Calculating Shear Force: Understanding Two Forces Acting in Opposite Directions

[Diptangshu]

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]

 

[chetzread]

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

 

[Diptangshu]

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...

 

[David Lewis]

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.

 

[David Lewis]

So what does the shear force diagram for the pin look like ?

 

[chetzread]

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 ?

 

[chetzread]

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 ?

 

[chetzread]

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 .

 

[David Lewis]

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).

 

[David Lewis]

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.

 

[chetzread]

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 ?

 

[chetzread]

5kN inertial reaction force acting on the left part (pointing to the right).

why ?

 

[David Lewis]

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.

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.

 

[chetzread]

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 ?

 

[David Lewis]

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.

 

[David Lewis]

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.