Strength of Materials |
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Page 65
... solution satisfying these three steps is unique ; i.e. , it is the only possible solution . In deriving the torsion formulas , we make the following assumptions . These assumptions may be proved mathematically , and some may be ...
... solution satisfying these three steps is unique ; i.e. , it is the only possible solution . In deriving the torsion formulas , we make the following assumptions . These assumptions may be proved mathematically , and some may be ...
Page 281
... solution ( b ) shown in Fig . 8-30 . As a start , reduce A and D to free ends by releasing them , applying the balancing moments of +1200 at A and + 1500 at D , and carrying over half these amounts with opposite signs to B and C as ...
... solution ( b ) shown in Fig . 8-30 . As a start , reduce A and D to free ends by releasing them , applying the balancing moments of +1200 at A and + 1500 at D , and carrying over half these amounts with opposite signs to B and C as ...
Page 309
... Solution ( a ) Regular procedure K = { 2x1 = 1 23x1 = DF 1 317 FEM -1200 Release A & D +1200 34 -1200-1500 -600 ㄋ ? 3 / t 1 -1500-1000 -1500 -750 +1500 Adjusted FEM 0 -1800-1500 -1500-1750 0 1st distribution Carry - over 2nd ...
... Solution ( a ) Regular procedure K = { 2x1 = 1 23x1 = DF 1 317 FEM -1200 Release A & D +1200 34 -1200-1500 -600 ㄋ ? 3 / t 1 -1500-1000 -1500 -750 +1500 Adjusted FEM 0 -1800-1500 -1500-1750 0 1st distribution Carry - over 2nd ...
Contents
SIMPLE STRESS | 1 |
RIVETED AND WELDED JOINTS | 39 |
TORSION | 65 |
Copyright | |
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acting AISC formula allowable stresses angle applied assumed axes axial load bending bending moment C₁ C₂ Carry-over centroidal column formula compressive stress Compute the maximum concrete critical load cross-section deflection deformation Determine the maximum diameter elastic curve element end moments equal equivalent Euler's formula factor of safety flange flexural stress flexure formula ft long ft-lb Hence ILLUSTRATIVE PROBLEMS in.¹ lb-ft³ lb/ft length maximum shearing stress maximum stress midspan modulus Mohr's circle moment of inertia moments of inertia neutral axis normal stress obtain P₁ plane plate principal stresses PROB product of inertia proportional limit R₁ R₂ radius reinforced resisting resultant rivet S₂ shaft shear center shear diagram shear flow shearing force shown in Fig simply supported slenderness ratio slope Solution span steel strain tangent tensile stress torsional vertical shear whence zero ΕΙ