This document provides information about riveted joints, including definitions of common riveted joint types like lap joints and butt joints. It describes important terminology used in riveted joints like pitch, back pitch, and margin. Potential failure modes of riveted joints like tearing of plates, shearing of rivets, and crushing are explained. The document also discusses the efficiency of riveted joints and provides steps for designing longitudinal butt joints for boilers according to Indian Boiler Regulations, including how to determine rivet diameter, pitch, row spacing, and strap thickness. Eccentrically loaded riveted joints are also addressed.

1. RIVETED JOINT
 Krushal kakadiya
 Dhruvit Kardani
 Tirth Lad
 Milan Lakhani
 Vishal Lapsiwala

2. INTRODUCTION
A rivet is a short cylindrical bar with head integral to it. The cylindrical
portion of the rivet is called shank or body and lower portion of shank
is known as tail, as shown in Fig. The rivets are used to make
permanent fastening between the plates such as in structural work,
ship building, bridges, tanks and shells. The riveted joints are widely
used for joining light metal. It’s permanent joint.
Head
Shank or
body
Ta
il
Fig. Rivet
Joint

3. Types of
Riveted Joint
Lap Joint Butt Joint

4. (a) Lap Joint (b) Butt Joint
• Lap Joint: A lap joint is that in which one plate overlaps the other and
the two plates are then riveted together.
• Butt Joint: A butt joint is that in which the main plates are kept in
alignment butting (i.e. touching) each other and a cover plate (i.e.
strap) is placed either on one side or on both sides of the main plates.
The cover plate is then riveted together with the main plates.

5. Important Terminology Used in Riveted Joints
Pitch. It is the distance from the centre of one rivet
to the centre of the next rivet measured parallel to
the seam as shown in Fig. It is usually denoted by p.
Back pitch. It is the perpendicular distance between
the centre lines of the successive rows as shown in
Fig. It is usually denoted by pb.
Diagonal pitch. It is the distance between the
centres of the rivets in adjacent rows of zig-zag
riveted joint as shown in Fig. It is usually denoted by
pd.
Margin or marginal pitch. It is the distance
between the centre of rivet hole to the nearest edge
of the plate as shown in Fig. It is usually denoted by
m.

6. Failures of a Riveted Joint
Tearing of
the plate at
an edge.
Tearing of
the plate
across a row
of rivets.
Shearing of
the rivets.
Crushing of
the plate or
rivets.

7. Tearing of the plate at an edge.
• A joint may fail due to tearing of the plate at an edge as shown in Fig.
This can be avoided by keeping the margin, m = 1.5d, where d is the
diameter of the rivet hole.
d
m

8. Tearing of the plate across a row of rivets.
• Tearing of the plate across a row of rivets. Due to the tensile
stresses in the main plates, the main plate or cover plates may tear off
across a row of rivets as shown in Fig. 9.14. In such cases, we consider
only one pitch length of the plate, since every rivet is responsible for
that much length of the plate only.
The resistance offered by the plate against tearing is known as
tearing resistance or tearing strength or tearing value of the plate.

9. Let p = Pitch of the rivets,
d = Diameter of the rivet hole,
t = Thickness of the plate, and
σt = Permissible tensile stress for the plate material.
Tearing area per pitch length, At = ( p – d ) t
∴ Tearing resistance or pull required to tear off the plate per pitch
length, Pt = At.σ t = (p – d)t.σ t
d
d
p

10. Shearing of the rivets.
• The plates which are connected by the rivets exert tensile stress on the
rivets, and if the rivets are unable to resist the stress, they are sheared
off as shown in Fig.

11. Let d = Diameter of the rivet hole,
τ = Safe permissible shear stress for the rivet material, and
n = Number of rivets per pitch length.
Shearing area, As =
𝜋
4
× d 2 ...(In single shear)
= 1.875 ×
𝜋
4
× d 2 ...(In double shear, IBR)
∴ Shearing resistance or pull required to shear off the rivet per pitch length,Ps
= n ×
𝜋
4
× d 2 × τ ...(In single shear)
= n × 1.875 ×
𝜋
4
× d2 × τ ...(In double shear, IBR)

12. Crushing of the plate or rivets.
• Sometimes, the rivets do not actually shear off under the tensile stress,
but are crushed as shown in Fig. 9.17. Due to this, the rivet hole
becomes of an oval shape and hence the joint becomes loose. The
failure of rivets in such a manner is also known as bearing failure. The
area which resists this action is the projected area of the hole or rivet
on diametric plane.
P P

13. Let d = Diameter of the rivet hole,
t = Thickness of the plate,
σc = Safe permissible crushing stress for the rivet or plate material
n = Number of rivets per pitch length under crushing.
Crushing area per rivet (i.e. projected area per rivet), Ac = d.t
Crushing resistance or pull required to crush the rivet per pitch length
(for total n no. of rivets),
Pc = n.d.t.σc

14. Efficiency of a Riveted Joint
• The efficiency of a riveted joint is defined as the ratio of the strength of
riveted joint to the strength of the un-riveted or solid plate.
η =
Least of Pt , Ps and Pc
p × t × σt
Where, p = Pitch of the rivets,
t = Thickness of the plate,
σt = Permissible tensile stress of the plate material.

15. Design of Longitudinal Butt Joint for a Boiler
• According to Indian Boiler Regulations (I.B.R), the following procedure should be
adopted for the design of longitudinal butt joint for a boiler.
1. Thickness of boiler shell. First of all, the thickness of the boiler shell is determined by
using the thin cylindrical formula, i.e.
t =
P. D
2 .σt..
η
l
+ 1 mm as corrosion allowance
2.Diameter of rivets.
(A) The diameter of the rivet hole (d) may be determined by using Unwin's empirical
formula, i.e.
d = 6√t …(when t is greater than 8 mm)
(B) Otherwise find diameter from relation Ps = Pc …(Refer to slide 8)

16. 3. Pitch of rivets. The pitch of the rivets is obtained by equating the tearing resistance
of the plate to the shearing resistance of the rivets. It may noted that
(a) The pitch of the rivets should not be less than 2d, which is necessary for the
formation of head.
(b) The maximum value of the pitch of rivets for a longitudinal joint of a boiler as per
I.B.R. is pmax = C × t + 41.28 mm
Use the minimum value from (a) and (b)
No.of rivet per pitch Lap Joint
Butt Joint
Single Double
1 1.31 1.53 1.75
2 2.62 3.06 3.5
3 3.47 4.05 4.63
4 4.17 - 5.52
5 - - 6
Value of C

17. 4. Distance between the rows of rivets. The distance between the rows of rivets as
specified by Indian Boiler Regulations is as follows :
For equal number of rivets in more than one row for lap joint or butt joint, the distance
between the rows of rivets ( pb) should not be less than
0.33 p + 0.67 d, for zig-zig riveting, and
2 d, for chain riveting.
5. Marginal distance.
m = 1.5 d

18. 6. Thickness of the butt strap t1 / t2
If t1 = t2 then, t1 = 1.125 t (single butt strap)
t1 = 0.625 t (double butt strap)
If t1 ≠ t2 then, t1 = 0.75 t
t2 = 0.625 t
where t is thickness of the boiler shell

19. Eccentric Loaded Riveted Joint
Let P = Eccentric load on the joint, and
d = Eccentricity of the load i.e. the distance between the line of action
of the load and the centroid of the rivet system i.e. G.
Step 1: First of all, find the centre of gravity G of the rivet
system.
𝑥 =
𝑥1
+𝑥2
+ …
𝑛
; 𝑦 =
𝑦1
+𝑦2
+ …
𝑛

20. Step 2 : Introduce two forces P1 and P2 at the centre of gravity ‘G’ of the rivet
system. These forces are equal and opposite to P as shown in Fig.

21. Step 3 : Assuming that all the rivets are of the same size, the effect of P1 = P is
to produce direct shear load on each rivet of equal magnitude. Therefore, direct
shear load on each rivet,
Ps =
𝑃
𝑛
acting parallel to the load P.

22. Step 4 : Equating the moment then,
P.e = F1.l1 + F2.l2 + F3.l3 + …
=
𝐹1
𝑙1
𝑙1
2 + 𝑙2
2 + …
From the relation,
𝐹1
𝑙1
=
𝐹2
𝑙2
=
𝐹3
𝑙3
= …
From the above two relation find F1, F2, F3, …

23. Step 5: The primary (or direct) and secondary shear load may be added
vectorially to determine the resultant shear load (R) on each rivet as shown in
Fig.
R = 𝑃 𝑠
2 + 𝐹2 + 2. 𝑃𝑠. 𝐹. cos 𝜃
Where θ = Angle between the primary or direct shear
load (Ps) and secondary shear load (F).

24. Then, Maximum resultant shear load (R) =
π
4
× d 2 × τ
From the above relation, we get the value of d in mm.

25. REFERENCE
 Machine Design by R.S Khurmi and Jk Gupta

26. THANK YOU