英汉双语弹性力学3

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σx 图
σx
y
y
l
h
x
1
h
Fig.3-2
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2 3.应力函数 = cy 能解决矩形板在 x 方向受均布拉力 (设 c > 0)或均布压力(设 c < 0 )的问题.如图3-1(c).
三,应力函数取三次式 对应的应力分量: 3-2所示的矩形梁.
= ay 3 σ x = 6ay, σ y = 0, τ xy = τ yx = 0
εx =
σx =
M y, σ y = 0, τ xy = τ yx = 0 I
1 (σ x σ y ) E 1 ε y = (σ y σ x ) E 2(1 + ) γ xy = τ xy E
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§3 - 2
位移分量的求出
以矩形梁的纯弯曲问题为例,说明如何由应力分量求出 位移分量. 一,平面应力的情况 将应力分量 σ x =
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σ x = 0, σ y = 0,τ xy = τ yx = b
= bxy
b y (b) Fig.3-1 ,stress components
y (c)
= ax 2 能解决矩形板在 y 方向受均布拉力 结论:应力函数
2a b
(设 a > 0)或均布压力(设 a < 0 )的问题.如图3-1(a).
M y, σ y = 0, τ xy = τ yx = 0 代入物理方程 I
1 ε x = (σ x σ y ) E 1 ε y = (σ y σ x ) E 2(1 + ) γ xy = τ xy E
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Then the distortion components are:
M M εx = y, γ xy = 0 y, ε y = EI EI
Then put (a) into the geometric equation:
(a)
so: Integrate the two equation above:
u M = y, = y, + =0 x EI y EI x y
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3.The Stress Function in the form of a Polynomial of the Third Degree The Corresponding Stress Components
= ay 3
σ x = 6ay, σ y = 0, τ xy = τ yx = 0
(a)
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= ax 2 + bxy + cy 2
§3 - 1
一,应力函数取一次多项式
多项式解答
= a + bx + cy
应力分量: 应力边界条件:
σ x = 0, σ y = 0,τ xy = τ yx = 0
X =Y =0
结论:(1)线性应力函数对应于无面力,无应力的状态. (2)把任何平面问题的应力函数加上一个线性函 数,并不影响应力. 二,应力函数取二次多项式
M σx = y, σ y = 0, τ xy = τ yx = 0 I
The result is same with corresponding part in material mechanics Note: : The result above is useful to the beam which the length l is greatly more than the depth h ; to the beam which the length l is equal to the depth h , this result is useless.
Elasticity
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Chapter 3 Two-dimensional Problem in Rectangular Coordinates
§3-1 Solution by Polynomials §3-2 Determination of Displacements §3-3 Bending of a Simply Supported Beam by Uniform Load §3-4 Loading of a Wedge by Gravity and Hydraulic Pressure §3-5 Solution by Series §3-6 Bending of a Simply Supported Beam by Arbitrary Lateral Load Exercises
2a
b
o
2a
b
x
o
b x 2c
o
2c
x
y (a)
2.Corresponding to
Conclusion: The stress function = bxy can used to solute the problem of uniformly distributed shearing stresses rectangular plate.(Fig3-1b)
o
2a
b
x
o
b y (b)
图3-1
b x 2c
o
2c
x
y (a)
y (c)
2.对应于 = bxy ,应力分量 σ x = 0, σ y = 0,τ xy = τ yx = b
.
结论:应力函数 = bxy 能解决矩形板受均布剪力问题.如 图3-1(b).
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3.The stress function = cy can be used to solute the problem of uniformly distributed tensile(if c < 0 ) or compressive(if c > 0 ) stresses on -axis of rectangular plate.(Fig3-1c)
u ε x = x v εy = y v u + γ xy = x y v M v u
(b)
M M 2 u= xy + f1 ( y ), v = y + f 2 ( x) EI 2 EI
(c)
f1 and f 2 Is the arbitrary function.put(c)into the third equation of(b)
2.The Stress Function in the form of a Polynomial of the Second Degree 1.Corresponding to = ax 2,the stress components s x = 0, s y = 2 a ,t xy =t yx = 0.
= a + bx + cy
The Stress Components: The Stress Boundary Condition:
σ x = 0, σ y = 0,τ xy = τ yx = 0
X =Y =0 Conclusion:(1)The linear stress function is corresponding to the state of no surface force and no stress.(2)There's no effect to the stress to add a linear function to any stress function of two-dimensional problem.
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1× h 3 因为梁截面的惯矩是 I = ,所以上式可改写为: 12 M σx = y, σ y = 0, τ xy = τ yx = 0 I
结果与材料力学中完全相同. 注意: 注意: 对于长度 l 远大于深度 h 的梁,上面答案是有实用价值 的;对于长度 l 与深度 h 同等大小的所谓深梁,这个解答是 没有什么实用意义的.
Conclusion:The stress function(a) can be used to solute the problem of pure bending of rectangular beam. The rectangular beam is shown in Fig3-2.
M
+
h
M
2
6a ∫ ydy = 0,6a ∫ y 2 dy = M
Put in x σ equation(a)into the formula above,then:
h 2 h 2 h 2 h 2
h 2 h 2
h 2 h 2
∫
σ x dy = 0, ∫ σ x ydy = M
The first one always can meet ,and the second one request that : 2M a= h3 Put into formula(a),then:
(a)
= ay 3 能解决矩形梁受纯弯曲的问题.如图 结论:应力函数
M
+
h
M
2 2
σx 图
σx
y
y
l
图3-2hxh Nhomakorabea1
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Specification as following: In fig3-2,considering the unit width beam ,named the moment of double force on per unit width M .The order of here is[force][length]/[length],the result is [force] M . On the left or right,the two-dimensional force should be combined to a double force,as the moment of double force is M ,this request that: