A cylindrical pressure vessel with a 1.5 m inside diameter is subjected to internal steam pressure of 1.5 MPa. It is made from steel plate by triple-riveted double-strap longitudinal butt joint with equal straps. The pitch of the rivets in the outer row is twice of the pitch of the rivets in the inner rows. The rivets are arranged in a zigzag pattern. The efficiency of the riveted joint should be at least 80%. The permissible stresses for the plate and rivets in tension, shear and compression are 80, 60 and 120 N/mm² respectively. Assume that the rivet in double shear is 1.875 times stronger than in single shear. Design the joint and calculate:
(i) thickness of the plate;
(ii) diameter of rivets;
(iii) pitch of rivets;
(iv) distance between the rows of rivets;
(v) margin;
(vi) thickness of the straps; and
(vii) efficiency of the joint.
Draw a neat sketch of the riveted joint showing calculated values of dimensions.
Question 8.21: A cylindrical pressure vessel with a 1.5 m inside diameter i...
\text { Given For vessel, } \quad D_{i}=1.5 m \quad P_{i}=1.5 MPa .
\eta=80 \% \quad \sigma_{t}=80 N / mm ^{2} \quad \tau=60 N / mm ^{2} .
\sigma_{c}=120 N / mm ^{2} .
The triple-riveted, double-strap butt joint with equal straps, is shown in Fig. 8.63.
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{1.5(1500)}{2(80)(0.8)}+2 .
= 19.58 or 20 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{20}=26.83 \text { or } 27 mm (ii).
Step III Pitch of rivets
The pitch of rivets is obtained by equating the shear strength of rivets with the tensile strength of 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. 8.63, the number of rivets per pitch length in various 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. It connects the inner strap, the shell plate and the outer strap. This results in double shear in the rivet in the outer row.
(b) Middle row There are two rivets per pitch length. They connect the inner strap, shell plate and outer strap. This results in double shear in the rivets in the middle row.
(c) Inner row There is one half-rivet on the left side, one complete rivet in the middle and one half-rivet on the right side. These make two rivets per pitch length. They connect the inner strap, shell plate and outer strap. This results in double shear in the rivets in the inner row.
Therefore, there are five rivets per pitch length for a triple-riveted butt joint. Since the straps are of equal width, all rivets are in double shear.
Therefore,
n_{1}=0 \quad n_{2}=5 .
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 5) \pi(27)^{2}(60)}{4(20)(80)}+27 .
= 228.29 or 230 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(27)=54 mm .
p_{\max }=C t+41.28=6(20)+41.28 .
= 161.28 mm
Since, p>p_{\max } .
it is assumed that,
p=p_{\max }=161.28 \text { or } 160 mm (iii).
The pitch of rivets in the outer row is 160 mm.
The pitch of rivets in the middle and inner rows is (160/2) or 80 mm.
Step IV Distance between rows of rivets
The number of rivets in the outer row is one-half of the number of rivets in the middle and inner rows.
Also, the rivets have a zig-zag pattern in middle and inner rows. From Eq. (8.52), the distance between the outer and middle rows is given by,
p_{t}=0.2 p+1.15 d (8.52).
p_{t}=0.2 p+1.15 d=0.2(160)+1.15(27) .
= 63.05 or 65 mm
From Eq. (8.53), the distance between middle and inner rows is given by,
p_{t}=0.165 p+0.67 d (8.53).
p_{t}=0.165 p+0.67 d=0.165(160)+0.67(27) .
= 44.49 or 50 mm (iv)
Step V Margin
From Eq. (8.54),
m=1.5 d (8.54).
m = 1.5d = 1.5 (27) = 40.5 or 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(20)\left[\frac{160-27}{160-2 \times 27}\right] .
= 15.68 or 16 mm (vi)
Step VII Efficiency of joint
The tensile strength of plate per pitch length in the outer row is given by,
P_{t}=(p-d) t \sigma_{t}=(160-27)(20)(80) .
= 212 800 N (a)
The shear strength of 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 5)\left[\frac{\pi}{4}(27)^{2}(60)\right] .
= 322 062.33 N (b)
The crushing strength of the plate is given by,
P_{c}=\left(n_{1}+n_{2}\right) d t \sigma_{c} .
= (0 + 5) (27) (20) (120)
= 324 000 N (c)
The tensile strength of the solid plate per pitch length is given by,
P=p t \sigma_{t}=160(20)(80)=256000 N (d).
From (a), (b), (c) and (d),
\eta=\frac{212800}{256000}=0.8313 \quad \text { or } \quad 83.13 \% (vii).
Figure 8.63 shows the sketch of longitudinal butt joint with calculated values of the dimensions.
