Machinery's Handbook, 31st Edition
1864
Riveted Joints
the rivet holes is 1 ∕ 16 inch larger than that of the rivets. The rivets are so placed that the main plates will not tear diagonally from one rivet row to the others nor will they tear or fail in shear out to their edges. The safe tensile load L and the efficiency η may be determined in the following way: Design stresses for 8500 psi for shear, 20,000 psi for bearing, and 10,000 psi for tension are arbitrarily assigned. a) The safe tensile load L based on double shearing of the rivets is equal to the number of rivets n times the number of shearing planes per rivet times the cross-sectional area of one rivet A r times the allowable shearing stress S s or . , L n A S 2 5 2 4 0 875 8500 51 112 pounds r s 2 # # # # # # π = = = ^ h b) The safe tensile load L based on bearing stress is equal to the number of rivets n times the projected bearing area of the rivet A b (diameter times thickness of plate) times the allowable bearing stress S c or L = n × A b × S c = 5 × (0.875 × 0.5) × 20,000 = 43,750 pounds. (Cover plates are not considered since their combined thickness is 1 ∕ 4 inch greater than the main plate thickness.) c) The safe tensile load L based on the tensile stress is equal to the net cross-sectional area of the plate between the two rivets in the outer row A p times the allowable tensile stress S t or L = A p × S t = 0.5[12 − 2(0.875 + 0.0625)] × 10,000 = 50,625 pounds. In completing the analysis, the sum of the load that would cause tearing between rivets in the three-hole section plus the load carried by the two rivets in the two-hole section is also investigated. The sum is necessary because if the joint is to fail, it must fail at both sections simultaneously. The least safe load that can be carried by the two rivets of the two-hole section is based on the bearing stress (see the foregoing calculations). 1) The safe tensile load L based on the bearing strength of two rivets of the two-hole sec tion is L = n × A b × S c = 2 × (0.875 × 0.5) × 20,000 = 17,500 pounds. 2) The safe tensile load L based on the tensile strength of the main plate between holes in the three-hole section is L × A p × S t = 0.5[12 − 3(0.875 + 0.0625)] × 10,000 = 45,938 pounds. The total safe tensile load based on this combination is 17,500 + 45,938 = 63,438 pounds, which is greater than any of the other results obtained. The safe tensile load for the joint would be the least of the loads just computed or 43,750 pounds and the efficiency η would be equal to this load divided by the tensile strength of the section of plate under consideration if it were unperforated or . , , . 05 12 10 000 43750 100 729 # # # η= = percent Formulas for Riveted Joint Design.— A riveted joint may fail by shearing through the rivets (single or double shear), crushing the rivets, tearing the plate between the rivets, crushing the plate or by a combination of two or more of the foregoing causes. Rivets placed too close to the edge of the plate may tear or shear the plate out to the edge but this type of failure is avoided by placing the center of the rivet 1.5 times the rivet diameter away from the edge. The efficiency of a riveted joint is equal to the strength of the joint divided by the strength of the unriveted plate, expressed as a percentage. In the following formulas, let d = diameter of holes t = thickness of plate t c = thickness of cover plates p = pitch of inner row of rivets P = pitch of outer row of rivets S s = shear stress for rivets S t = tensile stress for plates S c = compressive or bearing stress for rivets or plates
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