1. Jan 15, 2009

### tommy100

does any one know an equation that can help me find the stress of a hollow circular tube standing up-right and fixed at the bottom, and also a force is pushing it from one side?
many thanks

2. Jan 15, 2009

### minger

The case is just a simple cantelivered beam; it doesn't matter which way it stands neglecting gravity). Exactly how and where is the force applied?

3. Jan 15, 2009

### tommy100

the force is actually a wind of 1.0KN/m^2
so it will be applied accross the whole length of the tube

4. Jan 15, 2009

### minger

The maximum bending stress is
$$\sigma_{max} = \frac{My}{I}$$
Where M is the maximum moment, y is your radius, and I is the moment of inertia. For a cantelivered case, the maximum moment is at the base (obviously). We can get an equivalent force of all the pressures combined, which will act at the center of applied pressure (the center of the beam). The total force is
$$F = pL$$
Since it is applied at the center of the beam, the moment it creates on the base is:
$$M = Fd = pL(L/2) = \frac{pL^2}{2}$$
The moment of inertia for a hollow cylinder is:
$$I = \frac{\pi r^4}{4}$$
So, your maximum bending stress is then (canceling out the r term)
$$\sigma_{max} = \frac{2pL^2}{\pi r^3}$$

By the way, how were you able to get such an accurate loading for wind?

Last edited: Jan 15, 2009
5. Jan 15, 2009

### stewartcs

For a hollow cylinder (i.e. tube or pipe) the moment of inertia is:

$$I = \frac{\pi}{4} \cdot (r_o^4 - r_i^4)$$

Which would make the max stress at the base:

$$\sigma_{max} = \frac{pL^2y}{2I}$$

CS

6. Jan 15, 2009

### nvn

I agree with the moment of inertia posted by stewartcs, but the maximum stress on the tube is sigma_max = p*[(ro*L)^2]/I, where ro = tube outside radius, and L = tube length.

7. Jan 15, 2009

### stewartcs

Not for a cantilevered beam with a uniform load it wouldn't.

CS

8. Jan 16, 2009

### tommy100

what does the p stand for in F=pL?

cheers

9. Jan 16, 2009

### minger

p is the uniform pressure in units of force/length

10. Jan 16, 2009

### nvn

tommy100: In my post, p is pressure. According to your second post, pressure p = 1.0 kPa. Pressure has units of force per unit area.

11. Jan 16, 2009

### minger

For beam calculations, we typically treat the problem as a 2d problem, and p (or w as its sometimes called) has units of force/length. Remember that multiplying that uniform loading by the length of the loading gives an equivalent force.

If you have an actual pressure, then you'll need to take into account the shape of the beam and integrate the pressure around to find the net force in the vertical direction.

12. Jan 16, 2009

### stewartcs

minger brings up a good point (I actually overlooked the wind loading and took p = w). So for a wind load the equivalent force on the column would be:

$$F = \rho A V^2 C$$

where,

$\rho$ is the air density
A is the area
V is the wind velocity
C is the shape factor

Which would then make the max stress:

$$\sigma_{max} = \frac{\rho A V^2 C L y}{2I}$$

CS

13. Jan 16, 2009

### nvn

tommy100 already gave the differential pressure, p = 1.0 kPa, in post 3. The answer in my post already takes that into account. For the given differential pressure, all he needs to do is plug pressure p = 1.0 kPa into the formula in my post.

14. Jan 16, 2009

### minger

But.....pressure always acts normal to the surface. When we just use pressure as the uniform loading, it's assuming that the loading is also uniform across the beam (into the paper). However, since the beam is curved, the vertical component of the pressure varies across the beam.

For the most part, it was just semantics. For a 2d beam, the loading has units of force/length, not force/area. If you have a variable that's force/area, you need to determine if this is applicable to be able to use. In this case, it's probably fine because at the very least it will be a conservative estimate.

Again though, the determination of the pressure from wind loading should be verified.

15. Jan 16, 2009

### nvn

I already took that into account. For the given pressure (post 3), all tommy100 needs to do is plug pressure p = 1.0 kPa into the formula in my post (post 6).