STRENGTH OF MATERIAL
5 SHEAR STRESS DISTRIBUTION IN BEAMS
SHEAR STRESS IN BEAM M P
m
P >
V
n x >
< V
Shear force V = P , M = P x M
n A B
-------------
<>
<>
m DC
acts on one of vertical face BC is balanced by that in AD. To balance this moment on horizontal face also acts.
Shear Stress at a cross section :-
= FQ AB >
>
Ib xxxxxxxxxxxxxxxxx dA
h/2 <
y yh
___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___ _ ___
>
M M + dM
<
Q = A1y1+ A2y2+---- CD
F = Shear force at section +d
= Shear stress at a distance y F2
A = Area above / below the point F Copyright : Ascent Gate Academy 51
1
y = Distance of Centroid of area from NA <>
I = M.I. about N.A.
F
3
b = Width
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MECHANICAL ENGINEERING
Average Shear Stress : -
= F b
avg cross section area
C
= F y<>
avg b.d
FAy d
Ib
=
max)NA
= F x bd x d
max)NA 24
bd3
xb
12
= 3F
2bd
3F 3
max 2 bd = 2 = 1.5
= (for rectangle & square & Triangle)
F
avg bd
= 50% more than for rectangle & square
max. avg.
SHEAR STRESS DISTRUBTION
i) Rectangular Section
b
dy 20 parabola<>
h ___ . ___ . ___ . __y_ . ___ . ___ . ___ . ___ . ___ .
= 6 F h2 y2 Parabolic distribution
bh3 4
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STRENGTH OF MATERIAL
(ii) Circular Section
= 4 F Cos 2 - Parabolic variation dy
3R2
4R <>
y = 3 ___ . ___ . ___ . ___) .___ . ___ . ___ max.
< >
At = 900 , = 0
At = 00 = 4F =
3R2 max
= F = 4 for circular section
mean R2 max 3
mean
Note : - In case of pure shear, a circular cross section (whose / is less than that of rectangle or
max avg
square ) is efficient in shear because of more area at N.A.
* In beams when both shear & bending acts, design is done for bending so beam are generally of rect-
angle section.
(iii) Triangular Section :-
* Shear stress at y from vortex is given by
12 F y(h-y) h <> <> y
b.h3 y= 2 < >< > >< > h/2
* Maximum shear stress exists
MAX
at the middle triangle and is given by 2/3h 3F
h A h/2 bh
3 F
max bh N <
* Average shear stress is given by h/3
D
F 2 F
avg bh
1 bh
2
Hence ,
max avg
2h
* Shear stress at Neutral Axis(y = 3 from top) is given by
8 F
n.a. avg 3 bh
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MECHANICAL ENGINEERING
(iv) Dimand Section :-
< aa >< > 3d
NA max 8
< max >
d = 2 a
1 d
d .__<_N._A _>_ 8
___ .___ .___ .___ .___ .___ .___ .___ .___ .___
9 .___ >< ><
max 8 = 1.125
= 1d
8
avg
< max >
= 12.5 % greater than a a 3d
max avg. 8
<
FAy F 1 2a x a 1.a
Ib 2 2 3 2
= =
N A
a4
12 2a)
= F
NA a2
= F
NA a2
Zsquare section = 2 Zsqure diagonal
= 1.
NA
avg
Flanged section : - Eg. : I, T channel angle etc.
Flange
Web
___ .___ .___ .___ .___ .___ .___ .__<_ .___ .___>.___ .___ .___ .___ .___
max.
shear stress distribution bending stress distribution
Note : * The bending stress distribution is constant irrespective of shape of cross sesction, shear stress
distribution changes with width of cross section.
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STRENGTH OF MATERIAL
___ .___ .___ .___ .___ .___ .___ .__<_ .___ .___> ___ .___ .___ .___ .___ .___ .___ .__<_ .___ .___>
max. max.
___ .___ .___ .___ .___ .__N_.A._.__ .__<_ .___ .___> ___ .___ .___ .___ .___ .__N_.A._.__ .__<_ .___ .___>
max. max.
max.
<>
___ .___ .___ .___ .___ .___ .___ .___ .___ .___ .___ .___ .___ .___ .___ .
< >
max.
___ .___ .___ .___ .___ .___< .___ >.___ .___ .___ .___ .___ .___ .___ .___ .
max. max.
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MECHANICAL ENGINEERING
6 TORSION OR TWISTING MOMENT
Assumptions :-
i) Plane cross section remains plane after application of torsion (Euler - Berunaullis assumption)As per
this assumption, there is no warping or twisting in the cross section. This assumption is not valid for
torsion of non circular members.
ii) The twist along the shaft is uniform
iii) Material is homogenous isotropic & follows hook’s law.
iv)All radius remain straight before & after torsion. (As per this assumption, there is no change in shape
of cross section after torsion)
v) Shaft is of uniform circular section throughout.
CC
A ___ .___ .___ .__)_.___ .___ .___ .___ .___ .___ .___ .___._(__ .___ .___ .
B BR
<l >
Torsion equation :-
T = = G OR q
JR l R= r
T = Torsional moment of resistance offered by cross section of shaft = Torque applied
J = Polar moment of inertia of section of shaft
- Shear stress at surface of the shaft
q - Shear stress at radius r
R - Radius of shaft
G - Modulus of rigidty of shaft
-Angle of twist
l - length of shaft
Polar modulus :-
T T = x J
= R
JR
or T = x Zp
Zp = JP = Polar modulus
R
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STRENGTH OF MATERIAL
Torsional Stiffness : -
Torsion required to produce unit angular twist (in rad) is called Torsional stiffness
KT = T
Torsional Rigidity :- (GJ) Torsion required to produce unit angular twist over unit length
T G Tl
= GJ =
Jl
or GJ = T
l
Unit : - N x mm4 = Nmm2 or kNm2
mm2
Significance :- More GJ More rigid in torsion less angle of twist
Shear Strain A’
= shear strain = shear angle
) (
= AA’ = R..
AR
ll
Shear stress distribution over cross section under Torsion
Note : - * If the shaft is subjected to torsion, all the extrme pts.
max
on the surface will have same maximum shear stress.
max
* If the same shaft is sub. to bending, the extreme max
top & bottom pts. are with maximum bending stress.
All the pts. on N.A. will have zero stress.
Power Transmission
2NT
P = watt ( N - rpm)
60
or P = 2NT ( N - rps or Hz )
1 kW = 1000 N-m/s.
1 Hp = 746 watt
T = Avg. Torque N - m
Note : * In power transmission there will be losses so we use average torque.
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MECHANICAL ENGINEERING
Arrnagement of shafts
1) Series
1 23
T
T1 = T2 = T3 = T
(Total angle of twist)
2) Parallel 1 1 2
T = T1 + T2. 2 T
T
Note : - * To transfer higher torsion capacities it is better to use shafts in parallel
COMPARISION OF HOLLOW & SOLID SHAFTS : -
Let hollow shaft and solid shafts have same material and length.
D0 = external diameter of hollow shaft
Di = Internal diameter of hollow shaft
D= Diameter of the solid shaft
Case (i) When the hollow and solid shafts have the same torsional strength
* In this case polar modulus section of two shafts would be equal.
* Weight of hollow shaft = 1 - n2
Weigth of solid shaft (1 - n4)2/3
USE : % Saving in weight can be calculated for same torsional strength.
Case (ii) : When the hollow and solid shafts are of equal weights
* In this case torsional strength is compared
* Thollow = 1 + n2
Tsolid 1 - n2
USE : ratio of strength for same weight can be calculated.
Case (iii) : When the diameter of solid shaft is equal to the external diameter of the hollow shaft
* Thollow = 1 - n4
T
solid
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STRENGTH OF MATERIAL
TORSION OF THIN CIRCULAR TUBE A A1 BB
Shear & Torsional Resilience : ---- 1
Shear resilience : - - - - - - -)- - - - - - - - - -
- - - - - - -)- - - - - - - - - -
Strain Energy stored (U) = 1 P.
2
U = 1 l2 or U = x vol. of block D
2 G 2G
This is energy stored by the block
Strain Energy stored / unit volume =
2G.
Torsional Resilience :-
In this case, shear stress due to torsion varies uniformly from zero at axis to maximum value at surface
s
Strain Energy stored / unit vol.
2 2 D2 + d2
U= U=
for Hollow shaft 4G D2
4G
Torsion of shafts of Non Circular section :-
i) Rectangular Section :-
Torsional resistance T = x2 y2 2s
3y + 1.8 x
x = Short side , y = Long side
Note :- The maximum shearing stress occurs at middle of longer side
= k y2 + x2 Tl ; K = 3.645 - 0.06 b
y3 + x3 G d
ii) Square Section :-
Torsonal resistance T = 0.208 x3
s
x = side of square
Note : The maximum shearing stress occurs at the middle pt of a side.
= 7.11 Tl
Gx2
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MECHANICAL ENGINEERING
SPRINGS
* A spring is used to absorb energy due to resilience which may be used as & when required
* The spring capable of storing greatest amount of energy for the given stress is the best one
Close Coiled Helical springs
Total shear stress = Direct shear + Torsional shear
= + .
= 4P + 16 PR L = 2RN
d2 d3
= 16 PR 1+ d d <>
m d3 4R
Walls equation :- (Considered due to curvature) <D >
P
16 PR 4m - 1 0.615
= d3 +
m 4m - 4 m
D 2R D = mean coil dia
m= = d = dia. of wire
m = spring index
d d. N = No. to trun
K = stiffness of spring
= R..
= Tl PR x 2Rn
GJ = G x d4
32
= 64PR3n
Gd4
Total strain Energy stored
U = 1 P..
2
32R3P2n
U=
Gd4
Spring const. K = P
K = Gd4
64 R3n.
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STRENGTH OF MATERIAL
Spring under impact load
Work done by following weight. = work stored in spring
W(h + ) = 1 W1. ( h = height from which weight is dropped)
2 ( = Deflection of spring )
= 64W1R3n (W = weight)
Gd4
W1 & can be determined.
Note : - If a closely coiled helical spring is subjected to Mo
axial couple or torsion, it behaves as a beam subjected to
pure bending. The stressees are bending stresses.
Arrangement of Springs :- Axial torque.
Mo
i) Springs in Series ii) Springs in Parallel
W = W1 + W
= + 2.2
1
W = W + W K = W W1 = W2 K K
ke k1 k2 1 ke = k1 k2 1 2
W
or Ke = K1 K2 K W= W.K1
K1 + K2 2 1 K.
P W.K2
K.
W2 =
W
W= (K +K )
K. 1 2
Ke = K1 + K2
iii) SNS (Spring in Spring)
Ke = K + K = W
12
= W
Ke
Ke is more is less
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MECHANICAL ENGINEERING
Laminated spring / Leaf spring / Carriage spring W/2
W/2
semi elleptical arch
W (axle of wheel)
Note : - i) This is the only eg. of beam of uniform strength
i.e. max = const.
ii) In laminated springs, the plates can move one over the other freely shear stress developed is zero.
iii) max = 0
M=Z
M Z
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STRENGTH OF MATERIAL
7 DEFLECTION AND SLOPE
METHOD FOR SLOPE AND DEFLECTION ATA SECTION
1. Double Intergration Method
* This method is suitable for simple loading in simplysupported beams and cantilevers with uniformly
distributed loads and traingular loadings.
* d2y M is +ive sagging
EI dx2 = M M is -ive hogging
After intergrating end conditions are applied for determination of constants of intergration.
2. Macaulay’s Method
* This is convenient method for beams subjected to point loading or in general discontinous loads or
beams subjected to couples (concentrated moments).
* This method is similar to the double integration method but speciality of this method lies in the
manner in which thebendingmoment atanysection is expressed and in themanner in which intergration
in carried out.
3. Moment Area Method
* This method is suitable for cantilevers and simply supported beams carrying symmetrical loadings
and beams fixed at both ends i.e. those beams for which the area and C.G. of area of B.M.D. can
be found easily. This means this methods is not suitable for triangular loading and irregular loading.
This can also be used for non prismatic bars.
* This method is suitable for
(i) Cantileverse (because slope at the fixed end is zero)
(ii) Simplysupported beams carrying symmetrical loading (slope at mid span is zero).
(iii) Beams fixed at both ends (slope at each end is zero).
Theorem 1 : -
- = Area of M diagram beetween C & D
D C EI
Note : If M/EI diagram is possitive angle is measured from C in anticlockwisse direction and if negative then
in CW direction.
Therom 2 :-
t D/C = vertical display of point D from tangent at C.
= I momentof area
=Ax
A = Area of M/EI diagram
x = taken from D
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MECHANICAL ENGINEERING
SLOPES & DEFLECTIONS OF DIFFERENT LOADINGS
Type of loading Slope Deflection
-Wl3
W Ymax = 3EI
1) A _ __B_ B<> = -Wl2
_ B 2EI
< a w b > -Wa2 -Wa2 (3l - a)
B 2EI 6EI
>< = Y=
max
2)
3) A w/unit run = -wl3 wl4
B 6EI YB = 8EI
B
4) A w/unit run B = -wa3 Ymax = - Wa4+ Wa3(l - a)
= 8EI 6EI
c B C 6EJ
c = -W (l3 - a3) W (3l4-4Ia3+a4)
B 6EJ Ymax = - 24EI
5) A B
6) A B = -Ml YB = -Ml2
B EJ 2EI
M<
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STRENGTH OF MATERIAL
Type of loading Slope Deflection
-Wl4
7) = -Wl3 Ymax = 30EI
B 24EI
w/unit trun
w = -Wl2 -Wl3
l/2 l/2 A 16EI YC = 48EI
8) A B
= +Wl2
B 16EI
w = -Wb (l2 - b2) Wb/(l2 - b2) 3/2
>< b > A 6EIl 83EIL
<a = +Wa (l2 - a2) Ymax =
9) A 6EIl
w/unit run = -Wl2 Y = -5Wl4
A 24EI max 384EI
A B = +Wl3
10) B 24EI
= -5Wl3 -Wl4
A 192EI 120EI
11) +5Wl3 YC =
192EI
=
B
= -7Wl3 Ymax = -2.5Wl4
A 360EI 384EI
12) B = +Wl3 (x = 0.519 l from A)
A B 45EI
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MECHANICAL ENGINEERING
IMPORTANT RELATIONS
Y = Deflection
dy
dx = Slope
d2y 1 =P = curvature
dx2 = R
or d3y = 1 = M (when EI = 1)
dx3 EI
d3y 1 dM E = F [ If EI = 1]
dx3 = EI dx = EI
dM = F = Shear Force
dx
d4y 1 dF W = W [ If EI = 1]
dx4 = EI dx = EI
dF
dx = W = Load
for M to be maximum
dm
dx = 0, i.e. F = 0
for yto be maximum
dy
= 0, i.e. Slope = 0
dx
Note :
* At the point of zero slope, deflection is maximum but at the point of zero deflection slope may not be
maximum.
* The tangent drawn at the point of maximum deflection should be better to axis of beam
Shear is 0.
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STRENGTH OF MATERIAL
Conjugate Beam Method
* Usefull mainlywhen EI varies Conjugate Beam
* Semi graphical method i) Shear Force at corresponding section
* Conjugate Beam is an imaginary beam
ii) Bending moment at corresponding section
Real Beam M
i) Slope at any section
ii) Deflection at any section iii) EI diagram of real beam becomes loading
iii) Loading the given system for conjugate beam
iv) Roller support iv) Hinged support
v) v)
vi) Free end vi)
vii) Fixed end vii) Free end
viii) Intermadiate hinge viii) Intermadiate support
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MECHANICAL ENGINEERING
Real Beam Conjugate Beam
ix) Intermediate support ix) Intermediate hinge
Slope exists, deflection = 0 Shear Force exists, Bending moment = 0
x) Possivite Bending Moment diagram x) M Loading diagram is possitive load is
(sagging) EI downwards
Negative Bending Moment (hogging) M diagram is negative loading is
EI upwards
xi)
xi)
xii) xii)
xiii) xiii)
Propped cantilever xiv)
xv)
xiv)
xv)
xvi) xvi)
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STRENGTH OF MATERIAL
Real Beam Conjugate Beam
xvii) xvii)
xviii) xviii)
Note :-
* A Conjugate Beam need not be a stable beam.
* M/EI diagram for Real Beams loading for Conjugate Beam.
* Shear Force at any point in Conjugate Beam gives slope at that point at Real Beam.
* Bending moment at any point of Conjugate Beam gives deflection at that point of Real Beam.
MAXWELL RECIPROCAL THEOREM A PB
. B
y
P A
A
AB
y
B
<>
<>
PA y = PB y - This is Maxwell theorem If PA = PB y =y
A B AB
P B A PB
A y y
B
A B A
<> PA y = PB y
<>A B
<>
<>
P P
B
Ay
B y
A
P y =P y
AA BB
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MECHANICAL ENGINEERING
8 COLUMNS AND STRUTS
A strut is a vertical member subjected to an axial compressive force. A strut may be horizontal,
inclined or even vertical. A vertical strut is called a column. When a column or strut is subjected to a
compressive load, compressive stress on it is given by,
= P
c A
When load on the column is increased, column fails bycrushing. The load corresponding to crushing
is knwon as crushing load. The load at which the column just backles is called buckling load, critical load or
crippling load. Euler has proposed theories for the columns.
Euler’s Formula
The following assumptions are made in the Euler’s Column theory applied for long columns :
* The column is perfectly straight and the load applied is axial.
* The column material is perfectly classic, homogeneous and isotropic and obeys Hooke’s law.
* The cross section of the column is unifrom though out its length.
* The length of the column is very large compared to its cross sectional dimensions.
* Column fails due to buckling only.
END CONDITIONS FOR DESIGNING THE COLUMNS P
1. Column with both ends hinged :
<>
Consider a columnAB of length / and uniform sectional areaA, <>
hinged of both ends A and B. Let P be the crippling load at
which the column has just buckled. ------------------
The load acting on the column is given by ly
2EI
<>
Load, P = l2
x
Fig : Columns with both ends hinged
2. Columns with one end fixed and the other B< ><>
free critical load on the column :
Consider a columnAB of length l and uniform ly ------------------
sectional areaA, fixed at one end A and free at B.
Let P be the crippling load at which the column <>
has just buckled.
The load acting on the column is given by x
2EI A
Load, P = 4l2 Fig : Columns with one end fixed and the
other free critical load on the column
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STRENGTH OF MATERIAL
3. Column with one end fixed and P<>
the other B
------------------
hinged critical load on the column : ly
Consider a columnAB of length l and uniform x<>
sectional aeaA, fixed at one endAand hinged
at end B. Let P be the crippling load at which the A MA
column has just buckled.
Fig : Column with one end fixed an the
The load acting on the column is given by, other hinged critical load on the column
2EI
Load, P = l2
4. Column with both ends fixed (critical load on the column) : MB<>
Consider a columnAB of length l and uniform
sectional area A, fixed at both ends A and B. ly ------------------
Let P be the crippling load at which the column >
has just buckled. A
Critical load on the column is given by , Fig : Column with both ends fixed
P = 2EI critical load on the column
l2
EQUIVALENT LENGTH (L) OFA COLUMN
Equivalent lengh (L) of a given column with given end conditoins is the length of an equivalent column of the
same material and cross section with hinged ends and having the value of crippling load equal to that of the
given column
If both ends are hinged, L = 1
If both ends are fixed, L = l/2
If one end is fixed and other end is free, L = 2l
If one end is fixed and the other is hinged, L = l/2
Slenderness ratio
Writing the general relation for crippling load,
P = 2EI
l2
Substituting, I =Ak2 where, Ais the cross-sectional area and k is the least radius of gyration of the section
2EA
=
1/k 2
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MECHANICAL ENGINEERING
RANKINE’S RELATION FOR A COLUMN
When the strut or columns are neither short nor long, the failure of the member will be due to the combined
effect of direct and bending stress. Rankine has proposed the following relation for the crushing load for all
types of columns.
If P = crippling load by Rankine’s relation
Pc = fcA
= Ultimate crushing load for the column
PE = 2EI
l2
= crippling load by Euler’s relation,
then
1= 1= 1
P Pc P
E
Or
P = fc A/{1+a(L/k)7}
where
a= fc
2E
JOHNSON’S RELATION FOR A COLUMN
Straight line formula due to Johnson is given by ,
Safe load , P = f-n L
K
where n = a constant whose value depends upon the column material
When a column is subjected to an eccentric load, safe axial load according to
* Rankine : 1+ e.yc fcA 2 d eP<>
Load, P = k2 1+a
L < >< >------------------
* Euler :
K B -------
MB
ly
Maximum stress, Pmax= P.esec1 P A
P EI Fig : Safe axial load
A+ Z
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STRENGTH OF MATERIAL
9 THIN CYLINDERS AND SPHERES
THIN CYLINDERS
- If the thickness of the cylinder is less than 1 = to 115of the diameter of the cylinder, it is
10
treated as the thin cylinder.
- It is assumed that the stresses are uniformly distributed through out the thickness of the wall.
- ‘Hoop stress’ or ‘Circumferential stree’ is given by
f1 = p2td(tensile)
- ‘Longitudinal stress is given by
f2 = p4td(tensile)
- The maximum shear stress is given by
qmax = f1 - f2 pd
2 8t
- Hoop strain is given by
= pd 2- 1
4 tE m
- Longitudinal strain is given by
= pd 1- 2
4 tE m
- Volumetric strain is given by
= pd 5- 4
V 4 tE m
Where
p = internal pressure
d = diameter of cylinder
t = thickness of the cylinder
= 1 poisson’s ratio
m=
- If ‘fa’be the permissible tensile stress for the shell material, then from strength point of view, the
major principle stress (f) should be less than or equal to fa
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MECHANICAL ENGINEERING
Hence f= pd fa
2t
pd
or t 2fa
THIN SPHERICAL SHELLS
- In case of thin spherical shells, longitudinal stress and circumferential stress are equal and are
given by
pd (tensile in nature)
f1 = f2 = 4t
- The maximum shear stress, qmax= f1 - f2 = 0
2
- The strain in any direction is given by
[ ]dpd1
d 4tE m
- The volumetric strain is given by
3pd 1
V 4tE m
CYLINDERS WITH HEMISPHERICAL ENDS
Let
tc = thickness of the cylinder
ts = thickness of the hemisphere
tc
-------------------
ts d -------------------
<l >
- Hoop stress in cylindrical part
fC1 = pd
2tC
- Hoop stress in hemispherical part
fs1 = pd
4tS
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STRENGTH OF MATERIAL
- Longitudinal stress in cylindrical part
- fC2 = pd in hemispherical part
Longitudinal st4rteCss
- CircumffSe2r=entialp4stdtSrain in hemispherical part
eS1 = pd 1 - 1
4t E m
S
- circumferential strain in cylindrical part
e= pd 1
C1 4tCE 2 - m
- For the condition of no distortion of the junction
e = eS1
C1
> pd 2 - 1 = pd 1 -m1
4tCE m 4tSE
or 2 -
1-
tc = 2m -1 =
ts m- 1
this means thickness of cylindried part should be more than the hemispherical part.
- For the condition of same maximum stress in cylindrical and hemispherical parts,
pd pd
or 2 tC = 4 tS
tC = 2
tS
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