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 27
... superposition . Since the strains Ep and ε yield the displace- ments Ap and △ , the deflection of the center of the beam , A , with both loads acting is also obtained by superposition . P W The principle of superposition of ...
... superposition . Since the strains Ep and ε yield the displace- ments Ap and △ , the deflection of the center of the beam , A , with both loads acting is also obtained by superposition . P W The principle of superposition of ...
Page 34
... superposition of deformations are met , the analysis of a stable , symmetric structure can be obtained as the superposition of the analyses of the symmetrically and an- tisymmetrically loaded halves of the structure . From the ...
... superposition of deformations are met , the analysis of a stable , symmetric structure can be obtained as the superposition of the analyses of the symmetrically and an- tisymmetrically loaded halves of the structure . From the ...
Page 177
... Superposition of load systems in frame member MA L RBq Mq ( x ) = -RAqX + q ( t ) ( x - t ) dt ( a ) Superposition of load systems in simply supported beam MMA ( x ) = MA- MAX L MMB ( X ) = MB- MBX L ( b ) Superposition of moment ...
... Superposition of load systems in frame member MA L RBq Mq ( x ) = -RAqX + q ( t ) ( x - t ) dt ( a ) Superposition of load systems in simply supported beam MMA ( x ) = MA- MAX L MMB ( X ) = MB- MBX L ( b ) Superposition of moment ...
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