Reaction Forces on a Frame With Roller Joint and Pin Joint

In summary, the conversation discussed a problem involving calculating the support reactions of a roof support frame. The frictionless pin joint at point A has both vertical and horizontal reaction forces, while the rollers at point B only have a vertical reaction. The conversation also mentioned the equation Moment = Force * Distance and the attempt to find the support reactions using this method. However, it was pointed out that the moments from horizontal forces on the columns must also be included in the calculations. The conversation then discussed the method of summing forces in the x direction to find the horizontal reaction at point A. Finally, the possibility of using the moments method as a check was mentioned.
  • #1
AliesAbawe
4
0
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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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
  • #5
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 unfortunatly :/
 
  • #6
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 unfortunatly :/
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??
 
  • #7
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!
 
  • #8
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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Related to Reaction Forces on a Frame With Roller Joint and Pin Joint

1. What are reaction forces on a frame with roller joint and pin joint?

Reaction forces are the forces exerted by a structure on its supports or joints in order to maintain equilibrium. In a frame with roller and pin joints, there are two types of reaction forces: axial and shear forces. Axial forces are parallel to the member while shear forces act perpendicular to the member.

2. How do you calculate reaction forces on a frame with roller joint and pin joint?

To calculate reaction forces on a frame with roller and pin joints, you must first draw a free body diagram of the structure and consider all external forces acting on it. Then, using the equations of equilibrium, you can solve for the unknown reaction forces at the joints.

3. What is the difference between a roller joint and a pin joint?

A roller joint allows a member to rotate freely while a pin joint allows a member to rotate and translate. This means that a roller joint only supports axial forces while a pin joint can support both axial and shear forces.

4. How do reaction forces affect the stability of a frame with roller joint and pin joint?

Reaction forces play a crucial role in the stability of a frame with roller and pin joints. These forces must be carefully calculated and distributed in order to prevent any unwanted movement or collapse of the structure. Improperly calculated reaction forces can lead to structural failure.

5. Are there any real-life applications of frames with roller joint and pin joint?

Yes, frames with roller and pin joints are commonly used in structural engineering and construction. They are often found in bridges, cranes, and other types of structures that require the transfer of loads and forces between different members. They are also used in mechanical systems such as trusses and frames.

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