I would like to start by saying I am not an engineer, so I apologize if this question seems simpleminded.

I am making a manual tarp reel from pvc pipe, and I am trying to decided whether to use Schedule 40 or Schedule 80. I need to know which one will sag, deflect?, the least amount under its own weight. The pipe will be 10ft long and in a fixed-fixed position.

Type Outer Diameter Inner Diameter Weight E
Schedule 40 1.315” 1.049” .32 lbs/ft 400,000 psi
Schedule 80 1.315” .957” .41 lbs/ft 400,000 psi

When I used $\frac{wL^4}{384EI}$ , I got 4.683 for Sch 40 and 5.233 for Sch 80. Is this even the correct formula? The Sch 40 doesn’t visually seem to sag 4 inches.

Any help and explanations are much appreciated! Thank you!

  • 1
    $\begingroup$ What about torsion? Go get a stick of 3" pvc DWV. (The thinwall will deflect less, but it will only take a tiny load (say a gerbil) for the thickwall to win out.) $\endgroup$ – Phil Sweet Mar 6 at 21:39

You probably have mixed up the units (i.e. you have 10 ft pipe and inches for the Moment of area)

The formulas I've used:

$$ d_{max}= \frac{5wL^4}{384EI} $$

enter image description here

$$I = \frac{\pi (d_o^4 − d_i^4)}{ 64}$$

the values I've used are the following:

Type Outer Diameter Inner Diameter Weight
Schedule 40 1.315” (0.0334m) 1.049” (=0.0266m) .32 lbs/ft (=4.67 N/m)
Schedule 80 1.315” (0.0334m) .957” (=0.0234m) .41 lbs/ft (=5.98 N/m)


  • $E=400,000 psi =2.758 [GPa] $
  • $L=3.048[m]$

In metric units what I got from the data you provided for:

  • Schedule 40 is about 5.23[cm]. (which is about half what you have calculated) ($I_{40}=36354[mm^4]=0.0873 [in^4]$)
  • Schedule 80 is about 5.55[cm] ($I_{80}=43957[mm^4] =0.1056 [in^4]$)

My comparison gives $\frac{I_{80}}{I_{40}}=\frac{43957}{36354}\approx 1.2091$ and the weight disadvantage $\frac{w_{80}}{w_{40}}\approx 1.28$, therefore according to my calculations due to the self weight the SCH 80 will deflect more.

regarding the formula - if you are supporting the pipe at the two ends- the correct formula is similar to the equation in the original post times 5 .

enter image description here

  • $\begingroup$ Can you show your working? You (and OP) seem to disagree with Kamran $\endgroup$ – Jonathan R Swift Mar 6 at 21:06
  • $\begingroup$ @JonathanRSwift probably the difference with kamran is that I am using the diameter and not radius. IMHO it is not correct the way he is using the equation (i.e using 1.315 as external radius), but he is a lot more experienced and US trained so I am most likely wrong (but I can't see where - units in imperial does not help). (Also since he got already upvoted, someone must have verified his calculation ). $\endgroup$ – NMech Mar 6 at 21:32
  • $\begingroup$ @JonathanRSwift if you can spot my mistake please let me know. $\endgroup$ – NMech Mar 6 at 21:37
  • $\begingroup$ @NMech The deflection equation used by you and the OP are different. $\endgroup$ – r13 Mar 7 at 0:19
  • 1
    $\begingroup$ Sorry for doubting you - also don't doubt yourself! I also saw the upvote on Kamran as a hint it had maybe been checked, was going to wait for comparison working before checking myself but seems it's all solved for now. $\endgroup$ – Jonathan R Swift Mar 7 at 9:21

The formula is correct, the answer is likely wrong.

Let's look at the two variables that can affect the outcome.

The weight and the I.

The schedule 80 I is.

$$I = π (r_o^4 − r_i^4) / 4 = \pi ( 1.315^4-0.975^4)/4= 1.898"^4$$ $$\delta= \omega*120"^4/384EI $$

The schedule 40 I is.

$$I = π (r_o^4 − r_i^4) / 4 = \pi ( 1.315^4-1.049^4)/4=1.397"^4$$

this gives a 1.359 advantage in I compared to 1.281 disadvantage in weight.

So we get less deflection for schedule 80.


I made a mistake using the diameter for radius to calculate the I. so the new ratio of I of schedule 80 to 40 is 1+ 0.359/16= 1.0224.

Therefore considering that the weight of the schedule 80 pipe is relatively more it will deflect more.

  • 1
    $\begingroup$ You used the diameter instead of the radius for your calculation of the inertia. $\endgroup$ – JohnHoltz Mar 6 at 22:00
  • $\begingroup$ @JohnHoltz, sorry, you are right. i have landed at Catalina island. i will fix it when i get back to LA. so the ratio of the 2 I's will be much less and the wigjt will be the dominant factor. $\endgroup$ – kamran Mar 6 at 23:48
  • $\begingroup$ Is it possible that the correct equation is $\frac{5wL^4}{384EI}$ instead of $\frac{wL^4}{384EI}$ that the OP claimed? I've checked and doublechecked, and I don't understand why the 5 is ommited in the OP. Because you are backing it, I am doubting myself. $\endgroup$ – NMech Mar 7 at 4:56
  • $\begingroup$ @NMech, yes of course it's true. my apologies. i should be more careful. but it doesn't change the fact that the fraction w/I is linear and the only variable that changes. I sometimes answer these questions while holding short on the tarmac. no good. I have to wait for departure clearance for some 20 mins. I don't know what is causing the congestion recently. $\endgroup$ – kamran Mar 7 at 6:01
  • $\begingroup$ @NMech-About the 5 being omitted, I wasn’t sure about the correct equation because I’m not even in engineering school, I’m just a commoner. I’m very interested in learning and understanding, so would you mind explaining to me when to use which? $\endgroup$ – Callie Mar 7 at 12:44

For schedule 40 pipe:

I = \Pi x (1.315^4 - 1.049^4)/64 = 0.0873 in^4

\delta = [0.32(lb/ft) x (10 ft)^4 x (12 in/ft)^3]/[384 x (400000 lb/in^2) x (0.0873 in^4)] = 0.4124"

Please check my math.

  • $\begingroup$ I suggest using mathjax to describe equations. $\endgroup$ – Mahendra Gunawardena Mar 7 at 23:17

Here's my solution in GNU Octave. Everything is in inches. The formula you used (fix-fix support) gives 0.412" deflection for S-40 and 0.437" deflection for S80.

clc, clear
format long
E = 400000 ;
D_1 = 1.315;
d_1 = 1.049;
d_2 = 0.957;
w_1  = 0.32/12;
w_2 = 0.41/12;
L = 10*12;

I_1 = pi/64 * (D_1^4 - d_1^4);
I_2 = pi/64 * (D_1^4 - d_2^4);

fprintf('I_1 = %f\n', I_1)
fprintf('I_2 = %f\n', I_2)

disp("Pin-Pin Support")
d_1 = w_1 * L/2 /(24*E*I_1) *(L^3 - 2*L*(L/2)^2 + (L/2)^2 )
d_2 = w_2 * L/2 /(24*E*I_2) *(L^3 - 2*L*(L/2)^2 + (L/2)^2 )

disp("Pin-Roller Support")
d_1 = 5*w_1*L^4 / (384*E*I_1)
d_2 = 5*w_2*L^4 / (384*E*I_2)

disp("Fix-Fix Support")
d_1 = w_1*(L/2)^2 / (24*E*I_1) * (L - L/2)^2
d_2 = w_2*(L/2)^2 / (24*E*I_2) * (L - L/2)^2


I_1 = 0.087343
I_2 = 0.105609
Pin-Pin Support
d_1 =  1.655542171849186
d_2 =  1.754296885580493
Pin-Roller Support
d_1 =  2.060840877820563
d_2 =  2.183772056739203
Fix-Fix Support
d_1 =    4.121681755641125e-01
d_2 =    4.367544113478407e-01

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