Strength of Materials |
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Page 344
... equivalent torque and equivalent moment must be remembered : T. = √M2 + T2 M2 = { ( M + T ) e ( 9-13 ) ( 9-14 ) For the numerical data given in this problem , the equivalent torque and equiva- lent moment are Ꭲ . Te = √M2 + T2 ...
... equivalent torque and equivalent moment must be remembered : T. = √M2 + T2 M2 = { ( M + T ) e ( 9-13 ) ( 9-14 ) For the numerical data given in this problem , the equivalent torque and equiva- lent moment are Ꭲ . Te = √M2 + T2 ...
Page 365
... equivalent section in terms of one material to which the theory can be applied . h h 012 hi nb K9 ( a ) Timber and steel section ( b ) Equivalent wood section ( c ) Equivalent steel section - Fig . 10-1 . Equivalent sections . To obtain ...
... equivalent section in terms of one material to which the theory can be applied . h h 012 hi nb K9 ( a ) Timber and steel section ( b ) Equivalent wood section ( c ) Equivalent steel section - Fig . 10-1 . Equivalent sections . To obtain ...
Page 382
... equivalent section of concrete shown in Fig . 10-10 . Thus we obtain V [ S. = A'T ] Ib Ss = V ( nAs ) ( d Ib ' - kd ) ( a ) where b ' is the effective width of the steel bars , equivalent to the sum of the perimeters of the steel bars ...
... equivalent section of concrete shown in Fig . 10-10 . Thus we obtain V [ S. = A'T ] Ib Ss = V ( nAs ) ( d Ib ' - kd ) ( a ) where b ' is the effective width of the steel bars , equivalent to the sum of the perimeters of the steel bars ...
Contents
APPENDIX B TABLES | 2 |
Shearing Stress | 10 |
SIMPLE STRAIN | 26 |
Copyright | |
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acting actual allowable angle applied assumed axial axis beam beam shown bending bolt carries caused centroid circle column compressive compressive stress compute concentrated concrete consider constant couple cross section deflection deformation determine developed diagram diameter direction distance effect elastic curve element equal equation equivalent expressed flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia joint lb/ft length limit load material maximum method midspan moments negative neutral axis normal obtain occurs plane plate positive Prob PROBLEMS produce R₁ radius reaction relation resisting respect resultant rivet segment shaft shearing stress shown in Fig shows simply slope Solution Solve span spring steel strain strength supported Table tangent tensile thickness torque torsional uniformly varies vertical wall yield zero ΕΙ