Example 9.12. A tie-bar in a bridge consists of flat 350 mm wide and 20 mm thick.
connected to a gusset plate of the same thickness by a double cover butt joint.
Design an economical
joint if the permissible stresses are :
t = 90 MPa, = 60 MPa and c
= 51.95 N/mm2 = 51.95 MPa
Since this stress is well below the given crushing stress of 120 MPa, therefore the
satisfactory.Riveted Joints 329
Example 9.15. The bracket as shown in Fig. 9.27, is to carry a load of 45 kN.
size of th
= 9 87 000 = 783 000 N
The strength of the joint is the least of Pt1, Pt2, Pt3, Pt4, Ps and Pc.
Strength of the joint
= 577 800 N along section 1-1
We know that the strength of the un-riveted plate,
t = 350 20 90 = 630 000 N
Efficiency of the joint
= P / n = 45 103 / 9 = 5000 N
The direct shear load acts parallel to the direction of load P, i.e. vertically
downward as shown
in the figure.
Turning moment produced by the load P due to eccentricity e
=P.e = 45 103 500 = 22.5 106 N-mm
This turning m
= = 0.75 or 75% Ans.
Since the efficiency of the designed joint is equal to the given efficiency of 75%,
design is satisfactory.Riveted Joints 307
Example 9.8. A pressure vessel has an internal
diameter of 1 m and is to
= 178 500 N along section 1-1
We know that strength of the un-riveted plate,
t = 20 12.5 80 = 200 000 N
Efficiency of the joint,
Strength of the joint 178 500
Strength of the unriveted plate 200 000
= 0.8925 or 89.25% Ans.
6. Pitch of r
We know that strength of the un-riveted plate
t = 200 10 112 = 224 000 N
Efficiency of the joint,
Strength of the joint 181 225
Strength of the un-riveted plate 224 000
= 0.809 or 80.9% Ans.
9.21 Eccentric Loaded Riveted Joint
When the line of ac
1 = 0.75 t = 0.75 20 = 15 mm Ans.320 A Textbook of Machine Design
5. Efficiency of the johint
First of all, let us find the resistances along the sections 1-1, 2-2, 3-3 and 4-4.
At section 1-1, there is only one rivet hole.
Resistance of the joint in tea
6 = Ps + F6 = 5000 + 14 800 = 19 800 N
The resultant shear load (R3 or R9) may be determined graphically as shown in
From above we see that the maximum resultant shear load is on rivets 3 and 9.
If d is the diameter of the rivet hole, then maxi
(34.5)2 60 = 476 820 N .(ii)
Equating equations (i) and (ii), we get
2100 ( p 34.5) = 476 820
p 34.5 = 476 820 / 2100 = 227 or p = 227 + 34.5 = 261.5 mm
According to I.B.R., the maximum pitch,
pmax = C.t + 41.28 mm
From Table 9.5, we find that for doub
3 + .
[(l1)2 + (l2)2 + (l3)2 + .]
From the above expression, the value of F1 may be calculated and hence F2 and F3
known. The direction of these forces are at right angles to the lines joining the
centre of rivet to the
centre of gravit
shown in Fig. 9.24.
The bracket plate is 25 mm thick. All rivets are to be of the same size. Load on the
P = 50 kN ; rivet spacing, C = 100 mm; load arm, e = 400 mm.
Permissible shear stress is 65 MPa and crushing stress is 1
resultant shear load (R) on each rivet as shown in Fig. 9.23 (c). It may also be
obtained by using the
R = ( ) 2 cos P F P F s s 2 2 + +
where = Angle between the primary or direct shear load (Ps) and
secondary shear load (F).
When the secondary
From Table 9.7, we see that according to IS : 19291982 (Reaffirmed 1996), the
diameter of rivet hole ( d ) is 29 mm and the corresponding diameter of rivet is 27
2. Number of rivets
Let n = Number of rivets.
We know that the maximum pull
The direct shear load acts parallel to the direction of load P i.e. vertically
downward as shown
in Fig. 9.25.
Turning moment produced by the load P due to eccentricity (e)
= P e = 50 103 400 = 20 106 N-mm
This turning moment is resisted by seven rivets
1 = 1.25 t = 1.25 10 = 12.5 mm Ans.
4. Efficiency of the joint
First of all, let us find the resistances along the sections 1-1, 2-2 and 3-3. At section
1-1, there is
only one rivet hole.
Resistance of the joint in tearing along section 1-1,
t1 = (b d
The number of rivets required for the joint may be obtained by the shearing or
resistance of the rivets.
t = Maximum pull acting on the joint. This is the tearing resistance
of the plate at the outer row which has only one rivet.
= ( b d )
8. Distance between the rows of rivets
= 2.5 d = 2.5 29 = 72.5 say 75 mm Ans.
Note : If chain riveting with three rows of three rivets in each is used instead of
diamond riveting, then
Least strength of the joint
= (b 3 d ) t t = (350 3 29) 20 90 = 473 40
112 MPa and 200 MPa respectively and shear stress of the rivets as 84 MPa. Show
the disposition of
the rivets for maximum joint efficiency and determine the joint efficiency. Take
diameter of rivet hole
as 25.5 mm for a 24 mm diameter rivet.
From Table 9.7, the standard diameter of the rivet hole ( d ) and the rivet diameter
specified, according to IS : 1929 1982 (Reaffirmed 1996).Riveted Joints 325
Notes : 1. In the solution of a problem, the primary and shear loads may be laid off
4. The effect of P
2 = P is to produce a turning moment of magnitude P e which tends to rotate
the joint about the centre of gravity G of the rivet system in a clockwise direction.
Due to the turning
moment, secondary shear load on each rivet is produced.
force to twist the joint about the centre of gravity in addition to direct shear or
Let P = Eccentric load on the joint, and
e = Eccentricity of the load i.e. the distance between the line of
action of the load and the centroid of the rivet