Stresses in structural steel angles : with special tables . al stresses will occurare b and d of the angle, but if the plane of loading is shiftedto OPr the points of critical stresses will be a and d. The most advantageous plane of loading is that forwhich there is the. greatest sec-tion modulus. For an angle usedas a purlin, for a vertical load,a study of Fig. 4 will show thatthe purlin should be set as indi-cated at a of Fig. 5, not as shownat b. It will be observed that for an angle in tension or com-pression, the application of the load at the gauge line is notfar from the most advantageo
Stresses in structural steel angles : with special tables . al stresses will occurare b and d of the angle, but if the plane of loading is shiftedto OPr the points of critical stresses will be a and d. The most advantageous plane of loading is that forwhich there is the. greatest sec-tion modulus. For an angle usedas a purlin, for a vertical load,a study of Fig. 4 will show thatthe purlin should be set as indi-cated at a of Fig. 5, not as shownat b. It will be observed that for an angle in tension or com-pression, the application of the load at the gauge line is notfar from the most advantageous point of loading. Art. 5. Neutral Axis For pure flexure the neutral axis will pass through thecenter of gravity of the section, assuming some direction, asfor example nn (Fig. 1), which direction will depend uponthe plane of loading and upon the section, but the neutralaxis will not necessarily be perpendicular to the plane ofloading. In Art. 2, the relation of a to 6 (see Eq. 11), wasobtained in the derivation of the expression for the 10 STRESSES IN STRUCTURAL STEEL ANGLES This equation and two of its equivalent forms are: Ix — J tan 6 , \ tan a = -——- -, (20) J-Iy tan 0 v J , Ix cot 0 — 7 e N tan a = - (21) J cot 6-Iy v J Ix cos 0 —/ sin 0 , N tana = - —:—. . (22) J cos B — Iy sin 0 For a combination of flexure and direct loading theneutral axis will not pass through the center of gravity of thesection, but it will be parallel to the position which it wouldhave for pure flexure, and the distance between the two par-allel lines will be, F (y cos a — x sin a) , N v=a 7 • • • • (23) in which v = distance from the neutral axis to the centerof gravity; F = total stress acting normal to the section;A =area of the section;/=unit stress at the extreme fiber due to flexure;x, y = coordinates of the extreme fiber;a = angle which the neutral axis makes with X-X. By the use of Eqs. (20) to (23) the position and the direc-tion of the neutral
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