梁的应力的深入讨论.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,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,resolve,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,.,The deformed axis lies in the plane the loads act,F,A,A,F,1,F,2,B,Symmetric axis,Longitudinal plane of,symmetry,The plane of bending,F,B,Sign Conventions for Bending Moments,x,z,y,M,y,M,z,On a positive face,the moments are positive when their vectors point in the 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,M,y,M,z,A,(,y,z,),C,n,n,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 passing through the,centroid,C,.,Bending Stresses and Neutral Axis,z,y,M,y,M,z,A,(,y,z,),C,n,n,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,M,y,M,z,C,n,n,The angle,b,between neutral axis and the,z,axis is,b,M,q,q,P,x,z,y,P,q,M,y,(,x,),=,P,sin,q,(,l-x,),M,z,(,x,),=,P,cos,q,(,l-x,),So usually the neutral axis is not perpendicular to the loading plane.,Special Cases(),z,y,M,y,M,z,C,n,n,b,M,q,q,P,(1)When the load lies in the,xy,plane and,z,is the neutral axis;,(2)When the load lies in the,xz,plane and,y,is the neutral axis;,(,3)When the principal moments of inertia are equal(,I,z,=,I,y,).,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 perpendicular 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,y,y,C,d,A,s,d,A,(a),+,-,(b),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,d,A,s,d,A,(a),+,-,(b),If,z,is the neutral axis,The moment resultants are,Only when,M,y,and,M,z,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,d,A,s,d,A,(a),+,-,(b),That is,an,unsymmetric,beam bends in the same manner as a symmetric beam provided that the,z,axis is a principal,centroidal,axis and the only bending moment is,M,z,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,centroidal,axis and the only bending moment is,M,y,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,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,d,A,s,d,A,(a),Procedures for Analyzing an,Unsymmtric,Bending,q,(a),z,y,M,y,C,n,n,b,M,M,y,=M,sin,q,M,z,=M,cos,q,The normal stress at any point is,The equation of the neutral axis,nn,is,The angle,b,between the neutral axis and,z,axis is,例,已知：,I,y,=28310,-,8,m,4,，,I,z,=1 93010,-,8,m,4,，,I,yz,=53210,-,8,m,4,，,s,=170,MPa,。,求,q,。,解：,将,M,z,=,M,C,沿形心主轴,y,0,，,z,0,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,y,0,，,z,0,的位置如图,d,所示。,z,y,0,y,z,0,D,(,d,),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,y,0,，,z,0,弯矩的分量分别为,D,点在,y,0,，,z,0,坐标中的坐标的绝对值分别为,z,y,0,y,z,0,D,(,d,),D,点的强度条件为,解得,z,y,0,y,z,0,D,(,d,),q,z,y,形心,M,*,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,I,z,与,I,y,，以及惯性积,I,yz,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,z,y,C,M,0,S,P,x,z,y,P,The Example of an Unbalanced I-Beam,z,y,C,M,0,S,P,P,A lateral load acting on a beam will produce bending without twisting only if it acts through the shear center.,z,y,C,M,0,S,P,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 symmetry,For an,unsymmetric,cross section,to locate the shear center needs more advanced methods.,y,z,.,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 sections,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 sections shown in fig a,b and c respectively?,z,(,b,),.,C,(,c,),C,(,a,),y,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,O,y,z,P,t,e,R,0,q,d,s,x,z,y,P,(,b,),O,y,z,P,t,e,R,0,q,d,s,(,b,),x,z,y,P,If the neutral axis is the,y,axis,6.8 Shear Stresses in Wide-Flange Beams,Vocabulary,The rear face,the front face,Shear Stresses in the Web,x,y,t,f,h,z,O,d,b,t,y,d,A,x,z,y,O,s,A,*,d,x,t,t,At the junctions of the web and flanges,At the neutral axis,z,y,O,t,max,t,w,max,t,w,min,(2)Shear stresses in the flanges,a.As there is no shear stress on the top and bottom surfaces of the flanges,the shear stresses 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,x,y,t,f,h,z,O,d,b,t,y,Obviously the shear stresses parallel to the axis,y,on the flanges are very small which usually are not taken into consideration in engineering.,x,y,t,f,h,z,O,d,b,t,y,2 The shear stresses perpendicular to the axis,y,d,s,t,1,t,1,x,y,t,f,h,z,O,d,b,t,s,2 The shear stresses perpendicular to the axis,y,x,y,t,f,h,z,O,d,b,t,s,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 derivations 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,”.,z,y,O,t,max,t,w,max,t,w,min,t,1max,Example A channel-shaped simply supported beam and the dimensions of its cross section is shown in the fig.Point,C,is the,centroid,.Known that the moment of inertia of the cross section about the neutral axis is,I,z,=1152,10,4,mm,4,.,Pls,plot the shear stress distribution in section,D,.,Solution:The reactions are,The shear force in section,D,0.6m,0.4m,A,D,F,=110 kN,B,D,F,B,F,A,220,10,10,y,z,C,103,47,140,10,The first moment of the two webs below the neutral axis about the neutral axis,The first moment of the web below point,a,about the neutral axis,The shear stress at point,a,y,220,a,10,10,z,C,103,47,140,10,The horizontal shear stress at point,d,The first moment of the web to the right of point,d,about the neutral axis,y,220,d,10,10,z,C,103,47,140,10,y,C,z,V,t,d,t,max,t,a,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，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 bending about both principal axes.,Channel Section,V,z,V,y,F,R,V,y,Angle Section,C,y,z,t,t,b,b,C,y,z,V,y,s,t,max,t,max,C,y,z,s,t,max,t,max,V,y,S,When,s,=,b,t,max,occurs:,F,So the resultant vertical force is equal to,V,y,and the shear center,S,is located at the junction of the two legs of the angle.,Sections Consisting of Two Intersecting Narrow Rectangles,V,y,S,F,S,S,S,S,Z-Section,y,z,C,This section has no axes of symmetry but is symmetric about the,centroid,C.,y,z,C,F,1,F,1,F,2,2,F,1,V,y,When a shear force,V,y,acts parallel to the,y,axis,the line of its action must be through the,centroid,.,When a shear force,V,z,acts parallel to the,z,axis,we can arrive at a similar conclusion.,So the shear center of the Z-section coincides with the,centroid,.,S,Example,O,(,a,),y,z,d,e,A,R,0,Solution:,(,b,),O,y,z,V,y,d,e,R,0,d,s,j,where,O,y,z,V,y,d,e,R,0,q,d,q,j,So,，,O,y,z,V,y,d,e,R,0,q,d,q,j,.,Location of the,Center of Flexure,of some cross sections,The intersection of the midlines of the two rectangles,The,centroid,