Strength of Materials |
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Page 198
... expressed as 0AB = 1 ΕΙ ( Area ) AB ( 6-4 ) This is the algebraic expression of Theorem I , which is stated as follows : Theorem I : The change in slope between tangents drawn to the elastic 1 curve at any two points A and B is equal to ...
... expressed as 0AB = 1 ΕΙ ( Area ) AB ( 6-4 ) This is the algebraic expression of Theorem I , which is stated as follows : Theorem I : The change in slope between tangents drawn to the elastic 1 curve at any two points A and B is equal to ...
Page 200
... expressed as a function of x . Our purpose here , however , is to discuss a method of dividing moment diagrams into parts whose areas and centroids are known ; this permits . simple numerical calculations to replace integrations . The ...
... expressed as a function of x . Our purpose here , however , is to discuss a method of dividing moment diagrams into parts whose areas and centroids are known ; this permits . simple numerical calculations to replace integrations . The ...
Page 536
... expression of the form Sp2 dA , where p is the perpendicular distance from dA to the axis of inertia . This integral appears so frequently that it has been named moment of inertia . * Moment of inertia applied to areas has no real ...
... expression of the form Sp2 dA , where p is the perpendicular distance from dA to the axis of inertia . This integral appears so frequently that it has been named moment of inertia . * Moment of inertia applied to areas has no real ...
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