Drawing Bending Moment & Shear Force for a Beam w/ Constant Moment

Click For Summary

Discussion Overview

The discussion revolves around drawing bending moment and shear force diagrams for a beam subjected to a constant moment. Participants explore the implications of a constant moment on the bending moment diagram and how it interacts with shear force calculations. The scope includes theoretical and mathematical reasoning related to structural analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks assistance in drawing the bending moment diagram, expressing uncertainty about applying the area under the shear force diagram rule due to the presence of a constant moment.
  • Another participant explains that a constant moment creates a "discontinuity" in the bending moment diagram, similar to how loads affect the shear diagram, and describes how to adjust the diagram based on the direction of the moment.
  • There is a question about whether a constant moment at a specific point should be added or subtracted in the diagram, indicating a need for clarification on this point.
  • A participant proposes a formula for calculating the bending moment under a uniformly distributed load and questions how to incorporate a constant moment into this equation.
  • Another participant confirms that the constant moment should be added to the bending moment calculation if it is clockwise.

Areas of Agreement / Disagreement

Participants express differing views on how to accurately represent the constant moment in the bending moment diagram. While some agree on the need to adjust the diagram based on the moment's direction, there is no consensus on the overall approach to integrating this into the bending moment calculations.

Contextual Notes

The discussion includes assumptions about the behavior of bending moments and shear forces, but these assumptions are not universally agreed upon. The application of the area under the shear force diagram rule in the context of a constant moment remains a point of contention.

frozen7
Messages
163
Reaction score
0
Can anyone draw me a correct Bending moment diagram and shear force diagram for this beam?
i have drawn the shear force diagram for this case but I do not know how to draw the bending moment disgram for this case because this question involve a constant moment. Normally the rule for bending moment is the value of bending moment at certain point is the area under shear force diagram from the starting point to that certain point. Can this rule applied on this question as well? I felt the bending moment diagram is quite strange if follow this rule for this question. Can anyone help me?
Thanks.
 

Attachments

  • beam.JPG
    beam.JPG
    29.8 KB · Views: 568
Physics news on Phys.org
Hello, the only thing a couple or moment does for the bending diagram is create a "discontinuity" (more like a derivative doesn't exist at said point), just like the loads do on the shear diagram. If the moment is clockwise it goes up (adds to the area [method]), and if its counterclockwise it goes down (substract to the area [method]).
 
Last edited:
Does it mean when there is a constant moment at point b, then we should either add or substract that value of constant moment in that point (either draw a straight line goes up or goes down)?
 
frozen7 said:
Does it mean when there is a constant moment at point b, then we should either add or substract that value of constant moment in that point (either draw a straight line goes up or goes down)?

exactly what i said above.
 
One more thing have to be confirmed.
For uniformly distributed load, Mx = (Ay)(x) - (wx^2)/2 (where x is the distance from staring point to x, Ay is the reaction force at the starting point and w is the force per unit length and Mx is the bending moment)
Let say if there is a constant moment which is 50N/m at point a ,then Mx become Mx = (Ay)(x) - (wx^2)/2 +50 ??
 
frozen7 said:
One more thing have to be confirmed.
For uniformly distributed load, Mx = (Ay)(x) - (wx^2)/2 (where x is the distance from staring point to x, Ay is the reaction force at the starting point and w is the force per unit length and Mx is the bending moment)
Let say if there is a constant moment which is 50N/m at point a ,then Mx become Mx = (Ay)(x) - (wx^2)/2 +50 ??

Yes +50 if its clockwise.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K