Pre-Owned Night of the Moonjellies (Hardcover) 1930900163 9781930900165

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1916 edition. Excerpt: ...the distortion of the angles is greatest on the flatter side of the section; hence the shearing unit-stress is there the greatest, and the torsion formula (90) does not apply to the elliptical bar; in fact, that formula applies only to circular sections, and it should not be used for other sections except for approximate investigations. The correct torsion formulas for elliptical and rectangular sections will now be deduced. Let Fig. 99a represent a section of an elliptical bar which is subject to a twisting moment Pp from a force P acting in a plane normal to the axis and at a distance p from that axis. Let m be the major axis and n the minor axis of the ellipse, let y and x be the vertical and horizontal coordinates of any point on the circumference of the ellipse with respect to m and - as coordinates axes. Let 5i and 52 be the shearing unit-stresses at the extremities of the minor and major axes, and 5 the shearing unit-stress at the point where cqordinates are y and x; these stresses are tangential to the circumference. Let 5' and 5" be the components of 5 parallel to 5i and 52, and let x be the angle which 5 makes with 5i; then 5' =5 cos and 5" =5 sin, whence S"/S' =tan . The equation of the ellipse is m2y2 ] n2x2 = niln2, and by differentiation there is found dy/dx=--n2x/m2y =tan; accordingly S"/S' = n2x/m2y and it thus appears that the components 5" and 5' are proportional to their distances from the coordinate axes n and m. When the elastic limit of the material is not exceeded, the same relation must hold between the components of the unit-stress at any point within the ellipse, for here, as in the circle, the unit-stresses along any radius vector vary proportionally as their distances from the center....