Strength of Materials |
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Page 61
... Shaft is loaded by twisting couples in planes that are perpendicular to the axis of the shaft . 5. Stresses do not exceed the proportional limit . 3-2 . Derivation of Torsion Formulas Fig . 3-1 shows two views of a solid circular shaft ...
... Shaft is loaded by twisting couples in planes that are perpendicular to the axis of the shaft . 5. Stresses do not exceed the proportional limit . 3-2 . Derivation of Torsion Formulas Fig . 3-1 shows two views of a solid circular shaft ...
Page 69
... shaft that will not twist through more than 3 ° in a 20 - ft length when subjected to a torque of 10,000 ft - lb ? What maximum shearing stress is developed ? G = 12 X 106 psi . Ans . d = 4.65 in .; S , 6080 psi 305. A solid shaft 16 ft ...
... shaft that will not twist through more than 3 ° in a 20 - ft length when subjected to a torque of 10,000 ft - lb ? What maximum shearing stress is developed ? G = 12 X 106 psi . Ans . d = 4.65 in .; S , 6080 psi 305. A solid shaft 16 ft ...
Page 70
... shaft at 5 ft from the right end . ( a ) Find the uniform shaft diameter so that the shearing stress will not exceed 8000 psi . ( b ) If a uniform diameter of 4 in . is specified , determine the angle by which one end of the shaft lags ...
... shaft at 5 ft from the right end . ( a ) Find the uniform shaft diameter so that the shearing stress will not exceed 8000 psi . ( b ) If a uniform diameter of 4 in . is specified , determine the angle by which one end of the shaft lags ...
Contents
APPENDIX B TABLES | 2 |
Shearing Stress | 10 |
SIMPLE STRAIN | 26 |
Copyright | |
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acting actual allowable angle applied assumed axial axis beam beam shown bending bolt carries caused centroid circle column compressive compressive stress compute concentrated concrete consider constant couple cross section deflection deformation determine developed diagram diameter direction distance effect elastic curve element equal equation equivalent expressed flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia joint lb/ft length limit load material maximum method midspan moments negative neutral axis normal obtain occurs plane plate positive Prob PROBLEMS produce R₁ radius reaction relation resisting respect resultant rivet segment shaft shearing stress shown in Fig shows simply slope Solution Solve span spring steel strain strength supported Table tangent tensile thickness torque torsional uniformly varies vertical wall yield zero ΕΙ