Opposing Moments on Levered Beam

In summary, the task at hand is to calculate the force F1 required to bend the beam AB to a radius of 7.5m. The beam is a steel tube with an outer diameter of 200mm and wall thickness of 20mm. The force acts on pivots at points C and D and is always in the x direction. Using the formula for second moment of area and Young's modulus for steel, the required moment was calculated to be 1.441 x 10^5 N, resulting in a force of 9.9 x 10^5 N at points C and D.
  • #1
nickbarker77
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Homework Statement



Calculate the force F1 to bend the beam AB to a radius of 7.5m. Assume the beam is a steel tube 200mm OD with wall thickness 20mm. The arm's AC and BD can be assumed rigid.

The force F1 acts on pivots at C and D and is always in the x direction.

Homework Equations



I = pi (do4 - di4) / 64



The Attempt at a Solution



My first thought was to assume the beam was fixed at one end and then work out the deflection required to achieve a a radius of 7.5m. From this I substituted into the deflection formula to work out a point load need on the end of the cantilevered beam to produce the deflection.

I think i should be calculating the moment at point C and D but I'm struggling a bit, any help would be greatly appreciated.
 

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  • #2
Why not just use

M/I = σ/y = E/R

E is given for the material and you know R. Just get M.
 
  • #3
Thanks rock.freak667.

i have since found this formula and came out with an answer of 1.25 x 10^6 N/m for the moment required.

this is using a value of 4.637 x 10^-5 for the second moment of area and a young's modulus of 200GPa for the steel.

As the arms at the end of the beam are 1.25m long then the required force at points C and D would be 9.9 x 10^5 N.

This seem a little high?
 
  • #4
When I calculated the second moment of area I got 2.701*10^-5 m^4 based on this:
[itex]I=\frac{\pi*\left[\left(200mm\right)^{4}-\left(200mm-20mm\right)^{4}\right]}{64}=2.701\times10^{-5}m^{4}[/itex]

This would change your results a little bit. Solving out the ratio before of M/I = E/R gives:

[itex]\frac{M}{I}=\frac{E}{R}[/itex]

[itex]M=\frac{EI}{R}[/itex]

[itex]M=1.441\times10^{5}N[/itex]

I would definitely double check what I did. 32,394 pounds of force seems reasonable to me though.
 
  • #5




As a scientist, the first step in solving this problem would be to clarify any uncertainties or missing information. It is important to know the material properties of the steel tube, such as its modulus of elasticity and yield strength, in order to accurately calculate the required force. Additionally, the location of the pivots at C and D should be specified in order to determine the lever arm and moment arm.

Assuming these values are known, the next step would be to calculate the moment required to bend the beam to a radius of 7.5m. This can be done by using the formula for the moment of inertia of a hollow cylinder (I = pi (do4 - di4) / 64) and substituting in the values for the outer and inner diameters of the steel tube. This moment can then be equated to the force F1 multiplied by the lever arm, which is the distance from the pivot to the point of force application.

Finally, the force F1 can be calculated by dividing the calculated moment by the lever arm. It is important to note that the force F1 will vary depending on the location of the pivots at C and D. If the pivots are closer to A, the required force will be larger, and if they are closer to B, the required force will be smaller.

In conclusion, to accurately calculate the force F1 required to bend the beam to a radius of 7.5m, we need to know the material properties of the steel tube, the location of the pivots, and the lever arm. These values can then be used to calculate the moment required and subsequently the force F1.
 

FAQ: Opposing Moments on Levered Beam

What is a levered beam?

A levered beam is a simple machine consisting of a rigid bar or beam that is able to rotate around a fixed point. It is used to amplify or redirect forces in order to make tasks easier to accomplish.

How does opposing moments occur on a levered beam?

Opposing moments on a levered beam occur when there are two equal and opposite forces acting on either side of the pivot point. This creates a torque or rotational force, causing the beam to rotate.

What factors influence the magnitude of opposing moments on a levered beam?

The magnitude of opposing moments on a levered beam is influenced by the distance between the forces and the pivot point, as well as the magnitude of the forces themselves. The longer the distance between the forces, the greater the opposing moments will be. Similarly, the larger the forces, the greater the opposing moments.

How is the equilibrium of a levered beam achieved when there are opposing moments?

In order for a levered beam to be in equilibrium when there are opposing moments, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments. This means that the opposing moments must be balanced and cancel each other out.

How does the placement of the pivot point affect opposing moments on a levered beam?

The placement of the pivot point on a levered beam can greatly impact the opposing moments. If the pivot point is closer to one of the forces, it will have a greater effect on the opposing moments and can potentially cause the beam to rotate more easily.

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