Strength of Materials |
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Page 178
... gives the force F to be resisted in a length e : F = S. ( eb ) - V Ib Q ( eb ) = Ve Q I The same result can be obtained more directly by using the concept of shear flow , which is the longitudinal shearing force developed per unit ...
... gives the force F to be resisted in a length e : F = S. ( eb ) - V Ib Q ( eb ) = Ve Q I The same result can be obtained more directly by using the concept of shear flow , which is the longitudinal shearing force developed per unit ...
Page 381
... gives M [ x = S / ] SI Mc = 600 ( 11,840 ) 6.16 = 1,154,000 in . - lb In terms of the concrete equivalent of the steel , the permissible limit is ƒ : 18,000 = 1500 psi , and the flexure formula gives = n 12 [ M - St ] SI = M = 8 1500 ...
... gives M [ x = S / ] SI Mc = 600 ( 11,840 ) 6.16 = 1,154,000 in . - lb In terms of the concrete equivalent of the steel , the permissible limit is ƒ : 18,000 = 1500 psi , and the flexure formula gives = n 12 [ M - St ] SI = M = 8 1500 ...
Page 537
... gives the moment of area ; when multiplied a second time by its moment arm it gives the moment of inertia . The term second moment of area is preferable to the expression moment of inertia ; the latter is confusing when applied to an ...
... gives the moment of area ; when multiplied a second time by its moment arm it gives the moment of inertia . The term second moment of area is preferable to the expression moment of inertia ; the latter is confusing when applied to an ...
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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