工程材料學實驗
王雲哲
國立成功大學土木工程學系
2/22/2026 ~ 6/26/2026
量測技術
（物理/實體）實驗
（數值/虛擬）實驗
創造材料
（物理/實體）實驗
（數值/虛擬）實驗
Uniaxial Deformation
σA = P = kx = k(ΔL)
σ/L = k(ΔL)/(AL) = (k/A)ε
σ = (kL/A)ε = Eε
kL
Young’s Modulus E =
A
E
Shear Modulus G =
2(1 + ν)
Poisson’s ratio
−1 < ν <
1
2
Bending
Bazant and Cedolin
σ=
Torsion
Mz My
=
Izz
Iyy
θ=
TL
GJ
τ=
Tr
J
γ=
J = I xx + I yy
θ γ max
=
L
a
Resonant Ultrasound
Tr
GJ
•RUS = Resonant ultrasound spectroscopy
•Two piezoelectric transducers
•Corner contact between transducers and solid
samples to reduce mechanical impedance
•Apply sinusoidal driving voltage to one
transducer
•Record vibration signal from the other sensor
Requirement on sample geometry:
Cube: f = 2 G
πL
ρ
1
G
Short cylinder (L=D): f =
2L ρ
Sphere:
Why? Analytical solutions
available.
If not available, use numerical
solutions
70w%Cu+30w%Zn
黃銅
Viscoelastic damping can be
determined from the width of
resonant peaks:
tanδ =
1 Δf
3 f
Wave Ultrasound
De ne three-dimensional longitudinal and transfer wave speed
ρcL2
cL =
E(1 − ν)
ρ(1 + ν)(1 − 2ν)
cT =
E
2ρ(1 + ν)
\rho
2
−1 < ν <
1
2
4
c ≈ 1.155cT
3 T
Equal when ν = − 1
cL ≥
\nu
∂
∂
∂
+j +k
∂x
∂y
∂z
∂u
E
E
ρ 2 =
∇2 u +
∇( ∇ ⋅ u)
∂t
2(1 + ν)
2(1 + ν)(1 − 2ν) \nabla
Boundary Conditions?
∂2u \mu 2 \lambda
ρ 2 = μ ∇ u + (λ + μ) ∇( ∇ ⋅ u) Initial Conditions?
Shear modulus Lamé constant
∂t
∂2
∂2
∂2
∇ = 2+ 2+ 2
∂x
∂y
∂z
2
Navier’s Equation
λ=
Eν
(1 − 2ν)(1 + ν)
∇=i
∂2u
= cT2 ∇2 u + (cL2 − cT2) ∇( ∇ ⋅ u)
2
∂t
E
3(1 − 2ν)
Bulk modulus
K=
ρu1,tt = μu1,xx + (λ + μ)(u1,x + u1,y + u1,y),x
ρu2,tt = μu2,yy + (λ + μ)(u2,x + u2,y + u2,z),y
ρu3,tt = μu3,zz + (λ + μ)(u3,x + u3,y + u3,z),z
fi
∂2u
E
E
ρ 2 =
∇2 u +
∇( ∇ ⋅ u)
∂t
2(1 + ν)
2(1 + ν)(1 − 2ν)
Poisson’s ratio
cL =
E(1 − ν)
ρ(1 + ν)(1 − 2ν)
cT =
E
=
2ρ(1 + ν)
G
ρ
Longitudinal
Transverse
u = uL + uT
3 unknowns
6 unknowns
∇ ⋅ uT = 0
∇ × uL = 0
solenoidal vector eld
irrotational vector eld
One eqn.
Three eqn.
Two indepnd.
·· + u
·· ) = c 2 ∇2 (u + u ) + (c 2 − c 2) ∇( ∇ ⋅ u )
(u
L
T
L
T
L
T
L
T
2 2
2
2
··
∇ ⋅ uL = cT ∇ ( ∇ ⋅ uL) + (cL − cT ) ∇ ⋅ ∇( ∇ ⋅ uL)
·· = c 2 ∇2 ( ∇ ⋅ u )
∇2
∇⋅u
L
L
L
2
2
2
∂2uL
∂
∂
∂
= cL2
+ 2 + 2 uL
2
2
∂t
∂x
∂y
∂z )
(
·· + u
·· ) = c 2 ∇2 (u + u ) + (c 2 − c 2) ∇( ∇ ⋅ u )
(u
L
T
L
T
L
T
L
T
·· = c 2 ∇2 ( ∇ × u ) + (c 2 − c 2) ∇ × [ ∇( ∇ ⋅ u )]
∇×u
T
T
L
T
L
T
·· = c 2∇ × ∇2 u + (c 2 − c 2) ∇[ ∇ ⋅ ( ∇ × u )]
∇×u
T
T
T
L
T
·· = c 2∇ × ∇2 u
∇×u
T
T
T
·· = c 2 ∇2 u
u
T
T
T
∂2f(x, y, z, t)
∂t 2
Resonant frequencies of a cube
Zero stresses on all surfaces. Di cult.
= c 2 ∇2 f(x, y, z, t)
Algebraic equations for solving U, V, W
For nontrivial solutions
on x =0 and a
i.e. Ω1 = Ω2 = 0 and Ω3 = -s.
on y = 0 and b
Resonant Modes of a Cube
L
on z = 0 and c
Easier boundary conditions
(i.e. ux = 0, σxy = σxz = 0)
On x = 0 and a
De ne wave numbers
On y = 0 and b
On z = 0 and c
Assume displacement elds
Assume
Lowest mode
=
In RUS, the 1st torsional resonant frequency (Hz) of the cube
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fi
fi
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