The Diameter and Pitch of Rivets


As for a single riveted seam, with the rivet in single shear, the shear strength at failure of a 20 mm diameter rivet is about 10 tonnes and, therefore, the shear strength of a rivet of any diameter is given by:


Fs = 10(d/20)2 If the shear strength of the steel plate is 0.028 t/mm2 tonnes (3) , then, with the rivet’s centreline 1.5 d


from the plate’s edge and the plate in double shear, the force required to shear the plate from the edge of the rivet hole to the plate’s edge in the wake of the rivet (the bearing surface) is given by: -


Fp = d/tp


= ≈


0.028 x 2 x dtp


The two forces are assumed to be equal, therefore: 10(d/20)2


0.028 x 2 x dtp 2 (4a)


That would mean that for a vessel plated in 6 mm thick mild steel, the rivets should be 12 mm in diameter and, when inspecting a riveted vessel, the marine surveyor should keep that fact in mind in those areas subject to the maximum shear forces. The length of the rivet before fitting is usually twice the diameter plus 30 if the hole is countersunk and 24 if not.


With double riveted seams the marine surveyor should consider a strip of a double zig-zag rivets shell plating butt as shown above in Figure 8 with a width equal to the rivet pitch p.


Then the resistance to tearing is: Rt


=


the resistance to shearing is: Rs


=


and the resistance to crushing is: - Rc


= (p – d)tp σt kg/cm2 (5) 2πd2σs/4 2dtp σc kg/cm2 (6) kg/cm2 (7) If the lap joint has n rows of rivets, the factor 2 in Formulae 7 should be changed to n.


A little thought would suggest that the three resistances given above in a good joint should be as near as possible equal. Realistically, however, it is only possible to equate Formulae (6) and (7). By equating those two Formulae to each other a useful relation for the diameter of the rivet to be used is obtained and that is:


dtpσc = in single shear d


and, in double shear d


d and = =


= =


1.274σc.tp 0.837σc Generally, τs may be taken as 60 N/mm2 .tp / τs /τs and σc 2.75tp 1.37tp as 130 N/mm2 mm mm giving:


in single shear mm in double shear mm


(10a) (10b)


As a common practice for plate thickness greater than 8 mm the diameter of rivet hole is often determined by Unwin’s formula which gives: -


d = tp½ mm 48 | The Report • June 2021 • Issue 96 (11) (9a) (9b) π/4 x d2τs Substituting the suggested relationships of σc and τs - given above that reduces to: (8) tonnes (4)


Page 1  |  Page 2  |  Page 3  |  Page 4  |  Page 5  |  Page 6  |  Page 7  |  Page 8  |  Page 9  |  Page 10  |  Page 11  |  Page 12  |  Page 13  |  Page 14  |  Page 15  |  Page 16  |  Page 17  |  Page 18  |  Page 19  |  Page 20  |  Page 21  |  Page 22  |  Page 23  |  Page 24  |  Page 25  |  Page 26  |  Page 27  |  Page 28  |  Page 29  |  Page 30  |  Page 31  |  Page 32  |  Page 33  |  Page 34  |  Page 35  |  Page 36  |  Page 37  |  Page 38  |  Page 39  |  Page 40  |  Page 41  |  Page 42  |  Page 43  |  Page 44  |  Page 45  |  Page 46  |  Page 47  |  Page 48  |  Page 49  |  Page 50  |  Page 51  |  Page 52  |  Page 53  |  Page 54  |  Page 55  |  Page 56  |  Page 57  |  Page 58  |  Page 59  |  Page 60  |  Page 61  |  Page 62  |  Page 63  |  Page 64  |  Page 65  |  Page 66  |  Page 67  |  Page 68  |  Page 69  |  Page 70  |  Page 71  |  Page 72  |  Page 73  |  Page 74  |  Page 75  |  Page 76  |  Page 77  |  Page 78  |  Page 79  |  Page 80  |  Page 81  |  Page 82  |  Page 83  |  Page 84  |  Page 85  |  Page 86  |  Page 87  |  Page 88  |  Page 89  |  Page 90  |  Page 91  |  Page 92  |  Page 93  |  Page 94  |  Page 95  |  Page 96  |  Page 97  |  Page 98  |  Page 99  |  Page 100  |  Page 101  |  Page 102  |  Page 103  |  Page 104  |  Page 105  |  Page 106  |  Page 107  |  Page 108  |  Page 109  |  Page 110  |  Page 111  |  Page 112  |  Page 113  |  Page 114  |  Page 115  |  Page 116  |  Page 117  |  Page 118  |  Page 119  |  Page 120  |  Page 121  |  Page 122  |  Page 123  |  Page 124  |  Page 125  |  Page 126  |  Page 127  |  Page 128  |  Page 129  |  Page 130  |  Page 131  |  Page 132  |  Page 133  |  Page 134  |  Page 135  |  Page 136