材料力学第七章应力状态分析
合集下载
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
(*)τα = Nhomakorabea2
sin 2α + τ xy cos 2α
(**)
(*) 2 + (**) 2
(σ α −
σ x +σ y
2
) + (τ α ) = (
2 2
σ x −σ y
2
2 ) 2 + τ xy
(7 - 6)
In a given problem, σx, σy, τxy are the three constants, σα,, τα are the variables. This equation is an expression for a circle of radius
σ x −α y
2
cos 2α − τ xy sin 2α
(7-1)
τα =
sin 2α + τ xy cos 2α
3. Principle Stresses in Two-dimensional Problems To find the plane for a maximum or a minimum normal stress, let σ x −α y dσ α = −2[ sin 2α + τ xy cos 2α ] = 0 = −2τ α 2 dα 2τ xy tg 2α1 = − σ x −σ y
σ'=
σ x +σ y
(7 - 5)
∴τ max = ±
min
σ1 − σ 2
2
Example 7-1 For the state of stress shown in the figure, (a) find the stresses acting on the inclined plane with θ=-22.5°; (b) find the principle stresses and shown their sense on a properly oriented element; and (c) find the maximum shear stresses with the associated normal stresses and show the results on a properly oriented element. Solution: For original state of stress σx=3 Mpa σy=1 MPa τxy= -2 Mpa (a) From Eq.(7-1)
7-5 Yield and Fracture Criteria 1. Failure criteria for a material or for an element :
工程构件丧失正常承载能力或工作能力统称为失效。 磨损、腐蚀和疲劳断裂是机械失效的三大原因。除此以外,
常温静载下工作的构件还会有屈服和断裂引起的强度失效, 由于过大的弹性变形引起的刚度失效以及稳定性不够而引 起的局部屈曲或整体坍塌。在高温下长期工作的构件会产 生蠕变失效和应力松驰失效。
(σ x' ) max = σ 1 or 2 =
min
σ x +σ y
2
± (
σ x −σ y
2
2 ) 2 + τ xy
(7 - 2)
至于σ1 , σ2 与α1 , α1+90°的对应关系可由(7-1)式确定.
4. Maximum shear stresses in two-dimensional problems dτ α σ x − σ y = 2cos2α - 2τ xy sin2α = 0 dα 2 (σ x − σ y ) / 2 tg 2α 2 = = −ctg 2α1 (7 - 3) τ xy α 2 = α1 ± 450 These locate two planes on which τ max act.
+ σ y dA sin α cos α + τ yx dA sin α sin α = 0
Noting that τxy = τyx, we have
σ α = σ x cos 2 α + σ y sin 2 α − 2τ xy sin α cos α
=
σ x +α y
2 σ x −α y 2
+
σα = τα = σ x +α y
2 σ x −α y 2 +
σ x −α y
2
cos 2α − τ xy sin 2α
sin 2α + τ xy cos 2α
Rewrite last two equations as:
σα − σ x +σ y
2 σ x −α y =
σ x −σ y
2
cos 2α − τ xy sin 2α
7-2 Transformation of stress in two-dimensional problems
1. The basic problem 2. Stresses on the inclined plane Sense of σ , τ , α : σ : tensile → "+"; compressive → "-". τ : τ xy making a moment clockwise → "+"
and counterclockwise → "-". α : counterclockwise → "+"; clockwise → "-". For in equilibrium
y σy
τyx n α
τxy=-τyx σx τxy x
σx
σα τα
dAcos α
τxy
α dA
∑ Fσ = 0 ∑ Fτ = 0
0⎞ ⎟ 0⎟ σ3 ⎟ ⎠
This state of stress is said to be : triaxial.
When σ3=0, the state of stress is said to be : biaxial.
When σ2=0, σ3=0, the state of stress is said to be : Uniaxial.
⎧ α1 α1 = ⎨ α1 + 900 ⎩ One of these locates a plane on which σ max acts; the other locates the corresponding plane for the σ min .
Conclusion : on plane on which maximum or minimum normal stresses occur, there are no shear stresses. These planes are called " principle planes" of stress, and the stresses acting on these plane - -σ max and σ min - -are called the " principle stresses"; acting directions of σ max and σ min are called the " principle directions" of stress - state of a point.
2. Construction of Mohr’s circle
σy α1
τyx σx τxy
(σα,τα) 2α 2α1
σ1
Procedure: ①选择合适比例尺, 建立σ- τ 坐标系. ②由(σx, τxy)确定A点, ((σx+ σy)/2, 0)确定C点. ③以C点为圆心, CA为半径作 圆--Mohr’ circle. ④σ1所在的主平面位于与应 力圆上转向相同的α1处.
τ max = ± (
min
σ x −σ y
2
2 ) 2 + τ xy
(7 - 4)
一般只需计算τmax 即可. 其位置可由(7-1)式确定. When α=α2 , from (7-1)
2 - - - - the normal stresses acting on the planes of τ max If σ x , σ y are the principle stresses, then τ xy = 0
dAsin α σy τyx − σ x dA cos α cos α + τ yx dA sin α cos α − σ y dA sin α sin α = 0
σ α dA + τ xy dA cos α sin α
τ α dA − τ xy dA cos α cos α − σ x dA cos α sin α
R= (
σ x −σ y
2 2 σ α − τ α coordinate. The circle is called Mohr' s circle of stress.
) +τ
2
2 xy
with its center at (
σ x +σ y
,0) corresponding to
Obviously, 2α is the referent variable of Mohr' s circle.
Considering stresses on inclined plane with α.
7-4 三向应力状态简介
工程实际问题中, 也存在三向应力状态. 如低碳钢拉伸实 验中“颈缩”部分处于三向拉伸应力状态; 滚珠轴承中的滚珠与 内环接触处为三向压应力状态.
σ2 σ3 σ2 σ1
σ1
σ3
Fig. 7 - 16
Expressing the state of stress in matrix :
All of them are equivalent.
7-3 Mohr’s circle of stress for two-dimensional problems 1. Circle of stress The stresses on the inclined plane (α):
(b) The σmax , σmin
From(7-2)
Location: ? Let θ=31°43’, from (7-1)
3 +1 3 −1 σα = cos 63o 26'−(−2) sin 63o 26' = 4.24 MPa + 2 2
(c) τmax
From (7-4)
The normal stress on τ max plane
1.Stress Tensor
用无限小的3组相互垂直的平行平面截取构件所得到的小 立方体称为单元体（element slice)。 作用其上的应力如图所 示。 Matrix representation of the stress components: 由于 τi,j 两下 标的任意性(i,j =x, y, z)，要完整 描述一点的应力需要9个分量。 因此，应力分量（stress components）可表示为如下的 张量形式：
2. State of stress The two-dimensional stress shown in the figure is referred to as plane stress. In matrix representation such a stress can be written as
⎛σ x ⎜ ⎜τ ⎜ 0 ⎝
τ σy
0
0⎞ ⎟ 0⎟ 0⎟ ⎠
x,y,z坐标是任意选择的直角坐标系. 当用其 它一组坐标系表示同一点的应力状态时, 称作Stress transformation. 可以证明, 总 有一组坐标, 使得一点的应力张量表示为 如下形式:
⎛σ 1 0 ⎜ ⎜ 0 σ2 ⎜0 0 ⎝
Chapter 7 Analysis of Stress State 7-1 Introduction In the first part of this chapter, a formal treatment for changing the components of the state of stress given in one set of coordinate axes to any other set of rotated axes is discussed. The criteria for the onset of yield or the occurrence of fracture is treated in the second part.
sin 2α + τ xy cos 2α
(**)
(*) 2 + (**) 2
(σ α −
σ x +σ y
2
) + (τ α ) = (
2 2
σ x −σ y
2
2 ) 2 + τ xy
(7 - 6)
In a given problem, σx, σy, τxy are the three constants, σα,, τα are the variables. This equation is an expression for a circle of radius
σ x −α y
2
cos 2α − τ xy sin 2α
(7-1)
τα =
sin 2α + τ xy cos 2α
3. Principle Stresses in Two-dimensional Problems To find the plane for a maximum or a minimum normal stress, let σ x −α y dσ α = −2[ sin 2α + τ xy cos 2α ] = 0 = −2τ α 2 dα 2τ xy tg 2α1 = − σ x −σ y
σ'=
σ x +σ y
(7 - 5)
∴τ max = ±
min
σ1 − σ 2
2
Example 7-1 For the state of stress shown in the figure, (a) find the stresses acting on the inclined plane with θ=-22.5°; (b) find the principle stresses and shown their sense on a properly oriented element; and (c) find the maximum shear stresses with the associated normal stresses and show the results on a properly oriented element. Solution: For original state of stress σx=3 Mpa σy=1 MPa τxy= -2 Mpa (a) From Eq.(7-1)
7-5 Yield and Fracture Criteria 1. Failure criteria for a material or for an element :
工程构件丧失正常承载能力或工作能力统称为失效。 磨损、腐蚀和疲劳断裂是机械失效的三大原因。除此以外,
常温静载下工作的构件还会有屈服和断裂引起的强度失效, 由于过大的弹性变形引起的刚度失效以及稳定性不够而引 起的局部屈曲或整体坍塌。在高温下长期工作的构件会产 生蠕变失效和应力松驰失效。
(σ x' ) max = σ 1 or 2 =
min
σ x +σ y
2
± (
σ x −σ y
2
2 ) 2 + τ xy
(7 - 2)
至于σ1 , σ2 与α1 , α1+90°的对应关系可由(7-1)式确定.
4. Maximum shear stresses in two-dimensional problems dτ α σ x − σ y = 2cos2α - 2τ xy sin2α = 0 dα 2 (σ x − σ y ) / 2 tg 2α 2 = = −ctg 2α1 (7 - 3) τ xy α 2 = α1 ± 450 These locate two planes on which τ max act.
+ σ y dA sin α cos α + τ yx dA sin α sin α = 0
Noting that τxy = τyx, we have
σ α = σ x cos 2 α + σ y sin 2 α − 2τ xy sin α cos α
=
σ x +α y
2 σ x −α y 2
+
σα = τα = σ x +α y
2 σ x −α y 2 +
σ x −α y
2
cos 2α − τ xy sin 2α
sin 2α + τ xy cos 2α
Rewrite last two equations as:
σα − σ x +σ y
2 σ x −α y =
σ x −σ y
2
cos 2α − τ xy sin 2α
7-2 Transformation of stress in two-dimensional problems
1. The basic problem 2. Stresses on the inclined plane Sense of σ , τ , α : σ : tensile → "+"; compressive → "-". τ : τ xy making a moment clockwise → "+"
and counterclockwise → "-". α : counterclockwise → "+"; clockwise → "-". For in equilibrium
y σy
τyx n α
τxy=-τyx σx τxy x
σx
σα τα
dAcos α
τxy
α dA
∑ Fσ = 0 ∑ Fτ = 0
0⎞ ⎟ 0⎟ σ3 ⎟ ⎠
This state of stress is said to be : triaxial.
When σ3=0, the state of stress is said to be : biaxial.
When σ2=0, σ3=0, the state of stress is said to be : Uniaxial.
⎧ α1 α1 = ⎨ α1 + 900 ⎩ One of these locates a plane on which σ max acts; the other locates the corresponding plane for the σ min .
Conclusion : on plane on which maximum or minimum normal stresses occur, there are no shear stresses. These planes are called " principle planes" of stress, and the stresses acting on these plane - -σ max and σ min - -are called the " principle stresses"; acting directions of σ max and σ min are called the " principle directions" of stress - state of a point.
2. Construction of Mohr’s circle
σy α1
τyx σx τxy
(σα,τα) 2α 2α1
σ1
Procedure: ①选择合适比例尺, 建立σ- τ 坐标系. ②由(σx, τxy)确定A点, ((σx+ σy)/2, 0)确定C点. ③以C点为圆心, CA为半径作 圆--Mohr’ circle. ④σ1所在的主平面位于与应 力圆上转向相同的α1处.
τ max = ± (
min
σ x −σ y
2
2 ) 2 + τ xy
(7 - 4)
一般只需计算τmax 即可. 其位置可由(7-1)式确定. When α=α2 , from (7-1)
2 - - - - the normal stresses acting on the planes of τ max If σ x , σ y are the principle stresses, then τ xy = 0
dAsin α σy τyx − σ x dA cos α cos α + τ yx dA sin α cos α − σ y dA sin α sin α = 0
σ α dA + τ xy dA cos α sin α
τ α dA − τ xy dA cos α cos α − σ x dA cos α sin α
R= (
σ x −σ y
2 2 σ α − τ α coordinate. The circle is called Mohr' s circle of stress.
) +τ
2
2 xy
with its center at (
σ x +σ y
,0) corresponding to
Obviously, 2α is the referent variable of Mohr' s circle.
Considering stresses on inclined plane with α.
7-4 三向应力状态简介
工程实际问题中, 也存在三向应力状态. 如低碳钢拉伸实 验中“颈缩”部分处于三向拉伸应力状态; 滚珠轴承中的滚珠与 内环接触处为三向压应力状态.
σ2 σ3 σ2 σ1
σ1
σ3
Fig. 7 - 16
Expressing the state of stress in matrix :
All of them are equivalent.
7-3 Mohr’s circle of stress for two-dimensional problems 1. Circle of stress The stresses on the inclined plane (α):
(b) The σmax , σmin
From(7-2)
Location: ? Let θ=31°43’, from (7-1)
3 +1 3 −1 σα = cos 63o 26'−(−2) sin 63o 26' = 4.24 MPa + 2 2
(c) τmax
From (7-4)
The normal stress on τ max plane
1.Stress Tensor
用无限小的3组相互垂直的平行平面截取构件所得到的小 立方体称为单元体（element slice)。 作用其上的应力如图所 示。 Matrix representation of the stress components: 由于 τi,j 两下 标的任意性(i,j =x, y, z)，要完整 描述一点的应力需要9个分量。 因此，应力分量（stress components）可表示为如下的 张量形式：
2. State of stress The two-dimensional stress shown in the figure is referred to as plane stress. In matrix representation such a stress can be written as
⎛σ x ⎜ ⎜τ ⎜ 0 ⎝
τ σy
0
0⎞ ⎟ 0⎟ 0⎟ ⎠
x,y,z坐标是任意选择的直角坐标系. 当用其 它一组坐标系表示同一点的应力状态时, 称作Stress transformation. 可以证明, 总 有一组坐标, 使得一点的应力张量表示为 如下形式:
⎛σ 1 0 ⎜ ⎜ 0 σ2 ⎜0 0 ⎝
Chapter 7 Analysis of Stress State 7-1 Introduction In the first part of this chapter, a formal treatment for changing the components of the state of stress given in one set of coordinate axes to any other set of rotated axes is discussed. The criteria for the onset of yield or the occurrence of fracture is treated in the second part.