Strength of Materials |
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Page 243
... developed in the beams . = 1400 lb , compute the maximum flexural stress Ans . S = = 1200 psi SUMMARY 1 M Starting with the relation . = developed in Art . 5-2 , two separate Ρ ΕΙ methods of determining slopes and deflections are ...
... developed in the beams . = 1400 lb , compute the maximum flexural stress Ans . S = = 1200 psi SUMMARY 1 M Starting with the relation . = developed in Art . 5-2 , two separate Ρ ΕΙ methods of determining slopes and deflections are ...
Page 244
... developed a method of drawing moment diagrams by parts ( ie . , in terms of equivalent cantilever loadings ) , which is equivalent to , and re- places , the conventional moment diagram . Deflections in cantilever beams ( Art . 6–5 ) are ...
... developed a method of drawing moment diagrams by parts ( ie . , in terms of equivalent cantilever loadings ) , which is equivalent to , and re- places , the conventional moment diagram . Deflections in cantilever beams ( Art . 6–5 ) are ...
Page 371
... developed between the steel and wood , and between the wood and aluminum . Express the results as a func- tion of the vertical shear V. Ans . 0.129V ; 0.124V 10-4 . Reinforced Concrete Beams Concrete is an excellent building material ...
... developed between the steel and wood , and between the wood and aluminum . Express the results as a func- tion of the vertical shear V. Ans . 0.129V ; 0.124V 10-4 . Reinforced Concrete Beams Concrete is an excellent building material ...
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allowable stresses aluminum angle applied assumed axial load beam in Fig beam loaded beam shown bending bending moment bolt bronze cantilever beam caused centroid column compressive stress Compute the maximum concentrated load concrete continuous beam cross section deformation Determine the maximum diameter elastic curve end moments equal equivalent Euler's formula factor of safety fibers flange flexure formula free-body diagram ft long ft-lb Hence Hooke's law horizontal ILLUSTRATIVE PROBLEMS inertia lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan deflection modulus Mohr's circle moment of area moment of inertia neutral axis obtain plane positive proportional limit R₂ radius reaction Repeat Prob resisting restrained beam resultant segment shaft shear diagram shearing force shown in Fig Solution Solve Prob span statically indeterminate steel strain tensile stress three-moment equation torque torsional uniformly distributed load vertical shear weld zero ΕΙ