Cyclopedia of Civil Engineering; Statics Materials Roof Trusses Cost Analysis

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. 1909 Excerpt: ...= 3,100 cos 55" + 6,200 cos 55" H 3,100 cos 55"= 7,113 pounds, Fig. 85. and from the third, K, = 12.2 + 3,100 x 24.4 = Substituting this value of R, in the second equation we find that R," =3,100 cos 35 + 6,200 cos 35 + 3,100 cos 35-3,782 = 10,150-3,782 = 0,374 pounds. If desired, the reaction R, can now be found by eompounding its two components R,' and Rj". (J) Using the second set of conditions of equilibrium stated in Art. 35 we obtain the following three "equilibrium equations": As in (1), resolving forces along the horizontal gives-It,' + 3,100 cos 55 + 6,200 cos 55 + 3,100cos 55 = 0, and taking moments about the left end, 6,200 X 12.2 + 3,100 X 24.4-K2 X 40 = 0. Taking moments about the right end gives R," X 40-3,100 X-6,200 X£6-3,100 X c6" = 0 Just as in (a), we find from the first and second equations the values of R,' and E3. To find It," we need values of the arms «6, 66, and d,. By measurement from a drawing we find that a6 = 32.7, "Mi = 20.5, and (T = 8.3 feet. Substituting these values in the third equation and solving for R," we find that 3,100X 32.7 + 6,200X20.5 + 3,100X8.3 c 0„ K, = = o,355 pounds. (c) Using the third set of conditions of equilibrium stated in Art. 35 we obtain the following three equilibrium equations: As in (6), taking moments about the right and left ends we get R," X 40-3,100 X 32.7-6,200 X 20.5-3,100 X 8. 3 = 0, and-R3 X 40 + 6,200 X 12.2 + 3,100 X 24.4 = 0. Choosing the peak of the truss as the origin of moments for the third equation we find that R,' X 14 + R," X 20-3,100 X 24.4-6,200 X 12.2 It, X 20 = 0. As in (6) we find from the first two equations the values of R,"and Ko. These values substituted in the third equation chang...