. The strength of materials; a text-book for engineers and architects. diameter of thecircle of the holt centres is lOJ inches (i. e. r = o25 inches) andthe coupling has 6 holts. The stress allowed is 2J tons per sq. in. In this case from equation (3) 120 X 33,000 X 12 1 = —c^ 1-r^ iri • lbs. 2 IT X 75 = 100,000 in. lbs. nearly. Also from equation (2) T = . X ozi^ m. tons 4 6x2-5x7rX6Z2x2,240 ^ ^^. „ = — —. 4 = 139,000 d~ in. lbs. nearly. , _ 100,000 •• ~ 139,000 , /i6o,oo() ^^. , ^ = Vr397000 = ^^^^^-^^^^^^ . •. Adopt bolts Y ^^ diameter. (2) ExA3iPLE OF Cleat.—We ivill now take t
. The strength of materials; a text-book for engineers and architects. diameter of thecircle of the holt centres is lOJ inches (i. e. r = o25 inches) andthe coupling has 6 holts. The stress allowed is 2J tons per sq. in. In this case from equation (3) 120 X 33,000 X 12 1 = —c^ 1-r^ iri • lbs. 2 IT X 75 = 100,000 in. lbs. nearly. Also from equation (2) T = . X ozi^ m. tons 4 6x2-5x7rX6Z2x2,240 ^ ^^. „ = — —. 4 = 139,000 d~ in. lbs. nearly. , _ 100,000 •• ~ 139,000 , /i6o,oo() ^^. , ^ = Vr397000 = ^^^^^-^^^^^^ . •. Adopt bolts Y ^^ diameter. (2) ExA3iPLE OF Cleat.—We ivill now take the case shown inFig. 140 of the cleat given in the Handbook of Messrs. Dorman,Long d; Co., Ltd., for a 16 in. hy 6 in. standard I beam with aminimum span of 18 ft., the rivets being of f in. diameter. The safe uniformly distributed load given for this span and TORSION AND TWISTING OF SHAFTS 315 beam is 25 tons, so that the reaction at each end will be 25 ^ = 12-5 tons, and half of this will be carried by each angle. or the load P will be 6-25 - 2i-^Z^ McVi 625 Fig. 140. First find the position of the centre of gravity of the is clearly on the horizontal line through the rivet 3, and itsdistance from the line 1, 3, 5 is obtained by moments thus— 5 ^ = 2 X 21 ^ 4-5 ^ . 1. e. a = -zr = 9 316 THE STRENGTH OF MATERIALS Then we tabulate the dimensions as follows— No. of Rivet. r r2 12345 4-58 2-62 -90 2-62 4-58 21-06 6-88 -81 6-88 21-06 2 7-2 = 56-69 Pa; _ 6^25 X 3^5•*• * ~ 2>2 - 56^69 = -348 ton. The moment load will be a maximum on rivets 1 and 5because they are farthest from x, and will be equal toTg = -348 X 4-58 = 1-59 tons. The direct load W on these rivets 6-25 = 1-25 tons. Therefore resultant load = R5 = 2*20 tons. [See Fig. 140.]Now bearing area of a |-in. rivet in a |-in. plate 9 _3 3 ~4 ^ 8 32 sq. m. 2-20 X 32Bearing stress on rivet = ^ = 782 tons per gq. in. Area of a |-in. rivet in section ^ -r x \~^] = 442 .4 2-20. •.
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