1550078481-Ordinary_Differential_Equations__Roberts_

306 Ordinary Differential Equations

To solve this differential equation and find the deflection of the beam, we need
to know the load, w(x), the moment of area of the cross section, I(x), and
the initial or boundary conditions. Unless I ( x) is an exponential function,
equation ( 43) will be a linear differential equation with variable coefficients.
The solution of the linear differential equations with variable coefficients is
discussed in chapter 7. Let us suppose I(x) is an exponential function, say,
I(x) = beax. Then I' = abeax, I" = a^2 beax and upon substitution (43)
reduces to

(44)

If a beam is simply supported at x = 0, then y(O) = 0 and y" (0) = 0.
The condition y(O) = 0 means the end of the beam is fixed- that is , it cannot
move vertically. The condition y" (0) = 0 means the beam can rotate about
x = 0 in the plane of deformation. A simply supported end of a beam is

sometimes called a pinned end- see Figure 6.12a. Of course, if a beam is

simply supported at x = L , then y(L) = 0 and y"(L) = 0.

If a beam is clamped at x = 0, then ·y(O) = 0 and y' (0) = 0. The condition
y' (0) = 0 means at x = 0 the slope of the beam is zero- that is , at x = 0
the beam is horizontal. A clamped beam is often a built-in beam as shown in
Figure 6.12b. If a beam is built-in at x = L , then y(L) = 0 and y' (L) = 0.

/y=O, y"=O

~run

a. Simply Supported Beam

or Pinned End

b. Clamped Beam or

Built-in Beam

Figure 6.12 T ypes of Beam Support at an End

Exercise 6. Find the equation for the deflection of a horizontal beam of
length L , assuming y' is small and I is constant (that is , solve equation (41))
under the following conditions:

a. Both ends are simply supported and the load is

i. uniform, w(x) = wo, where w 0 is constant ii. w(x) = w 0 sin(7rx/L)

iii. w(x) = w 0 cos (7rx/ L)

b. The end at x = 0 is simply supported, the end at x =Lis clamped, and

the load is

i. w(x) = wo ii. w(x) = wosin(7rx/L) iii. w(x) = w 0 cos(7rx/L)