Ps
= =
τs
πdh2
/4 nτs πdh2 /4
N/mm2
We may also write if n is the number of rivets per pitch length: Ps
N/mm2
(24) (25)
If the rivet is in double shear as in Figure 11 (b) the effective area over which failure occurs is twice that as in Figure 11 (a) and the allowable stress in double shear is 1.75 times that in single shear. Hence in double shear:
Ps Ps = ≥ = 1.75τsπdh2/4
Failure will not occur in either case in this mode if: - Ps
≥ P Rivet Head P P/2 P P P/2 Rivet po (a) Single Shear Figure 12 Rivet Failure in Mode 2 Figure 11 Rivet Failure in Mode 2 Failure Mode 3 - Crushing of Both Plate and Rivet
Due to the rivet being compressed against the inner surface of the hole, there is the possibility that either the rivet or surface of the hole may be crushed. The area, which resists this action, is the projected area of hole or rivet on a diametric plane. The area per rivet is:
Ac where Ac
= =
dh tp mm2 area of rivet resisting crushing
If the allowable crushing or bearing stress of rivet or plate is σc in tones per square mm the crushing strength of the joint or load carrying capacity of the joint against crushing is: -
Pc = ≥ σc dh tp
Then failure in this mode will not occur if: Pc
P N (29) (30)
where P is applied as a load per pitch length and there is one rivet per pitch. If the number of rivets in a pitch length is n then the right hand side in Formula (29) should be multiplied by n. Thus, this type of failure is sometimes found at the vessel’s quarter lengths and at about half her moulded depth. It is found by tap testing the rivet heads on both sides of the vessel.
tp P P Figure 12 Rivet Failure Mode 3 Figure 12 Rivet Failure Mode 3 Figure 13 Rivet Failure Mode 4 Figure 14 Rivet Failure Mode 4
The modes of failure discussed above are primary in nature and in certain cases they have to be considered uniquely. One such case is when rivets are arranged in lozenge or diamond shape. In writing down the above equations for strength of the joint certain assumptions have been made and it is worthwhile to know and remember them (see below). Most importantly it should be remembered that most direct stresses have been assumed to be induced in rivet and plate which may not be the case.
However, ignorance of the actual state of the stress and its replacement by the most direct stress is compensated for by lowering the allowable values of stresses σt
, τs and σc, i.e. by increasing factor of
safety or, as the more cynical call it, the factor of ignorance. As the rivets are driven hot, it is customary, when the holes are drilled to multiply the diameter used in the above calculation by 1.05 and, if the holes are punched by 1.06. A multiplier of 1.25 is also used to allow for the countersink.
...Continued in Issue 97.
(28)
Failure in this case will not occur if: - Pes
(b) Double Shear ≥ P e
Failure in this case will not occur if: Pes
≥ P 1.75τs πdh2 /4
Failure will not occur in either case in this mode if: Ps
P N/mm2 N/mm2 N N (26) (26) (27) (27)
Failure Mode 4 - Shearing of the Plate Edge near the Rivet Hole
Figure 13 shows this mode of failure which can take place in both either or both inner and outer plate and in which the plate edge can shear along planes marked in red. If the length of the edge set is e, then the area resisting this failure is:
Aes
where Aes
= = 2etp mm2 (31) the area resisting such shear
If the allowable shearing stress of the plate is τs in newtons per mm2 then the load carrying capacity of the joint against the shearing of the edge is:
Pes = 2et.τs N/mm2 (32)
The Report • June 2021 • Issue 96 | 51
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