problems with integration

[Claudio L.]

I don't have a Prime, but here's my take on the problem itself:

You have several piecewise functions (your load functions), call them w1, w2, w3..., separated at points a, b, c...

You want to integrate each piecewise function a number of times to obtain shear, moment, rotation and deflection diagrams.

Any CAS will always give you the "family" of curves, and you must supply the boundary conditions for each piecewise function to determine the value of integration constants. The CAS may or may not add the constants explicitly (in most cases they don't).

Your steps to integrate this properly are (same as doing it by hand):

1. Compute the indefinite integral (integrate as many times as needed)
2. Add the correction constants to the expression, as a polynomial. If you integrated 'n' times, you'll need to add C1*x^(n-1)+C2*x^(n-2)+...+Cn. For example, if you integrate 3 times, the first integral will have +C1, which integrated will become +C1'*x, and a new C2 has to be added, and the next step you'll end up with C1''*X^2+C2'*x+C3.
3. Repeat the two previous steps for all your piecewise components
4. Now you have a system of equations with 'n' unknowns (the Cn constants), where you need to apply your boundary and continuity conditions. If you do deflection (let's call the resulting functions D1, D2, D3...), you can write D1(0)=0, D2(a)=D1(a), D3(b)=D2(b)... This basically does continuity of deflections at all points in the beam (and zero deflection at left support, assuming x=0 is the support). You'll need additional equations, so you can write some for continuity of the rotations, or points of zero moment, etc. (bear in mind that you'll have to derive the deflection functions with the added constants to obtain rotation and moment equations that have the SAME constants you want to solve for). When you have enough equations, you solve for all your constants and will have a complete family of equations that work the way they are supposed to be.
   

The key in all this is to obtain the proper boundary conditions to get the right constants. The solution proposed by Han to do multiple integration in one shot will only help you get a nice and clean expression at the first step of my list above, but you can't forget that you have to add a constant at each integration step, which ends up in a polynomial expression in x. His solution quickly dismisses those constants by applying an implicit boundary condition Fn(a)=0 at each integration step (resulting in Cn=-Fn(a)=0). This does not apply to your problem. In the beam these constants will exist and not necessarily be zero, it depends heavily on the configuration of the beam and loads.

Here's a full example using two constant distributed loads w1 and w2, that change at point 'a' in a simply supported beam:

w1(x)= w1 (it's a constant) from 0 to a
w2(x)= w2 (constant) from a to b (b=right support)

Step 1: integrate 4 times to get deflection:

V1(x)=w1*x ( +C1 ) --> Shears
M1(x)=w1*x^2/2 (+C1*x+C2 ) --> Moments
r1(x)=w1*x^3/6 ( +C1/2*x^2 + C2*x + C3 ) --> Rotations
d1(x)=w1*x^4/24 ( +C1/6*x^3 + C2/2*x^2 + C3*x + C4 ) --> Deflections

We can do the same integration for the other function:
[[ skipping the most obvious ]]
d2(x)=w2*x^4/24 ( +C5/6*x^3 + C6/2*x^2 + C7*x + C8 ) --> Deflections


Side note: Han's program would give you just the main term:
d1(x)=w1*x^4/24
d2(x)=w2*(x-a)^4/24
Also notice that d2() in this case would be "shifted by a" in the x direction, so the constants would end up with a different numeric value, but the end function would be the same after applying the correct boundary conditions. While it may look better, it doesn't really save you any work versus the more generic approach I'm following here. Especially because at a given point 'a' in the middle of a beam, in general neither deflection, rotation, moment or shears are zero, so all your constants will have a numeric value.

Now we have 8 unknown constants, here are the boundary conditions:

Deflections are zero at supports:
d1(0)=0 --> C4=0
d2(b)=0

Zero moments at supports:
M1(0)=0 --> C2=0
M2(b)=0

Material continuity at the point where load changes:
d1(a)=d2(a)
r1(a)=r2(a)

Equilibrium:
V1(0)+V2(b)=0
Continuity of shear at point a:
V1(a)=V2(a)

8 equations with 8 constants:
2 are immediately eliminated (C2 and C4 =0)
2 are easy to find from the last 2 shear equations (C1 and C5)

This was a simple example, I don't need to state that these boundary conditions only apply to this very simple example and need to be adapted for every problem.

The Prime CAN and WILL help you do all these individual steps, which involve not only the symbolic integration, but also valuing your functions at various points for the boundary conditions AND also solving the linear system of equations to get your constants. You just have to follow a very methodical solution path to the problem and it will get you there.

EDIT: If your instructor is asking for the integration constants, he is asking for trouble because those constants will change depending on how you integrate (not just a CAS may use different techniques, but each student could use a different technique too). But all final polynomial expressions (including the numeric constants) will have the exact same coefficients after taking them to the same canonical form. This could help you "check" if your constants are correct and prove your instructor that you are right.

Claudio

PS: My apologies to everybody for the excessive length of this post.