874 STRENGTH OF MATERIALS RIVETED JOINTS 875
(jj) Double riveted lap joint having chain riveting. This is shown in Fig. 20.8. In case of Sol. (t) Single riveted buttjoint with a single cover plate. In case of a butt joint, the edges
double riveted lap joint, there should be two rows of rivets as shown in plan. For a chain of the two plates to be joined together butt (i.e., touch) against each other and a cover plate is
riveted joint, every rivet of a row should be opposite to the other rivet of the other row as placed on one side of the two plates. Rivets are passed through the main plates and cover
shown in Fig. 20.8. The distance between the two rows of the rivets should be at least = 2D + 6, plates. For a single riveted butt joints, one row of rivets is used on each side of the joint. Hence
in total, there will be two rows of rivets as shown in Fig. 20.10. For a single cover plate, the
Dwhere is the diameter of the rivet. thickness of the cover plate is taken as equal to t or 1.125 1, where t is the thickness of the main
plate to be joined by butt joint. The distance between the two rows of rivets should be equal to
The plan and elevation of the double riveted lap joint having chain riveting is shown in
D D3 where is the diameter of the rivet. The plan and elevation of the single riveted butt joint
Fig. 20.8.
with single cover plate is shown in Fig. 20.10.
(Hi) Double riveted lap joint having zig-zag riveting. This is shown in Fig. 20.9. There
are two rows of rivets as shown in plan. For a zig-zag riveted joint, the spacing of the rivets is
staggered in such a way that every rivet is in the middle of the two rivets of the opposite row as
shown in Fig. 20.9. The distance between the two rows of rivets should be atleast = 2D, where
D is the diameter of the rivet. The plan and elevation of the double riveted lap joint having zig-
zag riveting is shown in Fig. 20.9.
Fig. 20.10. Single riveted butt joint with Fig. 20.11. Single riveted butt joint
a single cover plate. with double cover.
Fig. 20.9. Double-riveted lap joint—Zig-zag riveting. (ii) Single riveted butt joint with double cover plates. In case of single riveted butt joint,
Problem 20.2. Draw neat sketches (Plan and Elevation) of the following riveted joints : there will be two rows of rivets as shown in Fig. 20.11. One row of rivets is used on each side of
the joints. There will be two cover plates, one on each side of the main plates. The thickness of
(i) Single riveted butt joint with a single cover plate each cover plate is taken as equal to 0.6 t to 0.8 t where t is the thickness of the plates to be
(ii) Single riveted butt joint with double cover plate joined together. The plan and elevation of the single riveted butt joint with double cover plates
(Hi) Double riveted butt joint with a single cover plate.
are shown in Fig. 20,11.
(AJV1IE, Summer 1985 and Winter 1981)
(Hi) Double riveted butt joint with a single cover plate. In case of double riveted butt
joints, two rows of rivets are used on each side of the joint and hence in total there will be four
rows of rivets as shown in Fig. 20.12. One cover plate is placed on one side of the rivets. The
thickness of the single cover plate is taken as equal to t to 1.125 1 where t is the thickness of the
main plate. The distance between the two rows of rivets which are one side of the joint is taken
, 376 STRENGTH OF MATERIALS
equal to 3D whereas the distance between two rows of rivets which are on either side of the
Ddnt is taken equal to 3D, where is the diameter of the rivet. The plan and elevation of the
double riveted butt joint with a single cover plate is shown in Pig. 20.12.
Fig. 20.13 Fig. 20.14
Let p = Pitch of the rivet
d = Diameter of the rivet
t - Thickness of the Plate
P = Tensile load acting on the plates
cr = Safe tensile stress in the plate
(
P = Safe tensile load that the plates can withstand for one pitch length
t
Then safe tensile load per pitch length is given by
P = Safe tensile stress x Area of plate per pitch length
t
= o x(p- d)xt ,..(20.1)
t
Fig. 20. 12. Double riveted butt joint with a single cover plate. If the value of tensile load (P) is more than the value of P given by equation (20.1), then
t
20.6. FAILURE OF A RIVETED JOINT
tearing of plate between the rivets will take place.
A riveted joint may fail in any one of the following ways :
20.6.3. Failure Due to Shearing of Rivet. If the diameter of the rivets is less than the
(i) Failure due to tearing of the plate between the rivet hole and the edge of the plate,
(ii) Failure due to tearing of the plates between rivets, required diameter, then the rivets will be sheared off as shown in Fig. 20.15 and Fig. 20.16.
{Hi) Failure due to shearing of rivet, Fig. 20.15 shows the rivet in a single shear whereas Fig. 20.16 shows the rivets in double
(iv) Failure due to crushing (or bearing) of rivet,
(a) Failure due to crushing of plate. shear.
20.6.1. Failure Due to Tearing of the Plate between the Rivet hole and the Fig. 20.15. Rivet in single shear. Fig. 20.16. Rivet in double shear.
xldge. If the distance between the centre of the rivet and the nearest edge of the plate (also The cross-sectional area resisting shear, when the rivet is in single shear is equal to
known as margin, m) is not sufficient, the tearing of the plate between the rivet hole and the
—Jl
dge of the plate will take place as shown in Fig. 20.13. This type of failure may be avoided if,
d2 whereas the cross-sectional area resisting shear, when the rivet is in double shear is
m = 1.5d
m'here = Margin i.e. , distance between the centre of rivet and nearest edge of the plate and equal to 2 x — d2 where d is the diameter of the rivet.
d = Diameter of rivet. In a lap joint and in a single cover butt joint, the rivets are in single shear as shown in
Fig. 20.17 and Fig. 20.18. But the rivets are in double shear in a double cover butt joint as
20.6.2. Failure Due to tearing of the Plate between Rivets of a Row. The plate shown in Fig. 20.19.
will tear between the rivet holes across a row if the tensile stress (due to tensile force P) on the
section corresponding to the line of rivets is having a large value as shown in Fig. 20.14. In
ouch cases, we consider only a pitch length of the plate.
STRENGTH OF MATERIALS RIVETED JOINTS 879
878
In case of a lap joint, the rivets are in single shear.
Fig. 20.17. Shearing off a rivet in a lap joint. Fig- 20.18. Shearing off a rivet in a Safe load which a rivet can withstand in single shear
single cover butt joint.
= ^d2 xx
4
Strength of joint per pitch length against shearing
Fig. 20. 19. Shearing off a rivet in a double cover butt joint. where n = Number of rivets covered per pitch length
Let x = Safe shear stress for the rivet material = 1 for a single riveted lap joint
d = Diameter of rivet
= 2 for a double riveted lap joint
A —Jl -n
= 3 for a triple riveted lap joint.
= Area of cross-section of rivet a"
20.6.4. Failure Due to Crushing (or bearing) of Rivet or Plate. In one of the plates
P = Safe load which a rivet can withstand against shearing
n - Number of rivets in one pitch length. in a joint is weaker than the other, the crushing of that plate (or rivet in contact with that
plate) will take place. Let the top plate of a lap joint shown in Fig. 20.20 is weaker than the
(i) Safe load per pitch length in case of lap joint. In case of lap joint the rivets are in
single shear. bottom plate. Now if the top plate is pulled by a load P, the crushing stress (or bearing stress)
P = 1 x — d2 x x for a single riveted lap joint will be induced between the top plate and the rivet. The plate or rivet will be crushed if these
stresses are having large values as shown in Fig. 20.21.
s4
Fig. 20.20 Fig. 20.21
= 2 x — d2 x x for a double riveted lap joint
Let a - Allowable crushing stress,
4 c
t= 3 x d2 x t for a triple riveted lap joint and so on. d = Diameter rivet,
4 t = Thickness of plate,
(ii) Safe load per pitch length in case of butt joint with single cover plates. The rivets are
in single shear. But there is one rivet in a single riveted butt joint on one side of the joint per P = Safe load which a rivet can withstand against crushing
pitch length (see Fig. 20.18). b
= cr. x projected area
—P - 1 x d2 x t for a single riveted butt joint
-a xd.t. (- Projected area = d.tj
s4 c
= 2 x — d2 x t for a double riveted butt joint :. Strength of the joint per pitch length of the joint against crushing •
4 = nxo xdxt
c
= 3 x — d 2 x x for a triple riveted butt joint and so on.
where n = Number of rivets covered per pitch length.
4
20.7. STRENGTH OF A RIVETED JOINT
(i«) Safe load per pitch length in case of butt joint with double cover plates. The rivets
are in double shear in case of butt joint with double cover plates as shown in Fig. 20.19. The maximum force, which a riveted joint can withstand without failure, is known as
.-. Safe load which a rivet can withstand in double shear the strength of the joint. The strength of the riveted joint is obtained as :
Let o = Allowable tensile stress in the plate,
—7t _
£
= 2 x d2 x x
4
.'. Strength ofjoint per pitch length against shearing
where n = Number of rivets covered per pitch length on one side of the joint
= 1 for a single riveted butt joint
= 2 for a double riveted butt joint
= 3 for a triple riveted butt joint.
880 STRENGTH OF MATERIALS RIVETED JOINTS
x = Allowable shear stress for the rivet material, Also rivets are in simple shear
a = Allowable crushing stress of rivet material. (Single riveted lap joint is shown in Fig. 20.7)
c
Then per pitch length, Consider the pitch length of the joint.
(i) The tearing strength is given by, (i) Tearing strength per pitch length is given by equation (20.1).
P - o (p - d) x t Using equation (20.1),
t
N NPt - = -
(it) The shearing strength (Pj is given by, = a (p d) x 120(50 x = 32640
t
t 16) 8
P = n x t x — d2 ... if the rivet is in single shear ...(20.2) (ii) Shearing strength per pitch length is given by equation (20.2),
s4 P_
= nxxx —dJt n
i
—= n x x x ^2 x d 2 ... if the rivet is in double shear ...(20.3) 4sst
where j = lxx x-d (v n = 1)
n = Number of rivets in one pitch length for lap joint
= Number of rivets in one pitch length on one side of the joint for butt joint. N= 1 x 90 x ~ x 162 = 18095.5
(Hi) The crushing strength (P ) or bearing strength (P ) is given by 4
c t
(iii) Crushing strength per pitch length is given by equation (20.4),
P = n x o x ti x f ...(20.4)
c c P = n x o xdxt
cc
where n - Number of rivets in one pitch length. N= 1 x 160 x 16 x 8 = 20480
The maximum force, which a riveted joint can transmit without failure, is the least
value of P P or P . Once the minimum of the three values is reached, the joint will fall. Hence .'. Least strength per pitch length
(, s c
the strength of the joint will be equal to the least value of P P and P .
(, s c - Least of P P and P
(, s c
20.8. EFFICIENCY OF A RIVETED JOINT = 18095.5 N
Strength of the solid plate per pitch length is given by equation (20.6),
The efficiency of a riveted joint is the ratio of the strength of the riveted joint to the NP = = x =
strength of the solid plate (i.e., strength of unriveted plate). Mathematically cr, p . t 120 50 x 8 48000
.
.•. Efficiency of the riveted joint
Efficiency, Strength of the riveted joint Least of P P and P c
-r\ Strength of solid plate ts
P PLeast of Strength of solid plate
P,. and c 18095.5 N
t,
...(20.5) N= ' = 0.3769 = 37.69%. Ans.
48000
where P - Strength of solid plate
Problem 20.4. If in problem 20.3, the plates are joined by a double riveted lap joint and
= arp.t. per pitch length ...(20.6) pitch = 8 cm determine :
Least of P P and P (i) strength of the riveted joint and (ii) efficiency of the riveted joint.
t, s c
Efficiency,
Sol. Given :
mmmmFrom problem 20.3, t = 8 mm, d = 16
mmProblem 20.3. Two plates 8 thick are joined by a single riveted lap joint. The and p = 8 cm = 80
mmdiameter of the rivets is 16 and pitch = 50 mm. If a - 120 N/mm2 t = 90 N/mm2 and o = 120 N/mm2 t = 90 N/mm2 and o = 160 N/mm2
t , (, c
N/mma - 160 2 determine the efficiency of the joint. Nature of the joint = Double riveted lap joint
c
,
Sol. Given : Number of rivets in one pitch length, n = 2
mmThickness of plates, t = 8 The rivets are in single shear.
Dia. of rivet, d = 16 mm (Double riveted lap joint is shown in Fig. 20.8 and Fig. 20.9)
Pitch of rivet, p = 50 mm Consider one pitch length of the joint.
Tensile stress, o = 120 N/mm2 (i) Tearing strength per pitch length is given by equation (20.1).
Shear stress, (
x = 90 N/mm2 P- x (P ~ rf)*
f
Crushing stress, c = 160 N/mm2 = 120 x (80 - 16) x 8 = 61440 N.
Nature of joint c
= Single riveted lap joint (ii) Shearing strength per pitch length is given by equation (20.2).
The number of rivets in one pitch length, n = 1 P = n xt x — d2
s4
:
STRENGTH OF MATERIALS RIVETED JOINTS
882
(b ) Shearing strength per pitch length is given by equation (20.2).
= 2 X 90 X — X 162 (v n = 2) P=rexxx — 2 = 2x90 x— x 20 2
=v( re 2) <f
4
*4 4
= 3619 N
= 56548 N.
(iii) Crushing strength per pitch length is given by equation (20.4).
(c) Crushing strength per pitch length is given by equation (20.4).
P =n xo xdxt
cc P = nx 0 xiixt <
= 2 x 160 x 16 x 8 £ t
= 40960 N
= 2 x 160 x 12 x 20 = 76800 N
Least of P P and P = 56548
t, s c
Strength of riveted joint of Pv P and P Efficiency of the joint is given by equation (20.5).
s c
PThe least of P and P gives the strength of riveted joint. But least - P PLeast of P, , s and c
t, s c
36191 N
Strength of riveted joint = 36191 N. Ans. where P = Strength of solid plate per pitch length
Efficiency of the riveted joint N= cr x p x t = 120 x 6 x 12 = 86400
(
PUsing equation (20.5), q = Least of P and P —56548
t, s c
.-. Efficiency of joint - 86400 = 0.6545 = 65.45%.
where P = Strength of solid plate per pitch length (ii) 2nd case
= or p . t = 120 x 80 x 8 = 76800 N Dia. of rivets, d = 3 cm = 30 mm
Pitch of rivets,
36191 %712 _ 47 12 m Ans. Thickness of plate, p = 8 cm = 80 mm
' 76800 mmt = 1.2 cm = 12
Problem 20.5. Double riveted lap joints are made in the following two ways : () Tearing strength per pitch length is given by equation (20.1).
=6(:i ) Diameter of rivets = 2 cm, pitch of rivets cm P = o x(p-d)xt= 120 x (80-30) x 12 = 72000 N
((
(U) Diameter of rivets = 3 cm, pitch of rivets = 8 cm
() Shearing strength per pitch length is given by equation (20.2).
Ifo = 120 N/mm2 x = 90 N/mm 2 and a N/mm= 160 2 find out which joint has higher ^P = re x t x d2 - 2 x 90 x x 302 ( v re = 2)
, c s
, ^4
t (AMIE, Winter 1982) 4
efficiency. *The thickness of the plates is 1.2 cm in each case.
Sol. Given = 127234.5 N.
Tensile stress,
Shear stress, o = 120 N/mm2 (c) Crushing or bearing strength is given by equation (20.4).
Crushing stress, t
t = 90 N/mm2 P = rexo xdxi=2x 160 x 30 x 12 = 115200 N
CC
cr. = 160 N/mm2
Strength of solid plate per pitch length,
NP= = =
a xp x t 120 x 80 x 12 115200
t
Nature of joint = Double riveted lap joint
Least of P P and P = 72000 N.
t, s c
Number of rivets in one pitch length,
.•. Efficiency is given by
re = 2 ———Least of P, P, and P
72000
n = Strength of solid plate = 115200 = 0.625 = 62.5%
The rivets are in single shear
(Double riveted lap joint is shown in Fig. 20.8 and Fig. 20.9) .-. The 1st joint has higher efficiency. Ans.
Consider one pitch length of the joint. Problem 20.6. In a double riveted lap joint, the pitch of the rivets is 7.5 cm, thickness of
(i) 1st Case the plate = 1.5 cm and rivet diameter = 2.5 cm. What minimum force per pitch length will rupture
Dia. of rivets, N mm mmthe joint when ultimate stresses are a - 400 I 2 2 and o = 640 N/mm2.
Pitch of rivets, t r
Thickness of plates d = 2 cm = 20 mm , % = 320 N/
p - 6 cm = 60 mm (AMIE, Winter 1981)
mmt = 1.2 cm = 12
Sol. Given :
(a) Tearing strength per pitch length is given by equation (20.1) Nature of the joint = Double riveted lap joint
P = a x (p - d) x t .'. Number of rivets in one pitch length,
tt
= 120 x (60 - 20) x 12 = 57600 N. re = 2
884 STRENGTH OF MATERIALS RIVETED JOINTS 885
The rivets are in single shear (iii) Crashing (or bearing) strength per pitch is given by equation (20.4).
Pitch of rivets, p = 7.5 cm = 75 mm P = n x o x d x / = 1 x 600 x 22 x 12 = 158400 N.
c c
mmt - 1.5 cm = 15
NLeast
Thickness of plate, of P, 7, P. and P = 114040
I S C
d = 2.5 cm = 25 mm
Dia. of rivets, Efficiency of the joint
o = 400 N/mm2 P P —tUtsi• ng
Ultimate tensile stress, ( .. - Least of -( , and P
Ultimate shear stress, x - 320 N/mm2
Ultimate Crushing stress, equation (20.5), r|
o = 640 N/mm2
c where P = strength of solid plate per pitch length
Consider one pitch length of the joint. = a x p x t = 450 x 50 x 12 = 270000 N
t
(i) Tearing strength per pitch length is given by equation (20.1) u = 114040 = 0„.4224 = 42.24%. Ans.
1 270000
P = a x(p-d)xS = 400 x (75 - 25) x 15 = 300000 N
ff AProblem 20.8. single riveted double cover butt joint is used to connect two plates
iii) Shearing strength per pitch length is given by equation (20.2). mm mm15 thick. The rivets are 26 in diameter and are provided at a pitch of 10 cm. The
P = nxxx^d2 = 2x 320 x^ x 252 N/mm mm N/mmallowable stresses in tension, shear and crushing are 130
(v n = 2) 2 75 N/ 2 and 150 2
,
s4
4 respectively, find :
= 314200 N. (i) Safe load per pitch length of the joint, and
(iii) Crushing strength per pitch length is given by equation (20.4).
P = n x a x d x f = 2 x 640 x 25 x 15 ( v n = 2) (,ii ) Efficiency of the joint.
cc
= 480000 N. Sol. Given :
N.-. Nature of joint = Single riveted double cover butt joint.
Minimum force, that will rupture the joint is the least of P, P and P, i.e., 300000
, s
or 300 kN. Ans. Number of rivets per pitch length on one side* of the joint.
mm mmProblem 20.7. A thin cylindrical shell 1500 n= 1.
in diameter is made of 12 plates.
mmThe circumferential joint is a single riveted lap joint with 22 diameter rivets at a pitch of As the joint is double cover butt joint, hence the rivets are in double shear.
N/mm50 mm. If the ultimate tensile stress in the plate is 450 2 and the ultimate shearing and Thickness of plates, t - 15 mm
N/mm N/mmcrushing stresses for the rivets are 300 2 respectively, calculate the
2 and 600 Dia. of rivets, d = 25 mm
efficiency of the joint. (AMIE, Winter 1974) Pitch of rivets, p = 10 cm = 100 mm
Sol. Given : Allowable tensile stress, o = 130 N/mm2
(
Dia. of shell, D - 1500 mm
Shear stress, x = 75 N/mm2
l
mmThickness of plates, t = 12 Crushing stress, o = 150 N/mm2
c
Dia. of rivets, d - 22 mm
Pitch of rivets, p = 50 mm Consider one pitch length of the joint.
Tensile stress, o = 450 N/mm2 (o) Tearing strength per pitch length is given an equation (20.1).
Shearing stress, (
P = a x (p - d) x t = 130 x (100 - 25) x 15 = 146250 N.
t = 300 N/mm2 t t
Crushing stress, o = 600 N/mm2 (b) Shearing strength per pitch length is given by equation (20.3).
c
Nature of the joint = Single riveted lap joint P. = n x t x 2 rivets are in double shear)
Number of rivets per pitch length,
n = 1.0 25 2
J
N= = 73631 N.
1 x 75 x [2 ^ x
The rivets are in single shear. x
Consider one pitch length of the joint. (c) Crushing strength per pitch length is given by equation (20.4).
(£) Tearing strength of the plate per pitch length is given by equation (20.1). P = n x a x d x t = 1 x 150 x 25 x 15 N = 56250 N.
cc
P = a (p - d) x t = 450(50 - 22) x 12 = 151200 N.
tt (i) Safe load per pitch length of the joint. The safe load per pitch length of the joint will
(«) Shearing strength per pitch length is given by equation (20.2). P Pbe P
the least of the three values of P s and . But the value of is least.
;,
c c
P=raxTX — d? = \ k 300 x — x 222 Safe load = 56250 N. Ans.
s4 (v «=1) * Please note that in case of butt joint the number of rivets on one side of joint per pitch length
4 are taken.
= 114040 N.
:)
STRENGTH OF MATERIALS RIVETED JOINTS
(ii) Efficiency of the joint 36815.5 Strength of solid plate = a .p.t.)
t
PPLfeast P v .p.t
of three values of t , and c t 18.88%. Ans.
36815.5
s
130 x 100 x 15
Using equation (20.5), ri = Strength of solid plate
56250
Strength of solid plate AProblem 20.10. single riveted double cover buttjoint in a structure is used for connecting
But strength of solid plate is given by equation (20.6) as mmtwo plates 12 thick. The diameter of the rivets is 24 mm. The permissible stresses are
N mm mm N mm120 V
NP = a xp x t = 130 x 100 x 15 = 195000 2 in tension, 100 N/ 2 in shear and 200 I 2 in bearing. Calculate the necessary
t
pitch and efficiency of the joint. (AMIE, May 1967)
"S' 0-2884 ' 28-84*' *“•
Sol. Given
Problem 20.9. If in problem 20.8, the single butt joint is having a single cover plate Nature of the joint = Single riveted double cover butt joint.
instead of double cover plates, find : (i) Safe load per pitch length of the joint and {ii) Efficiency
.-. Number of rivets per pitch length on one side of the joint,
of the joint.
n= 1
Sol. Given : As the butt joint is having double cover plates, the rivets will be in double shear.
Nature of the joint = Single riveted butt joint with a single cover plate. mmThickness of plates, t = 12
Number of rivets per pitch length on one side of the joint, mmDiameter of rivets, d = 24
.
n - 1. Tensile stress, a = 120 N/mm2
Shearing stress, t
As the butt joint is having a single cover plate, the rivets will be in single shear. x = 100 N/mm2
The other data from problem 20.8 are : Bearing stress, o = 200 N/mm2
c
t = 15 mm, d = 25 mm, p - 100 mm, Let p = Pitch of the joint and
N/mm2 x = 75 N/mm2 and o N/mm 2 . i] = Efficiency of the joint.
, c
a = 130 = 150 Consider one pitch length of the joint.
t
(i) Tearing strength per pitch length is given by equation (20.1).
Consider one pitch length of the joint.
P -a x{p-d)xt
() Tearing strength per pitch length is given by equation (20.1). tt ... ii
= 120 x(p- 24) x 12 = 1440 {p - 24) N.
NP = o x {p - d ) x t = 130 x (100 - 25) x 15 = 146250 N.
t( (ii) Shearing strength per pitch length is given by equation (20.3).
() Shearing strength per pitch length is given by equation (20.2).
= n x x x — a,2 (v Rivets are in single shear) P = n x ax — dn ,2^1 rivets are in double shear)
\4 ; x,
;
4
X 25 2
Ny: 1 X 75 X (v n = 1) = 1 x 100 x 2 x -- x 24 2
14 ; (v n = 1) V4
= 36815.5 N. = 90500 N.
(c) The crushing strength per pitch length is given by equation (20.4). (iii) Bearing (or crashing) strength per pitch length is given by equation (20.4).
P -nxa xdxt P = nx a x d x t - lx 200 x 24 x 12 N = 57600 N
cc cc
N= 1 x 150 x 25 x 15
PEquating P Pto the lesser of the forces we get
t c,
and
s
= 56250 N. 1440(p - 24) = 57600
(i) Safe load per pitch length of the joint 57600 m_ 40 + 24 = 64 m.
1440
The of three values Pv P and P the load on the But the y _ Ans.
s c
least of is safe joint. least of
Pthree values is of i.e., 36815.5 N. Now strength of solid plate
£
- a xp xt
Safe load per pitch length of the joint t
= 36815.5 N. Ans. = 120 x 64 x 12 = 92160 N
(ii) Efficiency of the joint Least force
Strength of solid plate
... and : 0.625 = 62.5%. Ans.
—Using
= Least of three values of P. , P. P,- Efficiency,
equation (20.5), i] 2
Strength of solid plate
.
888 STRENGTH OF MATERIALS RIVETED JOINTS 889
mmProblem 20.11. Two plates of 12 thickness are connected by a double riveted cover Sol. Given :
mmbutt joint using 18 diameter rivets at a pitch of 8 cm. If the ultimate tensile stress in plate Nature of the joint = Double riveted double cover butt joint
N N/mmand shearing and bearing stresses in the rivets are 460 /mm2 320 2 and 640 N/mm2 Number of the rivets per pitch length on one side of the joint,
,
respectively, find the pull per pitch length at which the joint will fail. (AMIE, Summer 1974) n = 2.
Sol. Given : The butt joint is having double cover plates and hence the rivets will be in double shear
mmThickness of plates, t = 12 Thickness of plates, mmt - 1.2 cm = 12
Nature of the point = Double riveted double cover butt joint. Diameter of rivets, d = 2.2 cm = 22 mm
Number of rivets per pitch length of one side of the joint, Tensile stress, o = 100 N/mm2
Shear stress, (
n = 2. x = 80 N/mm2
As the butt joint if having double cover plates, the rivets will be in double shear. Crushing or bearing stress, o = 160 N/mm2
c
Dia. of rivets, d = 18 mm Let p = Pitch of the joint and
Pitch, p = 8 cm = 80 mm r| = Efficiency of the joint
Tensile stress, a = 460 N/mm2 Consider one pitch length of the joint.
Shearing stress, t
t = 320 N/mm2
Bearing stress, a = 640 N/mm2
c
Consider on pitch length of the joint.
(i) Tearing strength per pitch length is given by equation (20.1).
P, = o x(p-d)xt = 460(80 - 18) x 12 N = 34,2200 N
t
(a) Shearing strength per pitch length is given by equation (20.3)
P =nxT* xT^ ] ('•' Rivets are in double shear)
s
-= 2 x 320 x I 2 x x 18^ (v n = 2)
= 325800 N
(iii) Bearing (or crushing) strength per pitch length is given by equation
N NP= n x a x d = x
c c
x t 2 x 640 18 x 12 = 276500
The pull per pitch length at which the joint will fall is the least of the above three forces.
But the least of the above three forces is 276500 N.
The joinbwill-faH-at a pull of 276500 N. Ans.
Efficiency of the joint
—— ——rUrsi• ng equat..i. on (20.5), n =
Least of the three forces of P. P. and P,
‘-, t—?
Strength of solid plate
276500 ('.' Strength of solid plate = a..p.t)
-
,
276500 Ans.
=
460 x 80 x 12
= 0.6262 = 62.62%.
Problem 20.12. A double riveted double cover butt joint is used for connecting plates
1.2 cm thick. The diameter of the rivets is 2.2 cm. The permissible stresses are 100 N/mm2 in Fig. 20.22
, tension, 80 N/mm2 in shear and 160 N/mm2 in bearing. Draw a neat sketch ofthp joint and
calculate the necessary pitch and efficiency of the joint. (AMIE, November 1966)
890 STRENGTH OF MATERIALS RIVETED JOINTS
( i ) Tearing strength per pitch length is given by equation (20.1). = 1 X 94.5 x - X 22 = 29688 N
P = a x (p - d) x t = 100(p - d) x 12 4
tt
N= 1200(p - 22) Crushing strength per pitch length is given by equation (20.4),
Hi) Shearing strength per pitch length is given by equation (20.3). P = nxa xdxt
cc
P —= n x x x ! 2 x d‘ = 1x212.5x20x10 - (/ n = 1)
b = 42500 N
( Rivets are in double shear)
= 2 x 80 x 2 x-x!22 N = 121600 N Tearing strength per pitch length is given by equation (20.1),
I4 P = (p - d) x t x a = (p - 20) x 10 x 150
t (
(iii) Bearing (or crushing) strength per pitch length is given by equation (20.4). Equating the tearing strength to the lesser of crushing and shearing strengths
P = n x a x d x t = 2 x 160 x 22 x 12 N = 84480 N W1500(p - 20) = 29688
c e + 20 = 398mm
Equating P to the lesser of the forces P and P we get
t s c,
1200( p- 22) = 84480
84480 = + 22 Generally the pitch should not be less than 3d i.e., 3 x 20 = 60 mm. Hence provide a
pitch of 60 mm. Ans.
P = ' + 22 70-4
1200
mm= 92.4
say 92.5 mm. Ans. Efficiency of the riveted joint
Now strength of the solid plate P P PLeast of t , s , c
N= a .p.t - 100 x 925 x 12 = 111000 Using equation (20.5), q - g^reng^.jj 0f soiid plate per pitch length
t
29688
.-. Efficiency of the joint,
Least force Strength of solid plate per pitch length
11 “ Strength of solid plate
84480 But strength of solid plate per pitch length
= mooo = a761 = 76- 1%- Ans' = o .p.t = 150 x 60 x 10 = 90000 N.
t
The neat sketch of the joint (plan and elevation) is shown in Fig. 20.22. 29688
q = qqq'qq = 0.33 = 33%. Ans.
.-. Efficiency,
Problem 20.13. Find the suitable pitch for a riveted lap joint for plates 1 cm thick if
safe working stresses in tension in the plates and crushing and shearing of the rivet material Hi) Double riveted lap joint
mm N N mmare respectively 150 N! Number of rivets per pitch length = 2
2 212.5 /mm2 and 94.5 V 2 in the following types ofjoints :
,
(i ) Single riveted and Hi) Double riveted. Find also the efficiency of the joint in the above two The rivets in double riveted lap joint are also in single shear. Consider one pitch length
cases. Take d = 6-Jt . (AMIE, Summer 1977) of the joint.
Sol. Given : Shearing strength per pitch length is given by equation (20.2).
Thickness of plate,
t - 1 cm = 10 mm P=nxTX — d2
*4
Safe tensile stress, a = 150 N/mm2
t
Safe crushing stress, o = 212.5 N/mm2 = 2 x 94.5 x 7- 202 (v >i = 2)
c 4
Safe shearing stress, t = 94.5 N/mm2 = 59376 N
Let d - Dia. of the rivet in mm.
(t) Single riveted lap joint Crushing strength per pitch length is given by equation (20.4).
No. of rivets per pitch length, n = 1. P = nxf xdxt = 2x 212.5 x 20 x 10 = 85000 N
c c
Dia. of rivet, mmd = 6 Jt = 6jlQ = 19 say 20 mm Tearing strength per pitch length is given by equation (20.1),
The rivets in lap joint are in single shear. P = a x (p - d) x t = 150 x (p - 20] x 10 = 1500(p - 20)
tt
Consider one pitch length of the joint.
Equating the tearing strength to the lesser of crushing and shearing strengths, we get
Shearing strength per pitch length is given by equation (20.2). 1500(p - 20) = 59376
P=nxa.x — d2 <Y >1 = 1) mmp = say 60 mm. Ans.
S {A + 20 = 59.6
- 1500
STRENGTH OF MATERIALS
Efficiency of the joint The rivet value is obtained by considering shearing strength and bearing strength of
one rivet. The minimum of the two gives the least rivet value.
Least of P P and P
Using equation (20.5), q : t, s c 4. Thickness of cover plates (in case of butt joints)
Strength of solid plate per pitch length The thickness of the cover plates is obtained from the relations :
59376 52376 Ans. t - 1.125/ for ordinary butt joint with a single cover plate. ...(20.13)
= 0.6597 = 65.97%. 1 for a butt joint with double cover plates ...(20.14)
o,x pxt = 0.625/
150 x 60 x 10
20.9. DESIGN OF A RIVETED JOINT where t = Thickness of the main plates and
Designing a riveted joint means to find the following quantity : t = Thickness of cover plates.
1. Diameter of rivets, x
mmProblem 20.14. Two plates 10 thick are joined by a single riveted lap joint. The
2. Pitch of rivets, plates are subjected to a load of 200 kN. If the permissible tensile, shear and bearing stresses
mm mm N/mmare 120 N/ 2 respectively, determine :
2 100 N! 2 and 160
,
3. No. of rivets required for the joint, and (i) diameter of the rivets, (ii) pitch of the rivet,
4. Thickness of the cover plates (in case of butt joints).
1. Diameter of rivets (d) (i ii) number of rivets, and (iv) efficiency of the joint.
The diameter of rivets is calculated by using the relation :
Sol. Given
d = 1.6 Jt ...(20.8) :
where d = Diameter of the rivets in cm, and t
t = Thickness of the main plates in cm. mmThickness of plates, t = 10
The diameter of the rivets can also be calculated from the relation :
Nature of the joint - Single riveted lap joint
Number of rivets per pitch length = 1
And rivets are in single shear
Load or pull through which the plates are subjected,
d = Q4i ...(20.9) P = 200 kN = 200,000 N
where d and t are in mm.
Permissible tensile stress, a, = 120 N/mm2
2. Pitch of rivets (p) Permissible shear stress, r = 100 N/mm2
(i) If the efficiency of the riveted joint is given, then pitch is obtained from the relation : Permissible crushing stress, fc = 160 N/mm2
p-d (j) The diameter of rivets is obtained by using equation (20.8).
0= ...(20.10) d = Qyft
P
mm= 6 x yflO = 18.97
In the above relation q and diameter id) are known hence the pitch (p) can be obtained. say 19.0 mm. Ans.
(ii) If the efficiency of the joint is not known, then pitch is obtained by considering the (ii) Pitch of the rivet
shearing strength and crushing strength of the joint per pitch length. The minimum of these First find the shearing strength and bearing strength of one rivet. The rivets are in
two values are equated with the tearing strength of the plate per pitch length. single shear.
Or Minimum of P or P = o, x (p - d) x t ...(20.11) Shear strength of one rivet
s c
where P = Shearing strength per pitch length, NP = -
s S
4 d2 x x= y4 x 19 2 x 100 = 28352.8
P = Bearing strength per
c pitch length,
a = Safe tensile stress, Crushing strength of one rivet,
t
P = d x t x a = 19 x 10 x 160
d = diameter of rivets, cc (fc = 160)
= 30400 N
t = Thickness of plate, and
p = pitch. .'. Least rivet value = 28352.8 N
The value of the pitch obtained from equations (20.10) or (20.11) should be between
Now using equation (20.11).
2.5 to 3.0 times the rivet diameter.
Minimum of P or P = a \p — d] x t
c s t
o(P
3. No. of rivet required for a joint where a - d\ x is the tearing strength of the plate per pitch length. But minimum c or
t
The number of rivets in a small joint are obtained from the relation [p t
P P = 28352.8 N
S
71
28352.8 = 120(p - 19) x 10
Least rivet value for shearing and bearing
—(20.12)
where P = Force or pull to be transmitted across the joint.
,
STRENGTH OF MATERIALS RIVETED JOINTS
894
28352.8 Ui)No. of rivets
First of all, let us find the rivet value, i.e., the pulls required for shearing and crushing
1200 of a rivet.
mm28352.8 .-. Shearing strength of one rivet,
+ 19 = 42.6
mmBut the pitch should not be less than 2.5 d i.e., less than 2.5 x 19 = 47.5 P = 2 x x x ~ d2 (v Rivets are in double shear)
Hence provide a pitch of 47.5 mm. Hence p = 47.5 mm. Ans. $4
(Hi) Number of rivets N= 2 x 90 x - x (24)2 = 81160
4
Let n = number of rivets.
Crashing strength of one rivet,
= ” _P
P = a xtxd = 200 x 15 x 24 = 72000 N
Using equation (20.12), n Least rivet vaiue for shearing and bearing CC
where P = Pull to be transmitted across the joint P P.'. Rivet value is the least of and .
sc
N= 200000 (given)
Nand least rivet value = 28352.8 NHence rivet value = 72000
.-. Number of rivets are given by equation (20.12) as
=^m8W 200000 =7-054Say8- AnS* n _~ P
(Please note that the number of rivets are always taken the next higher integer). Rivet value
330000 (( v P from ee quation (ii) - 330000 N)
=
(iv) Efficiency of the joint
72000
‘”Using equation P P and P 28352.8
Least of t s c 120 x 47.5 x 10 = 4.6
= ,
(20.7), q a - Let us provide 5 rivets. Ans.
= 0.4974 = 49.74%. Ans. (iii) Pitch of the rivets (p)
Problem 20.15. Design a double cover butt joint to connect two plates 1.5 cm thick and- Using equation (20.11)
N/mm N/mm N mm20 cm wide. The safe stresses are a = 125 2 Minimum of P or P = ft [p -d]xt
2 x = 90 2 and a = 200 / Also c s
c . '
t 72000 = 125 (p - 24) x 15
determine the efficiency of the joint. ^p ~ 24 = l^Ii 01 ^ = Ii
Sol. Given : +24 = 62-4mm
mmThickness of the plate, t = 1.5 cm = 15
Nature of the joint = Double cover butt joint. But the pitch should not be less than 2.5 d
.-. The rivets are in double shear mmor 2.5 x 24 = 60
Width of the plates, b = 20 cm = 200 mm .-. Let us provide a pitch of 62.5 cm. Ans.
The safe tensile stress, o = 125 N/mm2 Thickness of cover plates (tf)
The safe shear stress, t
t = 90 N/mm2 mmUsing equation (20.14), = 0.625 x t = 0.625 x 15 = 9.37
The safe crushing stress, a = 200 N/mm2 mmLet us provide 10 thick cover plates. Ans.
c
In this joint, the pull or load transmitted across the joint is not given. This is obtained by Efficiency of the joint
considering the plates to be weakened by a single hole in the last row. Using equation (20.10), q = p — d
.-. Tearing strength of the plate across the last row
= a (b - d) x t -(i) 62.5 - 24 38.5
t = 0.616 = 61.6%. Ans.
= 125(200 - d) x 15 N
(i) Diameter of the rivet (d)
Using equation (20.8), d = 1.9 Jt
= 1.9 /L5 = 2.33 cm say 2.4 or 24 mm. Ans. HIGHLIGHTS
Substituting the value of d in equation (i), we get the safe strength of the plate as 1. Riveted joints are mainly of two types i.e., lap joint and butt joint.
2. In case of lap joint, the edges of the plates to be joined together overlap each other whereas in
P = 125(200 - 24) x 15 = 33000 N -(H)
case of butt joint the edges 'ofjthe plates butt (i.e., touch) against each other.
STRENGTH OF MATERIALS RIVETED JOINTS g97
The lap joints may be : (t) Single riveted lap joint, (ii) double riveted lap joint and (Ui ) triple 16. Efficiency of a riveted joint is the ratio of the strength of the joint to the strength of the solid
plate. Mathematically,
riveted lap joint. Pn ~__ Strength of riveted joint _ Least value of t , Ps and Pr
Strength of solid plate p
In case of single riveted lap joint, only one row of rivets is used for connecting two plates whereas
in case of double riveted lap joint, two rows of rivets are used for the connection. where P = a x p x t.
t
In case of a butt joint, atleast two rows of rivets one on each side of the joint, are required.
Butt joints are also classified as : (i) Single riveted butt joint, (ii) double riveted butt joint, and 17. Design of a riveted joint means to find : (£) diameter of rivets, (ii) pitch of rivets, (Hi) number of
(sit) tripple butt joint. rivets required for the joint, and (iv) thickness of the cover plates (in case of butt joints).
In case of a single riveted butt joint, one row of rivets is used on each side of the joint. Hence in 18. Diameter of rivets is given by
total there are two rows of rivets. d = 19 -Jt when d and t are in cm
In case of a chain riveted joint, every rivet of a row is opposite to the other rivet of the other row. = 6 -It when d and t are in mm.
In case of a zig-zag riveted joint, every rivet is in the middle of the two rivets of the opposite row. 19. The pitch of rivets is obtained either from the relation
The failure due to tearing of the plate between the rivet hole and the edge may be avoided if
margin is equal to 1.5d where d = diameter of rivet.
The safe tensile load per pitch length is given by
P = a x (jd - d) x t
tt
where p - Pitch, or from the relation.
d = Diameter of rivet, and Minimum of P or P - o x (p - d) X t.
s c (
t = Thickness of the plate. 20. The number of rivets in a small joint are obtained from
In a lap joint and in a single cover butt joint, the rivets are in single shear. But in case of double
cover butt joint, the rivets are in double shear. Force transmitted across the Saint
Safe shear load per pitch length in case of lap joint is given by, Least rivet value for shearing and bearing
21. Thickness of cover plates in case of butt joints is given by
P = 1 X —% dn2 x x t = 1.1254 for single cover butt joint
1 for double cover butt joint.
s4 ... For a single riveted lap joint
= 0.625t
=2x K d2„ x x ... For a double riveted lap joint
-7
4 EXERCISE 20
= 3 x —K d2 x t ... For a triple riveted Sap joint.
4 (A) Theoretical Questions
Safe shear load per pitch length in case of butt joint with double cover plates is given by,
P = n x ^2 x — d" x 1. Define the terms : riveted joint, lap joint and butt joint.
t 2. What are the different types of lap joint and butt joint ? Explain clearly with neat sketches.
where n = Number of rivets covered per pitch length on one side of the joint 3. Explain the terms : Chain riveted joint, zig-zag riveted joint and diamond riveted joint.
4. Draw neat sketches (plan and elevation) of the following riveted joints.
= 1 ... For a single riveted butt joint
= 2 ... For a double riveted butt joint and so on. (£) Single riveted lap joint,
The minimum force, which a riveted joint can withstand without failure, is known as the strength (ii) Double riveted lap joint having chain riveting, and
P P Pof the joint. It is equal to the least value of where (fit) Double riveted lap joint having zig-zag riveting.
and
t, c,
s
P = ct, x (p - d) x t, 5. Describe the different types of failure of a riveted joint.
t
6. What should be the value of margin so that there is no failure due to tearing of the plate between
n xxx — d2 ... if the rivet is in single shear , the rivet hold and the edge of the plate.
4 7. Prove that the safe tensile load per pitch length is given by
2 x-d' ... if the rivet is double shear Pt = a x (p - d) x t
t
4
where a = Safe tensile stress in the plate,
nxa xdxt t
c
and P = p = Pitch of the rivet,
c
where n - Number of rivets in one pitch length. d = Diameter of the rivet, and
t - Thickness of the plate.
8. Prove the statement that in a lap joint the rivets are in single shear whereas in double cover butt
joint, the rivets are in double shear.
)
9. Find an expression for the safe shear load per pitch length in case of a butt joint with double 11 . Find the suitable pitch for a riveted lap joint for plates 1.2 thick if safe working stresses '
cover plates. in tension in the plates and crushing and shearing of the rivet materials are respectively"
10. What do you mean by strength of a riveted joint ? Find an expression for the tearing strength, _J40N/nun2 200 N/mm2 and 90 N/mm2 in the following types of joints : (i) single riveted and
,
shearing strength and bearing strength of a riveted joint.
(ii) double riveted. Find also the efficiency of the joint in the above two cases. Take d = lSjt
11. Define the efficiency of a riveted joint. How will you find the efficiency of a riveted joint ?
12. What do you mean by ‘Design of a riveted joint’ ? While designing a riveted joint, how will you [Ans. (i) 5.5 cm, 33.73% (ii) 6 cm, 61.84%]
find different quantities ? mm12. Two plates 10 thick are joined by a single riveted lap joint. The plates are subjected to a load
(B) Numerical Problems of 180 kN. If the permissible tensile shear and crushing stresses are 125 N/mm2 90 N/mm2 and
,
N/mm170 2 respectively. Determine : (i) diameter of rivets, (ii) pitch of the rivet, (iii) number of
mm1. Two plates 10 thick are joined by single riveted lap joint. The diameter of the rivets is rivets and (iv) efficiency of the joint. [Ans. 1.9 cm ; 4.75 cm ; 8 ; 42.97%]
mm20 and pitch = 60 mm. If a, = 125 N/mm! x = 80 N/mmz and a t 160 N/mm2 determine the 13. Design a double cover butt joint to connect two plates 1.2 cm thick and 18 cm wide. The safe
, c ,
efficiency of the joint. [Ans. 33.51%] stresses are o = 120 N/mm2 x = 90 N/mm2 and ct = 180 N/mm2 . Also determine the efficiency of
{ , c
I
2. If in problem 1, the plates are joined by a double riveted lap joint and pitch = 80 mm, determine : the joint. [Ans. d = 2.1 cm, n = 5, p = 6.5 cm, t = 0.75 cm and q = 67.7%]
x
(i) strength of the riveted joint and (ii) efficiency of the riveted joint.
[Ans. (i) 50.265 kN, (ii) 50.26%]
3. Double riveted lap joints are made in the following two ways :
(i) Diameter of rivets = 2.5 cm, pitch of rivets = 7.5 cm
(ii) Diameter of rivets = 3.5 cm, pitch of rivets = 8.5 cm
If a, = 120 N/mmz x = 90 N/mmz and o = 160 N/mm2 find out which joint has higher efficiency.
, c ,
The thickness of the plates in each case is 1.3 cm. [Ans. 1st joint]
4. In a double riveted lap joint, the pitch of the rivets is 9.0 cm, thickness of the plate = 1.6 cm and
rivet diameter = 2.6 cm. What minimum force per pitch length will rupture the joint when ulti-
mate stresses are a, = 450 N/mm2 t - ‘320 N/mm2 and o = 640 N/mm 2 . [Ans. 33.98 tonnes]
, c
mm mmA5. thin cylindrical shell 1600 plates. The circumferential
in diameter is made of 13.5
mmjoint in a single riveted lap joint with 24 diameter rivets at a pitch of 60 mm. If the ultimate
tensile stress in the plate is 400 N/mm 2 and ultimate shearing and crushing stresses for rivets
300 N/mm 2 and 600 N/mm2 respectively, calculate the efficiency of the joint. [Ans. 37.04%]
A mm6. single riveted double cover butt joint is used to connect two plates 60 thick. The rivets are
mm25 in diameter and are provided at a pitch of 10 cm. The allowable stresses in tension, shear
and crushing are 160 N/mm2 90 N/mm2 and 180 N/mm2 respectively, find :
,
( i Safe load per pitch length of the joint, and (ii) Efficiency of the joint.
10. [Ans. (i) 72 kN, (ii) 28.125%]
7. If in the above problem, the single riveted butt joint is having a single cover plate instead of
double cover plates, find : (t) Safe load per pitch length of the joint and (ii) Efficiency of the joint.
[Ans. (i) 44.178 kN, 17.25%]
mmA8. single riveted double cover butt joint in a structure is used for connecting two plates 15
thick. The diameter of the rivets is 25 mm. The permissible stresses are 125 N/mm2 in tension
90 N/mm2 in shear and 180 N/mm2 in crushing. Calculate the necessary pitch and efficiency of
the joint. [Ans. 6.1 cm, 59.01%]
mm9. Two plates of 15 thickness are connected by a double riveted double cover butt joints using
mm20 diameter rivets at a pitch of 10 cm. If the ultimate tensile stress in plate and shearing
and crushing stresses in the rivets are 450 N/mm2 30 N/mm2 and 600 N/mm 2 respectively, find
,
the pull per pitch length at which the joint will fail. [Ans. 36 kN]
A double riveted double cover butt joint is used for connecting plates 1.5 cm thick. The diameter
of the rivets is 2.5 cm. The permissible stresses are 120 N/mm2 in tension, 90 N/mm2 in shear
and 180 N/mm2 in crushing. Draw a neat sketch of the joint and calculate the necessary pitch
and efficiency of the joint. [Ans. 10 cm, 75%]
WELDED JOINTS 901
4.
21 Since no provision is kept for expansion or contraction in the frame, therefore, crack
Welded Joints may develop in it.
21.1. INTRODUCTION 21.3.
The process of permanently joining two or more metal parts by the fusion of the edges of TYPES OF WELDED JOINTS
the metals with or without the application of pressure and a filler material, is known as weld-
ing. If the pressure is used for joining the two parts, the process is known as forge welding The two important types of welded joints are : ...(21.1)
whereas if the two parts are joined without any pressure but with a separate weld metal, the
process is known as fusion welding. The joints so formed are known as welded joints. 1. Butt weld and
The heat of melt for the weld metafis generally obtained by gas welding or by electric 2. Fillet weld or lap joint.
are welding. 21.3.1. Butt Weld Joint. If the edges of
the two members butt (i.e., touch) against each
21.2. ADVANTAGES AND DISADVANTAGES OF WELDED CONNECTIONS other and the two members are joined by weld-
ing, then the joint so formed is known as butt-
The advantages and disadvantages of welded connections over riveted connection are
given below : weld joint. Fig. 21.1 shows the isometric view
of a butt-weld joint.
Advantages
1. The welded structure are comparatively lighter than corresponding riveted structure. Let l = Length of the weld. It is equal to
2. The welded joint has greater strength as compared to the riveted joint. Hence efficiency the width of plate.
of a welded joint is more than that of a riveted joint. t - Depth of weld. It is equal to the
3. Addition and alternations can be easily made in the existing welding structure more
thickness of the plate.
easily than in riveted structure.
F = Tensile force.
A4. welded structure has a better finish and appearance than a corresponding riveted
a = Allowable tensile stress in the
member. The maintenance and painting cost for a welded structure is less than for the riveted t
weld.
structure.
Then the tensile force is given by
5. Welding takes less time than riveting.
6. In welded connections the tension members are not weakened as in the riveted joints. F = Tensile stress x Area
7. Members of such shapes, which are difficult for riveting, can be easily welded. = cr xlxf
8. It is possible to weld any part of a structure at any point. But riveting requires enough
(
clearance.
Equation (21.1) is also used to calculate the strength of the butt-weld joint.
' Disadvantages The following types of butt-weld joints are mostly used :
. 1. Single V-butt joint [Fig. 21.2 (o)j
1. Welding requires skilled labour and supervision. 2. Single U-butt joint [Fig. 21.2 (6)1
2. As there is an uneven heating and cooling during welding, the members may get 3. Double V-butt joint [Fig. 21.2 (c)]
distorted or additional stresses may develop.
4. Double U-butt joint [Fig. 21.2 (d)].
3. Testing a weld joint is difficult. An X-ray examination alone can enable us to study
(c) Double V-Butt joint (d) Double U-Butt joint
the quality of the connection.
Fig. 21.2. Different types of butt-weld joints.
900
21.3.2. Fillet Weld or Lap Joint. When the two members overlap each other and
they are joined by welding, then the joint so formed is known as lap joint or fillet weld
joint. Fig. 21.3 (a) shows the isometric view of a fillet weld joint.
902 STRENGTH OF MATERIALS
Fig. 21.3 (a)
Let t = Thickness of the plate. This is also known as the size of the weld
l = Length of the weld
a = Allowable tensile stress for weld metal Fig. 21.4. Single fillet lap joint.
t
F = Tensile strength of double fillet lap joint.
Then throat thickness AD [See Fig. 21.3 (6)] 1. Single fillet lap joint [Fig. 21.4]
ABC= AB sin 45° ( v is an isosceles right angled triangle) Area of fillet weld = Length of weld x Throat thickness
1 = l x 0.707t ...(21.4)
...(21.5)
where ; = Thickness of plate and
l = Length of weld.
2. Double fillet lap joint [Fig. 21.5]
Area of fillet weld = l x Throat thickness + l x Throat thickness
= 2lx Throat thickness
= 2lx 0.707;
= 1.414 xixf.
Fig. 21.3 (6)
F
Then tensile strength of the double fillet lap is given by,
F = Tensile stress x Area of double fillet weld
= o x 2 x length of weld x Throat thickness
(
= o,x2xlx 0.707 x t (v Throat thickness = 0.7070
= 1.414 x a x l x t ...(21.2)
t
If the fillet weld is a single fillet lap joint, then strength is given by,
F = a x Area of single fillet weld
t
= ct x Length of weld x Throat thickness
(
= qt x l x 0.707i (v Throat thickness = 0.7070
= 0.707 x a x l x t ...(21,3) Fig. 21.5. Double fillet lap joint.
t
3. Parallel fillet weld [Fig. 21.6]
The following types of fillet weld joints are mostly used : Let l ~ Length of parallel weld
1. Single fillet lap joint, [Fig. 21.4] ; = Thickness of plate.
2. Double fillet lap joint, [Fig. 21.5]
3. Parallel fillet joint, [Fig. 21.6],
1 WELDED JOINTS 905
904 STRENGTH OF MATERIALS
Let l = Length of the weld.
Now area of parallel fillet weld is given by
equation (21.6).
Area of parallel fillet weld
= 1.414xZ xf
mm= 1.414 x l x 10 2
The load carried by parallel fillet weld is
given by Fig. 21.8
Load = Stress x Area of parallel fillet weld
(v Area = 1.414 xlx 10)
Fig. 21.6. Parallel fillet weld. or 50000 = 55 x 1.414 x Z x 10
Then, 50000 = 64.3 mm. Ans.
Area of parallel fillet weld 55 x 1.414 x 14
= l x Throat thickness + 1 x Throat thickness Problem 21.3. Two plates of width 12 cm and thickness 1.15 cm are welded by a single
= Z x 0.707 + Z x 0.707Z N/mmV-butt joint. If the safe stress in the weld is 140 * find the permissible load earned by the
= 1.414 xlxt. ...(21.6) plates. •
Notes 1 . The throat thickness of depth of weld in case of butt weld joint is equal t o the thickness Sol. Given : b = 12 cm = 120 mm
of plate (i.e., t) whereas the throat thickness for lap joint or fillet weld joint is equal to 0.707f. Width of plates,
2. The single and double fillet lap joints are designed for tensile strength whereas the parallel Thickness of the plates,
fillet weld is designed for shear strength.
mmf = 1.25 cm = 12.5
Problem 21,1. Two steel plates 10 cm wide and 1.25 cm thick are to be joined by
double lap weld joint. Find the length of the weld if the maximum tensile stress is not to Safe stress, o, = 140 N/mm-
exceed 75 N/mm2 and maximum tensile load carried by the plates is 100 kN.
FLet = Permissible load carried by the plates.
Sol. Given : In case of a butt joint, the width of the plate is equal
Width of the plate, b - 10 cm = 100 mm to the length of the weld.
mmThickness of the plate, t = 1.25 cm = 12.5 mm.-. Length of weld, l = b = 120
Maximum tensile stress, a = 75 N/mm2 Now using equation (21.1),
t
F=a .l .t = 140 x 120 x 12.5
Minimum tensile load, F = 100 kN = 100000 N t
= 210000 N = 210 kN. Ans.
Let l - Length of the weld.
The tensile strength of the double fillet lap joint is 21.4. ANALYSIS OF A COMPOUND WELD
given by equation (21.2).
Using equation (21.2), we have Fig. 21.7 Fig. 21.10 shows a combination of parallel
F = 1.414 x a x l x t fillet weld and a single fillet lap weld. This type
t of a weld is known as a compound weld. The weld
or 100000 = 1.414 x 75 x Z x 12.5 AB and CD are parallel fillet weld whereas the
100000 BCweld is a single fillet lap weld.
75.4 cm. Ans.
FLet = load carried by single fillet lap
1.414 x 75 x 12.5
AProblem 21.2. steel plate 10 cm wide and 1 cm thick is to be joined by parallel filler { weld (i.e., by weld BC)
welds to another plate. The plates are subjected to a load of 50 kN. Find the length of the weld F - Load carried by parallel fillet
, I 2 weld (i.e. by welds AB and CD)
N/mmif the maximum shear stress does not exceed 55
2 ,
.
Sol. Given : i F = Total load carried by the plates
Width of plates, b = 10 cm = 100 mm = F + F ~(i)
l 2
mmThickness of plates, t = 1.0 cm = 10
Zj - Length of single fillet lap weld
Load, F = 50 kN = 50000 N Z, = Length of the parallel fillet welds
Max. shear stress, x = 55 N/mm2 t = Thickness of the plates
’ .
STRENGTH OF MATERIALS WELDED JOINTS
a, - Maximum tensile stress in the weld 90000 = 63630 + 933.24 x l2 Alls.
r = Maximum shear stress in the weld. 90000 - 63630 = 933.24 x l2
We know that the fillet lap weld is designed for tensile stress whereas the parallel fillet
^_= 90000 - 63630 _ 28.25 mm.
welds are designed for shear stress.
2 933.24
The load carried by a single fillet lap weld is given by equation (21.3). 21.5. ANALYSIS OF UNSYMMETRICAL WELDED SECTIONS WHICH ARE LOADED
Using equation (21.3)
AXIALLY
Load = 0.707 x o x Length of weld x Thickness of plate
( Fig. 21.12 shows an angle section welded to a plate and carries an axial load F. Such
welded sections are known as unsymmetrical welded section.
or ~ 0.707 x o x^xt ...(21.7)
t
Now the area of parallel fillet weld is given by equation (21.6) as
Area of fillet parallel weld = 1,414 xi2 x t.
The load carried by parallel fillet weld is given by.
Load = Area of parallel fillet weld x -c
or F = 1.414 xl xt xx ...(21.8)
2 2
Substituting equations (21.7) and (21.8) in equation (i), we get
F = 0.707 xo xljX(+ 1.414 x l2 x t x x ...(21.9)
(
Problem 21.4. A plate 10 cm wide and 1.20 cm thick is joined with another plate by a
single fillet lap weld and a double parallel fillet weld as shown in Fig. 21.11. The maximum
N mm N mmtensile and shear stresses are 75 I
2 and 55 ! 2 respectively. Find the length of each
parallel fillet if the joint is subjected to a total load of 90000 N. Fig. 21.12
Sol. Given : The axial load F is applied along the axis which passes through the C.G. of the
mmWidth of plane, b = 10 cm = 100 ..... V xnsymmetrical section, in order to avoid the effect of eccentricity. Now the lengths of the weld
1
mmThickness of plate, t = 12 T j and l2 should be so proportioned that the sum of the resisting moments of the welds about
I he centre of the gravity axis is zero. Or welds
Maximum tensile stress, F I F in other words, the C.G. of the lengths of f and
I
a = 75 N/mm2 * | F, lies on the line of action of the load
(
10 cm 2
Minimum shear stress,
1
t = 55 N/mm2
—*— Let x = Allowable shear stress in the weld
h— /j = Length of weld at the top
Length of single fillet lap weld, H H l2 - Length of weld at the bottom
l\, = width of pelate Fig, 21.11 l = Total length of the weld = +l
= 100 mm. 2
a = Distance of top edge of the angle section from gravity axis
Let t = Length of each parallel fillet weld b = Distance of bottom edge of the angle section from gravity axis
2
Fj = Load carried by single fillet lap weld F - Axial load on the angle
F2 = Load carried by parallel fillet weld s = Resistance offered by the weld per unit length
Total load carried by the plates, F = Resistance of the top weld
l
F = 90000 N. - Resistance of weld per unit length x Length of top weld
The load carried by single fillet lap weld is given by equation (21.7). = sxl
t
Fl = 0.707 x o x/jXf
t F0 = Resistance of the bottom weld
= 0.707 x 75 x 100 x 12 = 63630 N = sxl
2
The load carried by double parallel fillet weld is given by equation (21.8).
F = Axial load on the angle
F2 = 1.414 x (2 x i x r F= + F (For the equilibrium of the welded section)
2
l
= 1.414 x l2 x 12 x 55 = 933.24 x l2 . Now the moment of the top weld about gravity axis
by the plates is given by
Now total load carried - Resistance of top weld x Distance of top weld from gravity axis
f=f +f = FjXa
12
zsx/jXa.
908 STRENGTH OF MATERIALS WELDED JOINTS 909
Similarly moment of the bottom weld about gravity axis N mmtop and bottom if the allowable shear stress in the weld is 102.5 / 2 . The distances between
F= 2 x b = s x Z2 x b. mm mmthe neutral axis and the edges of the angle section are 144.7
Since the sum of moment of the weld about the gravity axis must be zero. Hence and 55.3 respectively.
s x Zj x a = s x Z2 x b
or Zj x a = l2 x 6
Now we know that total length of the weld is equal to the sum of the lengths of top weld
and bottom weld.
l = l + l2 ...(it)
t
—L x a „ a( ... Z, x ")
= Zj + From —vI (t), ,
., =
Z2
equation
j
- £= 6Zt + Zj x a = Zj (a + b)
z, = —6 x Z r ...(21.10) Fig. 21.13
1
a + b
Substituting the value of Z, in equation (it), we get Sol. Given : = 200 mm x 150 mm x 10 mm
Dimension of angle
. 6 xZ . b x Z + Z2 (a + b)
mmThickness of angle = 10
a+b (a + b)
t = 10 mm
or l(a + b) - b x l + (a + b) .-. Size of weld,
l 2
F — 200 kN = 200000 N
or l x a + l x b = b x l + l2 (a + b) Axial load,
or lxa + lxb T bxl = l (a + b) Allowable shear stress, x = 102.5 N/mm2
2 Distance of the top edge of the angle section from neutral axis,
or lx a = Z 2(a + 6) mma = 144.7
axZ -.<.( 221L.1111 )> Distance of the bottom edge of the angle section from neutral axis,
or ** =
b = 55.3 mm
(flT6)
From equations (21.10) and (21.11), the length of the top weld and length of bottom weld
Fcan be calculated. Then the resistance of the top weld ( i.e. , ) and resistance of bottom weld Let Z = Length of weld at the top
;
y
F(i.e., 2 ) can be easily obtained as given below : l = Length of the weld at the bottom
2
F =s x l Z = Total length of weld = Z + Z2
yy L
bxl Now using equation (21.14), we have
SX
v From equation (2110), l = - F- x x (Z + Z2) x 0.707 x t
(a + b) x x
bxF 200000 = 102.5 x (Zj + Z2 ) x 0.707 x 10
a+b (v s x l = Total resistance = F) ...(21.12) mm200000
w ~+ =(/ 1in02o..K5 *x n07.f70V77x“x"1it0n: = 276
i
ax F Z 2>)
Similarly F,, = ...(21.13) I - 276 mm —(".* Z Zj t* Zg)
(<2 + O )
Axial load in terms of shear stress Now using equation (21.10),
The axial load IF) in terms of shear stress is given by : II = bxi " 55.3 x 276 =~ 76.3 mm. Ans.
(144.7 + 55.3)
Axial load = Shear stress x Area of weld oTb
Ff-e., = x x Area of weld Substituting the value of Zj in equation (i), we get
But area of weld = (Total length of weld) x Throat thickness 76.3 + Z 2 = 276
= 276 - 76.3 = 199.7 mm. Ans.
x 50 x8 carrying
= (Zj + Z2) x 0.707 x t (•.' Throat thickness = 0.707 x Z) mm mm mmor Z2 angle 125 kN to be
where t = Size of weld or thickness of plate
Problem 21.6. A 90 a load of is
F = r x (Z, + Z2 ) x 0.707 x t connected to a gusset plate by welding. If the size of the weld is 6 mtn and maximum allowable
shear stress the lengths of the weld at the top and bottom. The
...(21.14) in the weld is 102.5 N/mm2 find
,
mm mm mmProblem 21.5. A 200
x 150 x 10 angle, carrying a load of 200 kN, is to be
welded to a steel plate by fillet welds as shown in Fig. 21.13. Find the lengths of the weld at the
:, r
STRENGTH OF MATERJA'LS WELDED JOINTS 911
2di8s.t7ances between the neutral axis and the bottom and the top edges of the angle sections are Shear stress in weld, x = 102.5 N/mm2
mm mmand 61.3 respectively. Let Zt = Length of top weld,
Z2 = Length of bottom weld.
SoL Given Z = Total length of weld.
mm mmDimensions of the angle = 90 x 8 mm The single is welded to the plate as shown in Fig. 21.14.
x 50
Gusset
NAxial load, F = 125 kN = 125000
mmSize of weld, t* - 6
Maximum allowable shear stress, x = 102.5 N/mm2
Distance between the neutral axis and bottom edge,
mmb = 28.7
Distance between the neutral axis and top edge,
mma = 61.3
Let Zj = Length of weld at the top,
Z2 - Length of weld at the bottom.
I = Total length of weld = Z + Z2
2
Now using equation (21.1), we have
F= x x (Z + Z2 ) x 0.707 x t
x
the angle
where t - size of the weld
125000 = 102.5 x (Zj + Z ) x 0.707 x 6 Fig. 21.14
2
++Z2 = 125000 Now using the equation (21.14), we have
Z 102.5x0.707x6 F = t x (Zj + Z2) x 0.707 x t
1
= 287.5 mm ,..(Z) or 200000 = 102.5 x (Z + Z2) x 0.707 x 6
t
mmor Z = 287.5
(v = +Z 200000
l Z2 )
x y-^l + _
102.5x 0.707x6
Now using equation (21.10),
mm. 6 x Z 28.7x287.5= = 459.97 mm say 460 mm ...(i)
= ' Ans‘
h (28.7 + 61.3) = 91 '68 mmor Z = 460
J^b)
(v Z + Z = Z)
1
2
Substituting the value of L in equation (Z), we get Using equation (21.10), we get
x
91.68 + Z = 287.5 —136.7 bxl
2
Z = 287.5 - 91.68 = 195.82 mm. Ans. i, =
2 1 £1 + 5
mm mm mmProblem 21.7. A 150 x 115 x 8 angle carrying a tensile force of200 kN is to 44.6 x 460 44.6 x 460
mmbe connected to a gusset plate by 6 ==
fillet welds at the extremities of the longer leg, the
(105.4 + 44.6) 150
mmshorter leg being outstanding. Find lengths of welds if the C.G. of angle is 44.6 22.7
from the top
= 136.7 mm. Ans.
of the shorter leg. Take permissible shear stress in the welds as 102.5 N/mm2.
Substituting this value of l in equation (t), we get
(AMIE, Winter 1983) t
Sol. Given : + Z = 460
2
mm mmmmDimensions of angle = 150 Z2 = 460 - 136.7 = 323.3 cm. Ans.
x 115 x8
mm mmProblem 21.8. Two angles 80 mmx 8 carrying a load of 220 kN are
Load carried by angle, F = 200 kN = 200 x 1000 - 20000 N x 80
Size of the weld, t = 6 mm mmto be connected to a gusset plate on either side of the plate by 6 fillet welds as shown
N/mmin Fig. 21.15. The permissible shear stress in the weld is 102.5 2 The distances
.
The distance of C.G. of the angle from the top of shorter leg, mmbetween the neutral axis and top and bottom edges of the angle sections are 57.3 and
b = 44.6 mm mm respectively. Find the lengths of the weld at the top and bottom.
Distance of the C.G. of the angle from the top edge of the angle i.e. SoL Given :
a = 150 - 44.6 = 105.4 mm. mm mmDimensions of two angles = 80 x 8 mm
x 80
‘Here thickness of the angle is 8 mm. But the size of weld is 6 mm. This means the thickness of Axial load, F = 220 kN = 220 x 1000 = 220000 N
the weld is 6 mm. Hence t = 6 mm.
:
912 STRENGTH OF MATERIALS WELDED JOINTS 913
Size of weld, t = 6 mm Sol. Given
Permissible shear stress, x = 102.5 N/mm2 Nature of welded joint = Lap joint
Distance between neutral axis and top edge of the angle, Size of tie bar = 15 cm x 1 cm
a - 57.3 mm A.'. Area of tie bar, = 15 x 1
Distance between neutral axis and bottom edge of the angle, = 15 cm2
mmb = 2.27 cm = 22.7 = 1500 mm2
Working stress in the tie bar cm cm ’em
= 150 N/mm2 Fig. 21.16
.-. Maximum load, the tie bar can resist
= Working stress in tie bar x Area of bar
Size of fillet weld, = 150 x 1500 = 225000 N
t = 8 mm
Throat thickness of weld
mm. = 0.707 xf = 0.707 x 8
Fig. 21.15 Safe stress in the weld = 102 N/mm2
Total length of the weld = Length AB + Length BC + Length DE
+ Length EF + Length FG
Let = Length of top weld on one angle But length AB = Length EF,
and
l = Length of bottom weld on one angle length DE = Length BC = Length FG - Length AH
2
l - Total length of weld for two angles Total length of the weld
= + )x2 (v There are two angles on either side of the plate) = 2 (Length AB) + 4 (Length AH)
(l L l 2
=2 + l 2). 2x5= + 4 2 + 62 ) (• AH = )
(^/s
The two angles are welded to the gusset plate as shown in Fig. 21.15. mm= 10 + *761 = 41.24 cm = 412.4
Now using equation (21.14), we have Strength of weld = Safe stress in weld x Area of weld
Axial load =tx Total area of weld = 102 x (Total length of weld x Throat thickness of weld)
For = t x Total length of weld x Throat thickness = 102 x (412.4 x 0.707 x 8) = 237918.5 N.
= t x 2(lj + l2 ) x 0.707 x t (v Throat thickness = 0.707 f) Since the strength of the weld is greater than the maximum load, the tie bar can resist.
or 220000 = 102.5 x2« + Z2 )x 0.707x 6.0 (v t = 6) Hence the joint is safe. Ans.
1
Problem 21.10. A welded lap joint is provided to connect two tie bars 12 cm x 1 cm as
a +n 220000
12 102.5 x 2 x 0.707 x 6 shown in Fig. 21.17.
= 253.0 mm ...(t) N/mmThe working stress in the tie bar is 150 2 Investigate the design, if the size of the
.
Now using equation (21.10), we get fillet is 8 mm. The safe stress for the end fillet weld and diagonal fdlet weld may be taken as
— —bx(l,+L) 22.7x253 102.5 N/mm2 and 80 N/mm2 respectively.
= =, = 22.7 x253 = 71.8„.
* Ans.
l,
Sol. Given :
1 (o + 6) 57.3 + 22.7 80
Substituting the value of (j in equation (i), we get Nature of joint = Lap welded joint
71.8 + l2 = 253 Size of tie bar = 12 cm x 1 cm
l = 253 - 71.8 = 181.2 mm. Ans. Area of tie bar = 12 x 1 = 12 cm 2
2
= 1200 mm2
Problem 21.9. A welded lap joint is provided to connect two tie bars 15 cm x 1 cm as
shown in Fig. 21.16. Working stress in tie bar
NlmmThe working stress in tie bars in 150 2 Investigate the design if the size of the fillet = 150 N/mm2
.
N/mmweld is 8 mm. The safe stress for the weld may be taken as 102 2
.
Fig. 21.17
STRENGTH OF MATERIALS WELDED JOINTS
Maximum load, the tie bar can resist 7. The strength of a single V-butt weld is given by
= Working stress in tie bar x Area of bar F ~ o xl xt
t
= 150 x 1200 = 180000 N
where a - Allowable tensile stress in the weld,
t
l = Length of the weld, and
mmSize of the fillet weld, t = 8 t = Depth of weld or thickness of the plate.
Throat thickness of weld mm= 0.707 x / = 0.707' x 8 8. When the two members overlap each other and they are joined by welding, then the joint so
formed is known as lap joint or fillet weld joint.
Safe stress in the end fillet weld = 102.5 N/mm2
Safe stress in the diagonal fillet weld = 80 N/mm2 9. The strength of a single lap weld joint is given by
Total length of end fillet welds P - a x l x 0.707 x t
t
= AB +EF
where a, = Allowable tensile stress,
mm= 5 + 5 = 10 cm = 100
l = Length of weld,
Total length of diagonal fillet welds t - Size of the weld.
= BC + DE + FG + AH 10. The strength of a double lap weld joint is given by
AH= 4 x = 4 x 2 "+ 42 = 4 x 4 x V2 cm F-o xlx 1.414 x t.
t
^4
11. The throat thickness of a fillet weld is given by
Throat thickness = 0.707 X t
mm= 16 x 42. cm = 160 x J2 where t = Size of the weld.
But strength of weld = Safe stress in weld x Area of weld may12. The fillet weld joint be : (i ) Single fillet lap joint, (ii) Double fillet lap joint, and (iii) Parallel
.•. Strength of end fillet weld
fillet joint.
= Safe stress in end fillet weld x Area of end fillet weld
= 102.5 x (Total length of end fillet welds x Throat thickness) 13. The area of fillet weld is given by, ... For a single fillet lap joint
N= 102.5 x'(100 x 0.707 x 8) = 57974 A = l x 0.707 t
Similarly strength of the diagonal fillet welds = 2 x l x 0.707 t ... For a double fillet tap joint
= Safe stress in the diagonal fillet weld
x Area of diagonal fillet welds = / x 0-707 t ... For a parallel fillet weld,
= 80 x (Total length of diagonal fillet welds x Throat thickness)
where l = Total length of weld
N= 80 x (160 x J2 x 0.707 x 8) = 102400
t = Size of the weld.
.-. Total strength of weld
= Strength of end fillet welds 14. The throat thickness in case of a butt-weld joint is equal to the thickness of the plate whereas in
+ Strength of the diagonal fillet welds case of fillet weld or lap weld the throat thickness is equal to 0.707f where t = Size of weld.
= 57974 + 102400 = 160374 N ,
Since the strength of the weld is less than the maximum load the tie bar can resist. 15. The single and double fillet lap joints are designed for tensile strength whereas the parallel fillet
weld is designed for shear strength.
Hence the joint is not safe. Ans.
A16. combination of a parallel fillet weld and a fillet lap weld is known as compound weld.
17. The strength of a parallel fillet weld is given by
F = x x l x 0.707 x t
where l = Total length of parallel weld,
x = Allowable shear stress in the weld, and
t - Size of the weld.
18. An angle section welded to a plate is an example of unsymmetrical welded sections.
HIGHLIGHTS 19. The lengths of the welds at the top and bottom of an unsymmetrical welded sections should be so
proportioned that the sum of the resisting moments of the welds about the centre of the gravity
1. Welding is a process ofjoining permanently two or more metal parts by the fusion of the edges of
the metals with or without the application of the pressure and a fillet material. axis is zero.
2. The joint formed by welding is known as welded joint. 20. The lengths of welds at the top and bottom of an unsymmetrical welded sections which are
3. The welded joint may be a butt weld joint or a fillet weld joint (i.e., iap joint). loaded axially are given by
4. If the edges of the two members butt (i.e., touch) against each other and the two members are
ax l
joined by welding, then the joint so formed is known as butt-weld joint.
5. The depth of weld in case of a butt weld is equal to the thickness of the plate. a+a
6. The butt-weid joint may be : (i) Single V-butt joint, (ii) Single U-butt joint, (iii) Dquble V-butt
where = Length of weld at the top,
joint, and (fu) Double U-butt joint.
l2 - Length of weld at the bottom,
l = Total length of the weld = l + Z
t 2
STRENGTH OF MATERIALS WELDED JOINTS 917
5.
a = Distance of the top edge of the angle section from gravity axis
b = Distance of the bottom edge of the angle section from gravity axis or neutral axis. N/mmN/mmIf in the above problem, the safe stress for the end fillet weld is 102.52 2 whereas the safe
stress for the diagonal fillet weld is 75
t , investigate the design. The other dimensions of
21. The axial load on the unsymmetrical welded section at the neutral axis is given by the joint are the same as given in Fig. 21.18. [Ans. Joint is not safe]
F = t x (Zj + Z2 ) x 0.7Q7Z.
n
(A) Theoretical Questions 6.
1. Define the terms : Welding, welded joints, forge welding and fusion welding. .
2. Explain the advantages and disadvantages of a welded joint,
Fig. 21.18
(AMIE, Winter 1983 and Summer 1985)
A plate 10 cm wide and 1.15 cm thick is joined with another plate by a single fillet lap weld
3. Name the two important types of a welded joint. Find an expression for the strength of a single
and a double parallel fillet welds as shown in Fig. 21.19. The maximum tensile and shear
V-butt weld joint.
4. With neat sketches, give the important types of butt-weld joint. stresses are 75 N/mm2 and 55 N/mm2 respectively. Find the length of each parallel fillet if the
5. Define a fillet-weld joint and a butt joint. What are the different type of fillet-weld joint ? joint is subjected to a total load of 80 kN. [Ans. 2.12 cm]
6. Prove that the strength of a single lap weld joint is given by
F=0'XCx 0.707 t
where a = Allowable tensile stress in the weld,
,
l - Length of weld, and
t = Size of the weld.
7. Define a compound weld. How will you find the strength of a compound weld consisting of paral-
lel fillet weld and a single lap weld.
8. What do you mean by unsymmetrical welded section ? How will you determine the lengths of
welds at the top and bottom of an unsymmetrical welded section, which is loaded axially ?
9. Prove that the lengths of welds at the top and bottom of an unsymmetrical welded section which
is loaded axially, are given by
axl—b x l . , Fig. 21.19
iff
slid
mm mm mm7 A 200 angle candying a load of 250 kN, is to be welded to a steel plate by
x 150 x 10
where i = Length of the weld at the top, fillet welds as shown in Fig. 21.20. Find the length of the weld at the top and bottom if the
l
N/mm 2 The and
l2 - Length of the weld at the bottom, allowable shear stress in the weld is 102.5 . distance between the neutral axis
mm mmthe edges of the angle section are 144.7
= length of the weld = + and 55.3 respectively.
l Total l l2
x
[Ans. /, = 9.54 cm and l2 = 24.96 cm]
a = Distance of the top edge of the angle section from gravity axis or neutral axis, and
b = Distance of the bottom edge of the angle section from neutral axis.
(B) Numerical Problems
1. Two steel plates 12 cm wide and 1.30 cm thick are to be joined by double lap weld point. Find the
length of the weld if the maximum tensile stress is not to exceed 72 N/mm2 and maximum tensile
load carried by the plate is 110 kN. [Ans. 8.55 cml
A2. steel plate 10 cm wide and 1.25 cm thick is to be joined by parallel fillet welds to another plate.
The plates are subjected to a load of 60 kN. Find the length of the weld if the maximum shear
stress does not exceed 56 N/mm2 . [Ans. 6.06 cm]
3. Two plates of width 15 cm and thickness 1.25 cm are welded by a single V-butt joint. If the safe
stress in the weld is 135 N/mm2 find the permissible load carried by the plates.
,
[Ans. 25312.5 N]
A4 . welded lap joint is provided to connect two tie bars 15 cm x 1 cm as shown in Fig. 21.18. Steel
plate
The working stress in the tie bar is 150 N/mm2 . Investigate the design if the size of the fillet weld
is 8 mm. The safe stress for the weld may be taken as 102.5 N/mm2 [Ans. Joint is safe] Fig. 21.20
.
STRENGTH OF MATERIALS
A mm mm mm8. 80 x 80 x 8 angle carrying a load of 100 kN is to be connected to a gusset plate
mmby welding. If the size of the weld is 6 and maximum allowable shear stress in the weld
N/mm2 the lengths of the weld at the top and bottom. The distances between ,
, bottom and top edges of the angle sections are 2.27 cm and
is 102.5 find 5.73 cm
the neutral axis and
respectively. h[A”8- l = 652 cm = 16 47 cro! 22
i-
Rotating Discs and Cylinders
x 8 angle carrying a tensile force of 150 kN is to be connected to a
mm mm mm9. A 150
x 115
mmgusset plate by 6 fillet welds at the extremities of the longer leg, the shorter leg being
outstanding. Find lengths of welds if the C.G. of the angle is 4.46 cm from the top of the shorter
leg. Take permissible shear stress in the welds as 102.5 N/mm2 .
[Ans. = 10.26 era, l = 24.14 cm]
2
22.1. INTRODUCTION
The stress developed due to rotation in a thin cylinder has already been derived in
article 17.12 of chapter 17. In that article it has been assumed that hoop stress (or circumfer-
ential stress) is uniform in thin cylinder and is given by
0 = p x z x r2 or p x o2
co
where co = Angular speed of rotation of thin cylinder
r = Mean radius of cylinder
p = Density of material of the cylinder and
v - Tangential velocity of cylinder = co x r
a = Hoop stress in the thin cylinder.
This chapter deals with the study of stresses developed due to rotation in circular discs
and cylinders. The machine members of the rotating type and bodies like circular discs, cylin-
ders, flywheels etc. invariably rotate at high speeds. Due to rotation, these members are
subjected to centrifugal forces. The stresses are set up in the material of these members due to
centrifugal forces.
Also in case of thin disc, the stress in axial direction is zero. But in case of thick disc (or
long cylinder) the stress in axial direction (i.e., longitudinal direction) will not be zero.
22.2. EXPRESSION FOR STRESSES IN A ROTATING THIN DISC
In case of thin disc, only two stresses namely circumferential and radial stresses are
existing.
A thin disc of inner radius r and outer radius r2 rotating about its axis is shown in
i
Fig. 22.1.
VConsider an element ABCD of the disc at a radius of radial width dr and thickness T.
Let the element subtends an angle d(i at the centre.
Let a = Radial stress on the face AB,
a + do = Radial stress on the face CD,
r r
a - Circumferential stress on faces BC and AD.
C
The forces due to these stresses along with centrifugal force acting on the element are
shown on the enlarged view of the element.
VNow consider the equilibrium of the element ABCD of radius radial width ‘dr’ and
Vthickness subtending an angle dd at the centre as shown in Fig. 22.1.
919
920 rotating discs and cylinders
STRENGTH OF MATERIALS
The forces acting on the element are :
AS(0 Radial force on face and equal to a x dQ x Cancelling dQ x t to both sides, we get
r
r t,
CD(ii) Radial force on face and equal to (o + do )(r + dr) x dQ x t, a x r + 2a x dr x = (o + da)(r + dr) + pr2 to2
r r r c r
| x dr
(Hi) Circumferential force on face BC = a xdrxt
°r a x r + a x dr = [o x + a x dr+rdo
AD(iv) Circumferential force on face or r c r r r + (da x dr] + pr2 2
= a x- dr x t r .) a> x dr
y
m(u) Centrifugal force = x w 2 x r where a x dr = a x dr + r da + pr2 2 x dr
c r r oi
= mass of element = p x volume of element [Neglecting the product of two /small quantities i.e., (dor) x dr]
Dividing by dr to both sides, we get
= p x (r x dQ x dr x t) da.
r + r —j—
" Centrifugal force = (p x r x dQ x dr x t) x 2 x r \-oc + pr2„ id2„
co
= pr 2 2 dQ dr xt
co
The forces in circumferential direction will be equal and opposite. or (a - or) = r + pr2 w2 .
c (}
There are two unknowns in equation (i). They are o and o. To find their values we
require one more equation. The c r
Hence first find circumferential
second equation is obtained from stress strain relationshivp
and radial strain.
When the disc is rotating at high speed, let the radius r becomes (r + u) and dr
becomes (dr + du).
Circumferential strain,
e _ Final circumference ~ Initial circumference
c Initial circumference
— ^ ^_ 2n(r + u)~ 2kr u
2nr ~ r
and radial strain, e= radl al - InitiaI radial dth
r Initial radial width
_ (dr + du) - (dr) _ du
dr dr
The radial and circumferential strains interms of stresses are also given by,
Fig. 22.1
Resolving the force in radial direction, we get cE E
—o Ee
r r
xr dQ x t + 2o x dr x t x sin E
c
2
Equating the two values of e and e we get
= (or + dar) 0" + dr) dQ x t + centrifugal force c r,
m mwhere centrifugal force =x 2 x = mass of element —Circumferential strain,
co r where - por = —u
Er
= p * volume of element e = E
= p x(rdQ xdrxt) — — —e = ar -
du
= (p x rdQ x dr x t) x oi 2 x r and radial strain, -=
r E E dr
- pr2 x a>2 x d8 x dr x t
In equations (ii) and (iii), u is the increase in radius r due to rotation and du is the
and . dd dQ increase in radial thickness dr. These two values u and du are unknowns. They can be elimi-
Sln 2 — 2 as dQ is very small nated from equations (it) and (iii) as given below :
Hence the above equation becomes as From equation (it), we have
o xrdQxt + 2a xdrxtxdYQ=(o + do (r + dr)dBxt + pr 2 o2 dQ
r c r r
) x x dr x eu = {°‘ ~ !lar)
t
Exu-or r(o - po )
c r
1"
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS
922 ...{iu
Differentiating the above equation w.r.t. r, we get — pxrW )+2 - r dcv
2 dr
du . f da c da,.^
,
da r =- “p x 2 x 2 n(3
—drr~ co r ^ +, Cj
2o + r + p)
r
But from equation (iit), we have 2
du _ JfHs.
Multiplying by r to both sides, we get
dr ~ E E
-d3dorr 2 x 3
E-du = a -„o„, co
r r
—2 . = C^p x v + xr
x r x a+ r2 r (3 + p) x
r
2
...(«
—a7 , 20 px 2 x 3 (3 + (x) + Cj x r
(r r
co
~Equating the two values of E x oj
dr r
given by equations (iv) and (o), we get 2
Integrating both sides, we get
(u - gor) + r - 1* ^f) = <V = - xp- x co —r .v 2 2
c
x +x (3 |x) •
(where C is constant of integration)
2
- 1 ^r) - °r + "« = °
(o - pa ) + r -— p)+y= x 2 x r2 Cl ^Cg.
c r
°r p to8 +r2
(fo3 + ,
a (l + |A)-o (l + |A) + r[-^£--lx -^r) =0 VC Cs 2 2
c r co r
2 2 p x x 22...(
.1)
“ (3 + p)
8
Equation (22.1) gives the expression for radial stress in terms of constants, C\ and C2. To
Cget C substitute equation (22.1) into equation (vii).
Substituting the value of (ac - or) from equation (i), into above equation (vi), we get the value of o in terms of constants t and
c z,
+= (1- Pi + C,
2 r2 8 2
~ ^r^p- —or x r x oi
2 =0 - Ci 0? p x co*" x r . - —p (/i1 +. p) +, rC*j
pr2 (1 + p) + r -pxrx o„ = -r- (3 +
+ xra y+ p)
— --? p-Ci
=
^ ^^ ^+ + pr2 0)2 + + r “ t1 x r x fa ~ 0 Co p x co x r (3 + 4 -. j, v
+ 4p)
2
2 r 8
it ^'+ * xr M- ><r "dr =0 C, C2 x 2 x r2
+ 'a xt°2(1 + x)+r p to (- 1 - 3p)
p,
'
^na uu„ pr2 2 + = 0
r + x o) Cl + p) r -rr 2
= P^.y .-.— (1 + 3u) ...(22.2)
VDividing by to both sides, we get 2 r2 8
The constants C and C are obtained from boundary conditions. In equations (22.1) and
x 2
dr + p1 x r x w2 (1 + si) + ~drr- = 0 L.H.S. is in N/m 2 hence every term on R.H.S. should be in N/m 2 Hence C and C will
x 2
(22.2), , .
be in N/m2 .
+ ^L=-pxrx<o2 (l + p) 22.2.1. Expression for Circumferential Stress and Radial Stress in a Solid Disc.
dr dr From equations (22.1) and (22.2), it is clear that the stresses set up in a rotating disc will
Integrating, we get become infinite at the centre of the disc where r = 0. But the stresses at the centre, can not be
—o + a = - x x 2 x (1 + p) + Cj infinite. Hence C should be zero. Substituting C = 0 in these equations, we get
r c oi 2 2
p
2 —a = Ci
r2 2 x 2
where C is a constant of integration. p x co ...(A)
t r
Subtracting equation (i) from equation ( vii ), we get
(3 + p)
8
x z 2 x (1 + (i) and Ci 22 ...(B)
r ac 2
p x id +, Cr, - r - pr2 X 2 p x oj x r (1 + 3p)
2a = co 8
r
2 1 dr
*L
924 STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS
Q„ _ C x 2 x r2
O)
+, 2 ®p
+ M>
8
~Y~Yand „_ C2 pxo 2 xr! ,
c % (1 + 3 f‘)
CTo find O and 2 , the boundary conditions are used Tf r s,r>R „ ,, •
i
then radial stress is zero at these radii. This means
Substituting these values, we get two equations as
Jr ^radii, a ° Uter
r = = r mldatr^
0 1
pxa 2 xr, 2
2 r2 W<3 + [Here a = 0 at r = rj ...(;)
r
§
— ^0 =Cl p x 2 x r2
^9 to 2
+ r~22 (3+p) [Here a = 0 at r = r2]
r
...(H)
Subtracting equation (ii) from equation (i), we get
—--^(3Cft = x 2
„r 2 oj
2 p x rf pxffl ! xr,2
§ (3 + p) + + p)
1 1 P X 2
| Q)
2 .2 [rj 2
.2 (3 + p) - r 2
2 ]
r2 - nr I (3 + p) [r^ - r/]
(3 + p)(r 2 - rj 2 )
2
• (3 + p) [Cancelling (r 2 - rf) to both sides]
2
2
<»
— «.°2 P x , r, ,z r 2,
g (3 + p) 2
Substituting the value of C in equation we get
2
(*),
0=t^° = + [-£^Lq- (.3„+lp,) 2r*,r2 x 1 X 2 X r2
01
—p
X (3 + Ji)
P x oi“ x r?2 x2 r,2
<i>
5
8
—^IL + , - Cp x
9 (3 p) (3 + p)
4
g
^L _ p x u> + [r 2 + rj 2]
2
<3 ip)
Cj - x (3 + p) (r 2 + rfi
2
Substituting the values of C and C is equations (22.1) and we
i 2
(22.2), get
° = p x co + +2 r 2 + p x 2 (3 + p)r 2 r22 1 p x co 2 x
r l) a>
4x2 ' (’'z ^
l (3 + p)
= £~x(3 + p) (rU| +rr?2 ))--^-r* ...(22.7)
|
r
STRENGTH OF MATERIALS
926
^a = ^Z~9~ (3 + X (r 2 + r 2) ' P^pxo)2 (3 + p^r.? ! _ 2 ^’
2l
(1 + 3p)
^( „-r= L t, 1
3 + 1 nir2 - (3 + p) j
|
Equations ( 22 .7 ) and (22 . 8 ) give the expression for radial stress and circumferential
V ^ ^ ^stress at any radius in a disc with a central o e.
circunlferential stress
of °c- -|
2 +ri2\)+ rl2r2 W (1 + 3p)
(^= r ^T"2 o+tf J
r-R2 (3+ ^)
^-g— ^+n= £p X o2 (® + S 2
2 -2 rr +
2
(3 4-11)
2 + p) - 2 (l + 3p)
r, (3 rt
2
(3 + p) 2 2 +
= £i^(3 + p)[2r| + -^-(3 + u-l-3p)
C3 + M5 2 tf + 7jfh) (2 - 2|i)
= (3 + p) j^2rj +2r 22 (1-H)
x1 (3 + p)
rfd-li)
(3 + p) x 2 r,22 +
(lzA n= 2 2 22 9...(
+ .)
r2
(3 + p) 3+p
4 ..
Equation (22.9) gives the maximum value of o.
c
If r -> r = r, then
2
x ^(3 +^rvCvoc);max = P a
+ p)[r2 3 + |i 2
j
L
2 (3 + p) + (1 - p)r 2
r
(3 + p) (3 + p)
x r2 (3 + p + 1 - p) = P * w2 * j
which is same as of a thin rotating cylinder.
, --
f
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS ,
P^(3 --+ n) + Also at the outer radius, a = 0. This means at r = 0.45 m,
r
(3 11
2r 2 + 2 ^- a =
1 r2 r 0. Substituting these values in equation (Hi), we get
8 3 + (x |_px ^0 =
2 x(0.45 ) 2
^(3= + ,) ^ + ^-2p) , (3 +
PX (°.45)2
(3 + p) or l=
pxw -Q f 1 LI f P^iiM5)!“'; Cor +(3 !i)
if +
4
3 + ji
T—7/n 2 22 12...( . ) 7800 2 2
re) (0.45) (3
(3 + p) ri x (100 x +
j 0.3)
= 128.6 x 106 N/m2
22.2.3. Expression for Maximum o and Maximum o for a Hollow Disc with a Substituting the value of C in equations and we
0 r x
—Pin Hole at the Centre. For a disc with a pin hole at the centre, 7 0. Hence substituting (iii) (iv), get
= 0 in the equations (22.9) and (22.11), we get the expression for a and o for a disc with pin 128.6 x 10 b 22
r c
p x o> x r
(3+ p)
hole at the centre.
—pxd, , 2 128.6 x 10 6 p x 2 x 2
2 to r
(or) ma* cu g
P>2 , 2 = p x x r2 g (1 + 3p)
= ) 2 (3 + p) ...(22.13)
(3 +
22 = 128.6 x 10 6 = .
10 6
=^0to.U At r = 0, MN/mo
p x (1) x r2 c 64.3 x N/m2 = 64.3 2 Ans.
o=0At r = 0.45,
.
+ p) r2 = (3 + p) ...(22.14)
2
The maximum circumferential stress in a solid disc is given by equation (22.5), as 128.6 x 10® (100 k)2 2
(0.45)
"2
—o =
c
" 7800 x x
(3 + u) r,2 . (1 + 3 x 0.3)
(a )max for solid disc = $
c = 64.3 x 10 s - 37.024 x 10s N/m2
If we compare the maximum circumferential stress in a solid disc and in a disc with a = 27.276 x 10 6 N/m2 = 27.276 MN/m 2 Ans.
.
.
pin hole at the centre, the maximum circumferential stress in a disc with a pin hole at the Alternate Method
centre is two times the maximum circumferential stress in a solid disc. (i) Stresses at the centre. The radial stress and circumferential stress at the centre of a
mmProblem 22.1. A steel disc of uniform thickness and of diameter 900 is rotating solid disc are maximum and are equal. They are given by equation (22.5) as
about its axis at 3000 r.p.m. Determine the radial and circumferential stresses at the centre and
outer radius. The density of material is 7800 kg/m2 and Poisson’s ratio = 0.3. tym** = <O = (3 + p) r2
c
)max 2
Sol. Given : where —mr = Outer radius = 0.45
2
900
.-. Radius of disc, r = —r- = 450
mm mm mDiameter = 900
= 0.45 g°X 03t)
NSpeed, = 3000 r.p.m. = (o ) max = (3 + 0.3) x (0.45)2
c
—Angular speed, o> = = 64.3 x 10® N/m 2 = 64.3 MN/m 2 Ans. -
2nN = 2k x 3000
.
= 100 at , (ii) Stresses at the outer radius
rad/s
bU bU The radial stress at the outer surface is zero.
Density, p = 7800 kg/m 3
Poisson’s ratio, p = 0.3 The circumferential stress at outer radius is given by equation (22.6) as
The radial and circumferential stresses are given by equations (22.1) and (22.2) as 2 2
22 ------fpxw“ x r2 7800 x (100 x) x--0.-4-5 -
-
—Ci- —c§2 p x cu x r (3 + p) (1-1*)= (1-0.3)
ar = 8
' 2 + r2 ...(i)
= 27.28 x 10® N/m2 = 27.28 MN/m2. Ans.
c, c2 p x 9 x 2 At any section, there are radial and circumferential stresses which are pendicular to
co each other and tensile in nature. There is no shear stress. Hence these stresses are also known
r
as principal stresses. Hence the above stresses are also principal stresses.
and “ '2 (1 + 3p) ...(ii)
r mmProblem 22.2. If for the problem 22.1, the disc is having a central hole of 150
2 8
diameter, then determine :
As the stresses cannot have infinite value at r - 0, hence C should be zero. Hence
2
equations (i) and (ii ) becomes as
C 22
„ i - p X O) X r
r 28 (3 + p) ...(iii) (i) circumferential stress at inner radius and outer radius,
• ...(iv)
2 2 (ii) radius at which radial stress is maximum, and
r
|jp p x o> x (iii) maximum radial stress.
and & -ii (1 + 3p)
0to
a
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS 931
930
Sol. Given : Sol. Given :
From Problem 22.1,
Outer radius. r = 0.45 m. Data from Problem 22.1
Angular speed, 2
N = 3000 r.p.m., u> = (100 it) rad/s
to = {100 Jt) rad/s
mp
= kg/m3 = = 7800 kg/m3 [.i - 0.3, outer radius, r = 0.45
, ,
7800 0.3 2
p p
-( i ) Circumferential stress ( )
mm mm: 75 c
Inner dia = 150 mm, Inner radius, r, = - = 0.075 YThe circumferential stress for a solid disc at any radius is given by equation (22.4) as
(i) Circumferential stress (ac) at inner and outer radii a = [(3 + n)r22 - (1 + 3p)r2]
c
The circumferential stress in a disc with central hole is maximum at the inner radius.
7800 x (100 2
This stress is given by equation (22.9) as it)
=P^(3 [(3 + 0.3) x 0.452 - (1 + 3 x 0.3) r2]
'(o/)man 4 + ^t) ^3 + (ll = 96.228 x 106 [0.66825 - 1.9 r2]
^
By substituting different values of r in the above equation, different values of a, are
7800 x (100 kY 2 (1-0.3) ‘ obtained as given below :
(3 + 0.3) 0.45 + (0.0 75)
3 + 0.3 At r = 0, a = 96.228 x 10 x 0.66825 = 64.3 x 106 N/m2 = 64.3 MN/m2
c
= 635.109 x 106 [0.2025 + 1.193 x 10“3 = 0.05 m, a = 96.228 x 10 s [0.66825 - 1.9 x 0.05 2] = 63.84 x 10« N/m2 = 63.84 MN/m 2
3 c
= 129.367 x 106 N/m2 = 129.367 MN/m2 . Ans. = 0.1 m, o = 96.228 x 106 [0.66825 - 1.9 x 0.1 2] = 62.47 x 108 N/m2 = 62.47 MN/m2
c
The circumferential stress at the outer radius is given by equation (22.12) as = 0.2 m, a = 96.228 x 10 6 [0.66825 - 1.9 x 0.2 2 ] = 56.99 x 106 N/m2 = 56.99 MN/m 2
c
ac= Ppx_w_2 (3 + ^, )^2 ^f_l--^ = 0.3 m, a = 96.228 x 10 6 [0.66825 - 1.9 x 0.3 2 ] = 47.85 x 10 8 N/m2 = 47.85 MN/m 2
c
= 0.4 m, a = 96.228 x 10 6 [0.66825 - 1.9 x 0.42 ] = 35.05 x 10 s N/m2 = 35.05 MN/m2
c
= 0.45 m, o = 96.228 x 10 6 [0.66825 - 1.9 x 0.45 2] = 27.28 x 10 6 N/m2 = 27.28 MN/m2
c
2 x (3 + 0.3) °'3 | x 0 45 2 J(ii ) Radial stress (a
it) VThe radial stress for a solid disc at any radius is given by equation (22.3) as
-7800 x
_ (100 n 07S 2 + f
L' 13 + 0.3J
= 635.109 x 106 [5.625 x 10~3 + 0.04295] + (T 2 -r2 )
2
MN/m 2 |0
= 30.85 x 10s N/m2 = 30.85 . Ans. ^ ^= l*00 100
(«) Radius at which radial stress is maximum (3 + 0 . 3 ) (0.452 - r2) N/m2
The radius at which radial stress is maximum is given by equation (22.10) as 8
= 317.5545 x 106 (0.2025 - r2) N/m2
r = = V0.075 x 0.45 = 0.1837 m. Ans. = 317.5545 (0.2025 - r2 ) MN/m 2
(iii) Maximum radial stress YBy substituting different values of in the above equation, different values of o are
The maximum radial stress is given by equation (22.11) as r
obtained as given below :
At r = 0, a = 317.5545 (0.2025 - 0) = 64.3 MN/m2
r = 0.05 m, r
<°An« = (3 + h) (f2 - rj2 r = 0.1 m,
r = 0.2 m, o = 317.5545 (0.2025 - 0.05 2 ) = 63.5 MN/m2
r = 0.3 m, r
r = 0.4 m,
7800 x (100 a) 2 (3 + 0.3) (0.45 - 0.075)2 r = 0.45 m, o = 317.5545 (0.2025 - 0.12) = 61.13 MN/m2
r
= 317.5545 x 106 (0.375)2
o = 317.5545 (0.2025 - 0.2 2 ) = 51.6 MN/m2
= 44.656 x 106 N/m2 = 44.656 MN/m2. Ans. r
Problem 22.3. For the data given in Problem 22.1, plot the variations ofcircumferential o = 317.5545 (0.2025 - 0.3 2 ) = 35.7 MN/m2
and radial stresses along the radius of the solid disc. Take radius along y-axis and stresses r
Yalong x-axis. Also plot these variations taking along x-axis and stresses along y-axis. a = 317.5545 (0.2025 - 0.42 ) = 13.5 MN/m2
r
a = 317.5545 (0.2025 - 0.452 ) = 0
r
The variation of o and a along the radius (taking radius along y-axis and stress along
cr
r-axis) are plotted as shown in Fig. 22.4.
1
Ji
STRENGTH OF MATERIALS, ROTATING DISCS AND CYLINDERS
(i) Circumferential stress (aj
The circumferential stress for a disc with a hole at the centre is given by equation (22 8)
pxw 2 r\ r£ - r 1 + 3m
+ +r,.2\ .
ro +, ,
-pr l377* )
dm*?- 7800 ( 3 + 0 3) 2 + 0.075 2 ) + 0- 075 1* P- 45 * _f r2
'
8 (0.45 2 J
rl 3 + 0.3
1.139 x 10' 3 2
r
: 317.55 x 10 6 0.208125 + -0.575 N/m 2
0 10 20 30 40 SO 60 = 317.55 0.208125 + 1139 * 10 3 - 0.575 r 2 MN/m 2
2
Stress r
By substituting different values of r in the above equation, different values of a are
c
mobtained. Let us start from inner radius i.e., from r = 0.075
Fig. 22.4
If radius is taken along s-axis and stresses along y-axis, then variation of cr, and o will
r
be as shown in Fig. 22.5. At r = 0.075 m, cr, = 317.55 jo.208125 + - 0.575 x 0.075 2
j
= 317.55 [0.208125 + 0.2025 - 0.00323] = 129.368 MN/m2
At r = 0.1 m, L139 x 10“ - 0.575 x 2
a = 317.55 0.208125 + 0.1
= 317.55 (0.208125 + 0.1139 - 0.00575) = 100.43 MN/m2
At r = 0.2 m, 1.139 x 10" 0.575 x0.2 2
a = 317.55 0.208125 +
j
= 317.55 (0.208125 + 0.0285 - 0.023)l) = 67.83 MNN//m2
At r= 0.3 m, a = 317.55 0.208125 + 1.1-39-x-j10— 0.575 x0.32
c
j
= 317.55 (0.208125 + 0.01265 - 0.05175) = 50.05 MN/m2
I13 0.575 x 2
At r = 0.4 m, a = 317.55 |o.208125 + 0.4
c
^
j
0 0.1 0.2 0.3 0.4 O.S = 317.55 (0.208125 + 0.007118 - 0.092) = 39.13 MN/m2
At r = 0.45 m, a = 317.55 0.208125 + L139 x 10- • 0.575 x 0.45
Fig. 22.5 c
Problem 22,4. For the problem 22.2, plot the variation of circumferential and radial = 317.55 (0.208125 + 0.005625 - 0.1164) = 30.9 MN/m2.
stresses along the radius of disc with a central hole. (it) Radial stress (a
The radial stress for a disc with a hole at the centre is given by equation (22.7) as
Sol. Given :
Data from Problem 22.2, ^(3arr= 8 >-#-r+ p)[(r| + r2
m mOuter 2
radius, r = 0.45 ; inner radius, r = 0.075 , ; r
l
2
N = 3000 r.p.m. ; w = 100 n rad/s ; p = 0.3, p = 7800 kg/m3
934 STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS 935
7800 x (100 it) 2 „2 —0.0752 x 0.45 2 2
0.075 ) 5 r
(3 + 0-3) 2 +
(0.45
8
1139 x 10' x 106 N/ra2
= 317.55 0.208125
= 317.55 0.208125- 1139 : 10--r- MN/m2
By substituting the different values of r in the above equation, different values of a are
r
mobtained. = and at r = 0.45 m. Also we have got in problem 22.2, that o
Also o = 0 at r 0.075 r
r
is maximum at r = = 70.075 x 0.45 = 0.1837 m. Let us start the calculations from inner
radius (= 0.075 m).
0.208125 - 1139 *A° 3 - 0.075 2 =0
At r = 0.075 m, o = 317.55 -
0.075 2 Fig. 22.6
1139x10 0.1 2 Fig. 22.7 shows the variation of o and a along the radius, when radius is taken along
cr
At r = 0.1 m, o = 317.55 0.208125- 2
r x-axis and stresses along y-axis.
0.1
= 317.55 [0.208125 - 0.1139 - 0.01] = 26.74 MN/m2
1139 x 10' -0.1837 2
At r = 0.1837 m, a = 317.55 0.208125- 0.1837 2
= 317.55 [0.208125 - 0.03375 - 0.03375] = 44.65 MN/m2
r - 0.2 m, a = 317.55 0.208125- 1139 *,1<r‘ -°.2*
0 22
.
= 317.55 [0.208125 - 0.02847 - 0.04] = 44.34 MN/m2
r = 0.3 m, a = 317.55 0.208125 0.3’
r
2
0.3
= 317.55 [0.208125 - 0.01265 - 0.09] = 33.49 MN/m2
r = 0.4 m, 0 = 317.55 0.208125 -l^i^-0.42
r
0.4 2
= 317.55 [0.208125 - 0.007118 - 0.16] = 13.02 MN/m2
r = 0.45 m, a = 317.55 0.208125 - —1139 x 1 - 0.45 2
! ,
0.45 2
= 317.55 [0.208125 - 0.005625 - 0.2025] = 0 MN/m2
.
The variation of a and o along the radius is shown in Fig. 22.6 taking radius along Fig. 22.7
cr
y-axis and stresses along x-axis.
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS
936
Problem 22.5. For the problems 22.1 and 22.2, find the maximum shear stress in the Centrifugal force
solid disc and in the disc with a central hole.
a \ 17, R* a
Sol. The maximum shear stress at any radius is given by, T =-(a - or )
c
roax
(i) Solid disc
In case of solid disc the stresses calculated in Problem 22.1 at the centre are 0 = a -
r c
64.3 MN/m2 whereas the stresses at outer radius are a = 0 and o = 27.276 MN/m2 .
, r c
Hence principal stresses at the centre are : 64.3 MN/m2 64.3 MN/m2 0
, ,
.-. Shear stress at centre = — (64.3 - 64.3) = 0
MN/mThe principal stresses at the outer radius are : 27.276 2 0, 0
,
Shear stress at outer radius = ~ (27.276 — 0) = 13.638 MN/m2
Maximum shear stress is at the outer radius andequal to 13.638 MN/m2 Fig. 22.8
MN/mxmux. = 13.638 2 Ans. The forces acting on the element are :
.
(ii) Disc with a central hole AS(i) Radial force due to radial stress a on face = a x rdQ x t
From Problem 22.2, the calculated values of stresses are : CD(ii) Radial force on face = a x (r + dr) dQ x (t + dt)
At the inner radius, a = 129.367 MN/m2 a = 0 BC( iii ) Circumferential force on face = a x dr x t
c , r
At the outer radius, a - 30.85 MN/m 2 a = 0 AD(iv) Circumferential force on face = a x dr x t
c , r
m(v) Centrifugal force on the element =
MN/m.-. Principal stresses at inner radius are : 129.367 3 0, 0 x 2 x r
co
,
MN/mPrincipal stresses at outer radius are : 30.85 2 0, 0. where m - Mass of element
,
Hence maximum shear stress will be at the inner radius. = p x Volume of element
= p x (rdO x dr x Z)
°max = | <°c - °r> .-. Centrifugal force = (p x rdQ x dr x /) x 2 x r = p x w2 x r2 x dQ x dr x Z.
to
= | (129.367 - 0) = 64.683 MN/m2 . Ans.
The forces in circumferential direction are equal and opposite. Considering the equilibrium
of the element, and resolving all forces in radial direction, we get
— —axrxdQxt + axdrxtx sin dQ + a x dr xt x sin dQ
22.3. DISC OF UNIFORM STRENGTH = a x (r + dr) dQ x (t + dt) + pxro2 xr2 xdQxdrxt
A disc which has equal values of circumferential and radial stresses at all radii, is known — —dQ dO
=
as a disc of uniform strength. Hence for a disc of uniform strength, a = o = constant a for all But sin = as dQ is very small.
r c
radii. The thickness of the disc of uniform strength will not be constant. It will be varying as Hence above equation becomes, as
shown in Fig. 22.8 (a).
Consider an element of the disc of uniform strength at a distance r from the axis^ of a x rdQ xt+2oxdrxtx^-=ax(r + dr) dfl x (t + dt) + pm2 x r2 x do x dr xt
rotation and of radial width dr as shown in Fig. 22.8 (a). The same element ABCD along with Cancelling dQ to both sides, we get
radial and circumferential stresses (o = a = a) is shown in Fig. 22.8 b( ).
c
r
Let t = Thickness of element at radius r axrxt + 2a x dr x — = o(r + dr) x (t + dt) + pto2r2 x dr x t
t+dt = Thickness of element at radius r + dr or axrx/4axdrxf = cj(rxt + rdt + tdr + dr x dt) + pto3r2 x dr x t
a = Stress in radial and circumferential direction - a (r x t + rdt + tdr) + pt»2r2 x dr xt
dO = Angle subtended by faces AD and BC with centre O (Neglecting the product of dr x dt i.e., product of two small quantity)
o> = Angular speed of rotation. or axrxt+axdrx/-CTxrxf- 0 x rdt - a x t x dr = p x a2 x r2 x dr x t
tn = Thickness of element at radius r = 0. or -axrxdt=px(o2 xr2 xdrxt
—
ROTATING DISCS AND CYLINDERS 939
938 STRENGTH OF MATERIALS
mm mLet t = Thickness of the rotor at r = 350
= 0.35
2 x r2 x dr p x (a x r Using equation (22.16), we get
pto
Integrating to both sides, we get m m(Here
p is in kg/m3 r in and a in N/m2 . If tn is in then t
,
22 will also be in m. If t0 is in mm, t will be in imp).
—log t = - £ * —— x
^ o2 + Constant - 8000 X (140 *)2 « (0.35)8
C C = 25 x e 2x85x106
The constant may be taken as = log where is also constant.
e
= 25 x e" 1115 = 8.19 mm. Ans.
22 ~,
+ log,
t = - p x to x r
- C, ms —where
1115 - -=-!-=• = = 0.328
l°g e'
c 2q
L e 3.049
22 mmProblem 22.7. The minimum thickness of a turbine rotor is 9 at a radius of300 mm.
= - p x ai x r
Clog - MN/mIf the rotor is to be designed for a uniform stress of 200 2
t log find the thickness of the rotor
e ,
O_ 2 2 mmat a radius of 25 when it is running at 9000 r.p.m. Take p = 8000 kg/m2.
to r
p x x Sol. Given
:
mm mm mThickness, t = 9
x when = 300 = 0.3
Uniform stress, a = 200 MN/m2 = 200 x 106 N/m2
Density, p = 8000 kg/m3
Speed,
N = 9000 r.p.m.
p x o) x r N —2it
or f = Cxc 20 w=
-.(22.15) .-. Angular speed., = 2it x 9000 = 300 it rad/s
60 60
mm mFind thickness when radius = 25
The value of C is obtained from boundary condition. = 0.025
At r - 0, i = t0 Using equation (22.15), we get
Substituting this condition in equation (22.15), we get
Cxe# = C t=Cx e 2a
i = mm mm m(t) When t - 9
0 radius, r = 300 = 0.3
Substituting C = t0 in equation (22.5), we get ;
Hence above equation becomes,
p x 2 x r2 2 x 2
id o> (0,3)
xi=t t0 - p x
20 ...(22.16) 9=Cx e 2a
.
Equation (22.16) gives the thickness of disc of uniform strength at any radius. The rotor - p x 2 x 0-Q9
to
of a steam turbine has constant strength throughout the radius and are designed according to
=Cx e Ta
equation (22.16).
AProblem 22.6. steam turbine rotor is running at 4200 r.p.m. It is to be designed for C- ---Vw
MN/m mmuniform strength for a stress of 85 - P y. g>~ x 0.09
2 If the thickness of the rotor at the centre is 25 e 2a
and density of its material is 800 kg/rn3 then find the thickness of the rotor at a radius of
,
350 mm. p x x 0.09
=9x e 2cr
Sol. Given : N = 4200 r.p.m. m(»} When thickness is t, radius is 0.025
Speed,
Substituting the above values in equation (t), we get
.-. Angular speed, —o> = 2 nN = 2it x—4200 = 140 it rad/s - p x m1 x 2
60 bO
(0.Q25)
t=Cx e 20
mmThickness at the centre, tQ = 25 px(l) 2 X 0.09 p x or x 0,000625
2o
Density, p = 8000 kg/m3 = 9 x e 20
Allowable stress, o = 85 x 106 N/m2
Rotor is to be designed for uniform strength. This means the radial and circumferential p X 2 X 0.09
CO
stresses should be equal at all radii.
From equation («), C ~ 9 x e
STRENGTH OF MATERIALS
p x 2 x 0.09 pxm^x 0.000625
id
j
j
2a
2a
J
{
px 2 (0,09 - 0.000625)
to
M2 8000 X (300 z X 0,089375
Jt)
* 0.089375
=9xeP 2 * 200 x 10 s
= 9 x e „2o
= 9x e 1 -587 = 9 x 4.892 = 44.03 mm. Ans.
22.4. LONG CYLINDERS
In case of rotating long cylinders, it is assumed that the longitudinal strain (e ) even at
f
high speeds of rotation is constant (this means that the cross-sections remain plane, which is
true for the sections away from the ends.)
Let a = Circumferential (or hoop stress)
a - Radial stress
r
Of = Longitudinal (or axial) stress
e = Circumferential strain
c
e ~ Radial strain,
r
e. = Longitudinal strain.
When the cylinder rotates at high speeds, let the radius r becomes (r+u ) and 'dr' becomes
(dr + du). Then circumferential strain,
2it (r + u)- 2nr _ u
e
° 2jrr r
err = (dr + du) -dr du
Radial strain, ~z ~~r-
dr dr
The stresses ac , o and a, are acting on any element of a section of the cylinder. The
r
strains produced by these stresses in circumferential direction, radial direction and longitudi-
nal direction are circumferential strain, radial strain and longitudinal strains respectively.
These strains in terms of stresses are given by.
OU = 14
+ a)
'
Circumferential strain. Y-I 7(o
'
or p du
Radial strain,
a, p
cainudu liAoMnigg.iiLtUuUd.iiini.acln Dsturaaiinn,, g o+(ycrr
Cc )
where u = Poisson’s ratio, and
E = Young’s modulus
In equations (i) and (ii), u is the increase of radius r due to rotation and du is the in-
crease in thickness dr. These two values u and du are unknowns. They can be eliminated as
given below :
From equation (t), we have
“= [o - i L (W +
,
or Eu = r [o - p(o + a )]
c r ;
ROTATING DISCS AND CYLINDERS 943
STRENGTH OF MATERIALS ar) equation we get
942
Substituting this value of {a - in (tnit),
c
Substituting this value of “f in equation (w), we get r + pr2(02 + r(l - p) - p x r x dr = 0
dr
( dac da
r
+|X
- ar) + hr ^rdar + p—r dr (1 - p) + pr2 cu2 + r (1 - p) ar = 0
(1 + p) (o r -5r
c
°P+ -(1 (xi) ((aa,, dac da z da c 2 jgr. _ 0
( +r r dr J
[ dr ^ dr
^ dr
^"a+io<o,-«X+ r (1 ""’ > do r + do^ __ 2Z _ pxr 2
[
pr ai xoi
]"|I <1 * ,‘) 0 r(l-p) (1-p)
Tfr' dr dr
~or (1 + M-)= 0 Integrating both sides, we get
{1 + ix)(oc - or) + r(l- 2 - (xx t — —q + o = - -
n)
- - x 2 x + Cj ...(i.T)
co 1
.Xl-|dai,)-^-^^(l^d)o^r =0
c (1-p) 2
(
or (l + (1)(o{ -or) + r(l + where C, is a constant of integration.
(o -o )^<f-|0^-l*xrx^=0 values of a and a the two equations iix) and iviu) are to be solved. As
el r c,
-(ou) To get the
or d°~ we should eliminate a, from these equations. This can be done
equation (viii) contains ,
[Cancelling (1 + p)l equilibrium of an element ABCD of radius r radial width d,r , by subtracting equation {viii) from equation (ix), then we get
Now considering the angle dO at the centre as shown in Pig. 22.9, , and
thickness r subtending an
we get
— -r~-x
2a, = - 2 x O + Cj - pr2*)2
co x
centrifugal force rfr
ya = (o + dar) (r + dr) dQ x i +
r x + 2a x dr x x sin r
x rde t c t
(Resolving the force in radial direction) ^2o +r = -(T^ x “2x T~ prW+Ci
r
m mwhere centrifugal force = x w2 x r where = mass of element
pxwo xr o + 2 +C
= p x volume of element 1
= p -x {rdQ x dr x t)
= (px rdQ x dr x t) x 2 x r
co
= pr2 x 2 x dQ x dr x t i x 2 x 2 fl±i-2p
to o) r
— —and dQ — d0 dO • very smaolill
sin as is
Z£ 2 x 2 3 - 2p
D r f
Hence the above equation becomes as
Z r + dar) (r + dr) dQ x t 4 pr2co2 x d6 x dr x t
xdrxixy=(oa Multiplying by r to both sides, we get
r
x rde x t + 2a
c
—2ra px(it 2 xr 3 ^ ~ ^9
Cancelling dQ x t to both sides, we get —+rn2 da r =-
f
-
i_ u
a x r + 2a x dr x = (a + dar) (r + dr) + pr 2 2 x dr
r c r <o
^ 2 3 (3 - 2p) '
r 1
^ (rX0 w 1-9or px x
d ,, =,
x pr2o>2 x dr 2
o xr + a y.dr = [o Y.r + o xdr + rda + (dar) dr] + ^)
or r c r r r
or a x dr = a x dr + rdo, + pr2 2 x dr Integrating the above equation, we get
co
c r (da x dr]
r
[Neglecting the product of two small quantities i.e., )
p x 2 x 4 C(3 - 2p) 2
to
Dividing by dr to both sides, we get 2^4= ~ ' r + xr
^r2 x 1
+ r —dop^_ 1-jT 2
a =o + pro o where C is another constant of integration.
cr 2
or
dr
r^ ^or ' -<«“>
-o(a = +P 2
c
r)
944 STRENGTH OF MATERIALS
p x 2 x r2 f 3 - 2p ~)
co
+
=y ^-L_Cor 2 Z f3-2p^
a Ci 2 o x CD x r
+
...(22.17)
r
J
The equation (22.17) gives the expression for radial stress set up in a rotating long
cylinder. To get the expression for circumferential stress (he., cr.), substitute the above value in
equation (ix).
[Cj C2 x 2 x r2 (3 - 2p)l 22
to
TT r~o ' +a to x r
;
—p p X + c.
=-
C C2 p x 2 x 2 (3 - 2p)
x at r
^T“7^C+, r, + I^TT
§
'
C C2 22
1 p x tu x r
8(1- p) [4 - (3 - 2p)]
CCi ~ 2 ~ p x 2 x 2
72 2 co r
” (1 + 2p)
8(1 - p)
|£i)
The equation (22.18), gives the expression for circumferential stress set up in a rotating
long cylinder.
22.4.1. Expression for Circumferential and Radial Stresses in a Solid Cylinder.
For a solid cylinder, at the centre r = 0, the stresses will become infinite. But the stresses
cannot be infinite at the centre. Hence in equations (22.17) and (22.18), the constant C should
2
be C Hence
zero he., 2 = 0. the stresses for solid cylinder will be as :
The constant C from these equations will be obtained from boundary condition.
x
Now for a solid cylinder on the free surface, o = 0. This means at r - r.,
r
a - 0 where r, = radius of the surface of the shaft.
r
2
From equation (x) in which o * 0 at r= r„ we get
r
c 1-9" 2 r22
x _ p X X ( 3 - 2p )
co
2 8l J
Cj p x 2 x 2 f 3 - 2p
o) r2
2 " 8 l 1-9
—r r
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS 947
946
subtract equation Uv) from equation (in), x 2 2 f 3 - 2p rf x r| 1 _ p x 2 x 2 ^ f 3 - 2p
co ^| <o
C from these equation, p p x co r
L
To eliminate and 2x4cl = (r 2 + r2 2 ) + ^
t
Pp ^x o<o_2 x l 1-1* . u-t* j
Ct pxo> 2 ((33--22li,) x)_C^2
~r x ri2x* n - u) + 8 2x
on = o
r? 2 r2 x 2
co f3-2p’ i r' 2
p x .2 + r„22 - r2 ...(22.23)
1 2
1 l] pxco 2 ^x f3^--^2|.L\[r 2 -r 2 8 l 1-1* ,
]
Equation. '22.22) gives the expression for circumferential stress and equation (22.23)
2 I J2. ,.2 l l-V* )
gives the expression for radial stress in a hollow long rotating cylinder.
(r£ - 2 2 f3 -r(T22 2 Maximum circumferential stress (oc) max
) px(i> ) From equation (22.22), it is clear that circumferential stress will be maximum when r’
x is minimum.
"2 22
r r2 Hence circumferential stress will be maximum at the inner radius where r is minimum
l
2 3 - 2p z - r 2
(r 2
+ p x co ( x )]
~“^Co [Cancelling
YFYF l i-i*
c = _ P xa)2 filial r 2 r. Hence substituting r = r L in equation (22.22), we get max. value of a,,
t
*4*
CSubstituting of as
value of C in equation (iii\ we get the value
2 1
this
mp 2 1 + 2p 2
„0 C 1 x CO f 3 - 2ix ^ r2 2"| _ px r'l + 2 2 - | ...(22.24)
x
= + -2 [ r
-TT
8 l-p
l.
_c , Maximum radial stress (o
t
r
y r"lTTJ=
3-22 i 2 _PfJgl x>- ) raax
p X a) t^ r ^ From equation (22.23), we cannot say that a will be maximum when r is minimum.
r
[ 28
Ct pxto 2 f3-2Y 2 p*jS - Hence for a to be maximum d(o r ) shoulTdJ be zero.
r
r
~2 ~ ~~8
i 1- P j 2 8 l 1-1* J
2 x r'22_
d 2 f 3 - 2)i rr 1 +, r2 _ r
l 2 2
P x f° l
p x 2 f 3 - 2p. +Ir,* '‘ *] dr 8 1- P
co
1
2 (3- 2p
p x co
P*«S(3^) 2 + r22] 8 l 1-P .
[ri
1 4 ^ 1 - M- ) 2r xrl-2r=0
l
and we get
values of C and C in equations (i) (ii), 3
the 1 2 r
Substituting
pxo> 2 f3-2u'l , , x 2 ("3 - 2p \ rfr£_1 px o)_j or 2r 2 x r 2 - 2r4 = 0
to _
9\ p _ 2
[ r4 = r 2 x r 2 r2 =
L or r x r
or
L 2
, ...(22.25)
2 r2 ( 1 + 2u or r = VJrr,i x r92
2 x Equation (22.25) gives the value of r at which radial stress will be maximum. Substitut-
2 p x co f 3 - 2p^ __ r xr2 p x (a
to ( 3 — 2p t
p x [r, 2 + 2 +
r 2
] 2 Ving this value of in equation (22.23), we get maximum value of radial stress.
r
2 3a CM 3 - 2 p'
( 3 - 2p (Y +> +a > XCL
p x o> f
CO
x 2 (3 - 2u. p x 2 , - r L]2
co 0)
If3-2p'
p 2 + - 2r irJ = [r
r2 2
22 22...( . ) t'Y 8 1-P
1
l
...(22.26)
Equation (22.26) gives the value of maximum radial stress.
:r
STRENGTH OF MATERIALS
AProblem 22.8. long cylinder is rotating at a speed of 3600 r.p.m. The diameter of the
mmcylinder is 500 and density of its material is 8000 kg/m3. If Poisson’s ratio = 0.3, then :
(i) Calculate the maximum stress in the cylinder and
( ii ) Plot the circumferential stress and radial stress along the radius.
Sol. Given : mm: 500
Diameter,
Radius, —r = = ——- = 250 mm = 0.25 m
2
Speed of rotation, N = 3600 r.p.m.
Ang*ular spreed, 2jxN 2jt x 3600
’ = 120 ji rad/s
60
60
Density, p = 8000 kg/m3 Poisson’s ratio, p = 0.3
,
(i) Maximum stress in the cylinder. Maximum circumferential and radial stresses occur
at the centre, where r - 0. It is given by equation (22.21) as
p x co 2 x 2 3 ~ 2p
r2
),r 8 l l-(t
8000 x (120 2 x 2 (3 -2 x 0.3
jt)
(0.25)
8 1-0.3
= 8000 x 14400 x —x 0.0625 x 2.4 .N./. m,2
8
0.7
= 30.45 x 10® N/m2 = 30.45 MN/m2. Ans.
(ii) Plot of the stresses :
Circumferential stress (a ) Circumferential stress is given by equation (22.20), as
a. = cp X O) 1 + 25p“ r_2
—„2 - ~
r2
- 2 \i)
8000 x (120 2 3 -2x0,3 1 + 2x0,3 1 2
ji) 3 -2x0.3 )
[
8 1-0.3
8000 x 14400 x 2 —0.0625 - 2
it
8 2.4
= 4.872 x 108 [0.0625 - 0.667 r3 ] N/m2
= 487.2 x 10 6 [0.0625 - 0.667 r2 ] N/m2
= 487.2 (0.0625 - 0.667 r2 ) MN/m2
r = 0, a = 487.2 x 0.0625 MN/m2 = 30.45 MN/m2
r = 0.05 m, c
r = 0.10 m,
r = 0.15 m, a = 487.2 (0.0625 - 0.667 x 0.05 2 ) = 29.63 MN/m2
r = 0.20 m, c
r = 0.25 m,
o = 487.2 (0.0625 - 0.667 x 0.1 2 ) = 27.2 MN/m2
c
a = 487.2 (0.0625 - 0.667 x 0.15 2 ) = 23.14 MN/m2
c
MN/mcr
c
= 487.2 (0.0625 - 0.667 x 0.22 ) = 17.45 2
a = 487.2 (0.0625 - 0.667 x 0.252 ) = 10.14 MN/m 2
c
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS
{ii) Radius at which radial stress is maximum. The radial stress is maximum at radius
given by,
mr = x r2 = -J0.1 x 0.25 = 0.158
(iii) Maximum radial stress. Maximum radial stress is given by equation (22.26) as
3-2
f
a ^(r, \
- P x0j2 (r _ r )2
( r)max g [ 1 - |A J 2 1
_ 8000 X (120 2 f3 - 2 x 0.3 (q ' 25 -
it) ^
0 1)2
x
0.15 8 1-0.3 J
Radius (r)
= 487.2 x 106 x 0.0225 = 10.962 x 10e N/m2
Fig. 22.11 = 10.962 MN/m2. Ans.
Problem 22.9. If in Problem 22.8, the long rotating cylinder is having a central hole of (iu) Variation of stresses along the radius
(a) Circumferential stress (aj. The circumferential stress at any radius is given by equa-
diameter 200 mm, then determine . tion (22.22) as
(j) Maximum circumferential stress pxto 2 f3- 2p.^| !" 2 + 2 + r xr2 f 1+2lO 2
(ii) Radius at which radial stress is maximum, i r
°‘ = B r2
^2 [ 3 - 2u j
Maximum radial stress,
( iii ) and radial stresses a ong era is nftu~ nvllnrlpr
(iv) The variation of circumferential
3|~ - 2 x 0.3 ]~ x 1+ 2
_f
] fl 2
Take all other data from Problem 22.8. 2 + 0 25 2 + r 2
r
8000 x (120 Jt) '
^=
“~ 8 L 1-0.3 J [ 1^3 - 2 x 0.3 J
Sol. Given : mmOuter radius. r2 - 250
- 0.25
mmOuter diameter = 500 - 0.1
mmInner diameter = 200 mm m/.
Inner radius, r = 100 = 487 2 x 10 6 x 0.01 + 0.0625 i 2 * ~ 0.667 x 2 N/m2
L r
NSpeed = 3600 r.p.m ; oi = 120 a rad/s ; p = 8000 kg/m , n 0.3
(i) Maximum circumferential (o The circumferential stress at any radius
c
stress ) max - °'00< 62 -
? r2 )
MN/m= - 2
487.2 x ( 0.0725 + 0.667 x
given by equation (22.22) a.s
2 r| f 1 + 2(1 YBy substituting the different values of in the above equation, we get different values
r,
of o . Let us start the calculation from inner radius {i.e., r = r = 0.1)
x
c
Vinner Hence at the °- 000325 2 MN/m2
circumferential stress occurs where radius is minimum. = - 0.667 x O.l = 62.5
The maximum = r,), the circumferential stress will be maximum, At r = 0.1, a 487.2 fo.0725 +
c ]
radius (i.e., at r L o.l J
2(3- 2ll 22 —1 + 2^.} 2 At r = 0.125, aa== 487.2 00..C0725 + - 0.667 x 0.125 2
2 'l r2 r c
. 3 - 2Vi J i
^c^max = 487.2 [0.0725 + 0.04 - 0.0104] = 49.73 MN/m2
1 + 2|1 ,2 At r = 0.15, n - 487 2 L0„.„07„25c + 0.000625 0n.c6c6n7 x 0.15
|1 5
2
3-2n J' L 0.15 J
ft*) = 487.2 [0.0725 + 0.0277 - 0.015] = 41.5 MN/m2
L8000 x (120 K'y f 3 - 2 x 0.3 ) iiliiM) 2 ^?p^-At + 0.667 x 0.175 2
2 2 x r = 0,175, a = 487.2 [0.0725
o.i c
l + 2 x (0.25)
3 - 2 X 0.3 )
8 "l 1-0.3 )[ ' = 487.2 [0.0725 + 0.0204 - 0.0204] = 35.32 MN/m2
~= 487.2 x 106 j^O.Ol + 0.1250 - 2 At r = 0.2, cr, = 487.2 [o- 0725 + - 0.667 x 2
x 0.1 0.2
j
= 62.5 x 106 N/m2 = 62.5 MN/m2. Ans. = 487.2 [0.0725 + 0.0156 - 0.02668] = 29.92 MN/m2
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS
At r = 0.25 m.
^000
At r = 0.225, 1.0725 + P- . _ 0.667 x 0.225 2
: 487.2 [0.0725 + 0.0123 - 0.0337] = 24.86 MN/m2
La
^c
0.000625 _
0.667x0.25
At r = 0.25, = 487.2 I 0.0725 +
q
= 487.2 [0.0725 + 0.01 - 0.04168] = 19.88 MN/m'2
V(6) Radial Stress (a ). The radial stress at any radius is given by equation (22.23) as
t
P^ ^ 140 = --A
x fil^l r r
\ rl-
8000 x (120 2 ( 3 - 2 x 0,3 ) 2 p n5 2 0,12 X °- 253 r2
jc ) q.i
, 2
,
(
= 487.2 x 10 s x O.OI + Oj062g _ 01)00625 r2 N/m2 ——0 0.05 0.1 0.15 0.2 0.25
z Radius (r)
r
Fig. 22. 12
0.000625 MN/m2
= 487.2 x 0.0725 The variation of circumferential and radial stresses are shown in Fig. 22.12.
VBy substituting the different values of in the above equation, we get different values
m-of a,.. Also ar is zero at r = 0.1 and r - 0.25 m. HIGHLIGHTS
At r - 0.1 m, a = 0 j[
r
- °-0006 5 0.125 2 1. The hoop stress (or circumferential stress) in a rotating thin cylinder is given by
At r = 0.125 m, o = 487.2 x fc.0725 2^ a = x 2 x r2 p x o2.
(o
L 0.125 p or
= 487.2 x [0.0725 - 0.04 - 0.0156] = 8.23 MN/m2 2. For a rotating solid disc, the stresses at any radius r are :
2
At r = 0.15 m, a = 487.2 x fo.0725 - °-000625 _ 0.15 2 ^= P - <0 r2 + 3p) r2 ]
2
L 0.15 i - [(3 + p) 2 - (1
= 487.2 [0.0725 - 0.0277 - 0.0225] = 10.86 MN/m2 and a=P 2
r
mWe have already calculated that a is maximum at r = 0. 158 maximum of a ~ (3 + u) [r 2 - r2 ]
r r
and value
is 10.962 MN/m2 where u - Circumferential stress,
c
.. At r = 0.158 m, a. = 10.962 MN/m 2 cr = Radial stress,
r
-Ppa 2aiV
° ,0 ° to = Angular velocity =
At r = 0.175 m, = 4587/..2Z|{U".oU.0/7Z2O5 - 0.175 52 - 0V.L1i7D5 .
L J b(J
487.2 x [0.0725 - 0.0204 - 0.0306] = 10.47 MN/m2 u = Poisson’s ratio,
At r = 0.2 m. cr = 487.2 0.0725 - 0.000625 - 0. 22 3« At the centre of rotating solid disc, the radial and circumferential stress are maximum and are
equal.
= 487.2 [0.0725 - 0.0156 - 0.04] = 8.23 MN/m2 KL —g—= m<<*r> ax = <3 + r22-
At r = 0.225 m. a = 487.2 0.0725 - 0.000625 0.225 2 4. The circumferential stress at outer radius of a solid disc is given by
= 487.2 [0.0725 - 0.0123 - 0.0506] = 4.66 MN/rri2 pxw„ 2 xr22
= (1 - p).
STRENGTH OF MATERIALS ROTATING DISCS AND CYLINDERS 955
954
5. For a rotating disc with a central hole p x 2 3 -
(i) The stresses at any radius are : f
o>
8 U-b
— ^o = P xt“2 ri _" ,2 11, For a rotating hollow cylinder :
r ^2 (i) The stresses at any radius are :
(3 + ji) ,2 + *2 > _ 1
~
Oi
3 - 2p ^|
f
2 f l+3p l.2 ^-l3^2 r r2 ( 1+
2 +. I|2)' +. Tt x cc i
£^( ^-^j^,)^ ^and f roi+r22 +
ac= 3+ 8 l 1-h
(ii) The circumferential stress is maximum at inner radius and is given by x 2 |j-2p^ „ .Q! Jr^k22 _ 2
0>
U-pJLp r2 + r| r2
P xt°2 \rr?2++ [{ 4i-r^-1 2 8
- r, ]
(„ \ - f3 + ul o(ii) will be maximum at inner radius and is equal to
c
(iii) The radial stress is maximum at the radius given by ,P1(oc)'max xm2g. 2 + Zr*2 2 Vl + 2p
r- x r2
^ 1- g 11\nL U(
(iu) The maximum radial stress is given by + 2pj
a(iii) is maximum at r = xr2 and is given by
r
£ir"<v ~arr/>mma«x = g (3 + iO Kr2 - r^2] x 2 (3 - 2p
to
The value of o at the outer radius is 1-P8 l 2-[(r r ),]-
c i
J
(y)
2 2 . f 1 + 1* 1 J-
EXERCISE 22
6. The maximum radial and circumferential stresses in a disc with pin hole at the centre are . (A) Theoretical Questions
p X 2 X rf ^ 1. Find an expression for the circumferential stress developed in a thin rotating cylinder.
co -'2. Find an expression for the circumferential and radial stresses developed in a rotating solid disc.
&r\nax =
( 3. Prove that the circumferential and radial stresses are maximum and are equal at the centre of a
8 rotating solid disc.
” p x 2 x rf 4. Prove that the circumferential stress at the outer radius of a rotating solid disc is given by
io 22
^c^max 4 -p x o) x r2
7. The maximum circumferential stress in a disc with a pin hole at the centre is two times the
maximum circumferential stress in a solid disc.
A8. disc which has equal values of a and a at all radii, is known as a disc of uniform strength.
c r
a “ a = a for all radii. 5. Prove that the radius at which radial stress is maximum in a rotating disc with a central hole is
rc
9. The thickness of a disc of uniform strength is given by
- p x « 2 * '*2 given by r= x r2 .
t— e 2a Also find an expression for the maximum radial stress.
where £0 = Thickness at r = 0. 6. Prove that the circumferential stress in a rotating disc with a pin hole at the centre is two times
10. For a rotating solid long cylinder
the maximum circumferential stress in a rotating solid disc.
( i ) stresses at any radius are
7. What do you mean by a disc of uniform strength ? Find an expression for the thickness of a disc
pxwg 2 3 - 2p. 2 _ ( 1+ ^ r2
2 of uniform strength.
(^
I, 1 - ii J [ 8. Prove that in case of a long solid rotating cylinder the maximum radial and circumferential
pxoi 2 3~ 2P p x 2 f 3 - 2p
f ) to
and- r_ 2 stresses at the centre are equal and they are I x r,“ in magnitude.
j
o(ii) is maximum at the centre and is given by 9. Prove that in case of a rotating hollow cylinder, the radial stress is maximum at radius = x r2
r
and is given by
('voV)max = - - xr. 2 f 3 - 2p
g ^ 1-p J =
;rj
(Hi) ct is maximum at the centre and is given by
c
, pxo, 2 fiz^il Xr
1-n J
) .
STRENGTH OF MATERIALS
956
XO. Prove that the thickness of a disc of uniform strength is given by
- p x w2 x 2 23
r
t = t0 e 2°
where /0 = Thickness at r = 0
o = Stress developed due to rotation. Bending of Curved Bars
(B) Numerical Problems 23.1.
1. Find the hoop stress developed in the thin rim of a wheel of radius 400 mm, when it is rotating INTRODUCTION
at a speed of 3000 r.p.m. The density of the material of wheel is 8000 kg/m 3 .
[Ans. 126.33 MN/m2]
A mm2. steel disc of uniform thickness and diameter 900 is rotating about its axis at 2400 r.p.m.
Determine : For a straight beam, the bending equation ^ has been derived in chapter 7.
(i) radial and circumferential stresses at the centre, and This equation can be applied, with sufficient accuracy? to the beams or bars having small
(ii) circumferential stress at the outer radius. ceitnhcia.tpiwtalheirccudhrevaaartseurwheia.tvhiHntoghweleabvreegnrediitnhnietgiamolfacccuuhrrivvnaeetdumrbeea,rmsbt,heewrhssiicsmhpulcaehrebaeshnacdvriiannnggeelhqaouroagktesi,ionnicthciaaailnnncoultrivnbakestuuarsene.dd ArTcithnuig-ss
MN/mMN/m[Axis. ( i 41.152 2
Take p = 7800 kg/m2 and Poisson’s ratios = 0.3 2 (ii) 17.46
]
mm3. If for the question 2, the disc is having a central hole of 200 diameter, then determine :
(i) circumferential stress at inner radius and outer radius. ally curved bars means the bars of large initial curvature.
(ii) radius at which radial stress is maximum and Generally, if the radius of curvature is more than 5 times the depth
is said to be having small initial curvature. But if radius of curvature is
(Hi) maximum radial stress beam of the beam the
(in) maximum shear stress. less than 5 times
the depth, the beam is said to be having large initial curvature. Hence for large initial curva-
MN/m MN/mMN/m MN/m[Ans. (i) 82.16 2
2 21.5 2 (ii) 0.212 m, (iii) 24.89 2 (iv) 41.58 ture, the radius of curvature is small. Also curved
, , , ] donot coincide.
for beams, the neutral and centroidal axes
4. For the data given in question 2, plot the variation of circumferential and radial stresses along
the radius.
5. For the data given in question 3, plot the variation of circumferential and radial stresses along 23.2. ASSUMPTIONS MADE IN THE DERIVATION OF STRESSES IN A CURVED BAR
the radius.
A6. steam turbine rotor is running at 4800 r.p.m. It is to be designed for uniform strength for a Before deriving the expression for the stresses developed in a curved bar when it is
mmstress subjected to some bending moments, the following assumption are made :
of 90 MN/m'2 If the thickness of the rotor at the centre is 30 and density of its material
.
[Ans. 5 mm]
is 8000 kg/m 3 find the thickness of the rotor at a radius of 400 mm. 1. Transverse sections which are plane before bending remain plane after bending.
,
mm7. The minimum thickness of a turbine rotor is 10 at a radius of 300 mm. If the rotor is to be 2. Hooks law is applicable. This means the working stresses are below the
designed a uniform stress of 150 MN/m2 find the thickness of the rotor at a radius of 30 mm, portionality. limit of Hnro-
,
for
[Ans. 62.84 mm]
when it is running at 8400 r.p.m. Take p = 8000 kg/m3 .
A mm180.. long cylinder is rotating at a speed of 4200 r.p.m. The diameter of the cylinder is 400 ^ The longi tudlnal fibres of the bar, parallel to centroidal axis exert no pressure on each
and
other. This means the distance between any longitudinal fibre from centroidal axis is same
density of its material is 7800 kg/m3 . If Poisson’s ratio = 0.3, then before and after bending.
\ (i) calculate the maximum stress in the cjdinder and 4. Each layer of the beam is free to expand or contract, independently of the layer above
(ii) plot the variation of circumferential and radial stresses along the radius. or below it.
[Ans. (i) 25.866 MN/m2
]
9. If in question 8 above, the long rotating cylinder is having a central hole of diameter 100 mm, 23.3. EXPRESSION FOR STRESSES IN A CURVED BAR
then determine :
(i ) maximum circumferential stress, Fig. 23.1 shows the two positions of a curved bar, one position lie.. Fig. 23.1 (a)] is before
(ii ) radius at which radial stress is maximum, Mbending whereas the second position [i.e., Fig. 23.1 (6)] is after bending when some moment.
(iii) maximum radial stress, LMis applied at the ends of the bar. The centroidal axis is shown by line and LM' in the two
(iv) variation of circumferential and radial stresses along the radius of the cylinder. Take all positions respectively.
data from question 8. [Ans. (i) 52.27 MN/m-, (ii) 0.1 m, (iii) 14.55 MN/m-] YConsider any fibre EF at a distance from the centroidal axis LM. This fibre takes the
Mposition of EF' when the moment is applied. Then the fibre EF' will be at a distance of y"
Calculate : (i) maximum and minimum circumferential stress (ii) maximum and minimum ra-
mm mmdial stress in a thin uniform disc of inner diameter 50 from the centroidal axis LM'
and outer diameter 250 when
rotating at 8000 r.p.m. Take u = 0.33 and p = 8000 kg/m3 .
[Ans. (i) 74.9 MN/m 2 17.9 MN/m 2 (ii) 23.8 MN/m2 0] 957
, , ,
' ',
STRENGTH OF MATERIALS
958
Refer to Fig. 23.1 (a). Here: .
LMR = radius of curvature of centroidal axis LAI, at the centre of curvature 0,
0 = angle subtended by centroidal axis
y = distance of fibre EF from centroidal axis LM.
Refer to Fig. 23.1 (6). Here :
M = uniform bending moment applied to the bar,
LMR'
0'
= radius of curvature of centroidal axis at the centre of curvature 0 ,
=
angle subtended by centroid axis LM'
y’ = distance of fibre EF' from centroidal axis LM'.
Let a = stress in the strained fibre EF' due to bending moment M, and
e - strain in the same fibre EF',
e 0 = strain in centroidal layer LM.
° ...... O'il Curved bar AB’CO. after bending
(a) Curved bar ABCD before bending
Fig. 23.1
Now, from Figs. 23. 1 (a) and 23.1 (6), we have
EF R-
EF = (R + y) x 0 and ( + y’) x 0 toN
Also LM = R x 0 and LM' ^ R' xfl’
Strain,
^EF' - EF R( + /) x 8‘ - (R + y) x 8
e = = (R + y) x 0
(R' + y') x 6' (R-»y)x0 _ (R‘ +y’> „ g'
(R + y) x 0 (R + y) x 0 (R + y) 6
=1 4- p —(R' + y') 6'X
(R + y) 0
LM' - LM R' x 8' - R x 9 _ R^
Also, strain, e =
0
R' 8' ...(H)
X
or
R0
960 STRENGTH OF MATERIALS BENDING OF CURVED BARS
Re 1+ v T 1 l\ 2
°i
y (l + e0 )
= E J e0 y dA + Ej
fl IV MM1 ,(1 1
eo)
W~R. + = Q + E(l+en
= en + = ea + l
7 ITT /T
4i + i+ £
The equation (iii) gives the expression for the strain produced in fibre EF under the
action of moment M.
Now the expression for stress produced in the fibre EF can be obtained. = E( 1 + en
We know that ———Strpss
=E
/^L-Ah2
Strain
Jf f, VI
Estress = x strain
a=Exe
[W~Ry(n1 + eo)if 11
where h2 = a constant for the cross-section of the bar
E= x e0 + M = £(1 + e0)
xAx/t 2 ...(of)
The above stress is produced due to bending moment M. Hence this stress is known as In the above equation e and R' are unknown.
0
bending stress.
Let us find the value of e and R'
Consider a small strip of area dA at a distance y from centroidal axis LM.
The force on the strip = stress x area of strip 0
= a x dA We have seen that force on the strip of area dA - a x dA
Moment of this force about centroidal axis = Force x y
Total force on the cross-section will be obtained by integrating the above equation.
= (o x dA) x y
The above expression gives the resisting moment offered by the strip. Total resisting F E.". Total force, = J cr x dA = J x e0
moment is obtained by integrating the above equation.
EFrom equation ( iv), a = R)y(l + e0 ) 1 _ J. V,
.'. Total resisting moment = J (a x dA x y )
For equilibrium, the total resisting moment must be equal to applied moment M. R'
x e0 +
M = f ct x dA xy or Joxyx dA
y(1+ n.('ll ,Eje0 dA + Ej y^fy^dA
= J E x e0 + ip-r yxdA H)
GH)1y(l + e0 ) where A = Area of cross-section of the beam.
EFrom equation (to), a = x e0 + - But for pure bending, there is no force on the section of the beam.
F=0
STRENGTH OF MATERIALS BENDING OF CURVED BARS 963
962
Jl lA h2 _e E A[Cancelling x to both sides]
0=Exe0 xA+E(l + e|))^. ^jj
(1+e°} {r; ~rj r° ...(uiif)
H)Let us now find the value of j ,.. , -Substituting the value of (1 + e Q) [^i - in equation (vi),
j
dA = Ah2 from equation (u). Hence let us use this value to find We get M — E x eQ x R x A x h
We know that —= E x eg x R x A
„ — V**/
1 0 ExRxA A
Prom equation (a) it is clear that e is constant for a given value of M, and R.
0
Now the expression for stress (a) can be obtained by substituting the value of (1 + eg )
2 „2 i — 1 from (viii) in equation (iv ).
R' R R' R)
'r]"' 1 ni Eo = x eQ + -
—Adding and subtracting to the numerator j
^v (1 + e0 = from equation (viii
)
j
K) iLl.-Xv dA —EX 1+
T * (‘* 5)1
iy
i—*=J dA^ X 7
-J fi T-, . .
-siJf 21 [• ydA = °] Prom equation (ix) e0 = -
J
f -2^— dA = AA 2 w J ...(23.1)
Bxa[ 1 i h 2
J H) + yj_
] The equation (23.1), gives the expression for tensile stress. The value of will be -ve,
on the downward side of centroidal axis LM. Hence stress will be opposite to the stress given
Substituting the value of f in equation <n£i>, we get by equation (23.1).
11 + 1-1
o(comPressive) = 1 + fr(^)]
mr r2 ...(23.2)
ior £(l + e°)(i7 "i) X xAh;i = E * eo*A ra[_ h2
OTQCMfSTU nc MATPRIA1 .Q ..
The expression for the stress given by equation (23.1) or (23.2) is known as
Winkler-Bach Formula. The distribution of the stress given by equation (23.1) or (23.2) is
hyperbolic (and not linear as in the case of straight beams), The stress distribution is shown in
Fig. 23.2.
Stress distribution
across the section
Fig. 23.2
Position of Neutral Axis. Let y 0 = distance of neutral axis from eentr'oidal axis.
The stress (a) is zero at the neutral axis. Hence position of neutral axis is obtained by
substituting o = 0 and y = y0 in equation (23.1).
MA R°„ “ fix i 0 or 1 +
1, +
h2
Ror
2 = - h2 x - h2 x y0
l? y 0
Ry0(R2 + h2 ) = - h 2 x or 2 h2 ...(23.3)
i?
+
yBut is the distance from centroidal axis , As y is -ve, this means the neutral axis will
ybe at a distance of below the centroidal axis. Fig. 23.2 (a) shows the position of neutral axis
and Fig. 23.2 (6) shows the stress distribution.
Sign Convention. The stress across the cross-section will be tensile or compressive
according to the following conventions :
M1. The bending moment will be positive if it increases the curvature of the beam and
negative if it decreases the curvature. (Increase of curvature of beam will decrease the radius
of curvature).
y2. The distance is positive if it is measured from the centroidal axis towards the
convex side of beam, and negative when measured towards the concave side (or towards the
centre of curvature).
3. The positive stress means tensile stress, whereas negative stress means compressive
stress.
V23.4. DETERMINATION OF FACTOR FOR VARIOUS SECTIONS
From equation (u), we have
H)
+
STRENGTH OF MATERIALS BENDING OF CURVED BARS 967
966
Now area of strip, dA = b x dy
Area of section, A = b x d
From equation (23.4), we have
R 3 +d/2 Ixjxrfy
p,
b x d -d/2 (fl + y)
( v dA = b x dy)
Q t2
3 x b dy-R2 = 3 r ~f - r2
fl
—I log (« +
bxd e y)\
+ y)
R^r (r dm -R2 ...(23.5)
d l°ge (R - d/2) 2j
R + d/2 (2 R + d)
R-dl2
(2R - d)
2R + d
2R-d
= fi2 1+-
3
...[23.5 (A)]
23.4.2. Triangular Section. A triangular section of a curved beam of width ‘b’ and
height ‘d’ is shown in Fig. 23.4. The centre of curvature of the curved beam is at 0. The radius
of curvature from the centroidal axis is R. Consider a strip of width b* and depth dy at a
distance ‘y’ from the centroidal axis and also at a distance ‘r’ from the axis of curvature.
Centre of curvature
Fig. 23.4
bH R
968 STRENGTH OF MATERIALS BENDING OF CURVED BARS 969
Trapezoidal Section. A trapezoidal section of a curved beam of widths b at -b2 )x
1 Rb2
r
23.4.3. i[El (6X - 62 ) x R^ _ (5 r dr -
t
A 2
the top and of height d shown in Fig. 25.5. The centre of curvature of the dxr dxr
the bottom, b at is
2
curved beam is at O. The radius of curvature from the centroidal axis is R. Consider a strip of
[Here the limits of integration wall be according to ‘dr’ in which r is the distance from the
width b and depth ‘dy’ at a distance ofy from the centroid axis and at a distance r from the axis
Raxis of curvature. Hence limits will be from 7?, to ? (i.e., distances of bottom and top layers of
of curvature.
trapezoidal section from axis of curvature.)]
r=R+y
-Q{ Tf>i - b2
and dr = 0 + dy = dy El ^}E‘,r h L d' -R2
A JJr,
Let R = distance of bottom layer of trapezoidal from axis of curvature )
1
& -£>1 br
R = distance of top layer of trapezoidal from axis of curvature w(bj-b2 ) |
2 d
A b2 + X ^2 H l°&e r
— —H 2
R R3 6o + (5X — b2 ) x 2
-RA
R% ~ 5g) (R2 - Ri) 2
R( b v d
b and are constants)
2, ?
hAEl + ---- b2) x iog -«>!- b2 ) - 2 ...(23.7)
- \f
(v R -R ^d)
2 1
A23.4.4. Circular Section. circular section of a curved beam of diameter ‘d’ is shown
in Fig. 23.6. The centre of curvature of the curved beam is at O. The radius of curvature from
centroidal axis is R. Consider a strip of width ‘b’ and depth ‘dy’ at a distance ‘y’ from the
centroidal axis.
Fig. 23.5
Area of trapezoidal (6j + b2 ) x d
WArea of strip, dA = b x dy where b - b2 + ~ ^2
•M CRa-r) dy = bv + d (R2 ~r) dr
( dy = dr)
Now from equation (23.4), we have
h2 ~ Fig. 23.6
Area of strip, dA = b x dy where b= 2 - 2 x2
-J(d/ 2)
y
dA = bn + bi ~b2 R( 2 ~ r)
-R2
R2 A it
Total area of section, - d2
4
R
STRENGTH OF MATERIALS BENDING OF CURVED BARS Q71
970 -y 2 \dy
From equation (23.4), we have ( + y)
A -R Rwhere =
= total area of T-section b x (R L) + 62 x (R - z)
l 2
s
r d/2
dA Ri _ 23.4.6. I-section. Fig. 23.8 shows the I-section of a curved beam. Consider a strip of
4 '-d* 2
A RJ-rf/2 ( + y) width b and depth dy at a distance y from centroidal axis and at a distance r from the axis of
curvature. Then
SR 3 t* d/2 V 4 y
dy-R2
nd 2 J-d/2 (# + y)
The above integral can be expanded by binomial expression and then integrated, we
will get
„U J. u. ...(23.8)
h ~ +
16 128 R 2
A23.4.5. T-section. T-section of a curved beam is shown in Pig. 23.7 . Consider a strip of
width ‘b’ and thickness ‘dy’ at a distance y from centroidal axis and at a distance r from axis of
curvature. Then r = R + y and dr = 0 + dy = dy
-H b2 «-
LUdy O
Centroidal axis f
Fig. 23.8
r=R +y where R = Radius of curvature i.e., distance
of centre of curvature from centroidal axis
]T dr = dy
| Area of strip, dA - b x dy = b x dr (v dy = dr )
R R, A R RTotal area of I-section, = + x
fejX <R 2 - i? ) b x (i? - + b (R -
Axis of curvature s 3 2) 4 3)
z a
To find A 2 use equation (23.4)
,
Fig. 23.7 R3 r*h bj x dr f Ra b2 xdr + b3 x dr
+ rJ
A 4[Jr,
Area of strip = b x dy r Jr, r
To find A 2 use equation (23.4),
,
1 t£A
r
J (R + ...(23.10)
:
R3 C®2 bL x dr Awhere = b x (f? - Rfi + b 2(.R 3 ~ fry + &3CR4 ~ ^3)-
I" f^i 4 2
b2 x dr
Problem 23.1. Determine ; (i) location of neutral axis, (ii) maximum and minimum
R R[-.- R stress, and (Hi) ratio of maximum and minimum stress, when a curved beam of rectangular
= x dr and between mm mm Ncross-section ofwidth 20
dA between 2 and 3 = b x dy b2 i and ofdepth 40 is subjected to pure bending ofmoment + 6'0O m.
2
Rand dA = b xdy = b x tfr] The beam is curved in a plane parallel to depth. The mean redius of curvature is SO mm. Also
2, 1 1
plot the variation of the stresses across the section.
J 1J
STRENGTH OF MATERIALS BENDING OF CURVED BARS
Sol. Given : : 2500 [0.0533 + 0.00512] = 2500 x .05842 = 146.05
Curved beam of rectangular section, - 50 x 146.05 - 50 x 146.05
Width, b = 20 mm, depth, d - 40 mm, 2646.05 — — 2.759 — — 2,76 mm. Ans.
Area, A = ixd = 20x 40 - 800 mm2 50 2 + 146.05
Radius of curvature, R = 50 mm. (ii) Maximum and minimum stresses. The stress at any layer in the cross-section at a
MBending moment, = + 600 Nm = 600 x 1000 = 600,000 Nmm ydistance from centroidal axis is given by equation (23.1) as
k-20 mm-H •-P-TIh 2 t,-R + yJJ
Centroidat For a given value of M, R and A, the stress will be maximum when y is -ve and maxi-
mum. Hence maximum stress will occur at the extreme bottom layer where y = - 20 mm.
600,000 50 2 ' -20 V!
°max ~ 50 x 800
(f
/r Oil C
- _ 146.05 ^50 - 20
,m 10
mm M[v A = 800
15 2 h2 = 146.05, = 600,000 Nmm]
,
£==d-20 = 15 1- 2500 x 20 1 = 15 [1 - 11.41] = 15 x (- 10.41)
156.17 .
R 1 50 mm (fa) Stress distribution [ 146.05 x 30 J
= - 156.17 N/mm 2 (compressive) Ans.
.
Axis of curvature Centre of curvature mmMinimum stress occurs at the extreme top layer where y = 20
_ jJ./.- '
o 600,000 x 20 = 15 [1 + 4.89]
mln 50 x 800
= 15 1 +
[_
146.05 ^50 + 20 JJ L 146.05i x 7700 J
Fig. 23.9 = 15 x 5.89 = 88.36 N/mm2. (tensile) Ans.
(i) Location of neutral axis. Let v,. = distance of neutral axis from centroidal axis using (Hi) Ratio of and amin
(
g max
equation (23.3), we get
«»** - 156.17 (Here take magnitude only)
-
yv .,- . ...(i) T^min 88.00
°
R2 + h2 = 1.767. Ans.
Let us find now the value of h 2 for rectangular section ( iv ) Plotting the stresses across the cross-section. Stress at any distance y’ from neutral
axis is given by equation (23.1) as
The value of A2 for rectangular section is given by equation (23.5) as
R3 (2R + d 2 x 50 + 40 h2 \R + y ' (50 + y j
2 x 50 - 40
h, 2 ~ l0g
d
* 2fi-d
= 3125 log e - 2500 = 3125 [ log (2.333)] - 2500 U-ll= 15 1 +17.H
e 1,50 + y)
By substituting the various values of y in equation (ii), different values of stress are
= 3125 x 0.847 - 2500 = 2646.8 - 2500 = 146.8
obtained.
_ 50 x (146.8)_ = _ 2 7?3 mm< Ans.
17.11 x 0
0 50 2 + (146.8) [1+^^0 = 15 = 15 x 1 = 15 N/mm2
At y = 0,
-ve sign means the neutral axis will be below central axis at a distance of 2.773 mm.
The value of h2 can also be obtained from equation [23.5 (A)] as
3UR) 5{2R At y = 5 mm, a - 15 1 + 11-11 x ^Tgj =38.33 N/mm2
- o50u 2, —1 x ( 40 At y = 10 mm, a =45 l+ 17-llx 50TT0 ) =57.77 N/mm2
3 I 2 X 50 At y = 15 mm, ct = 15 [l + 17.11 x = 74.22 N/mm2
— x 0.16 + —
)
|_3 5
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