Given w = 250 mm t = 20 mm.
\sigma_{t}=80 N / mm ^{2} \quad \tau=60 N / mm ^{2} \quad \sigma_{c}=120 N / mm ^{2} .
Step I Diameter of rivets
The diameter of the rivet is given by,
d=6 \sqrt{t}=6 \sqrt{20}=26.83 \quad \text { or } \quad 27 mm (i).
Step II Number of rivets
The shear resistance of one rivet in double shear is given by,
P_{s}=1.875\left[\frac{\pi}{4} d^{2} \tau\right] .
=1.875\left[\frac{\pi}{4}(27)^{2}(60)\right]=64412.47 N (a).
Crushing resistance of one rivet is given by,
P_{c}=d t \sigma_{c}=27(20)(120)=64800 N (b).
\text { From (a) and (b), } P_{s}<P_{c} .
Therefore, shear strength of the rivet is the criterion of design. It is assumed that rivets are so arranged that there is only one rivet in the outer row. The tensile strength of the plate in the outer row is given by,
P_{t}=(w-d) t \sigma_{t}=(250-27)(20)(80) .
= 356 800 N (c)
Equating the strength of the plate with the shear strength of n rivets
356 800 = n (64 412.47)
∴ n = 5.54 or 6 rivets (ii).
Step III Arrangement of rivets
The arrangement of six rivets in three rows is illustrated in Fig. 8.61. This type of arrangement in diamond shape is called Lozenge joint. The dimensions given in the figure have the following numerical values,
m = 1.5d = 1.5 (27) = 40.5 or 45 mm
p_{t}=2 d=2(27)=54 \text { or } 55 mm .
t_{1}=0.625 t=0.625(20)=12.5 mm .
From the figure,
m + p + p + m = 250 or 45 + 2p + 45 = 250
∴ p = 80 mm.
Step IV Efficiency of joint
In order to calculate the efficiency of the joint, we will proceed from the outer row of rivets to the inner row and fi nd out the weakest cross-section of the joint. Let us consider the failure behaviour of three cross-sections denoted by AA, BB and CC.
(i) The joint can fracture along the section-AA without affecting the rivets in the middle and inner rows.
(ii) The joint cannot fracture along the section- BB without shearing one rivet in double shear. This is the rivet in the outer row.
(iii) The joint cannot fracture along the section- CC, without shearing three rivets in double shear. These are the rivets in outer and middle rows.
Based on these assumptions the strength of the joint along three sections is calculated.
Along the section-AA,
\text { strength }=(w-d) t \sigma_{t}=(250-27)(20)(80) .
356 800 N (a)
Along the section-BB,
\text { strength }=(w-2 d) t \sigma_{t}+1 \times P_{s} .
= (250 – 2 × 27) (20) (80) + 64 412.47
= 378 012.47 N (b).
Along the section-CC,
\text { strength }=(w-3 d) t \sigma_{t}+3 \times P_{s} .
= (250 – 3 × 27)(20)(80)
+ 3 × 64 412.47
= 463 637.41 N (c).
\text { Shear resistance of all rivets }=6 \times P_{s} .
= 6 × 64 412.47 = 386 474.8 N (d)
From (a), (b), (c), and (d), the lowest strength of the joint is along the section-AA.
\text { Strength of solid plate }=w t \sigma_{t}=250(20)(80) .
= 400 000 N
Therefore
\eta=\frac{356800}{400000}=0.892 \quad \text { or } \quad 89.2 \% (iv).
