Transcript ME16A - Faculty of Engineering

ME16A: CHAPTER ONE
STATICALLY
DETERMINATE
STRESS SYSTEMS
INTRODUCTION

A problem is said to be
statically determinate if the
stress within the body can be
calculated purely from the
conditions of equilibrium of the
applied loading and internal
forces.
2.1 AXIALLY LOADED BARS,
STRUT OR COLUMN
2.1 AXIALLY LOADED BARS, STRUT OR COLUMN
The external force applied at the ends of the member is balanced by
internal force which is average stress x cross sectional area.
F
F
X
If the axially loaded bar is cut perpendicular to the axis into two:
F =  x A i . e x 
F
A
( tensile stress)
If the bar is cut at an angle to the axis, two components of
stress will be created: one normal to the plane,  h and the other
parallel to the plane,  s .
 h  s.
 s.
h
2.1.1. Principle of St. Venant

It states that the actual distribution
of load over the surface of its
application will not affect the
distribution of stress or strain on
sections of the body which are at
an appreciable distance (> 3 times
its greatest width) away from the
load
Principle of St. Venant Contd.





e.g. a rod in simple tension may have
the end load applied.
(a) Centrally concentrated
(b)
Distributed
round
the
circumference of rod
(c) Distributed over the end crosssection.
All are statically equivalent.
Principle of St. Venant Concluded
F
F
Uniform stress – stress distribution
not affected by distribution of load but
by its resultant.
Alternatively:
The principle states that the stress distribution at sections far removed
from the point of application of concentrated forces depends on stress resultants and not
on the actual distribution of forces.
Example




The piston of an engine is 30 cm in diameter
and the piston rod is 5 cm in diameter. The
steam pressure is 100 N/cm2.
Find (a) the stress on the piston rod and
(b) the elongation of a length of 80 cm when
the piston is in instroke.
(c) the reduction in diameter of the piston
rod (E = 2 x 107 N/cm2; v = 0.3).
Solution
p = 100 N/cm2
x
F
Piston rod
(dia. = 5 cm)
Piston
(a) For horizontal equilibrium of forces
 x 52
 x (302  52 )
2
x
 100 N / cm x
4
4
 x   rod 
(b) Elongation =
y 
(c)
 / 4(100)( 302  52 )
 3500 N / cm2
2
 /4 x 5
FL  L 3500 N / cm2 x 80 cm


 0.014 cm
AE
E
2 x 107 N / cm2
Change in dia 1
1
 [ y    x ] 
[0  0.3(3500)]
Original dia E
2 x 107
  5.25 x 105
Change in diameter = 5.25 x 10 -5 x 5 = 0.0002625 cm
30 cm
2.2 THIN-WALLED PRESSURE
VESSELS


Cylindrical and spherical pressure
vessels are commonly used for storing
gas and liquids under pressure.
A thin cylinder is normally defined as
one in which the thickness of the metal
is less than 1/20 of the diameter of the
cylinder.
THIN-WALLED PRESSURE
VESSELS CONTD


In thin cylinders, it can be assumed that
the variation of stress within the metal
is negligible, and that the mean
diameter, Dm is approximately equal to
the internal diameter, D.
At mid-length, the walls are subjected
to hoop or circumferential stress, and
a longitudinal stress, .
Hoop and Longitudinal Stress
2.2.1 Hoop stress in thin cylindrical
shell
Hoop stress in thin cylindrical shell
Contd.


The internal pressure, p tends to
increase the diameter of the cylinder
and this produces a hoop or
circumferential stress (tensile).
If the stress becomes excessive, failure
in the form of a longitudinal burst would
occur.
Hoop stress in thin cylindrical shell
Concluded
Consider the half cylinder shown. Force due to internal pressure, p is balanced by the
force due to hoop stress,  h .
i.e. hoop stress x area = pressure x projected area
h x 2 L t = P x d L
 h = (P d) / 2 t
Where: d is the internal diameter of cylinder; t is the thickness of wall of cylinder.
2.2.2. Longitudinal stress in thin
cylindrical shell
Longitudinal stress in thin cylindrical
shell Contd.
The internal pressure, P also produces a tensile stress in
longitudinal direction as shown above.
Force by P acting on an area
longitudinal stress,  L
dt
4
is balanced by
acting over an approximate area,
(mean diameter should strictly be used). That is:
L x d t  P x
L 
 d2
Pd
4t
 d2
4
Note



1. Since hoop stress is twice longitudinal
stress, the cylinder would fail by tearing
along a line parallel to the axis, rather than
on a section perpendicular to the axis.
The equation for hoop stress is therefore
used to determine the cylinder thickness.
Allowance is made for this by dividing the
thickness obtained in hoop stress equation
by efficiency (i.e. tearing and shearing
efficiency) of the joint.
Longitudinal stress in thin cylindrical
shell Concluded
Example

A cylindrical boiler is subjected to an
internal pressure, p. If the boiler has a
mean radius, r and a wall thickness, t,
derive expressions for the hoop and
longitudinal stresses in its wall. If
Poisson’s ratio for the material is 0.30,
find the ratio of the hoop strain to the
longitudinal strain and compare it with
the ratio of stresses.
Solution
Hoop stress will cause expansion on the lateral direction and is
equal to  y while the longitudinal stress is  x
Hoop stress,  h

Longitudinal stress,
p d p x 2r p r


2t
2t
t
L 
ie y
p d p x 2r p r


i. e. x
4t
4t
2t
(a) Stress ratio = 2
(b)  x 
1
1 pr
pr
0.2 pr
[ x    y ]  [  0.3 ] 
( Longitudinal strain)
E
E 2t
t
E t
1
1 pr
pr
0.85 pr
[ y    x ]  [  0.3 ] 
( Hoop strain)
E
E t
2t
E t
Hoop strain
0.85
Ratio of strains 

 4.25
Longitudinal strain 0.2
y 
2.2.3 Pressure in Spherical Vessels
2.2.3 Pressure in Spherical Vessels
Problems dealing with spherical vessels follow similar solutions to that for thin cylinders
except that there will be longitudinal stresses in all directions. No hoop or circumferential
stresses are produced.
i.e
L 
Pd
4t
2.3 STRESSES IN THIN ROTATING
RINGS



If a thin circular ring or cylinder, is rotated
about its centre, there will be a natural
tendency for the diameter of the ring to be
increased.
A centripetal force is required to maintain a
body in circular motion.
In the case of a rotating ring, this force can
only arise from the hoop or circumferential
stress created in the ring.
STRESSES IN THIN ROTATING
RING
STRESSES IN THIN ROTATING
RINGS CONTD.

Consider a thin ring of mean radius, r, density,
and
having a cross-sectional area, A, to be rotating about
centre O with an angular velocity, w (rad/s).
For an elemental length which sustends an angle d at O,
as shown in Fig. (a).
Circumferential length of element =
Volume of element =
r A d
Mass of element =
r A d

r d

Centripetal force to maintain circular motion = mass x w2 r =
=

w2 r2 A d
r A d w2 r
STRESSES IN THIN ROTATING
RINGS CONTD.
If the hoop stress created in the ring is  h
Force F acting on cross-section =  h . A (see diagram b)
Radial component of the force, F = 2 ( h . A) sin d 2 =
2 (  h . A) d 2 ( for small d )
This radial component of forces, F supplies the required
centripetal force to maintain the element in circular motion. Thus:
2 (  h . A) d 2 = 
i.e.
w2 r2 A d
 h =
Putting velocity, V = wr ;
w2 r2
 h =
V2
STRESSES IN THIN ROTATING
RINGS CONCLUDED


Hence: Hoop stress created in
a thin rotating ring, or cylinder
is independent of the crosssectional area.
For a given peripheral speed,
the stress is independent of the
radius of the ring.
EXAMPLE


A thin steel plate having a tensile
strength of 440 MN/m2 and a density of
7.8 Mg/m3 is formed into a circular
drum of mean diameter 0.8 m.
Determine the greatest speed at which
the drum can be rotated if there is to be
a safety factor of 8. E = 210 GN/m2.
SOLUTION
Greatest stress to be applied =
440 MN / m 2
 55 MN / m 2
Factor of safety (8)
  7.8 Mg / m 3  7800 kg / m 3
hoop stress,  h
=
V2 =
w2 r2
h
55 x 106 N / m2
w

 209.9 rad / s
2
3
2
2
r
7800 kg / m x 0.4 ( m )

rad = 1800,
rad = 57.2960
In 3600 (1 rev), we have 360/57.296 = 6.283 rad
i.e 209.9 rad /s = 33.407 rev/s = 2004.4 rev/min
2.4 STATICALLY INDETERMINATE
STRESS SYSTEMS

There is the need to assess the
geometry of deformation and
link stress and strain through
modulus and Poisson’s ratio for
the material.
2.4.1 Volume Changes

Example: A pressure cylinder, 0.8 m
long is made out of 5 mm thick steel
plate which has an elastic modulus of
210 x 103 N/mm2 and a Poisson’s ratio
of 0.28. The cylinder has a mean
diameter of 0.3 m and is closed at its
ends by flat plates. If it is subjected to
an internal pressure of 3 N/mm2,
calculate its increase in volume.
SOLUTION
Hoop stress,  h = (P d) / 2 t =
3 N / mm2 x 300 mm
 90 N / mm2
2 x 5 mm
Longitudinal stress,  L = (P d) / 4 t = 45 N/mm2
Longitudinal strain,
1
1
 L  [ L    h ] 
[ 45  0.28 x 90]  0.00009429
3
2
E
210 x 10 N / mm
SOLUTION CONCLUDED
Hoop strain,
h 
1
1
[ h    L ] 
[90  0.28 x 45]  0.0003686
3
2
E
210 x 10 N / mm
Volumetric strain =
 L  2  h  0.00083134 (See Section 1.4)
Original volume of cylinder is equal to :
 x 3002
x 800  56.5487 x 10 6 mm3
4
Increase in volume  56.5487 x 10 6 x 0.00083134  47009 mm3
Example

The dimensions of an oil storage tank
with hemispherical ends are shown in
the Figure. The tank is filled with oil
and the volume of oil increases by
0.1% for each degree rise in
temperature of 10C. If the coefficient
of linear expansion of the tank material
is 12 x 10-6 per 0C, how much oil will
be lost if the temperature rises by 100C.
SOLUTION
For 100C rise in temperature:
Volumetric strain of oil = 0.001 x 10 = 0.01
Volumetric strain of tank =
3
T
= 3 x 12 x 10-6 x 10 = 0.00036
Difference in volumetric strain = 0.01 - 0.00036 = 0.00964
Volume of tank =

10000


x 102 x 100 + 4/3 x

x 103 =
+ 1333.33

= 11333.33
m3

Volume of oil lost = strain difference x volume of tank = 0.00964 x 11333.33
= 343.2 m3.
m3
2.4.2 IMPACT LOADS
L
W
h
x = dl
IMPACT LOADS CONTD.
Consider a weight, W falling through a height, h
on to a collar attached to one end of a uniform bar.
The other end of the bar is fixed.
Let dl be the maximum extension caused
and

be the stress set up.
Let P be the equivalent static or gradually applied load
which would cause the same extension, dl
Strain energy in the bar at this instant = 1/2 P. dl
Neglecting loss of energy at impact:
L
W
h
x = dl
IMPACT LOADS CONTD.
Loss of potential energy of weight, W on impact = Gain of strain energy of bar
i.e. W (h + dl) = 1/2 P dl
Recall that :
i.e
dl 
PL
AE
PL
1 P2 L
W (h 
) (
)
AE
2 AE
PL 1 P 2 L
Wh  W

AE 2 AE
1 P 2 L WPL
i. e.

 Wh  0
2 AE
AE
AE
Multiplying by
L
P2
WhAE
 WP 
0
2
L
L
W
h
x = dl
IMPACT LOADS CONTD.
Recall the quadratic equation formula:
b  b 2  4ac
P
2a
i. e. P  W  W 2  4 x
1 WhAE
2
L
L
W
Using only the positive root:
h
P  W  W  2WhAE / L
2
 W [1  1 
From which
2hAE
]
WL
PL
dl 
AE
x = dl
and

P

A
can be obtained.
IMPACT LOADS CONTD.


Note: 1. For a suddenly applied load ,
h = 0 and P = 2 W i.e the stress
produced by a suddenly applied load is
twice the static stress.
2. If there is no deformation, ‘ x’ of the
bar, W will oscillate about, and come to
rest in the normal equilibrium position.
IMPACT LOAD CONCLUDED



3. The above analysis assumes that the
whole of the rod attains the same value of
maximum stress at the same instant.
In actual practice, a wave of stress is set up
by the impact and is propagated along the
rod.
This approximate analysis, however, gives
results on the “safe” side.
EXAMPLE


A mass of 100 kg falls 4 cm on
to a collar attached to a bar of
steel, 2 cm diameter, 3 m long.
Find the maximum stress set
up. E = 205,000 N/mm2.
SOLUTION CONCLUDED
 x 202
 314.2 mm2
Area of bar =
4
W = 100 x 9.81 = 981 N
2 x 40 x 314.2 x 205000
P  981[1  1 
]  42029.65 N
981 x 3000
P 42029.65 N
2
Stress  

133
.
77

134
N
/
mm
A 314.2 mm2
ALTERNATIVE SOLUTION
Using Alternative Equation:
 '    ( 2 


2 Eh 1/ 2
)
L
By Benham and Crawford (1987)
100 x 9.81
981
2
2


3122
.
N
/
mm
;

 9.748
2
20
314.2
x
4
 '  3122
.
 9.748 
2 x 3122
.
x 205000 x 40
3000
= 133.80 = 134 N/mm2.