214
6 Free Vibration of Multiple Degrees of Freedom System
when x = 0, θ b = θ b i which yields
C 1 = θ b i
(6.82)
From Eqs. (6.81) and (6.82), the relation between the bending slopes at stations i
and (i + 1) can be obtained as
(θ b ) i + 1 =
1
(E I ) i
M i L i +
M i + 1 − M i
2
L i
+ (θ b ) i
or
(θ b ) i + 1 = (θ b ) i + (M i + M i + 1 )
1
(E I ) i
L i
2
(6.83)
Similarly, the bending deflection at a distance x from station i is
y b = y b i +
θ b dx
(6.84)
Substituting the value of θ b from Eq. (6.81) into Eq. (6.83) and by performing the
integration, we get
y b = y b i +
1
(E I ) i
M i
x
2
2
+
M i + 1 − M i
L i
x
3
6
+ θ b i x
(6.85)
The relation of the bending deflections between stations i and (i + 1) is as follows:
y b i + 1 = y b i + (θ b ) i L i +
M i
3
+
M i + 1
6
L
2
i
(E I ) i
(6.86)
Lastly, we can write
y i + 1 = y b i + 1 + y s i + 1
(6.87)
Sequence of calculations given by Eqs. (6.77), (6.78), (6.79), (6.83), (6.86) and
(6.87) is performed for an assumed value of p
2 .
A beam has two boundary conditions at each end. Therefore, out of four conditions
required in a beam, two are known. The calculation is started by assuming a trial
frequency p. The value of p that satisfies simultaneously both end conditions of the
beam is its correct value.
We start the computation from one end. A value of the frequency p is to be
assumed. Let it be p
. Let us take the case of a simply supported beam. At both ends,
y = M = 0, but the slope and the shear force are not zero at that end. We assume a
slope θ
, at the left hand end (for convenience, it is taken as 1.0), and shear force at
6 Free Vibration of Multiple Degrees of Freedom System
when x = 0, θ b = θ b i which yields
C 1 = θ b i
(6.82)
From Eqs. (6.81) and (6.82), the relation between the bending slopes at stations i
and (i + 1) can be obtained as
(θ b ) i + 1 =
1
(E I ) i
M i L i +
M i + 1 − M i
2
L i
+ (θ b ) i
or
(θ b ) i + 1 = (θ b ) i + (M i + M i + 1 )
1
(E I ) i
L i
2
(6.83)
Similarly, the bending deflection at a distance x from station i is
y b = y b i +
θ b dx
(6.84)
Substituting the value of θ b from Eq. (6.81) into Eq. (6.83) and by performing the
integration, we get
y b = y b i +
1
(E I ) i
M i
x
2
2
+
M i + 1 − M i
L i
x
3
6
+ θ b i x
(6.85)
The relation of the bending deflections between stations i and (i + 1) is as follows:
y b i + 1 = y b i + (θ b ) i L i +
M i
3
+
M i + 1
6
L
2
i
(E I ) i
(6.86)
Lastly, we can write
y i + 1 = y b i + 1 + y s i + 1
(6.87)
Sequence of calculations given by Eqs. (6.77), (6.78), (6.79), (6.83), (6.86) and
(6.87) is performed for an assumed value of p
2 .
A beam has two boundary conditions at each end. Therefore, out of four conditions
required in a beam, two are known. The calculation is started by assuming a trial
frequency p. The value of p that satisfies simultaneously both end conditions of the
beam is its correct value.
We start the computation from one end. A value of the frequency p is to be
assumed. Let it be p
. Let us take the case of a simply supported beam. At both ends,
y = M = 0, but the slope and the shear force are not zero at that end. We assume a
slope θ
, at the left hand end (for convenience, it is taken as 1.0), and shear force at