At the end of this lesson, the students should be able to understand:

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Intructional Objective: At the end of thi leon, the tudent hould be able to undertand: Baic failure mechanim of riveted joint. Concept of deign of a riveted joint. 1. Strength of riveted joint: Strength of a riveted joint i evaluated taking all poible failure path in the joint into account. Since rivet are arranged in a periodic manner, the trength of joint i uually calculated conidering one pitch length of the plate. There are four poible way a ingle rivet joint may fail. a) Tearing of the plate: If the force i too large, the plate may fail in tenion along the row (ee figure 10..1). The maximum force allowed in thi cae i 1 = t ( p d) t where t = allowable tenile tre of the plate material p = pitch d = diameter of the rivet hole t = thickne of the plate Failure path in tenion Figure 10..1: Failure of plate in tenion (tearing) Verion ME, IIT Kharagpur

b) Shearing of the rivet: The rivet may hear a hown in figure 10... The maximum force withtood by the joint to prevent thi failure i = ( for lap joint, ingle trap butt joint d ) where = ( d ) for double trap butt joint =allowable hear tre of the rivet material. Figure 10..: Failure of a rivet by hearing c) Cruhing of rivet: If the bearing tre on the rivet i too large the contact urface between the rivet and the plate may get damaged. (ee figure 10..3). With a imple aumption of uniform contact tre the maximum force allowed i = dt 3 c where c =allowable bearing tre between the rivet and plate material. Figure 10..3: Failure of rivet by Verion ME, IIT Kharagpur

d) Tearing of the plate at edge: If the margin i too mall, the plate may fail a hown in figure 10... To prevent the failure a minimum margin of m= 1.5d i uually provided. Figure 10..: Tearing of the plate at the edge. Efficiency: Efficiency of the ingle riveted joint can be obtained a ratio between the maximum of, and and the load carried by a olid plate which i pt t. Thu 1 3 min{ 1,, 3} efficiency ( η)= pt t In a double or triple riveted joint the failure mechanim may be more than thoe dicued above. The failure of plate along the outer row may occur in the ame way a above. However, in addition the inner row may fail. For example, in a double riveted joint, the plate may fail along the econd row. But in order to do that the rivet in the firt row mut fail either by hear or by cruhing. Thu the maximum allowable load uch that the plate doe not tear in the econd row i = ( p d) t+ min{, } 3. t Further, the joint may fail by (i) hearing of rivet in both row (ii) cruhing of rivet in both row (iii) hearing of rivet in one row and cruhing in the other row. Verion ME, IIT Kharagpur

The efficiency hould be calculated taking all poible failure mechanim into conideration. 3. Deign of rivet joint: The deign parameter in a riveted joint are d, p and m Diameter of the hole ( d ): When thickne of the plate ( t ) i more than 8 mm, Unwin formula i ued, Otherwie d = 6 t mm. d i obtained by equating cruhing trength to the hear trength of the joint. In a double riveted zigzag joint, thi implie However, d t c = d (valid for t < 8 mm) hould not be le than t, in any cae. The tandard ize of i tabulated in code IS: 198-1961. itch ( p ): itch i deigned by equating the tearing trength of the plate to the hear trength of the rivet. In a double riveted lap joint, thi take the following form. ( ) ( t p d t = d ) But p d in order to accommodate head of the rivet. Margin ( m ): m= 1.5d. In order to deign boiler joint, a deigner mut alo comply with Indian Boiler Regulation (I.B.R.). p : uually 0.33p + 0.67d mm) ( b d Review quetion and anwer: Q. 1. Two plate of 7 mm thickne are connected by a double riveted lap joint of zigzag pattern. Calculate rivet diameter, rivet pitch and ditance between row of rivet for the joint. Aume = 90Ma, = 60Ma, = 10Ma. t c An. Since t = 7mm< 8mm, the diameter of the rivet hole i elected equating hear trength to the cruhing trength, i.e., Verion ME, IIT Kharagpur

d = dt c yielding d = 17.8mm. According to IS code, the tandard ize i d = 19 mm and the correponding rivet diameter i 18 mm. itch i obtained from the following ( ) = (, where d = 19 mm t p d t d ) p = 5 + 19 = 73mm [Note: If the joint i to comply with I.B.R. pecification, then pmax = ct. + 1.8 mm, where c i a contant depending upon the type of joint and i tabulated in the code.] The ditance between the two rivet row i p d p = + d = 37 mm. 3 3 Q.. A triple riveted butt joint with two unequal cover plate join two 5 mm plate a hown in the figure below. Figure: 10..5 The rivet arrangement i zigzag and the detail are given below: itch = cm in outer row and 11 cm in inner row, Rivet diameter = 33 mm Calculate the efficiency of the joint when the allowable tree are 75 Ma, 60 Ma and 15 Ma in tenion, hear and cruhing, repectively. Verion ME, IIT Kharagpur

An. From code it may be een that the correponding rivet hole diameter i 3.5 mm. To find trength of the joint all poible failure mechanim are to be conidered eparately. kn (a) Tearing reitance of the plate in outer row: = ( p d ) t = (0-3.5) X 5 X 75 = 37.81 1 hole T (b) Shearing reitance of the rivet: = d S + d S = 61.86 kn Note that within a pitch length of cm four rivet are in double hear while one rivet in ingle hear. row (c) Cruhing reitance of the rivet 3 = 5 d t C = 515.6 kn (d) Shear failure of the outer row and tearing of the rivet in the econd = ( p d hole ) tt + d S = 33. kn Note that in econd row there are rivet per pitch length and the rivet in outer row undergoe ingle hear. There are other mechanim of failure of the joint e.g. tearing along the innermot row and hearing or cruhing of rivet in other two row etc., but all of them will have higher reitance than thoe conidered above. Hence the efficiency of the joint i min{ 1,, 3, } η = = 0.8108 pt T or when expreed in percentile 81.08 %. Q.3. How i a rivet joint of uniform trength deigned? An. The procedure by which uniform trength in a riveted joint i obtained i known a diamond riveting, whereby the number of rivet i increaed Verion ME, IIT Kharagpur

progreively from the outermot row to the innermot row (ee figure below). A common joint, where thi type of riveting i done, i Lozenge joint ued for roof, bridge work etc. Figure 10..6: Diamond riveting in tructural joint Q.. Two mild teel tie rod having width 00 mm and thickne 1.5 mm are to be connected by mean of a butt joint with double cover plate. Find the number of rivet needed if the permiible tree are 80 Ma in tenion, 65 Ma in hear and 160 Ma in cruhing. An. A dicued earlier for a tructural member Lozenge joint i ued which ha one rivet in the outer row. The number of rivet can be obtained equating the tearing trength to the hear or cruhing trength of the joint, i.e., from the equation b d t = n d [Double hear] ( ) T 1( ) or ( b d) tt = n ( dt) c where b and t are the width and thickne of the plate to be joined. In the problem b = 00 mm, t = 1.5mm, = 80Ma, = 160Ma, T c = 65Ma and d i obtained from Unwin formula d = 6 t mm= 1.mm. According to IS code, the tandard rivet hole diameter i 1.5 mm and correponding rivet diameter i 0 mm. The number of rivet required i the minimum of the number calculated from the above two expreion. It may be checked that i found out to be 3.89 while n i.16. Therefore, at leat 5 rivet n1 are needed. Verion ME, IIT Kharagpur