How to Calculate Vertical Reactions and UDLs for a Beam?

Click For Summary

Discussion Overview

The discussion focuses on calculating vertical reactions at points A and C for a beam subjected to a uniformly distributed load (UDL). Participants explore methods for solving the equations of equilibrium, particularly the application of moments about specific points.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant states the need to calculate vertical reactions at points A and C, noting the challenge of having two variables in the equations.
  • Another participant suggests taking moments about points A and C to solve for the unknown forces, emphasizing the general rule of equilibrium.
  • A participant questions the correctness of their moment equation for vertical reaction at C, presenting a specific equation involving various forces and distances.
  • Another participant confirms the correctness of the moment equation but points out a minor error regarding the sign of one term.
  • One participant proposes a method for calculating the moment created by the UDL, suggesting it should be based on the load multiplied by the distance and the halfway point, leading to a different moment value than previously calculated.
  • A later reply acknowledges the oversight in the units and confirms the need to multiply the load by the length to obtain the correct moment value.

Areas of Agreement / Disagreement

Participants express some agreement on the methods for calculating moments and vertical reactions, but there are differing interpretations regarding the calculations and the application of the UDL, indicating that the discussion remains unresolved.

Contextual Notes

Participants reference specific distances and forces without fully resolving the implications of their calculations, and there are indications of potential misunderstandings regarding units and moment calculations.

Ry122
Messages
563
Reaction score
2
I have to calculate the vertical reactions at point A and C.
I know that the sum of all vertical forces is equal to zero but i can't solve the equation since there's two variables.
http://users.on.net/~rohanlal/Q3.jpg
Bv=50
Cv=55
Dv=45
w=80
 
Last edited by a moderator:
Physics news on Phys.org
Ry122 said:
I have to calculate the vertical reactions at point A and C.
I know that the sum of all vertical forces is equal to zero but i can't solve the equation since there's two variables.

Hi Ry122! :smile:
general rule … when you have a unknown force you can't get rid of, take moments about a point (strictly, an axis) through the line of that force.​
In this case, take moments about A and about C.

(Alternatively, find the point about which the moment of all the given forces is zero :wink:)
 
Is this correct for the vertical reaction at C?
0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)
 
Ry122 said:
Is this correct for the vertical reaction at C?
0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)

Hi Ry122! :smile:

Yes … you've taken moments about A …

and everything's correct (except you need a minus in front of the 15/2*80 :wink:) :smile:

(I suggest you take moments about C also, just for practice … but of course, as I expect you've noticed, you don't need to, because you already know the total reaction for A and C together :wink:)
 
for the UDL wouldn't the moment created about point A be the load (80) multiplied by the distance it covers (15m) then multipied by the half way point (7.5m) giving you a moment of 9000 KN/M about point a?

rather than what you have with 7.5 x 80 giving you a moment of 600 KN/M about point a?
 
Welcome to PF!

badeany said:
for the UDL wouldn't the moment created about point A be the load (80) multiplied by the distance it covers (15m) then multipied by the half way point (7.5m) giving you a moment of 9000 KN/M about point a?

rather than what you have with 7.5 x 80 giving you a moment of 600 KN/M about point a?

Hi badeany! Welcome to PF! :smile:

ooh … you're right!

I didn't notice the small print ("kN/m") on the diagram! :redface:

Yes, 80 is the force-per-length, so it does have to be multiplied by the length. :wink:

Thanks, and very well spotted! :biggrin:
 

Similar threads

Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
Replies
11
Views
144K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K