Strength of Materials |
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Page 553
... axes , marked X ' and Y ' and rotated 90 ° counterclockwise from the original set of axes , the area is in the fourth quadrant . The new coordinates of dA are x ' y and y ' with respect to the new axes , the product of inertia is Px'y ...
... axes , marked X ' and Y ' and rotated 90 ° counterclockwise from the original set of axes , the area is in the fourth quadrant . The new coordinates of dA are x ' y and y ' with respect to the new axes , the product of inertia is Px'y ...
Page 556
... axes . Solution : The angle section can be considered composed of a 5 by 1 in . rec- tangle plus an 8 by 1 in . rectangle . For the first rectangle , the centroidal axes parallel to the X and Y axes are axes of symmetry ; hence , from ...
... axes . Solution : The angle section can be considered composed of a 5 by 1 in . rec- tangle plus an 8 by 1 in . rectangle . For the first rectangle , the centroidal axes parallel to the X and Y axes are axes of symmetry ; hence , from ...
Page 558
... Axes In some cases , it is necessary to determine the moment of inertia with respect to axes which are inclined to the usual axes . The moment of inertia in such cases can be obtained by formal integration , but a general formula is ...
... Axes In some cases , it is necessary to determine the moment of inertia with respect to axes which are inclined to the usual axes . The moment of inertia in such cases can be obtained by formal integration , but a general formula is ...
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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