, solve for the unknown reactions, and then determine the total reactions for the combined loading.

https://engineering.purdue.edu/~ce474/Docs/Fixed End Moments.pdf

3. Jun 25, 2014

### soronemus

Thanks for the reply, I was looking at the pdf you linked for a while before I posted here and was having trouble following along as they seem to skip steps. In particular, in example 10-4 (which if I can work through is very similar to part of my problem) they go straight to an equation R_a= (Pb/L) + (M_a/L) - (M_b/L) which seems like they are summing the forces in the y direction to get R_a but they are summing moments about two points at once. I am not really sure what they are doing there because I have only ever summed moments about a single point. Along with this they are multiplying the point load by a dimensionless number. Any clarification of that equation would be very much appreciated.

4. Jun 25, 2014

### SteamKing

Staff Emeritus
I understand your confusion, but write the equations of static equilibrium for the beam as shown in Ex. 10-4.

The first equation is simply the sum of the forces acting on the beam:

$∑F = 0$

$R_{A}+P+R_{B} = 0$

The second equation is the sum of the moments acting on the beam. We'll pick point A as the point about which to calculate the moments (CCW moments are positive):

$∑M_{A} = 0$

$M_{A}-M_{B}-Pa+R_{B}L = 0$

Then, solving for $R_{B}$:

$R_{B}=Pa/L-M_{A}/L+M_{B}/L$

$R_{A}$ can then be found by substituting the expression for $R_{B}$ back into the first equation of static equilibrium.

5. Jun 26, 2014

### soronemus

Sweet! I knew it had to be something simple like that. If you have enough patience for another question, where are the 'force-displacement relations' coming from? I have tried google searching but wasn't having much luck on that specific subject as it relates to the form these equations are in. I have worked through several online examples using the differentiation/integration of EIv=bla to get load, sheer, moment, ext, but i'm having trouble with how they arrived at these equations even though it must have something to do with said examples.

6. Jun 26, 2014

### SteamKing

Staff Emeritus
Once you have established the reactions and fixed-end moments for the beam, you can draw the complete shear and moment diagram for the loading. The slope and deflection of the beam as a function of position can then be calculated, by integration along the length of the beam if desired. The force-displacement relations come from solving the fixed-fixed beam with a single loading, say a point load or a distributed load along the full length of the beam.

These functions or formulas used to be compiled into tables which could be found in most strength of materials handbooks like Roarks, but it is hard to find a complete table of both slope and deflections on the internet nowadays, because so few people do hand calculations for beams anymore. The best and most complete table I found was in a Roarks Formulas for Stress and Strain from the mid 1960s; the later editions of this work have tables which are inferior, IMO.

If you scan down from Example 10-4 to the top of p. 11, the formula:

$δ_{C}=PL^{3}/(192EI)$

is the central deflection of a fixed-fixed beam loaded with a single point load as shown at the bottom of the previous page. This formula can be derived by integration, as described above.

7. Jun 30, 2014

### soronemus

The problem i'm running into is that I cannot construct the shear and moment diagram with the information given after they sum the forces and the moments, it is statically indeterminate. I can get equations for the reactions but I have too many unknowns. Are you saying to just construct the shear/bending moment diagram with the equations instead of with actual values? Here is a link to the problem description and a depiction of me trying to break up the problem into separate parts, maybe someone can look at it and show me what i'm doing wrong. The only thing I really need to solve for is the maximum bending moment so I can determine how large of i-beam I need. http://imgur.com/AqpUG5M

8. Jun 30, 2014

### SteamKing

Staff Emeritus
If you want to construct the shear and bending moment diagrams for the beam, you can either determine the end reactions and fixed end moments by integration (which is a lot of calculation), or you can use the tables of Fixed End Moments here:

https://engineering.purdue.edu/~ce474/Docs/Fixed End Moments.pdf

to determine the total end moments on the beam by superposition. Once you have the total fixed end moments, and you know the loads on the beam, you can still write two equations of statics, from which you can determine the unknown reaction forces at the ends of the beam.

Once all of this information has been compiled, you can then construct the SF and BM diagrams for the fixed-end beam, and then carry out your stress analysis.

9. Jun 30, 2014

### soronemus

For my case, (http://imgur.com/XmgNq11) how do I find the equivalent force (P)? I was thinking sum the moments about one of the supports to find a force and distance that gives an equal moment but I would have to use the reaction moment/force at the other end, which I don't know. Is there a simpler way to find the equivalent force? Or a method to combine different loading conditions from the chart you linked? Because there is no case dictated to this loading condition on that page.

10. Jun 30, 2014

### SteamKing

Staff Emeritus
If I understand the beam loading from your post #7, you have a 1000 N force located at 1.9 m from the left end of the beam, a UDL of 950 N/m which extends from the left end of the beam to 1.9 m, and another UDL of 800 N/m extending from 1.9 m to the right end of the beam.

In order to calculate the Fixed End Moments using the tables provided, you do not need to calculate equivalent loads. You select the particular loading (single point load, full UDL, partial UDL, etc.), calculate the FEM for that load case, and repeat the process for the additional load cases. The total FEM is found by superposing the FEMs from all of the individual load cases.

For the case of the 1000 N load, the FEM at A is [1000*(1.9)*(5-1.9)^2]/5^2 N-m
[Case No. 2 from the top left hand side in this link: https://engineering.purdue.edu/~ce474/Docs/Fixed%20End%20Moments.pdf] [Broken]

In the image you linked to in Post #9, the FEM formulas are located adjacent to the curved arrows, while the fixed-end reaction formulas are located on the bottom.

There may be additional load cases from the same source from which you obtained the image in Post #9. If so, you would have the means to solve your problem. A note of caution: the fixed end moments and reactions due to concentrated loads are different from those produced by UDLs, which is why you shouldn't use equivalent loads to calculate these quantities.

Last edited by a moderator: May 6, 2017
11. Jul 3, 2014

### soronemus

When combining the fixed end reactions using superposition, are you just adding them? For example the moment at A would be the sum of the moments at A for each of the three loads? If so I think I have finished. Here is a link to a google doc containing the shear and bending moment diagram I came up with.

I summed the forces and moments and they all came to zero which makes me think I did it correctly. Thanks again for all your help, it was extremely appreciated. I wish I knew how to come up with the equations for determining the reactions given in the chart you linked but for now I will settle for trusting the chart lol.

12. Jul 3, 2014

### SteamKing

Staff Emeritus
Yep, superposition means you add reactions, moments, deflections, whatever, together. As long as your beam is statically determinate with the fixed-end reactions and moments (I checked your calcs.), you have solved it.

The fixed-end reactions and moments can be determined by integration, and this problem is treated in most basic introductory strength of materials texts.

Here are a couple of slide shows illustrating the integration method. In the first slide show, Example 3, slide 22, shows how to calculate FEM for a fixed-fixed beam with a UDL:

http://am.hit.edu.cn/courses/mechmat2012/Courseware_files/19_bendsi_print.pdf [Broken]

http://www.assakkaf.com/courses/enes220/lectures/lecture18.pdf

By using a different loading on the beam, you can create (or check) your own table of fixed-end reactions and moments.

Last edited by a moderator: May 6, 2017