Analysis and Behavior of StructuresOffering students a presentation of classical structural analysis, this text emphasizes the limitations required in creating mathematical models for analysis, including these used in computer programs. Students are encouraged to use hand methods of analysis to develop a feel for the behaviour of structures. |
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Page 61
... Loading diagram for general distributed load Q = area under loading diagram ( b ) Sum of forces with distributed load b ΣΜΟ ( x ) a x Centroid of loading diagram - xo ) dx = q ( x ) x dx - xoq ( x ) dx = XQ - x0Q = ( x - xo ) Q Figure ...
... Loading diagram for general distributed load Q = area under loading diagram ( b ) Sum of forces with distributed load b ΣΜΟ ( x ) a x Centroid of loading diagram - xo ) dx = q ( x ) x dx - xoq ( x ) dx = XQ - x0Q = ( x - xo ) Q Figure ...
Page 136
... loading diagram . In the third step , the varia- tion of the shear and moment from A to B is second and third degree , re- spectively . The diagrams can be drawn correctly if the relationships defining the slopes between moment and ...
... loading diagram . In the third step , the varia- tion of the shear and moment from A to B is second and third degree , re- spectively . The diagrams can be drawn correctly if the relationships defining the slopes between moment and ...
Page 149
... diagrams for any applied loading , as illustrated in Example 4.7 . Example 4.7 The member shown is loaded with a distributed axial load . Once the reactive force , R , is obtained by equilibrium in step 1 , the loading diagram is easily ...
... diagrams for any applied loading , as illustrated in Example 4.7 . Example 4.7 The member shown is loaded with a distributed axial load . Once the reactive force , R , is obtained by equilibrium in step 1 , the loading diagram is easily ...
Common terms and phrases
acting action analysis applied applied loads assumed assumptions axial force axis beam behavior bending calculation caused Chapter column components Compute condition constant continued create curvature defined deflection deformations developed direction displacement distribution Draw elastic end moments energy equal equations equilibrium equilibrium equations established Example expression Figure fixed force system frame free body function geometric gives hinge horizontal indeterminate structure influence line integration internal joint length limitations linear load magnitude material mathematical matrix maximum member forces ments method Note obtained occur plane positive presented principle Problem provides reaction relation relative rotation shear shown in Fig simple slope solution solve statically determinate STEP stiffness strain stresses structure symmetric Table tion truss unit load unknown vertical virtual yields zero ΕΙ