Essentials in the theory of framed structures . braic sum of the quantities in column 8 is 31,916; and the deflection of point 4 is ^1,016,000 A4 = ^-^^—- = m. 29,000,000 Similarly the deflection of point 6 is 22,68;,ooo o . Ae = —-— = m. 29,000,000 Since the design and loading of the truss in this problem areboth symmetrical about the center line, it is obvious that thevertical deflections at the points 8 and 10 are respectively thesame as at points 4 and 2. The horizontal displacement of any point may be obtained 288 THEORY OF FRAMED STRUCTURES Chap. VII in a similar manner. Suppose
Essentials in the theory of framed structures . braic sum of the quantities in column 8 is 31,916; and the deflection of point 4 is ^1,016,000 A4 = ^-^^—- = m. 29,000,000 Similarly the deflection of point 6 is 22,68;,ooo o . Ae = —-— = m. 29,000,000 Since the design and loading of the truss in this problem areboth symmetrical about the center line, it is obvious that thevertical deflections at the points 8 and 10 are respectively thesame as at points 4 and 2. The horizontal displacement of any point may be obtained 288 THEORY OF FRAMED STRUCTURES Chap. VII in a similar manner. Suppose that the horizontal displacementof the point i in Fig. 176 is reqmred, when the truss is held fastat the left support and rests on rollers at the right support; asshown in Fig. 175. The loads are assumed, as in Fig. 176; PIand the values of j of column 4, Table I, are therefore appli-cable. Place I lb. at point i, acting horizontally either to theright or left, let us say to the right (as in Fig. 179); compute. Fig. 179. the reactions and tabulate the resulting ratios in a column PuJ,marked Mi; and make the extensions —r-- The horizontal displacement of the point i will be Ai = S AE Since the i lb. load was taken as acting to the right, a positivevalue for the summation will indicate a displacement to theright; and vice versa. Sec. II. Graphic Method 181. Williot Diagrams.—The graphic method of finding thedeflections of a truss is accompUshed by drawing a Williotdiagram; after the strain in each member has been determined,as described in Sec. I. In order to compare the results of thealgebraic method with the graphic, the latter will be explainedin connection with the problem illustrated in Fig. 176. Sincethe modulus of elasticity is constant for all members, thequantities in column 4 of Table I will be taken to represent thestrains; and the factors 1,000 and E = 29,000,000 lb. per Sec. n DEFLECTION OF TRUSSES 289 square inch will be introduced at the end
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