\text { Given For vessel, } D_{i}=1 m \quad P_{i}=2.5 MPa .
\eta=70 \% \quad \sigma_{t}=80 N / mm ^{2} \quad \tau=60 N / mm ^{2} .
A double-riveted double-strap butt joint, with equal straps, is shown in Fig. 8.64.
Step I Thickness of plate From Eq. (8.43),
t=\frac{P_{i} D_{i}}{2 \sigma_{t} \eta}+ CA (8.43).
t=\frac{P_{i} D_{i}}{2 \sigma_{t} \eta}+ CA =\frac{2.5(1000)}{2(80)(0.7)}+2 .
= 24.32 or 25 mm (i)
Step II Diameter of rivets
t > 8 mm
From Eq. (8.45),
d=6 \sqrt{t} (8.45).
d=6 \sqrt{t}=6 \sqrt{25}=30 mm (ii).
Step III Pitch of rivets
The pitch of rivets is obtained by equating the shear strength of the rivets to the tensile strength of the plate. From Eq. (8.47),
p=\frac{\left(n_{1}+1.875 n_{2}\right) \pi d^{2} \tau}{4 t \sigma_{t}}+d (8.47).
p=\frac{\left(n_{1}+1.875 n_{2}\right) \pi d^{2} \tau}{4 t \sigma_{t}}+d .
As shown in Fig. 8.64, the number of rivets per pitch length in two rows are as follows:
(a) Outer row There is one half-rivet on the left side and one half-rivet on the right side.
These make one rivet per pitch length.
(b) Inner row There are two rivets per pitch length.
Adding these values, the total number of rivets per pitch length is 3. Since the straps are of equal width, all rivets connect the inner strap, the shell plate and the outer strap. This results in double shear in all the rivets. Therefore,
n_{1}=0 \quad n_{2}=3 .
Substituting values in Eq. (8.47),
p=\frac{\left(n_{1}+1.875 n_{2}\right) \pi d^{2} \tau}{4 t \sigma_{t}}+d (8.47).
p=\frac{(0+1.875 \times 3) \pi(30)^{2}(60)}{4(25)(80)}+30 .
= 149.28 or 150 mm
From Eqs (8.48) and (8.49),
p_{\min .}=2 d (8.48).
p_{\max }=C t+41.28 (8.49).
p_{\min .}=2 d=2(30)=60 mm .
p_{\max }=C t+41.28=4.63(25)+41.28 .
= 157.03 mm
The pitch of 150 mm is within the limits from 60 mm to 157.03 mm. Therefore,
p = 150 mm (iii)
The pitch of rivets in the outer row is 150 mm.
The pitch of rivets in the inner row is (150/2) or 75 mm.
Step IV Distance between inner and outer rows
The number of rivets in the outer row is one-half of the number of rivets in the inner row. Also, the rivets are arranged in a zig-zag pattern. From Eq. (8.52), the distance between inner and outer rows is given by,
p_{t}=0.2 p+1.15 d (8.52).
p_{t}=0.2 p+1.15 d=0.2(150)+1.15(30) .
= 64.5 or 65 mm (iv)
Step V Margin
From Eq. (8.54),
m = 1.5d (8.54).
m = 1.5d = 1.5(30) = 45 mm (v)
Step VI Thickness of straps
The straps are of equal width and every alternate rivet in the outer row is omitted. From Eq. (8.57),
t_{1}=0.625 t\left[\frac{p-d}{p-2 d}\right] (8.57).
t_{1}=0.625 t\left[\frac{p-d}{p-2 d}\right] .
=0.625(25)\left[\frac{150-30}{150-2 \times 30}\right] .
= 20.83 or 21 mm (vi).
Step VII Efficiency of joint
The tensile strength of the plate per pitch length in the outer row of rivets is given by,
P_{t}=(p-d) t \sigma_{t}=(150-30)(25)(80) .
= 240 000 N (a)
The shear strength of the rivets per pitch length is given by,
P_{s}=\left(n_{1}+1.875 n_{2}\right)\left[\frac{\pi}{4} d^{2} \tau\right] .
=(0+1.875 \times 3)\left[\frac{\pi}{4}(30)^{2}(60)\right] .
= 238 564. 69 N (b)
It is assumed that the joint does not fail in crushing. Also, the tensile strength of the solid plate per pitch length is given by,
P=p t \sigma_{t}=150(25)(80)=300000 N (c)
From (a), (b), and (c),
\eta=\frac{238564.69}{300000}=0.7952 \quad \text { or } \quad 79.52 \% (vii).
Figure 8.64 shows the sketch of the longitudinal butt joint with calculated values of dimensions.
