梁的应力的深入讨论.ppt

1、Chapter 6 Stresses in Beams(Advanced Topics),6.1 Introduction,Vocabulary Composite beams , beams with inclined loads, unsymmetric beams, shear stresses in thin-walled beams, shear centers,6.4 Doubly Symmetric Beams with Inclined Loads,Vocabulary Inclined loads, a doubly symmetric cross section, reso

2、lve, components, superpose,Stresses under Symmetric bending,In symmetric bending the deformed axis lies in the same plane that all loads act, so it belongs to planar bending.,Sign Conventions for Bending Moments,x,z,y,My,Mz,On a positive face, the moments are positive when their vectors point in the

3、 positive directions of the corresponding axes, and the right-hand rule for vectors gives the directions of rotation.,Bending Stresses and Neutral Axis,z,y,Mz,A(y,z),C,The normal stress at point A is,The equation to determine the neutral axis is,It is shown that the neutral axis is a straight line p

4、assing through the centroid C.,Bending Stresses and Neutral Axis,z,y,Mz,A(y,z),C,The angle b between neutral axis and the z axis is,And the maximum stresses occur at points located farthest from the neutral axis.,b,Relationship Between the Neutral Axis and the Inclination of the Loads,z,y,My,Mz,C,Th

5、e angle b between neutral axis and the z axis is,b,M,q,q,P,x,z,y,P,q,My(x)=Psinq (l-x) Mz(x)= Pcosq (l-x),So usually the neutral axis is not perpendicular to the loading plane.,Special Cases ( ),z,y,My,Mz,C,b,M,q,q,P,(1)When the load lies in the xy plane and z is the neutral axis; (2)When the load l

6、ies in the xz plane and y is the neutral axis; (3)When the principal moments of inertia are equal (Iz =Iy).,In case (3), all axes through the centroid are principal axes and all have the same moment of inertia. Whatever direction of the plane of loading, it is always a principal plane and perpendicu

7、lar to the neutral axis.,Examle 6-4,Vocabulary Roof purlin, roof sheathing, top chords,Example 6-4,D,E,D,E,Maximum stresses (pnts D and E),Neutral axis:,(c),Points where maximum stresses occur,To be found: Maximum tensile stresses.,Known:,6.5 Bending of Unsymetric Beam,Vocabulary abondon,（b）,（a）,z,z

8、,y,y,C,dA,s dA,(a),If z is the neutral axis,Since it is pure bending,z axis be a centroidal axis.,Similarly, if y is the neutral axis,y axis be a centroidal axis.,So the origin of the y and z axes should be placed at the centroid.,z,z,y,y,C,dA,s dA,(a),If z is the neutral axis,The moment resultants

9、are,Only when My and Mz are in the ratio in the equations, an arbitrary directed z axis will be a neutral axis; If the z axis is selected as a principal axis, bending takes place in the xy plane and is similar to a symmetric beam.,z,z,y,y,C,dA,s dA,(a),That is, an unsymmetric beam bends in the same

10、manner as a symmetric beam provided that the z axis is a principal centroidal axis and the only bending moment is Mz about the same axis.,Similarly, if y is the neutral axis, we can arrive that:,An unsymmetric beam bends in the same manner as a symmetric beam provided that the y axis is a principal

11、centroidal axis and the only bending moment is My about the same axis.,Note: if either axis is a principal axis, the other is automatically a principal axis.,When an unsymmetric beam is in pure bending, the plane in which the bending moment acts is perpendicular to the neutral axis only if the z and

12、 y axes are principal centroidal axes of the cross section and the bending moment acts in one of the two principal planes.,z,z,y,y,C,dA,s dA,(a),Procedures for Analyzing an Unsymmtric Bending,q,(a),z,y,My,C,b,M,My =M sinq Mz = M cosq,The normal stress at any point is,The equation of the neutral axis

13、 nn is,The angle b between the neutral axis and z axis is,例 已知：Iy=28310-8 m4，Iz=1 93010-8 m4，Iyz=53210-8 m4，s=170 MPa。求q。,解：将Mz = MC沿形心主轴y0，z0方向分解(图d)，分别计算两个平面弯曲时的正应力，然后进行叠加。由材料力学()附录()的(13)式确定形心主轴的位置，即,形心主轴y0，z0的位置如图d所示。,由材料力学()附录()的(14)式，得形心主惯性矩分别为,沿形心主轴y0，z0弯矩的分量分别为,D点在y0，z0坐标中的坐标的绝对值分别为,D点的强度条件为

14、,解得,*思考： 图示一等截面直梁的横截面，它是Z字形的，该梁受纯弯，材料服从虎克定律，且截面的惯性矩Iz与Iy，以及惯性积Iyz均为已知。假定中性轴垂直于截面的腹板，即与y轴相重合，试确定弯矩向量与y轴之夹角。,答案：,6.6 The Shear-Center Concept,Vocabulary shear centers, the center of flexure Singly symmetric cross section, an unbalanced I-beam,The Example of an Unbalanced I-Beam,A lateral load acting o

15、n a beam will produce bending without twisting only if it acts through the shear center.,The location of the shear center,For doubly symmetric cross section, the shear center coincides with the centroid For singly symmetric cross section, both the shear center and the centroid lie on the axis of sym

16、metry For an unsymmetric cross section, to locate the shear center needs more advanced methods.,. Location of the Center of Flexure of some cross sections,The intersection of the midlines of the two rectangles,The centroid,The significance of shear center,For the beams of thin-walled open cross sect

17、ions, which are very weak in twisting, it is especially important to locate their shear centers.,Question: A cantilever beam supports a concentrated force F at the free end. If the force F acts through the centroid C of the cross section, what kind of deformations will take place for the three cross

18、 sections shown in fig a, b and c respectively?,6.7 Shear Stresses in Beams of Thin-Walled Open Cross Sections,Vocabulary structure section, profile section, centerline, tacitly,For symmetric bending,Shear Stresses in Beams of Thin-Walled Open Cross Sections,q,ds,q,ds,If the neutral axis is the y ax

19、is,6.8 Shear Stresses in Wide-Flange Beams,Vocabulary The rear face, the front face,Shear Stresses in the Web,At the junctions of the web and flanges,At the neutral axis,z,y,O,(2) Shear stresses in the flanges,a. As there is no shear stress on the top and bottom surfaces of the flanges, the shear st

20、resses parallel to the axis y equal zero at the top and bottom sides of the flanges; b. Calculations show that the shear force supports by the web is,1 The shear stresses parallel to the axis y,Obviously the shear stresses parallel to the axis y on the flanges are very small which usually are not ta

21、ken into consideration in engineering.,2 The shear stresses perpendicular to the axis y,2 The shear stresses perpendicular to the axis y,At s = b/2,Which shows that the shear stresses perpendicular to the axis y on the flanges vary linearly with the distance s from the axis y .,Through analogous der

22、ivations we can know that the directions of the shear stresses in the upper , lower flanges and the web of the I-shaped beam form a “continuous shear stress flow”.,Example A channel-shaped simply supported beam and the dimensions of its cross section is shown in the fig. Point C is the centroid. Kno

23、wn that the moment of inertia of the cross section about the neutral axis is Iz=1152104mm4. Pls plot the shear stress distribution in section D.,Solution: The reactions are,The shear force in section D,The first moment of the two webs below the neutral axis about the neutral axis,The first moment of

24、 the web below point a about the neutral axis,The shear stress at point a,The horizontal shear stress at point d,The first moment of the web to the right of point d about the neutral axis,y,The shear stress distribution in section D is shown above.,图示箱形截面简支梁用四块木板胶合而成，受三个集中力作用如图所示。已知横截面对中性轴的惯性矩,材料为红松

25、，其许用弯曲正应力,许用顺纹切应力,胶合缝的许用切应力,试校核梁的强度。,=,6.9 Shear Centers of Thin-Walled Open Sections,Only beams with singly symmetric or unsymmetric cross sections will be considered.,Procedures to locating the shear centers Calculating the shear stresses when bending occurs about one of the principal axes; Determining the resultant of those stresses. As the shear center is located on the line of action of the resultant, we can determine the position of the shear center by considering bend