Strength of Materials |
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Page 29
... aluminum bar , A = = 10 X 10 psi . = 30 X 10 psi . For 0.60 sq in . and E Ans . SAC 10,000 psi ; SAB = 2500 psi ; SAD = 1670 psi 148. Referring to the data in Prob . 147 , determine the maximum value of W that will not exceed an aluminum ...
... aluminum bar , A = = 10 X 10 psi . = 30 X 10 psi . For 0.60 sq in . and E Ans . SAC 10,000 psi ; SAB = 2500 psi ; SAD = 1670 psi 148. Referring to the data in Prob . 147 , determine the maximum value of W that will not exceed an aluminum ...
Page 74
... aluminum shaft of 3 in . outer diameter and 2 in . inner diameter is slipped over a solid steel shaft 2 in . in diameter and of the same length as the hollow shaft . The two are then fastened rigidly together at their ends . Determine ...
... aluminum shaft of 3 in . outer diameter and 2 in . inner diameter is slipped over a solid steel shaft 2 in . in diameter and of the same length as the hollow shaft . The two are then fastened rigidly together at their ends . Determine ...
Page 447
... Aluminum , cast 12.8 X 10-6 10 × 106 Aluminum Alloy 17ST 12.8 x 10 21,000 56,000 10.3 X 106 Brass , rolled ( 70 % Cu ) 10.4 X 10-25,000 15,000 55,000 b 14 X 106 ( 30 % Zn ) Y Y Weight Flange AXIS X - X AXIS Y. Weight Temp . Coeff . of ...
... Aluminum , cast 12.8 X 10-6 10 × 106 Aluminum Alloy 17ST 12.8 x 10 21,000 56,000 10.3 X 106 Brass , rolled ( 70 % Cu ) 10.4 X 10-25,000 15,000 55,000 b 14 X 106 ( 30 % Zn ) Y Y Weight Flange AXIS X - X AXIS Y. Weight Temp . Coeff . of ...
Contents
RIVETED AND WELDED JOINTS | 39 |
TORSION | 65 |
SHEAR AND MOMENT IN BEAMS | 91 |
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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