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 562
... column . This yields the moments at the base of the columns as the product of the shear in the column and half - height of the column . There are no intermediate loads on the column , so the variation of the moment is linear , which ...
... column . This yields the moments at the base of the columns as the product of the shear in the column and half - height of the column . There are no intermediate loads on the column , so the variation of the moment is linear , which ...
Page 564
... column and each beam and that the shear in each story is distributed to the bays in that story in relation to the column fixities . Bay shears are distributed to columns by the proportions shown in Figs . 14.3 to 14.6 . For most frames ...
... column and each beam and that the shear in each story is distributed to the bays in that story in relation to the column fixities . Bay shears are distributed to columns by the proportions shown in Figs . 14.3 to 14.6 . For most frames ...
Page 571
... columns . A free body taken by cutting through the columns and beams at their centers and sum- ming forces vertically will yield the beam shears . The sequence of compu- tation of beam shear can proceed either vertically along a column ...
... columns . A free body taken by cutting through the columns and beams at their centers and sum- ming forces vertically will yield the beam shears . The sequence of compu- tation of beam shear can proceed either vertically along a column ...
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