Strength of Materials |
From inside the book
Results 1-3 of 85
Page 110
... zero and the corresponding slope of the shear diagram is zero ( a horizontal line ) . We may conclude therefore that the shear diagram consists of straight hori- zontal lines for intervals in which the load intensity is zero , and of ...
... zero and the corresponding slope of the shear diagram is zero ( a horizontal line ) . We may conclude therefore that the shear diagram consists of straight hori- zontal lines for intervals in which the load intensity is zero , and of ...
Page 113
... zero , which means that the slope is hori- zontal . The concentrated load of 120 lb at E reduces the shear abruptly to zero . Before we locate the positions of zero shear at F and G on the shear diagram , consider the effect of ...
... zero , which means that the slope is hori- zontal . The concentrated load of 120 lb at E reduces the shear abruptly to zero . Before we locate the positions of zero shear at F and G on the shear diagram , consider the effect of ...
Page 553
... zero value . When in this position , the axes are known as the principal axes of the area . Their application is discussed in Art . A - 12 . A - 8 . Product of Inertia Is Zero with Respect to Axes of Symmetry If an area has an axis of ...
... zero value . When in this position , the axes are known as the principal axes of the area . Their application is discussed in Art . A - 12 . A - 8 . Product of Inertia Is Zero with Respect to Axes of Symmetry If an area has an axis of ...
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