ADVANCED STRENGTH OF MATERIALS - Ki??isel Sayfalarkisi.deu.edu.tr/emine.cinar/ASM16-Rotational...
Transcript of ADVANCED STRENGTH OF MATERIALS - Ki??isel Sayfalarkisi.deu.edu.tr/emine.cinar/ASM16-Rotational...
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ROTATIONAL STRESSES
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INTRODUCTION
High centrifugal forces are developed in machine components rotating at a
high angular speed of the order of 100 to 500 revolutions per second
(rps). High centrifugal force produces high hoop and radial stresses in
machine components such as rotors and blades of steam and gas turbines.
It is almost necessary to analyse these stresses in such components
because these are subjected to fatigue loading and creep strain. Many a
times a component fails due to crack developed by fatigue and creep
during operation. In this part we will analyse hoop and radial stresses in
components like thin disc, long cylinder, or a disc of uniform strength
rotating at high angular speeds.
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ROTATING RING
Consider a ring of mean radius R, rotating about its axis O, with circular
speed ω as shown in Figure 1. Say the section of the ring is rectangular with
breadth b and thickness t as shown in the figure. Take a small element abcd,
subtending an angle dθ at the centre, at angular displacement θ from x−x
axis.
Volume of small element = Rdθbt
if weight density of ring = ρ
weight of small element = ρRbtdθ
Figure 1 Rotating ring
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where ρ = weight density,
ω = angular velocity in rad/s, and
g = acceleration due to gravity.
dF, centrifugal force acting on this element =
Vertical component of centrifugal force =
Horizontal component of
Horizontal component of centrifugal force will be cancelled when we
consider another small element a′b′c′d′ in second quadrant at an angle
θ, but the vertical component of dF will be added.
maF
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Total vertical component or the bursting
force on ring in vertical direction acting on
horizontal diameter xx:
Say, σc is the hoop stress developed in horizontal section along xx.
Area of cross section resisting the force = 2bt
Resisting force (Eql.in y direction)
Circumferential or hoop stress developed in section of ring,
where V = linear velocity of ring at radius R = ωR.
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Example 8.1
To develop substantial stress in ring, ω has to be very high.
Let us take a ring of mild steel with weight density of 7644 kg/m3 or
76.44 × 103 N/m3.
Say the yield strength (yield point) of mild steel is 280 N/mm2 and let us
take
σyp = σc, developed in ring.
If we take the ring radius equal to 0.5 m, then angular speed required to develop hoop stress of the order of 280 N/mm2 is
or 3620 revolutions per minute.
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STRESSES IN A THIN ROTATING DISC
Consider a thin disc of inner radius R1 and outer radius R2, rotating at an angular
speed ω about its axis O. The thickness t of the disc is small and it is assumed
that stresses in the disc along the thickness do not vary and there is no axial
stress developed in the disc; in other words the disc is under plane stress
conditions. Take a small element abcd at a radius r, of thickness dr, subtending
an angle dθ at the centre of the disc (Figure 2).
Figure 2 Thin rotating disc
When the disc is rotating at a
high speed, say radius r
changes to r + u and the radius
r + dr changes to r + u + dr +
du. In other words, change in
radius r is u and change in
radial thickness dr is du. Say
circumferential stress developed
is σc at radius r.
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Radial stress at radius r is σc.
Say ρ = weight density of thin disc.
weight of the small element = ρ(rdθ)tdr.
Centrifugal force on small element,
Circumferential forces on faces ad and bc = σc dr t.
Radial force on face ab = σrr dθ t.
Radial force on face cd = (σr + dσr)(r + dr) dθ t.
Resolving the forces along the vertical direction OCF as shown in Figure (b):
But angle dθ → 0 is very small
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Moreover, r dθ is common on both sides; simplifying and neglecting the term
dσr dr, we get
Dividing this equation throughout by dr, we get
or
Now considering the strains in circumferential and radial directions
Circumferential strain, (Hooke’s law)
******
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where E is Young’s modulus and v is Poisson’s ratio.
Radial strains,
or
Differentiating with respect to r, we get
Equating these last two equations, we get
rCE
ru
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After the simplification of this equation
Putting the value of (σc − σr) from into above eq.
Simplifying this equation
or
Integrating this we get
where A is constant of integration.
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From
Putting this value in we get
or
Multiplying this by r throughout, we get
Integrating this equation,
where B is another constant of integration
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or
Radial stress,
From
Putting the value of σr, we get the hoop stress as
Let us take constant, and another constant.
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Expressions for stresses will now be: Radial stress
Constants A and B can be determined by using boundary conditions of a thin hollow disc: Radial stress σr = 0 at r = R1; and at r = R2
(normal stress on free surfaces is zero), therefore
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From these two equations, values of constants are
Final expressions for stresses for hollow disc are
where
For radial stress to be maximum, or
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or
From the expression for σc, it can be seen that the maximum
circumferential stress will occur at r equal to a minimum value,
i.e. r = R1, (inner radius).
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Solid Disc
In a solid disc, at r = 0, i.e. at the centre, stresses cannot be infinite, therefore
constant B has to be zero otherwise will become infinite at the centre.
So, stresses are
Say outer radius of solid disc = R. At outer surface, radial stress σr = 0. Therefore,
Constant,
Stresses at any radius r,
obviously both the stresses are maximum at the centre, i.e. at r = 0
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Example 8.2 A thin uniform steel disc of diameter 500 mm is rotating about its axis at 3000 rpm. Calculate the maximum principal stress and maximum in plane shear stress in the disc. Draw the circumferential stress and radial stress distribution along the radius of the thin disc. Given Poisson’s ratio, = 0.3, r = 7700 kg/m3 g = 9.81 m/s2
Radius, R = 250 mm
Constants
Maximum principal stress σr, σc at centre
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Expression for radial stress
Hoop stress
Distribution of radial and hoop stresses is shown in the figure. Maximum in plane shear stress occurs at the outer radius
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Example 8.3 A thin uniform disc of inner radius 50 mm and outer radius 200 mm is rotating at 6000 rpm about its axis. What are the maximum hoop and radial stresses? Draw the distribution of hoop and radial stresses along the radius of the disc. Given r = 7800kg/m3, = 0.3, N = 6000rpm. R1, inner radius = 50 mm R2, outer radius = 200 mm
Angular speed,
Constants,
Weight density, ρ = 7800 × 9.81 × 10−9 N/mm3 = 7.65 × 10−5 N/mm3
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Let us calculate also
Radial stress
Maximum σr occurs at
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Variation of hoop and radial stresses along the radius of the disc
Circumferential Stress
The maximum circumferential stress occurs at the inner radius and minimum circumferential stress occurs at the outer radius as is obvious from the figure.
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DISC OF UNIFORM STRENGTH
Rotors of steam or gas turbines, on the periphery of which blades are
attached, are designed as disc of uniform strength, in which the stress
developed due to centrifugal forces are equal and constant independent of the
radius. In order to achieve the objective of uniform strength throughout,
thickness of the disc is varied along the radius. Consider a disc of radius R,
rotating at angular speed ω about its axis, as shown in Figure 3. Take a small
element abcd subtending on angle dθ at the centre of the disc. Disc is of
uniform strength. Stress on faces ab, bc, cd, and da of the small element is σ
as shown. Thickness of the element at radius r is t and say at radius r + dr
thickness is t + dt.
Figure 3 Disc of uniform strength
t + dt
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Volume of small element = rdθtdr
Mass of the element =
Centrifugal force on small element,
Radial force on face ab = rdθtσ
Radial force on face cd = (r+dr) dθ(t+dt)σ
Forces on faces bc and da = rdθtσ
where σ is the uniform stress.
These forces are inclined at an angle to horizontal direction as shown.
Resolving all these forces along the radial direction OCF
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but dθ is very small, so
in all these terms dθ is common, simplifying the equation we can write
neglecting dr × dt as negligible.
or
or
Integrating this equation, we get
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or
or
or
where ln A is a constant of integration,
At the center, r = 0, thickness t = t0
So, t0 = Ae0 = A
or constant
A = t0
Equation for thickness becomes
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Example 8.4 A steel disc of a turbine is to be designed so that the radial and circumferential stresses are to be the same throughout the thickness and radius of disc and is equal to 80 MPa, when running at 3500 rpm. If the axial thickness at the centre is 20 mm, what is the thickness at the radius of 500 mm? ρ for steel = 0.07644N/cm3, g = 9810mm/s.
Thickness at centre, t0 = 20 mm
radius r = 500 mm
Angular velocity,
Constant strength, σ = 80MPa
density = 0.07644 × 10−3N/mm3
Thickness, t = t0 e−1.636 = 20 × 0.1947 = 3.895 mm
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STRESSES IN ROTATING LONG CYLINDERS
The analysis of stresses in long cylinders is similar to that of a thin disc. The
only difference is that length of the cylinder along the axis of rotation is large
as compared to its radius and axial stress is also taken into account, i.e. at any
radius r of the cylinder there are three stresses, i.e. σa, axial stress; σr radial
stress, and σc the circumferential stress. While developing the theory for long
rotating cylinders, the following assumptions are made:
1.Transverse sections of the cylinder remain plane at high speeds of rotation.
This is true only for sections away from the ends.
2.At the central cross section of the cylinder, shear stress is zero due to
symmetry and there are three principal stresses, i.e. σr, σa, and σc as stated
above.
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If E is the Young’s modulus and v is the Poisson’s ratio, then,
Radial strain,
Axial strain,
We have derived expression for (σc − σr) for a small element subtending an angle dθ at the centre of disc, the equations of equilibrium give
But circumferential strain,
Radial strain,
Hoop strain,
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Differentiating with respect to r, we get
Equating and the above equation and after
simplification, we get
As per the first assumption, the transverse sections remain plane after the long cylinder starts rotating at high angular speeds; it is implied that the axial strain is constant, therefore,
In this E and are elastic constants, therefore
σa − (σc + σr) = a constant
Differentiating the above equation with respect to r, we get
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or
Substituting the value of in ,
After further simplification
But from , we get
Substituting this value of (σc − σr) in
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or
Integrating this, we get
where A is constant of integration. But,
From and , we get
Multiplying throughout by r, we get
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Integrating this equation, we get
where B is another constant of integration.
Or radial stress,
But from
From Eqs and , we get the value of σc
or hoop stress,
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Solid cylinder
In the case of solid long cylinder, stress at the centre cannot be infinite; therefore
constant B = 0, expression for stresses will be
at r = R, outerradius
σr = 0,
∴
or constant
At the outer free surface, radial stress σr has to be zero, so
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Expression for stresses will now be
If we put constant
Stresses will be
σr and σc are maximum at the centre of the cylinder
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radial stress,
hoop stress,
Hollow cylinder
Stresses in a hollow cylinder are
But radial stress is zero at inner and outer free surfaces of cylinder, i.e.
So,
From these equations
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Finally the expressions for stresses are
Obviously σc will be maximum when r is minimum, i.e. at inner radius, R1
To obtain the value of σr max, let us put
or
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Putting this value of r in expression for σr
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Example 8.5
A solid long cylinder of diameter 600 mm is rotating at 3000 rpm. Calculate (i)
maximum and minimum hoop stresses and (ii) maximum radial stress.
Given ρ = 0.07644 N/cm3, g = 9.8 m/s2, v = 0.3
Radius R = 300 mm
ρ = .07644 × 10−3 N/mm3
g = 9800 mm/s2, v = 0.3
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Minimum hoop stress occurs at outer radius
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Example 8.6
A long cylinder of inside diameter 50 mm and outside diameter 250 mm is
rotating at 5000 rpm, calculate (i) minimum and maximum hoop stresses
and (ii) maximum radial stress and at what radius it occurs?
Given ρ = 0.07644 ×−3/mm3, v = 0.3, g = 9.8 m/s2
Inner radius, R1 = 25 mm
Outer radius, R2 = 125 mm.
Constants
Angular speed,
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occurs at
Circumferential stress,
At the inner radius,
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At the outer radius, at r = R2
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Exercise 8.1
A solid long cylinder of diameter 400 mm is rotating about its axis at an
angular speed of 300 rad /s. Determine the maximum and minimum values of
radial and hoop stresses.
Given ρ = 0.07644 × 10 3 N/mm3, =0.3, g = 9.81m/s2
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Exercise 8.2
A long cylinder of steel of outer diameter 750 mm and inner diameter 250 mm
is rotating about its axis at 4000 rpm. Determine the radial stress and
circumferential stress along the radius of cylinder.
Given ρ = 0.078 N/cm3, = 0.3, g = 980 m/s2
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THERMAL STRESSES IN A THIN DISC
Consider a thin disc rotating at a high angular speed ω and subjected to
temperature variation at the same time as shown in Figure 4. Say the
stresses are σc, circumferential, and σr, radial, and a is the coefficient of
thermal expansion of the material, T is the change in temperature.
Consider a small element abcd subtending an angle dθ at the centre,
radial thickness dr. At high speed
Figure 4 Rotating thin disc
r → changes to r + u
dr → changes to dr + du
Circumferential strain,
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Radial strain,
The relationship between the circumferential and radial stresses (page 9):
From
Equating and the above equation,
or
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Putting the value of (σc – σr) from
Integrating this equation
where A is a constant of integration. From and the above eq., eliminating σc
Multiplying throughout by r
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Integrating this, we get
where B is another constant of integration.
Radial stress,
putting the value of σr in , we get
Constants A and B can be determined by using the boundary conditions. For a solid disc, constant B = 0, because stress cannot be infinite at the centre of the disc.
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E = 208 kN/mm2, r = 0.07644 N/cm3; = 0.3,
g = 9.80 m/s2
Given a = 11 × 10–6/°C
Example 8.7 A thin disc of outer radius 300 mm and inner radius 100 mm is rotating about its axis at 3500 rpm. Temperature in the disc has a linear variation of temperature with temperature 80°C at outer radius and 20°C at inner radius, calculate the maximum stress.
Angular speed,
Temperature variation
T = 20 + 0.3(r – 100) at r = 100 mm, T = 20°C r = 300 mm, T = 80°C Radial stress, σr = 0 at r = 100 mm, r = 300 mn
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Taking temperature 20°C, constant from center to inner radius of disc
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Putting second boundary condition,
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From Eqs (i) and (ii)
Maximum stress occurs at the inner radius at r = 100 mm.
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maximum stress i.e. circumferential stress occurs at the inner radius.