土木工程专业英语教学.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,Statically Determinate Structures,Structures are said to be,statically determinate,when the forces and,reactions,produced by a given loading can be calculated using only the equations of equilibrium.,Statically determinate structure,Statically Determinate Structures,The,simply supported beam,shown in Figure 1.3 is statically determinate.We can solve for the three unknown reactions using the equations of equilibrium and then calculate the,internal forces,such as,bending moment,shear force,and,axial force,at any given location along the length of the beam.,Statically determinate structure,Statically Indeterminate Structures,Four unknown reactions,Three independent equilibrium equations,Statically Indeterminate Structures,Statically indeterminate beam,Satically indeterminate frame,Satically indeterminate,composite Structures,Force Method 力法,The,force method,(also called the flexibility method)is used to calculate internal forces and reactions in statically indeterminate structures due to loads and imposed deformations.,The steps in the force method,This is accomplished by,releasing external support conditions,creating internal,hinges,.,The system thus formed is called the,primary system,.Number the released constraints from 1 to,n,.,primary system,q,q,q,A,A,B,B,C,C,l,l,X,1,X,1,b)基本体系,a)一次超静定结构,3For a given released constraint,j,introduce an unknown,redundant,ri5dQndEnt,force,X,j,corresponding to,the type and direction,of the released constraint.,多余未知力,redundant force,基本体系沿多余未知力方向的位移应与原结构位移相同,F,P,F,P,F,Ax,F,Ay,M,A,X,1,基本体系,A,A,B,B,C,C,q,q,q,A,A,B,B,C,C,l,l,X,1,X,1,X,1,4Apply the given loading or imposed deformation to the primary system.Calculate,displacements,due to the given loading,at each of the released constraints in the primary system.These displacements are called,1P,2P,nP,.,.,F,P,A,A,B,B,C,1P,X,1,A,B,X,1,=1,d,11,11,=,d,11,X,1,F,P,A,B,C,EI,l,/2,l,/2,X,1,基本体系,(,1,=,B,=0),=,+,5For a given released constraint,j,apply a,unit load,X,j,=1,to the primary system.Calculate displacements due to,X,j,=1,at each of the released constraints in the primary system.These displacements are called ，,A,B,X,1,=1,d,11,F,P,A,B,C,EI,l,/2,l,/2,X,1,基本体系,(,1,=,B,=0),=,F,P,A,A,B,B,C,1P,X,1,11,=,d,11,X,1,+,unit load,单位力,.,.,6.Solve for redundant forces x,1,through x,n,by imposing the,compatibility conditions,of the original structure.These conditions transform the,primary system,back to the original structure by finding the combination of redundant forces that make displacement at each of the released constraints equal to zero.,Calculate force,S,at a given location in the structure using,Forcecan be bending moment,shear,axial force,or reaction.,The Classical Displacement Method,The force method:,unknowns are force quantities(the redundant forces),based on geometrical conditions(compatibility conditions at the location of each redundant force).,The classical displacement method,unknowns are displacement quantities,based on statical conditions(equilibrium conditions).,=,Z,1,R,21,1,2,3,4,1,3,4,P,R,2P,1,2,2,3,4,R,22,R,12,R,11,R,1P,Z,2,The steps in the,classical displacement method,1For a given structure and loading,consider the joints to,be fully fixed,against rotation.,P,L,1,2,3,4,EI=常数,Z,1,Z,2,(a),(b)基本结构,1,2,3,4,=,Z,1,Z,2,R,1,=0,=0,P,R,2,The steps in the,classical displacement method,2Calculate the moments in each member of the structure due to the given loads,R,1P,R,2P,assuming,full fixity,at the joints.These moments are called,fixed-end moments,.,3.Calculate moments,r,11,、r,12,at the ends of each member due to,unit displacements,of the joints.,4Express the total moment at each end of a given member as the sum of the,fixed-end moments,M,P,and the product of,unknown joint displacements,z,1,z,2,times the moments,M,1,、M,2,produced by,unit joint displacements Z,1,Z,2,calculated in Step 3.,5Generate an equation of moment equilibrium at each joint.,基本结构在荷载等外因和结点位移的共同作用下，每一个附加联系中的附加反力矩或反力都应等于零(静力平衡条件),。,6Solve the system of equations for the,unknown joint displacements.,7Calculate the member end moments using the expressions derived in Step 4 and the values of joint displacements calculated in Step 6.,r,11,Z,1,+r,12,Z,2,+R,1P,=0,r,21,Z,1,+r,22,Z,2,+R,2P,=0,r,11,Z,1,+r,12,Z,2,+R,1P,=0,r,21,Z,1,+r,22,Z,2,+R,2P,=0,R,1,=R,11,+R,12,+R,1P,=0,R,2,=R,21,+R,22,+R,2P,=0,8Calculate all remaining forces in the structure(shear forces and axial forces).,Force method,can solve all statically indeterminate structures.,computational complexity,prohibitive,for structures with more than three unknown forces.,Classical displacement method,allows a solution based on a member-by-member procedure,rather than one that requires consideration of the structure as a whole;,based on the pre-solution of standard cases of,intermediate,媒介物 load and displacement,reduce the number of unknowns in a given solution.,Moment Distribution Method,力矩分配法,The,moment distribution method,is used for statically indeterminate beams and frames by simple hand calculations.,This is basically an iterative,5itErEtiv,迭代的 process.,The procedure involves artificially restraining temporarily all the joints against rotation and writing down the,fixed end moments,for all the members.,restrain all the joints against rotation,Write down the,fixed end moments,The joints are then released one by one in succession,sEk5seFEn,连续.At each released joint the unbalanced moments are distributed to all the ends of the members meeting at that joint.,A certain fraction of these distributed moments are carried over to the far ends of members.The released joint is again restrained temporarily before proceeding to the next joint.,The same set of operations are carried out at each joint till all the joints are completed.This completes one cycle of operations.The process is repeated a number of times or cycles till the values obtained are within the desired accuracy.,Member distribution factor,弯矩分配系数,Given a unit moment applied to joint A,what moments are produced in each of the members?,For a unit rotation the moment in each member at joint A is just its stiffness,Joint stiffness:,Member distribution factor,Distribution factor,of member,i,at joint A:,member distribution factor=,member stiffness/joint stiffness,The center joint is fixed(the rotation is set to zero)which gives the so called,fixed-end moment,solution for beam on the right and no response in the beam on the left.This solution is valid except that it requires an external moment to be applied to the center joint.The final solution is constructed by releasing or balancing the center joint which is equivalent to applying a clockwsie moment of to this joint.,center joint is fixed,fixed-end moment solution,releasing or balancing the center joint,