Strength of Materials |
From inside the book
Results 1-3 of 91
Page 115
... equal but of opposite sign , thereby producing equal slopes oppositely directed . The moment curve between A and B , however , is not symmetrical about F because the shear ordinates do not have equal values at equal distances on either ...
... equal but of opposite sign , thereby producing equal slopes oppositely directed . The moment curve between A and B , however , is not symmetrical about F because the shear ordinates do not have equal values at equal distances on either ...
Page 160
... equal compressive forces C3 and T3 cancel out . We conclude that equal shear resistances are developed at layers fg and hk . However , this requires that the areas from the neutral axis to the equidistant layers be symmetrical with ...
... equal compressive forces C3 and T3 cancel out . We conclude that equal shear resistances are developed at layers fg and hk . However , this requires that the areas from the neutral axis to the equidistant layers be symmetrical with ...
Page 227
... equal to 28 , or twice the actual midspan deflection in Fig . 6-27a . Since we are considering the deviation of C from the midspan tangent drawn at B , we need the moment diagram of only half the beam . This M - diagram may be drawn by ...
... equal to 28 , or twice the actual midspan deflection in Fig . 6-27a . Since we are considering the deviation of C from the midspan tangent drawn at B , we need the moment diagram of only half the beam . This M - diagram may be drawn by ...
Other editions - View all
Common terms and phrases
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