3.15 Equations of Compatibility for Strains
73
And take the derivative of γ zx with respect to y and x.
Thus,
∂
2
γ zx
∂ x ∂ y
=
∂
3 u
∂ x ∂ y ∂z
+
∂
3 w
∂ x 2 ∂ y
(3.44)
Now, adding Eqs. (3.42) and (3.44) and subtracting Eq. (3.43), we get
−
∂
2
γ yz
∂ x 2
+
∂
2
γ xz
∂ x∂ y
+
∂
2
γ xy
∂ x∂z
=
2∂
3 u
∂ x∂ y∂z
(3.45)
From the above, we get
2∂
2
ε x
∂ y∂z
=
∂
∂ x
−
∂γ yz
∂ x
+
∂γ xz
∂ y
+
∂γ xy
∂z
(3.46)
Similarly, we can get
2∂
2
ε y
∂ x∂z
=
∂
∂ y
−
∂γ zx
∂ y
+
∂γ yx
∂z
+
∂γ yz
∂ x
(3.47)
2∂
2
ε z
∂ x∂ y
=
∂
∂z
−
∂γ xy
∂z
+
∂γ yz
∂ x
+
∂γ zx
∂ y
(3.48)
Thus, for a three-dimensional situation, the following are the six equations of
compatibility of strain.
∂ 2 ε x
∂ y 2 +
∂ 2 ε y
∂ x 2 =
∂ 2 γ xy
∂ x∂ y
∂ 2 ε y
∂z 2 +
∂ 2 ε z
∂ y 2 =
∂ 2 γ yz
∂ y∂z
∂ 2 ε z
∂ x 2 +
∂ 2 ε x
∂z 2 =
∂ 2 γ zx
∂z∂ x
(3.49)
2∂ 2 ε x
∂ y∂z =
∂
∂ x
−
∂γ yz
∂ x +
∂γ xz
∂ y +
∂γ xy
∂z
2∂ 2 ε y
∂z∂ x =
∂
∂ y
∂γ yz
∂ x −
∂γ zx
∂ y +
∂γ xy
∂z
2∂ 2 ε z
∂ x∂ y =
∂
∂z
∂γ yz
∂ x +
∂γ zx
∂ y −
∂γ xy
∂z