# Calculte max force that can be applied to a Beam flange?

1. Mar 9, 2009

### em07189

Hi everyone!

I need to design a beam for a monorail system, like in the pictures above.

http://img90.imageshack.us/img90/2629/monorailperfil.png [Broken]

http://img111.imageshack.us/img111/2987/perfiltransversal.png [Broken]

But the question, is how to know the maximum force that a beam flange can handle?

Thanks.

Last edited by a moderator: May 4, 2017
2. Mar 9, 2009

### minger

How will the monorail be attached to the beam (e.g. 4 'x'-spaced wheels)?

3. Mar 9, 2009

### em07189

beam lenght 3 meters.
distance between wheels 0.6 meters.

I'm concerned with weight loads, and find the beam with flanges that will handle with the weight load.

How can i do this calculation?

http://img154.imageshack.us/img154/2333/trolley.png [Broken]

Last edited by a moderator: May 4, 2017
4. Mar 9, 2009

### mooktank

Find the bending stress at the point where the flange meets the web (you'd need to assume some supports for the cross section, top diagram).

You should also find the bending stress of the entire beam. (bottom diagram)

Then calculate shear stresses.

Or plug the whole thing into your FEA software of choice.

5. Mar 10, 2009

### minger

If the trolley is stationary, then you do a simple beam analysis. You have vertical forces at the wheels. You draw a shear/moment diagram to find the location of the maximum moment and shear.

I would think that the worst case would be the trolley 'almost' to the end of the beam, such that the wheel is loading just before the support. This should generate the highest shears in your problem. Then, sum moments about either support to calculate the reaction force at the opposite support.

Then sum forces to calculate the final reaction force. From here, draw a shear diagram (if you need more info, let us know). Then, integrate the shear diagram to get the moment diagram. The shear stress will be:
$$\sigma_{shear} = \frac{F}{A}$$
Where F is the shear force, and A will be an area that it acts upon.
The bending stress will be:
$$\sigma_{bend} = \frac{My}{I}$$
Where y is the distance from the centerline to any point, and I is the area moment of inertia. For maximum stress location, y is the outer fiber of the beam (think OD).

However, your maximum stress will probably be an equivalent stress located at the top of the flange. If the web is either welded or brazed on, then there may be additional stress factors as well.

6. Mar 10, 2009

### Q_Goest

Hi em. Minger has provided the general method for determining stresses in beams, though the worst case stress is when the load is in the middle. Assume simply supported ends unless you have sufficient anchoring at the beam support points to assume otherwise.

Generally, trolleys have a load rating and that load rating is also dependent on the beam used. So your first option should be to check with the trolley manufacturer for minimum beam recommendations and start from there. If you don't have that or if you're designing your own trolley, I'd suggest applying the beam bending equations to the flanges themselves. Imagine the flange being two cantilevered beams in bending, one on each side of the web. This defines the length of the beam (half the width of the flange) and thickness of the beam (thickness of the flange). You also need width. I'd suggest for each contact point (wheel) you use a width equal to the length of the beam (half the width of the flange). This should give you a safe / conservative stress analysis for the flange itself. Don't forget to apply a safety factor - I'd suggest 2 to yield and 4 to ultimate.

WARNING: If this is for overhead work in a commercial application, the 'crane' must meet OSHA standards and should be designed & tested by a certified crane testing outfit.

7. Mar 10, 2009

### minger

ah whoops, sorry 'bout that. Good eye

8. Mar 10, 2009

### em07189

Hi minger!

Do i have to apply the von-misses equation for the shear and bend, right?

Or can i calculate shear and stress individually?

Thanks.

9. Mar 10, 2009

### Whatthe

There are standards for this type of design.

10. Mar 11, 2009

### minger

There are a few different equivalent stress formulations, Von Mises is just one; you'll have to decide which one to use.

However, I'd have to side with a few of the posts here and say that for a common part such as this, and its application, there should be standards that you can follow for proper design and safety.

11. Mar 11, 2009

### nvn

em07189: Maximum moment on the I cross section occurs underneath the wheels closest to the beam midspan when one axle is 140 mm from the beam midspan; and this moment is My = 1215(2*F1), where My = moment about the I-beam cross section lateral (strong) axis (N*mm), and F1 = downward load applied by one wheel (N). Therefore, the beam longitudinal bending stress, at the bottom face of the bottom flange, is sigma_x = My*(0.5*h)/I, where h = I-beam cross section height. The flange bending moment about the x axis, per Q_Goest, is Mx = -0.5*b*F1; therefore, the flange lateral bending stress, at the bottom face of the bottom flange, is sigma_y = -6*F1/t^2, where t = I-beam flange thickness. Notice sigma_x is positive, and sigma_y is negative. Combine sigma_x and sigma_y using von Mises, which will give sigma_vm. Using yield factor of safety FSy = 2.0, suggested by Q_Goest, ensure sigma_vm < Sty/FSy, where Sty = beam material tensile yield strength.

Transverse shear stress on a free face is zero; therefore, you can check peak shear stress at the flange midplane separately. Using the model suggested by Q_Goest, peak shear stress is tau = 1.50*F1/A = 3*F1/(b*t), where b = I-beam total flange width. Ensure tau < 0.577*Sty/FSy.

Use consistent units for all quantities; e.g., N, mm, MPa. Feel free to increase FSy if you think the design code requires a higher FSy. Follow a design code if it varies from the above.

12. Mar 12, 2009

### em07189

HI nvn!

Thanks for your help, but what is "1215" and why, in "My=1215(2*F1)"?

13. Mar 12, 2009

### nvn

em07189: Place your trolley at the position indicated in the first sentence of post 11. I.e., place one axle at 140 mm from the beam midspan, and place the other axle at 460 mm from the beam midspan. Now use statics to solve for the maximum moment in your beam. The answer, in N*mm, will turn out to be My = 1215(2*F1), if F1 is in units of N.