Strength of Materials |
From inside the book
Results 1-3 of 93
Page 116
Ferdinand Leon Singer. 500 lb 1000 lb 200 100 lb / ft 300 lb / ft 두 누구 lb / ft 5 ' 10 ' 10 ' R1 R2 Fig . P - 427 . 20 ' Fig . P - 428 . Ans . Max . M = 5000 ft - lb 427. Beam loaded as shown in Fig . P - 427 . 428. In the overhanging ...
Ferdinand Leon Singer. 500 lb 1000 lb 200 100 lb / ft 300 lb / ft 두 누구 lb / ft 5 ' 10 ' 10 ' R1 R2 Fig . P - 427 . 20 ' Fig . P - 428 . Ans . Max . M = 5000 ft - lb 427. Beam loaded as shown in Fig . P - 427 . 428. In the overhanging ...
Page 119
Ferdinand Leon Singer. w lb / ft R1 22 + Fig . P - 441 . 2/2 w lb / ft R2 442. Beam loaded as shown in Fig . P - 442 . 443. Beam loaded as shown in Fig . P - 443 . Ans . Max . M = 1267 ft - lb Ans . Max . M 4720 ft - lb 1200 lb 120 lb / ft ...
Ferdinand Leon Singer. w lb / ft R1 22 + Fig . P - 441 . 2/2 w lb / ft R2 442. Beam loaded as shown in Fig . P - 442 . 443. Beam loaded as shown in Fig . P - 443 . Ans . Max . M = 1267 ft - lb Ans . Max . M 4720 ft - lb 1200 lb 120 lb / ft ...
Page 284
... lb / ft . At the midpoint of the second span , there is a concentrated load of 2000 lb. Solve for the moments over the supports and check your answers , using Probs . 820 and 821 . Ans . M2 = -5880 ft - lb ; M3 = -8020 ft - lb 825. See ...
... lb / ft . At the midpoint of the second span , there is a concentrated load of 2000 lb. Solve for the moments over the supports and check your answers , using Probs . 820 and 821 . Ans . M2 = -5880 ft - lb ; M3 = -8020 ft - lb 825. See ...
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