求英文翻译:A和B同款，A取消了Piping，那么相应的，B款是否也应取消?_百度知道
求英文翻译:A和B同款，A取消了Piping，那么相应的，B款是否也应取消?
求英文翻译:A和B同款，A取消了Piping，那么相应的，B款是否也应取消?
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A和B同款，A取消了Piping，那么相应的，B款是否也应取消?A and B are in the same paragraph. A cancels Piping. So should the corresponding paragraph B be cancelled?
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baiduA和B同款?翻译成英语是：A and B are the same，A取消了Piping，那么相应的，B款是否也应取消://d.jpg" target="_blank" title="点击查看大图" class="ikqb_img_alink"><img class="ikqb_img" src="http://d.hiphotos.hiphotos.baidu.com/zhidao/pic/item/b999ae4eea540e087bf40bd1cbcf.com/zhidao/wh%3D600%2C800/sign=095c097f392ac65c67506e75cbc29e29/b999ae4eea540e087bf40bd1cbcf.jpg" esrc="http://d.hiphotos.baidu, A canceled Piping, so the corresponding, whether the B should be cancelled<a href="http
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服装加工工序英文翻译及解析
1.& &修边&&trimming pieces&&修剪毛坯裁片。
2.& &打线丁&&making tailor’s tack 用白棉纱线在裁片上做出缝制标记.
3.& &剪省缝&&slashing dart 毛呢服装省缝剪开。
4． 环缝&&overcasting stitches&&毛呢服装剪开的省缝用环形针法绕缝。
5． 缉省缝&&sewing darts 省缝折合机缉缝合。
6． 烫省缝&&pressing dart open 省缝坐倒熨烫或分开熨烫。
7． 推门&&blocking front piece 将平面前衣片推烫成立体衣片。
8． 缉衬省&&stitching dart of interlining 机缉衬布省道。
9． 缉衬&&stitching interlining&&机缉前身衬布。
10．烫衬&&pressing interlining 熨烫缉好的胸衬。
11. 敷胸衬&&attaching interlining&&前衣片敷上胸衬。
12．压衬&&pressing interfacing&&用粘合机将衣片和粘合衬进行热压粘合。
13．纳驳头&&pad-stitching lapel 手工或机扎驳头。（同义词：驳头）
14．敷止口牵条&&taping front edge 牵条布敷上驳口部位。
15．敷驳口牵条&&taping lapel roll line&&牵条布敷上驳口部位。
16．拼袋盖里&&matching flap facing 袋盖里拼接。
17． 做袋盖&&making flap&&袋盖面和里机缉缝合。
18． 翻袋盖&&making over flap&&袋盖正面翻出。
19． 做插笔口&&making an opening for pen on the flap 在小袋盖上口做插笔开口。
20． 滚袋口&&binding pocket mouth 毛边袋口用滚条包光。
21． 缉袋嵌线&&stitching bound pocket&&袋嵌线料缉在开袋上。
22． 开袋口&&cutting pocket mouth&&将缉好袋嵌线的袋口剪开。
23.&&封袋口&&stitching ends of pocket mouth&&袋口两头机缉封口。
24．缉转袋布&&stitching pocket bag&&袋布双层缝合。
25． 拼接挂面&&stitching facing 手工或机缝拼接挂面。
26． 拼接耳朵皮&&stitching flange&&手工或机缝拼接耳朵皮。
27． 敷挂面&&attaching facing&&挂面敷在前衣片止口部位，即敷过面。
28． 合止口&&joining front edge 门里襟止口机缉缝合。
29． 修剔止口&&trimming front edge&&止口缝毛边剪窄，剔薄。
30． 扳止口&&whipstitching front edge 止口缝毛边剪窄，剔薄。
31． 扎止口&&basting front edge 翻出的止口扎一道临时固定线。
32． 叠挂面&&basting front facing&&将挂面和大身扎在一起。
33． 合背缝&&joining center back seam&&背缝机缉缝合。
34.& &归拔后背&&blocking back seam&&将平面后衣片按体型归烫成立体衣片。
35． 敷袖窿牵条&&taping armhole 牵条布缝上后衣片的袖窿部位。
36． 敷背叉牵条&&taping back vent&&牵条布缝上背叉沿口部位。
37． 封背叉&&bartacking back vent&&背叉封结。
38． 合摆缝&&joining side seam 摆缝机缉缝合。
39． 分烫摆缝&&pressing open side seam&&摆缝缉缝分开熨烫。
40． 扣烫底边&&folding and pressing hem 衣边折转熨烫。
41．&&叠底边&&basting hem 底边扣烫后扎一道临时固定线。
42． 叠摆缝&&basting side seam&&将里子和面子的摆缝扎在一起。
43． 倒钩袖窿&&back-stitching armhole 沿袖窿用倒钩针法缝扎。
44． 合肩缝&&joining shoulder seam 肩缝机缉缝合。
45． 分烫肩缝 pressing open shoulder seam&&肩缝缉缝分开熨烫。
46． 叠肩缝&&slip-stitching shoulder seam&&肩缝缝头与衬扎牢。
47． 做垫肩&&making shoulder pad&&用布和棉花等做成垫肩。
48． 装垫肩&&setting shoulder pad 垫肩装在袖窿肩头部位。
49．倒钩领窝&&back-stitching neckline&&沿领窝用倒钩针法缝扎。
50．拼领衬&&joining collar interlining&&领衬拼缝机缉缝合或粘合搭拼。
51．拼领里&&applying interlining to collar 领里拼缝机缉缝合。
52．缉领里&&top-stitching under collar 机缉领里。
53. 归拔领里&&blocking under collar 领里归拔熨烫。
54. 归拔领面&&blocking top collar&&领面归拔熨烫。
55.敷领面&&attaching under collar to top collar 领面敷上领里。
56.合领子&&joining under collar and top collar 领面、里机缉缝合。
57. 翻领子&&turning over collar 领子正面翻出。
58．做领舌&&making collar band ends 做中山服底领探出的里襟。
59．包底领&&stitching collar band&&中山服底领包转机缉。
60. 绱领子&&sewing collar on and down 领子装上领窝。
61．分烫绱领缝&&pressing open collar seam&&绱领缉缝分开熨烫。
62．分烫领串口&&slip-stitching gorge seam 领串口缉缝分开熨烫。
63.&&叠领串口&&slip-stitching gorge seam 领串口缝与绱领缝扎牢。
64. 包领面&&turning over top collar seam allowances and catch-stitching it 西装、大衣领面外口包转，用三角针与领面绷牢。
65.上下领缝合&&attaching band to collar&&中山服或衬衫领上下结合。
66．压烫领脚&&pressing collar 装领完毕后，将前后领圈放平用熨斗压烫绱领缝子部位。
67．归拔偏袖&&blocking sleeve&&偏袖部位，归拔熨烫。
68．合袖缝&&joining sleeve seam&&袖缝机缉缝合。
69．分烫袖缝&&pressing open sleeve seam 袖缝缉缝分开熨烫。
70．挢袖叉&&slip-stitching sleeve slit 袖叉边与袖口贴边挢牢。
71．叠袖里缝&&matching and stitching sleeve lining seam allowance&&袖子面、里缉缝对齐扎牢。
72．翻袖子&&turning over sleeve&&袖子正面翻出，即缝袖山头吃势。
73．收袖山&&easing sleeve cap 收缩袖山松度或缝吃头。
74． 绱袖&&setting in sleeve&&袖子装在袖窿上。
75.&&挢袖窿&&slip-stitching sleeve lining to garment lining 挢袖窿里子。
76．压烫袖窿&&pressing armhole 装袖完毕，将袖窿按圆形放平（分段）用熨斗压烫绱袖缝子部分。
77．滚袖窿&&binding armhole 用滚条将袖窿毛边包光。
78．挢领钩&&attaching hook to collar band&&底领领钩开口处用手工挢牢。
79．挢领下口&&slip-stitching collar to garment 领里下口与大身挢牢。
80．叠暗门襟&&slip-stitching facing 暗门襟眼距间用暗针缝牢。
81．挢底边&&blindstitching hem 底边与大身挢牢。
B.&&女毛呢服装
1.合刀背缝&&stitching princess line&&刀背缝机缉缝合。
2.烫刀背缝&&pressing princess line&&刀背缝缉缝坐倒或分开熨烫。
3．定眼位&&marking button position&&划准扣眼位置。
4．滚扣眼&&bounding buttonhole 用滚眼料把扣眼毛边包光。
5．开扣眼&&cutting buttonhole&&扣眼剪开。
6. 滚挂面&&bias binding facing 挂面里口毛边用滚条包光。
7．合领面&&joining top collar&&领面拼缝机缉缝合（合领里与此同）
8．做袋爿&&making welt pocket 袋爿缝扣转挢上里子。
9．做袋&&making pocket 做各种袋。
10.绱袋&&attaching pocket to garment&&口袋装在袋位上。
11．翻小袢&&turning over tab&&小袢正面翻出。
12．绱袖袢&&attaching sleeve tab 袖袢装在袖口上规定的部位。
13．合腰带&&stitching waistband and lining&&腰带机缉缝合。
14．翻腰带&&turning over waistband&&腰带正面翻出。
15．坐烫里子缝&&pressing lining seam rolling to underside 里子缉缝坐倒熨烫。
16．合大身面里&&stitching garment and lining together 大身面里机缉缝合。
17．翻里子&&turning over lining&&面、里正面翻出。
18．敷里子&&attaching lining to garment 里子敷上大身。
19．挢扣眼&&slip-stitching buttonhole 扣眼毛边折光挢牢。
1. 角薄膜定位&&positioning collar stay&&领角薄膜在领衬上定位。
2．热缩领面&&pressing top collar for preshrinking&&领面防缩熨烫。
3．粘翻领&&fusing interlining&&领衬与领面三边沿口上浆粘合。
4．压领角&&pressing coller point&&上领翻出后领角热定型。
5.夹翻领&&attaching coller to band 翻领夹进底领机缉合。
6.扣烫过肩&&folding and pressing back yoke&&过肩毛边折转扣烫。
7．绱过肩&&setting back yoke 过肩缉在衣片上。
8．绱明门襟&&attaching facing&&门襟装在前衣片止口上。
9．镶边&&making bias binding as a decorative trim&&用镶边料装在衣片边上。
10．镶嵌线&&piping and binding&&用嵌线料镶在衣片上。
11．缉明线&&top stitching 机缉服装表面线迹。
12．合袖头&&joining sleeve bottom&&袖头里面的缝合。
13．翻袖头&&turning over cuff&&将兜好的袖头面翻
14. 绱袖叉条&&stitching placket tape to sleeve 袖叉条装在袖叉& & 位。
15.封袖叉&&top stitching close to placket&&袖开叉机缉封牢。
16．绱袖头&&attaching cuff to sleeve&&袖头与袖子缝合。
17．绱橡筋&&attaching elastic 橡筋装在部件上。
18．定扣位&&marking button position&&标出钮扣位置。
19．锁扣眼&&sewing buttonhole 扣眼毛边用线锁光，分机锁和手工锁眼。
20．钉扣&&sewing button&&钮扣钉在钮位上。
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reading material 1anal alysis static analysis of beamsa bar that is subjected to forces acting transverse to its axis is called &&&&a beam．in this section we will consider only a few of the simplest types of beams， such as those shown in fig． 2． every 1． in instance it is assumed that the beam has a plane of symmetry that is parallel to the plane of the figure itself．thus，the cross section of the beam has a vertical axis of symmetry．also，it is assumed that the applied loads act in the plane of symmetry，and hence bending of the beam occurs in that plane．later we will consider a more general kind of bending in which the beam may have an unsymmetrical cross section． (a)a simple supported beam (b) a cantilever beam (c) a beam with an overhang fig．1．2 types of beams the beam in fig．1．2(a)，with a pin support at one end and a roller support at the other，is called a simply supported beam， a simple beam． essential feature of a simple beam is that or the both ands of the beam may rotate freely during bending, but they cannot translate in the lateral direction. also, one end of the beam can move freely in the axial direction (that is, horizontally). the supports of a simple beam may sustain vertical reactions acting either upward or downward. the beam in fig. 1.2(b) which is built in or fixed at one end and free at the other end, is called a cantilever beam. at the fixed support the beam can neither rotate nor translate, while at the free end it may do both. the third example in the figure shows a beam with an overhang. this beam is simply supported at a and b and has a free end at c. loads on a beam may be concentrated forces, such as p1 and p2 in fig. 1. 2(a) and (c),or distributed loads, such as the load q in fig. 1.2(b). distributed loads are characterized by their intensity, which is expressed in units of force per unit distance along the axis of the beam. for a uniformly distributed load, illustrated in fig. 1.2(b), the i a varying load, on the other hand, is one in which the intensity varies as a function of distance along the axis of the beam. the beams shown in fig. 1.2 are statically determinate because all their reactions can be determined from equations of static equilibrium. for instance, in the case of the simple beam supporting the load pl[fig. 1.2(a)], both reactions are vertical, and their magnitudes can be found by summing mo thus, we find ra=p1 (l--a)/ l rb =p1a/ l the reactions for the beam with an overhang [fig. 1. 2 (c)] can be found in the same manner. for the cantilever beam [fig. 1.2 (b)], the action of the applied load q is equilibrated by a vertical force ra and a couple ma acting at the fixed support, as shown in the figure. from a summation of forces in the vertical direction, we conclude that ra =q b and, from a summation of moments about point a, we find ma =q b(a+b/2) the reactive moment ma acts counterclockwise as shown in the figure. the preceding examples illustrate how the reactions(forces and moments)of statically determinate beams may be calculated by statics．the determination of the reactions for statically indeterminate beams requires a consideration of the bending of the beams，and hence this subject will be postponed．the idealized support conditions shown in fig．1．2 are encountered only occasionally in practice. as an example, long—span beams in bridges sometimes are constructed with pin and roller support at the ends. however, in beams of shorter span, there is usually some restraint against horizontal movement of the supports．under most conditions this restraint has little effect on the action of the beam and can be neglected． however， the beam is very flexible, and if if the horizontal restraints at the ends are very rigid， may be necessary to consider their effects. it example* find the reactions at the supports for a simple beam loaded as shown in fig．1．3(a) neglect the weight of the beam． solution the loading of the beam is already given in diagrammatic form．the nature of the supports is examined next and the unknown components of these reactions are boldly indicated on the diagram．the beam，with the unknown reaction components and all the applied forces，is redrawn in fig． 3(b)to deliberately emphasize this important step in constructing a free—body 1． diagram．at a，two unknown reaction components may exist，since the end is pinned．the reaction at b can only act in a vertical direction since the end is on a roller．the points of application of all forces are carefully noted．after a free body diagram of the beam is made，tile equations of statics are applied to obtain the solution． fig．1，3 a simple beam ∑fx=0， rax=0 ∑ma=o+，)+160(15)-rb(20)=0，rb=+2700 lb↑ ∑mb=o+，ray(20)+)-160(5)=0，ray=-10lb↓ check：∑fy=0↑+，-10-100-160+270=0 note that ∑fx=0 uses up one of the three independent equations of statics．thus only two additional reaction components may be determined from statics．if more unknown reaction components or moments exist at the support，the problem becomes statically indeterminate． note that the concentrated moment applied at c enters only into the expressions for the summation of moments．the positive sign of rb indicates that the direction of rb has been correctly assumed in fig．1．3(b)．the inverse is the case of ray, and the vertical reaction at a is downward． note that a check on the arithmetical work is available if the calculations are made as shown． (selected from stephen p. timoshenko and james m. gere,mechanics of materials* van nostrand reinhold company ltd. ,1978. * selected from egor p. popov, introduction to mechanics of solids*prentice-hall inc. ,1968. )材料 1横梁的静力分析一条受到由截面向轴心的力的棒子称为横梁。 在这文章中我们将讨论几种最简单的受力 图，如图 1.2 所示。每种情况下，我们假设横梁水平对称即与其平整的外形对称。因此,横梁 截面梁垂直对称轴。同时假定施加载荷于对称平面，同时弯曲也发生在那个平面。之后，我们考虑一种更普遍的弯曲，可能有非对称截面。 （a）简单的支撑横梁 （b）悬臂梁 （c）外伸梁 图 1.2 横梁的种类 在图 1.2（a）中杆一端由固定支座支撑另一端由一个滚动支座支撑，被称为简支梁或者 单体梁。 简支的一个基本的特征是在弯曲的时候杆的两端可以自由旋转， 但是不能横向移动。 杆一端能够在轴向自由移动（意思是水平方向） 。简支梁的支撑可能垂直向上或者向下。如 图 1.2（b）杆一端固定，另一端自由的称作悬臂梁。在固定的一端既不能旋转也不能移动， 但另一端两者均可。 图中第三个例子展示的是带有悬垂部分的外伸梁。 梁在 ab 处受到支撑， 有一个自由端 c。杆上的载荷可能集中在一点，例如图 1.2（a）的 p1 和 1.2（c）的 p2，或 者分散分布，例如,图 1.2（b）的 q. 具有分布式荷载强度,主要表现为一个单位距离力沿杆 的主轴。对于均匀呢分布的载荷，例如 1.2（b）中，其强度是不变的；非均匀分布的载荷其 强度是随沿其杆轴线方向距离的一个函数。 图 1.2 的横梁是静定的，因为所以的反应都可以从静力平衡方程中得出。例如，图 1.2 （a）中简支梁受的力 p1，两个方向的受力都是垂直的，并且其大小也可以通过总结受力完 成瞬间得出，因此，我们发现 ra=p1（l-a）/l rb=ap1/l 图 1.2（c）的外 伸梁的受力也可以用同种方法获得。 对于图 1.2（b）的悬臂梁，所受应用载荷 q 与一个垂直力和固定端的力偶平行，如图 所示。从以上垂直方向的合力，我们总结得出 ra = qb 。从 a 点的合力，我们发现 ma=qb(a+b/2)，作用力力偶是逆时针的如图所示。 前面的例子说明了静定力能够由静力学方程求出来。 静力不确定横梁的力需要考虑梁的 变形，这个在以后会做讨论。 图 1.2 中的理想化条件在实际中只是偶尔遇到。例如，大桥中的长距离横梁两端有时候 需要铰链和滚动结构。当然，在短距离横梁中经常会有一些对支撑水平移动的制约力。在多 数情况下，这种约束力对横梁的影响很小，可以忽略不计。但是，如果横梁的韧性横好，而 两端的水平制约力非常的明显，那么就必须要考虑这个的影响了。 例子 找出图 1.3(a)简单横梁受力下的反作用力。忽略横梁的重力。 解决方案 横梁的受力已经在图中给出。 支撑力的性质接下来就会测出， 组成部分未知的反作用力 也将会在图中大胆的指出。 含有未知力部分和所有应用力的横梁被重新展示在图 1. 3(b)中用 来着重强调构建一个自由图示的步骤。在 a 点，可能存在两个未知的反作用力，因为两端 被约束。b 点的约束反力只能在垂直方向起作用，因为 b 端在滑动支座上。力的作用点被仔 细的标记出来。当横梁的自由结构图示制作出来，就可以利用静力学方程得出求解。 ∑fx=0 ,rax=0 ∑ma= 0 +,)+160(15)-rb(20)=0,rb=+2700lb↑ ∑mb=0+,ray(20)+)-160(5)=0,ray=-10lb↓ 检查 :∑fy=0↑+,-10-100-160+270=0 注意，∑fx= 0 既是三个静力学独立方程的一个，所以只有两个额外的反作用力可以从 静力学中求得。 如果出现更多的反作用力部分或者更多的支持瞬间存在， 问题就会变成静力 学不确定。 注意， 集中作用与 c 点的力只有一个合力瞬间的表达式。 的正方向表明 rb 已经在图 rb 1.3（b）中准确的假设了。ray 的情况相反，a 点的垂直作用力向下。注意假如计算过程如 上所示，那么计算的验证是有效的。 （选自 stephen p. timoshenko and james m. gere， 材料机械， 诺斯特兰德莱有限公司， 1978。选自 egor p. popov，固体机械介绍，普伦蒂斯-霍尔公司，1968）reading material 2 shear force and bending moment in beamslet us now consider，as an example，a cantilever beam acted upon by an inclined load p at its free end [fig.1.5(a)].if we cut through the beam at a cross section mn and isolate the left-hand part of the beams as free body [fig.1.5(b)],we see that the action of the removed part of the beam (that is, the right-hang part) upon the left-hand part must be such as to hold the left-hand part in the equilibrium. the distribution of stresses over the cross section mn is not known at this stage in our study，but we do know that the resultant of these stresses must be such as to equilibrate the load p. it is convenient to resolve the resultant into an axial force n acting normal to the cross section and passing through the centroid of the cross section，a shear force v acting parallel to the cross section，and a bending moment m action in the plane of the beam. the axial force, shear force, and bending moment acting at a cross section of a beam are known as stress resultants. for a statically determinate beam, the stress resultants can be determined from equations of equilibrium for the cantilever beam pictured in fig.1.5, we may write three equations of statics for the free-body diagram shown in the second part of the figure. from summations of forces in the horizontal and vertical directions we find, respectively, n=p cosβ v=p sinβ and, from a summation of moments about an axis through the centroid of cross section mn, we obtain m=pxsinβ where x is the distance from the free end to section mn. thus through the use of a free-body diagram and equations of static equilibrium, we are able to calculate the stress resultants without difficulty. the stresses in the beam due to the axial force n acting alone have been discussed in the text of unit.2; now we will see how to obtain the stresses associated with bending moment m and the shear force v. the stress resultants n, v and m will be assumed to be positive when they act in the directions shown in fig.1.5 (b). this sign convention is only useful, however, when we are discussing the equilibrium of the left-hand part of the beam. if the right-hand part of the beam is considered, we will find that the stress resultants have the same magnitudes but opposite directions [see fig.1.5(c)]. therefore, we must recognize that the algebraic sign of a stress resultant does not depend upon its direction in space, such as to the left or to the right, but rather it depends upon its direction with respect to the material against which it acts. toillustrate this fact, the sign conventions for n,v and m are repeated in fig.1.6, where the stress resultants are shown acting on an element of the beam. we see that a positive axial force is directed away from the surface upon which it acts(tension), a positive shear force acts clockwise about the surface upon which it acts, and appositive bending moment is one that compresses the upper part of the beam. example a simple beam ab carries two loads, a concentrated force p and a couple mo, acting as shown in fig.1.7 (a). find the shear force and bending moment in the beam at cross sections located as follows: (a) a small distance to the left of the middle of the beam and (b) a small distance to the right of the middle of the beam. solution the thirst step in the analysis of this beam is to find the reactions ra and rb. taking moments about ends a and b gives two equations of equilibrium, from which we find ra=3p/4-mo/l bb=p/4+mo/l next, the beam is cut at a cross section just to the left of the middle, and a free-body diagram is drawn of either half of the beam. in this example we choose the left-hand half of the beam, and the corresponding diagram is shown in fig.1.7 (b). the force p and the reaction ra appear in this diagram, as also do the unknown shear force v and bending moment m, both of which are shown in their positive directions, the couple mo does not appear in the figure because the beam is cut to the left of the point where mo is applied. a summation of forces in the vertical direction gives v=ra-p=-p/4-mo/l which shows that the she hence, it acts in the opposite direction to that assumed in fig.1.7(b). taking moments about an axis through the cross section where the beam is cut [fig.1.7 (b)] gives m=ral/2-pl/4=pl/8-mo/2 depending upon the relative magnitudes of the terms in this equation, we see that the bending moment m may be either positive or negative. to obtain the stress resultants at a cross section just to the right of the middle, we cut the beam at that section and again draw an appropriate free-body diagram [fig.1.7(c)]. the only difference between this diagram and the former one is that the couple mo now acts on the part of the beam to the left of the cut section. again summing forces in the vertical direction, and also taking moments about an axis through the cut section, we obtain v=-p/4-mo/l m=pl/8+mo/2 we see from these results that the shear force does not change when the cut section is shifted from left to right of the couple mo, but the bending moment increases algebraically by an amount equal to mo. (selected from:stephen p. timosheko and james m. gere, mechanics of materials,van nostrand reinhold company ltd. , 1978. )材料 2梁所受剪力和弯矩 梁所受剪力和弯矩现在我们考虑，例如图 1.5（a）,悬臂梁的自由端受到斜向载荷 p。如果我们从截面 mn 将梁截开， 将左边的部分孤立成自由体如 1.5 （b） 所示， 我们可以发现被移去一部分的梁 （即 右边部分）在维持左边部分的平衡上是必须的。在截面 mn 上的力的分布在我们学习的现阶 段不可知， 但是我们知道应力的合力必须是与载荷 p 平衡。 当带入一个沿轴向过截面轴心的 力 n 和一个与截面平行的剪切力 v，一个作用与横梁的弯矩。 截面上的轴向力，剪切力和弯矩都是应力的结果。对于静力平衡的横梁，我们能够通过 静力平衡方程知道合应力。对于图 1.5.9 的悬臂梁，从第二部分图示的自由图解中我们可以 列出 3 个静力平衡方程。水平方向和垂直方向的合力，各自为： n=p cosβ v=p sinβ 从通过截面 mn 轴心轴向的合力，我们得：m=pxsinβ。x 是自由端与截面 mn 的距离。因此， 通过利用自由体图示和静力平衡方程，我们很容易的可以求出合应力。横梁的轴向应力 n 单独作用的情况我们已经在第 2 单元讨论过了。现在我们将要看到如何联合弯矩 m 和剪切 力 v 共同求得。 如图 1.5(b)所示，应力 n，v 和 m 需假定正方向。这样一般性假定只在讨论横梁左边部 分时才有用。 如果把右边部分加以考虑的话， 我们将得到同样大小但是方向相反的力。 因此， 我们必须确定数学标定不是依靠其空间的方向， 例如向左或者向右， 而是依靠力作用其材料 上产生的反作用力的方向。为了说明这种情况，对于 n，v 和 m 的符号规定在图 1.6 中再次 展示。 我们可以看到，正向轴力方向远离截面表面表现为拉伸，正向剪切力沿顺时针方向，正 弯矩压紧横梁的上部分。 简单横梁 ab 受到两个载荷的作用，集中力 p 和力偶 mo，作用如图 1.7（a）所示。我 们发现横梁的剪切力和弯矩位于断面如下： （a）距离横梁的左端 1/4 个横梁的距离。 （b）距 离中右端一端距离。 分析横梁的第一步是找出反作用力 ra 和 rb。对于 a，b 两端我们列出两个平衡方程， 从中我们发现：ra=3p/4-mo/l rb=p/4+mo/l 然后，从中间将横梁截断成两个如图所示 的自由体结构，我们以左半部分为例，相应受力图解如 1.7（b）所示。力 p 和反作用力 ra， 还有我们先假定正向的剪切力 v 和弯矩 m 都在图中画出。力偶 mo 没有出现，因为它应用 到了被分开的横梁的左边部分去了。 垂直方向上的合力： v=ra-p=-p/4-mo/l 在垂直方 向上的合力是负数，因此与它图 1.7（b）所假设的方向相反。他图 1.7（b）给出的沿截面 轴向的合力： m=lra/2-pl/4=pl/8-mo/2 根据方程中的大小关系，我们可以发现弯矩可能 既不正向也不反向。 为了获得作用在右边中间截面上合力，我们将横梁沿截面砍断，再画一个自由结构示意 图如 1.7（c） 。这份图表和和之前的的差别在，现在力偶 mo 作用于横梁的左截面。我们再 2 假设垂直方向的力和通过截面轴心的力， 我们获得： v=-p/4-mo/l m = pl/8+mo/2 从 结果分析，有以下结论：力矩 mo 在梁上左右移动时，剪切力并没有改变，但弯矩和 mo 成线 性比例关系。 (选自 stephen p. timosheko and james m. gere，材料力学，范·诺斯特兰德·莱茵霍尔德股 份有限公司，1978)reading material 3 theories of strength1、 principal stresses 、 the state of stress at a point in a structural member under a complex system of loading is described by the magnitude and direction of the principal stresses. the principal stresses are the maximum values of the normal s which act on planes on which the shear stress is zero. in a two-dimensional stress system,fig.1.11,the principal stresses at any point are related to the normal stresses the x and y directions σx and σy and the shear stress, xy at the point by the following equation: principal stresses,σ1
1 1 2 (σ y
σ x ) 2 + 4τ xy
= (σ y + σ x ) ± σ2 2 2(1.7)the maximum shear stress at the point is equal to half the algebraic difference between the principal stresses: maximum shear stress , τmax=1 （σ1—σ2） 2(1.8)compressive stresses are conventional tensile as positive. 2、 classification of pressure vessels 、 for the purposes of design and analysis, pressure vessels, pressure vessels are sub-divided into two classes depending on the ratio of the wall thickness to vessel diameter: thin-walled vessels, with a thickness ratio of less than 1/10,and thick-walled above this ratio. the principal stresses acting at a point in the wall of a vessel, due to a pressure load, are shown in fig.1.12.if the wall is thin, the radial stressσ3 will be small and can be neglected in comparison with the other stresses, and the longitudinal and circumferential stressesσ1 and σ 2 can be taken as constant over the wall thickness. in a thick wall, the magnitude of the radial stress will be significant, and the circumferential stress will vary across the wall. the majority of the vessels used in the chemical and allied industries are classified as thin-walled vessels. thick-walled vessels are used for high pressures. fig.1.11 two-dimensional stress system fig.1.12 principal stress in pressure-vessel wall3、 allowable stress 、 in the first two sections of this unit equation were developed for finding the normal stress and average shear in a structural member. these equations can also be used to select the size of a member if the member’s strength is known. the strength of a material can be defined in several ways, depending on the material and the environment in which it is to be used. one definition is the ultimate strength or the stress. ultimate strength is the stress at which a material will rupture when subjected to a purely axial load. this property is determined from a tensile test of the material. this is a laboratory test of an accurately prepared specimen which usually is conducted on a universal testing machine. the load is applied slowly and is continuously monitored. the ultimate stress or strength is the maximum load divided by the original cross-sectional area. the ultimate strength for most engineering materials has been accurately determined and is readily available.if a member is loaded beyond its ultimate strength it will fail—rupture. in most engineering structures it is desirable that the structure not fail. thus design is based on some lower value called allowable stress or design stress. if, for example, a certain steel is known to have an ultimate strength of 110000psi,a lower allowable stress would be used for design, say 55000psi.this allowable stress would allow only half the load the ultimate strength would allow. the ratio of the ultimate strength to the allowable stress is known as the factor of safety: (1.9) factor of safety=ultimate strength/allowable stress or n=su/sa we use s for strength or allowable stress andσfor the actual stress in a material. in a design: σ≤sa this so-called factor of safety covers a multitude of sins. it includes such factors as the uncertainty of the load, he uncertainty of the material properties, and the inaccuracy of the stress analysis. it could more accurately be called a o factor f ignorance! in general, the more accurate, extensive, and expensive the analysis, the lower the factor of safety necessary. 4、 theories of failure 、 the failure of a simple structural element under unidirectional stress (tensile or compressive ) is easy to relate to the tensile strength of the material , as determined in a standard tensile test ,but foe components subjected to combined stresses (normal and shear stress) the position is not so simple ,and several theories of failure have been proposed .the three theories most commonly used are described below: maximum principal stress theory :which postulates that a member will fail when one of the principal stresses reaches the failure value in simple tension, σ’e. the failure point in a simply tension is taken as the yield-point stress, or the tensile strength of the material divided by a suitable factor of safety. maximum shear stress theory: which postulates that failure will occur in a complex stress system when the maximum shear stress reaches the value of the shear stress at failure in simple tension. for a system of combined stresses there are three shear stresses maxima: τ1=±σ1-σ2/2, τ2=±σ2-σ3/2, τ3=±σ3-σ1/2 (1.10) in the tensile test, τe=σ’e/2 (1.11) the maximum shear stress will depend on the sign of the principal stresses as well as their magnitude, and in a two-dimensional stress system, such as that in the wall of a thin-walled pressure vessel, the maximum value of the shear stress may be given by putting σ3 =0 in equations 1.10. the maximum shear stress theory is often called tresca’s , or gueat’s, theory. maximum strain energy theory: which postulates that failure will occur in a complex stress system when the total strain energy per unit volume reaches the value at which failure occurs in simple tension. the maximum shear-stress theory has been found to be suitable for predicting the failure of ductile materials under complex loading and is the criterion normally used in the pressure-vessel design. (selected from r. k. sinnott chemical engineering* vol.6, 2nd edition, pergamon press, 1996. * selected from raymond f. neathery ， statics and applied strength of materials, john wildey &. sons inc. , 1985. )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，σ1
1 1 2 (σ y
σ x ) 2 + 4τ xy
= (σ y + σ x ) ± σ2 2 2(1.7)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=1 （σ1—σ2）……………………………………(1. 8) 2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 &thick shell” 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)， 19) and (1. 24) (1. 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, 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 = 12 + (025) = 2.47 oz. in. to be removed2θ1 = tan 11 = 24 o 0&#39; clockwise from plane of first unbalance 2.2523 c 2 =
+ 12 = 1.25 oz. in. to be removed 4θ1 = tan 11 = 53o clockwise from plane of first unbalance 0.75( william t. thomson, vibration theory and applications, prentice-hall inc. ,1965. )材料 5旋转体的静态和动态平衡单个圆盘在它的两个平行刀刃的车轴上旋转可以检测到圆盘失去平衡， 如图 1.22 所示。 圆盘会旋转最终会偏停在重的一面。 这类的不平衡称为静态平衡， 因为它能够通过静力学的 方法检测。 通常，一般的转子都会安装在前轴上，例如电动机电枢或者汽车发动机机轴。一个类似 上述的测试表明，这种零件处在静态平衡，但是当系统旋转起来后就表现出不平衡了。 举例，考虑带两个圆盘的轴，如图 1.23。如果两个不等重的圆盘处于平衡，质心位于轴 的两侧且两质心成 180°，系统会沿轴处于静态平衡。然而，当系统旋转，不平衡的圆盘将 会受到旋转离心力会摇晃轴承上的轴。因为这种不平衡的类型我们称作动态失衡。 图 1.24 展示了一个一般情况下地动态和静态失衡。通过在旋转的任意两平行面增加正 确的重量不平衡力 p 和 q 能够被消除。 首先考虑失衡力 p,它能够被两个平行的力 q *c/l 和 q *d/l 代替。 类似的， 可以被 q *c/l q 和 q *d/l 代替。这两个不同平面的力可以合为单个的合力就可以被单个的校正重量平衡如 所展示。两个校正重量 c1 和 c2 用以平衡 p 和 q，系统式静态和动态平衡。着重强调动态平 衡体系也是静态平衡的。相反，静态平衡体系就是动态平衡体系就不一定是正确的。 在中心面 举例 一个 4 英寸的转子在距离最左边一英寸的一个平面上有 3 盎司的重量， 上有 2 盎司的重量，从左边看去第一个不平衡点顺时针方向两者成 90°。校正重量分别在 两个端面上，角度处于两不平衡点的角度里 解答 3 盎司不平衡重量相当于左边平面 2.24 盎司的重量和右边的 0.75 盎司的重量， 如图 1.25 所示。中部 2 盎司的重量显然等于两端 1 盎司的重量之和。 分别合成两端的不平衡量，得校正量： 左端：c1=1 + ( 2 . 25 )212= 2 . 47 盎司ο &#39;θ = tan11 = 2 . 2524 0第一个不平衡点的顺时针方向右端：cθ2=3 ( ) +1 4tan122= 1 . 25 盎司2=1 = 53 ° 第一个不平衡点的逆时针方向 0 . 75(摘自：william t. thomoson, 《振动理论及应用》普伦蒂斯霍尔公司 ,1965.）reading material 6stainless steelstainless steels do not rust in the atmosphere as most other steels do. the term “stainless” implies a resistance to staining, rusting, and pitting in the air, moist and polluted as it is, and generally defines a chromium content in excess of 11% but less than 30%. and the fact that the stuff is “steel” means that the base is iron. stainless steels have room-temperature yield strengths that range from 205mpa (30kis) to more than 1725mpa (250 ksi). operating temperatures around 750c (1400f) are reached. at the other extreme of temperature some stainless steels maintain their toughness down to temperatures approaching absolute zero. with specific restrictions in certain types, the stainless steels can be shaped and fabricated in conventional ways. they can be produced and used in the as- shapes can be produced by power-m cast ingots can be rolled or forged (and this accounts for the greatest tonnage by far). the roll product can be drawn, bent, extruded, or spun. stainless steel can be further shaped by machining, and it can be joined by soldering, brazing, and welding. it can be used as an integral cladding on plain carbon or low-alloy steels. the generic term “stainless steel” covers scores of standard composition as well as variations bearing company trade names and special alloys made for particular applications. stainless steels vary in their composition from a fairly simple alloy of, essentially, iron with 11% chromium, to complex alloys that include 30% chromium, substantial quantities of nickel, and half a dozen other effective elements. at the high-chromium, high-nickel end of the range they merge into other groups of heat-resisting alloys, and one has to be arbitrary about a cutoff point. if the alloy content is so high that the iron content is about half, however, the alloy falls outside the stainless family. even with these imposed restrictions on composition, the range is great, and naturally, the properties that affect fabrication and use vary enormously. it is obviously not enough to specify simply a “stainless steel.” classification the various specifying bodies categorize stainless steels according to chemical composition and other properties. however, all the stainless steels, whatever specifications they conform to, can be conveniently classified into six major classes that represent three distinct types of alloy constitution, or structure. these classes are ferritic, martensitic, austenitic, manganese-substituted austenitic, duplex austenitic ferritic, and precipitation-hardening. each class is briefly described below. (1)ferrous stainless steels: this class is so named because the crystal structure of the steel is the same as that of iron at room temperature. the alloys in the class are magnetic at room temperature and up to their curie temperature (about 750 1400f). common alloys in the ferrous class contain between 11% and 29% chromium, no nickel, and very little carbon in the wrought condition. (2) martensitic stainless steels：stainless steels of this ： class，which necessarily contain more than 11% chromium，have such a great hardenability that substantial thickness will harden during air cooling，and nothing more drastic than oil quenching is ever required. the hardness of the as-quenched martensitic stainless steel depends on its carbon content. however the development of mechanical properties through quenching and tempering is inevitably associated with increased susceptibility to corrosion. (3)austenitic stainless steels： the traditional and familiar austenitic stainless steels have a composition that ：contains sufficient chromium to offer corrosion resistance，together with nickel to ensure austenite at room temperature and bellow. the basic austenitic composition is the familiar 18% chromium，8% nickel alloy. both chromium and nickel contents can be increased to improve corrosion resistance, and additional elements (most commonly molybdenum) can be added to further enhance corrosion resistance. (4)manganese-substituted austenitic stainless steels：the ： austenitic structure can be encouraged by elements other than nickel，and the substitution of manganese and nitrogen produces a class that we believe is sufficiently different in its properties to be separated from the chromium-nickel austenitic class just described. the most important difference lies in the higher strength of the manganese-substituted alloys. (5)duplex austenitic-ferrous stainless steels: the structure of these steels is a hybrid of the structures of f and the mechanical properties likewise combine qualities of each component steel type. the duplex steels combine desirable corrosion and mechanical properties, and their use is as a result increasing in both wrought and cast form. (6)precipitation-hardening stainless steels：stainless steels can be designed so that their composition is amenable to ： precipitation hardening. this class cuts across two of the other classes， give us martensitic and to austenitic precipitation-hardening stainless steels. in this class we find stainless steels with the greatest useful strength as well as the highest useful operating temperature. properties in selection of stainless steels，three kinds of properties have to be considered： (1)physical properties: density, thermal conductivity, electrical resistivity, (2) mechanical properties: strength, ductility, hardness, creep, resistance, fatigue, and (3) corrosion-resistant properties. note that properties of stainless steels are substantially influenced by chemical composition and microstructure. hence specifications include chemical composition, or, more, correctly, an analysis of the most important elements (traces of unreported elements also may be present) as well as a heat treatment that provides the optimum structure. applications since stainless steels were first used in cutlery industry, the number of applications has increased dramatically. the relative importance of the major fields of applications for flat and long stainless steel products is shown in table 1. chemical and power engineering is the largest market for both long and flat products. it began in about 1920 with the nitric acid industry. today, it includes an extremely diversified range of service conditions, including nuclear reactor vessels, heat exchangers, oil industry tubulars, components for the chemical processing and pulp and paper industries, furnace parts, and boilers used in fossil fuel electric power plants.application industrial equipment chemical and power engineering food and beverage industry transportationpercentageapplication architecture consumer goods domestic application，house utensils small electric and electronic applicationpercenta ge 5 28 634 18 9(selected from- stainless steels, materials park, asm international, 1994. )材料 6不锈钢不锈钢如其他一些钢材一样在大气中不会生锈。 “不锈”意味着能够抵抗着色，生锈， 空气中腐蚀，潮湿和污染，一般规定铬的含量在 11%到 30%。实际上钢铁材料意思是组成基 础成分是铁。 不锈钢有室温的屈服应力从 205 mpa (30 kis) 到超过 1725 mpa (250 ksi)。调整温度到 750 摄氏度（1400 华氏度）左右就可以达到。在温度接近绝对 0 度时一些不锈钢获得良好 的韧性。 一些类型的特殊限制条件， 不锈钢能够通过惯常的方式被塑造和焊接。 他们能够被生产 和使用在铸造条件下，形状可以通过冶金技术获得，铸锭可以滚压或者锻造获得（归功于大 型的吨压机） 。轧制的产品能够被拉拔，弯曲，挤压或者旋转。不锈钢能够机械造型，能够 通过焊接，铜焊等方法连接。它能够被用于整体的电镀于碳素和低合金钢。 一般的术语 “不锈钢” 有着标准的组成成分就好像变轴承公司交易名称和应用于特殊条 件的特殊合金。不锈钢的构成多种多样，一种相当简单的合金，实质上，含铬 11%，更复杂 一点的合金那种含 30%的铬，含大量的镍，和 6 种其他的有效元素。高铬，高镍他们合并为 一种高抗性合金和一种截止点任意的合金。如果合金的含量太高，铁含量大概一半，这个合 金就不是不锈家族的了。尽管在构成物上强加限制条件，不锈钢的分布任然是很大范围的， 实际上，合金的使用非常的多。明显简单说不锈钢是不足够的。 分类 各种说明文根据化学成分和性能来划分不锈钢。然而，所有的不锈钢，不管是说 明样的规格， 它都可以简单的划分为 6 种等级代表了三种特有的合金构造或者结构。 这些等 级分别是铁素体， 马氏体， 奥氏体， 锰代铬奥氏体， 奥氏体铁素体双相和沉淀硬化型不锈钢。 每种的简单描述如下： （1）亚铁不锈钢：这种命名是因为钢的晶体结构类似于铁在室温下的 铁。这个等级的合金在室温下有磁性且能够接近居里温度(大概 750 1400 f)普通的合金在 亚铁等级包含 11%到 29%的铬，无镍，锻造条件下含有非常少的碳。 （2）马氏体不锈钢， ： 这个等级的不锈钢必须包含超过 11%的铬，有很强的硬度，空冷得时候非常的坚硬。淬火马 氏体的硬度取决于其含碳量。 然而， 通过淬火和回火的机械性能的发展必然与腐蚀系数增加 相关联。 （3）奥氏体不锈钢：传统和常见的奥氏体不锈钢的结构成分包含了足够提供腐蚀抗 力的铬和保证在室温下奥氏体在较低含量的镍。基础的奥氏体组成成分接近 18%的铬，8% 的镍合金。 增加铬和镍的含量用以提高抗腐蚀能力， 增加其他元素也能大大增强抗腐蚀能力。 （4）锰代铬奥氏体不锈钢：奥氏体结构能够受到镍以外的元素的鼓舞，锰的替代物和氮生 产出一个等级我们相信它的性能与上述所说的铬镍奥氏体充分的不一样。 最重要的不同点是 它是高强度的锰代铬合金。 （5）奥氏体铁素体双相不锈钢：这种钢的结构师铁素体与奥氏体 的混合物， 这个的机械性能也一样是这两种钢型的结合。 这种双相钢包含了令人满意的腐蚀 和机械性能，它们在锻造和铸造上的使用也一直在增加。 （6）沉淀硬化型不锈钢：不锈钢可 以被设计成它们的成分服从沉淀硬化。 这等级近似与其它的两个等级， 给我们马氏体和奥氏 体沉淀硬化型不锈钢。在这等级我们发现最大使用强度的不锈钢也有最高使用温度。 我们必须要考虑的三个性能： 物理性能： （1） 密度 导热性 电 性能 在不锈钢的选择上， 阻等等（2）机械性能：强度 韧性 硬度 延伸性 抗性 疲劳等等（3）抗腐蚀性能。注意不 锈钢的性能影响是化学成分和微观结构的结果。 因此说明书上要包含化学成分或者大部分重 要的元素的分析和适合结构的温度治疗方法。 （选自，不锈钢 r.a.lula 美国自然五金，1986） 不锈钢最初是应用在餐具行业，应用量戏剧般的增加。一些相关的重要的应用 应用 领域和不锈钢的生产如图 1 所示。化学制品和力量工程应用市场最大。它开始于 1920 年和 硝酸工业。今天，它包括及其多样的服役条件，包括原子反应堆容器，换热器和炼油器。应用 工业设备 化工和电力工程 饮食业 运输业百分数应用 农业 生活用品 家庭用品，家庭器皿 小型电气化应用百分数 5 28 634 18 9（选自不锈钢，材料公园，美国金属协会，1994）reading material 7 standard mechanical tests to summarize the previous discussion, it is very important to know the strength of a material, both for its eventual use and also to determine the forces required to shape it. since it is impracticable to test every article after it has been designed and made, several simple general tests are used to measure the mechanical properties of the stock material before, during and after manufacture of the final product. (1) tensile tests the simplest and most widely accepted tensile test requires a cylindrical (or flat) bar with enlarged ends. this tensile specimen is subjected to a steadily increasing tensile force along its axis, and the extension of a gauge length is accurately measured as the load-extension curve , according to the appropriate standard. the results usually required are the maximum tensile stress, the yield stress, the percentage elongation to fracture and the reduction of cross-sectional area at fracture. in addition, the young’ s modulus of elasticity, or young modulus may be measured. (2) compression tests it is important for metal forming calculations to know the yield stress at much higher strains than can be obtained in tension. axial compression of a short cylinder may be used, with suitable correction for the frictional resistance on the flat ends, but a more accurate result is obtained by the transverse plane strain compression of a well-lubricated strip. (3) hardness testing tensile and compression tests are destructive of the sample, but it is often important to check the strength properties of stock material or finished components, without destruction. there are several types of hardness test for this purpose, which make only a small indentation in the surface. the oldest and best known hardness tests in the u.k are the brinell test in which a standard ball (usually 10 mm dia. ) is pressed into a metal under a prescribed load, typically 3000 kgf (= 29. 42 kn or 6615 lbf), and the vickers test. the brinell hardness number (bhn or hb) is defined as the load in kgf divided by the actual spherical surface area of the indentation in mm2. likewise, the vickers hardness number (vhn or hv) is the load in kgf divided by the pyramidal surface area(again in mm2) of the indentation. in the u. s. a. , the rockwell test is favored. in that test the depth of the indentation is measured whilst the load is still being applied (rather than the lateral dimensions). the rockwell hardness number is designated as hr. (4) fatigue tests a very important phenomenon is called fatigue. it has been recognized for many years that static tensile or compressive testing is not adequate for predicting the strength of components subjected to vibration or repeated loading. these can fail at much lower stress levels, and there is a general relationship (due to goodman) which shows the allowable oscillating stress level for a given mean stress. fatigue testing needs considerable time, since each point on the final graph of applied stress s against the number n of cycles to failure requires a new specimen and n is usually between 106 and 108. for many non-ferrous alloys the s-n curve falls steadily, but for steels there is often a leveling-off after some 106 to 107 cycles. if the stress does not exceed this endurance limit, the specimen will last indefinitely. another very important failure phenomenon is that of high-stress low-cycle fatigue which is potentially dangerous in materials as disparate as animal bone and aerospace components. (5) impact testing another important subject is that of the behavior of relatively brittle materials such as cast iron, which may fail under even a single impact. since it may be very important to avoid this type of fracture, impact tests have been devised in which a notched specimen is hit by a heavy pendulum. the energy absorbed is measured from the height of follow-through of the pendulum. (6) high-temperature tests at high temperature the plastic deformation of materials is dominated by diffusion processes which, for metals, become evident above about 2/3 of the absolute melting temperature tm. tensile, compression or hardness tests may all be used at elevated temperatures. (7) creep tests an important feature of the hot tensile deformation of metals and alloys is that, at sufficiently high temperatures, extension will continue at a very slow rate under very low loads. this phenomenon, termed creep, is very important in gas turbines and many other high-temperature components. creep tests are conducted over long periods, typically from 1000 to 10000 hours. because of the length of time involved in creep testing, a shorter method is often used in which only approximate measurements of strain are made during the test, the main purpose of which is to determine the time to rupture at a given temperature and stress. these stress-rupture tests can be further speeded up by testing a string of specimens in series in a long furnace. the specimens are all subjected to the same end load but differing temperatures (which must be accurately measured, of course). (8) fracture toughness in recent years much attention has been given to the fracture toughness of certain brittle materials, which is related to the ease with which a crack, once started, will propagate. a simple view of this process is that the opening of a crack releases elastic deformation energy but also requires the supply of surface energy to the two newly created areas of crack surface. if, in a brittle material, the released strain energy u is sufficient for this, the crack will propagate. (9) plastic anisotropy in sheet metal forming in particular it is important to recognize that the properties of rolled sheetmay differ substantially in the rolling and transverse directions as well as in the &through-thickness& direction. this feature can be measured in terms of the now well-known so-called 7-value, which is the ratio of the transverse to the longitudinal strain in a tensile test on wide, flat strip using techniques described by hosford and caddell. volume is always approximately conserved in plastic deformation, so the thickness strain is also dependent upon the 7-value. ( j. m. alexander and j. s. gunasekera, strength of materials, vol. 2: advanced theory and applications,ellis harwood ltd, 1991. )材料 7机械测试标准总结前面的讨论， 我们知道了解材料的强度是非常重要的， 对于最终的使用和使其变形 所需要的力的确定都非常重要。 因为产品设计和制造后的测试是非常重要的， 在以前库存材 料都会经过几个简单通用的测试才会生产为最终的产品。 （1）拉力试验 最简单和广泛应用的拉力测试需要一个两端扩大的圆柱棒。 拉力试样受到沿其轴线的稳 定增加的拉力，拉伸的长度在适当的标准下在应力-应变曲线上精确的表示出来。结果通常 需求的最大拉伸应力,屈服应力、断裂伸长率及断面处的收缩率。此外，还要求杨氏弹性模 量或者说是杨氏模量。 （2）抗压试验 在金属成形计算中，获得工件在比所受拉力更高的力的情况下的屈服强度是非常重要 的。可以使用一个轴向压缩圆筒进行测试，测试前要适当的修正摩擦阻力，但是一个比较正 确的结果是横的应变加上轴线所的压缩应变得来的 （3）硬度测试 拉力和抗压测试是破坏样本的， 但是在没有破坏的情况下测试库存材料或是完成件的力 学性能是非常重要的。 在联合国最原始和最为人知的硬度测试是布氏硬度测试和维氏硬度测 试，布氏硬度测试是在试验中标准棒（通常直径是 10 毫米）在规定搞得压力下被压入金属， 典型的是 3000kgf(等于 29.42kn 或者是 6615lbf)。布氏硬度数值（bhn 或者 hb）的测试是在 2 毫米压痕时的力除以被压球表面面积。同样的，维氏硬度值得测量是在 2 毫米压痕时的力 除以被压锥面面积。 在美国，洛克威尔测试是受人追捧的。在这个试验中，压痕的深度是要测量的，同时载 荷是要应用的。hr 特指洛克威尔硬度值。 （4）疲劳测试 一个非常重要的现象就是疲劳。 静力拉压测试不能够充分的预测承受震动或是反复载荷 作用的组件的强度的事已经被认可多年了。 这种可以在很低的应力水平下失效， 对于在给定 的平均应力下允许的振动应力水平的一般关系。 疲劳试验需要尝试很多的次数， 因为每个点 n 都承受应力 s 循环破坏 n 次直到最后实效，需要很多新的试样做试验。通常 n 是在 106 和 108 之间。对于许多无铁的合金 s-n 曲线稳定地下降。如果是钢，则在约 n 为 106 到 107 个循环之后常有个平衡点。 认为应力作用不允许超过这一个疲劳限制， 否则试样可能不会正常工作。 另外一个很重要地事实是，在高载荷低循环下，如动物骨骼运动和太空用品，在这种情 况下工件有很大潜在危险。 （5）碰撞试验 另外的重要测试是对铸铁那样的脆性材料的测试， 在很小的碰撞下铸铁会被破坏， 这种 测试能避免试样被破坏，它用锥形压件压入试样，锤击的大小用锤下落的高度来测试。 （6）高温测试 在高温下，材料容易发生塑性变形，对于金属，温度高于熔化温度 tm 的 2/3 就变的容 易伸展了。拉伸、压缩或者硬度试验可能要全部在高温下重复测试。 （7）蠕变试验 金属和合金受热变形的重要特征是， 在足够高温度下， 伸长将会在非常低的负载下以一 个非常慢的速度连续发生变化。 这个现象在力学上称为蠕变， 在气轮机和其它的许多在高温 下工作的工件是非常重要的。蠕变测试需要很长时间，一般需要 1000 到 10000 小时。 由于蠕变时间过长，我们采用一个更简单的方法，用一个设定的应力来测试，主要目的 是测定试样在一定温度和应力下裂开所用的时间， 那些应力测试在一个炉子里完成的， 试样 受同样的应力。 （当然，这点必须被正确的测量） （8）断裂韧性 近年来我们更关注脆性材料，它伴随强烈的松弛，一旦开始，将会一直蔓延，我们看一 个简单的例子，裂开时试样将释放弹性能。在脆性材料中，如果释放的应变能 u 足够，裂 纹就会蔓延。 （9）塑性的向导性 在金属片中， 认为钢板性能在旋转和横向移动时和穿过厚度的方向上的金属性能时不一 样时很重要的。这一特点可以从现在人所共知的被测量的 γ－型值知道，这是一个拉伸试验 在纵向应变中的定量，平面裂纹被 hosford 和 caddell 用技术性的语言描述出来，大量的断 裂通常被认为是塑性变形，所以厚度的应变也是根据 γ－型值决定的。 （选自：j.m.alexander and j.s.gunasekera，材料强度，第二卷：先进的理论和应用，埃利斯 哈伍德有限公司，1991。 ）reading material 8 examples of manufacturing processes (continued) casting casting can be characterized as: mass conserving, fluid state of material, mechanical primary basic process—filling of the die cavity. casting is one of the oldest manufacturing methods and one of the best known processes. the material is melted and poured into a die cavity corresponding to the desired geometry. the fluid material takes the shape of the die cavity and this geometry is finally stabilized by the solidification of the material. the stages or steps in a casting process are the making of a suitable mold, the melting of the material, the filling or pouring of the material into the cavity, and the solidification. depending on the mold material, different properties and dimensional accuracies are obtained. equipment used in a casting process includes furnaces, mold-making machinery, and casting machines. turning turning can be characterized as : mass reducing, solid state of work material, mechanicalprimary basic process—fracture. the turning process, which is the best known and most widely used mass-reducing p}