2. Riveting applications
Rivets
Types of riveted joints
Failure of riveted joints
3. Riveting applications:
Pressure vessels, boilers
tanks
Bridges
Hulls of ships
Airplanes
Cranes
buildings
Machinery in general
4. Rivets
head
A rivet is a round bar consisting of
head
shank.
The rivet blank is heated to a red glow
Inserted into the holes; Shank
The head is held firmly against the Point
plate
The projecting end is formed into a
second head, called the point,
8. Failure of riveted joints
Bending of rivet or plate
F F
•In Lap connections the offset creates a moment
• M=Ft/2
•Bending moment causes complex deformations
and stresses
•In most cases this offset moment is neglected
•A suitable factor of safety is used.
9. b) Shearing of the rivets:
Single shear
FF
F
Double shear
F/2
F
F/2
10. Joint strength in shear
2
πd S s
Fs (2 n 2 n1 )
4
Where,
n1: :number of rivets in single shear
n2: number of rivets in double shear
Ss: allowable shear stress
d : diameter of rivet
11. c) Crushing of the rivets or the plates
Crushing of margin
F F
Crushing of the rivet or plate occurs due to the pressure
on the cylindrical surface of the rivet and the plate
The resistance to crushing of rivets is,
Fc (n 2 h 1 n 1 h 2 ) dS c
where
h1 h2: plate thickness
Sc: allowable crushing stress
12. d) Rupture of plate by tension:
F L F
Rupture of plate occurs at the section between the
rivets
The resistance of rupture can be obtained from the
expression:
Where: Ft ( L nd ) hS t
L: plate width
h: plate thickness:
St : allowable tensile stress
n : number of rivet holes at the section
13. Undrilled
Section
F F
L
In the undrilled section:
Ft LhS t
Where
L : width of plate
14. d) Tearing and shearing of the margin
Tearing of the margin
F F
Margin
Shearing of the margin
F F
For the riveted joint to resist tearing and
shearing of the margin, the margin (m) =
1.5 d for double shear
2d for single shear.
15. Example:
Fig.(3) shows a lap riveted joint, consists of two Rolled steel plates,
SAE 1020, of 0.5 in thickness. The plates are riveted together with
four rivets 0.375 inch in diameter of low carbon steel, SAE 1010.
Estimate the maximum value of the force F that the joint can stand
while considering a factor of safety equals 2 and the rivets are driven
by hand hammer. 0.5 in
F
F
4 in. 2
1
1
16. Solution
Shearing of the rivets:
d = 0.375
fs = 2
π
2
n1 = 4 F (2 n2 n1)
d Ss
s
n2 =0 4 fs
π 0 .375 Rivet-driving
2
Load carrying member Type of stress 10000 Rivets acting in Rivets acting in
Fs 4 power single shear double shear
Rolled steel, SAE 1020 Tension 4 2 ….. 18000 18000
Shear Power 13500 13500
Rivets, SAE1010
FS = Shear lb
2209 Hand 10000 10000
Crushing Power 24000 30000
Crushing Hand 16000 20000
17. Crushing of the rivets:
(n 2 h1 n 1 h 2 ) dS c
Fc
fs
Load carrying member Type of stress
4 0 . 375 Rivet-driving Rivets acting in
0 . 5 16000 shear Rivets acting in
Fc power single double shear
Rolled steel, SAE 1020 Tension
Shear
2 …..
Power
18000
13500
18000
13500
Shear Hand 10000 10000
Rivets, SAE1010
Crushing Power 24000 30000
F = 6000 lb
C Crushing Hand 16000 20000
18. Rupture of plate by tension
(L nd ) hS t
Ft
fs
Load carrying member Type of stress Rivet-driving Rivets acting in Rivets acting in
(4 2 0 . 375 ) 0 . 5
power
18000
single shear double shear
Rolled steel, SAE 1020
Ft Tension ….. 18000 18000
Shear 2
Power 13500 13500
Shear Hand 10000 10000
Rivets, SAE1010
Ft = 14625 lb
Crushing
Crushing
Power
Hand
24000
16000
30000
20000
19. Crushing of the plates
(n 2 h1 n 1 h 2 ) dS c
Fc
fs
Load carrying member Type of stress Rivet-driving Rivets acting in Rivets acting in
4 0 .5 0 . 375
power 36000
single shear double shear
Rolled steel, SAE 1020 Fc Tension ….. 18000 18000
Shear 2
Power 13500 13500
Shear Hand 10000 10000
Rivets, SAE1010
FC = 13500 lb
Crushing
Crushing
Power
Hand
24000
16000
30000
20000
Maximum value of F = 2209 lb
20. Design procedure for structural joints
The following sequence of steps applies to the calculations
for structural joints:
• The load on each member is determined analytically or
graphically
• The shape and size of each member is determined based
on the load
• The diameter of the rivets is determined by the thickness
of the structural shapes apply the equation:
1
d 2h 16
21. •The number of rivets required is based upon the
shearing or crushing stress which ever determine the
cause of failure.
•The rivets in the joint are spaced in order to utilize the
material economically
•avoiding eccentric loading as far as possible
•The margin of the edge parallel to the load
m1 1 .5 d
•The margin of the edge normal to the load
m2 2d
•Pitch limit: 16 h p 3d
where h is the thickness of the thinnest plate
used in the joint
