Strength of Materials |
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Page 126
... Neutral surface Ss dA ( b ) N.A. Neutral axis S dA FIG . 5-2 . — Forces acting upon any element of the cross - section of a beam . any exploratory section satisfy the conditions EX = 0 , V = V ,, and M = M ,. The pictorial sketch1 ( Fig ...
... Neutral surface Ss dA ( b ) N.A. Neutral axis S dA FIG . 5-2 . — Forces acting upon any element of the cross - section of a beam . any exploratory section satisfy the conditions EX = 0 , V = V ,, and M = M ,. The pictorial sketch1 ( Fig ...
Page 144
... neutral axis . Because flexure stresses vary directly with distance from the neutral axis which is the centroidal axis such beam sections are de- sirable for materials that are equally strong in tension and compression . However , for ...
... neutral axis . Because flexure stresses vary directly with distance from the neutral axis which is the centroidal axis such beam sections are de- sirable for materials that are equally strong in tension and compression . However , for ...
Page 293
... neutral axis with the X and Y axes respec- tively are found by substituting first y = 0 and then x = 0 in Eq . ( a ) . This gives 2 u = ― ky2 ex kz and v = Cy ( b ) The neutral axis passes through the quadrant which is opposite to that ...
... neutral axis with the X and Y axes respec- tively are found by substituting first y = 0 and then x = 0 in Eq . ( a ) . This gives 2 u = ― ky2 ex kz and v = Cy ( b ) The neutral axis passes through the quadrant which is opposite to that ...
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