Reaction Forces on a Frame With Roller Joint and Pin Joint

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SUMMARY

The discussion focuses on calculating reaction forces in a roof support frame with a frictionless pin joint at point A and a roller joint at point B. The vertical reaction forces were determined to be 5.4 kN at point B and 8.6 kN at point A. The horizontal reaction force at point A was calculated to be 0.5 kN directed to the left. Participants emphasized the importance of summing forces in both x and y directions to ensure equilibrium in the frame.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of reaction forces in structural analysis
  • Familiarity with moment calculations in mechanics
  • Experience with frame analysis techniques
NEXT STEPS
  • Study the method of joints in frame analysis
  • Learn about calculating horizontal reaction forces in structures
  • Explore the application of the moment method in structural mechanics
  • Review examples of static equilibrium problems involving pin and roller supports
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Civil engineers, structural analysts, and students studying mechanics of materials will benefit from this discussion, particularly those focused on calculating reaction forces in framed structures.

AliesAbawe
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1. Hi, The problem:
A roof support frame has various leadings on it. The frictionless pin joint at points A has vertical and horizontal reaction forces. The rollers at point B allow free movement in the horizontal direction and has a vertical reaction.

a/Calculate The total Horizontal reactions in the pin at point A in KN, and the direction as indicated by the arrow in the answer?

b/Describe the vertical reaction at point B in KN?

c/ What is the vertical reaction at point A in KN?



2. Moment = Force * Distance



3. I have attempted this question, but only as if it were a beam, and for the Vertical Force at B i got 5.4KN, and for The Vertical force at A i got 8.6KN. I don't know what to do with the horizontal forces and distances.
 

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AliesAbawe said:
1. Hi, The problem:
A roof support frame has various leadings on it. The frictionless pin joint at points A has vertical and horizontal reaction forces. The rollers at point B allow free movement in the horizontal direction and has a vertical reaction.

a/Calculate The total Horizontal reactions in the pin at point A in KN, and the direction as indicated by the arrow in the answer?

b/Describe the vertical reaction at point B in KN?

c/ What is the vertical reaction at point A in KN?



2. Moment = Force * Distance



3. I have attempted this question, but only as if it were a beam, and for the Vertical Force at B i got 5.4KN, and for The Vertical force at A i got 8.6KN. I don't know what to do with the horizontal forces and distances.
as long as the roof beam is rigidly connected to the columns, you can calculate the support reactions as a frame by summing moments about a point = 0 and summing forces in x and y directions = 0. You are missing a distance for the 4 kN force from the left end, so I can't check your work. Please show how you arrived at your numbers. Indicate support reaction directions.
 
PhanthomJay said:
as long as the roof beam is rigidly connected to the columns, you can calculate the support reactions as a frame by summing moments about a point = 0 and summing forces in x and y directions = 0. You are missing a distance for the 4 kN force from the left end, so I can't check your work. Please show how you arrived at your numbers. Indicate support reaction directions.

The distance of the 4KN from Point A is 0m.

The support reaction direction have been indicated on the attachment i posted.

Working out for the Vertical Forces:

Taking moments about point A, taking anti clockwise as positive:

By*10-2*9-5*6-3*2-4*0 = 0

By = (18+30+6+0)/10 = 5.4KN

Taking moments about point B, taking clockwise as positive:

Ay*10-4*10-3*8-5*4-2*1 = 0

Ay = (40+24+20+2)/10 = 8.6KN
 
AliesAbawe said:
The distance of the 4KN from Point A is 0m.

The support reaction direction have been indicated on the attachment i posted.

Working out for the Vertical Forces:

Taking moments about point A, taking anti clockwise as positive:

By*10-2*9-5*6-3*2-4*0 = 0

By = (18+30+6+0)/10 = 5.4KN

Taking moments about point B, taking clockwise as positive:

Ay*10-4*10-3*8-5*4-2*1 = 0

Ay = (40+24+20+2)/10 = 8.6KN
You forgot to include the moments from the horizontal forces on the columns. It is essential that these be included in your calcs for the support reactions.
Redo the work, and check results for the vert reaction forces by summing forces in y direction = 0. What about the horizontal reaction force at A?
 
PhanthomJay said:
You forgot to include the moments from the horizontal forces on the columns. It is essential that these be included in your calcs for the support reactions.
Redo the work, and check results for the vert reaction forces by summing forces in y direction = 0. What about the horizontal reaction force at A?

The problem in the first place was, i don't know how to calculate the horizontal forces, what distances to use etc in order to find the reaction force at A, I want to learn how to do it. i have no knowledge on doing horizontal reaction forces unfortunately :/
 
AliesAbawe said:
The problem in the first place was, i don't know how to calculate the horizontal forces, what distances to use etc in order to find the reaction force at A, I want to learn how to do it. i have no knowledge on doing horizontal reaction forces unfortunately :/
The only knowledge you need is that the sum of all forces in the x direction must be equal to 0. Since there is no horizontal force reaction at B ( which is a roller support free to slide and which cannot provide horizontal force support), then all of the horizontal force reaction must be at A, which is ? KN in the ? direction??
 
PhanthomJay said:
The only knowledge you need is that the sum of all forces in the x direction must be equal to 0. Since there is no horizontal force reaction at B ( which is a roller support free to slide and which cannot provide horizontal force support), then all of the horizontal force reaction must be at A, which is ? KN in the ? direction??

ohhhh i get it, must be equal to 0.5KN since 2-1.5= 0.5KN at ---> direction, is that right?, and can the moments method be used aswell? and if so could you explain. thanks!
 
AliesAbawe said:
ohhhh i get it, must be equal to 0.5KN since 2-1.5= 0.5KN at ---> direction, is that right?, and can the moments method be used aswell? and if so could you explain. thanks!
Well yes it is 0.5 kN to the left...show it as
0.5 kN <--- rather than -0.5 kN --->, the minus sign is already too confusing in itself. You can sum moments about say the top left corner of the frame to solve for it, now that you know B_y, but it is more work and errors are more likely, but give it a shot if you want it as a check.
 
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