Roark's Formulas for Stress and Strain - pipe span

[Master slacker]

In pulling a tube bundle from its shell, we need to have a vertical support so the bundle won't just fall to the deck when yanked out. We'll do this with a steel pipe spanning structural steel on the deck above the exchanger and dropping a come-along and choker to the bundle.

Referencing Roark's formulas, Table 13.3, Case 8a (page 636), I'm getting membrane stress 17000 PSI and bending stress of 101000 PSI for a 3 in. Sch 80 pipe and a 3800 lb load. (With a 6 in. Sch 40 I get 6500 and 23000).

When assuming the pipe is a beam, I calculate bending stress of 73000 PSI and max shear of 2100 PSI. (73000 and 1100)

It's not often that I do actual engineering calculation work and this is the first time I've used Roark at all. My only intent is to ensure that the pipe to be used for vertical support won't crush / fail.

Am I looking at this correctly? :huh:

 

[MA_PE]

What is the span? why do you need a pipe section? Why not an I-section. simple beam bending, so why use Roark's just grab a steel manual and use the tables.

Are you adding any dynamic factor to the weight of the bundle when you compute the load?

Are you fixing the ends of the pipe to the structural steel?

 

[Master slacker]

The span is roughly 5 ft. We typically use pipe because we won't have to worry about kinking or cutting chokers on sharp edges. Also, we have lots of pipe and not so much beam. The load will be "purely" static. The pipe will lie across the span on top of the grating with no rigid fixing and the structural steel is W14 so I am not worrying about that.

This isn't the first time this pipe / choker method has been used in bundle extraction, but I'm going through this exercise to determine how safe it is for each instance.

 

[MA_PE]

for bending stress use Fb = M/S where M is moment and S = section modulus for the pipe.

6 in. sch 40 pipe wall t = 0.280 in., S = 8.50 in^3

for yield say 36 ksi

simple span moment with concentrated load at the center M = PL/4 = L=60 in.

substitute M --> FB = PL/4/S for 6500 psi - I calculate your load as ~3.7kips. Is that correct?

this gives 36/6.5 = safety factor of ~5.5 against yield for bending

Is that what your're figuring?

 

[Master slacker]

(sigh) I'm an idiot. I figured it was a silly mistake because I used Mc/I for my beam calcs. So, like I did when I worked problems for three months in prep for the PE exam, I wrote my units out. I noticed "I" had units of "in2" and not "in4". Fixed that issue and I'm getting 18223 PSI for the 3" sch. 40 and 3707 PSI for the 6" sch. 40 (offset load from center). All is right with the world again.

Today's lesson: CHECK YOUR UNITS! :brick:

:lmao:

 

[MA_PE]

so you're load is 2100lbs?

 

[Master slacker]

Nope. 3800 lbs. M=Pab/L (MERM A-119). a ~ 10 in.

M = 31448 lb-in

c = 1.75 in (or 3.3125 in)

I = 3.02 in4 (or 28.1 in4)

sb = 18223 PSI (or 3707 PSI)

 

[Lumber Jim]

Haven't gone through the calc but have some experience with "pipe" sections in bending. Localized buckling can occur in thin walled vessels even when the bending stress checks out. This is the reason why pipes kink instead of bending smoothly. (Leading to a catastrophic failure) You have checked both the compressive stress on the top of the pipe and the tensile stress on the bottom of the pipe but you might want to check buckling at the point at which you are applying the load. Some people limit deflection and there are some equations to use out there for buckling. I've not had to calculate this specific application (only larger diameter thin wall with stiffeners) but After a quick google search, Eng tips forum has addressed this to a certain extent. Don't have any references sitting on my couch otherwise I'd look some stuff up but I'm sure you'll be able to find what you need...

2 cents.