Strength of Materials |
From inside the book
Results 1-3 of 81
Page 28
... Limit . From the origin O to a point called the proportional limit , Fig . 2-1 shows the stress - strain diagram to be a straight line . From this we deduce the well - known relation , first postulated by Robert Hooke * in 1678 , that ...
... Limit . From the origin O to a point called the proportional limit , Fig . 2-1 shows the stress - strain diagram to be a straight line . From this we deduce the well - known relation , first postulated by Robert Hooke * in 1678 , that ...
Page 397
... limit is 30,000 psi . 1109. Select the lightest WF section that will act as a column 40 ft long with fixed ends and support an axial load of 150,000 lb with a factor of safety of 2. Assume that the proportional limit is 30,000 psi ...
... limit is 30,000 psi . 1109. Select the lightest WF section that will act as a column 40 ft long with fixed ends and support an axial load of 150,000 lb with a factor of safety of 2. Assume that the proportional limit is 30,000 psi ...
Page 531
... limit load for this possibility , we draw the free - body diagrams as shown in part ( b ) where the wall moments become the limit moments ML . Expressing these moments in terms of the common contact force P , we have [ MA = ( 2M ) R ] ...
... limit load for this possibility , we draw the free - body diagrams as shown in part ( b ) where the wall moments become the limit moments ML . Expressing these moments in terms of the common contact force P , we have [ MA = ( 2M ) R ] ...
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