Mechanics of Materials
CHAPTER 7 ANALYSIS OF THE STATE OF
STRESS AND STRAIN
§ 7–4 PRINCEPAL STRESSES AND THEIR TRAJECTORIES OF
THE BEAM § 7–5
ANALYSIS OF TRIAXIAL STRESSED STATE—METHOD
OF STRESS CIRCLE
§ 7–6 ANALYSIS OF STRAIN IN A PLANE
§ 7–7 RELATION BETWEEN STRESS AND STRAIN UNDER
COMPLEX STRESSED STATE—（ GENERALIZED HOOKE’S LAW ）
§ 7–8 STRAIN -ENERGY DENSITY UNDER COMPLEX
STRESSED STATE
§ 7–1 CONCEPTS OF THE STATE OF STRESS
§ 7–2 ANALYSIS OF THE STATE OF PLANE STRESS —
ANALYTICAL METHOD
§ 7–3 ANALYSIS OF THE STATE OF PLANE
STRESS — GRAPHYCAL METHOD
第七章 应力状态与应变状态分析
§ 7–1 应力状态的概念
§ 7–2 平面应力状态分析 ——解析法
§ 7–3 平面应力状态分析 ——图解法
§ 7–4 梁的主应力及其主应力迹线
§ 7–5 三向应力状态研究 ——应力圆法
§ 7–6 平面内的应变分析
§ 7–7 复杂应力状态下的应力 -- 应变关系
——（ 广义虎克定律 ）
§ 7–8 复杂应力状态下的变形比能
§ 7–１ CONCEPTS OF THE STAFE OF STRISS
1,Forward
1),Investigation on the tensile,compressive and torsional test of cast iron
and low-carbon steel
M
Low-carbon
steel
Cast iron
P
P Cast iron in
tension P
Cast iron in
compression
2),How will the member rupture in
combined deformations？
M
P
§ 7–１ 应力状态的概念
一、引言
1、铸铁与低碳钢的拉、压、扭试验现象是怎样产生的？
M
低碳钢
铸铁 P
P 铸铁拉伸
P
铸铁压缩
2、组合变形杆将怎样破坏？
M
P
4,Expression of stresses in general case
3,Element,?Element— Delegate of a point in the member,It is a infinitesimal
geometric body enveloping the studied point,In common use it is a correctitude cubic
body,
?Properties of an element—a,Stresses are distributed
uniformly in the sections；
b,The stresses in two planes that are
parallel to each other are equal,
2,State of stress at a point,
There are countless sections through a point,The gathering of stresses in all
sections is called the state of stress at this point,
x
y
z
s x
sz
s y
txy
四、普遍状态下的应力表示
三、单元体, ?单元体 ——构件内的点的代表物，是包围被研究点
的无限小的几何体，常用的是正六面体。
?单元体的性质 ——a、平行面上，应力均布；
b、平行面上，应力相等。
二、一点的应力状态,
过一点有无数的截面，这一点的各个截面上应力情况的集合，
称为这点的应力状态（ State of Stress at a Given Point）。
x
y
z
s x
sz
s y
txy
x
y
z
s x
sz
s y
txy
5,Theorem of conjugate shearing stress,
Shearing stresses on perpendicular planes are equal in magnitude and have
directions such that both stresses point toward,or both point away form,the line
of intersection of the faces,
0 ?? zM
0d)dd(d)dd( ?? yxzxzy yxxy tt
yxxy tt ??
Provement, The element is in
equilibrium,
x
y
z
s x
sz
s y
txy
五、剪应力互等定理（ Theorem of Conjugate Shearing Stress),
过一点的两个正交面上,如果有与相交边垂直的剪应力分
量,则两个面上的这两个剪应力分量一定等值、方向相对或相
离。
0, ?? zM单元体平衡证明
0d)dd(d)dd( ?? yxzxzy yxxy tt
yxxy tt ??
6,Original element（ known element）,
Example 1 Plot the known elements of point A,B,C shown in
the following figures,
tzx
P P A
A
s x s x
M
P
x
y
z
B
C
s x s x B
txz C
t xy
t yx
tzx
六、原始单元体（已知单元体）,
[例 1] 画出下列图中的 A,B,C点的已知单元体。
P P A
A
s x s x
M
P
x
y
z
B
C
s x s x B
txz C
t xy
t yx
7,Principal element,principal planes,principal stresses,
? Principal element,
The element in which the shearing stresses in side
planes are all zero,
? Principal Planes,
The planes on which the shearing stresses are zero,
? Principal stresses,
Normal stresses acting on the principle planes,
?convention of the order for three principal stresses,
In magnitude of the algebraic value,
321 sss ??
s1
s2
s3
x
y
z
sx
sy
sz
七、主单元体、主平面、主应力,
?主单元体 (Principal bidy),
各侧面上剪应力均为零的单元体。
?主平面 (Principal Plane),
剪应力为零的截面。
?主应力 (Principal Stress ）,
主平面上的正应力。
?主应力排列规定：按代数值大小,
321 sss ??
s1
s2
s3
x
y
z
sx
sy
sz
? State of the uniaxial stress,
State of stress that one principal stress is not equal to zero,
? State of the biaxial stress,
State of stress that one principal stress is equal to zero,
? State of the triaxial Stress,
State of stress that all the three principal stresses are not equal to zero,
A
s x s x
tzx
s x s x B
txz
?单向应力状态（ Unidirectional State of Stress）,
一个主应力不为零的应力状态。
?二向应力状态（ Plane State of Stress）,
一个主应力为零的应力状态。
?三向应力状态（ Three—Dimensional State of Stress）,
三个主应力都不为零的应力状态。
A
s x s x
tzx
s x s x B
txz
§ 7–2 ANALYSIS OF THE STATE OF PLANE STRESS —
ANALYTICAL METHOD
equivalent
sx t
xy
sy
x
y
z x
y
sx
txy
sy
O
§ 7–2 平面应力状态分析 ——解析法
等价
sx t
xy
sy
x
y
z x
y
sx
txy
sy
O
Stipulate,?s? is positive if its direction is the
same with one of the external normal line of the
section； ?t ? is positive if it make the element
rotate clockwise； ? A countclockwise angle ?
is considered to be positive,
Fig.1
Assume that area of the inclined section is S,
According to the equilibrium of the free body we get,
? ? F n 0
0c o ss i ns i n
s i nc o sc o s
2
2
???
??
??t?s
??t?ss ?
SS
SSS
yxy
xyx
1,Stresses acting in arbitrary inclined
plane
x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
Fig.2
规定,?s? 截面外法线同向为正；
?t ?绕研究对象顺时针转为正；
??逆时针为正。
图 1
设：斜截面面积为 S,由分离体平衡得,
? ? F n 0
0c o ss i ns i n
s i nc o sc o s
2
2
???
??
??t?s
??t?ss ?
SS
SSS
yxy
xyx
一、任意斜截面上的应力
x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
图 2
Fig.1 x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
Fig.2
?t?sssss ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
Considering conjugate of shearing stresses
and trigonometric identities we get,
Similarly,
图 1 x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
图 2
?t?sssss ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
考虑剪应力互等和三角变换，得,
同理,
? ? 02c o s22s in,00
0
?????
?
?t?ss
?
s
??
?
xyyxd
dLe t
2,The extreme values for the stress
yx
xy
ss
t
?
?
??
2
2tg 0）：、（
20101
??? ?
00 ??t
）
22
22 xy
yxyx
m in
m ax tssss
s
s
?
?
±
?
?
?
?
?
（
x
y
sx
txy
sy
O
and two extremums
Thus we can get two stationary
points
Extreme normal stresses are principal stresses,
? ? 02c o s22s i n,00
0
?????
?
?t?ss
?
s
??
?
xyyxd
d令
二、极值应力
yx
xy
ss
t
?
?
??
2
2tg 0和两个极值：）、（
由此得两个驻点：
2
0101
?
?? ?
！极值正应力就是主应力?? 00?t
x
y
sx
txy
sy
O
）
22
22 xy
yxyx
m in
m ax tssss
s
s
?
?
±
?
?
?
?
?
（
s1 is in the quadrant for the shearing stress to
point and lean to the larger of both sx and sy
0
d
d
:et
1
?
? ???
t ?
L
xy
yx
t
ss
?
2
2tg 1
?
?
2 2
2 x y
y x
min
max t s s
t
t ? ? ± ?
? ?
?
） （
,
410
??? ??
m a x2m a x1 ; ssss ??
x
y
sx
txy
sy
O
Main
elemeut
2s
1s
That is the angle between the planes in
which shearing stresses reach extremums
and the principal planes is, 045
s1在剪应力相对的象限内,
且偏向于 sx 及 sy较 大的一侧。
0
d
d
:
1
?
? ???
t ?
令
xy
yx
t
ss
?
2
2tg 1
?
?
0
10 45,4 面成即极值剪应力面与主平
??? ??
2 2
2 x y
y x
min
max t s s
t
t ? ? ± ?
? ?
?
） （
x
y
sx
txy
sy
O
主
单元体
2s
1s
m a x2m a x1 ; ssss ??
Example 2 Analyze the failure of the circular shaft in torsion,
Solution,?Determine the critical
point and plot the original element,
?Determine the extreme-value stress
0?? yx ss
P
n
xy W
M
?? tt
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（
tt ???? 2xy
t xy C
t yx
M
C
x
y
O
txy
tyx
[例 2] 分析受扭构件的破坏规律。
解,?确定危险点并画其原
始单元体
?求极值应力
0?? yx ss
P
n
xy W
M
?? tt
tt ???? 2xy
t xy C
t yx
M
C
x
y
O
txy
tyx
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（
?Analysis of failure,
tt
ss
t
t
???
?
??
?
?
? 22
m i n
m a x
2 xy
yx ）（
tssts ???? 321 ;0; ?
45
2
2tg 00 ????
?
?? ?
ss
t
?
yx
xy
0
11 0022tg ???
?
? ?
t
ss
?
xy
yx
M P a2 0 0;M P a2 4 0 ?? ss ts
M P a300~198;M P a960~640
M P a280~98
??
?
byb
Lb
ts
s
Low-carbon
steel
Cast iron
Low-carbon steel,
Cast iron,
?破坏分析
tssts ???? 321 ;0; ?
45
2
2tg 00 ????
?
?? ?
ss
t
?
yx
xy
0
11 0022tg ???
?
? ?
t
ss
?
xy
yx
M P a2 0 0;M P a2 4 0,?? ss ts低碳钢
M P a300~198;M P a960~640
M P a280~98:
??
?
byb
Lb
ts
s灰口铸铁
低碳钢
铸铁
tt
ss
t
t
???
?
??
?
?
? 22
m i n
m a x
2 xy
yx ）（
§ 7–3 ANALYSIS OF THE STATE OF STRESS —
GRAPHYCAL METHOD
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?t?
ss
t
?t?
ssss
s
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
xy
yxyx
2
2
2
2
22 xy
yxyx tsstsss
?? ???
?
?
??
?
? ?
????
?
?
??
?
? ?
?
Eliminating the parameter 2? from
the above equation,we get,
1,Stress Circle
x
y
sx
txy
sy
O
sy
txyx sx
s?
t?
?
x
y
O t
n curve of this equation is a circle—stress circle
（ or Mohr’s circle,introduced by German
engineer Otto Mohr)
§ 7–3 平面应力状态分析 ——图解法
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?t?
ss
t
?t?
ssss
s
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
xy
yxyx
2
2
2
2
22 xy
yxyx tsstsss
?? ???
?
?
??
?
? ?
????
?
?
??
?
? ?
?
对上述方程消去参数（ 2?），得,
一、应力圆（ Stress Circle）
x
y
sx
txy
sy
O
此方程曲线为圆 —应力圆（或莫尔圆，
由德国工程师,Otto Mohr引入）
sy
txyx sx
s?
t?
?
x
y
O t
n
?Set up a stress coordinate system as shown
in the following figure,
（ Pay attention to the selection of the scale ）
2,Method to plot the stress circle
?Plot the point A(s x,txy) and the point
B(sy,tyx) in the coordinate system,
?Inclined line AB intersects the axis sa
at the point C,This point is the center
of the stress circle,
?Plot a circle—stress circle with the
center C and the radius AC,
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
?建立应力坐标系，如下图所示,
（注意选好比例尺）
二、应力圆的画法
?在 坐标系内画出点 A(s x,txy)和
B(sy,tyx)
?AB与 s? 轴的交点 C便是圆心。
?以 C为圆心，以 AC为半径画
圆 ——应力圆；
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
3,Corresponding relation between the
element and stress circle
?Stress(s ?,t ?) in ? plane
A point (s ?,t ?) on the stress
circumference
?Normal line of ? plane
Radius of the stress circle
?Angle ? between two sections
Angle 2? between two radiuses；
And the direction of rotation is the
same,
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
三、单元体与应力圆的对应关系
??面上的 应力 (s ?,t ?)
应力圆上一点 (s ?,t ?)
??面的法线 应力圆的半径
?两面夹角 ? 两半径夹角 2? ；
且转向一致。
22
3
1
22
xy
yxyx
r a d i u s
ROC
t
ssss
s
s
?
?
?
?
???
?
?
?
）（
4,Mark extreme stresses on the circumference of
the stress circle
22
m i nm a x
m i n
m a x
2
2
xy
yx
r a d i u s
R
t
ss
ss
t
t
?
?
?
?
????
?
?
?
）（
O C s
?
t?
A(sx,txy)
B(sy,tyx)
x
2?1
mint
maxt
2?0
s1 s2 s3
22
3
1
22
xy
yxyx
ROC
t
ssss
s
s
?
?
?
?
???
?
?
?
）（
半径
四、在应力圆上标出极值应力
22
m i nm a x
m i n
m a x
2
2
xy
yx
R
t
ss
ss
t
t
?
?
?
?
????
?
?
?
）（
半径
O C s
?
t?
A(sx,txy)
B(sy,tyx)
x
2?1
mint
maxt
2?0
s1 s2 s3
s3
Example 3 Determine principal stresses and orientation of principal
planes of the element as shown in the figure.(unit,MPa)
45
325
325
95
150
°
A
B
s 1
s2
Method 1-graphical method,
? Stress coordinate
system is shown in the figure,
?Intersection C of the
perpendicular bisection line of
AB and the axis sa is the center
of the circle,Plot the circle with
the center C and the a radius
AC—stress circle,
?0
s1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
)325,45(B
?Plot the point and )325,95(A
s3
例 3 求图示单元体的主应力及主平面的位置。 (单位,MPa)
45
325
325
95
150
°
A
B
s 1
s2
解法 1——图解法,
?主应力坐标系如图
?AB的垂直平分线与
s? 轴的交点 C便是
圆心，以 C为圆心，
以 AC为半径画
圆 ——应力圆
?0
s1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
)325,45(B
)325,95(A
?在 坐标系内画出点
s3 s
1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
?Principal stresses and principal
planes as shown in the figure
M P a 0
M P a 20
M P a 120
3
2
1
?
?
?
s
s
s
?30
0 ???
45
325
325
95
150
°
s 1
?0 s2
A
B
s3 s
1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
?主应力及主平面如图
?30
0 ???
45
325
325
95
150
°
s 1
?0 s2
A
B
M P a 0
M P a 20
M P a 120
3
2
1
?
?
?
s
s
s
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
45
325
325
95
150
°
Method 2— analytical method,Analysis—set up the coordinate as
shown in the figure,
xyyx
y
tt
s
???
?
M P a325
M P a45
?xs
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（
60°
M Pa325
M Pa95
0
0
60
60
?
?
t
s
x
y
O
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
45
325
325
95
150
°解法 2—解析法：分析 ——建立坐标系如图
xyyx
y
tt
s
???
?
M P a325
M P a45
?xs
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（60°
x
y
O
M Pa325
M Pa95
0
0
60
60
?
?
t
s
§ 7–4 PRINCIPAL STRESSES AND THEIR
TRAJECTORIES OF THE BEAM
z
z
xy Ib
QS ?
?t
z
x I
My?s
1
2
3
4
5
P1 P2 q
Element,
22
3
1
22 xy
xx tss
s
s
???
?
?
?
）（
As shown in the figure,the beam
produced the shear bending（
transversal bending）,where, M、
Q>0 in the beam,Try to determine the
magnitude of the principal stresses and
the position of the principal planes of
each point in the section,
§ 7–4 梁的主应力及其主应力迹线
z
z
xy Ib
QS ?
?t
z
x I
My?s
1
2
3
4
5
P1 P2 q 如图，已知梁发生剪切弯
曲（横力弯曲），其上 M、
Q>0，试确定截面上各点主
应力大小及主平面位置。
单元体,
22
3
1
22 xy
xx tss
s
s
???
?
?
?
）（
1
s1
s3
s3
3
s1
s3
s1
s1
s3
5
?0
–45°
?0
s
t
A1 A2 D2 D1
C O
s
A2
D2
D1
C
A1
O
t
2?0
s
t
D2
C
D1
O
2?0= –90°
s
D2
A1 O
t
2?0
C
D1
A2
s
t
A2 D2 D1
C
A1
O
1
s1
s3
s3
3
s1
s3
s1
s1
s3
5
?0
–45°
?0
s
t
A1 A2 D2 D1
C O
s
A2
D2
D1
C
A1
O
t
2?0
s
t
D2
C
D1
O
2?0= –90°
s
D2
A1 O
t
2?0
C
D1
A2
s
t
A2 D2 D1
C
A1
O
Tensile
force
Compressive
force
Principal stress trajectories,
Envelopes of the direction lines of principal stresses—tangent at
each
point on the curve indicates the orientation of the tensile(or
compressive)
principal stress at the same
point,
Solid lines express the tensile
principal stress trajectories；
dashed lines express the
compressive principal stress
trajectories,
s1 s3
拉力
压力
主应力迹线（ Stress Trajectories),
主应力方向线的包络线 ——曲线上每一点的切线都指示
着该点的拉主应力方位（或压主应力方位）。
实线表示拉主应力迹线；
虚线表示压主应力迹线。
s1 s3
x
y
Method to plot principal
stress trajectories,
1
1
2
2
3
3
4
4
i
i
n
n
b a
c
d
q
s3
s1
Section section section section section section
q
x
y
主应力迹线的画法,
1
1
截面
2
2
截面
3
3
截面
4
4
截面
i
i
截面
n
n
截面
b a
c
d
s3
s1
§ 7–5 ANALYSIS OF TRIAXIAL STRESSED STATE—
METHOD OF STRESS CIRCLE
s2
s1
x
y
z
s3 1s2s3s
?s
?t
1,Spatial stressed state
§ 7–5 三向应力状态研究 ——应力圆法
s2
s1
x
y
z
s3
1s
2s3s
?s
?t
1、空间应力状态
2,Analysis of the
triaxial stress
Fig.a Fig.b
?The maximum shearing stress inside the
whole element is,
t max
2
31
m a x
ss
t
?
?
s2
s1
x
y
z
s3
1s
2s3s
?s
?t
?Elastic theory proved that stresses on any plane passing through a point in by the
element shown in Fig.a may be corresponding to the coordinates of a point on the
circumference of the stress circle or in the shadow region,
2、三向应力分析
?弹性理论证明，图 a单元体内任意一点任意截面上的应
力都对应着图 b的应力圆上或阴影区内的一点。
图 a
图 b
?整个单元体内的最大剪应力为,
t max
2
31
m a x
ss
t
?
?
s2
s1
x
y
z
s3
1s
2s3s
?s
?t
Example 4 Determine the principal stresses and the maximum shearing stress
of the element shown in the figure.(MPa)
Solution,?From
the element sketch we
know plane yz is a
principal plane,
501 ?s
?Set up stress coordinates
as shown in the figure,
Plot the stress circle and
locate the point s1,get,
27
50
58
3
2
1
??
?
?
s
s
s
44m ax ?t
50 40
x
y
z
30
10
(MPa)
s?
（ MPa ） t?
A
B
C
A
B
s1 s2 s3
t max
[例 4] 求图示单元体的主应力和最大剪应力。（ MPa）
解,?由单元
体图知,y
z面为主平
面
501 ?s
?建立应力坐标系如
图，画应力圆和
点 s1,得,
27
50
58
3
2
1
??
?
?
s
s
s
44m ax ?t
50 40
x
y
z
30
10
(M Pa)
s?
（ M Pa ） t?
A
B
C
A
B
s1 s2 s3
t max
§ 7–6 ANALYSIS OF STRAIN IN A PLANE
x
y
O
?
1,Determine the expression of strain analysis by the method of superposition
??? ? c o sd 11 xaDD ??
??? ? 21 c o sx?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
s i n/
c o s
1
x
xx
a
a
b
b
B O EA O D
??
???
?????
a b
c d
?
A
O
B
Shearing strain,Incremental
quantity of the right angle！（ Only
like this the contents in the preceding
section and the following section Can
correspond to each other,）
?
D
D1
E
E1
? ? ?
§ 7–6 平面内的应变分析
x
y
O
?
一, 叠加法求应变分析公式
??? ? c o sd 11 xaDD ??
??? ? 21 c o sx?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
s i n/
c o s
1
x
xx
a
a
b
b
B O EA O D
??
???
?????
a b
c d
?
A
O
B
剪应变,直角的增大量！
（只有这样，前后才对应）
?
D
D1
E
E1
? ? ?
??? ? s i nd 22 ycDD ??
??? ? 22 s iny?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
/ c o s
s i n
2
y
yy
c
c
c
c
B O EA O D
??
???
?????
x
y
O
a b
c d
?
A
O
B D
D2
E
E2
? ?
??? ? s i nd 22 ycDD ??
??? ? 22 s iny?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
/ c o s
s i n
2
y
yy
c
c
c
c
B O EA O D
??
???
?????
x
y
O
a b
c d
?
A
O
B D
D2
E
E2
? ?
??? ? c o sd 33 xycADd ???????
???? ? s o cxy s i n3 ??
? ????
?
??
?
??
?
?
22
33
s i nc o s
/ c o s
c o s
s i n/
s i n
??
???
?????
xy
xyxy
c
c
c
c
B O EA O D
D
D3
E
E3
?
xy?
xy?
x
y
O
a b
c d
?
A
O
B
??? ? c o sd 33 xycADd ???????
???? ? s o cxy s i n3 ??
? ????
?
??
?
??
?
?
22
33
s i nc o s
/ c o s
c o s
s i n/
s i n
??
???
?????
xy
xyxy
c
c
c
c
B O EA O D
D
D3
E
E3
?
xy?
xy?
x
y
O
a b
c d
?
A
O
B
????????? ?? c o ss i ns i nc o s 22
3
1
xyyx
i
i ???? ?
?
? ?????????? ?? 22
3
1
s i nc o s2s i n2s i n ????? ?
?
xyyx
i
i
?t?
ssss
s ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ????
?
?
???
????
? ? 2s i n
2
1
2c o s
22 xy
yxyx ?????
???
???
? 2c o s
2
1
2s i n
22 xy
yx ????
?
?
????????? ?? c o ss i ns i nc o s 22
3
1
xyyx
i
i ???? ?
?
?t?
ssss
s ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ????
?
?
? ?????????? ?? 22
3
1
s i nc o s2s i n2s i n ????? ?
?
xyyx
i
i
???
????
? ? 2s i n
2
1
2c o s
22 xy
yxyx ?????
???
???
? 2c o s
2
1
2s i n
22 xy
yx ????
?
?
2),Plot the strain circle according
to strain( )of a point,
xyyx ???,,
2,Graphical method of strain analysis—strain circle
22 ; 2 ; ??t?s? ???? ???
1),Analogic relation of stress circle and strain circle
?Set up strain coordinates as shown in
the figure
?locate the point A(?x,?xy/2) and
B(?y,-?yx/2) in the coordinate system
?intersection of AB and axis ?? is the
center of the circle,
?plot circle by the center C with a radius AC —strain circle,
??
??/2
A
B
C
2、已知一点 A的应变（ ），画应变圆
xyyx ???,,
二、应变分析图解法 ——应变圆 （ Strain Circle）
22 ; 2 ; ??t?s? ???? ???
1、应变圆与应力圆的类比关系
?建立应变坐标系如图
?在 坐标系内画出点
A(?x,?xy/2)
B(?y,-?yx/2)
?AB与 ?? 轴的交点 C便是圆心
?以 C为圆心，以 AC为半径画圆 ——应变圆。
??
??/2
A
B
C
??
??/2
3,Corresponding relation between the strain in ? direction
and strain circle
?max ?min
2?0
D(??,??/2)
2?
n
? Strain in ?direction(? ?,? ?/2)
A point on the strain circle(??,? ?/2)
?The direction line of ?
Radius of the strain circle
?Angle ? between two directions
Angle 2? between the two
radiuses and the direction of
rotation is the same,
A
B
C
??
??/2
三, ?方向上的 应变与 应变圆的对应关系
?max ?min
2?0
D(??,??/2)
2?
n
??方向上的 应变 (? ?,? ?/2)
应变圆上一点 (??,? ?/2)
?? 方向线 应变圆的半径
?两方向间夹角 ?
两半径夹角 2? ；且转向一致。
A
B
C
4,Values and orientation of principal strains
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
22 ;
2; ??t?s? ???? ???
22
m i n
m a x
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（ yx
xytg
ss
t
??
?
??
2
2 0
yx
xy
??
?
?
?
?
?02tg
四, 主应变数值及其方位
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
22 ;
2; ??t?s? ???? ???
yx
xy
??
?
?
?
?
?02tg
22
m i n
m a x
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（ yx
xytg
ss
t
??
?
??
2
2 0
Example 5 Knowing three strain s??1,??2 and ??3 in
directions ?1,?2 and ?3 at a point in some plane,determine the
principal strains in this plane,
Solution,
1,2,3 )(i c o ss i ns i nc o s 22 ???? iixyiyixi ???????? ?
First find out ? x,? y,? x y by solving the above three
equations then determine the principal strains,
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
[例 5] 已知一点在某一平面内的 ?1,?2,?3 方向上的线 应变分
别为 ??1,??2,??3，,求该面内的主应变。
解：由
iixyiyixi ???????? ? c o ss i ns i nc o s
22 ???
i =1,2,3这三个方程求出 ? x,? y,? x y；然后再求主应变。
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
Example 6 Determine the principal strains of the point after
three linear strains at this point are tested by the strain foil of 45°,
x
y u
45o ?0 ?max
? ?][2)(
2
1 22
m a x ）（）（ yuuxyx ??????? ??????
? ?][2)(
2
1 22
m i n ）（）（ yuuxyx ??????? ??????
yx
yxu
??
???
?
?
??
?
2
2tg 0
[例 6] 用 45° 应变花测得一点的三个线应变后，求该点的主应变。
x
y u
45o ?0 ?max
? ?][2)(
2
1 22
m a x ）（）（ yuuxyx ??????? ??????
? ?][2)(
2
1 22
m i n ）（）（ yuuxyx ??????? ??????
yx
yxu
??
???
?
?
??
?
2
2tg 0
§ 7–7 STRESS — STRAIN RELATION UNDER THE COMPLEX
STRESSED STATE—（ GENERALIZED HOOKE’S LAW ）
1,Stress-strain relation in uniaxial tension
2,Stress-strain relation in pure shear
G
xy
xy
t
? ?
E
x
x
s? ?
xy E s
?? ??
xz E s
?? ??
) 0 x,y,z( i,jij ???
)( 0 x,y,zii ???
0?? zxyz ??
x
y
z
sx
x
y
z
t x y
§ 7–7 复杂应力状态下的应力 -- 应变关系
——（ 广义虎克定律 ）
一、单拉下的应力 --应变关系
二、纯剪的应力 --应变关系
G
xy
xy
t
? ?
)( 0 x,y,zii ???
0?? zxyz ??
x
y
z
sx
x
y
z
t x y
E
x
x
s? ?
xy E s
?? ??
xz E s
?? ??
) 0 x,y,z( i,jij ???
3,Stress-strain relation in complex stressed state
According to
superposition we get,
? ?? ?
zyx
zyx
x
E
EEE
ss?s
s
?
s
?
s
?
???
???
1
? ?? ?xzyy E ss?s? ??? 1
? ?? ?yxzz E ss?s? ??? 1
G
xy
xy
t
? ?
G
yz
yz
t
? ?
G
zx
zx
t? ?
? ?? ?zyxx E ss?s? ??? 1
?
?
?
?
?
?
?
?
?
?
?
x
y
z
sz
sy
txy
sx
三、复杂状态下的应力 --- 应变关系
依叠加原理,得,
? ?? ?
zyx
zyx
x
E
EEE
ss?s
s
?
s
?
s
?
???
???
1
? ?? ?xzyy E ss?s? ??? 1
? ?? ?yxzz E ss?s? ??? 1
G
xy
xy
t
? ?
G
yz
yz
t
? ?
G
zx
zx
t? ?
? ?? ?zyxx E ss?s? ??? 1
?
?
?
?
?
?
?
?
?
?
?
x
y
z
sz
sy
txy
sx
Principal stress—principal strain relation
4,Stress-strain relation
under the state of plane
stress
0??? zxyzz tts
The directions are the same
02tg
2
?
ss
t
?
?
??
yx
xy
yx
xy
??
?
?
?
?
??02tg
? ?? ?1322 1 ss?s? ??? E
? ?? ?1233 1 ss?s? ??? E
? ?? ?3211 1 ss?s? ??? E
xyxy G ?t ?
? ?yxx E ???
?
s ?
?
? 2
1
? ?xyy E ???
?
s ?
?
? 2
1
主应力 --- 主应变关系
四、平面状态下的应力 ---应变关系,
0??? zxyzz tts
方向一致
? ?? ?1322 1 ss?s? ??? E
? ?? ?1233 1 ss?s? ??? E
? ?? ?3211 1 ss?s? ??? E
xyxy G ?t ?
? ?yxx E ???
?
s ?
?
? 2
1
? ?xyy E ???
?
s ?
?
? 2
1
02tg
2
?
ss
t
?
?
??
yx
xy
yx
xy
??
?
?
?
?
??02tg
The directions of the principal stress and the principal
strain are the same,
0
2
0 2tg
)()]1)([(
1
22
2tg ?
??
?
???
?
?
ss
t
? ??
?
??
??
?
?
?
?
?
?
yx
xy
yx
xy
yx
xy
E
G
主应力与主应变 方向一致。
0
2
0 2tg
)()]1)([(
1
22
2tg ?
??
?
???
?
?
ss
t
? ??
?
??
??
?
?
?
?
?
?
yx
xy
yx
xy
yx
xy
E
G
5,Relation between the volumetric strain and stress components,
321 aaaV ?
)1()1()1( 3322111 ??? ???? aaaV
321
1 ??? ??????
V
VV
Volumetric strain,
)(
21
)(
21
321
zyx
E
E
sss
?
sss
?
??
?
?
??
?
??
Relation between the volumetric
strain and stress components,
五、体积应变与应力分量间的关系
321 aaaV ?
)1()1()1( 3322111 ??? ???? aaaV
321
1 ??? ??????
V
VV
体积应变,
)(
21
)(
21
321
zyx
E
E
sss
?
sss
?
??
?
?
??
?
??
体积应变与应力分量间的关系,
Example 7 A structure member is subjected to some forces,Two principal
strainsat a point on the free surface of the member are ?1=240?10-6 ?2=–160?10-6，
modulus of elasticity is E=210GPa,Possion’s ratio is ?=0.3,Try to determine the
principal stresses and another principal strain at this point,
03 ?s
? ? M P a3.4410)1603.0240(
3.01
10210
1
6
2
9
2121 ?????
???
?
?? ????
?
s E
,Then this point is in the stat
? ?
M P a3.2010)2403.0160(
3.01
10210
1
6
2
9
1222
??????
?
?
?
?
?
?
?
???
?
s
E
1s?
2s?
Solution,On the free surface
Of plane stress.,
[例 7] 已知一受力构件自由表面上某一点处的两个面内主应变分别
为,?1=240?10-6,?2=–160?10-6，弹性模量 E=210GPa，泊松比
为 ?=0.3,试求该点处的主应力及另一主应变 。
03, ?s自由面上解
所以，该点处为平面应力状态
1s?
2s?
? ? M P a3.4410)1603.0240(
3.01
10210
1
6
2
9
2121 ?????
???
?
?? ????
?
s E
? ?
M P a3.2010)2403.0160(
3.01
10210
1
6
2
9
1222
??????
?
?
?
?
?
?
?
???
?
s
E
? ? 669132 103.3410)3.443.22(
10210
3.0 ???????
?
????? ss??
E;M P a3.20;0;M P a3.44 321 ???? sss
? 3 34 2, ? ? × 10-6
? ? 669132 103.3410)3.443.22(
10210
3.0 ???????
?
????? ss??
E;M P a3.20;0;M P a3.44 321 ????? sss
? 3 34 2, ? ? × 10-6
Example 8 A thin-walled container subjected to inside pressure is shown in
Fig.a,In order to determine the value of the inside pressure the hoop strain tested
on the surface of the container with strain foil is ? t =350× l06,If the mean diameter
of the container is D=500 mm,thickness of its wall is ?=10 mm,E=210GPa，
?=0.25,Try to, 1.derive the expressions of stress in the lateral and longitudinal
sections of the container, 2.calculate the inside pressure of the container,
p
p
p
x
st
sm
L
p
O D
x
A B
y
Fig.a
[例 8] 图 a所示为承受内压的薄壁容器。为测量容器所承受的内
压力值，在容器表面用电阻应变片测得环向应变 ? t =350× l06，
若已知容器平均直径 D=500 mm，壁厚 ?=10 mm，容器材料的
E=210GPa,?=0.25，试求,1.导出容器横截面和纵截面上的正应
力表达式； 2.计算容器所受的内压力。
p
O D
x
A B
y
图 a
p
p
p
x
st
sm
L
1)Longitudinal stresses
Solution,Expressions of the circumferential and longitudinal stress of
the container reservoir
Cutting the container along a section shown and in Fig.b
considering the equilibrium of the right part,we get
? ? 42DpDm ???s ??
?s 4
pD
m ?
p
sm
sm
x D
Fig.b
1、轴向应力,(longitudinal stress)
解：容器的环向和纵向应力表达式
用横截面将容器截开，受力如图 b所示,根据平衡方程
? ? 42DpDm ???s ??
?s 4
pD
m ?
p
sm
sm
x D
图 b
Imagine cut off the container along
longitudinal section,and take the upper
pant as the study object,The free body
diagram is shown in Fig.c
2,Circumferential stress
? ? Dlplt ??? ?s 2
?s 2
pD
t ?
3,Determine the inside pressure
（ by stress-strain relation）
? ? ? ????ss? ???? 241 EpDE mtt
M P a36.3
)25.02(5.0
1035001.0102104
)2(
4
69
?
??
?????
?
?
?
?
?
??
D
E
p
t
st
sm External
surface
y
p s
t s t D
q
dq )d2( q?? Dlp
z
Fig,c
O
用纵截面将容器截开，取长为 L的一部
分为研究对象，受力如图 c所示
2、环向应力,(hoop stress)
? ? Dlplt ??? ?s 2
?s 2
pD
t ?
3、求内压（以应力应变关系求之）
? ? ? ????ss? ???? 241 EpDE mtt
M P a36.3
)25.02(5.0
1035001.0102104
)2(
4
69
?
??
?????
?
?
?
?
?
??
D
E
p
t
st
sm
外表面
y
p s
t s t D
q
dq )d2( q?? Dlp
z
图 c
O
§ 7－ 8 STRAIN-ENERGY DENSITY OF
COMPLEX STRESSED STATE
332211 2
1
2
1
2
1 ?s?s?s ???u
)(31 321 ssss ???m
s2
s3
s 1
Fig,a
Fig,c
s3 -sm
s 1 -sm
s2 -sm
ba E ????
??? )(21
321 sss
? 0??
c
? ?? ?312321232221 22 1 ssssss?sss ?????? E
sm
Fig,b
sm
sm
§ 7－ 8 复杂应力状态下的变形比能
332211 2
1
2
1
2
1 ?s?s?s ???u
)(31 321 ssss ???m
s2
s3
s 1
图 a
图 c
s3 -sm
s 1 -sm
s2 -sm
ba E ????
??? )(21
321 sss
? 0??
c
? ?? ?312321232221 22 1 ssssss?sss ?????? E
sm
图 b
sm
sm
? ? ? ? ? ?? ?213232221
6
1 ssssss? ???????
E
u x
Where ux is called the strain-energy density corresponding to the
distortion,
Fig,c
s3 -sm
s 1 -sm
s2 -sm
Strain energy of the element shown in Fig.c is
? ? ? ? ? ?? ?213232221
6
1 ssssss? ???????
E
u x
:单元体的应变能为图 c
称为形状改变比能或歪形能。
图 c
s3 -sm
s 1 -sm
s2 -sm
Example 9 Prove the relations between three elastic constants by
the energy method,
G
u
22
1 2tt? ??
?Specific strain energy of the element in pure shear is,
? Expression of the specific strain energy of the
element in pure shear by principal stresses is,
? ?? ?312321232221 2
2
1 ssssss?sss ??????
E
u
? ?? ?tt?tt )(002)(02 1 22 ???????? E
21 t?
E
??
? ???? 12
EG
txy
A
s1
s3
[例 9] 用能量法证明三个弹性常数间的关系。
G
u
22
1 2tt? ??
?纯剪单元体的比能为,
?纯剪单元体比能的主应力表示为,
? ?? ?312321232221 2
2
1 ssssss?sss ??????
E
u
? ?? ?tt?tt )(002)(02 1 22 ???????? E
21 t?
E
??
? ????? 12
EG
txy
A
s1
s3
Chapter 7 Exercises
1,As shown in the figure,a certain point is in biaxial stress,
Try to determine when the two principal stresses are
both tensile stresses,
Solution,
xyt
? ? 0222 40402 4040
m in
m a x ???? ??
xyts
s
40??? xyt
4040 ???? xyt
第七章 练习题
一、某点应力状态为图示二向应力状态。试求
该点的两个主应力均为拉应力时 的取值范围。
解,
xyt
? ? 0222 40402 4040
m in
m a x ???? ??
xyts
s
40??? xyt
4040 ???? xyt
2,The stress state of an element is shown in the figure,
Knowing E andμof the material,Try to determine the
maximum shearing strain of this element,
Solution,
max?
oss ?1 02 ?s o
ss ??3
os
sst ???
2
31
m a x
? ?
oEG s
?t? ??? 12m a x
m a x
二、图示单元体的应力状态,已知材料的 E和 μ,
试求该单元体的最大剪应变 。
解,
max?
oss ?1 02 ?s oss ??3
os
sst ???
2
31
m a x
? ?
oEG s
?t? ??? 12m a x
m a x
3,The stresses on sections ac and bc at a certain point
of a force-acting member are shown in the figure,Try to
determine the magnitude and the directions of the stresses
on section ab by the stress circle and plot them on an
element,
三、受力构件某点处两截面 ac和 bc上的应力
如图所示,试利用应力圆确定 ab面上应力的大小和
方向，并画在单元体上。
CHAPTER 7 ANALYSIS OF THE STATE OF
STRESS AND STRAIN
§ 7–4 PRINCEPAL STRESSES AND THEIR TRAJECTORIES OF
THE BEAM § 7–5
ANALYSIS OF TRIAXIAL STRESSED STATE—METHOD
OF STRESS CIRCLE
§ 7–6 ANALYSIS OF STRAIN IN A PLANE
§ 7–7 RELATION BETWEEN STRESS AND STRAIN UNDER
COMPLEX STRESSED STATE—（ GENERALIZED HOOKE’S LAW ）
§ 7–8 STRAIN -ENERGY DENSITY UNDER COMPLEX
STRESSED STATE
§ 7–1 CONCEPTS OF THE STATE OF STRESS
§ 7–2 ANALYSIS OF THE STATE OF PLANE STRESS —
ANALYTICAL METHOD
§ 7–3 ANALYSIS OF THE STATE OF PLANE
STRESS — GRAPHYCAL METHOD
第七章 应力状态与应变状态分析
§ 7–1 应力状态的概念
§ 7–2 平面应力状态分析 ——解析法
§ 7–3 平面应力状态分析 ——图解法
§ 7–4 梁的主应力及其主应力迹线
§ 7–5 三向应力状态研究 ——应力圆法
§ 7–6 平面内的应变分析
§ 7–7 复杂应力状态下的应力 -- 应变关系
——（ 广义虎克定律 ）
§ 7–8 复杂应力状态下的变形比能
§ 7–１ CONCEPTS OF THE STAFE OF STRISS
1,Forward
1),Investigation on the tensile,compressive and torsional test of cast iron
and low-carbon steel
M
Low-carbon
steel
Cast iron
P
P Cast iron in
tension P
Cast iron in
compression
2),How will the member rupture in
combined deformations？
M
P
§ 7–１ 应力状态的概念
一、引言
1、铸铁与低碳钢的拉、压、扭试验现象是怎样产生的？
M
低碳钢
铸铁 P
P 铸铁拉伸
P
铸铁压缩
2、组合变形杆将怎样破坏？
M
P
4,Expression of stresses in general case
3,Element,?Element— Delegate of a point in the member,It is a infinitesimal
geometric body enveloping the studied point,In common use it is a correctitude cubic
body,
?Properties of an element—a,Stresses are distributed
uniformly in the sections；
b,The stresses in two planes that are
parallel to each other are equal,
2,State of stress at a point,
There are countless sections through a point,The gathering of stresses in all
sections is called the state of stress at this point,
x
y
z
s x
sz
s y
txy
四、普遍状态下的应力表示
三、单元体, ?单元体 ——构件内的点的代表物，是包围被研究点
的无限小的几何体，常用的是正六面体。
?单元体的性质 ——a、平行面上，应力均布；
b、平行面上，应力相等。
二、一点的应力状态,
过一点有无数的截面，这一点的各个截面上应力情况的集合，
称为这点的应力状态（ State of Stress at a Given Point）。
x
y
z
s x
sz
s y
txy
x
y
z
s x
sz
s y
txy
5,Theorem of conjugate shearing stress,
Shearing stresses on perpendicular planes are equal in magnitude and have
directions such that both stresses point toward,or both point away form,the line
of intersection of the faces,
0 ?? zM
0d)dd(d)dd( ?? yxzxzy yxxy tt
yxxy tt ??
Provement, The element is in
equilibrium,
x
y
z
s x
sz
s y
txy
五、剪应力互等定理（ Theorem of Conjugate Shearing Stress),
过一点的两个正交面上,如果有与相交边垂直的剪应力分
量,则两个面上的这两个剪应力分量一定等值、方向相对或相
离。
0, ?? zM单元体平衡证明
0d)dd(d)dd( ?? yxzxzy yxxy tt
yxxy tt ??
6,Original element（ known element）,
Example 1 Plot the known elements of point A,B,C shown in
the following figures,
tzx
P P A
A
s x s x
M
P
x
y
z
B
C
s x s x B
txz C
t xy
t yx
tzx
六、原始单元体（已知单元体）,
[例 1] 画出下列图中的 A,B,C点的已知单元体。
P P A
A
s x s x
M
P
x
y
z
B
C
s x s x B
txz C
t xy
t yx
7,Principal element,principal planes,principal stresses,
? Principal element,
The element in which the shearing stresses in side
planes are all zero,
? Principal Planes,
The planes on which the shearing stresses are zero,
? Principal stresses,
Normal stresses acting on the principle planes,
?convention of the order for three principal stresses,
In magnitude of the algebraic value,
321 sss ??
s1
s2
s3
x
y
z
sx
sy
sz
七、主单元体、主平面、主应力,
?主单元体 (Principal bidy),
各侧面上剪应力均为零的单元体。
?主平面 (Principal Plane),
剪应力为零的截面。
?主应力 (Principal Stress ）,
主平面上的正应力。
?主应力排列规定：按代数值大小,
321 sss ??
s1
s2
s3
x
y
z
sx
sy
sz
? State of the uniaxial stress,
State of stress that one principal stress is not equal to zero,
? State of the biaxial stress,
State of stress that one principal stress is equal to zero,
? State of the triaxial Stress,
State of stress that all the three principal stresses are not equal to zero,
A
s x s x
tzx
s x s x B
txz
?单向应力状态（ Unidirectional State of Stress）,
一个主应力不为零的应力状态。
?二向应力状态（ Plane State of Stress）,
一个主应力为零的应力状态。
?三向应力状态（ Three—Dimensional State of Stress）,
三个主应力都不为零的应力状态。
A
s x s x
tzx
s x s x B
txz
§ 7–2 ANALYSIS OF THE STATE OF PLANE STRESS —
ANALYTICAL METHOD
equivalent
sx t
xy
sy
x
y
z x
y
sx
txy
sy
O
§ 7–2 平面应力状态分析 ——解析法
等价
sx t
xy
sy
x
y
z x
y
sx
txy
sy
O
Stipulate,?s? is positive if its direction is the
same with one of the external normal line of the
section； ?t ? is positive if it make the element
rotate clockwise； ? A countclockwise angle ?
is considered to be positive,
Fig.1
Assume that area of the inclined section is S,
According to the equilibrium of the free body we get,
? ? F n 0
0c o ss i ns i n
s i nc o sc o s
2
2
???
??
??t?s
??t?ss ?
SS
SSS
yxy
xyx
1,Stresses acting in arbitrary inclined
plane
x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
Fig.2
规定,?s? 截面外法线同向为正；
?t ?绕研究对象顺时针转为正；
??逆时针为正。
图 1
设：斜截面面积为 S,由分离体平衡得,
? ? F n 0
0c o ss i ns i n
s i nc o sc o s
2
2
???
??
??t?s
??t?ss ?
SS
SSS
yxy
xyx
一、任意斜截面上的应力
x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
图 2
Fig.1 x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
Fig.2
?t?sssss ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
Considering conjugate of shearing stresses
and trigonometric identities we get,
Similarly,
图 1 x
y
sx
txy
sy
O
sy
tyx sx
s?
t?
?
x
y
O t
n
图 2
?t?sssss ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
考虑剪应力互等和三角变换，得,
同理,
? ? 02c o s22s in,00
0
?????
?
?t?ss
?
s
??
?
xyyxd
dLe t
2,The extreme values for the stress
yx
xy
ss
t
?
?
??
2
2tg 0）：、（
20101
??? ?
00 ??t
）
22
22 xy
yxyx
m in
m ax tssss
s
s
?
?
±
?
?
?
?
?
（
x
y
sx
txy
sy
O
and two extremums
Thus we can get two stationary
points
Extreme normal stresses are principal stresses,
? ? 02c o s22s i n,00
0
?????
?
?t?ss
?
s
??
?
xyyxd
d令
二、极值应力
yx
xy
ss
t
?
?
??
2
2tg 0和两个极值：）、（
由此得两个驻点：
2
0101
?
?? ?
！极值正应力就是主应力?? 00?t
x
y
sx
txy
sy
O
）
22
22 xy
yxyx
m in
m ax tssss
s
s
?
?
±
?
?
?
?
?
（
s1 is in the quadrant for the shearing stress to
point and lean to the larger of both sx and sy
0
d
d
:et
1
?
? ???
t ?
L
xy
yx
t
ss
?
2
2tg 1
?
?
2 2
2 x y
y x
min
max t s s
t
t ? ? ± ?
? ?
?
） （
,
410
??? ??
m a x2m a x1 ; ssss ??
x
y
sx
txy
sy
O
Main
elemeut
2s
1s
That is the angle between the planes in
which shearing stresses reach extremums
and the principal planes is, 045
s1在剪应力相对的象限内,
且偏向于 sx 及 sy较 大的一侧。
0
d
d
:
1
?
? ???
t ?
令
xy
yx
t
ss
?
2
2tg 1
?
?
0
10 45,4 面成即极值剪应力面与主平
??? ??
2 2
2 x y
y x
min
max t s s
t
t ? ? ± ?
? ?
?
） （
x
y
sx
txy
sy
O
主
单元体
2s
1s
m a x2m a x1 ; ssss ??
Example 2 Analyze the failure of the circular shaft in torsion,
Solution,?Determine the critical
point and plot the original element,
?Determine the extreme-value stress
0?? yx ss
P
n
xy W
M
?? tt
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（
tt ???? 2xy
t xy C
t yx
M
C
x
y
O
txy
tyx
[例 2] 分析受扭构件的破坏规律。
解,?确定危险点并画其原
始单元体
?求极值应力
0?? yx ss
P
n
xy W
M
?? tt
tt ???? 2xy
t xy C
t yx
M
C
x
y
O
txy
tyx
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（
?Analysis of failure,
tt
ss
t
t
???
?
??
?
?
? 22
m i n
m a x
2 xy
yx ）（
tssts ???? 321 ;0; ?
45
2
2tg 00 ????
?
?? ?
ss
t
?
yx
xy
0
11 0022tg ???
?
? ?
t
ss
?
xy
yx
M P a2 0 0;M P a2 4 0 ?? ss ts
M P a300~198;M P a960~640
M P a280~98
??
?
byb
Lb
ts
s
Low-carbon
steel
Cast iron
Low-carbon steel,
Cast iron,
?破坏分析
tssts ???? 321 ;0; ?
45
2
2tg 00 ????
?
?? ?
ss
t
?
yx
xy
0
11 0022tg ???
?
? ?
t
ss
?
xy
yx
M P a2 0 0;M P a2 4 0,?? ss ts低碳钢
M P a300~198;M P a960~640
M P a280~98:
??
?
byb
Lb
ts
s灰口铸铁
低碳钢
铸铁
tt
ss
t
t
???
?
??
?
?
? 22
m i n
m a x
2 xy
yx ）（
§ 7–3 ANALYSIS OF THE STATE OF STRESS —
GRAPHYCAL METHOD
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?t?
ss
t
?t?
ssss
s
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
xy
yxyx
2
2
2
2
22 xy
yxyx tsstsss
?? ???
?
?
??
?
? ?
????
?
?
??
?
? ?
?
Eliminating the parameter 2? from
the above equation,we get,
1,Stress Circle
x
y
sx
txy
sy
O
sy
txyx sx
s?
t?
?
x
y
O t
n curve of this equation is a circle—stress circle
（ or Mohr’s circle,introduced by German
engineer Otto Mohr)
§ 7–3 平面应力状态分析 ——图解法
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?t?
ss
t
?t?
ssss
s
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
xy
yxyx
2
2
2
2
22 xy
yxyx tsstsss
?? ???
?
?
??
?
? ?
????
?
?
??
?
? ?
?
对上述方程消去参数（ 2?），得,
一、应力圆（ Stress Circle）
x
y
sx
txy
sy
O
此方程曲线为圆 —应力圆（或莫尔圆，
由德国工程师,Otto Mohr引入）
sy
txyx sx
s?
t?
?
x
y
O t
n
?Set up a stress coordinate system as shown
in the following figure,
（ Pay attention to the selection of the scale ）
2,Method to plot the stress circle
?Plot the point A(s x,txy) and the point
B(sy,tyx) in the coordinate system,
?Inclined line AB intersects the axis sa
at the point C,This point is the center
of the stress circle,
?Plot a circle—stress circle with the
center C and the radius AC,
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
?建立应力坐标系，如下图所示,
（注意选好比例尺）
二、应力圆的画法
?在 坐标系内画出点 A(s x,txy)和
B(sy,tyx)
?AB与 s? 轴的交点 C便是圆心。
?以 C为圆心，以 AC为半径画
圆 ——应力圆；
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
3,Corresponding relation between the
element and stress circle
?Stress(s ?,t ?) in ? plane
A point (s ?,t ?) on the stress
circumference
?Normal line of ? plane
Radius of the stress circle
?Angle ? between two sections
Angle 2? between two radiuses；
And the direction of rotation is the
same,
sx
txy
sy
x
y
O
n
s?
t?
?
O
s?
t?
C
A(sx,txy)
B(sy,tyx)
x
2?
n D( s?,t??
三、单元体与应力圆的对应关系
??面上的 应力 (s ?,t ?)
应力圆上一点 (s ?,t ?)
??面的法线 应力圆的半径
?两面夹角 ? 两半径夹角 2? ；
且转向一致。
22
3
1
22
xy
yxyx
r a d i u s
ROC
t
ssss
s
s
?
?
?
?
???
?
?
?
）（
4,Mark extreme stresses on the circumference of
the stress circle
22
m i nm a x
m i n
m a x
2
2
xy
yx
r a d i u s
R
t
ss
ss
t
t
?
?
?
?
????
?
?
?
）（
O C s
?
t?
A(sx,txy)
B(sy,tyx)
x
2?1
mint
maxt
2?0
s1 s2 s3
22
3
1
22
xy
yxyx
ROC
t
ssss
s
s
?
?
?
?
???
?
?
?
）（
半径
四、在应力圆上标出极值应力
22
m i nm a x
m i n
m a x
2
2
xy
yx
R
t
ss
ss
t
t
?
?
?
?
????
?
?
?
）（
半径
O C s
?
t?
A(sx,txy)
B(sy,tyx)
x
2?1
mint
maxt
2?0
s1 s2 s3
s3
Example 3 Determine principal stresses and orientation of principal
planes of the element as shown in the figure.(unit,MPa)
45
325
325
95
150
°
A
B
s 1
s2
Method 1-graphical method,
? Stress coordinate
system is shown in the figure,
?Intersection C of the
perpendicular bisection line of
AB and the axis sa is the center
of the circle,Plot the circle with
the center C and the a radius
AC—stress circle,
?0
s1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
)325,45(B
?Plot the point and )325,95(A
s3
例 3 求图示单元体的主应力及主平面的位置。 (单位,MPa)
45
325
325
95
150
°
A
B
s 1
s2
解法 1——图解法,
?主应力坐标系如图
?AB的垂直平分线与
s? 轴的交点 C便是
圆心，以 C为圆心，
以 AC为半径画
圆 ——应力圆
?0
s1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
)325,45(B
)325,95(A
?在 坐标系内画出点
s3 s
1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
?Principal stresses and principal
planes as shown in the figure
M P a 0
M P a 20
M P a 120
3
2
1
?
?
?
s
s
s
?30
0 ???
45
325
325
95
150
°
s 1
?0 s2
A
B
s3 s
1 s2
B A
C
2s0
s?
t?
(MPa)
(MPa)
O
20MPa
?主应力及主平面如图
?30
0 ???
45
325
325
95
150
°
s 1
?0 s2
A
B
M P a 0
M P a 20
M P a 120
3
2
1
?
?
?
s
s
s
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
45
325
325
95
150
°
Method 2— analytical method,Analysis—set up the coordinate as
shown in the figure,
xyyx
y
tt
s
???
?
M P a325
M P a45
?xs
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（
60°
M Pa325
M Pa95
0
0
60
60
?
?
t
s
x
y
O
?t?
ss
t ? 2c o s2s i n
2 xy
yx ???
45
325
325
95
150
°解法 2—解析法：分析 ——建立坐标系如图
xyyx
y
tt
s
???
?
M P a325
M P a45
?xs
22
2
1
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（60°
x
y
O
M Pa325
M Pa95
0
0
60
60
?
?
t
s
§ 7–4 PRINCIPAL STRESSES AND THEIR
TRAJECTORIES OF THE BEAM
z
z
xy Ib
QS ?
?t
z
x I
My?s
1
2
3
4
5
P1 P2 q
Element,
22
3
1
22 xy
xx tss
s
s
???
?
?
?
）（
As shown in the figure,the beam
produced the shear bending（
transversal bending）,where, M、
Q>0 in the beam,Try to determine the
magnitude of the principal stresses and
the position of the principal planes of
each point in the section,
§ 7–4 梁的主应力及其主应力迹线
z
z
xy Ib
QS ?
?t
z
x I
My?s
1
2
3
4
5
P1 P2 q 如图，已知梁发生剪切弯
曲（横力弯曲），其上 M、
Q>0，试确定截面上各点主
应力大小及主平面位置。
单元体,
22
3
1
22 xy
xx tss
s
s
???
?
?
?
）（
1
s1
s3
s3
3
s1
s3
s1
s1
s3
5
?0
–45°
?0
s
t
A1 A2 D2 D1
C O
s
A2
D2
D1
C
A1
O
t
2?0
s
t
D2
C
D1
O
2?0= –90°
s
D2
A1 O
t
2?0
C
D1
A2
s
t
A2 D2 D1
C
A1
O
1
s1
s3
s3
3
s1
s3
s1
s1
s3
5
?0
–45°
?0
s
t
A1 A2 D2 D1
C O
s
A2
D2
D1
C
A1
O
t
2?0
s
t
D2
C
D1
O
2?0= –90°
s
D2
A1 O
t
2?0
C
D1
A2
s
t
A2 D2 D1
C
A1
O
Tensile
force
Compressive
force
Principal stress trajectories,
Envelopes of the direction lines of principal stresses—tangent at
each
point on the curve indicates the orientation of the tensile(or
compressive)
principal stress at the same
point,
Solid lines express the tensile
principal stress trajectories；
dashed lines express the
compressive principal stress
trajectories,
s1 s3
拉力
压力
主应力迹线（ Stress Trajectories),
主应力方向线的包络线 ——曲线上每一点的切线都指示
着该点的拉主应力方位（或压主应力方位）。
实线表示拉主应力迹线；
虚线表示压主应力迹线。
s1 s3
x
y
Method to plot principal
stress trajectories,
1
1
2
2
3
3
4
4
i
i
n
n
b a
c
d
q
s3
s1
Section section section section section section
q
x
y
主应力迹线的画法,
1
1
截面
2
2
截面
3
3
截面
4
4
截面
i
i
截面
n
n
截面
b a
c
d
s3
s1
§ 7–5 ANALYSIS OF TRIAXIAL STRESSED STATE—
METHOD OF STRESS CIRCLE
s2
s1
x
y
z
s3 1s2s3s
?s
?t
1,Spatial stressed state
§ 7–5 三向应力状态研究 ——应力圆法
s2
s1
x
y
z
s3
1s
2s3s
?s
?t
1、空间应力状态
2,Analysis of the
triaxial stress
Fig.a Fig.b
?The maximum shearing stress inside the
whole element is,
t max
2
31
m a x
ss
t
?
?
s2
s1
x
y
z
s3
1s
2s3s
?s
?t
?Elastic theory proved that stresses on any plane passing through a point in by the
element shown in Fig.a may be corresponding to the coordinates of a point on the
circumference of the stress circle or in the shadow region,
2、三向应力分析
?弹性理论证明，图 a单元体内任意一点任意截面上的应
力都对应着图 b的应力圆上或阴影区内的一点。
图 a
图 b
?整个单元体内的最大剪应力为,
t max
2
31
m a x
ss
t
?
?
s2
s1
x
y
z
s3
1s
2s3s
?s
?t
Example 4 Determine the principal stresses and the maximum shearing stress
of the element shown in the figure.(MPa)
Solution,?From
the element sketch we
know plane yz is a
principal plane,
501 ?s
?Set up stress coordinates
as shown in the figure,
Plot the stress circle and
locate the point s1,get,
27
50
58
3
2
1
??
?
?
s
s
s
44m ax ?t
50 40
x
y
z
30
10
(MPa)
s?
（ MPa ） t?
A
B
C
A
B
s1 s2 s3
t max
[例 4] 求图示单元体的主应力和最大剪应力。（ MPa）
解,?由单元
体图知,y
z面为主平
面
501 ?s
?建立应力坐标系如
图，画应力圆和
点 s1,得,
27
50
58
3
2
1
??
?
?
s
s
s
44m ax ?t
50 40
x
y
z
30
10
(M Pa)
s?
（ M Pa ） t?
A
B
C
A
B
s1 s2 s3
t max
§ 7–6 ANALYSIS OF STRAIN IN A PLANE
x
y
O
?
1,Determine the expression of strain analysis by the method of superposition
??? ? c o sd 11 xaDD ??
??? ? 21 c o sx?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
s i n/
c o s
1
x
xx
a
a
b
b
B O EA O D
??
???
?????
a b
c d
?
A
O
B
Shearing strain,Incremental
quantity of the right angle！（ Only
like this the contents in the preceding
section and the following section Can
correspond to each other,）
?
D
D1
E
E1
? ? ?
§ 7–6 平面内的应变分析
x
y
O
?
一, 叠加法求应变分析公式
??? ? c o sd 11 xaDD ??
??? ? 21 c o sx?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
s i n/
c o s
1
x
xx
a
a
b
b
B O EA O D
??
???
?????
a b
c d
?
A
O
B
剪应变,直角的增大量！
（只有这样，前后才对应）
?
D
D1
E
E1
? ? ?
??? ? s i nd 22 ycDD ??
??? ? 22 s iny?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
/ c o s
s i n
2
y
yy
c
c
c
c
B O EA O D
??
???
?????
x
y
O
a b
c d
?
A
O
B D
D2
E
E2
? ?
??? ? s i nd 22 ycDD ??
??? ? 22 s iny?
??
?
??
?
??
?
?
2s i n
/ c o s
s i n
/ c o s
s i n
2
y
yy
c
c
c
c
B O EA O D
??
???
?????
x
y
O
a b
c d
?
A
O
B D
D2
E
E2
? ?
??? ? c o sd 33 xycADd ???????
???? ? s o cxy s i n3 ??
? ????
?
??
?
??
?
?
22
33
s i nc o s
/ c o s
c o s
s i n/
s i n
??
???
?????
xy
xyxy
c
c
c
c
B O EA O D
D
D3
E
E3
?
xy?
xy?
x
y
O
a b
c d
?
A
O
B
??? ? c o sd 33 xycADd ???????
???? ? s o cxy s i n3 ??
? ????
?
??
?
??
?
?
22
33
s i nc o s
/ c o s
c o s
s i n/
s i n
??
???
?????
xy
xyxy
c
c
c
c
B O EA O D
D
D3
E
E3
?
xy?
xy?
x
y
O
a b
c d
?
A
O
B
????????? ?? c o ss i ns i nc o s 22
3
1
xyyx
i
i ???? ?
?
? ?????????? ?? 22
3
1
s i nc o s2s i n2s i n ????? ?
?
xyyx
i
i
?t?
ssss
s ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ????
?
?
???
????
? ? 2s i n
2
1
2c o s
22 xy
yxyx ?????
???
???
? 2c o s
2
1
2s i n
22 xy
yx ????
?
?
????????? ?? c o ss i ns i nc o s 22
3
1
xyyx
i
i ???? ?
?
?t?
ssss
s ? 2s i n2c o s
22 xy
yxyx ?????
?t?
ss
t ? 2c o s2s i n
2 xy
yx ????
?
?
? ?????????? ?? 22
3
1
s i nc o s2s i n2s i n ????? ?
?
xyyx
i
i
???
????
? ? 2s i n
2
1
2c o s
22 xy
yxyx ?????
???
???
? 2c o s
2
1
2s i n
22 xy
yx ????
?
?
2),Plot the strain circle according
to strain( )of a point,
xyyx ???,,
2,Graphical method of strain analysis—strain circle
22 ; 2 ; ??t?s? ???? ???
1),Analogic relation of stress circle and strain circle
?Set up strain coordinates as shown in
the figure
?locate the point A(?x,?xy/2) and
B(?y,-?yx/2) in the coordinate system
?intersection of AB and axis ?? is the
center of the circle,
?plot circle by the center C with a radius AC —strain circle,
??
??/2
A
B
C
2、已知一点 A的应变（ ），画应变圆
xyyx ???,,
二、应变分析图解法 ——应变圆 （ Strain Circle）
22 ; 2 ; ??t?s? ???? ???
1、应变圆与应力圆的类比关系
?建立应变坐标系如图
?在 坐标系内画出点
A(?x,?xy/2)
B(?y,-?yx/2)
?AB与 ?? 轴的交点 C便是圆心
?以 C为圆心，以 AC为半径画圆 ——应变圆。
??
??/2
A
B
C
??
??/2
3,Corresponding relation between the strain in ? direction
and strain circle
?max ?min
2?0
D(??,??/2)
2?
n
? Strain in ?direction(? ?,? ?/2)
A point on the strain circle(??,? ?/2)
?The direction line of ?
Radius of the strain circle
?Angle ? between two directions
Angle 2? between the two
radiuses and the direction of
rotation is the same,
A
B
C
??
??/2
三, ?方向上的 应变与 应变圆的对应关系
?max ?min
2?0
D(??,??/2)
2?
n
??方向上的 应变 (? ?,? ?/2)
应变圆上一点 (??,? ?/2)
?? 方向线 应变圆的半径
?两方向间夹角 ?
两半径夹角 2? ；且转向一致。
A
B
C
4,Values and orientation of principal strains
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
22 ;
2; ??t?s? ???? ???
22
m i n
m a x
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（ yx
xytg
ss
t
??
?
??
2
2 0
yx
xy
??
?
?
?
?
?02tg
四, 主应变数值及其方位
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
22 ;
2; ??t?s? ???? ???
yx
xy
??
?
?
?
?
?02tg
22
m i n
m a x
22 xy
yxyx tssss
s
s
?
?
?
?
?
?
?
?
）（ yx
xytg
ss
t
??
?
??
2
2 0
Example 5 Knowing three strain s??1,??2 and ??3 in
directions ?1,?2 and ?3 at a point in some plane,determine the
principal strains in this plane,
Solution,
1,2,3 )(i c o ss i ns i nc o s 22 ???? iixyiyixi ???????? ?
First find out ? x,? y,? x y by solving the above three
equations then determine the principal strains,
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
[例 5] 已知一点在某一平面内的 ?1,?2,?3 方向上的线 应变分
别为 ??1,??2,??3，,求该面内的主应变。
解：由
iixyiyixi ???????? ? c o ss i ns i nc o s
22 ???
i =1,2,3这三个方程求出 ? x,? y,? x y；然后再求主应变。
? ?? ?22
m i n
m a x
2
1
xyyxyx ?????
?
?
?????
?
?
?
）（
Example 6 Determine the principal strains of the point after
three linear strains at this point are tested by the strain foil of 45°,
x
y u
45o ?0 ?max
? ?][2)(
2
1 22
m a x ）（）（ yuuxyx ??????? ??????
? ?][2)(
2
1 22
m i n ）（）（ yuuxyx ??????? ??????
yx
yxu
??
???
?
?
??
?
2
2tg 0
[例 6] 用 45° 应变花测得一点的三个线应变后，求该点的主应变。
x
y u
45o ?0 ?max
? ?][2)(
2
1 22
m a x ）（）（ yuuxyx ??????? ??????
? ?][2)(
2
1 22
m i n ）（）（ yuuxyx ??????? ??????
yx
yxu
??
???
?
?
??
?
2
2tg 0
§ 7–7 STRESS — STRAIN RELATION UNDER THE COMPLEX
STRESSED STATE—（ GENERALIZED HOOKE’S LAW ）
1,Stress-strain relation in uniaxial tension
2,Stress-strain relation in pure shear
G
xy
xy
t
? ?
E
x
x
s? ?
xy E s
?? ??
xz E s
?? ??
) 0 x,y,z( i,jij ???
)( 0 x,y,zii ???
0?? zxyz ??
x
y
z
sx
x
y
z
t x y
§ 7–7 复杂应力状态下的应力 -- 应变关系
——（ 广义虎克定律 ）
一、单拉下的应力 --应变关系
二、纯剪的应力 --应变关系
G
xy
xy
t
? ?
)( 0 x,y,zii ???
0?? zxyz ??
x
y
z
sx
x
y
z
t x y
E
x
x
s? ?
xy E s
?? ??
xz E s
?? ??
) 0 x,y,z( i,jij ???
3,Stress-strain relation in complex stressed state
According to
superposition we get,
? ?? ?
zyx
zyx
x
E
EEE
ss?s
s
?
s
?
s
?
???
???
1
? ?? ?xzyy E ss?s? ??? 1
? ?? ?yxzz E ss?s? ??? 1
G
xy
xy
t
? ?
G
yz
yz
t
? ?
G
zx
zx
t? ?
? ?? ?zyxx E ss?s? ??? 1
?
?
?
?
?
?
?
?
?
?
?
x
y
z
sz
sy
txy
sx
三、复杂状态下的应力 --- 应变关系
依叠加原理,得,
? ?? ?
zyx
zyx
x
E
EEE
ss?s
s
?
s
?
s
?
???
???
1
? ?? ?xzyy E ss?s? ??? 1
? ?? ?yxzz E ss?s? ??? 1
G
xy
xy
t
? ?
G
yz
yz
t
? ?
G
zx
zx
t? ?
? ?? ?zyxx E ss?s? ??? 1
?
?
?
?
?
?
?
?
?
?
?
x
y
z
sz
sy
txy
sx
Principal stress—principal strain relation
4,Stress-strain relation
under the state of plane
stress
0??? zxyzz tts
The directions are the same
02tg
2
?
ss
t
?
?
??
yx
xy
yx
xy
??
?
?
?
?
??02tg
? ?? ?1322 1 ss?s? ??? E
? ?? ?1233 1 ss?s? ??? E
? ?? ?3211 1 ss?s? ??? E
xyxy G ?t ?
? ?yxx E ???
?
s ?
?
? 2
1
? ?xyy E ???
?
s ?
?
? 2
1
主应力 --- 主应变关系
四、平面状态下的应力 ---应变关系,
0??? zxyzz tts
方向一致
? ?? ?1322 1 ss?s? ??? E
? ?? ?1233 1 ss?s? ??? E
? ?? ?3211 1 ss?s? ??? E
xyxy G ?t ?
? ?yxx E ???
?
s ?
?
? 2
1
? ?xyy E ???
?
s ?
?
? 2
1
02tg
2
?
ss
t
?
?
??
yx
xy
yx
xy
??
?
?
?
?
??02tg
The directions of the principal stress and the principal
strain are the same,
0
2
0 2tg
)()]1)([(
1
22
2tg ?
??
?
???
?
?
ss
t
? ??
?
??
??
?
?
?
?
?
?
yx
xy
yx
xy
yx
xy
E
G
主应力与主应变 方向一致。
0
2
0 2tg
)()]1)([(
1
22
2tg ?
??
?
???
?
?
ss
t
? ??
?
??
??
?
?
?
?
?
?
yx
xy
yx
xy
yx
xy
E
G
5,Relation between the volumetric strain and stress components,
321 aaaV ?
)1()1()1( 3322111 ??? ???? aaaV
321
1 ??? ??????
V
VV
Volumetric strain,
)(
21
)(
21
321
zyx
E
E
sss
?
sss
?
??
?
?
??
?
??
Relation between the volumetric
strain and stress components,
五、体积应变与应力分量间的关系
321 aaaV ?
)1()1()1( 3322111 ??? ???? aaaV
321
1 ??? ??????
V
VV
体积应变,
)(
21
)(
21
321
zyx
E
E
sss
?
sss
?
??
?
?
??
?
??
体积应变与应力分量间的关系,
Example 7 A structure member is subjected to some forces,Two principal
strainsat a point on the free surface of the member are ?1=240?10-6 ?2=–160?10-6，
modulus of elasticity is E=210GPa,Possion’s ratio is ?=0.3,Try to determine the
principal stresses and another principal strain at this point,
03 ?s
? ? M P a3.4410)1603.0240(
3.01
10210
1
6
2
9
2121 ?????
???
?
?? ????
?
s E
,Then this point is in the stat
? ?
M P a3.2010)2403.0160(
3.01
10210
1
6
2
9
1222
??????
?
?
?
?
?
?
?
???
?
s
E
1s?
2s?
Solution,On the free surface
Of plane stress.,
[例 7] 已知一受力构件自由表面上某一点处的两个面内主应变分别
为,?1=240?10-6,?2=–160?10-6，弹性模量 E=210GPa，泊松比
为 ?=0.3,试求该点处的主应力及另一主应变 。
03, ?s自由面上解
所以，该点处为平面应力状态
1s?
2s?
? ? M P a3.4410)1603.0240(
3.01
10210
1
6
2
9
2121 ?????
???
?
?? ????
?
s E
? ?
M P a3.2010)2403.0160(
3.01
10210
1
6
2
9
1222
??????
?
?
?
?
?
?
?
???
?
s
E
? ? 669132 103.3410)3.443.22(
10210
3.0 ???????
?
????? ss??
E;M P a3.20;0;M P a3.44 321 ???? sss
? 3 34 2, ? ? × 10-6
? ? 669132 103.3410)3.443.22(
10210
3.0 ???????
?
????? ss??
E;M P a3.20;0;M P a3.44 321 ????? sss
? 3 34 2, ? ? × 10-6
Example 8 A thin-walled container subjected to inside pressure is shown in
Fig.a,In order to determine the value of the inside pressure the hoop strain tested
on the surface of the container with strain foil is ? t =350× l06,If the mean diameter
of the container is D=500 mm,thickness of its wall is ?=10 mm,E=210GPa，
?=0.25,Try to, 1.derive the expressions of stress in the lateral and longitudinal
sections of the container, 2.calculate the inside pressure of the container,
p
p
p
x
st
sm
L
p
O D
x
A B
y
Fig.a
[例 8] 图 a所示为承受内压的薄壁容器。为测量容器所承受的内
压力值，在容器表面用电阻应变片测得环向应变 ? t =350× l06，
若已知容器平均直径 D=500 mm，壁厚 ?=10 mm，容器材料的
E=210GPa,?=0.25，试求,1.导出容器横截面和纵截面上的正应
力表达式； 2.计算容器所受的内压力。
p
O D
x
A B
y
图 a
p
p
p
x
st
sm
L
1)Longitudinal stresses
Solution,Expressions of the circumferential and longitudinal stress of
the container reservoir
Cutting the container along a section shown and in Fig.b
considering the equilibrium of the right part,we get
? ? 42DpDm ???s ??
?s 4
pD
m ?
p
sm
sm
x D
Fig.b
1、轴向应力,(longitudinal stress)
解：容器的环向和纵向应力表达式
用横截面将容器截开，受力如图 b所示,根据平衡方程
? ? 42DpDm ???s ??
?s 4
pD
m ?
p
sm
sm
x D
图 b
Imagine cut off the container along
longitudinal section,and take the upper
pant as the study object,The free body
diagram is shown in Fig.c
2,Circumferential stress
? ? Dlplt ??? ?s 2
?s 2
pD
t ?
3,Determine the inside pressure
（ by stress-strain relation）
? ? ? ????ss? ???? 241 EpDE mtt
M P a36.3
)25.02(5.0
1035001.0102104
)2(
4
69
?
??
?????
?
?
?
?
?
??
D
E
p
t
st
sm External
surface
y
p s
t s t D
q
dq )d2( q?? Dlp
z
Fig,c
O
用纵截面将容器截开，取长为 L的一部
分为研究对象，受力如图 c所示
2、环向应力,(hoop stress)
? ? Dlplt ??? ?s 2
?s 2
pD
t ?
3、求内压（以应力应变关系求之）
? ? ? ????ss? ???? 241 EpDE mtt
M P a36.3
)25.02(5.0
1035001.0102104
)2(
4
69
?
??
?????
?
?
?
?
?
??
D
E
p
t
st
sm
外表面
y
p s
t s t D
q
dq )d2( q?? Dlp
z
图 c
O
§ 7－ 8 STRAIN-ENERGY DENSITY OF
COMPLEX STRESSED STATE
332211 2
1
2
1
2
1 ?s?s?s ???u
)(31 321 ssss ???m
s2
s3
s 1
Fig,a
Fig,c
s3 -sm
s 1 -sm
s2 -sm
ba E ????
??? )(21
321 sss
? 0??
c
? ?? ?312321232221 22 1 ssssss?sss ?????? E
sm
Fig,b
sm
sm
§ 7－ 8 复杂应力状态下的变形比能
332211 2
1
2
1
2
1 ?s?s?s ???u
)(31 321 ssss ???m
s2
s3
s 1
图 a
图 c
s3 -sm
s 1 -sm
s2 -sm
ba E ????
??? )(21
321 sss
? 0??
c
? ?? ?312321232221 22 1 ssssss?sss ?????? E
sm
图 b
sm
sm
? ? ? ? ? ?? ?213232221
6
1 ssssss? ???????
E
u x
Where ux is called the strain-energy density corresponding to the
distortion,
Fig,c
s3 -sm
s 1 -sm
s2 -sm
Strain energy of the element shown in Fig.c is
? ? ? ? ? ?? ?213232221
6
1 ssssss? ???????
E
u x
:单元体的应变能为图 c
称为形状改变比能或歪形能。
图 c
s3 -sm
s 1 -sm
s2 -sm
Example 9 Prove the relations between three elastic constants by
the energy method,
G
u
22
1 2tt? ??
?Specific strain energy of the element in pure shear is,
? Expression of the specific strain energy of the
element in pure shear by principal stresses is,
? ?? ?312321232221 2
2
1 ssssss?sss ??????
E
u
? ?? ?tt?tt )(002)(02 1 22 ???????? E
21 t?
E
??
? ???? 12
EG
txy
A
s1
s3
[例 9] 用能量法证明三个弹性常数间的关系。
G
u
22
1 2tt? ??
?纯剪单元体的比能为,
?纯剪单元体比能的主应力表示为,
? ?? ?312321232221 2
2
1 ssssss?sss ??????
E
u
? ?? ?tt?tt )(002)(02 1 22 ???????? E
21 t?
E
??
? ????? 12
EG
txy
A
s1
s3
Chapter 7 Exercises
1,As shown in the figure,a certain point is in biaxial stress,
Try to determine when the two principal stresses are
both tensile stresses,
Solution,
xyt
? ? 0222 40402 4040
m in
m a x ???? ??
xyts
s
40??? xyt
4040 ???? xyt
第七章 练习题
一、某点应力状态为图示二向应力状态。试求
该点的两个主应力均为拉应力时 的取值范围。
解,
xyt
? ? 0222 40402 4040
m in
m a x ???? ??
xyts
s
40??? xyt
4040 ???? xyt
2,The stress state of an element is shown in the figure,
Knowing E andμof the material,Try to determine the
maximum shearing strain of this element,
Solution,
max?
oss ?1 02 ?s o
ss ??3
os
sst ???
2
31
m a x
? ?
oEG s
?t? ??? 12m a x
m a x
二、图示单元体的应力状态,已知材料的 E和 μ,
试求该单元体的最大剪应变 。
解,
max?
oss ?1 02 ?s oss ??3
os
sst ???
2
31
m a x
? ?
oEG s
?t? ??? 12m a x
m a x
3,The stresses on sections ac and bc at a certain point
of a force-acting member are shown in the figure,Try to
determine the magnitude and the directions of the stresses
on section ab by the stress circle and plot them on an
element,
三、受力构件某点处两截面 ac和 bc上的应力
如图所示,试利用应力圆确定 ab面上应力的大小和
方向，并画在单元体上。
