Strength of Materials |
From inside the book
Results 1-3 of 84
Page 168
... shearing stress is = [ S. - YA'T ] Ib S & = 1800 72 ( 4 ) ( 4 X 3 ) ( 1.5 ) = 112.5 psi If desired , Eq . ( 5-6 ) may be used . As noted on page 165 , this equation determines the maximum shearing stress on any rectangular section . V 2 ...
... shearing stress is = [ S. - YA'T ] Ib S & = 1800 72 ( 4 ) ( 4 X 3 ) ( 1.5 ) = 112.5 psi If desired , Eq . ( 5-6 ) may be used . As noted on page 165 , this equation determines the maximum shearing stress on any rectangular section . V 2 ...
Page 329
... maximum or minimum normal stresses are found by differentiating Eq . ( 9-5 ) with respect to 0 and setting the derivative equal to zero , whence 2STV tan 20 = St S Similarly , the planes of maximum shearing stress ... maximum and minimum ...
... maximum or minimum normal stresses are found by differentiating Eq . ( 9-5 ) with respect to 0 and setting the derivative equal to zero , whence 2STV tan 20 = St S Similarly , the planes of maximum shearing stress ... maximum and minimum ...
Page 461
Ferdinand Leon Singer. This indicates that the maximum tensile or compressive stress alone is not sufficient to ... shearing stress equals the maximum shearing stress developed at yielding in simple tension . Since the maximum shearing ...
Ferdinand Leon Singer. This indicates that the maximum tensile or compressive stress alone is not sufficient to ... shearing stress equals the maximum shearing stress developed at yielding in simple tension . Since the maximum shearing ...
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