Cyclopedia of architecture, carpentry, and building : a general reference work . a push; hence the member isunder a compression of 2,308 pounds. The section might have been passed so as to cut members %,fg^ and/d, giving Fig. 41 as the left part, and the desired forcemight be obtained from the system of forces acting on that part(3,000, 2,000, Fi, Fo, and Fg.) To compute Fj we take momenteabout the intersection of F2 and Fg, thus 200 STATICS G7 Fj X - 2,000 X 8 + 3,000 X 10 = 0. This 13 the same equation as was obtained in the first solution,and hence leads to the same result. 4. It is r
Cyclopedia of architecture, carpentry, and building : a general reference work . a push; hence the member isunder a compression of 2,308 pounds. The section might have been passed so as to cut members %,fg^ and/d, giving Fig. 41 as the left part, and the desired forcemight be obtained from the system of forces acting on that part(3,000, 2,000, Fi, Fo, and Fg.) To compute Fj we take momenteabout the intersection of F2 and Fg, thus 200 STATICS G7 Fj X - 2,000 X 8 + 3,000 X 10 = 0. This 13 the same equation as was obtained in the first solution,and hence leads to the same result. 4. It is required to determine the stress in the member 12of the truss represented in Fig. 20, due to the-loads shown. Passincr a section cutting; members 12 and li. and consider-ing the left part, we get Fig. 42. To determine Fj we may writea moment equation preferably with origin at joint 4, thus: Fj X + 4,000 X 10 == 0*; 4,000 X 10 or. Fi = r= _ S,94S poimds. the minus sign meaning that the stress is compressive. Fj might be determined also by writing the algebraic sum of. Fiff. 43. the vertical components of all the forces on the left part equal tozero, thus: Fj sin 2G 34 + 4,000 = 0;- 4,000 - 4,000 or Fi - 8,948, sin 2(1^ 34 with the first result. EXAMPLES FOR PRACTICE. 1. Determine by the method of sections the stresses in mem-bers 23, 25, and 45 of the truss represented in Fig. 26, due to theloads shown. ( Stress in 23 = - 5,000 pounds,Ans. -^ Stress in 25 = — 3,350 in 45 = + 8,000 pounds * The arm of Fi with respect to (4) is feet. 201 68 STATICS 2. Determine the stresses in the members 12,15, 84, and 56of the truss represented in Fig. 27, due to the loads shown. ( Stress in 12 =- 11,170 pounds, Stress in 15 = + 10,000 pounds,I Stress in 84 = - 8,940 pounds,I Stress in 56 = + 6,000 pounds. Ans. d.(i)
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