Strength of Materials |
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Page xvii
... solution , values are substituted in the respective order in which the symbols appear in the equation . This procedure enables the reader to follow readily the various steps of the solution without continually referring to the body of ...
... solution , values are substituted in the respective order in which the symbols appear in the equation . This procedure enables the reader to follow readily the various steps of the solution without continually referring to the body of ...
Page 296
... Solution : This problem was solved on page 288 by double integration . Its solution here by the area - moment method shows the close similarity of the two methods and demonstrates the direct use of the moment diagram . We begin by ...
... Solution : This problem was solved on page 288 by double integration . Its solution here by the area - moment method shows the close similarity of the two methods and demonstrates the direct use of the moment diagram . We begin by ...
Page 355
... solution ( b ) shown in Fig . 8-23 . As a start , reduce A and D to free ends by releasing them , applying the balancing moments of -8000 at A and +7200 at D , and carrying over half these amounts with the same signs to B and C as 6000 ...
... solution ( b ) shown in Fig . 8-23 . As a start , reduce A and D to free ends by releasing them , applying the balancing moments of -8000 at A and +7200 at D , and carrying over half these amounts with the same signs to B and C as 6000 ...
Common terms and phrases
allowable stresses aluminum angle applied load assumed axial load beam carrying beam in Fig beam loaded beam shown bending moment bolt bronze cantilever beam centroid column compressive stress concentrated load continuous beam cross section deformation Determine the maximum diameter elastic curve end moments equal equilibrium equivalent exploratory section factor of safety fibers free-body diagram GN/m² Hence Hooke's law horizontal ILLUSTRATIVE PROBLEMS kN.m kN·m kN/m length load diagram loaded as shown M₁ M₂ maximum shearing stress midspan deflection mm² MN/m² Mohr's circle moment of inertia N/m² neutral axis obtain P₁ P₂ plane plate positive principal stresses proportional limit R₁ R₂ R2 Figure radius reaction resisting restrained beam resultant rivet segment shaft shear diagram shearing force shown in Fig simply supported beam Solution span statically indeterminate steel strain tangent drawn tensile stress three-moment equation torque torsional uniformly distributed load vertical shear weld ΕΙ