Engineers' and firemen's license law . y A, and thequotient will be the efficiency of a butt and double strap joint, triple-riveted, as shown in Fig. 5. T. S. = 55,000 pounds. a= .5185 square inches. t = f = .375. s = 42,000 pounds, b = 5/16 = .3125. S = 78,000 pounds. P = 6|/r = c = 95,000 = || =.8125. Number of rivets in single shear in a unit of length of joint = of rivets in double shear in a unit of length of joint = 4. A = X -375 X 55,000 = 134,062. B = ( —.8125) .375 X 55,000 = 117,304. C = 4 X 78,000 X .5185 + 1 X 42,000 X .5185 = 183,549. D = (—2 X • 8
Engineers' and firemen's license law . y A, and thequotient will be the efficiency of a butt and double strap joint, triple-riveted, as shown in Fig. 5. T. S. = 55,000 pounds. a= .5185 square inches. t = f = .375. s = 42,000 pounds, b = 5/16 = .3125. S = 78,000 pounds. P = 6|/r = c = 95,000 = || =.8125. Number of rivets in single shear in a unit of length of joint = of rivets in double shear in a unit of length of joint = 4. A = X -375 X 55,000 = 134,062. B = ( —.8125) .375 X 55,000 = 117,304. C = 4 X 78,000 X .5185 + 1 X 42,000 X .5185 = 183,549. D = (—2 X • 8125) .375 X 55,000 + 1 X 42,000 X .5185 = 122, = ( — 2 X .8125) .375 X 55,000 + .8125 X .3125 X 95,000 = 124, = 4 X .8125 X .375 X 95,000 + 1 X .8125 X .3125 X 95,000 = 139, = 4 X -8125 X • 375X 95,000 + 1 X 42,000 X .5185 = 137,558. !o!,8°4!^ = -875, Efficiency of ,062 (A) J J 18 Part II. — Section 7. Butt 6. Example.—Butt and double strap joint, quadruple- quadruple- • i jriveted. Fig. 6. A = Strength of solid plate = P X t X T. S. B = Strength of plate between rivet holes in the outer row = (P—d)t XT. = Shearing strength of eight (8) rivets in double shear, plus the shearing strength of three (3) rivets in single shear = N X S X a + n X s X = Strength of plate between rivet holes in the second row, plus the shearing strength of one (1) rivet in single shear in the outer row = (P—2d) t X T. S. + n X s X = Strength of plate between rivet holes in the third row, plus the shearing strength of two (2) rivets in the second row in single shear and one (1) rivet in single shear in the outer row = (P—4d) t X T. S. + n X s X = Strength of plate between rivet holes in the second row, plus the crushing strength of butt strap in front of one (1) rivet in the outer row = (P—2d) t X T. S. + d X b X cG = Strength of plate between rivet holes in the third row, plus the crushing strength of butt strap in front of tw
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