A cylindrical steam pressure vessel of 1 m inside diameter is subjected to an internal pressure of 2.5 MPa. Design a double-riveted, double-strap longitudinal butt joint for the vessel. The straps are of equal width. The pitch of the rivets in the outer row should be twice of the pitch of the

Question

A cylindrical steam pressure vessel of 1 m inside diameter is subjected to an internal pressure of 2.5 MPa. Design a double-riveted, double-strap longitudinal butt joint for the vessel. The straps are of equal width. The pitch of the rivets in the outer row should be twice of the pitch of the rivets in the inner row. A zigzag pattern is used for rivets in inner and outer rows.
The efficiency of the riveted joint should be at least 70%. The permissible tensile stress for the steel plate of the pressure vessel is 80 N/mm². The permissible shear stress for the rivet material is 60 N/mm². Assume that the rivets in double shear are 1.875 times stronger than in single shear and the joint do not fail by crushing. Calculate:

(i) thickness of the plate;
(ii) diameter of the rivets;
(iii) pitch of the rivets;
(iv) distance between inner and outer rows of the rivets;
(v) margin;
(vi) thickness of the straps; and
(vii) efficiency of the joint.
Make a neat sketch of the joint showing all calculated values of dimensions.

Accepted Answer

[latex] ext { Given For vessel, } D_{i}=1 m \quad P_{i}=2.5 MPa [/latex].

[latex] \eta=70 \% \quad \sigma_{t}=80 N / mm ^{2} \quad au=60 N / mm ^{2} [/latex].

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),

[latex] t=\frac{P_{i} D_{i}}{2 \sigma_{t} \eta}+ CA [/latex] (8.43).

[latex] t=\frac{P_{i} D_{i}}{2 \sigma_{t} \eta}+ CA =\frac{2.5(1000)}{2(80)(0.7)}+2 [/latex].

= 24.32 or 25 mm (i)
Step II Diameter of rivets
t > 8 mm
From Eq. (8.45),

[latex] d=6 \sqrt{t} [/latex] (8.45).

[latex] d=6 \sqrt{t}=6 \sqrt{25}=30 mm [/latex] (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),

[latex] p=\frac{\left(n_{1}+1.875 n_{2} ight) \pi d^{2} au}{4 t \sigma_{t}}+d [/latex] (8.47).

[latex] p=\frac{\left(n_{1}+1.875 n_{2} ight) \pi d^{2} au}{4 t \sigma_{t}}+d [/latex].

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,

[latex] n_{1}=0 \quad n_{2}=3 [/latex].

Substituting values in Eq. (8.47),

[latex] p=\frac{\left(n_{1}+1.875 n_{2} ight) \pi d^{2} au}{4 t \sigma_{t}}+d [/latex] (8.47).

[latex] p=\frac{(0+1.875 imes 3) \pi(30)^{2}(60)}{4(25)(80)}+30 [/latex].

= 149.28 or 150 mm
From Eqs (8.48) and (8.49),

[latex] p_{\min .}=2 d [/latex] (8.48).

[latex] p_{\max }=C t+41.28 [/latex] (8.49).

[latex] p_{\min .}=2 d=2(30)=60 mm [/latex].

[latex] p_{\max }=C t+41.28=4.63(25)+41.28 [/latex].

= 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,

[latex] p_{t}=0.2 p+1.15 d [/latex] (8.52).

[latex] p_{t}=0.2 p+1.15 d=0.2(150)+1.15(30) [/latex].

= 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),

[latex] t_{1}=0.625 t\left[\frac{p-d}{p-2 d} ight] [/latex] (8.57).

[latex] t_{1}=0.625 t\left[\frac{p-d}{p-2 d} ight] [/latex].

[latex] =0.625(25)\left[\frac{150-30}{150-2 imes 30} ight] [/latex].

= 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,

[latex] P_{t}=(p-d) t \sigma_{t}=(150-30)(25)(80) [/latex].

= 240 000 N (a)
The shear strength of the rivets per pitch length is given by,

[latex] P_{s}=\left(n_{1}+1.875 n_{2} ight)\left[\frac{\pi}{4} d^{2} au ight] [/latex].

[latex] =(0+1.875 imes 3)\left[\frac{\pi}{4}(30)^{2}(60) ight] [/latex].

= 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,

[latex] P=p t \sigma_{t}=150(25)(80)=300000 N [/latex] (c)
From (a), (b), and (c),

[latex] \eta=\frac{238564.69}{300000}=0.7952 \quad ext { or } \quad 79.52 \% [/latex] (vii).

Figure 8.64 shows the sketch of the longitudinal butt joint with calculated values of dimensions.

Question 8.22: A cylindrical steam pressure vessel of 1 m inside diameter i...

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