Statically Indeterminate Structures |
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Page 2
... shown in Fig . 1b , there will be five reactive A FIG . 1b . D f B forces and , consequently , there will be two redundant forces . The question may well arise as to which two reactive forces should be classi- fied as redundant . In ...
... shown in Fig . 1b , there will be five reactive A FIG . 1b . D f B forces and , consequently , there will be two redundant forces . The question may well arise as to which two reactive forces should be classi- fied as redundant . In ...
Page 41
... shown to be extremely small as compared with the length of the member , the same movements can be obtained with sufficient accuracy by using perpendiculars as shown in Fig . 246 instead of arcs . Moreover , it is now apparent that the ...
... shown to be extremely small as compared with the length of the member , the same movements can be obtained with sufficient accuracy by using perpendiculars as shown in Fig . 246 instead of arcs . Moreover , it is now apparent that the ...
Page 217
... Fig . 1286 , in which the ordinate is equal to and the abscissa is unity . y ds EI y ds x The angle de has or Ay L EI as ordinate and unity as a base . " Δαπ and , L Therefore , if these angles are drawn as shown in Fig . 1286 , the dif ...
... Fig . 1286 , in which the ordinate is equal to and the abscissa is unity . y ds EI y ds x The angle de has or Ay L EI as ordinate and unity as a base . " Δαπ and , L Therefore , if these angles are drawn as shown in Fig . 1286 , the dif ...
Contents
CHAPTER PAGE | 1 |
FUNDAMENTAL PRINCIPLES OF STRUCTURAL MECHANICS | 9 |
CONTINUOUS BEAMS AND FRAMES WITH STRAIGHT PRISMATIC MEM | 48 |
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analysis angle applied load arch assumed auxiliary force system axial load axial stress axis bending C₁ calculated carry-over factors change in length columns computed concentrated load conjugate beam constant continuous beam continuous truss correction cross section dead load deformation determined distribution factors elastic curve end moments end rotations equilibrium Example expressed in terms fixed-end moments flange force F ft-kips funicular girders HARDY CROSS Hinge Hooke's law horizontal displacement influence diagram joint kip per foot live load max.d MF ba MFab MFba moment of inertia moments in Fig numerical obtained P₁ panel problem procedure R₁ reactions reciprocal theorem redundant forces shearing force shearing stress shown in Fig simply supported slope-deflection equations solution span bc statically indeterminate stiffness strain energy structure symmetrical tangent tion uniform load values vertical displacement Vierendeel truss Williot diagram ΕΙ ΕΙΟ θα ΣΟΚ