Calculating Shear Force: Understanding Two Forces Acting in Opposite Directions

AI Thread Summary
The discussion centers on understanding shear forces in a system with multiple forces acting in opposite directions. It emphasizes that for a system in equilibrium, the total forces must balance, resulting in no net external force, but internal shear forces can still exist. Participants clarify that shear force is present regardless of whether the system is accelerating or at rest, as long as there are opposing forces acting on components. The conversation also highlights the importance of analyzing load paths in mechanical assemblies to understand how forces interact. Ultimately, grasping these concepts is crucial for studying engineering statics and mechanics.
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Homework Statement


in this diagram , we could see two forces acted to left , one force acted to right , total of two forces acting to left = force acting to right .

Homework Equations

The Attempt at a Solution


i'm wondering if the total of two forces on the left are not the same with the force on the right , then , how to do the question ? We know that tau = P/ A , tau = stress , P = force . let's say one of the force is 30kN to left , another one is 40kN to left , total is 70kN to the left , but on the right , the force is 60kN to right ... so , the shear stress = 10kN to right divided by area ? [/B]
 

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Why my post is deleted?
 
Not deleted from my end, check again.

The system is in equilibrium, so the applied force of 60kN to the right must be balanced by the two 30 kN forces on the left. You can't have it any other way for equilibrium And if you thus have 60 kN to the right and 60 kN to the left, the shear force on the bolts is not 0, is it? Follow the solution explanation.
 
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PhanthomJay said:
Not deleted from my end, check again.

The system is in equilibrium, so the applied force of 60kN to the right must be balanced by the two 30 kN forces on the left. You can't have it any other way for equilibrium And if you thus have 60 kN to the right and 60 kN to the left, the shear force on the bolts is not 0, is it? Follow the solution explanation.
Do you mean if the system is not in equilibrium, then shear force can't exist?
 
chetzread said:
Do you mean if the system is not in equilibrium, then shear force can't exist?
If the system is not in equilibrium, it is accelerating, and you don't even want to think about that when studying engineering statics and mechanics for bodies at rest.
 
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Can anyone try to answer?
 
PhanthomJay said:
If the system is not in equilibrium, it is accelerating, and you don't even want to think about that when studying engineering statics and mechanics for bodies at rest.
Can you explain further? If the body is not in equilibrium, why there's no shear force?
 
PhanthomJay said:
if you thus have 60 kN to the right and 60 kN to the left, the shear force on the bolts is not 0, is it
it's in equlibrium , so shear force = 0
 
One of the first things that an engineer looks at when analysing stresses in a mechanical assembly is the load path .

Load path describes how forces interact sequentially between individual components of an assembly . The complete load path starts with the applied forces , goes through the assembly and terminates with the balancing forces .

Load path can be a very simple linear chain with simple forces at each end or very complex with multiple sub paths and multiple forces .

What do you think that the load path is for the forces acting in your assembly ?
 
  • #10
Nidum said:
One of the first things that an engineer looks at when analysing stresses in a mechanical assembly is the load path .

Load path describes how forces interact sequentially between individual components of an assembly . The complete load path starts with the applied forces , goes through the assembly and terminates with the balancing forces .

Load path can be a very simple linear chain with simple forces at each end or very complex with multiple sub paths and multiple forces .

What do you think that the load path is for the forces acting in your assembly ?
Load path can be start from the 2 30m forces or the single 60m forces
 
  • #11
chetzread said:
it's in equlibrium , so shear force = 0
It is in equilibrium, so the net external force acting on the system is zero. But that does not mean that the shear force or stress on the bolts is zero. Follow the solution explanation.
 
  • #12
PhanthomJay said:
It is in equilibrium, so the net external force acting on the system is zero. But that does not mean that the shear force or stress on the bolts is zero. Follow the solution explanation.
I don't understand why the shear force not =0 when the whole structure in equilibrium? When it's in equilibrium, there's no net force acting on it, right?
 
  • #13
Pin2.jpg


Think about a slightly simpler problem with just one pin .

What forces are acting on the pin ?
 
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  • #14
chetzread said:
I don't understand why the shear force not =0 when the whole structure in equilibrium? When it's in equilibrium, there's no net force acting on it, right?
systems in equilibrium have no net external force acting on them, per Newton's first law. You apply a 60 N force on the right plate, and the end plate reactions are 30 kN apiece or 60 kN to the left.. But you are not looking at the internal forces acting within the system. Imagine the right plate is your head and the left plates are your legs, and someone pulled on your head with a force of 400 N and another person was pulling on your legs and you didn't move. Then the person pulling on your legs also exerts a force of 400 N in the other direction, since net external force must be zero. But if you think there is no force in your body, think again, because you will the pain as your body is being stressed. 400N force internal. If you draw a FBD of the right part of your body (cutting you in half in an imaginary sense), then the internal force of 400N is pulling you leftward and the applied force of 400 N is pulling ' you rightward, no net force, but you sure feel the 400 N force, it is quite real and certainly non-zero.
 
  • #15
PhanthomJay said:
If the system is not in equilibrium, it is accelerating, and you don't even want to think about that when studying engineering statics and mechanics for bodies at rest.
Why when the structure is accelerating , there can't be shear force acting?
 
  • #16
Nidum said:
View attachment 106974

Think about a slightly simpler problem with just one pin .

What forces are acting on the pin ?
2 F/2 act to left, and one F act to right?
 
  • #17
So what does the shear force diagram for the pin look like ?
 
  • #18
chetzread said:
Why when the structure is accelerating , there can't be shear force acting?
I did not say that . I ask you to focus instead on systems in equilibrium where net forces are zero.
 
  • #19
chetzread said:
I'm wondering if the total of two forces on the left are not the same with the force on the right, then how to do the question? Let's say one of the forces is 30kN to left, another one is 40kN to left (total is 70kN to the left), but on the right, the force is 60kN to right.
Then you multiply the mass of the right portion by the acceleration of the body, and add it to the 60kN force. For example, if the mass of the right portion is negligible, shear force would be 60 kN.
 
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  • #20
chetzread said:

Homework Statement


in this diagram , we could see two forces acted to left , one force acted to right , total of two forces acting to left = force acting to right .

Homework Equations

The Attempt at a Solution


i'm wondering if the total of two forces on the left are not the same with the force on the right , then , how to do the question ? We know that tau = P/ A , tau = stress , P = force . let's say one of the force is 30kN to left , another one is 40kN to left , total is 70kN to the left , but on the right , the force is 60kN to right ... so , the shear stress = 10kN to right divided by area ? [/B]
My apology for the 1st deleted post.

I don't know Whether Force imbalance is Possible unless there is Deformation of the Bolt.

But when the whole system is in Acceleration, Force P in both Segments of Bolted Joint is still the Same.

See the Figure below.
 

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  • #21
Diptangshu said:
I don't know Whether Force imbalance is Possible unless there is Deformation of the Bolt.
If the force on one side is zero, there will be no load on the shear pin other than inertial forces due to acceleration.
 
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  • #22
Diptangshu said:
My apology for the 1st deleted post.

I don't know Whether Force imbalance is Possible unless there is Deformation of the Bolt.

But when the whole system is in Acceleration, Force P in both Segments of Bolted Joint is still the Same.

See the Figure below.
So, when the whole structure is in acceleration, is the shearing possible?
 
  • #23
David Lewis said:
Then you multiply the mass of the right portion by the acceleration of the body, and add it to the 60kN force. For example, if the mass of the right portion is negligible, shear force would be 60 kN.
How if the mass is not negligible, then how the shearing force is?
 
  • #24
David Lewis said:
If the force on one side is zero, there will be no load on the shear pin other than inertial forces due to acceleration.
Do you mean if the force on one side is 0, then, the shear force will be just the inertia force? How if the 2 sides are subjected to different forces?
 
  • #25
chetzread said:
So, when the whole structure is in acceleration, is the shearing possible?
Of course... the figure shows that the bolted joint has force P on right and left, which generates the Shear Stress.
 
  • #26
David Lewis said:
If the force on one side is zero, there will be no load on the shear pin other than inertial forces due to acceleration.

Yes...but real life physical problem doesn't make mass on one side zero...i think.

I would love, if you give me insight on real life cases with force imbalance, Sir.
 
  • #27
Diptangshu said:
Of course... the figure shows that the bolted joint has force P on right and left, which generates the Shear Stress.
Then, why someone said that when the structure is in acceleration ( not in equilibrium), then the shearing is not possible?
 
  • #28
Diptangshu said:
Of course... the figure shows that the bolted joint has force P on right and left, which generates the Shear Stress.
But, the whole structure is in equilibrium, right? There's no acceleration...
 
  • #29
chetzread said:
But, the whole structure is in equilibrium, right? There's no acceleration...
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.
 
  • #30
Nidum said:
So what does the shear force diagram for the pin look like ?
i really have no idea . i haven't learn the shear force diagram
 
  • #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 ?
 
  • #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 ?
 

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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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