CNC二維工作平臺(tái)設(shè)計(jì)【說明書+CAD】,說明書+CAD,CNC二維工作平臺(tái)設(shè)計(jì)【說明書+CAD】,cnc,二維,工作,平臺(tái),設(shè)計(jì),說明書,仿單,cad 外文翻譯 （從選出：史蒂芬.Timoshenko 和詹姆士M.蓋爾，材料力學(xué)，NostrandReinhold廂式客貨兩用車有限公司，1978） Shear Force and Bending Moment in Beams Let 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 beam as free body [Fig.1.5(b)], we see that the action of the removed part of the beam (that is , the right-hand part)upon the left-hand part must as to hold the left-hand in equilibrium. The distribution of stresses over the cross section mn is not known at this stage in our study , but wee do know that the resultant of these stresses must be such as to equilibrate the load P. It is convenient to resolve to the resultant into an axial force N acting normal to the cross section and passing through the centriod of the cross section , a shear force V acting parallel to the cross section , and a bending moment M acting 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. Thus , for the cantilever beam pictured in Fig.1.5, we may writer three equations of stactics 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=Pcosβ V=Psinβ 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 stress 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 the 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 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 is acts(tension), a positive shear force acts clockwise about the surface upon which it acts , and a positive 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 first 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 RB=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 bean, 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=R – P= -P/4-M0/L Which shown that the shear force is negative; 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 force 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 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.) 平衡梁的剪力和彎矩 讓我們來共同探討像圖1.5(a)所示懸梁自由端在傾斜拉力P的作用下的問題。如果將平衡梁在截面mn處截?cái)嗲覍⑵渥筮叢糠肿鳛楦綦x體（圖1.5（b）?？梢钥闯龈綦x體截面（右邊）的作用國(guó)必須和左邊的作用力平衡，截面mn處應(yīng)力的分布情況我們現(xiàn)階段是不知道的，但我們知道這些應(yīng)力的合力必須和拉力P平衡。按常規(guī)可將合力分解成為通過質(zhì)點(diǎn)作用于橫截面的軸向應(yīng)力N、平行于截面的剪切力V和作用在平衡梁平面中的彎矩M。 作用在截面上的軸向應(yīng)力、剪切力和彎曲應(yīng)力就是應(yīng)力的合成力。比如靜止的固定梁合成力可由平衡方程得出，如圖1.5所示懸臂梁結(jié)構(gòu)。這樣就可以得到圖形另一部分中的圖示自由部分的三個(gè)平衡方程式。由水平合力和垂直合力的方向，可得： N=Pcosβ如果將平衡梁在截面mn處截?cái)嗲覍⑵渥筮叢糠肿鳛楦綦x體（圖1.5（b）?？梢钥闯龈綦x體截面（右邊）的作用國(guó)必須和左邊的作用力平衡，截面mn處應(yīng)力的分布情況我們現(xiàn)階段是不知道的，但我們知道這些應(yīng)力的合力必須和拉力P平衡。按常規(guī)可將合力分解成為通過質(zhì)點(diǎn)作用于橫截面的軸向應(yīng)力N、平行于截面的剪切力V和作用在平衡梁平面中的彎矩M。 作用在截面上的軸向應(yīng)力、剪切力和彎曲應(yīng)力就是應(yīng)力的合成力。比如靜止的固定梁合成力可由平衡方程得出，如圖1.5所示懸臂梁結(jié)構(gòu)。這樣就可以得到圖形另一部分中的圖示自由部分的三個(gè)平衡方程式。由水平合力和垂直合力的方向，可得： N=Pcosβ V=Psinβ 如果將平衡梁在截面mn處截?cái)嗲覍⑵渥筮叢糠肿鳛楦綦x體（圖1.5（b）。可以看出隔離體截面（右邊）的作用國(guó)必須和左邊的作用力平衡，截面mn處應(yīng)力的分布情況我們現(xiàn)階段是不知道的，但我們知道這些應(yīng)力的合力必須和拉力P平衡。按常規(guī)可將合力分解成為通過質(zhì)點(diǎn)作用于橫截面的軸向應(yīng)力N、平行于截面的剪切力V和作用在平衡梁平面中的彎矩M。 作用在截面上的軸向應(yīng)力、剪切力和彎曲應(yīng)力就是應(yīng)力的合成力。比如靜止的固定梁合成力可由平衡方程得出，如圖1.5所示懸臂梁結(jié)構(gòu)。這樣就可以得到圖形另一部分中的圖示自由部分的三個(gè)平衡方程式。由水平合力和垂直合力的方向，可得： N=Pcosβ V=Psin β 由通過截面mn質(zhì)心的軸向總彎矩，可得 M=Pxsinβ 其中力是自由端到截面mn的距離。因此，通過隔離體圖解和靜態(tài)平衡方程，可簡(jiǎn)單地計(jì)算出各合成力。屬于單獨(dú)作用的軸向應(yīng)力N的應(yīng)力已經(jīng)在第二單元討論過了，在這里我們將討論怎樣解出與這些應(yīng)力有關(guān)的彎矩M和剪切力V。 假設(shè)如圖1.5(b)所示合成力N、V和彎矩M的作用方向?yàn)檎?，?dāng)我們?cè)谟懻摿鹤蟀氩糠质芰ζ胶鈺r(shí)，符號(hào)很重要的。如果考慮到右半部分時(shí)我們會(huì)發(fā)現(xiàn)合成力大小相等且方向相反，如圖1.5(c)，然而，我們必須意識(shí)到應(yīng)力的代數(shù)符號(hào)不是取決于應(yīng)力的空間方向，如左、右之類而更取決于與其作用的材料有關(guān)的方向。為了說明事實(shí)，應(yīng)力N、V和彎矩M的規(guī)定方向在圖1.6平衡梁微元中反復(fù)使用。 大家知道軸向應(yīng)力：以彎矩壓縮梁的上部為正，從它作用的面指向外為正（拉伸）剪切力是其作用面內(nèi)順時(shí)鐘作用為正。 例題 剪支梁AB受集中應(yīng)力P和彎矩M0的作用。如圖1.7（a）所示，在下面條件下在梁截面中求剪切力和彎曲應(yīng)力， (a)距中心左側(cè)微小距離 (b) 距中心點(diǎn)右側(cè)微小距離 解：首先分析平衡梁，求出支反力RA 和RB。由AB兩點(diǎn)的彎矩得兩平衡方程。 由下式求得 RA=3P/4 - M0/L RB=P/4+M0/L 梁中點(diǎn)截面左側(cè)，梁兩側(cè)自由體圖解已給出，此外我們選擇梁的左側(cè)詳細(xì)圖如圖1.7(b)所示。此圖中應(yīng)力P、支反力RA，還有未知剪力V和彎矩M，這兩個(gè)力是反方向的，彎矩M0 沒有標(biāo)出，因?yàn)槠胶饬簞偤帽粡腗0作用點(diǎn)截開。垂直方向的總力為 V=R - P= -P/4-M0/L 這個(gè)方向表明剪力V是反方向的，因此，它的作用方向如圖1.7(b)所假設(shè)。由切割處的軸向彎矩可得： M = RAL/2-PL/4=PL/8-Mo/2 由方程中兩項(xiàng)的大小關(guān)系可以看出彎矩M子可能是負(fù)的。為了得到截面右側(cè)的應(yīng)力合力，將平衡梁用如上方法切開，其隔離體如圖1.7(c)所示，此圖和前者的維一不同之處是彎矩M0作用在這物體左側(cè)截面處，再由垂直方向的合力和截面處的軸向彎矩得： V= - P/4- Mo/L M=PL/8+Mo/2 綜上所述，剪彩應(yīng)力不會(huì)隨著截面從左側(cè)到有M0 作用的右側(cè)的改變而改變，但是彎矩代數(shù)值增加到了與M0相等。 （從中選出來：史蒂芬蓋爾，P. Timosheko 和詹姆士M.材料力學(xué)，reinhold Nostrand 廂式客貨兩用車有限公司，1978 。）