Articoli correlati a The application of graphic methods to the design of...
The application of graphic methods to the design of structures, specially prepared for the use of engineers - Brossura
Sinossi
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. 1896 Excerpt: ...here, as they will be found advantageous in deducing the stresses in other indeterminate forms. The direction of z p is known, and the problem will be completely solved, if we can find geometrically the point p, or if the magnitude of z p be round. From the symmetry of the figure, p will occupy a similar position in the triangle p, q, r to that of the point z in the triangle mzn. The point q is not yet found; hence, assume any point qu in eq produced, and through q1 draw q1 r1, intersecting f r in r1, this being parallel to Q R. Then through q1 and r1 draw q1 pu and r p1 parallel to Q P and RP respectively, intersecting in pl. Next select any other point q2, and draw the triangle p2, q2, r2. The line pi. pz-vvill contain the vertices of all the triangles, p, q, r for every possible value of e q; and hence the intersection of pi p2 with z p gives the required point p, and the stress diagram may now be completed. We may, if we like, resolve all the forces acting at the point EDNZPQ in two directions, along and perpendicular respectively to E Q. The components parallel to E Q cannot affect the stress in P Zi, because they are at right angles to it; therefore consider only the perpendicular components. By the first law of equilibrium the sum of these components must be zero. The stress in N Z is tensile, and from symmetry we should expect to find that in P Q tensile also. That it is tensile may be easily shown thus: Because P S and R S are in the same straight line, there can be no resultant force acting at the hinge joining them in a direction perpendicular to either of them, and consequently the stress in P Q must be of the opposite kind to that in Flo. 64. R; that is, P Q is in tension. Also the resolved part of F perpendicular to F R equals the component of...
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