User:Tiato/CE 130N

Based on the 1/23 and 1/28 lectures. This information is accurate to the best of my knowledge and understanding of the material, which isn't very much, so please watch out for errors.

Introduction
In CE 30, we were able to determine the deflection of a beam due to arbitrary transverse loads (point loads, distributed loads, and moments). By repeatedly integrating a differential equation defined by the loading conditions, we obtained an elastic curve describing the shape of the deflected beam. Specifically, from a distributed load function w(x), we integrated the fourth-order differential equation $$ EI \frac{d^4 y}{dx^4} = w(x) $$ four times in order to find an equation y(x), with four constants of integration that we determine using boundary conditions. This method is described in chapter 15 of Beer and Johnston's Statics and Mechanics of Materials.

In this section, we use a similar method in order to determine the axial deformation of a beam. In contrast with transverse deformation, we begin with a second-order differential equation; its solution contains only two constants of integration, which we determine using boundary conditions.

Basic form
We begin with the relation between internal axial force R(x) and the distributed axial force b(x), where b(x) has dimensions of force per length. A positive b(x) means force pulling towards the right. An external loading b(x) causes parts of the beam to undergo internal tension or compression R(x). (Remember to keep internal and external forces distinct, because the two play different roles in solving our problem.) The differential relation between R(x) and b(x) is readily derived by considering force equilibrium on a thin slice of the beam. The result is

Our objective is to find a function u(x) that gives the axial displacement for any given cross-section of the bar. This can be accomplished if we can relate R(x) to u(x), which we can do if we remember a few definitions.

Substitutions
We first recall the relation between force and stress:

We then recall the definition of strain, the relative displacement of particles in a body relative to length:

Finally, we will use the constitutive relationship between stress and strain for linear elastic materials

where E is the elastic modulus (Young's modulus).

Using these three definitions, we can rewrite Equation ($$) in terms of u(x).

Integration
By integrating twice, we obtain the general solution for u(x) in terms of b(x):

Note that, unlike in Professor Li's examples, I kept EA on the left side of the equation until the end. I do not think there is any reason to divide both sides by EA at the beginning (as long as EA is constant) as it will just make the integration of b(x) more complicated. Your results for C1 and C2 will be different depending on when you move EA.

Boundary conditions
We can solve for the two constants of integration by using two boundary conditions.

Fixed supports
For example, if both ends of the bar are fixed, then u(0) = 0 and u(L) = 0 (where L is the length of the bar). Substituting each pair of x and u into Equation ($$) will yield values for C1 and C2.

End loads
To find the constants of integration using known end loads, note that

which is the left side of Equation ($$). This is how we establish boundary conditions for known forces at the ends of the beam. For example, if the right end of the beam is free, then the internal force there R(L) is zero.

If there is an end-load F at the right end of the bar, we know the internal force there R(L) is F. Note the difference between distributed loads and end loads. If the internal force at either end of the bar R(0) or R(L) is nonzero, then this internal force is the result of either an end-load or a reaction from a fixed support. By itself, a distributed load cannot produce an internal force at the end of a bar because there is no length for the distributed load to act upon.

Review of singularity functions
To deal with point loads applied to the interior of a beam, we will take advantage of singularity functions. Again, these may be familiar from beam deflection in CE 30.

The Dirac delta function is zero everywhere except for a "spike" at zero; the area (integral) of this spike is defined as one. In engineering, we write it as $$\langle x \rangle^{-1}$$. The function can be scaled and translated in the form $$P \langle x - a \rangle^{-1}$$; this function is zero everywhere except for a "spike" at a with area P.

Its primary value to us is in its antiderivative, the Heaviside step function, which is zero when x < 0 and one when x > 1. We write its scaled, translated form as $$P \langle x - a \rangle^{0}$$; this function is zero to the left of a and P to the right of a.

We will also need the antiderivative of the Heaviside step function, the ramp function. It is zero when x < 0 and equal to x when x > 0. We write its scaled, translated form as $$P \langle x - a \rangle^{1}$$; this function is zero to the left of a and Px to the right of a.

Applying singularity functions
To model a point load P located at position x = a and pointing to the right, we use the Dirac delta function in place of b(x):

Integrating:

Consider the interpretation of Equation ($$). As we move along the bar in the x-direction, the internal force is constant (namely, C1) until we reach x = a, at which point the internal force decreases by P. If the ends of the bar are fixed, then the portion of the bar left of x = a is in tension, and the portion to the right is in compression. (Remember that tension is positive and compression is negative.)

The principle of superposition applies, so if you need to model both distributed loads and point loads, you can write b(x) as a sum of distributed load functions and Dirac delta functions.

Numerical solution
In order to solve for u(x) numerically, the second-order differential equation must be written as a system of two first-order differential equations:

These can then be written in matrix form as

The matrix on the left is named y with elements y1 and y2; it is the solution we are trying to find. The matrix on the right is a function that we define f(y, x). To set R equal to y2, replace it with a reference to the second element in y.