Fixed end moments with a beam with udl

In summary, the conversation discusses a beam with fixed ends and a UDL of "W" per unit length. The question arises about the end moments, with one person suggesting they should be W L^2/12 while the other suggests W L^2/4. The conversation also touches on the difference between fixed end beams and simply supported beams, and a request for a solution or link to solving a similar problem. An excerpt from Schaum's Outline of Strength of Materials is also mentioned.
  • #1
date.chinmay
10
0
I came across a reference where there is a beam (lets say length L)
both ends fixed... with a UDL of say "W" per unit length... i want to know how the end moments are W L^2/12 ...shudnt they be W L^2/4??


Its pretty simple mechanics. but it got me wondering,,,,
 
Physics news on Phys.org
  • #2
In the fixed end beam, both the slope and deflection at the ends must equal zero. In a simply supported beam, there are different end conditions and the moments are typically zero, unless there is also an applied couple.
 
  • #3
nice,,,
i wondered if u cud provide a solution or a link to the actual stepwise solving of one...
 
  • #4
Here is an excerpt from a Schaum's Outline of Strength of Materials (prob. 11.6):

This problem shows a beam which is partially loaded with a UDL. The technique can be used when the UDL is extended the full length of the beam.
 

Attachments

  • Schaum's Outlines - Strength of Materials300-01.pdf
    267.2 KB · Views: 883
  • #5


I can provide a response to this content. The end moments in a beam with both ends fixed and a uniform distributed load (UDL) of W per unit length are indeed W L^2/12. This is because in this scenario, the beam experiences a negative bending moment at both ends, which cancels out the positive bending moment created by the UDL. This results in a net end moment of W L^2/12 at each end of the beam.

On the other hand, if the beam had only one fixed end and one pinned end, the end moments would be W L^2/8. This is because in this case, the fixed end would experience a larger negative bending moment due to the UDL, while the pinned end would experience a smaller positive bending moment. The difference between these two moments results in an end moment of W L^2/8 at the fixed end.

In summary, the end moments in a beam with both ends fixed and a UDL of W per unit length are W L^2/12, while the end moments in a beam with one fixed end and one pinned end are W L^2/8. This is a result of the distribution of bending moments along the beam, which is influenced by the type of support at each end. I hope this explanation helps to clarify your understanding.
 

Related to Fixed end moments with a beam with udl

1. What is a fixed end moment?

A fixed end moment is a bending moment that occurs at the fixed ends of a beam. It is caused by the reaction forces from the supports and any applied loads on the beam.

2. How is a fixed end moment calculated?

A fixed end moment is calculated by multiplying the reaction force at the support by the distance from the support to the point of interest on the beam. This is then repeated for each support and the moments are summed together.

3. What is a beam with udl?

A beam with udl (uniformly distributed load) is a type of load that is evenly distributed along the length of the beam. This can be represented by a straight line on a load diagram.

4. How do fixed end moments affect a beam with udl?

Fixed end moments can cause the beam to bend and experience stress at the points where they occur. This can have an impact on the overall deflection and strength of the beam.

5. Can fixed end moments be negative?

Yes, fixed end moments can be either positive or negative, depending on the direction of the applied loads and the orientation of the beam. Negative moments will cause the beam to bend in the opposite direction compared to positive moments.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
8K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
138K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
8K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
Back
Top