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Fall 2005 (Week 13b) D. Beck
Name: _______________
More Second Order Differential Equations
- Warmup: Letโs finish the beam bending problem we started some time ago. Recall that the differential equation that governs the shape of the beam, y ( x ), is
p ( ) x dx
EI d y A (^) 4 =
4 .
We defined the shear and moment distributions via the differential equations ( ) (^) ( )
( ) (^) ( )
( ) ( ) (^) ( ).
2
2 p x dx
dV x dx
d M x
or
V x dx
dM x
px dx
dV x
With p ( ) x x 5
= 3 +^6 , the solution for M ( x ) is
( ) 2 3 5
M x L^3 L โ x + x + x โ
=โ โ^ +
a) Write the (homogeneous) second-order differential equation governing y ( x ) in terms of M ( x ) for the case where E and I (^) A are independent of x. ( ) (^) ( ) ( ) ( )
( ) (^) ( ).
2
2
2
2 2
2 4
4 2
2
M x dx
EI d yx
dx
d yx dx
EI d dx
px EI d yx dx
d M x
A
A A
=
b) Solve this equation in steps, first for dy / dx , and then for y ( x ) using the fact that y (0) = y ( L ) = 0.
Fall 2005 (Week 13b) D. Beck
( ) (^) ( )
( )
( )
( )
( )
( ). 300
(^34534)
1 3 3 4 3 4
(^451) 3
2
(^4512) 3
(^341) 2
2 3 2
2
โ + + +โ^ +
= โ โ^ +
=โ^ +
= =โ โ^ +
=โ โ^ +
=โ โ^ +
= =โ โ^ +
L L x x x L L x EI
y x
c L L L L L L
yL L L L L L cL
y c
EI yx L L x x x cx c
L L x x x c dx
EI dyx
M x L L x x x dx
EI d yx
A
A
A
A
c) Describe in a sentence what is represented by dy / dx. What are the values of dy / dx at x =0 and x = L? What signs would you expect for dy / dx at the two ends of the beam? Which direction (up or down) have we taken to be the positive direction for y ( x )? What sets this positive direction?
The quantity dy / dx is the slope of the line tangent to the beam at a given point x. ( )
( )
( )
= โก^ โ โ
= โก^ โ โ + + + +
= โก^ +
= โ โ^ +
=
=
3 4
3 4 3 4 3 4
3 4 0
(^23434)
L L
EI
L L L L L L
dx EI
dyx
L L
dx EI
dyx
L L x x x L L dx EI
dyx
A
xL A
x A
A
We would expect the signs of dy / dx to be opposite at opposite ends of the beam. The positive direction for y ( x ) is down and is set by taking p ( x ) positive in the downward direction. Therefore, we expect dy / dx to be positive at x = 0 and negative at x = L and that is what we find.
- In an earlier session, we talked about the motion of a pendulum, where the force, F , is proportional to the displacement and always opposed to it
2 dt
F =โ kx = md x
There is another very important class of problem that looks very similar and that we can also solve most efficiently by guessing the answer. Suppose the force is again proportional to the displacement, but that it is strictly proportional, i.e. no minus sign.
Fall 2005 (Week 13b) D. Beck
( ) ( )
( ) ( ) ( )
( )
โ
โ
t m
k k
t v m m
x k
e
k
x v m e
k
x v m xt
k
A B v m
A B v m
vt aA B k
xt A B x
xt Ae Be
kmt mkt
at at
1 1
0 1 0
1
0 0 1
0 0
1
0
(^10)
0
cosh sinh
1 1
You should note the similarity of this solution to that for the harmonic oscillator problem.
f) Suppose instead we know that the initial value of x is x 0 , and that (^) lim t โโ x ( ) t = 0. What is
the solution, x ( t ), in this case? ( ) ( ) ( )
( )
โโโ
โ โโโ โโ
โ โโ
โ
mk^ t
at at t
at at
xt xe
xt Ae Be A
xt A B x
xt Ae Be
1 0
0 (^0) lim 0
