Strength of Materials |
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Page 445
... weld is F = 9600t 9 = 9600 X 32 9 = 2700 lb / in . Hence the required lengths of weld are [ ] and = Li = 30,000 2700 = 11.1 in . L2 = 12,000 2700 = 4.44 in . These values should be increased by in . to provide for starting and stopping ...
... weld is F = 9600t 9 = 9600 X 32 9 = 2700 lb / in . Hence the required lengths of weld are [ ] and = Li = 30,000 2700 = 11.1 in . L2 = 12,000 2700 = 4.44 in . These values should be increased by in . to provide for starting and stopping ...
Page 451
... welds to be specified . What maximum . value of P can be applied vertically as shown if - in . welds are used ? 1251. Referring to Fig . P - 1250 , find the maximum force per inch of weld if a transverse fillet weld is added along the ...
... welds to be specified . What maximum . value of P can be applied vertically as shown if - in . welds are used ? 1251. Referring to Fig . P - 1250 , find the maximum force per inch of weld if a transverse fillet weld is added along the ...
Page 453
... weld . The maximum size of weld that may be ap- plied to a square edge is in . less than the thickness of the edge , and for rounded edges it is three - fourths of this thickness . Eccentrically loaded weld groups , in which the applied ...
... weld . The maximum size of weld that may be ap- plied to a square edge is in . less than the thickness of the edge , and for rounded edges it is three - fourths of this thickness . Eccentrically loaded weld groups , in which the applied ...
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