Example 2

Consider a bar of uniform cross-section shown in Fig. 3.4. Besides the point force P acting at the end x = L , there is an additional point force F acting at the midpoint, i.e. at the point .

Figure 3.4

We choose the same numerical values of the parameters EA , L and P as in example 1. Those values are given by equation (3.24). We choose the value of force F as

First, we chose the two-term approximation for the solution which is the same as in example 1. It is given by equations (3.25) and (3.26). Since the value of EA and the expressions for , i = 1, 2 are the same as in example 1, the coefficient matrix is the same as before (equation 3.28). For the evaluation of the right side vector, we use expression (3.53). Since, there is no distributed force, i.e. f = 0, and the force F acts at the midpoint , i.e. , equation (3.53) becomes

(3.55)

Substituting the expressions for (equation 3.26) and the values of P and F (equations 3.24 and 3.54 ), we get

Now, the matrix form of the algebraic equations (equation 3.16) becomes

(3.57)

The solution of this equation is given by

(3.58)

Therefore, the approximate solution (equation 3.25) becomes

(3.59)

Next, we choose the three-term approximation for this solution given by equations (3.33) and (3.34). Then the coefficient matrix is given by equation (3.36). Substituting the expression for (equation 3.34) and the values of P and F (equations 3.24 and 3.54), the expression (3.55) for { F } becomes:

	(3.60)

Now, the matrix form of the algebraic equations (equation 3.16) becomes

(3.61)

The solution of this equation is given by

(3.62)

Therefore, the approximate solution (equation 3.33) becomes

(3.63)

Thus, there is no improvement in the solution when the third term is added.

For comparison purpose, we find the exact solution. Substituting , f = 0 and the values of EA , L , P and F (equations 3.24 and 3.54), the boundary value problem (equations 3.45a, 3.45b, 3.45c, 3.45d) becomes

DE:		0 < x < 0.5, 0.5 < x <1;	(3.64a)
BC:		at x = 0,	(3.64b)
	(ii)	at x = 1,	(3.64c)
	(iii)		(3.64d)

The solution of this problem is given by

(3.65)

The graphical representation of the three solutions is shown in Fig. 3.5

Figure 3.5

Whereas the exact solution is piecewise linear, the approximate solution is quadratic over the whole domain. The maximum value of the displacement (i.e. u at x = 1) predicted by the approximate solution matches with the exact solution. However, the maximum error of the approximate solution is -8.33% which occurs at x = 0.5 (the point of application of F ).

Equations (3.59) and (3.65) give the following values of the displacement derivatives of the approximate solution and the exact solution at x = 1 :

	at x = 1,
	at x = 1.	(3.66)

Further, at the point of application of F (i.e., at x = 0.5) :

	at x = 0.5;
	at ,
	at	(3.67)

Thus, the approximate solution is neither able to provide a good approximation to the derivate nor able to capture the jump in the derivative.