# How to reduce a distributed load to a resultant force

by benanderson08
Tags: distributed, force, load, reduce, resultant
 P: 3 Hi everyone, So I am trying to study how different types of slurry will deflect a PVC pipe. The first step in my understanding is to reduce distributed load into a resultant force. I'm assuming that the resultant force will be placed in the center of the pipe (assuming a homogeneous slurry), but the magnitude eludes me. It's a 31in pipe, 4in diameter and the slurry (simple for the sake of calculation) is water @ 75F. The PVC is schedule 40, although that would matter for the reduction of the distributed load. Would anyone be able to point me in the right direction?
 Homework Sci Advisor HW Helper Thanks P: 12,739 You take the integral of the load along the pipe as usual. This means you need to know how the load is distributed over the pipe.
 P: 3 Although the integral method works for all loadings, it cannot be used if you look at the distributed load along the length of the pipe, because it is not uniformly distributed along the "b" distance of the pipe. So would I have to start off by looking down the center of the pipe i.e. a circle and then apply the integral method that way? so (integral) (b)(r^2-x^2)^0.5 would be the proper equation maybe???
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,466 How to reduce a distributed load to a resultant force If the cross-section of the slurry in the pipe varies over the length, you don't have to do a calculus integral, you can do a numerical integral with Simpson's Rule or equivalent. http://www.dummies.com/how-to/conten...sons-rule.html
 Sci Advisor HW Helper P: 2,124 benanderson08: Let's say the pipe has an inside diameter of d1. Let's say the slurry inside the pipe has a depth of h at the pipe centerline, and the slurry depth is constant along the pipe length. Therefore, the pipe inside radius is r = 0.5*d1. Looking down the center of the pipe, the cross-sectional area of the slurry is, A1 = (h - r)[h*(2*r - h)]^0.5 + (r^2){0.5*pi + asin[(h - r)/r]},where pi = 3.1416. Therefore, the total weight of the slurry in the pipe is, W1 = (rho1*g)*A1*L,where rho1*g = 9810 N/m^3 = 9.81e-6 N/mm^3, and L = pipe length. Therefore, to reduce the slurry uniformly-distributed load to a pipe midspan point load, place a force W1 at the pipe midspan. Let's try an example. In post 1 you have, d1 = 101.6 mm, and L = 787.4 mm. Therefore, r = 50.8 mm. And let's say the slurry depth in the pipe is, h = 70 mm. Therefore, A1 = (70 - 50.8)[70(2*50.8 - 70)]^0.5 + (50.8^2){0.5*3.1416 + asin[(70 - 50.8)/50.8]} = 19.20(47.032) + 2580.6{1.5708 + asin(0.3780)} = 19.20(47.032) + 2580.6{1.5708 + [(22.210 deg)/(57.2958 deg/rad)]} = 19.20(47.032) + 2580.6(1.5708 + 0.3876) = 903.01 + 5053.8 = 5957 mm^2.Therefore, the slurry total weight is, W1 = (9.81e-6 N/mm^3)(5957 mm^2)(787.4 mm) = 46.01 newtons. Therefore, the slurry can be represented as a pipe midspan force of W1 = 46.01 N. Did you want to include the pipe wall (self) weight? If so, compute the pipe self weight, W2, and add it to W1.
 Homework Sci Advisor HW Helper Thanks P: 12,739 ... and of course there is the possibility of doing a surface integral and/or exploiting what you know of the symmetry. ... and there you have it. Enjoy.