Strength of Materials |
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Page 314
... resultant stress at A is equal to the superposition of the two separate effects . Thus , the resultant force at A is the vector sum of the collinear forces Sa dA and Sf dA . Dividing this by the area dA gives the resultant stress S SaS ...
... resultant stress at A is equal to the superposition of the two separate effects . Thus , the resultant force at A is the vector sum of the collinear forces Sa dA and Sf dA . Dividing this by the area dA gives the resultant stress S SaS ...
Page 341
... resultant shearing stress S 12,000 psi or a resultant normal stress Sp = 16,000 psi . = Solution : The bending moment produces a maximum flexural stress at the top or bottom of the shaft . Its value is Mc S = [ s , - ME - AM ] I S = 4 ...
... resultant shearing stress S 12,000 psi or a resultant normal stress Sp = 16,000 psi . = Solution : The bending moment produces a maximum flexural stress at the top or bottom of the shaft . Its value is Mc S = [ s , - ME - AM ] I S = 4 ...
Page 342
... resultant shearing stress S , and the maximum resultant normal stress S , in terms of T , M , and the radius r of the shaft . By means of these relations , determine the proper diameter of a solid shaft to carry simultaneously T 900 ft ...
... resultant shearing stress S , and the maximum resultant normal stress S , in terms of T , M , and the radius r of the shaft . By means of these relations , determine the proper diameter of a solid shaft to carry simultaneously T 900 ft ...
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