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 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 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 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 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
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2诺斯特兰德2莱茵霍尔德股份有限公司，1978)
Reading Material 3
Theories of Strength
1、 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 stresses at the point; 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,
?1?1?(???2?2y??x)?12(?y??x)?4?xy (1.7)
22The maximum shear stress at the point is equal to half the algebraic difference between the principal stresses:
Maximum shear stress , τmax=
12（σ1—σ2） (1.8)
Compressive stresses are conventionally taken as negative; 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 σ2 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 wall 3、 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:
Factor of safety=ultimate strength/allowable stress or n=Su/SA (1.9) 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. )