Introduction to Mechanics of Deformable Solids |
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Page 127
... Substitution of ( 7.3 : 3 ) for en ( see also Fig . 7.3c ) gives One = -Enynoe L = де - Enyn Ī and with ( 7.3 : 7 ) де Pn = - EnAnyn L and after substitution for P , in the equation of equilibrium ( 7.3 : 1 ) де M = + EnAnyn2 ΤΣΕnAnyn 2 ...
... Substitution of ( 7.3 : 3 ) for en ( see also Fig . 7.3c ) gives One = -Enynoe L = де - Enyn Ī and with ( 7.3 : 7 ) де Pn = - EnAnyn L and after substitution for P , in the equation of equilibrium ( 7.3 : 1 ) де M = + EnAnyn2 ΤΣΕnAnyn 2 ...
Page 371
... substitution of ( 14.9 : 4 ) , Μι = +11022 σ12 dA - M2 = - a L1 y2 da b f yz da Μι = a S1 yz dA + b f1 z2 dA ( 14.9 ... substituted in Eq . ( 14.9 : 4 ) , give the bending stress σ ; substituted in Eq . ( 14.9 : 5 ) , they give the ...
... substitution of ( 14.9 : 4 ) , Μι = +11022 σ12 dA - M2 = - a L1 y2 da b f yz da Μι = a S1 yz dA + b f1 z2 dA ( 14.9 ... substituted in Eq . ( 14.9 : 4 ) , give the bending stress σ ; substituted in Eq . ( 14.9 : 5 ) , they give the ...
Page 411
... Substitution for M and for v ; from Eq . ( 16.3 : 6 ) gives ( 16.3 : 7 ) d2 dx2 ( πX vo sin = — ī Pv EI or d2v P dx2 ΕΙ + v = πX 12 vo sin ( 16.3 : 8 ) We no longer have an eigenvalue or buckling problem but simply a deflection problem ...
... Substitution for M and for v ; from Eq . ( 16.3 : 6 ) gives ( 16.3 : 7 ) d2 dx2 ( πX vo sin = — ī Pv EI or d2v P dx2 ΕΙ + v = πX 12 vo sin ( 16.3 : 8 ) We no longer have an eigenvalue or buckling problem but simply a deflection problem ...
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actual addition angle answer applied assemblage axes axial axis beam behavior bending circle circular column compatibility components compression compressive stress Consider constant creep cross section curve cylinder deflection deformation determined diameter direction displacement effect elastic equal equation equilibrium example Figure Find force given gives homogeneous idealization increase initial interior isotropic length limit linear linear-elastic load material maximum Maxwell modulus moment needed nonlinear normal obtained outer plane plastic positive pressure principal Prob problem produced pure radius range ratio replaced requires response result rotation shaft shear stress shell shown shows simple sketch solid solution steel strain stress-strain relations structural Suppose surface symmetry temperature tensile tension thickness thin-walled torsion tube twisting uniform unloading viscous yield zero