Strength of Materials |
From inside the book
Results 1-3 of 41
Page 226
... midspan tangent is equal to the midspan deflection . For simple beams that are unsymmetrically loaded , the midspan deflec- tion can be found as easily as for a symmetrically loaded beam . All that need be done is to add a symmetrically ...
... midspan tangent is equal to the midspan deflection . For simple beams that are unsymmetrically loaded , the midspan deflec- tion can be found as easily as for a symmetrically loaded beam . All that need be done is to add a symmetrically ...
Page 227
... midspan tangent drawn at B is 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 ...
... midspan tangent drawn at B is 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 ...
Page 229
... midspan value of EId for the beam loaded as shown in Fig . P - 680 . 681. Show that the midspan value of EId is wb ( L3 48 2Lbb3 ) for the beam in part ( a ) of Fig . P - 681 . Then use this result to find the midspan E18 of the loading ...
... midspan value of EId for the beam loaded as shown in Fig . P - 680 . 681. Show that the midspan value of EId is wb ( L3 48 2Lbb3 ) for the beam in part ( a ) of Fig . P - 681 . Then use this result to find the midspan E18 of the loading ...
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