Tension, shear and torque acting on a bolt and cantilever.

[Willy Cooper]

I am having some difficulty getting my head around the terms tension (stress/strain), shear and torque in a bolt and cantilever.

Lets say we have a cantilever 1m long bolted to the wall, with the bolt in the middle of the cantilever and a 10N force hanging and/or directly attached to the end.
(I understand that a 90 degree angle bracket acts as a lever depending on the bolt placement)

I also understand that though many web sites refer to the tension (stress/strain) and shear force acting on the bolt as N/m that this is incorrect because Tension (strain/stress) and shear are not torques but rather load which is measured in N or N/mm - not Nm

1) So if the cantilever attached at 90 degrees to a wall.
- Torque acting on the beam = 10 N/m?
- Tension (strain stress) in the bolt = 10N? (I am assumingthere is no horizontal component to the force only vertical)
- Shear = 10N/diameter of the bolt?

a) What I don't quite understand is that if I double the length of the beam to 2m then shouldn’t the Shear acting on the bolt also double? But from reading this is not the case?


2) Secondly if the beam is now attached to the roof (Hanging straight down).
The Tension (Strain/ Stress) in the bolt is the same as the above cantilever?
- Torque in the beam = 0 Nm
- Tension (strain/stress) in the bolt = 10N?
- Shear = 0 N


3) Thirdly if the cantilever is now attached at a 30 degree down angle out from the wall. (MesuringThe inside angle between the cantilever and the wall)

a) Torque acting on the beam.
If the weight is directly attached to the end of the beam, thus the force is being fully applied (perpendicularly at right angles) to the end of the beam.
([Sine 60 degrees * 1m] * 10N = 9.8 Nm). (I’m pretty sure this is right)

b) Torque acting on the beam.
If the weight is now suspended by a wire (which would cause the wire to form a 150 degree angle straight down from the end of the beam) has the torque acting on the beam now changed from '3a' above of 9.8N/m? I think so, and if so, how to calculate it?


4) Tension in the bolt. (this sounds silly but I have to ask).
To calculate the tension being exerted on the bolt (and threads) all that we have to consider is the tension exerted by the cantilever.
I.e we don’t add the initial bolt torque and cantilever tension together, right?


5) My final question is a about the yield point tension and torque.

a) - If we have a round metal bar with a yield of 10N and we twist it around on itself (perpendicular to its central axis) with a torque of 12N/m then we are applying the equivalent of 12N of tension (stress / strain) to the metal bar? So it will permanently deform.
- Also if we hold on end of the bar in a vice and push down with the same torque it will permanently bend?


b) So going back to my third question of a cantilever attached at a 30 degree angle.

- The standard way to calculate the tension in the bar is to calculate the horizontaland vertical forces
I.E.. Tension = (10 N) / [ sine 60 (degrees) ]

- However why can't we just calculate the torque instead?
I.E..Some correlation like 10 Nm of torque = 10N of tension?...Or 10 Nm X 2m (the actual length of the cantilever) = 20N of tension?

 

[PhanthomJay]

I am having some difficulty getting my head around the terms tension (stress/strain), shear and torque in a bolt and cantilever.

Lets say we have a cantilever 1m long bolted to the wall, with the bolt in the middle of the cantilever and a 10N force hanging and/or directly attached to the end.
(I understand that a 90 degree angle bracket acts as a lever depending on the bolt placement)

I also understand that though many web sites refer to the tension (stress/strain) and shear force acting on the bolt as N/m that this is incorrect because Tension (strain/stress) and shear are not torques but rather load which is measured in N or N/mm - not Nm

1) So if the cantilever attached at 90 degrees to a wall.
- Torque acting on the beam = 10 N/m?
- Tension (strain stress) in the bolt = 10N? (I am assumingthere is no horizontal component to the force only vertical)
- Shear = 10N/diameter of the bolt?

a) What I don't quite understand is that if I double the length of the beam to 2m then shouldn’t the Shear acting on the bolt also double? But from reading this is not the case?


2) Secondly if the beam is now attached to the roof (Hanging straight down).
The Tension (Strain/ Stress) in the bolt is the same as the above cantilever?
- Torque in the beam = 0 Nm
- Tension (strain/stress) in the bolt = 10N?
- Shear = 0 N


3) Thirdly if the cantilever is now attached at a 30 degree down angle out from the wall. (MesuringThe inside angle between the cantilever and the wall)

a) Torque acting on the beam.
If the weight is directly attached to the end of the beam, thus the force is being fully applied (perpendicularly at right angles) to the end of the beam.
([Sine 60 degrees * 1m] * 10N = 9.8 Nm). (I’m pretty sure this is right)

b) Torque acting on the beam.
If the weight is now suspended by a wire (which would cause the wire to form a 150 degree angle straight down from the end of the beam) has the torque acting on the beam now changed from '3a' above of 9.8N/m? I think so, and if so, how to calculate it?


4) Tension in the bolt. (this sounds silly but I have to ask).
To calculate the tension being exerted on the bolt (and threads) all that we have to consider is the tension exerted by the cantilever.
I.e we don’t add the initial bolt torque and cantilever tension together, right?


5) My final question is a about the yield point tension and torque.

a) - If we have a round metal bar with a yield of 10N and we twist it around on itself (perpendicular to its central axis) with a torque of 12N/m then we are applying the equivalent of 12N of tension (stress / strain) to the metal bar? So it will permanently deform.
- Also if we hold on end of the bar in a vice and push down with the same torque it will permanently bend?


b) So going back to my third question of a cantilever attached at a 30 degree angle.

- The standard way to calculate the tension in the bar is to calculate the horizontaland vertical forces
I.E.. Tension = (10 N) / [ sine 60 (degrees) ]

- However why can't we just calculate the torque instead?
I.E..Some correlation like 10 Nm of torque = 10N of tension?...Or 10 Nm X 2m (the actual length of the cantilever) = 20N of tension?

You are confusing tension and shear, in N, with tensile and shear stresses, in N/m^2. And you don't have a stable cantilever with just one bolt in the wall...Vertical loads produce shear or tension in a bolt, depending on orientation of beam or bolts...the torque (moment) produces tensile and compressive bending stresses in the beam, per Mc/I...you might want to rethink this...

 

[MKRAH]

Can someone tell me how to check the bearing capacity of plate and bolt when threads are considered in shear plane.