Calculating Shear Force: Understanding Two Forces Acting in Opposite Directions

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Discussion Overview

The discussion revolves around the calculation of shear force in a system with multiple forces acting in opposite directions. Participants explore concepts related to equilibrium, shear stress, and the implications of force imbalance in mechanical assemblies. The scope includes theoretical reasoning, technical explanations, and conceptual clarifications related to engineering statics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how to approach the problem if the total of two forces acting to the left is not equal to the force acting to the right, suggesting a scenario with 70 kN to the left and 60 kN to the right.
  • Another participant asserts that for the system to be in equilibrium, the forces must balance, indicating that the shear force on the bolts cannot be zero if there is a net force acting on them.
  • Some participants discuss the concept of load path, describing how forces interact within a mechanical assembly and questioning what the load path is for the forces in the given scenario.
  • There is a debate over whether shear force can exist in a system that is not in equilibrium, with some arguing that acceleration implies the presence of shear forces while others maintain that equilibrium means no net external force.
  • One participant uses an analogy involving forces acting on a body to illustrate that internal forces, such as shear forces, can exist even when the net external force is zero.
  • Another participant expresses confusion about why shear force would not equal zero in an equilibrium state, prompting further discussion on internal forces and their implications.
  • Some participants suggest that if the system is accelerating, the forces in the bolted joint remain the same, raising questions about force imbalance and deformation of the bolt.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the relationship between equilibrium and shear force, with multiple competing views on whether shear forces can exist in a system that is not in equilibrium. The discussion remains unresolved as participants explore different perspectives and analogies.

Contextual Notes

Participants express uncertainty regarding the conditions under which shear forces exist, particularly in relation to equilibrium and acceleration. There are references to specific forces and their interactions, but the discussion does not reach a consensus on the implications of these conditions.

  • #31
PhanthomJay said:
The system as given in the original posting is in equilibrium. But you have kept asking about what would happen if it was not, and instead, was accelerating under a net unbalanced force. Now you ask about equilibrium again. You are confusing yourself. Once again please focus on understanding shear forces on the pin for the equilibrium condition. Answer post 17 for the equilibrium condition. Once you understand that, then you can ask about the accelerated system if your inquisitive mind so desires. But you can't understand the advanced topics without first understanding the basic concepts.

Sorry , i misunderstood post #5 . I misunderstood that there's no shearing force when the whole structure is accelerating .

So, the correct concept is there's also shearing force when the object is accelerating ?
 
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  • #32
chetzread said:
Sorry , i misunderstood post #5 . I misunderstood that there's no shearing force when the whole structure is accelerating .

So, the correct concept is there's also shearing force when the object is accelerating ?
Yes, there is shearing force on the pin when it is at rest and there is shearing force on the pin when it is accelerating. But please focus on the given problem and its given solution, and confine your questions for now to that problem.
 
  • #33
chetzread said:
i really have no idea . i haven't learn the shear force diagram

I think that shear force diagrams have featured in some of your other threads ?

No matter . Can you sketch any sort of diagram showing the forces acting on the pin ?
 
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  • #34
PhanthomJay said:
Yes, there is shearing force on the pin when it is at rest and there is shearing force on the pin when it is accelerating. But please focus on the given problem and its given solution, and confine your questions for now to that problem.
ok , thanks . So , now i can understand about the head and leg that is pulled , shear stress exist happen there . How about when the object is accelerating ,. the object is subjected to force of total 70 kN to the left and total 60kN force to the right ? How to find the magnitude of shear force in this case ?
 
  • #35
chetzread said:
ok , thanks . So , now i can understand about the head and leg that is pulled , shear stress exist happen there .
that example I gave was a tensile stress on your body not a shear stress. I just wanted to show that when there is no net force on an object that there is still force and stress within that object. Tensile stresses act normal to the cross section plane, shear stresses act parallel to the cross section plane.
How about when the object is accelerating ,. the object is subjected to force of total 70 kN to the left and total 60kN force to the right ? How to find the magnitude of shear force in this case ?
oh darn chetzread there you go again back to the accelerating case! It would be best for you to first find the magnitude of the shear stress in the pin when you have 60 kN to the right and 2(30) = 60 kN to the left. Refresh your shear diagram knowledge or follow the problem solution.
Now when you have the accelerating case of 70 kN left and 60 kN right then the net force of 10 kN left will accelerate the system left from Newton 2nd law and you need then to know the mass of the system before finding the shear force on the pin and again I urge you to not dwell on this case at this time.
 
  • #36
chetzread said:
ok , thanks . So , now i can understand about the head and leg that is pulled , shear stress exist happen there . How about when the object is accelerating ,. the object is subjected to force of total 70 kN to the left and total 60kN force to the right ? How to find the magnitude of shear force in this case ?

chetzread, I already posted answer in Post #20... that is what the case is when the Joint (connecting the Carts) is accelerating.

But, your basics are the Most important parts. So, as PhanthomJay said you already, Clear your Static Problem first... there are Much Complicated Bolted joints you will Find as interesting to Solve. [PLAIN]https://www.physicsforums.com/members/phanthomjay.58874/[/PLAIN]
 
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  • #37
I can understand it, but it seems that your diagram doesn't suit my case, coz, I have 2 parts of the plane, one plane is subjected to 2 total smaller force, and one bigger force to right
 
  • #38
chetzread said:
I can understand it, but it seems that your diagram doesn't suit my case, coz, I have 2 parts of the plane, one plane is subjected to 2 total smaller force, and one bigger force to right

Well, my Friend. You Should listen to What other Advisors said to You...
 
  • #39
chetzread said:
How if the mass is not negligible, then how the shearing force is?
If the mass of the left part is equal to the mass of the right part then the shear force will be 65kN.

You have 70kN pulling to the left, and 60kN pulling to the right.
Acceleration will be 10kN/mass of the free-body (pointing to the left).
Inertial reaction force on the right part will be acceleration x mass of the right part.
The mass of the right part equals half the mass of the free-body:
Acceleration x mass/2 = (10kN/mass) x (mass/2) = 5kN.
 
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  • #40
Nidum said:
So what does the shear force diagram for the pin look like ?
 

Attachments

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  • #41
David Lewis said:
If the mass of the left part is equal to the mass of the right part then the shear force will be 65kN.

You have 70kN pulling to the left, and 60kN pulling to the right.
Acceleration will be 10kN/mass of the free-body (pointing to the left).
Inertial reaction force on the right part will be acceleration x mass of the right part.
The mass of the right part equals half the mass of the free-body:
Acceleration x mass/2 = (10kN/mass) x (mass/2) = 5kN.
if we consider the force from left side , the shear force would be = 70-5= 65kN ?
 
  • #42
David Lewis said:
If the mass of the left part is equal to the mass of the right part then the shear force will be 65kN.

You have 70kN pulling to the left, and 60kN pulling to the right.
Acceleration will be 10kN/mass of the free-body (pointing to the left).
Inertial reaction force on the right part will be acceleration x mass of the right part.
The mass of the right part equals half the mass of the free-body:
Acceleration x mass/2 = (10kN/mass) x (mass/2) = 5kN.
I don't understand the working , can you explain about it ? why the shear force is 60+5 = 65kN ?
 
  • #43
David Lewis said:
If the mass of the left part is equal to the mass of the right part then the shear force will be 65kN.

You have 70kN pulling to the left, and 60kN pulling to the right.
Acceleration will be 10kN/mass of the free-body (pointing to the left).
Inertial reaction force on the right part will be acceleration x mass of the right part.
The mass of the right part equals half the mass of the free-body:
Acceleration x mass/2 = (10kN/mass) x (mass/2) = 5kN.
is there snything to do with the shear force diagram ? btw , i have tried to sketch and here is it .
 

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  • #44
chetzread said:
If we consider the force from left side, the shear force would be = 70-5= 65kN ?
Yes. There is a 65kN shear force acting on the pin (pointing to the left), and there is a 5kN inertial reaction force acting on the left part (pointing to the right).
 
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  • #45
chetzread said:
Is there anything to do with the shear force diagram? BTW, I have tried to sketch and here it is:
That is not what I came up with. The pin does not experience any shear force except at the faces where the parts are joined. The SFD looks like the Greek letter π except one vertical leg is pointing up.

When the left side of a member is pushing up, and the right side is pushing down, shear force is considered positive, by convention. Look at the SFD attached to post #40.
 
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  • #46
David Lewis said:
Yes. There is a 65kN shear force acting on the pin (pointing to the left), and there is a 5kN inertial reaction force acting on the left part (pointing to the right).
can i explain in this way ? Since the acceleration for each block is 5kN( 5kN act to the right for left block) , so considering the left block , the shear force 70-5 = 65kN ?
 

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  • #47
David Lewis said:
5kN inertial reaction force acting on the left part (pointing to the right).
why ?
 
  • #48
If you have a body floating in deep space, tie a rope to it and pull it to the left, the body will exert an equal and opposite inertial reaction force to the right.
chetzread said:
Since the acceleration for each block is 5kN...

Remember to divide force by mass to get acceleration.
The physical quantity will work out to distance/time2.
 
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  • #49
David Lewis said:
If you have a body floating in deep space, tie a rope to it and pull it to the left, the body will exert an equal and opposite inertial reaction force to the right.

Remember to divide force by mass to get acceleration.
The physical quantity will work out to distance/time2.
do you mean for the block on the left , the acceleration force one block is 5kN to the left , so opposite inertia force is 5kN to the right ?

But , i am curious about the block on the right , from the diagram i sketched in post #46 , the 5kN act to the left for the right block , so , there will also be an opposite inertia force of 5kN act to the right ?
 
  • #50
Both 5kN inertial force vector arrows (green?) should point to the right since the body is accelerating to the left.
If the blue vector arrow represents acceleration in your diagram, you could label it acceleration or 10kN/mass.
 
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  • #51
When you make your force arrangement diagram, it is customary for the point of application of the inertial force vector to coincide with the center of gravity of the part on which the inertial force is acting.
 

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