# Homework Help: Torque problem

1. Mar 10, 2013

### Martin_79

1. The problem statement, all variables and given/known data
Hi All,

I am new to this forum and have a question for which I need some assistance. My knowledge about this subject is limited, but I need te following:

For a project we need to specify the torque required to uprighten a structure when rotated to a certain angle. The structure is located at a distance where the torque can be applied.
I know how to calculate the torque exterted by the structure at an angle. However, I presume that, when the counteracting torque is to be applied at a certain distance, the calculation would be different. Additionally, the pipe will twist as a result of the structure rotation, but how does this influence the required torque to uprighten the structure?

Regards,
Martin

2. Relevant equations
W(torque) = F (force) * a (arm)
a = sin (α) * distance CoG (relative to pipe centreline)

3. The attempt at a solution
F = 16 tonnes (156.96 kN)
Distance between CoG and pipe centreline = 0.5 m
α = 15 degrees (0.2618 radians)
a = sin (15) * 0.5 = 0.1294 m

W = 156.96 kN * 0.1294 m = 20.3 kNm

From this point onwards I do not know how and if additional calculations are required which include the twisting pipeline. The properties of the pipeline are:

Outer diameter = 300 mm
Wall thickness = 20 mm
Material = steel (E-modulus = 207 GPa)
Poisson ratio = 0.3
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### Torque problem.jpg
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2. Mar 10, 2013

### nvn

Martin_79: For the structure shown, the applied (required) torque will be the same. The torque calculation at a distance is not different. You already computed it correctly. Nice work. The pipe twist does not affect the required torque. There is no additional calculation required.

Last edited: Mar 10, 2013
3. Mar 11, 2013

### CWatters

The twist doesn't change the maximium torque but if the angle is specified at the torque end (rather then the load) end the curve of torque vs angle will move so max torque will occur at a different point.

4. Mar 11, 2013

### Martin_79

CWatters, could you explain your reply a bit more? If I am correct, you mean that when torque is applied and there is a ristriction in the angle at which the torque can be applied, the maximum torque will not necessarily occur at the location of the structure to uprighten it? If so, how can I calculate the torque required if the angle of application is different?

5. Mar 12, 2013

### CWatters

No I'm not talking about a restriction.

The equation for the torque will be..

T = mgB Sin(θ)

m=mass
g=9.81 m/s^2
B = Length of the arm. (eg Distance B on your diagram.)
θ = The angle of rotation measured at the end with the arm/COG.

The max torque occurs when Sin(θ) = 1 (eg when the θ = 90, so the arm is horizontal).

The twist does not change the maximium torque. Nor does it change the angle of the arm at which max torque occurs.

That might be all you need to know. Problem solved.

However IF you need to know the angle of the other end (the end where the torque is actually applied) you have to take into account the twist. This might matter if you are trying to limit how far the arm rotates by controling how far the other end rotates.

Lets call the angle of the Arm end θa and the angle of the end where the torque is applied θt. Both measured from the vertical.

In the max torque position θa=90. However θt will be less due to the twist.

θt = θa - ∠twist

unfortunately ∠twist isn't a constant. It depends on the torque which will vary with θa.

Also when the arm end is vertical θa=0 so the torque must reduce to zero to stop the arm moving. So when the arm is vertical the twist will be zero.

So the angle that the torque end moves will be less than 90 degrees.