# Opposing Moments on Levered Beam

1. Mar 22, 2012

### nickbarker77

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

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.

2. Relevant equations

I = pi (do4 - di4) / 64

3. 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.

#### Attached Files:

• ###### IMG_20120322_193451.jpg
File size:
13 KB
Views:
205
2. Mar 22, 2012

### rock.freak667

Why not just use

M/I = σ/y = E/R

E is given for the material and you know R. Just get M.

3. Mar 22, 2012

### nickbarker77

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. Mar 23, 2012

### HyperSniper

When I calculated the second moment of area I got 2.701*10^-5 m^4 based on this:
$I=\frac{\pi*\left[\left(200mm\right)^{4}-\left(200mm-20mm\right)^{4}\right]}{64}=2.701\times10^{-5}m^{4}$

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

$\frac{M}{I}=\frac{E}{R}$

$M=\frac{EI}{R}$

$M=1.441\times10^{5}N$

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