Strength of Materials |
From inside the book
Results 1-3 of 91
Page 114
... lb and acts at the centroid of the triangular area , i.e. , at of 6 ft from F. Thus we obtain = [ M ( ΣΜ ) , ] MF = 240 ( 6 ) 240 ( 9 ) = 960 ft - lb Similarly , the moment at B where x = 9 ft is found to be [ M = ( ΣΜ ) Σ ) MB = 240 ...
... lb and acts at the centroid of the triangular area , i.e. , at of 6 ft from F. Thus we obtain = [ M ( ΣΜ ) , ] MF = 240 ( 6 ) 240 ( 9 ) = 960 ft - lb Similarly , the moment at B where x = 9 ft is found to be [ M = ( ΣΜ ) Σ ) MB = 240 ...
Page 212
... lb -720 ft - lb -120x2 = -240 ft - lb 30 lb / ft C M = -120 × 6 = -720 ft - lb BOB B / C V = 120 lb ( a ) 8 ' M = 120x8 = 960 ft - lb V = 120 lb 4 ' 4 ' -240 ft - lb -720 ft - lb 8 ' M = -720 ft - lb M = -720 ft - lb 18 ' ( b ) 30 lb / ft 4 ...
... lb -720 ft - lb -120x2 = -240 ft - lb 30 lb / ft C M = -120 × 6 = -720 ft - lb BOB B / C V = 120 lb ( a ) 8 ' M = 120x8 = 960 ft - lb V = 120 lb 4 ' 4 ' -240 ft - lb -720 ft - lb 8 ' M = -720 ft - lb M = -720 ft - lb 18 ' ( b ) 30 lb / ft 4 ...
Page 345
Ferdinand Leon Singer. A = 250 lb к E 400 lb 1.5 100 lb C Dh = 750 lb 1.5 ' D1 = 575 lb B 1.5 ' A 100 lb 400 lb ( a ) Loading on line shaft 700 lb 100 lb A1 = 725 lb -1450 ft - lb -1150 ft - lb ( b ) Bending moments in vertical plane B 500 ...
Ferdinand Leon Singer. A = 250 lb к E 400 lb 1.5 100 lb C Dh = 750 lb 1.5 ' D1 = 575 lb B 1.5 ' A 100 lb 400 lb ( a ) Loading on line shaft 700 lb 100 lb A1 = 725 lb -1450 ft - lb -1150 ft - lb ( b ) Bending moments in vertical plane B 500 ...
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