应力平衡方程的推导
对于一点，已知其应力状态有：
σ x , τ x y , τ x z \sigma_x,\tau_{xy},\tau_{xz} σx,τxy,τxz
则其附近点的应力状态为：
σ
x
+
∂
σ
x
∂
x
d
x
,
τ
x
y
+
∂
τ
x
y
∂
x
d
x
,
τ
x
z
+
∂
τ
x
z
∂
x
d
x
\sigma_x+\frac{\partial \sigma_{x}}{\partial x}dx,\tau_{xy}+\frac{\partial \tau_{xy}}{\partial x}dx,\tau_{xz}+\frac{\partial \tau_{xz}}{\partial x}dx
σx+∂x∂σxdx,τxy+∂x∂τxydx,τxz+∂x∂τxzdx
之所以如此的原因：
对于附近的一点：
Δ
y
+
d
y
d
x
Δ
x
\Delta y+\frac{dy}{dx} \Delta x
Δy+dxdyΔx
也可由泰勒公式的二阶导知道：
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) 1 ! ( x − x 0 ) + o n ( x − x 0 ) f(x)=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+o^n(x-x_0) f(x)=f(x0)+1!f′(x0)(x−x0)+on(x−x0)
于是乎，x方向上的面积为 d y d z dydz dydz
根据外力平衡得：
σ x d y d z + τ x y d x d z + τ x z d x d y = ( σ x + ∂ σ x ∂ x d x ) d y d z + ( τ x y + ∂ τ x y ∂ x d x ) d x d z + ( τ x z + ∂ τ x z ∂ x d x ) d x d y \sigma_xdydz+\tau_{xy}dxdz+\tau_{xz}dxdy=(\sigma_x+\frac{\partial \sigma_{x}}{\partial x}dx)dydz+(\tau_{xy}+\frac{\partial \tau_{xy}}{\partial x}dx)dxdz+(\tau_{xz}+\frac{\partial \tau_{xz}}{\partial x}dx)dxdy σxdydz+τxydxdz+τxzdxdy=(σx+∂x∂σxdx)dydz+(τxy+∂x∂τxydx)dxdz+(τxz+∂x∂τxzdx)dxdy
化简得：
∂
σ
x
∂
x
d
x
d
y
d
z
+
∂
τ
x
y
∂
y
d
y
d
x
d
z
+
∂
τ
x
z
∂
z
d
z
d
x
d
y
=
0
\frac{\partial \sigma_{x}}{\partial x}dxdydz+\frac{\partial \tau_{xy}}{\partial y}dydxdz+\frac{\partial \tau_{xz}}{\partial z}dzdxdy=0
∂x∂σxdxdydz+∂y∂τxydydxdz+∂z∂τxzdzdxdy=0
∂ σ y ∂ y d x d y d z + ∂ τ y x ∂ x d y d x d z + ∂ τ y z ∂ z d z d x d y = 0 \frac{\partial \sigma_{y}}{\partial y}dxdydz+\frac{\partial \tau_{yx}}{\partial x}dydxdz+\frac{\partial \tau_{yz}}{\partial z}dzdxdy=0 ∂y∂σydxdydz+∂x∂τyxdydxdz+∂z∂τyzdzdxdy=0
∂ σ z ∂ z d x d y d z + ∂ τ x z ∂ x d y d x d z + ∂ τ y z ∂ x d z d x d y = 0 \frac{\partial \sigma_{z}}{\partial z}dxdydz+\frac{\partial \tau_{xz}}{\partial x}dydxdz+\frac{\partial \tau_{yz}}{\partial x}dzdxdy=0 ∂z∂σzdxdydz+∂x∂τxzdydxdz+∂x∂τyzdzdxdy=0
其中： d x d y d z dxdydz dxdydz可以约掉。
当有外力作用
F
F
F时，（这个外力是作用于微元体上的外力），其分量在三个方向上满足：
∂
σ
x
∂
x
+
∂
τ
x
y
∂
y
+
∂
τ
x
z
∂
z
+
F
x
=
0
\frac{\partial \sigma_{x}}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}+F_x=0
∂x∂σx+∂y∂τxy+∂z∂τxz+Fx=0
∂ σ y ∂ y + ∂ τ y x ∂ x + ∂ τ y z ∂ z + F y = 0 \frac{\partial \sigma_{y}}{\partial y}+\frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yz}}{\partial z}+F_y=0 ∂y∂σy+∂x∂τyx+∂z∂τyz+Fy=0
∂ σ z ∂ z + ∂ τ x z ∂ x + ∂ τ y z ∂ x + F z = 0 \frac{\partial \sigma_{z}}{\partial z}+\frac{\partial \tau_{xz}}{\partial x}+\frac{\partial \tau_{yz}}{\partial x}+F_z=0 ∂z∂σz+∂x∂τxz+∂x∂τyz+Fz=0
当考虑时间变化时，在 x , y , z x,y,z x,y,z三个方向上的位移 u , v , w u,v,w u,v,w时：
∂ σ x ∂ x + ∂ τ x y ∂ y + ∂ τ x z ∂ z = ρ ∂ 2 u ∂ t 2 \frac{\partial \sigma_{x}}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}=\rho \frac{\partial^2 u}{\partial t^2} ∂x∂σx+∂y∂τxy+∂z∂τxz=ρ∂t2∂2u
∂ σ y ∂ y + ∂ τ y x ∂ x + ∂ τ y z ∂ z = ρ ∂ 2 v ∂ t 2 \frac{\partial \sigma_{y}}{\partial y}+\frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yz}}{\partial z}=\rho \frac{\partial^2 v}{\partial t^2} ∂y∂σy+∂x∂τyx+∂z∂τyz=ρ∂t2∂2v
∂ σ z ∂ z + ∂ τ x z ∂ x + ∂ τ y z ∂ x = ρ ∂ 2 w ∂ t 2 \frac{\partial \sigma_{z}}{\partial z}+\frac{\partial \tau_{xz}}{\partial x}+\frac{\partial \tau_{yz}}{\partial x}=\rho \frac{\partial^2 w}{\partial t^2} ∂z∂σz+∂x∂τxz+∂x∂τyz=ρ∂t2∂2w
标签：tau,方程,frac,应力,推导,xz,xy,partial,sigma From： https://blog.csdn.net/qq_36980284/article/details/140362508