# Beam Deflection in a Tapered Pipe

1. Jun 18, 2013

### jrmainia

I need help comparing the deflection of pipe with two different cases. Case 1 is a regular non-tapering pipe. The case 2 is a tapered pipe that should deflect less than the straight pipe. I am working in MathCad and can change the value of my variables, so I have set the two cases to be equal to one another (by saying that the variable D.w is equal to D) for the sake of comparison. The answers should be equal to one another so I know that there is a problem but I am not sure where. I think that it is within setting up the double integral since I checked the case 1 pipe deflection against a different pipe deflection calculator. Any help would certainly be appreciated.

View attachment Mathcad - Tapered Pipe Bending.pdf

2. Jun 18, 2013

### SteamKing

Staff Emeritus
The problem is that 100 ft. of 6" pipe which is cantilevered is well outside of the elastic range for which the beam equations are valid. Your stress calculation of 113 ksi. should have tipped you off. Grade A pipe has a tensile yield of 30 ksi and Grade B pipe goes to 35 ksi. You are trying to analyze a piece of spaghetti.

3. Jun 18, 2013

### jrmainia

4. Jun 18, 2013

### SteamKing

Staff Emeritus
I think the problem is with your definition of the bending moment in the tapered beam calculation.

For a cantilever beam, the bending moment is a maximum at the fixed end (x = 0 in your coordinate system) and zero at the free end (x = L). By defining the bending moment as W*x in your MathCad integral, the BM is zero at the fixed end and W*L at the free end. If you change the BM to W*(L-x), this should solve your problem.

5. Jun 19, 2013

### jrmainia

That makes sense.

I changed my coordinate system picture in Case 2 so that the distance measured for the moment is zero at the free end.

I added equation y3. It is only to show that using a constant value for I, the deflection found through double integration is still incorrect, so this makes me think that the issue lies in the way I have the integrals set up.

D.w still is equal to D so all the deflection should still be equal to each other.

View attachment Mathcad - Tapered Pipe Bending.pdf

View attachment Tapered Pipe Bending.xmcd

6. Jun 19, 2013

### SteamKing

Staff Emeritus
In your latest MathCad calculation, you have introduced a new variable x1, which you set equal to zero.
Not surprisingly, your integral also evaluates to zero.

Instead of using x1, I think if you had defined the moment as W*(L-x) instead of W*x in your integral, the calculation would have worked out.

7. Jun 21, 2013

### jrmainia

In an effort to simplify the calc and let mathcad do the work, I defined (x=0) @ the free end and (L=x) at the fixed end and set up functions in terms of x. I also included graphs for the diameter, moment, moment of inertia, deflection angle, and deflection all in terms of x.

I think that the functions for diameter, moment and moment of inertia are correct. The deflection graph seems opposite what I expect in that @ x=L the angle should be 0deg. Could that be due to the definite integration range? Should this change to something else?

The deflection graph crosses 0 so I know that can't be right.

I still think that I have an error in how I define the integrals.

View attachment Alt Tapered Pipe Bending.xmcd
View attachment Mathcad - Alt Tapered Pipe Bending.pdf

Thanks for the help!

Last edited: Jun 21, 2013
8. Jun 21, 2013

### SteamKing

Staff Emeritus
I don't understand why you don't realize that your definition of the bending moment as a function of x is in error, and this is why your MathCad calculation of deflection fails to agree with your initial calculation.

Look, the bending moment M(x) must equal -1000 lbs times 120 in = -120000 in-lbs. at x = 0
and M(x) = 0 at x = 120 in. If M(x) = W*(L-x) is used in the MathCad calculation, then the results will agree using both methods.

9. Jun 22, 2013

### nvn

jrmainia: Your attached pdf file in post 3 is fine, as-is, except just change M*x in your two double-integral equations therein to W*(L - x).