Stress Analysis
Daiyao Xu
December 6, 2022
1
Introduction
This project mainly focus on work through the content of stress analysis
2
2.1
Stress
设计零件基本原则
1. evaluate the most likely modes of failure of the member 2.
2.2
Condition of Equilibrium
The equilibrium of the forces is the state in which the forces applied on a body are in balance.
2.3
Definition and components of stress
a free body diagram is simply a sketch of a body with all the appropriate forces bothn known and
unknown action on it.decompose the external force
dFx
∆Fx
=
∆A
dA
∆Fy
dFy
∆Fz
τxy = lim
=
, τxz = lim
=
∆A→0 ∆A
∆A→0 ∆A
dA

 
τxx τxy τxz
σx τxy τxz
[τij ] =  τyx τyy τyz  =  τyx σy τyz
τzx τzy τzz
τzx τzy σz
σx = lim
∆A→0
2.4
dFz
dA


variation of stress within one a body
近似条件：考虑足够小的二维微圆（应力可以uniformly分布在每个面上）Fx 和Fy 相互独立。（这里
的力是力比上单位体积）假设σx 施加在x面的负方向。对于应力关于位置的变化可以用泰勒级数的前
两项描述：
σx +
∂σx
dx
∂x
1
2.5
守恒公式
整理上式可得二维应力表达式
2.6
∂σx
∂τxy
+
+ Fx = 0
∂x
∂y
∂σy
∂τxy
+
+ Fy = 0
∂y
∂x
平面应力转化
px = σx cos θ + τxy sin θ
py = τxy cos θ + σy sin θ
σx′ = px cos θ + py sin θ
τx′ y′ = py cos θ − px sin θ
σx′ = σx cos2 θ + σy sin2 θ + 2τxy sin θ cos θ
τxy′ = τxy cos2 θ − sin2 θ + (σy − σx ) sin θ cos θ
σy′ = σx sin2 θ + σy cos2 θ − 2τxy sin θ cos θ
Equations (1.17) can be converted to a useful form by introducing the following trigonometric identities:
cos2 θ =
1
1
(1 + cos 2θ), sin θ cos θ = sin 2θ
2
2
1
sin2 θ = (1 − cos 2θ)
2
The transformation equations for plane stress now become
1
1
(σx + σy ) + (σx − σy ) cos 2θ + τxy sin 2θ
2
2
1
τx′ y′ = − (σx − σy ) sin 2θ + τxy cos 2θ
2
1
1
σy′ = (σx + σy ) − (σx − σy ) cos 2θ − τxy sin 2θ
2
2
σx′ =
2.7
最 大 应 力 principle stress and 最 大 平 面 内剪切力 maximum in-plane shear
stress（
（mohr’s circle 在 二维 压 力 下 ）
maximum or minimum σx′ the necessary condition dσx′ /dθ = 0
− (σx − σy ) sin 2θ + 2τxy cos 2θ = 0
tan 2θp =
2τxy
σx − σy
tan 2θp =
2τxy
σx − σy
τx′ y′ = 0 on a principal plane.
σmax,min = σ1,2 =
σx + σy
±
2
2
s
σx − σy
2
2
2
+ τxy
maximum shearing stress. Thus, setting dτx′ y /dθ = 0, we now have (σx − σy ) cos 2θ + 2τxy sin 2θ = 0
or
tan 2θs = −
σx − σy
2τxy
extreme values of shearing stress as follows:
s
2
1
σx − σy
2 = ± (σ − σ )
+ τxy
τmax = ±
1
2
2
2
maximum shearing stress, designated τmax . Normal stresses acting on the planes of maximum shearing
stress can be determined by substituting the values of 2θs from Eq. (1.21) into Eqs. (1.18a) and (1.18c)
tan 2θs = −
σx′ =
2.8
构件中的主应力
2.9
三维应力变换
σx − σy
2τxy
1
1
(σx + σy ) + (σx − σy ) cos 2θ + τxy sin 2θ
2
2
1
σ ′ = σave = (σx + σy )
2
cos α = cos(n, x) = l
cos β = cos(n, y) = m
cos γ = cos(n, z) = n
l2 + m2 + n2 = 1
The other two areas are similarly obtained. Collectively, we have
AQAB = Al,
AQAC = Am,
AQBC
= An
˙
Next, from the equilibrium of x, y, z-directed forces together with Eq. (a), we obtain, after canceling
A,
px = σx l + τxy m + τxz n
py = τxy l + σy m + τyz n
pz = τxz l + τyz m + σz n
Consider now a Cartesian coordinate system x′ , y ′ , z ′ , in which x′ coincides with n and y ′ , z ′ lie on
an oblique plane. The x′ y ′ z ′ and xyz systems are related by the direction cosines: l1 = cos (x′ , x) , m1 =
cos (x′ , y), and so on. The notation corresponding to a
x′
y′
z′
x
l1
l2
l3
y
m1
m2
m3
3
z
n1
n2
n3
complete set of direction cosines is shown in Table 1.2. The normal stress σx′ is found by projecting
px , py , and pz in the x′ direction and adding
σx′ = px l1 + py m1 + pz n1
σx′ = σx l12 + σy m21 + σz n21 + 2 (τxy l1 m1 + τyz m1 n1 + τxz l1 n1 )
Similarly, by projecting px , py and pz in the y ′ and z ′ directions, we obtain, respectively,
τx′ y′ =σx l1 l2 + σy m1 m2 + σz n1 n2 + τxy (l1 m2 + m1 l2 )
+ τyz (m1 n2 + n1 m2 ) + τxz (n1 l2 + l1 n2 )
τx′ z′ =σx l1 l3 + σy m1 m3 + σz n1 n3 + τxy (l1 m3 + m1 l3 )
+ τyz (m1 n3 + n1 m3 ) + τxz (n1 l3 + l1 n3 )
σy′ =σx l22 + σy m22 + σz n22 + 2 (τxy l2 m2 + τyz m2 n2 + τxz l2 n2 )
σz′ =σx l32 + σy m23 + σz n23 + 2 (τxy l3 m3 + τyz m3 n3 + τxz l3 n3 )
τy′ z′ =σx l2 l3 + σy m2 m3 + σz n2 n3 + τxy (m2 l3 + l2 m3 )
+ τyz (n2 m3 + m2 n3 ) + τxz (l2 n3 + n2 l3 )
From Table 1.2, observe that Eqs. (1.30a) are the sums of the squares of the cosines in each row,
and Eqs. (1.30b) are the sums of the products of the adjacent cosines in any two rows.
2.10
三维空间下最大应力
σx′ = σx l2 + σy m2 + σz n2 + 2 (τxy lm + τyz mn + τxz ln)
找到 σx′ 的极值. l, m, and n 并不互相独立t, （l2 + m2 + n2 = 1）, 只有 l and m may be regarded as
independent variables. Thus,
∂σx′
=0
∂m
∂σx′
= 0,
∂l
Differentiating Eq. (a) as indicated by Eqs. (b) in terms of the quantities in Eq. (1.26), we obtain
px + pz
∂n
= 0,
∂l
py + pz
∂n
=0
∂m
替换 n2 = 1 − l2 − m2 , we have ∂n/∂l = −l/n and ∂n/∂m = −m/n. Introducing these into Eq. (c),
the following relationships between the components of p and n are determined:
px
py
pz
=
=
l
m
n
这些比例关系表明，应力结果必须平行于单位法线，因此，不包含剪切成分。因此，在σx′具有极
值或主值的平面上，即主平面，剪应力消失。
So far, we have shown that three principal stresses and three principal planes exist. Denoting the
principal stresses by σp , Eq. (d) may be written as
px = σp l,
py = σp m,
4
pz = σp n
These expressions, together with Eq. (1.26), lead to
(σx − σp ) l + τxy m + τxz n = 0
τxy l + (σy − σp ) m + τyz n = 0
τxz l + τyz m + (σz − σp ) n = 0
A nontrivial solution for the direction cosines requires that the characteristic stress determinant vanish:
σx − σp
τxy
τxz
τxy
σy − σp
τyz
τxz
τyz
σz − σp
=0
Expanding Eq. (1.32) leads to
σp3 − I1 σp2 + I2 σp − I3 = 0
where
I1 = σx + σy + σz
2
2
2
I2 = σx σy + σx σz + σy σz − τxy
− τyz
− τxz
I3 =
σx
τxy
τxz
τxy
σy
τyz
τxz
τyz
σz
The three roots of the stress cubic equation (Eq. 1.33) are the principal stresses. Three corresponding
sets of direction cosines, along with these stresses, are used to establish the relationship of the principal
planes to the origin of the nonprincipal axes. The principal stresses are the characteristic values or
eigenvalues of the stress tensor τij . Since the stress tensor is a symmetric tensor whose elements are
all real, it has real eigenvalues. That is, the three principal stresses are real [Ref. 1.7]. The direction
cosines l, m, and n are the eigenvectors of τij .
3
STRAIN
3.1
定义应变
normal or longitudinal strains are:
εx =
∂u
,
∂x
εy =
∂v
∂y
假设 αx ≈ tan αx ；AB ≈ A′ B ′ . 可以得到, αx ≈ ∂v/∂x, 逆时针为正，同时可以得到 −αy ≈ ∂u/∂y.
The total angular change of angle DAB− is defined as the shearing strain and denoted by γxy :
γxy = αx − αy =
∂u ∂v
+
∂y
∂x
Lagrangian Description
在这个描述中，每个粒子都用它的初始位置以及时间来描述，此外，位移矢量被指定为初始位置
矢量的函数,它们被视为四个独立变量。参照图1，设为粒子A的原始坐标矢量，为同一粒子变形后的
5
坐标矢量，为位移矢量。对于三维情况，上述三个向量各有三个分量，在二维情况下，每个分量只有
两个。每个粒子的位置可以被指定为其初始坐标和时间的函数。
Figure 1: Lagrangian Description
a′ = a′ (x, y, z, t) = a′ (x, y, z)
for statics or equivalently x′ = x′ (x, y, z, t) = x′ (x, y, z)
statics y ′ = y ′ (x, y, z, t) = y ′ (x, y, z)
for statics z ′ = z ′ (x, y, z, t) = z ′ (x, y, z)
for
for statics Thus, the
displacement vector {u} is obtained as
{u} = {a′ } − {a}
or equivalently, for a three-dimensional case
ux = x′ − x = ux (x, y, z, t) = ux (x, y, z)
uy = y ′ − y = uy (x, y, z, t) = uy (x, y, z)
uz = z ′ − z = uz (x, y, z, t) = uz (x, y, z)
3.2
for statics
for statics
for statics
应变的定义
随着固体从一种构型转变为另一种构型，每一点附近的物质都会发生平移和旋转，从而产生应变。一
般来说，在一个固体体中，应变不会是均匀的，它将是身体的不同点，因为位移在粒子之间是不同
的。因此，位移分量被认为是粒子坐标的函数，进一步说，这些函数相对于他们的坐标的导数认为是
连续的。当位移和导数不再是很小的时候产生了应变。 被称为”有限应变”。现在，应变的表达式将被
开发。参照图2，设B和C是实体中变形前无限接近的两个点，变形后它们的新位置是B′和C′。
6
Figure 2:
The Cartesian co-ordinates of these four points can be written as
B : x, y, z
B ′ : x′ , y ′ , z ′ (or) x + ux , y + uz , z + uz
C : x + dx, y + dy, z + dz
C ′ : x′ + dx′ , y ′ + dy ′ , z ′ + dz ′ (or)
[x + dx + ux + dux ] , [y + dy + uy + duy ] , [z + dz + uz + duz ]
(dS)2 = (dx)2 + (dy)2 + (dz)2
2
2
2
B C之间距离
2
(dS ′ ) = (dx′ ) + (dy ′ ) + (dz ′ )
B ′ C ′ 之间距离
Expanding (x′ + dx′ ) using equation (2.2), employing Taylor’s series expansion and neglecting terms
involving higher powers of ∂x, ∂y and ∂z, we get
dx′ = (∂x′ /∂x) dx + (∂x′ /∂y) dy + (∂x′ /∂z) dz
Based on this result and similar results for dy ′ and dz ′ , we can write
  ∂x′ ∂x′ ∂x′  


 dx′ 
 dx 
∂x
∂y
∂z
′
′
′

∂y
∂y 
dy ′
dy
=  ∂y
∂x
∂y
∂z  


′
′
′ 
∂z
∂z
∂z ′
dz
dz
∂x
∂y
∂z
′
or briefly, {da } = [F ]{da}
where, [F ] is the matrix containing ”Spatial Deformation Gradients” ∂x′ /∂x, ∂y ′ /∂x and so
on. 用新的表达方法重写
(dS)2 = (dx)2 + (dy)2 + (dz)2
可得
2
(dS ′ ) = [dxdydz][F ]T [F ]{dxdydz}T
其中 [F ]T [F ] 叫做”Green’s strain tensor” and is denoted by the matrix [G].
2
(dS ′ ) − (dS)2 = [dxdydz][F ]T [F ] − [I]{dxdydz}T
7
where, [I] is the identity matrix. The so-called ”Lagrangian finite strain tensor” is defined by
[L] = 1/2 [F ]T [F ] − [I]
It is possible and convenient to re-write the above expression using the displacement gradients instead
of differentials of position vector, as follows. First, the expression for dx′ is written, by substituting
(dx + dux ) for dx′ and employing Taylor’s series expansion, as
dx′ =
∂ (x + ux )
∂ (x + ux )
∂ (x + ux )
dx +
dy +
dz
∂x
∂y
∂z
Similarly, the expressions for dy ′ and dz ′ can be re-written. Now, if we substitute these expressions
in equations (2.9), (2.10) and (2.11), the equivalent expression for Lagrangian finite strain tensor is
obtained to be
[L] = 1/2 [J] + [J]T + [J]T [J]
where,


[J] = 
3.3
∂ux
∂x
∂uy
∂x
∂uz
∂x
∂ux
∂y
∂uy
∂y
∂uz
∂y
∂ux
∂z
∂uy
∂z
∂uz
∂z



Infinitesimal strain 小 形变
If the displacement gradients, ∂ux /x, ∂uy /y and so on, are small compared to unity, then the finite
strain components given by equation (2.12) reduces to infinitesimal strain components. Thus, the
resulting equations represent small deformations. In equation (2.15), if the displacement gradients are
small compared to unity, theproducts of these displacement gradients become negligible and hence we
can write the Lagrangian infinitesimal strain tensor as
[L] = (1/2) [J] + [J]T
If both the displacement components are very small, the squares and products of the partial derivatives
become negligible and so on
Lxx = ∂ux /x
Lxy = (∂ux /∂y + ∂uy /∂x) /2
在连续介质力学中，无穷小应变理论是一种描述固体变形的数学方法，其中假定材料粒子的位移远小
于（实际上是无穷小）物体的任何相关尺寸；从而可以假设其几何形状和材料的本构属性（例如密度
和刚度）在空间的每个点都不会因变形而改变。 限应变理论形成对比。
无穷小应变理论在普遍用于对由相对坚硬的弹性材料（如混凝土和钢）建造的结构进行应力分
析，因为此类结构设计的共同目标是在典型载荷下将其变形降至最低。然而，这种近似在薄柔性体的
情况下需要谨慎，例如易受明显旋转影响的杆、板和壳，从而使结果不可靠。
8
3.4
Strain Components
假设由于某些加载的结果，只产生法向应力。可以预计，最初的矩形平行四边形将保持其矩形形状，
而作为刚性体的平移和旋转可能已经发生。因
因此 ， 可 能 只 有 平 行 四 边 形 的 边 长 发 生 了 变 化 ， 产 生 了
称为 ”扩
扩张 ”的
的体积 变 化 。这种类型的变形只产生法向应变“dilatation”，用εxx , εyy and εzz
Figure 3:
如果我们认为这个无限小的元素受到了剪切应力的作用，那么可以说，除了刚体的平移和旋转之
外，最初的矩形平行四边形的边的方向可能已经改变，而体积可能没有变化。这种类型的变形被称为
“distortion”，它可以通过平行四边形的两条边与初始直角的角度变化来衡量。这些角度变化的测量
方法被称为 “engineering shear strains”，它们用 γxy , γyz , γzx , γyx , γzy and γxz . 表示。
因此，九个应变（三个法向应变和六个工程剪切应变）对于描述相邻点沿三个正交轴的相对运动
是必要的。三个法向应变产生扩张（体积的变化），但形状没有变化，六个工程剪切应变产生变形，
但体积可能没有变化。这些应变在直角坐标系中被写为


εxx γxy γxz
 γyx εyy γyz 
γzx γzy εzz
cylindrical co-ordinate system written as

εrr
 γθr
γzr
γrθ
εθθ
γzθ
9

γrz
γθz 
εzz
3.5
几何角度
Figure 4:
εxx =
A′ C ′′ − AC
[ux + (∂ux /∂x) dx + dx − ux ] − dx
∂ux
=
=
AC
dx
∂x
In a similar manner, the other strain components can be shown to be
[uy + (∂uy /∂y) dy + dy − uy ] − dy
∂uy
A′ B ′′ − AB
=
=
AB
dy
∂y
∂uz
=
∂z
εyy =
εzz
Now, the angular deformation is determined. To this end, consider the rotation of edge AC with
respect to A.
tan β =
εxx =
[uy + (∂uy /∂x) dx] − uy
C ′ C ′′
∂uy /∂x
=
=
′
′′
AC
[ux + (∂ux /∂x) dx + dx] − ux
1 + ∂ux /∂x
A′ C ′′ − AC
[ux + (∂ux /∂x) dx + dx − ux ] − dx
∂ux
=
=
AC
dx
∂x
In a similar manner, the other strain components can be shown to be
[uy + (∂uy /∂y) dy + dy − uy ] − dy
∂uy
A′ B ′′ − AB
=
=
AB
dy
∂y
∂uz
=
∂z
εyy =
εzz
Now, the angular deformation is determined. To this end, consider the rotation of edge AC with
respect to A.
tan β =
C ′ C ′′
[uy + (∂uy /∂x) dx] − uy
∂uy /∂x
=
=
A′ C ′′
[ux + (∂ux /∂x) dx + dx] − ux
1 + ∂ux /∂x
When only infinitesimal deformations are considered, then ∂ux /∂x is very small compared to unity
and so,
tan β =
Similarly,
∂uy
≈β
∂x
∂ux /∂y ≈ α Thus, the change in angle between sides AB and AC due to deformation is
given as,
γxy = α + β = ∂ux /∂y + ∂uy /∂x
10
In a similar manner it can be shown that
γyz = ∂uy /∂z + ∂uz /∂y
γzx = ∂ux /∂z + ∂uz /∂x
Equations (2.22 − 2.24) and (2.28 − 2.30) together constitute the ”strain displacement relations”,
cylindrical co-ordinate system can be shown to be
εrr = ∂ur /∂r
εθθ =
1 ∂uθ
ur
+
r
r ∂θ
εzz = ∂uz /∂z
1 ∂ur
∂uθ
uθ
+
−
r ∂θ
∂r
r
∂ur
∂uz
=
+
∂z
∂r
1 ∂uz
∂uθ
+
=
∂z
r ∂θ
γrθ =
γrz
γθz
3.6
compatibility equations
Six strain components can be uniquely obtained fromstrain displacement relations. However, the
reverse situation is not that simple,
即从六个应变分量确定三个位移分量是不容易的。造成这种困难的原因是，为了获得三个未知
数ux , uy 和uz ，有六个方程需要解决。但这种（相反的）情况对实际分析问题是非常重要的! 因此，很
明显，如果位移分量的值是唯一解和连续的，显然六个应变分量之间必须存在某些关系，称为 ”应变
相容方程” “strain compatibility equations”。
在物理学上，这可以从以下方面观察到：如果我们在拉伸一个物体之前考虑一个三角形ABC，如
果指定一个任意的应变场，同一个三角形在拉伸后可能占据该图右侧所示的两个可能会出现间隙或重
叠。
The compatibility conditions can be derived by eliminating the displacement components ux , uy
and uz from the strain-displacement equations.
εxx = f (x, y)
εyy = g(x, y)
γxy = h(x, y)
Let us now examine (i) how consistent are these data and (ii) can we compute the displacements
uniquely from these strains.
εxx = ∂ux /∂x
= f (x, y)
εyy = ∂uy /∂y
= g(x, y)
γxy = ∂ux /∂y + ∂uy /∂x = h(x, y)
11
From the above three equations, the following partial derivative equations can be obtained.
∂ 3 ux
∂2f
=
∂y 2
∂x∂y 2
2
∂ h
∂ 3 ux
=
∂x∂y
∂x∂y 2
∂2h
∂2f
+
=
∂x∂y
∂y 2
∂2g
∂ 3 uy
=
∂x2
∂y∂x2
3
∂ uy
+
∂y∂x2
∂2g
∂x2
;
The experimental data must satisfy the above equation (2.41) in order to get consistent displacements. This type of equation is known as the ”equation of compatibility”. In the three-dimensional
situation, there are six equations of compatibility similar to the equation (2.41):
∂ 2 εxx
∂ 2 εyy
∂ 2 γxy
+
=
2
2
∂y
∂x
∂x∂y
∂ 2 εzz
∂ 2 γyz
∂ 2 εyy
+
=
∂z 2
∂y 2
∂y∂z
Further, from equations (2.38), (2.22 − 2.24) and (2.28 − 2.30), the following equations can be derived:
∂ 3 ux
∂ 2 εxx
=
∂y∂z
∂x∂y∂z
∂γyz
∂ 2 uy
∂ 2 uz
=
+
∂x
∂x∂z
∂x∂y
2
∂γzx
∂ uz
∂ 2 ux
=
+
∂y
∂x∂y ∂y∂z
∂γxy
∂ 2 ux
∂ 2 uy
=
+
∂z
∂y∂z
∂x∂z
From these three equations, the following condition is derived:
∂
∂γzx
∂γxy
∂γyz
∂ 2 εxx
+
+
−
=2
∂x
∂x
∂y
∂z
∂y∂z
∂γzx
∂γxy
∂ ∂γyz
∂ 2 εyy
−
+
=2
∂y ∂x
∂y
∂z
∂z∂x
∂ ∂γyz
∂γzx
∂γxy
∂ 2 εzz
+
−
=2
∂z ∂x
∂y
∂z
∂x∂y
The equations (2.42) and (2.44) are called as the ”compatibility equations”, given in the Cartesian coordinates. In cylindrical co-ordinates, the compatibility equations can be derived following the above
procedure and they are,
∂ 2 εrr
∂ 2 εzz
∂ 2 γrz
+
=
∂z 2
∂r2
∂r∂z
2
∂ εrr
∂εrr
∂ 2 (rεθθ )
∂ 2 (rγrθ )
−r
+r
=
2
2
∂θ
∂r
∂r
∂r∂θ
2
2
∂
ε
∂ε
∂
ε
∂γ
∂ 2 γθz
zz
θθ
zz
rz
r2
+
r
+
−
r
=
r
2
2
∂z
∂r
∂θ
∂z
∂z∂θ
∂ ∂ (rγrθ )
∂ ∂ (rγθz ) ∂γrz
∂ ∂ (rεθθ )
+
−
= 2r
− εrr
∂z
∂θ
∂θ
∂r
∂θ
∂z
∂r
2
2
∂ 1 ∂ (rγθz ) ∂γrz
∂ (rγrθ )
∂ (γεrr )
r2
−
−
=
∂r r
∂r
∂θ
∂z∂r
∂z∂θ
∂ ∂γrθ
∂ γθz
1 ∂γrz
∂ 1 ∂εzz
−r
−
= −2
∂z ∂z
∂r r
r ∂θ
∂r r ∂θ
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
 x = ρ sin φ cos θ
y = ρ sin φ sin θ

z = ρ cos φ
再用Jacobi矩阵 a :

∂(x, y, z) 
=
∂(ρ, φ, θ)
∂x
∂ρ
∂y
∂ρ
∂z
∂ρ
∂x
∂φ
∂y
∂φ
∂z
∂φ
∂x
∂θ
∂y
∂θ
∂z
∂θ


sin φ cos θ
 
sin φ sin θ
=
cos φ
ρ cos φ cos θ
ρ cos φ sin θ
−ρ sin φ

−ρ sin φ sin θ
ρ sin φ cos θ 
0
对应的Jacobi行列式 Q 的值为 : ρ2 sin φ, 此时的体积微元 Q 为: dV = ρ2 sin φdρdφdθ
13