材料3
力学理论
1、主应力
通过主应力的大小和方向来描述复杂系统下的点的受力状态。主应力是这点上所受正应力的最大值，此平面剪切力是0。在二维应力系统中，图1.11任意一点的主应力和正应力在x y方向上的σx 和σy和在此点的剪切力τxy有如下方程：
主应力，
?1?1?(???2?2y??x)?12(?y??x)?4?xy (1.7)
22此点最大剪切力数值上等于主应力之差的一半
最大剪切力：τmax=
12（σ1—σ2）……………………………………(1. 8)
压缩应力是常见的但是不是很明显，拉伸力则是很明显的。 2、压力容器的分类
为了设计和分析的目的，压力容器根据壁厚与容器直径的比例分为两类：薄壁容器，厚铉比小于1/10和厚壁容器，比大于1/10。
主应力对于容器壁的作用，就是压力载荷，如图1.12所示。如果壁比较薄，放射状应力σ3将会非常小，与其他的应力相比则可以忽略不计，纵向的和圆周的应力σ1σ2可以看做是常数。对于薄壁容器，放射状的应力的大小将会产生很大影响，圆周的应力将会绕容器壁变化。大多数的应用于化工和联合工业的压力容器分为薄壁容器。厚壁容器应用于高压条件下。
3、允许压力
在本章的最先两节，方程主要是为了找出结构中的正应力和平均剪切应力。这些方程也可以在构件强度已知的条件下用来选择结构尺寸。材料的强度可以在几个方面定义，材料本身和材料的使用环境。一种定义是最大的强度或压力，最大的强度是在其受到轴向纯粹的压力而断裂时的压力。这个性质是在拉伸试验中获得的。这是一项利用普遍的测试仪器对精确的试样进行的试验。应用载荷缓慢增加，始终处于监督之下。最大的压力或者说强度是可以在原始区域分开的最大载荷。对于多数的工程材料来讲，最大的强度是被精确的给出的。 如果所受载荷超出了最大强度值就将会发生断裂。在多数的工程结构中我们么所希望的是不要发生断裂。所以设计基于最低值也称为允许应力或设计应力。举个例子，一个确定的钢铁最大强度是110000磅每平方英寸，一个更低的允许应力将会被用于设计，比如55000磅每平方英寸。允许应力将可能是最大强度的一半。最大强度与允许应力的比值就作为安全系数： 安全系数=最大应力/允许应力 或者 n=Su/SA
我们用S表示强度或者允许应力用σ表示材料的实际应力。在设计上：σ≤Sa
这个所谓的安全系数包含了很多的缺点。包括了载荷的不确定性，材料性能的不确定性，力的分析的不精确性。它被称为忽略系数可能更准确。通常，分析的更准确，更广泛更昂贵，安全系数的必要性就越小。
4、失败理论
在单向力作用下的简单结构因素的失败很容易就联想到材料的拉伸强度，像标准的拉伸实验。但是对于部件受到复合应力（正应力和剪切应力），位置就不那么简单了。几个失败的理论已经被提出了。三个常用的理论描述如下：
最大的正应力理论：当一个正应力达到简单拉伸力的失败值σe’时假设就失败了。简单张力的失败点被称作屈服应力，或者材料的拉伸强度，被一个合适的安全系数划分。
最大剪切力理论：当最大剪切力达到简单拉伸失败的剪切力的值时失败就会出现在复杂
的系统中。对于一个有联合作用力的系统，存在以下三个最大剪切力：
τ1=±(σ1-σ2)/2 ,τ2=±(σ2-σ3)/2 , τ3=±(σ3-σ1)/2
在张力测试中：τe=σe’/2
最大剪切应力取决于主应力的方向和大小，在二维应力系统中，例如在薄壁压力容器壁，最大的剪切应力可以将σ3 = 0代入方程1.10中求出。最大剪切应力理论常被称为特雷斯卡理论或者盖斯特理论。
最大应变能理论:当单位体积的应变能达到了在简单拉伸力中失败的值时，假设的负载系统就会崩溃。
最大剪切力理论已经被发现适合于预测延伸材料在复杂的载荷作用下的失败，通常应用于压力容器的设计。
（选自R. K. Sinnott 化学动力第6卷第二版，帕加马出版社，1996
选自Raymond F. Neathery，静力学与应用强度材料，约翰威力父子出版公司，1985）
Reading Material 4
Stresses in Cylindrical Shells due to Internal Pressure
The classic equation for determining stress in a thin cylindrical shell subjected to pressure is obtained from Fig. 1. 16. Summation of forces perpendicular to plane ABCD gives
PL * 2r =2σθLt or σθ=PL/t (1.17)
where P=pressure, L=length of cylinder,σθ=hoop stress ,r=radius ,t = thickness the strain εθ is defined as
εθ=(final length-original length)/original length
and from Fig.1.7., εθ=[2π(r + W)-2πr r]/ 2πr r or εθ=W/r (1.18) also εr=d W /d r (1. 19)
The radial deflection of a cylindrical shell subjected to internal pressure is obtained by substituting the quantity into Eq. (1. 18). Hence for thin cylinders
W=Pr^2/Et (1.20) where W = radial deflection, E =modulus of elasticity.
Equations (1. 17) and (1. 20) give accurate results when r/t>10. As r/t decreases, however, a more accurate expression is needed because the stress distribution through the thickness is not uniform. Recourse is then made to the \ theory first developed by Lame. The derived equations are based on the forces and stresses shown in Fig. 1. 18. The theory assumes that all shearing stresses are zero due to symmetry and a plane that is normal to the longitudinal axes before pressure is applied remains plane pressurization. In other words , εl is constant at any cross section
A relationship between σr and σθcan be obtained by taking a free-body diagram of ring dr as shown in Fig. 1. 18b. Summing forces in the vertical direction and neglecting higher-order terms, we then have
σθ-σr =dσr /d r (1.21)
A second relationship is written as
σθ=E[εθ(1-μ)+ μ(εr+εl)]/[(1+μ)(1-2μ)]
σr=E[εr(1-μ)+ μ(εθ+εl)]/[(1+μ)(1-2μ)] (1. 22) σl=E[εl (1-μ)+ μ(εθ+εr)]/[(1+μ)(1-2μ)]
Substituting Eqs. (1. 18) and (1. 19) into the first two expressions of Eq. (1.22) and substituting the result into Eq. (1. 21) results in
d^2w/dr^2 + dw/rdr – w/r^2=0
A solution of this equation is
w=A r + B /r (1.23)
where A and B are constants of integration and are determined by first substituting Eq. (1 23) into the first one of Eq. (1. 22) and then applying the boundary conditions σr = -pi at r = ri and σr= -po at r=ro
Expression (1, 23) then becomes
w = -μrε1+1[r^2(1-μ-2μ^2)(Piri^2-Poro^2)+ri^2ro^2(1+μ)(Pi-Po))/Er(ro^2-r1^2) (1. 24) Once w is obtained, the values of σθdetermined from Eqs. (1. 18), and (1. 19), and (1.22) and expressed for thick cylinders as
σθ=[Piri^2-Poro^2+(Pi-Po)(ri^2 ro^2/r^2)]/(ro^2-ri^2) (1.25) σr=[ Poro^2-Piri^2 +(Pi-Po)(ri^2 ro^2/r^2)]/(ro^2-ri^2)
where σr = radial stress σθ = hoop stress Pi =internal pressure P0=external pressure ri=inside radius ro=outside radius r = radius at any point
The longitudinal stress in a thick cylinder is obtained by substituting Eqs. (1. 18)，(1. 19) and (1. 24) into the last expression of Eqs. (1. 22) to give σl = Eε1+[ 2μ(Piri^2-Poro^2)]/(ro^2-ri^2)
This equation indicates thatσl is constant throughout a cross section because εl is constant and r does not appear in the second term. Thus the expressionσl can be obtained from statics as σl = (Piri^2-Poro^2)]/(ro^2-ri^2) (1.26)
With σl known, Eq. (1. 24) for the deflection of a cylinder can be expressed as
w={ r^2(Piri^2-Poro^2)(1-2μ )+(Pi-Po)Ri^2ro^2(1+μ )}/Er(ro^2-ri^2) (1. 27)
(Selected from Maan H. Jawad and James R. Farr,Structural Analysis and Design of Process Equipment,
John Wiley & Sons Inc. , 1984. )
材料4
内部压力引起的柱状壳体应力
定义遭受压力的薄圆柱外壳应力的经典方程式是从图1.16中获得的。ABCD垂直面商的合力如下：
PL * 2r =2σθLt or σθ=PL/t (1.17) P是压力，L是圆柱体长度，σθ是环向应力，r是半径，t 是厚度，张力定义如下：
εθ=（最终长度-原始长度）/原始长度
从图1.17得：
εθ=[2π(r + W)-2πr r]/ 2πr r or εθ=W/r (1.18)
同时也有：εr=d W /d r (1. 19)
圆柱外壳遭受外压力产生的径向偏移获自替代方程1.18。因此，对于薄壁圆柱体有： W=Pr^2/Et (1.20)
W是径向变形偏移，E是弹性模量。
方程式1.17和1.20给出了当r/t>10时的精确结果。当r/t增加时，由于应力在厚度层上的分布不均匀所以需要更精确的表达式。救生索被应用于厚壳体理论。方程式的推导基于图1.18所示的作用力和应力。这个理论由于对称性而假设所有的剪切应力都为0。一个平面的轴向压力在施加压力前保持着增加。换句话说，εl在任何截面上是不变的。
σr 和σθ之间的关系可以通过图1.18的图解获得。在垂直方向的合力忽略高阶条件的话我们得：
σθ-σr =dσr /d r (1.21)
第二种关系如下写：
σθ=E[εθ(1-μ)+ μ(εr+εl)]/[(1+μ)(1-2μ)]
σr=E[εr(1-μ)+ μ(εθ+εl)]/[(1+μ)(1-2μ)] (1. 22) σl=E[εl (1-μ)+ μ(εθ+εr)]/[(1+μ)(1-2μ)]
将方程1.18和1.19代入方程1.22的前两个式子，再将结果代入方程1.21得结果：
d^2w/dr^2 + dw/rdr – w/r^2=0
方程的一个解是：
w=A r + B /r (1.23)
A和B是综合的常数，由式1.23代入1.22决定，应用边界条件 σr = -pi at r = ri and σr= -po at r=ro 表达式1.23变成
w = -μrε1+1[r^2(1-μ-2μ^2)(Piri^2-Poro^2)+ri^2ro^2(1+μ)(Pi-Po))/Er(ro^2-r1^2) (1. 24) w一旦获得，从1.18,1.19和1.22得出σθ对于厚钢瓶的价值如下：
σθ=[Piri^2-Poro^2+(Pi-Po)(ri^2 ro^2/r^2)]/(ro^2-ri^2) (1.25) σr=[ Poro^2-Piri^2 +(Pi-Po)(ri^2 ro^2/r^2)]/(ro^2-ri^2)
σr=径向应力 σθ=环向应力 Pi =内部压力 P0=外部压力 ri=内径 ro=外径 r = 任意半径 将方程式1.18,，1.19，1.24代入1.22的最后一个表达式可得厚钢管的纵向应力： σl = Eε1+[ 2μ(Piri^2-Poro^2)]/(ro^2-ri^2)
这个方程表示通过截面的σl是常数，因为εl是常数，r在第二时期没有出现。所以表达式σl能从静力学中获得：
σl = (Piri^2-Poro^2)]/(ro^2-ri^2) (1.26)
然后σl已知，圆柱的偏差方程式1.24可以表达为：
w={ r^2(Piri^2-Poro^2)(1-2μ )+(Pi-Po)Ri^2ro^2(1+μ )}/Er(ro^2-ri^2) (1. 27)
(选自Maan H. Jawad and James R. Farr，结构分析和工艺设备设计，约翰威力父子出版社1984) Reading Material 5
Static and Dynamic Balance of Rotating Bodies
The unbalance of a single disk can detected by allowing the disk to rotate on its axle between two parallel knife-edges, as shown in Fig. 1, 22. The disk will rotate and come to rest with the heavy side on the bottom. This type of unbalance is called static unbalance, since it can be detected by static means.
In general, the mass of a rotor is distributed along the shaft such as in a motor armature or an automobile-engine crankshaft. A test similar to the one above may indicate that such parts are in static balance, but the system may show a considerable unbalance when rotated.
As an illustration, consider a shaft with two disks, as shown in Fig. 1. 23. If the two unbalance weights are equal and 180 deg. apart, the system will be statically balanced about the axis of the shaft. However, when the system is rotated, each unbalanced disk would set up a rotating centrifugal force tending to rock the shaft on its bearings. Since this type of unbalance results only from rotation we refer to it as dynamic unbalance.
Fig. 1. 24 shows a general case where the system is both statically and dynamically unbalanced. It will now be shown that the unbalanced forces P and Q can always be eliminated by the addition of two correction weights in any two parallel planes of rotation.
Consider first the unbalance force P, which can be replaced by two parallel forces P*a/l and P *b/l In a similar manner Q can be replaced by two parallel forces Q *c/l and Q *d/l . The two forces in each plane can then be combined into a single resultant force that can be balanced by a single correction weight as shown. The two correction weights C1 and C2 introduced in the two parallel planes completely balanced P and Q, and the system is statically and dynamically balanced. It should be further emphasized that a dynamically balanced system is also statically balanced. The converse, however, is not always true; a statically balanced system may be dynamically unbalanced.
Example A rotor 4 in. long has an unbalance of 3 oz. in. in a plane 1 in. from the left end, and 2 oz. in. in the middle plane. Its angular position is 90 deg. in the clockwise direction from the first unbalance when viewed from the left end. Determine the corrections in the two end planes, giving magnitude and angular positions.
Solution. The 3-oz. in. unbalance is equivalent to2*1/4 oz. in. at the left end and 3/4 oz. in. at the right end,as shown in Fig. 1. 25. The 2 oz. in. at the middle is obviously equal to 1 oz. in. at the ends.
Combining the two unbalances at each end, the corrections are： Left end： C1?1??0252?1?2?2.47oz. in. to be removed
o?1?tan12.252?240'clockwise from plane of first unbalance
C2??3?2???1?1.25oz. in. to be removed ?4??1?1?tan10.75?53clockwise from plane of first unbalance
o(Selected from; William T. Thomson, Vibration Theory and Applications, Prentice-Hall Inc. ,1965. )