Assignment of Theory of Vibratiobns

1. Advanced Theory of Vibrations (AE 721)
Assignment No. 4
November 21, 2020
1 Problem Statement
Given a simply supported beam with constant E, I, and L acted upon by an external force
f (x, t) = sin
πx
L
sin (Ωt), find w (x, t), M (x, t) and σxx (x, t). The initial conditions of
the problem are w (x, 0) = w0 (x) = 0 and ˙w (x, 0) = ˙wx (x) = 0.
2 Natural Frequencies and Mode Shapes
The free vibration response of a simply supported beam made with constant E and I is
governed by
∂2
w (x, t)
∂t2
+ c2 ∂4
w (x, t)
∂x4
= 0
c2
=
EI
ρA
(1)
The boundary conditions for a simply supported beam in terms of w (x, t) are given by:
w (x = 0, t) = 0
∂2
w
∂x2
(x=0,t)
= 0
w (x = L, t) = 0
∂2
w
∂x2
(x=L,t)
= 0
(2)
The free vibration solution can be found using the method of separation-of-variables as
w (x, t) = W (x) T (t) (3)
In terms of W (x), the boundary conditions become
1

2. W (x = 0) = 0
d2
W
dx2
x=0
= 0
W (x = L) = 0
d2
W
dx2
x=L
= 0
(4)
Substituting the assumed solution w (x, t) = W (x) T (t) in to the governing differential
equations leads to
c2
W
d4
W
dx4
= −
1
T
d2
T
dt2
= −ω2
(5)
Here the choice of separation constant, ω2
, is made based on the experience that the nat-
ural frequencies comes from the temporal equation. Separating the spatial and temporal
equations yields
d4
W
dx4
− β4
W = 0
d2
T
dt2
+ ω2
T = 0
where, β4
=
ω2
c2
=
ρAω2
EI
(6)
The solution of temporal equation is given by
T (t) = A cos ωt + B sin ωt (7)
and A and B are constants that will eventually be determined by the specified initial
conditions after being combined with spatial solutions using orthogonality of mode shapes.
The solution of the spatial equation can be expressed as
W = C1 cos βx + C2 sin βx + C3 cosh βx + C4 sinh βx (8)
Applying the first two boundary condition (first two expressions of Equation 4) on spatial
solution gives
C1 + C3 = 0
−β3
C1 + β3
C3 = 0
(9)
which implies that C1 = C3 = 0. The spatial solution now becomes
W = C2 sin βx + C4 sinh βx (10)
Applying the last two boundary condition (last two expressions of Equation 4) on spatial
solution gives
2

3. C2 sin (βL) + C4 sinh (βL) = 0
−C2β2
sin (βL) + C4β2
sinh (βL) = 0
(11)
Adding the above equations gives
C4 1 + β2
sinh βL = 0 (12)
Since sinh βL and (1 + β2
) cannot be zero for positive values of β and L, C4 must be zero.
Subtracting the two expressions in Equation 11 gives
C2 1 + β2
sin βL = 0 (13)
For non-trivial solutions of governing differential equation, sin βL must vanish. The zeros
of sin (βL) are nπ, with n = 1, 2, 3, · · · .
(βL)n = nπ n = 1, 2, 3, · · ·
βn =
nπ
L
n = 1, 2, 3, · · ·
(14)
Using Equation 6, we can find the natural frequencies in terms of βn.
ωn =
n2
π2
L2
EI
ρA
(15)
Note that C1 = C3 = C4 = 0. If the value of C2 corresponding to βn is denoted as (C2)n,
we have
Wn (x) = (C2)n sin βnx (16)
where, Wn (x) are the mode shapes of pinned-pinned beam. The solution corresponding
to n−th value of βn is
wn (x, t) = (C2)n sin βx (An cos ωnt + B sin ωnt)
wn (x, t) = A∗
n sin βx cos ωnt + B∗
n sin βx sin ωnt
(17)
The general or total solution can be expressed as
w (x, t) =
∞
n=1
A∗
n sin βL cos ωnt + B∗
n sin βL sin ωnt (18)
Orthogonality of Mode Shapes
The principle of orthogonality of mode shapes states that



l
0
WnWmdx = 0 n = m
l
0
WnWmdx = L
2
n = m
(19)
3

4. Summary
Natural frequencies: ωn =
n2
π2
L2
EI
ρA
Mode shapes: Wn (x) = sin βnx
3 Forced Response
The forced response of a beam can be determined using the mode summation principle.
For this, the deflection of the beam is assumed as
w (x, t) =
∞
n=1
Wn (x) qn (t) (20)
where Wn (x) is the n−th mode shape which satisfies the spatial differential equation
and qn (t) is the generalized coordinate in the n−th mode. By substituting the assumed
solution in to the forced vibrations equation, we obtain
ρA
∂2
w (x, t)
∂t2
+ EI
∂4
w (x, t)
∂x4
= f (x, t)
ρA
∂2
∂t2
∞
n=1
Wn (x) qn (t) + EI
∂4
∂x4
∞
n=1
Wn (x) qn (t) = f (x, t)
∞
n=1
ρAWn (x)
d2
qn (t)
dt2
+
∞
n=1
qn (t) EI
d4
Wn (x)
dx4
= f (x, t)
(21)
From Equation 6, EI
d4
Wn (x)
dx4
= ρAω2
nWn (x). Substituting in to the above equation, we
have
∞
n=1
ρAWn (x)
d2
qn (t)
dt2
+
∞
n=1
qn (t) ρAω2
nWn (x) = f (x, t) (22)
By multiplying the above equation by Wm (x), integrating from 0 to L, and using principle
of orthogonality of mode shapes, we have
d2
qm
dt2
+ ω2
qm =
2
ρAL
L
0
f (x, t) Wm (x) dx (23)
Since m is a dummy symbol, we can replace it any other symbol, say n.
d2
qn
dt2
+ ω2
qn =
2
ρAL
L
0
f (x, t) Wn (x) dx = Qn (t) (24)
Substituting the expressions for Wn (x) and f (x, t) in to Qn (t), we have
4

5. Qn =
2
ρAL
L
0
sin
πx
L
sin Ωt sin
nπx
L
dx
Qn =
2
ρAL
sin Ωt
L
0
sin
πx
L
sin
nπx
L
dx
(25)
Because of orthogonality principle, the integral in the above expression will be equal
L
2
when n = 1, else it will be zero. So the above equation becomes



d2
qn
dt2 + ω2
nqn = 1
ρA
sin Ωt n = 1
d2
qn
dt2 + ω2
nqn = 0 n = 2, 3, 4, · · ·
(26)
This is essentially the same as the equation of motion of an undamped single degree-of-
freedom system. Its total solution can be expressed as
qn (t) = An cos ωnt + Bn sin ωnt + Particular Solution (27)
where the first two terms on the right-hand side represent the transient or free vibra-
tions (resulting from the initial conditions) and the third term denotes the steady-state
vibration (resulting from the forcing function).
For n = 2, 3, 4, · · · , both forcing function and initial conditions are zero, therefore, intu-
itively, we know that qn = 0 for n = 2, 3, 4, · · · . However, we will derive it mathematically
in the following.



qn (t) = An cos ωnt + Bn sin ωnt + 1
ρA(ω2
n−Ω2)
sin Ωt n = 1
qn (t) = An cos ωnt + Bn sin ωnt n = 2, 3, 4, · · ·
(28)
The final step is determine arbitrary constants An and Bn using initial conditions. From
Equation 20, the forced response of a beam is given by
w (x, t) =
∞
n=1
Wn (x) qn (t) (29)
The next step is to convert the initial condition specified on w (x, t) to initial condition
on q (t).
Displacement Initial Condition
The displacement initial condition is given as:
w (x, 0) = w0 (x) = 0 (30)
Substituting in Equation 29 gives:
w (x, 0) =
∞
n=1
Wn (x) qn (0) = 0 (31)
5

6. By multiplying the above equation by Wm (x) , integrating from 0 to L, and using principle
of orthogonality of mode shapes, and replacing dummy variable m with n we have
L
0
w (x, 0) Wm (x) dx =
∞
n=1
L
0
Wm (x) Wn (x) qn (0) dx
L
0
w (x, 0) Wm (x) dx =
L
2
qn (0)
qn (0) =
2
L
L
0
w (x, 0) Wn (x) dx
(32)
Since w (x, 0) = 0, qn (0) = 0.
Velocity Initial Condition
The velocity initial condition is given as:
˙w (x, 0) = w0 (x) = 0 (33)
Substituting in Equation 29 gives:
˙w (x, 0) =
∞
n=1
Wn (x) ˙qn (0) = 0 (34)
By multiplying the above equation by Wm (x) , integrating from 0 to L, and using principle
of orthogonality of mode shapes, and replacing dummy variable m with n we have
L
0
˙w (x, 0) Wm (x) dx =
∞
n=1
L
0
Wm (x) Wn (x) ˙qn (0) dx
L
0
˙w (x, 0) Wm (x) dx =
L
2
˙qn (0)
˙qn (0) =
2
L
L
0
w (x, 0) Wn (x) dx
(35)
Since ˙w (x, 0) = 0, ˙qn (0) = 0.
Total Response
Now we have initial conditions on qn (t) (qn (0) = 0 and ˙qn (0) = 0), we can solve Equation
28. Using these initial conditions, we have



qn = 2
ρAL(ω2
n−Ω2)
sin Ωt − Ω
ωn
sin ωnt n = 1
qn (t) = 0 n = 2, 3, 4, · · ·
(36)
6

7. Using the above values, the total response of the beam is given by
w (x, t) =
∞
n=1
Wn (x) qn (t)
w (x, t) = W1 (x) q1 (t)
w (x, t) =
2
ρAL (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin β1x
w (x, t) =
2
ρAL (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin
πx
L
(37)
The bending moment distribution is given by:
M (x, t) = EI
∂2
w
∂x2
M (x, t) = −
2π2
EI
ρAL3 (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin
πx
L
(38)
The bending stress is given by:
σxx (x, y, t) = −Ey
∂2
w
∂x2
σxx (x, y, t) = −
2Eyπ2
ρAL3 (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin
πx
L
(39)
Solution
w (x, t) =
2
ρAL (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin
πx
L
M (x, t) = −
2π2
EI
ρAL3 (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin
πx
L
σxx (x, y, t) = −
2Eyπ2
ρAL3 (ω2
1 − Ω2)
sin Ωt −
Ω
ω1
sin ω1t sin
πx
L
ω1 =
π2
L2
EI
ρA
(40)
7