Strength of materials |
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Page 31
A convenient variation of Hooke's law is obtained by replacing S by its P , S
equivalent — and replacing e by — , so that Eq. (2-3) becomes A L or PL _ SL 6 -
AE - E (2-4) Eq. (2-4) expresses the relation among the total deformation 5, the ...
A convenient variation of Hooke's law is obtained by replacing S by its P , S
equivalent — and replacing e by — , so that Eq. (2-3) becomes A L or PL _ SL 6 -
AE - E (2-4) Eq. (2-4) expresses the relation among the total deformation 5, the ...
Page 50
Thermal Stresses It is well known that changes in temperature cause bodies to
expand or contract, the amount of the linear deformation, St, being expressed by
the relati0D 8t=*L(AT) (2-14) in which a is the coefficient of linear expansion, ...
Thermal Stresses It is well known that changes in temperature cause bodies to
expand or contract, the amount of the linear deformation, St, being expressed by
the relati0D 8t=*L(AT) (2-14) in which a is the coefficient of linear expansion, ...
Page 62
Slice (2) will rotate past slice (1) until the elastic fibers joining them are deformed
enough to create a resisting torque which balances the applied torque. When this
happens, slices (1) and (2) will act as a rigid unit and transmit the torque to ...
Slice (2) will rotate past slice (1) until the elastic fibers joining them are deformed
enough to create a resisting torque which balances the applied torque. When this
happens, slices (1) and (2) will act as a rigid unit and transmit the torque to ...
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allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bolt cantilever beam caused centroid CN CN column compressive stress Compute the maximum concentrated load concrete cover plate 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 hinged Hooke's law horizontal ILLUSTRATIVE PROBLEMS lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan midspan deflection modulus Mohr's circle moments of inertia neutral axis obtain plane plastic positive product of inertia proportional limit radius ratio reaction Repeat Prob resisting restrained beam resultant segment shaft shear center shear diagram shearing force shown in Fig Solution Solve Prob span static steel strain tensile stress thickness torque torsional uniformly distributed load vertical shear weld zero