Strength of MaterialsSimple stress, simple strai, torsion, shear and moment in beams, beam deflections, continuous beams, combined stresses. |
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Page 72
Andrew Pytel, Ferdinand Leon Singer. 600 lb.ft 1000 lb.ft 900 lb.ft B 5 ft 500 lb.ft A -2 in . dia . 3 ft D 4 ft 600 lb.ft 1000 lb.ft 900 lb.ft 500 lb.ft D C B A TDC = 600 TBC == -400 TAB = 500 Figure 3-5 Angular deformations . TAB ...
Andrew Pytel, Ferdinand Leon Singer. 600 lb.ft 1000 lb.ft 900 lb.ft B 5 ft 500 lb.ft A -2 in . dia . 3 ft D 4 ft 600 lb.ft 1000 lb.ft 900 lb.ft 500 lb.ft D C B A TDC = 600 TBC == -400 TAB = 500 Figure 3-5 Angular deformations . TAB ...
Page 117
... ft The section of zero shear at G is found from the fact that the upward reaction applied over the interval CG must total 300 lb in order to reduce the shear of -300 lb at C to zero at G. Since the reaction is distributed at 105 lb / ft ...
... ft The section of zero shear at G is found from the fact that the upward reaction applied over the interval CG must total 300 lb in order to reduce the shear of -300 lb at C to zero at G. Since the reaction is distributed at 105 lb / ft ...
Page 226
Andrew Pytel, Ferdinand Leon Singer. 80 lb 60 lb / ft 6 ft- R1 4 ft . R2 ( a ) Original loading 4 ft A 60 lb / f 160 lb 2 ft ▽ 2 ft ↓ ↓ ↓↓ 60 lb / ft 4 ft tA / B 320 lb 28 IB 320 lb ( b ) Transformation to symmetry 6 ft 960 lb.ft 240 ...
Andrew Pytel, Ferdinand Leon Singer. 80 lb 60 lb / ft 6 ft- R1 4 ft . R2 ( a ) Original loading 4 ft A 60 lb / f 160 lb 2 ft ▽ 2 ft ↓ ↓ ↓↓ 60 lb / ft 4 ft tA / B 320 lb 28 IB 320 lb ( b ) Transformation to symmetry 6 ft 960 lb.ft 240 ...
Common terms and phrases
allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bending moment bolts cantilever beam centroid column compressive stress Compute the maximum concentrated load connector cross section deformations Determine the maximum diameter elastic curve element equal equivalent Euler's formula fibers flange flexural stress flexure formula free-body diagram Hence Hooke's law horizontal Illustrative Problem kips kN·m kN/m lb.ft lb/ft length loaded as shown M₁ M₂ maximum shearing stress maximum stress method midspan mm² Mohr's circle moment of inertia neutral axis obtain P₁ plane product of inertia proportional limit R₂ R2 Figure radius reaction resisting resultant rivet segment shaft shear center shear diagram shearing force shown in Fig slope Solution span steel strain tensile stress thickness torque torsional U.S. Customary Units uniformly distributed load vertical shear weld zero ΕΙ