Strength of Materials |
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Page 353
... hinged column of the same length . One other type of column is hinged at one end and built in at the other , as in Fig . 11-6 . For it , the point of inflection can be shown to be nearly 0.7 L from the hinged end . Hence substituting an ...
... hinged column of the same length . One other type of column is hinged at one end and built in at the other , as in Fig . 11-6 . For it , the point of inflection can be shown to be nearly 0.7 L from the hinged end . Hence substituting an ...
Page 354
... hinged Both ends hinged 421 4 2 One end fixed , the other free } L 0.7 L L 2 L 11-4 . Limitations of Euler's Formula A column always tends to buckle in its most limber direction . For this reason , and since flexural resistance varies ...
... hinged Both ends hinged 421 4 2 One end fixed , the other free } L 0.7 L L 2 L 11-4 . Limitations of Euler's Formula A column always tends to buckle in its most limber direction . For this reason , and since flexural resistance varies ...
Page 356
... hinged ends , Euler's formula is P = EIT2 L2 = PL2 ET2 whence 300,000 ( 22 X 12 ) 2 I = 70.7 in.1 ( 30 × 106 ) π2 = = The lightest WF section with this least value of I is the 10 WF 49 section , with least I 93.0 in . and least k 2.54 ...
... hinged ends , Euler's formula is P = EIT2 L2 = PL2 ET2 whence 300,000 ( 22 X 12 ) 2 I = 70.7 in.1 ( 30 × 106 ) π2 = = The lightest WF section with this least value of I is the 10 WF 49 section , with least I 93.0 in . and least k 2.54 ...
Contents
RIVETED AND WELDED JOINTS | 39 |
TORSION | 65 |
SHEAR AND MOMENT IN BEAMS | 91 |
Copyright | |
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acting actual allowable angle applied assumed axes axial axis beam bending carries caused centroidal circle column compressive compressive stress compute concrete consider constant cross-section deflection deformation Determine developed diagram diameter direction distance distributed effect elastic curve element equal equation equivalent expressed flange forces formula ft-lb given gives Hence horizontal ILLUSTRATIVE indicates inertia joint lb/ft length limit load material maximum method moments neutral axis normal obtain occurs plane plate position principal PROB PROBLEMS produced R₁ radius reaction reduces reference reinforced relation represents resisting respect resultant rivet shaft shearing stress shown in Fig shows simple slope Solution span spring steel strain strength Substituting supported Table tangent tensile thickness unit varies vertical wall weight whence zero