Let’s carefully compute the natural frequency of a fixed-clamped (F-C) beam using the one-term
Galerkin approximation and compare it with the exact solution. We'll go step-by-step in detail.
Step 1: Governing Equation and Boundary Conditions
The governing equation for the transverse vibration of an Euler-Bernoulli beam is:
𝐸𝐼
∂4 𝑤(𝑥, 𝑡)
∂2 𝑤(𝑥, 𝑡)
+
𝜌𝐴
= 0,
∂𝑥 4
∂𝑡 2
where:





𝑤(𝑥, 𝑡): Transverse displacement,
𝐸: Young's modulus,
𝐼: Area moment of inertia,
𝜌: Mass density per unit volume,
𝐴: Cross-sectional area.
Assume a separable solution:
𝑤(𝑥, 𝑡) = 𝑊(𝑥)𝑇(𝑡),
where 𝑊(𝑥) is the spatial mode shape, and 𝑇(𝑡) = sin(𝜔𝑡) represents harmonic motion.
Substituting this into the governing equation gives:
𝐸𝐼𝑊 (4) (𝑥) − 𝜌𝐴𝜔2 𝑊(𝑥) = 0.
For a fixed-clamped beam, the boundary conditions are:
1. At 𝑥 = 0 (fixed end): 𝑊(0) = 0, 𝑊 ′ (0) = 0,
2. At 𝑥 = 𝐿 (clamped end): 𝑊(𝐿) = 0, 𝑊 ′ (𝐿) = 0.
Step 2: Exact Solution
The exact solution involves solving the transcendental equation derived from the boundary
conditions:
cos(𝛽𝐿)cosh(𝛽𝐿) = 1,
where 𝛽𝑛 𝐿 are the roots of this equation. For the fundamental frequency (𝑛 = 1), the first root is:
𝛽1 𝐿 ≈ 4.73004.
The natural frequency is then:
𝜔Exact = 𝛽12 √
𝐸𝐼
.
𝜌𝐴
Substituting 𝛽1 𝐿 = 4.73004, we get:
4.73004 2 𝐸𝐼
𝜔Exact = (
) √ .
𝐿
𝜌𝐴
Numerically:
𝜔Exact ≈
22.373 𝐸𝐼
√ .
𝐿2
𝜌𝐴
Step 3: One-Term Galerkin Approximation
In the Galerkin method, we approximate the mode shape 𝑊(𝑥) using a trial function that satisfies
the geometric boundary conditions. For the F-C beam, choose:
𝑥 2
𝑥 3
𝑥 4
𝑊(𝑥) = 𝑎 [( ) − 2 ( ) + ( ) ].
𝐿
𝐿
𝐿
This trial function satisfies:
 𝑊(0) = 0,
 𝑊 ′ (0) = 0,
 𝑊(𝐿) = 0,
 𝑊 ′ (𝐿) = 0.
Weak Form of the Governing Equation:
The weak form of the governing equation is:
𝐿
∫ [𝐸𝐼𝑊 ′′ (𝑥)𝜙 ′′ (𝑥) − 𝜌𝐴𝜔2 𝑊(𝑥)𝜙(𝑥)] 𝑑𝑥 = 0,
0
where 𝜙(𝑥) = 𝑊(𝑥) is the weighting function.
Substituting 𝑊(𝑥) into the equation, we compute the stiffness term (𝐾) and mass term (𝑀):
Step 4: Stiffness Term (𝐾)
The stiffness term is:
𝐿
𝐾 = ∫ 𝐸 𝐼[𝑊 ′′ (𝑥)]2 𝑑𝑥.
0
First, compute 𝑊 ′′ (𝑥):
𝑥 2
𝑥 3
𝑥 4
𝑊(𝑥) = 𝑎 [( ) − 2 ( ) + ( ) ].
𝐿
𝐿
𝐿
Differentiating twice:
2𝑥 6𝑥 2 12𝑥 3
𝑊 ′ (𝑥) = 𝑎 [ 2 − 3 + 4 ],
𝐿
𝐿
𝐿
𝑊
′′ (𝑥)
2 12𝑥 36𝑥 2
= 𝑎 [ 2 − 3 + 4 ].
𝐿
𝐿
𝐿
Now square 𝑊 ′′ (𝑥):
4 48𝑥 216𝑥 2
[𝑊 ′′ (𝑥)]2 = 𝑎2 [ 4 − 5 +
].
𝐿
𝐿
𝐿6
Integrate over [0, 𝐿]:
𝐿
4 48𝑥 216𝑥 2
𝐾 = 𝐸𝐼 ⋅ 𝑎2 ∫ [ 4 − 5 +
] 𝑑𝑥.
𝐿
𝐿6
0 𝐿
Integrating term by term:
𝐿
4
4𝐿
4
𝑑𝑥 = 4 = 3 ,
4
𝐿
𝐿
0 𝐿
∫
𝐿
−48𝑥
−48 𝐿2 −24
𝑑𝑥
=
⋅ = 3 ,
𝐿5
𝐿5 2
𝐿
0
∫
𝐿
216𝑥 2
216 𝐿3 72
∫
𝑑𝑥 = 6 ⋅ = 3 .
𝐿6
𝐿
3
𝐿
0
Combine terms:
4 24 72
12
𝐾 = 𝐸𝐼 ⋅ 𝑎2 ⋅ ( 3 − 3 + 3 ) = 𝐸𝐼 ⋅ 𝑎2 ⋅ 3 .
𝐿
𝐿
𝐿
𝐿
Thus:
𝐾=
12𝐸𝐼
.
𝐿3
Step 5: Mass Term (𝑀)
The mass term is:
𝐿
𝑀 = ∫ 𝜌 𝐴[𝑊(𝑥)]2 𝑑𝑥.
0
Square 𝑊(𝑥):
𝑥 4
𝑥 5
𝑥 6
𝑥 7
𝑥 8
[𝑊(𝑥)]2 = 𝑎2 [( ) − 4 ( ) + 6 ( ) − 4 ( ) + ( ) ].
𝐿
𝐿
𝐿
𝐿
𝐿
Integrate term by term:
𝐿
𝑥 4
𝐿
∫ ( ) 𝑑𝑥 = ,
5
0 𝐿
𝐿
𝐿
𝑥 5
𝐿
∫ ( ) 𝑑𝑥 = ,
6
0 𝐿
𝐿
𝑥 6
𝐿
∫ ( ) 𝑑𝑥 = ,
7
0 𝐿
𝐿
𝑥 7
𝐿
∫ ( ) 𝑑𝑥 = ,
8
0 𝐿
𝑥 8
𝐿
∫ ( ) 𝑑𝑥 = .
9
0 𝐿
Combine terms:
1 4 6 4 1
𝑀 = 𝜌𝐴 ⋅ 𝑎2 ⋅ 𝐿 ⋅ ( − + − + ).
5 6 7 8 9
Simplify numerically:
1
= 0.2,
5
4
= 0.6667,
6
6
≈ 0.8571,
7
4
= 0.5,
8
1
≈ 0.1111.
9
𝑀 = 𝜌𝐴 ⋅ 𝑎2 ⋅ 𝐿 ⋅ (0.2 − 0.6667 + 0.8571 − 0.5 + 0.1111) = 𝜌𝐴 ⋅ 𝑎2 ⋅ 𝐿 ⋅ 0.0015.
Thus:
𝑀 = 𝜌𝐴 ⋅ 𝑎2 ⋅
𝐿
.
315
Step 6: Natural Frequency
The natural frequency is given by:
𝐾
𝜔Galerkin = √ .
𝑀
Substitute 𝐾 and 𝑀:
12𝐸𝐼
3
𝜔Galerkin = √ 𝐿
.
𝐿
𝜌𝐴 ⋅
315
Simplify:
3780𝐸𝐼
𝜔Galerkin = √
.
𝜌𝐴𝐿4
Numerically:
𝜔Galerkin ≈
61.48 𝐸𝐼
√ .
𝐿2
𝜌𝐴
Step 7: Comparison
Exact Solution:
𝜔Exact ≈
22.373 𝐸𝐼
√ .
𝐿2
𝜌𝐴
Galerkin Approximation:
𝜔Galerkin ≈
61.48 𝐸𝐼
√ .
𝐿2
𝜌𝐴
The Galerkin approximation overestimates the exact solution. Estimate vibration frequency of the
uniform F-C beam by using one-term Galerkin approximation method and compare the result with
exact solution the answer is correct for this question