. Tentative revised boiler safety orders. plate in front of two rivets, plus the shearing strengthof one rivet m single shear=A^Xc^XiXc+^XsXa £f= strength of butt straps between rivet holes in the inner row = (P—2d) 2b X TS. This method of failure is not possible for thicknesses of butt straps required by these Rules and the computation need only be made for old boilers in which thin butt straps have been used. For this reason this method of failure will not be considered in other joints. Divide B, C, D, E, F,GoxH (whichever is the least) by A, and the quotient will 100 REPORT OF BOILER CODE C
. Tentative revised boiler safety orders. plate in front of two rivets, plus the shearing strengthof one rivet m single shear=A^Xc^XiXc+^XsXa £f= strength of butt straps between rivet holes in the inner row = (P—2d) 2b X TS. This method of failure is not possible for thicknesses of butt straps required by these Rules and the computation need only be made for old boilers in which thin butt straps have been used. For this reason this method of failure will not be considered in other joints. Divide B, C, D, E, F,GoxH (whichever is the least) by A, and the quotient will 100 REPORT OF BOILER CODE COMMITTEE, be the efficiency of a butt and double strap joint, double-riveted, as shown inFig. 27. r^ = 55,000 lb. per sq. in. a= sq. in. ^ t= ^ in. = in. 5=44,000 lb. per sq. in. h= ^ in. = in. ^=88,000 lb. per sq. in, P=47^ c=95,000 lb. per sq. in, d= 7^ in. = in. Number of rivets in single shear in a unit length of joint = of rivets in double shear in a unit length of joint = Fig. 28 Example of Butt and Double Strap Joint, Triple-Riveted A =,000 = 100,547 B = (—) X55,000 = 82,500 C =2 X88,000 + 1 X44,000 = 132,286 Z) = (—),000+1X44,,910 .£; = (—),000+,000=90,429 F=2 X95,000+ X95,000 = 88,320 0=,000+1X44,,800 82,500 (B) 100,547 {A) = = efficiency of joint 414 Example: Butt and double strap joint, = strength of solid plate = PXtXTS J5=strength of plate between rivet holes in the outer row = (P—d) tXTSC= shearing strength of four rivets in double shear, plus the shearing strength of one rivet in single shear=i\rx>SXa+nXsXaZ)=strength of plate between rivet holes in the second row, plus the shearing strength of one rivet in single shear in the outer row = (P—2d) txTS •jrnXsXa APPENDIX 101 £^ = strength of plate bet
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