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7/17/2019 Solids Andstructures Mehanics IV Notes
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EMT 2407 SOLID AND STRUCTURAL ENGINEERING IV
COURSE NOTES
INNO ODIRA (BSc., MSc., PhD Ongoing)
(Mechanical Engineering)
JOMO KENYATTA UNIVERSITY OFAGRICULTURE AND TECHNOLOGY
c2012
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2
COURSE OUTLINE
EMT 2407 SOLID AND STRUCTURAL MECHANICS IV
Bending of curved beams with plane loading: Winkler’s analysis. Shear: Shear stress due to
torsion. Shear stress distribution due to torsion of thin-walled non-circular closed cross-section:
single cell and multi-cell cross-section. Shear deflection of beams - the slope and energy methods.
Total deflection of beams. Struts: Euler’s crippling load for struts with different end constraints,
struts with initial curvature, struts with eccentric loading, struts with transverse loading and
empirical strut formulae. Beam columns; Rigorous method and approximate engineering methods,
modified methods of superposition. Bending due to thermal stresses : Thermostats; commercial
practice, design concepts of thermostats, strip deflection constant and strip force constant, concept
of minimum volume thermostats. Rotating discs and cylinders: Stresses and strains, rotation of
shrink fit assemblies, disc with varying thickness and thermal effects. Plates: Simple concepts of
the general plate problem, cylindrical and spherical bending. Bending of circular plates - simple
cases. Introduction to stress functions and application to plate bending.
Reference Books
1. P. P. Benham, R. J. Crawford & C. G. Armstrong (1999) Mechanics of Engineering Mate-
rials , Longman, 2rd Ed.
2. Ferdinand P. Beer & Johnston, R. E. (1985) Mechanics of Materials , McGraw-Hill, student
Ed.
3. J. M. Gere & S. P. Timoshenko (1999) Mechanics of Materials , Stanley Thornes (Publishers)
Ltd, 4th Ed.
4. Arthur, P. Boresi & Sidebottom O. M. (1985) Advanced Mechanics of Materials , John Wiley
& Sons inc., 4th Ed.
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TABLE OF CONTENTS 3
TABLE OF CONTENTS
COURSE OUTLINE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
TABLE OF CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Columns and Struts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1 Euler’s Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.1 Struts with Pinned Ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.2 Strut with One Free End . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.3 Strut with one fixed end and the other end pinned . . . . . . . . . . . . . . 8
1.1.4 Both ends fixed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.1.5 Effective Length Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2 Limitation of Euler Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3 Rankine-Gordon Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4 Struts with Eccentric Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.5 Struts with Initial Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.6 Tutorial 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
General Solution of the Torsion Problem . . . . . . . . . . . . . . . . . . . . 19
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2 Torsion of Non-circular Solid Members . . . . . . . . . . . . . . . . . . . . . . . . 19
2.3 Torsion of Hollow Thin-walled Non-circular Members . . . . . . . . . . . . . . . . 21
2.4 Calculation of the Angle of Twist . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5 Single Cell X-sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.6 Multi-cell x-section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.7 Tutorial 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
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TABLE OF CONTENTS 4
Deflection Due to Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.2 Deriving the Shear Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.2.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.2.2 Distribution of Shear Stresses in a Rectangular Beam . . . . . . . . . . . . 37
3.2.3 Shear Stress Distribution in Beams with Flanges . . . . . . . . . . . . . . . 38
3.3 Shear Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Bending of Thin Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.1 Simple Concepts of the General Plate Problem . . . . . . . . . . . . . . . . . . . . 44
4.1.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.2 Rectangular Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.2.1 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending . . . . . . . 48
4.3.1 Relationship between Load, Shear Force and Bending Moment . . . . . . . 50
4.4 Tutorial 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
Bending of Curved Beams with Plane Loading . . . . . . . . . . . . . . . . . 56
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
5.2 Stresses and Strains in Curved Beams - Winkler Bach Analysis . . . . . . . . . . . 56
5.2.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.2.2 Case 1 - Slender Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
5.2.3 Case 2: Deeply Curved Beams . . . . . . . . . . . . . . . . . . . . . . . . . 58
5.3 Position of the Neutral Axis for a Deeply Curved Beam . . . . . . . . . . . . . . . 58
5.4 Bending Moment on the Cross-section . . . . . . . . . . . . . . . . . . . . . . . . 59
5.4.1 Terminologies used with Curved Beams . . . . . . . . . . . . . . . . . . . . 60
5.4.2 To determine R1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
5.5 Combined Direct and Bending Stresses . . . . . . . . . . . . . . . . . . . . . . . . 64
5.6 Tutorial 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
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TABLE OF CONTENTS 1
Bending due to Thermal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
6.3 Stresses in Bimetallic Strips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.3.1 Bending Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.3.2 Direct Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.3.3 Combined Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.4 Types of Thermostats Bimetallic Strips . . . . . . . . . . . . . . . . . . . . . . . . 74
6.4.1 Simply Supported Beam Type . . . . . . . . . . . . . . . . . . . . . . . . 74
6.4.2 Cantilever Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
6.5 Minimum Volume Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
6.6 Tutorial 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
Rotating Discs and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
7.2 Circumferential and Radial Strains . . . . . . . . . . . . . . . . . . . . . . . . . . 82
7.2.1 Solid Disc with Unloaded Boundaries . . . . . . . . . . . . . . . . . . . . . 84
7.2.2 Maximum Speed for Initial Yielding . . . . . . . . . . . . . . . . . . . . . . 85
7.2.3 Increase in Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
7.2.4 Change in Thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
7.3 Disc with Central Hole and Unloaded Boundaries . . . . . . . . . . . . . . . . . . 87
7.4 Disc Shrunk onto a Shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
7.5 Disc with Loaded Outer Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . 90
7.6 Disc of Uniform Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
7.7 Tutorial 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
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2
Chapter 1
Columns and Struts
A column is a long vertical slender bar or structural member subjected to an axial compressive
load and fixed rigidly at both ends. A strut refers to a long slender bar or structural member
in any position other than vertical, subjected to a compressive load. The strut may have one or
both ends fixed rigidly or hinged or pin-jointed. Examples of struts are: piston rods, connecting
rods for mechanisms, side-links in forging machines etc.
The primary concern in the analysis and design of struts and columns is the ability of the structure
to support a specified axial compressive load without undergoing unacceptable deformations1
Columns fail by crushing when the yield stress is exceeded while struts fail by buckling before the
the yield stress is reached.2
Buckling may occur due to a number of reasons:
1. the applied load may be higher than the critical load,
2. the strut may not be perfectly straight,
3. the load may not be applied exactly along the strut axis.
1.1 Euler’s Theory
A theory of buckling for slender struts under axial compression was developed by Leonhard Euler.
1.1.1 Struts with Pinned Ends
Consider a strut with pinned ends as shown in Figure 1.1.
Assumptions
i. strut is slender (l >> d)
ii. axial load P is applied through the centroid of the cross-section, aligned with the longitudinal
axis
iii. strut is initially perfectly straight
iv. material is obeys hooks law
v. bending is in a single plane (planar bendig)
1
stability of a structure is the ability of the structure to support a given load without undergoing/ experiencingsudden change in its configuration.
2Buckling is a failure mode characterized by sudden failure of a structural member subjected to high compressivestressed when the stress at the point of failure is less than the yield stress of the material in compression.
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EMT 2407: 1.1 Euler’s Theory 3
Figure 1.1: Strut with pinned ends
Loading conditions
1. A small load P is applied: the strut remains straight and experiences axial stresses only
(stable equilibrium). If a small lateral load is applied, the strut bends slightly but straightens
when the load is removed.
2. P is increased (P=Pcr: The strut may have a bent shape. If a small lateral load is applied,
the bent shape remains when the load is removed. The strut experiences a neutral or staticequilibrium in either straight or bent position.
3. P is increased further (P>Pcr): we have unstable equilibrium and the strut collapses by
bending.
Consider a section of the beam of length x shown in Figure 1.2.
Figure 1.2: Section of the strut
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EMT 2407: 1.1 Euler’s Theory 4
The displacement at A is v and the bending moment M = P v. Using the differential equation for
the deflection curve derived for beams,
EI d2v
dx = −M = −P v (1.1)
where,
I second moment of area of the section.
E Young’s Modulus of elasticity
∴d2v
dx +
P
EI v = 0 (1.2)
letk2 = P
EI (1.3)
equation 1.2 becomes
d2v
dx + k2v = 0
which is a homogeneous ode whose solution is:
v = C 1 sin kx + C 2 cos kx (1.4)
where C 1 and C 2 are constants which depend on the boundary conditions. Applying the boundary
conditions,
at x = 0, v = 0
at x = L, v = 0
0 = 0 + C 2 ⇒ C 2 = 0
0 = C 1 sin kL
⇒ C 1 = 0 or sin kL = 0
If C 1 = 0, we get a trivial case of the un-deflected strut which is of no importance in the analysis.
The condition sin kL = 0 leads to the solution kL = 0, π, 2π,...
Taking kL = 0 ⇒ P = 0 which is of no interest.
Thus kL = nπ, n=1,2,3...
Buckling first occurs for n=1 from which the critical load is given by:
kL = π ⇒ k =
π
L (1.5)
∴P crEI
= k2 =π
L
2or P cr =
π2EI
L2 (1.6)
P cr is the critical load which is also known as Euler Load or Euler Crippling Load.
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EMT 2407: 1.1 Euler’s Theory 5
The corresponding buckled mode shape is given by:
v = C 1 sin kx = C 1 sin
π
L x
Figure 1.3: mode shapes
To maintain same x-sectional area of the beam and achieve high second moment of area, we use,
1. hollow sections
2. I-sections
Figure 1.4: If I 1 > I 2, use the smaller value, I 2 in the equation
Optimum Columns/struts
1. Prismatic column/strut - same cross-section throughout
2. X-section is made larger in regions where B.M is maximum (constant strength column/strut)
3. Reinforced prismatic column/strut over part of the length.
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EMT 2407: 1.1 Euler’s Theory 6
Figure 1.5: mode shapes
The value of the compressive stress corresponding to the critical stress is called Critical stress and
is denoted by σcr.
Setting I = Ar2, where A is the cross-sectional area and r is its radius of gyration, we have:
σcr = P cr
A =
π2EAr2
AL2 =
π2Er2
L2 (1.7)
σcr = π2E
(L/r)2 (1.8)
The quantity L/r is called the slenderness ratio of the column.
1.1.2 Strut with One Free End
Figure 1.6: Strut with one free end
Consider a section of the strut, a distance x from the free end. The bending moment is given by:
ΣM x = 0 ⇒ M + P (δ − v) = 0∴ M = P v − P δ
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EMT 2407: 1.1 Euler’s Theory 7
Substituting the expression for M in the differential equation for the deflection curve,
EI d2v
dx2 = −M = −P (v − δ )
d2v
dx2 +
P
EI v =
P
EI δ let
P
EI = k2
∴d2v
dx2 + k2v = k2δ (1.9)
The solution to the differential equation 1.9 consists of a complementary function and a particular
integral.
Complementary function:
v1 = C 1 sin kx + C 2 cos kx (1.10)
Particular integral:
let v2 = C 3d2v2dx2
= 0
substituting in DE
0 + k2C 3 = k2δ
⇒ C 3 = δ
The solution to the D.E becomes,
v = v1 + v2 = C 1 sin kx + C 2 cos kx + δ (1.11)
Applying boundary conditions,
at x = 0, v = 0
∴ C 2 + δ = 0 ⇒ C 2 = −δ
at x = 0, dv
dx = 0
dv
dx = C 1k cos kx + δk sin kx
C k = 0 = since k = 0 ⇒ C 1 = 0
∴ v = −δ cos kx + δ = δ (1 − cos kx)
at x = L, v = δ
⇒ cos kL = 0
kL = nπ
2 , n = 1, 2, 3,...
P cr = k2
EI =
n2π2
4L2 EI
Buckling first occurs at n = 1
P cr = π2
4L2EI (1.12)
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EMT 2407: 1.1 Euler’s Theory 8
Figure 1.7: Fixed-pinned ends
1.1.3 Strut with one fixed end and the other end pinned
The bending moment at section x is:
M = P v − Rx
from which,
d2v
dx2 +
P
EI v =
R
EI x (1.13)
let P
EI
= k2, the solution to de becomes (1.14)
v = C 1 sin kx + C 2 cos kx + R
P x (1.15)
The boundary conditions are: v = 0 at x = 0, v = 0 at x = L and dvdx
= 0, at x = L from which,
C 2 = 0 C 2 = − R
P k cos kL = − RL
P sin kL∴ tan kL = kL (1.16)
The smallest value of kL to satisfy equation 1.16 is kL = 4.4934 (by trial and error) or
P cr = k2EI =
4.49342
L2
EI (1.17)
P cr = 20.19EI
L2 (1.18)
1.1.4 Both ends fixed
P cr = 4π2EI
L2 (1.19)
1.1.5 Effective Length Concept
This method relates the critical load for struts of varying support conditions to pinned end
columns.
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EMT 2407: 1.1 Euler’s Theory 9
Figure 1.8: Effective lengths
For Fixed-free ends, Le = 2L
P cr = π2EI
L2e
= π2EI
4L2 (1.20)
For Fixed-fixed ends, Le = L2
P cr = π2EI
L2e
= 4π2EI
L2 (1.21)
For Fixed-pinned ends,
P cr = π2EI
L2e
= 20.19EI
L2 =
2.046π2EI
L2 (1.22)
L2e =
L2
2.046 ⇒ Le ≈ 0.7L (1.23)
Example 1.1.1. A straight steel rod 9mm diameter is rigidly built into a foundation, the free end
protruding 0.5m normal to the foundation. An axial load is applied to the free end of the rod which
deflects as shown in Figure 1.9 . Determine the following,
i. Euler’s Buckling load
ii. The deflection at the free end of the rod when the total compressive stress reaches the elastic
limit.
Assume that E=200GPa and yield stress is 300MPa.
Solution
i.
P cr = π2EI
4L2
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EMT 2407: 1.2 Limitation of Euler Theory 10
Figure 1.9: Fixed-free strut
For a solid bar of diameter D,
I = πD4
64 =
π × (9 × 10−3)4
64 = 3.221 × 10−10m4
P cr = π2 × 200 × 109 × 3.22 × 10−10
4 × 0.52 = 636N
ii.
σc = P cr
A +
P vmaxy
I (1.24)
where:
σc compressive stress
y distance from the centroid to the outermost fibres
when the compressive stress reached the elastic limit, σc = σy, so that:
300 × 106 = 636
π × (4.5 × 10−4)2 +
636 × (4.5 × 10−3)vmax3.221 × 10−10
300 × 106 = 9.997 × 106 + 8.885 × 109vmax
vmax = 300 × 106 − 9.997 × 106
8.885×
109 = 32.6mm
1.2 Limitation of Euler Theory
Recall that
P cr = π2EI
L2 (1.25)
σcr = P
A =
π2E
(L/r)2 (1.26)
For long slender beams, (L/r) is large and σcr
is small
For short thick beams, (L/r) is small and σcr increases as L/r reduces.
However, σcr is normally limited by the materials yield stress such that after yielding, failure will
occur by plasticity. Therefore, Euler,s theory works only with long slender columns.
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EMT 2407: 1.3 Rankine-Gordon Method 11
Buckling becomes the limiting mode of failure when the critical stress is less than or equal to the
yield strength, i.e.
σy ≥ π
2
E (Le/r)2
⇒ Le
r ≥ π
E
σy
If a factor of safety, f is used,
Le
r
c
= π
f E
σy
The quantityLer
c
is known as the transition ratio.
Figure 1.10: Euler hyperbola
1.3 Rankine-Gordon Method
Is based on experimental results.
For a very short column or strut, collapse will result from direct crushing and the crippling load
is: P c = σcA, where
σc maximum compressive stress
A is the cross-sectional area
For a long strut, Euler formula applies. The Rankine hypothesis is;
1
P R =
1
P c +
1
P cr (1.27)
P R is the actual crippling load
P cr is Euler’s crippling load=π2EI L2e
= π2EA(L/r)2
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EMT 2407: 1.4 Struts with Eccentric Loading 12
Equation 1.27 can be rewritten as:
1
P R=
1
σcA +
(Le/r)2
π2EA (1.28)
A
P R=
1
σc+
(Le/r)2
π2E (1.29)
P RA
= 11σc
+ (Le/r)2
π2E
= σc
1 + σc(Le/r)2
π2E
(1.30)
P = σcA
1 + σcπ2E
(L/r)2 =
σcA
1 + aLer
2 (1.31)
Le is used to cater for the different end constraints. For any given material, σc and a are constants.
Other approximate engineering methods are:
1. Johnson’s Parabolic Formula
2. Straight-Line Formula
1.4 Struts with Eccentric Loading
In practice, struts or columns are rarely loaded exactly along the centroidal axis as the Euler
analysis assumes. Consider an eccentrically loaded strut illustrated in Figure ??.
Figure 1.11: Eccentrically loaded beam
M = P v + P e = P (v + e)
Substituting this in the differential equation,
d2v
dx2 =
−M
EI =
−P
EI (v + e)
d2v
dx2
+ P
EI v =
−P
EI e, Let
P
EI = k2
d2v
dx2 + k2v = −k2e
whose solution is v = C 1 sin kx + C 2 cos kx − e
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EMT 2407: 1.4 Struts with Eccentric Loading 13
Applying boundary conditions,
at x = 0, v = 0 at x = L, v = o
0 = C 2 − e ⇒ C 2 = eC 1 sin kL + e cos kL − e = 0 ⇒ C 1 sin kL = e(1 − cos kL)
but sin kL = 2 sin kL
2 cos
kL
2 and 1 − cos kL = 2 sin2 kL
2
∴ 2 sin kL
2 cos
kL
2 = 2e sin2 kL
2
C 1 = e tan kL
2
and v = e(tan kL
2 sin kx + cos kx − 1)
The value of the maximum deflection is obtained by setting x = L2
vmax = e(tan kL
2 sin
kL
2 + cos
kL
2 − 1)
sin2 kL2
cos kL2
+ cos kL
2 =
sin2 kL2
+ cos2 kL2
cos kL2
= 1
cos kL2
= sec kL
2
vmax = e(sec kL
2 − 1) = e(sec
L
2
P
EI − 1)
vmax becomes infinity when L2
pEI
= π2
. This implies that the defection becomes unacceptably
large when this condition is satisfied.
∴ P cr = π2EI
L2 (1.32)
The maximum stress occurs in the section of the column/strut where the B.M is maximum i.e.
σmax = P
A +
M maxc
I
M max = P (vmax + e) = P e(sec kL
2 − 1 + 1) = P e sec
kL
2
σmax = P
A + P e sec
kL
2 · c
Ar2
σmax = P A
1 + ec
r2 sec kL
2
= P
A
1 +
ec
r2 sec
L
2
P
EI
= P
A
1 +
ec
r2 sec
Le
2r
P
EA
Where, ec
r2= eccentricity ratio and Le
r = slenderness ratio
Le is used to make the formula applicable to various end conditions.
Example 1.4.1. A tubular strut is 60mm external diameter and 48mm internal diameter. It is 2.2m long and has hinged ends. The load is parallel to the axis of the axis of the strut but
eccentric. Find the maximum compressive stress for a crippling load of 0.75 of the Euler value
and an eccentricity of 4.5mm. Take E=207GPa.
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EMT 2407: 1.5 Struts with Initial Curvature 14
Solution
I = π(D4 − d4)64
= 3.756 × 10−7 m4
P cr = 0.75π2EI
L2 = 0.75
π2 × 109 × 3.756 × 10−7
2.22 = 118.91 kN
k =
P
EI = 0.75
π2
L2 =
0.75 × π2
2.22 = 1.237
M max = P e sec kL
2 = 118.91 × 103 × 0.0045 × sec
1.276 × 2.2
2 = 2565.74 Nm
σc = P
A +
M maxc
I , A =
π(0.062 − 0.0482
4 = 1.0179 × 10−3 m2
= 118.91 × 103
1.0179−3 +
2565.74 × 0.03
3.756 × 10−7
= 116.82 × 106 + 204.93 × 106 = 321.75 MPa
1.5 Struts with Initial Curvature
In some cases, the strut may not be perfectly straight before loading. This will influence the
stability of the strut. The initial shape of the beam may be assumed circular, parabolic of
sinusoidal but the most convenient form is the sinusoidal of the form:
vo = V sin πx
L (1.33)
where,
vo deflection at distance x from one end.
V the amplitude of the deflection or the initial maximum deflection.
Consider the strut with initial curvature as shown in Figure 1.12. On application of load P, the
deflection is increased by v and the B.M. at a section xx is:
M = P (v + vo) (1.34)
Substituting M in the differential equation of the deflection curve,
EI d2v
dx2 = −M = −P (v + V sin
πx
L (1.35)
d2v
dx2 +
P
EI v = − P
EI V sin
πx
L (1.36)
Let P EI
= k2 whichleadsto : (1.37)
d2v
dx2 + k2v = −k2V sin
πx
L (1.38)
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EMT 2407: 1.5 Struts with Initial Curvature 15
Figure 1.12: Strut with initial curvature
whose general solution is given by:
v = C 1 sin kx + C 2 cos kx − k2V
k2
− π2
L2
sin πx
L (1.39)
Applying the Boundary conditions,
at x = 0, v = 0
⇒ 0 = 0 + C 2 + 0 ∴ C 2 = 0
v = C 1 sin kx − k2V
k2 − π2
L2
sin πx
L
at x = L, v = 0
⇒ 0 = C 1 sin kL + 0B = 0
∴ v = − k2V
k2 − π2
L2
sin πx
L =
k2V π2
L2 − k2
sin πx
L
substituting back P EI
= k2
v =P EI
π2
L2 − P
EI
V sin πx
L (1.40)
= P
π2
EI L2 − P
V sin πx
L (1.41)
If we denote π2EI L2
= P cr, then
v = P
P cr − P V sin
πx
L (1.42)
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EMT 2407: 1.6 Tutorial 1 16
Thus the effect of the load is to increase the initial deflection by a factor P P cr−P
.
As P → P cr, the deflection tends to infinity.
We can obtain an expression of v in terms of the direct compressive stress.
Let σ = P A ⇒ P = σA and P cr = σcr
A ⇒ P = σcrA. Substituting these expressions in equation
1.42,
v = σ
σcr − σV sin
πx
L (1.43)
The total deflection at x is given by:
v = v + vo = σ
σcr − σV sin
πx
L + V sin
πx
L
= σ
σcr − σ + 1V sin
πx
L=
σ + σcr − σ
σcr − σ V sin
πx
L
v = σcrσcr − σ
V sin πx
L
The maximum deflection occurs at x = L2
vmax = σcrσcr − σ
V (1.44)
and the maximum bending moment is:
M max = P
σcrσcr − σ
V
The maximum compressive stress is:
σc =P V σcr
σcr−σy
I +
P
A (1.45)
=P V σcr
σcr−σy
Ar2 + σ (1.46)
= σV σcr
σcr − σ
y
r2 + σ (1.47)
1.6 Tutorial 1
1. A straight slender column of height 2.77 m is fixed at the lower end and is entirely free at
the upper end. The design criterion is to limit the compressive strain prior to buckling to
0.0008. Determine the required least radius of gyration. [Ans: 50 mm]
2. A column is made of two rolled steel joists of I-section and two thick plates as shown in Fig.
1.13. Determine by Rankine’s Formula, the safe load the column of 4m length, with both
ends fixed, can carry with a factor of safety of 3.
Take: a = 1
7500, and σc = 320 MN/m2
[Ans: 931.56 kN]
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EMT 2407: 1.6 Tutorial 1 17
Figure 1.13:
3. Show that, for and eccentrically loaded strut, the maximum resultant compressive stress is
given by:
σc = P
A
1 +
ec
r2 sec kL
where P is the applied load, A = cross-sectional area of the strut, e = eccentricity, k = P
A,
E = Young’s modulus of elasticity, c = distance from neutral axis to outermost fibres under
compression. Before application of the load, the strut has the configuration shown in Fig.
1.14
Figure 1.14:
4. A slender strut is built in at one end and an eccentric load is applied at the free end. Working
from first principles, find the expression for the maximum length of column such that the
deflection at the free end does not exceed the eccentricity of loading. The deflected strut is
shown in Fig. 1.15
Ans:
L =
π
3 EI
P
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EMT 2407: 1.6 Tutorial 1 18
Figure 1.15:
5. A slender column of straight circular section of length L has pinned ends. It carries an
axial load P and also a horizontal lateral load W applied at the mid-length. Show that themaximum deflection and maximum bending moment are given by:
vmax = W
2kP tan
kL
2 − W L
4P
M max = W
2k tan
kL
2 , where k =
P
EI
(a) In the case of a strut in question 5, the magnitude of P = P cr4
where P cr is Euler load
for the strut. Determine the ratio of the maximum deflection produced by P and the
lateral load acting together, to that produced by W acting alone. [Ans: 1.33]
(b) If the strut is made of steel 25mm diameter and 1.25 m long with an axial load of 16
kN applied, determine the value of W which would cause collapse if the yield stress is
280 MN/m2 and E= 206 GN/m2.
[Ans: W= 559.4 N]
6. A circular hollow steel column has a length of 2.44m, an external diameter of 101mm and
an internal diameter of 89mm with its ends pinned. Assuming the centreline is sinusoidal in
shape, with a maximum displacement at the mid-length of 4.5mm, determine the maximum
stress due to an axial compressive load of 10kN. Take E=205 GN/m2. [Ans: σc =6.72MN/m2]
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19
Chapter 2
General Solution of the Torsion Problem
2.1 Introduction
Torsion is a common engineering mode of deformation in which a solid or tubular member is
subjected to torque about its longitudinal axis resulting in twisting deformation.
Recall: Torsion equation for circular sections:
T
J =
τ
r =
Gθ
L (2.1)
where:
τ = shear stress at radius r
T = applied torque
J = polar second moment of area
G = shear modulus of elasticity
θ = angle of twist
L = length of member
Some of the assumptions used in deriving the torsion formula are:
1. Cross-sections which are plane remain plane after twisting (undistorted).
2. Radial lines remain radial during twisting
3. Deformation is by rotation of one cross-section plane relative to the next and planes remain
normal to the axis of the shaft.
2.2 Torsion of Non-circular Solid Members
There are some applications in machinery for non-circular cross-section members and shafts wherea regular polygonal cross-section is useful in transmitting torque to a gear or pulley that can have
an axial change in position. Because no key or keyway is needed, the possibility of a lost key is
avoided.
The assumption that plane sections remain plane For non-circular bars e.g square bars, due to
lack of axisymmetry1, lines drawn in the cross-section will deform when the bar is twisted and the
x-section will be warped out of its original shape. Thus the torsion equation cannot be used for
non-circular members.
1the appearance of the x-section remains the same when viewed from a fixed position and rotated about its axisthrough an arbitrary angle.
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EMT 2407: 2.2 Torsion of Non-circular Solid Members 20
Figure 2.1: Rectangular bar in torsion
Saint Venant (1855) showed that the maximum shear stress in a rectangular b × c section bar
occurs in the middle of the longest side b and is of magnitude
τ max = T
αbc2 (2.2)
= T
bc2
3 +
1.8b/c
(2.3)
where b is the longer side, and α is a factor that is a function of the ratio b/c as shown in Table
2.1.
The angle of twist is given by:
angle of twist θ = T L
βbc3G (2.4)
where β is a function of the ratio b/c as shown in
Equations 6.2 and 6.3 are only valid within the elastic limit.
Figure 2.2: Rectangular bar in torsion
Table 2.1: Coefficients for rectangular bar in torsionbc
1.0 1.2 1.5 2.0 2.5 3.0 4.0 5.0 10.0 ∞α 0.208 0.219 0.231 0.246 0.258 0.267 0.282 0.291 0.312 0.333β 0.1406 1661 0.1958 0.02290 0.2490 0.2630 0.2810 0.2910 0.3120 0.3330
Example 2.2.1. A rectangular brass bar of uniform cross-section is subjected to a torque T as
shown in Figure 2.4. If the allowable shear stress is τ max = 40 MN/m 2, determine the largest
torque which may be applied. Determine the angle of twist at this torque. Take G = 40 GN/m 2
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EMT 2407: 2.3 Torsion of Hollow Thin-walled Non-circular Members 21
Figure 2.3: Equivalent dimensions for other sections
Figure 2.4: Rectangular bar in torsion
Solution
Determine ratio ab
:a
b =
64
25 = 2.56
Interpolating:
α = 0.218 + (0.267 − 0.258)(3.0 − 2.5)
(2.56 − 2.5) = 0.259
β = 0.229 + (0.249 − 0.229)
(3.0 − 2.5) (2.56 − 2.5) = 0.2314
τ max = T
αbc2 ⇒ T = τ maxβbc2
T = 0.259 × 40 × 106 × 0.064 × 0.0252 = 414.4 Nm
θ = T L
c2bc3G =
414.4×0.2314 × 0.064 × 0.0253 × 40 × 109
= 0.0895 rad
2.3 Torsion of Hollow Thin-walled Non-circular Members
In some applications such as aeroplane structural members, the shear stress distribution due to
torsion of non-circular cross-sections is an important design factor.
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EMT 2407: 2.3 Torsion of Hollow Thin-walled Non-circular Members 22
The method used to find the shear stresses and angle of twist is simplified by assuming uniform
shear stress distribution across the wall of the section.
Consider a tube of non-circular cross-section with varying thickness shown in Figure 2(a). Let thecross-section be constant throughout the length of the tube/shaft.
Assume that the applied torque T acts about the longitudinal axis XX and it induces shearing
stresses over the end of the tube. These stresses have a direction parallel to that of a tangent to
the centerline of the wall of the tube.
A shearing stress of magnitude τ acting at any point in the circumference has a complementary
shear stress of the same magnitude acting in a longitudinal direction.
(a) Non-circular tube (b) Element of the tube
Figure 2.5:
Consider a small element ABCDEFGH of the tube and assume that the shear stress τ is constant
throughout the wall thickness t. The shearing force along the thin edge AB is τ t per unit length
of the tube. For longitudinal equilibrium of the element, this force must be equal to that on the
thin edge CD. It follows that τ t is constant for all parts of the tube. The quantity τ t = q is called
the shear flow and is the internal shearing force per unit length of the circumference of the section
of the thin tube.
The force on face ADEH (along the tangential direction) is given by:
dF = τtds (2.5)
The moment of force about the X-X axis of the tube is given by:
dT = τtds · r (2.6)
T =
τtrds (2.7)
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EMT 2407: 2.3 Torsion of Hollow Thin-walled Non-circular Members 23
The
means that the integral extends over the whole circumference.
T = q rds
= q · 2
dA = q · 2A
Figure 2.6:
From Figure 2.6,
rds = 2dA ⇒
rds = 2
dA = 2A
where A is the total plane area enclosed by the centreline of the wall of the tube.
∴ T = q (2A)⇒
q = T
2A
shear stress τ = q
t =
T
2At
Thin Rectangular x-section
(a) X-section (b) Shear stressdistribution
Figure 2.7:
A = a1a2 q = T
2A =
T
2a1a2
τ 1 = q
t1τ 2 =
q
t2τ 3 =
q
t3τ 4 =
q
t4
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EMT 2407: 2.4 Calculation of the Angle of Twist 24
Circular hollow section
Compute the shear stress for a thin-walled circular hollow shaft shown in Figure 2.8 and compare
the results with those obtained from the torsion equation.
Figure 2.8: Circular hollow section
A = πR2 q = T
2A =
T
2πR2
τ = q
t =
T
2πR2t
From the torsion equation,
T J = τ r = GθL
⇒ τ = T r
J J = 2πR3t for a thin section
∴ τ = T R
2πR3t =
T
2πR2t
Therefore the formulae for the torsion of non-circular tubes is fairly accurate.
2.4 Calculation of the Angle of Twist
The angle of twist can be determined from the strain energy stored in the tube. Consider a strip
of length L, thickness t and width ds as shown in Figure 2.9
Strain energy per unit volume is given by:
U s = τ 2
2G
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EMT 2407: 2.4 Calculation of the Angle of Twist 25
Figure 2.9: Elemental strip
general case
Figure 2.10: Elemental cube
If the bottom face is fixed and a force P is applied as shown, the elemental cube will shear.
P = shear force = τdxdz
tan γ = δdy
γ For small angles
shear strain energy =
1
2 P δ
= 1
2τdxdzγdy =
1
2τγdV
but G = τ
γ ∴ U s =
1
2
τ 2
G × Volume
strain energy per unit volume = τ 2
2G
strain energy stored in the element is given by:
dU s = τ 2
2GLtds
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EMT 2407: 2.5 Single Cell X-sections 26
Total strain energy stored in the tube is given by:
U s = τ 2
2GLtds
but τ = T
2At ∴
τ 2
2G =
T 2
2G · 22 · A2 · t2 =
T 2
8A2t2G
∴ U s =
T 2
8A2t2GLtds =
T 2L
8A2G
ds
t
But the stored energy is equal to the work done in twisting the tube,
U s = 1
2T θ
∴1
2T θ =
T 2L
8A2G ds
t (2.8)
θ = T L
4A2G
ds
t but T = 2Aq (2.9)
θ = 2AL
4A2G
q
tds =
L
2AG
qds
t (2.10)
If the thickness t is constant around the circumference, then
θ = T L
4A2Gt
ds, but
ds = S
=
T LS
4A2Gt
2.5 Single Cell X-sections
Figure 2.11: single celled x-sections
For a single cell x-section (c),
q = T
2A= τ 1t1 = τ 2t2
θ = L2AG
qds
t
= L
2AG
qπR
t1+
qπR
t2
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EMT 2407: 2.6 Multi-cell x-section 27
2.6 Multi-cell x-section
Figure 2.12: multi-celled x-sections
q 1 = T 12A1
, q 2 = T 22A2
, q 3 = q 1 − q 2
Applied torque, T = T 1 + T 2
For compatibility, θ1 = θ2 = θ
θ1 = L
2A1G
q 1a1
t1+
q 1a5
t5+
(q 1 − q 2)a3
t3
θ2 =
L
2A2G
q 2a2
t2
+ q 2a4
t4 −
(q 1 − q 2)a3
t3
Example 2.6.1. An aluminium-alloy structural member for a light aircraft has a cross-section
shown in Figure 2.13. If the shear stress is not to exceed 30MN/m 2 and the applied torque is 134
Nm, determine:
(i) the required thickness t of the metal
(ii) the angle of twist.
Take G=28 GN/m 2.
m3
Figure 2.13: Non-circular tube
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EMT 2407: 2.6 Multi-cell x-section 28
Figure 2.14: Non-circular tube
Solution
A = 0.02 × 0.08 + π × 0.022
2 = 2.2283 × 10−3m2
S 1 = π × 0.02 = 0.6283m
S 2 = S 3 =√
0.022 + 0.082 = 0.0825m
q = T
2A =
134
2 × 2.2283 × 10−3 = 30067.76N/m
τ 1 = q
2t ⇒ t =
30067.76
30 × 106 × 2 = 0.5 × 10−3m
τ 2 = q
t ⇒ t =
30067.76
30 × 106 = 1.0 × 10−3m
Angle of twist is given by:
θ = L
2AG
qds
t
= Lq
2AG
S 12t
+ S 2
t +
S 3t
=
3 × 30067.76
2 × 2.2283 × 10−3 × 28 × 109
0.0628
2 × 10−3 + 2
0.0825
1 × 10−3
= 0.142rad
Example 2.6.2. Figure 2.15 shows a two-celled tube with a cross-section as indicated. t1 = 4mm,
t2 = 5.5mm and t3 = 2.3mm. If a torque of magnitude 12KNm is applied and given G=80GPa,determine:
i. the shear flow distribution,
ii. the shear stress in all the walls
iii. the overall angle of twist per unit length.
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EMT 2407: 2.6 Multi-cell x-section 29
Figure 2.15: Example
Solution
Let q 1, q 2 and q 3 be the shear flow in the walls 1,2 and 3 respectively and S 1 − S 6 be the lengths
of the various walls. The total torque is given by:
Figure 2.16: Example
T = T 1 + T 2
where T 1 = 2q 1A1 torque in cell 1T 2 = 2q 2A2 Torque in cell 2
∴ T = 2q 1A1 + 2q 2A2
S 1 = S 5 = S 6 = 80
sin60 = 92.376mm
S 2 = S 4 =
(
92.376
2 − 30)2 + 802 = 81.621mm
S 3 = πR = π × 30 = 94.248mm
A1 = 12 × 80 × 92.376 = 3695.04mm2
A2 = 80
2 (92.376 + 60) +
π × 302
2 = 7508.76mm2
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EMT 2407: 2.6 Multi-cell x-section 30
Substituting the values of A1 and A2 in the torque equation gives,
12000 = 2(3695.04)q 1 + 2(7508.76)q 2
12000 × 106 = 7390.08q 1 + 15017.513q 2 (2.11)
From compatibility relation, we have
θ1 = θ2 = θ = L
2AiG
q ids
ti
For cell 1,
θ1 = L
2A1G
q 1ds
t1+
q 3ds
t3 = L
2A1G
(S 1 + S 5)q 1
t1+ (q 1 − q 2)S 6
t3
=
L
2A1G
46.188q 1 + 40.163q 1 − 40.163q 2
θ1 =
L
2A1G(86.351q 1 − 40.163q 2) (2.12)
For cell 2,
θ2 = L
2A2G(S 2 + S 3 + S 4)q 2
t2− (q 1 − q 2)
S 6t3
= L
2A2G
2 × 81.621 + 94.248
5.5 q 2 +
92.376
2.3 q 2 − 92.376
2.3 q 1
θ2 =
L
2A2G(86.98q 2 − 40.163q 1) (2.13)
Equating 2.12 and 2.13 gives,
L
2A1G(86.351q 1 − 40.163q 2) =
L
2A2G(86.98q 2 − 40.163q 1)
86.351q 1−
40.163q 2 = A1
A2
(86.98q 2−
40.163q 1)
= 3695.04
7508.76(86.98q 2 − 40.163q 1)
= 42.793q 2 − 19.7644q 1
⇒ 82.956q 2 = 106.115q 1
or q 2 = 1.279q 1 (2.14)
Substituting the value of q 2 in the torque equation gives,
12000
×106 = 7390.08q 1 + 15017.573(1.279q 1) = 26597.556q 1
∴ q 1 = 451169.275 N/m
q 2 = 1.279q 2 = 577045.503 N/m
q 3 = q 1 − q 2 = −125876.228 N/m
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EMT 2407: 2.7 Tutorial 2 31
Figure 2.17: Shear flow distribution
the -ve sign means that the direction of shear flow in wall 3 is opposite that assumed in the
diagram.
Shear stresses:
τ 1 = q 1t1
= 451169.275
4 × 10−3 = 112.79 MPa
τ 2 = q 2t2
= 577045.503
5.5 × 10−3 = 104.9 MPa
τ 3 = q 3t3
= 125873.228
2.3 × 10−3 = 54.727 MPa
The angle of twist can be given by either equation 2.12 or 2.13:
θ = θ1 = L
2A1G(86.351q 1 − 40.163q 2)
= L
2 × (3695.04 × 10−6) × 80 × 109[86.351(451169.275) − 40.163(577045.503)]
⇒ θ
L = 0.0267 rad/m
2.7 Tutorial 2
1. Fig. 2.18 shows a two-celled tube constructed from a steel plate of uniform thickness t with
the cross-section dimensions shown. The overall length of the tube is L and it is subjected
to a twisting moment T .
(a) Derive the expression for the shear stresses in all the walls.
(b) If a =96mm, t=6mm and the maximum shear stress is limited to 166MN/m2, calculate
the acceptable value of T
[Ans: τ 1 = 0.142T
a2
t
, τ 2 = 0.099T
a2
t
, τ 3 = 0.0433T
a2
t
; T =64.64kNm]
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EMT 2407: 2.7 Tutorial 2 32
Figure 2.18:
2. Fig. 2.19 shows a structural member for a marine vessel. t1 = 4.5mm, t2 = 6mm, t3 = 3mm,
and the dimensions of the cross-section are as shown. The member is subjected to a torque
of 15kNm. If G=80GPa, determine:
(a) the shear flow distribution,
(b) the shear stress in all the walls
(c) the overall angle of twist per unit length.
Figure 2.19:
[Ans: q 1=341536N/m, q 2=394132N/m, q 3=52597N/m; τ 1 = 75.9MPa, τ 2 = 65.69MPa,
τ 3 = 17.53MPa; θL
= 0.01257rad/m]
3. Fig. 20(a) shows the cross-section of a structural member for a light aircraft. The member
is made of aluminium alloy with the properties shown in Fig. 20(b). If the member is
subjected to a torque of 200Nm, and a factor of safety of 2.0 is to be used, determine the
allowable thickness t of the material. If the length of the member is L =4m, determine the
angle of twist.
[Ans: t = 1mm; θ=0.0898 rad]
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EMT 2407: 2.7 Tutorial 2 33
(a) Cross-section
6106.5 ×
4102 −
×
(b) Aluminiumproperties
Figure 2.20:
4. Fig. 2.21 shows the cross-section of a closed non-circular tube. The tube is 2.5m long and
is subjected to a torque of 155kNm. If G=28GPa, determine:
(a) the shear stress in all the walls,
(b) the overall angle of twist.
Figure 2.21:
[Ans: τ 1 = 34.8MPa, τ 2 = 38.2MPa, τ 3 = 28.1MPa; θ = 0.0192rad ]
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34
Chapter 3
Deflection Due to Shear
3.1 Introduction
Most beams are subjected to loads that produce both bending moments and shear forces (non-
uniform loading). In these cases, both bending (normal) stresses and shear stresses are developed
in the beam.
However in most cases, the deflection of a beam is calculated by taking into account the bending
moment only. Most structural members are normally subjected to non-uniform bending where
the bending moment varies introducing shear forces given by:
V = dM
dxAs a result of these shear forces, transverse sections will slip with respect to the adjacent sections
resulting to a deflection due to shear. The deflection due to shear can be calculated by use of
strain energy method. The strain energy due to shear is given by:
U s = τ 2
2G × V ol
3.2 Deriving the Shear Formula
The shear formula in a beam relates the shear force (V) and the shear stress (τ ).
3.2.1 Assumptions
1. The shear stresses acting on a cross-section are parallel to the shear force i.e. parallel to the
vertical sides of the cross-section as shown in figure 3.1(a).
2. The shear stresses are uniformly distributed across the cross-section of the beam, although
they may vary over the height as shown in figure 3.1(a).
The shear stresses acting on one side of an element are accompanied by shear stresses of equal
magnitude acting on perpendicular faces of the element as shown in figure 3.1(b).
At any point in the beam, these complementary shear stresses are equal in magnitude.
Consider an element of the beam of length dx subjected to non-uniform bending. Because of the
bending moments and shear forces, the element is subjected to normal and shear stresses on both
cross-sectional faces. Only the normal stresses are needed in the derivation of the shear stresses
by considering horizontal equilibrium.
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EMT 2407: 3.2 Deriving the Shear Formula 35
Figure 3.1:
Figure 3.2:
The normal bending stresses from the flexural formula are:
section mn : σ1 = My
I (3.1)
section m1n1 : σ1 = (M + dM )y
I (3.2)
We now isolate a subelement mm1 pp1, a distance y1 from the neutral surface as shown in figure
3.2(b).
The top surface is free from shear stresses. Its bottom surface, a distance y1 from the neutral
surface is acted upon by a shear stress τ .
Consider the cross-section of the beam at m1 and take an elemental area, dA, a distance y from
the neutral axis. The force acting on the element dA is given by
F = σdA (3.3)
The forces acting on the section of the beam are as shown in figure 3.3(b). From equation 3.3,
F 1 =
σ1dA (3.4)
F 2 =
σ2dA (3.5)
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EMT 2407: 3.2 Deriving the Shear Formula 36
Figure 3.3:
Considering equilibrium of forces in the horizontal direction,
F 3 = F 2 − F 1 (3.6)
=
M + dM
I ydA −
M
I ydA (3.7)
=
M
I ydA +
dM
I ydA −
M
I dA (3.8)
=
dM
I ydA (3.9)
At any given cross-section, dM and I are constants therefore,
F 3 =
dM
I
ydA (3.10)
Force F 3, is also given by:
F 3 = τbdx (3.11)
where bdx = area of bottom face of element.
⇒ τbdx = dM
dx
ydA
τ = dM
dx
1
Ib
ydA (3.12)
From the relationship between B.M. and S.F., the term
dM
dx = V (shear force).Equation 3.12 becomes:
τ = V
Ib
ydA (3.13)
ydA is the first moment of area above the level at which the shear stress τ is being evaluated
and is denoted by Q. Therefore the shear stress becomes:
τ = V Q
Ib (3.14)
Equation 3.14 is known as the shear formula. For a particular cross-section, the shear force, V ,
moment of inertia I and width b are constant. However, the first moment of area Q is not andhence the shear stress τ varies with the distance y1 form the neutral axis.
The sign convention for V and Q are ignored and the terms in the shear formula are treated as
positive quantities. We determine the direction of shear stress by inspection.
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EMT 2407: 3.2 Deriving the Shear Formula 37
3.2.2 Distribution of Shear Stresses in a Rectangular Beam
Consider the cross-section of a rectangular beam as shown in figure 6.10
Figure 3.4:
The first moment of area of the shaded area, Q is given by:
Q = Ay =
h2 − y1
b h2 − y12 + y1
(3.15)
= b
2
h
2 − y1
h
2 − y1 + 2y1
(3.16)
= b
2
h
2 − y1
h
2 + y1
(3.17)
= b
2
h2
4 − h
2y1 +
h
2y1 − y21
(3.18)
= b
2
h2
4 − y2
1
(3.19)
Substituting the expression for Q into the shear formula;
τ = V
2I
h2
4 − y2
1
(3.20)
Shear stresses in a rectangular beam vary quadratically with distance y1 from the N.A.
at y1 = ±h2
, τ = 0
τ max occurs at y1 = 0 (neutral axis)
I = bh3
12 (3.21)
τ max = V h2
8I = V h2
8 · 12
bh3 = 3V
2bh (3.22)
= 3V
2A (3.23)
The distribution of the shear stress along the height is shown in figure 6.10(b).
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EMT 2407: 3.2 Deriving the Shear Formula 38
3.2.3 Shear Stress Distribution in Beams with Flanges
Figure 3.5:
Consider the area between e − f and the bottom edge of the cross-section shown in figure 3.5. It
consists of two rectangle:
1st rectangle - flange
Af = bh
2 − h1
2
(3.24)2nd rectangle - part of the web
Aw = th1
2 − y1
(3.25)
Web: 1st moment of area Q of this area is given by:
Q = bh
2 − h1
2
h1
2 +
h2 − h1
2
2
+ th1
2 − y1
y1 +
h12 − y1
2
(3.26)
Simplifying:
Q = b
8h2
−h21+
t
8h2
1
−4y1 (3.27)
Substituting Q in the shear formula,
τ = V
8It
b(h2 − h2
1) + t(h21 − 4y21)
(3.28)
at y1 = 0, τ max = V
8It
b(h2 − h2
1) + th21
(3.29)
at y1 = ±h1
2 , τ min =
V b
8It
h2 − h2
1
(3.30)
Flange:
Q = bh
2 − y1
y1 +
h2
−y1
2
(3.31)
= b
8
h2 − 4y2
1
(3.32)
τ = V
8I
h2 − 4y2
1
(3.33)
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EMT 2407: 3.2 Deriving the Shear Formula 39
At the junction where the flange meets the web,
at y1 = ±h1
2 , τ =
V
8I h2 − h2
1 (3.34)
or τ = tb
τ min (3.35)
Example 3.2.1. An I section beam shown in figure 3.6 carries a shear force of 100KN . Sketch
the shear stress distribution across the section.
Figure 3.6:
Solution
I = bh3
12 − (b − t)h3
1
12 (3.36)
= 200 × 3403
12 − (200 − 10)3003
12 (3.37)
= 2.276 × 10−4 (3.38)
Flange:
τ = V
8I
h2 − 4y21
(3.39)
at surface y1 = h
2 τ = 0 (3.40)
at y1 = 0.3 τ = 100 × 103
8 × 2.276 × 10−4
0.342 − 0.32
(3.41)
= 1.406M P a (3.42)
Web:
At the junction with the flange, the shear stress suddenly increases from 1.406M P a to 20010
1.406 =
28.12MP a
τ = V
8It
b(h2
− h21) + t(h
21 − 4y
21)
(3.43)
τ max = 100 × 103
8 × 2.276 × 10−4 × 0.01
0.2(0.342 − 0.32) + 0.01 × 0.32
(3.44)
= 33.062MP a (3.45)
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EMT 2407: 3.3 Shear Deflection 40
3.3 Shear Deflection
Consider a section of a beam of length dx and a rectangular elemental strip of height dy, a distance
y from the neutral axis shown in Figure 3.7.
Figure 3.7: Portion of a beam subjected to non-uniform bending
The shear stress at a distance y from the neutral axis is given by:
τ = V Q
Ib (3.46)
where:
V shear force
Q First moment of area of the plane area above the point where τ is being evaluatedI Second moment of area of the cross-section
b Thickness of the section
For a rectangular section,
Q = b
2
h2
4 − y2
I =
bh3
12∴ τ =
6V
bh3
h2
4 − y2
The strain energy in the strip of height dy is given by:
strain energy = τ 2
2Gb · dx · dy
Substituting the expression for τ ,
strain energy = 1
2Gdx
36V
b2h6h4
16 −h2y2 + y4bdy
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EMT 2407: 3.3 Shear Deflection 41
The total strain energy for a portion of the beam of length dx is given by:
dU s = 18V 2dx
bh6G h2
−h
2
h4
16 − h2y2 + y4
dy
= 18V 2dx
bh6G
h4
16y − h2y3
3 +
y5
5
h2
−h2
= 18V 2dx
bh6G
h5
32 − h5
48 +
h5
160
−
− h5
32 +
h5
48 − h5
160
=
18V 2dx
bh6G · h5
30 =
3
5
V 2dx
bhG (3.47)
The strain energy stored in the whole beam can be obtained by integrating equation 3.47 with
respect to x.
Example
Consider a cantilever beam with point load at the free end shown in Figure 8(a).
(a) Cantilever beam (b) section of the beam, a distance x fromfree end
Figure 3.8: Cantilever with point load at free end
From the free body diagram of a section of the beam, Figure 8(b),
F v = 0 ⇒ P − V = 0
∴ P = V M x = 0 ⇒ M + P x = 0
∴ M = −P x
The strain energy for the whole beam of length L is:
U s = 3P 2
5bhG
L0
dx = 3P 2L
5bhG (3.48)
Equating the strain energy to the work done by P in deflecting the beam,1
2P vs =
3P 2L
5bhG (3.49)
⇒ vs = 6P L
5bhG (3.50)
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EMT 2407: 3.3 Shear Deflection 42
where vs is the deflection due to shear.
The deflection due to bending can be obtained from the differential equation of the deflection
curve,
EI d2vbdx2
= −M = P x
EI dvbdx
= P x2
2 + C 1
EI vb = P x3
6 + C 1x + C 2
Applying the boundary conditions,
at x = L, dvb
dx
= 0
⇒ C 1 =
−
P L2
2at x = L, vb = 0 ⇒ C 2 =
P L3
3
∴ vb = P x3
6EI − P L2
2EI x +
P L3
3EI
At the free end, x = 0 and the deflection due to bending is:
vb = P L3
3EI (3.51)
The total deflection due to bending and shear is:
v = vb + vs (3.52)
= P L3
3EI +
6P L
5bhG but I =
bh3
12 (3.53)
∴ v = 4P L3
Eh3b
1 +
3E
10G
h
L
2 (3.54)
or v = P L 4
Eh3bL2 +
6
5bhG
(3.55)
The moduli of elasticity in tension and shear (E and G) of a material are related by:
G = E
2(1 + ν ) (3.56)
where ν is Poisson’s ratio. For steel, ν ≈ 0.3 ⇒ G = E 2.6
and 3E 10G
≈ 0.78
For most materials, 0.5 < 3E 10G
< 1 and the contribution of shear to the total deflection depends
onhL
2i.e. deflection is only important for deep beams.
Assignment 1
(a) A cantilever beam of length L carries a uniformly distributed load of intensity q as shown in
Figure 3.9. The beam has a rectangular cross-section with b and h as the breadth and depth
respectively. Show that the deflection at the free end due to shear is given by:
vs = 3qL2
5Gbh
where G is the shear modulus of elasticity.
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EMT 2407: 3.3 Shear Deflection 43
Figure 3.9: Cantilever with uniformly distributed load
(b) If q = 20kN/m, L = 4m, b = 150mm, and h = 200mm, determine the total deflection at the
free end. Take G = 80GPa and E =200GPa
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44
Chapter 4
Bending of Thin Plates
4.1 Simple Concepts of the General Plate Problem
A plate is a three-dimensional structural element, with one of the dimensions (the plate thickness
h or Lz) being small compared to the in-plane dimensions Lx and Ly. The load on the plate is
applied perpendicular to the center plane of the plate (supports lateral loads).
Figure 4.1: Plate
The purpose of plates in engineering is to cover rectangular or circular area and to support
concentrated or distributed loading normal to the plane of the plate.Examples of application of plates:
• manhole covers
• sewage hole covers
• Pressure diaphragm1 e.g in safety or control devices
The analysis of plates involves the determination of shear forces, bending moments, stresses anddeflections.
4.1.1 Assumptions
The following assumptions are made in the derivation of the governing equations:
• The unloaded plate is thin, flat with uniform thickness h
• The material of the plate is homogenous, isotropic and linearly elastic.
• The middle plane of the plate is stress free, i.e. the middle plane is the neutral surface.
1a diaphragm is a sheet of a flexible material anchored at its periphery and most often round in shape. It serveseither as a barrier between two chambers, moving slightly up into one chamber or down into the other dependingon differences in pressure.
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EMT 2407: 4.2 Rectangular Plate 45
• The normal stresses in the direction transverse to the plate can be neglected i.e. σz = 0
• Points that lie on a line perpendicular to the center plane of the plate remain on a straight
line perpendicular to the center plane after deformation.
• The deflection w of the plate is small compared to the plate thickness. The curvature of the
plate after deformation can then be approximated by the second derivative of the deflection
w.
• Loads are applied in a direction perpendicular to the center plane of the plate.
4.2 Rectangular Plate
Consider an element of the material shown in Figure 4.2,
(a) Plate (b) Element of the plate
Figure 4.2: Plate under pure bending
M y and M x are the bending moments per unit length and are assumed to be positive as drawn
and act about the middle of the plate.
Above the neutral surface, the material is in a state of biaxial compression and below it in biaxial
tension. The curvature of the mid-plane are denoted by:1
Rx→ (x-z plane)
1
Ry
→ (y-z plane)
At a depth z below the neutral surface (N.S), the strains in the x and y directions of a lamina
such as abcd are:
εx = z
Rx
and εy = z
Ry
(4.1)
The general stress-strain relationships for plane stress state are:
εx = σx
E − ν
σyE
= z
Rx
εy = σy
E − ν
σxE
= z
Ry(4.2)
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EMT 2407: 4.2 Rectangular Plate 46
Multiplying equation 4.1 by ν , and adding to equation 4.2 gives,
σyE
(1 − ν 2) = z 1
Ry+
ν
Rx∴ σy =
Ez
1 − ν 2
1
Ry+
ν
Rx
(4.3)
and σx = Ez
1 − ν 2
1
Rx+
ν
Ry
(4.4)
Equations 4.3 and 4.4 show that the bending stresses are a function of the plate curvatures and
are proportional to the distance from the neutral surface.
Equating the internal resisting moments to the applied moments,
M xdy = h
2
−h2
σxdydz · z
M ydx =
h2
−h2
σydxdz · z
Substitute σy and σx from equations 4.3 and 4.4 and integrate,
M x = E
1 − ν 2
1
Rx+
ν
Ry
h2
−h2
z 2dz = Eh3
12(1 − ν 2)
1
Rx+
ν
Ry
= D
1
Rx+
ν
Ry
(4.5)
M y = E 1 − ν 2
1Ry
+ ν Rx
h2
−h2
z 2dz = Eh3
12(1 − ν 2)
1Ry
+ ν Rx
= D
1Ry
+ ν Rx
(4.6)
where D = Eh3
12(1 − ν 2) is known as plate constant or flexural rigidity
From equations 4.3, 4.4, 4.5 and 4.6,
σx = Ez
1 − ν 2
1
Ry
+ ν
Rx
=
M xD
Ez
1 − ν 2
= M x
Ez
1 − ν 2 · 12(1
−ν 2)
Eh3 = 12
M xz
h3
(σx)max = 12M x
h2
h3 =
6M xh2
similarly, σy = 12M yz
12
Relating the plate theory to beam theory,
d2w
dx2 = − M
EI = − 1
R for beam
∴
d2w
dx2 = − 1
Rx for plate
d2w
dy2 = − 1
Ryfor plate
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EMT 2407: 4.2 Rectangular Plate 47
where w is the deflection in the z direction. The relationship between the applied moments and
curvatures are:
M x = −Dd2w
dx2 + ν d2w
dy2
M y = −Dd2w
dy2 + ν
d2w
dx2
in matrix form
M xM y
= −D
1 ν ν 1
d2wdx2d2wdy2
(4.7)
To determine the expressions for the deflections, we pre-multiply both sides of equation 4.7 by
− 1D
1 ν ν 1
−1
and obtain:
d2wdx2d2wdy2
=
−1
(1 − ν 2)D
1 −ν −ν 1
M xM y
(4.8)
4.2.1 Special Cases
1. For a square plate with M x = M y, d2wdx2
= d2wdy2
and a state of spherical bending is said to
occur.
2. Total Bending moment M is applied to opposite ends of a rectangular plate of width b as
shown in Figure 4.3,
Figure 4.3: Bending in a single plane
M x = M
b M y = 0
Equation 4.8 reduces to:
d2
wdx2
= −1(1 − ν 2)D
M x = −12(1 − ν 2
)Eh3(1 − ν 2)
· M b
= −M EI
−1
Rx=
−M
EI ⇒ M
E =
I
Rx
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EMT 2407: 4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending 48
Therefore the plate reduces to the beam theory. The other curvature is:
d2w
dy2 =
−1
(1−
ν 2)D(−νM x) =
νM
EI (due to poisson’s effect)
1
Ry=
νM
EI
3. Cylindrical Bending:
If the plate is constrained so that displacement varies in only one direction, a state of
cylindrical bending is said to occur.
Figure 4.4: Cylindrical Bending
Assuming that w varies with x only so that d2w
dy2 = 0, the moment-curvature relation becomes:
M x = −Dd2w
∂x2 =
−Eh3
12(1 − ν 2)
−1
Rx
M y = −νDd2w
∂y2 = νM x
M x = M
b ⇒ M = M xb
∴ M =
Eh3b
12(1 − ν 2)Rx =
EI
(1 − ν 2)Rx
Beam stiffness = EI
Plate stiffness = EI
1 − ν 2
4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending
When loading on the surface of a circular plate is symmetrical about a perpendicular central axis,
the deflection surface is also symmetrical about that axis.
The curvature in the diametral plane r − z is:
1
Rr
= −d2w
dr2
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EMT 2407: 4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending 49
Figure 4.5: Circular Plate
For small values of w, the slope at any point is:
φ = −dw
dr ⇒ dφ
dr = −d2w
dr2
1
Rr
= −d2w
dr2 =
dφ
dr
sin φ = r
Rh
for small angles, sin φ = φ
∴ φ = r
Rh⇒ 1
Rh
= φ
r = −1
r
dw
dr
Consider an element of the plate subjected to bending moments along the edges:
Figure 4.6: Circular Plate
• M r - Bending moment per unit length of arc
• M h - Bending moment per unit length of radius
This element can be analyzed in the same manner as for rectangular plate:
M r = Eh3
12(1 − ν 2)
1
Rr
+ ν
Rh
= −D
d2w
dr2 +
ν
r
dw
dr
M r = Eh3
12(1 − ν 2)
1
Rh
+ ν
Rr
= −D
1
r
dw
dr + ν
d2w
dr2
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EMT 2407: 4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending 50
The relationship between the bending stresses and bending moments are:
σr = Ez
1
−ν 2
1
Rr+
ν
Rh = Ez
1
−ν 2
· 12(1 − ν 2)M rEh2
= 12M rz
h3
σh = Ez 1 − ν 2
1Rh
+ ν Rr
= 12M hz
h3
The maximum stress occurs at z = ±h2
(σr)max = 12M r · h
2
h2 =
6M rh2
(σh)max = 6M h
h2
4.3.1 Relationship between Load, Shear Force and Bending Moment
Consider an element of a plate under the action of a uniformly distributed load P per unit area
and resulting shear forces Q per unit length as shown in Figure 4.7. Due to symmetry, there are
no shear forces on the radial sides of the element.
Figure 4.7: Element of a plate
For vertical equilibrium,
Qrdθ + P rdrdθ −
Q + dQ
dr dr
(r + dr)dθ = 0
Qrdτθ + P rdθ − Qrdθ − dQ
dr dr2dθ − Qdrdθ − dQ
dr rdrdθ = 0
Simplifying and ignoring higher powers of small quantities, we obtain:
Prdr − Qdr − rdQ
dr dr = 0
or dQ
dr +
Q
r = P (4.9)
For moment equilibrium,M o = 0 ⇒
M r +
dM rdr
dr
(r + dr)dθ − M rrdθ − 2M hdr sin dθ
2 + Qrdθdr = 0
M rrdθ + rdM r
dr drdθ + M rdrdθ +
dM rdr
dr2dθ − M rrdθ − M hdrdθ + Qrdrdθ = 0
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EMT 2407: 4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending 51
Simplifying and ignoring higher powers of small quantities, we obtain:
rdM r
dr + M r − M h + Qr = 0 (4.10)
but M h = D
φr
+ ν dφdr
and M r
dφdr
− ν φr
dM r
dr = D
d2φ
dr2 + ν
r dφdr
− φ
r2
= D
d2φ
dr2 +
ν
r
dφ
dr − ν
φ
r2
Substituting for M r, M h and dM r
dr in equation 4.10, we obtain:
d2φ
dr2 +
dφ
dr − φ
r =
−Q
D (4.11)
Equation relates the slope at any radius to the shear force per unit length.
Recall 1
Rr= −d2w
dr2 =
dφ
dr
⇒ d2φ
dr2 = −d3w
dr3
and φ = −dw
dr ⇒ φ
r = −1
r
dw
dr
Substituting these equations in equation 4.11,
− d3w
dr3 − d2w
dr2 +
1
r
dw
dr = −Q
D
or d3w
dr3 +
d2w
dr2 − 1
r
dw
dr =
Q
D (4.12)
Equation 4.12 expresses the variation of the deflection w with the radius r.
Equation 4.11 can be written in a more convenient form as:
d
dr1
r
d
dr (rφ)
=
d2φ
dr2 +
dφ
dr − φ
r = −Q
D (4.13)
similarly d
dr
1
r
d
dr
r
dw
dr
=
Q
D (4.14)
From equation 4.9, the shear force Q is a function of the applied load P . Multiplying equation
4.9 by rdr gives:
rdQ + Qdr = Prdr
or d(Qr) = Prdr
Integrating, Q = 1r
Prdr
∴ Q = 1
r
Prdr
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EMT 2407: 4.3 Circular Plate Under Symmetrical Bending (Axisymmetrical Bending 52
I Plate subjected to uniform pressure P per unit area
Total force acting at radius r = P · πr2
Shear force per unit length of arc,
Q = π2r2P
2πr =
P r
2
∴d
dr
1
r
d
dr
r
dw
dr
P r
2DIntegrating with respect to r,
1
r
d
dr
r
dw
dr
=
P r2
4D + C 1
d
drrdw
dr = P r3
4D + C 1r
rdw
dr =
P r4
16D
C 1r2
2 + C 2
dw
dr =
P r3
16D
C 1r
2 +
C 2r
w = P r4
64D +
C 1r2
4 + C 2 ln r + C 3
The constants C 1, C 2 and C 3 can be evaluated from the boundary conditions.
Plate with Fixed Edges
Bondary conditions are:
i. at r = R, w = 0
ii. at r = 0, dwdr
= 0
iii. at r = R, dwdr
= 0
From ii, dwdr
→ ∞, since this is not practical, we set C 2 = 0.
∴
dw
dr =
P r3
16D +
C 1r
2
0 = P R3
16D + C 1R
2 ⇒ C 1 = −P R2
8D
w = P r4
64D − P R2
32Dr2 + C 3
0 = P R4
64D − P R4
32D + C 3 ⇒ C 3 =
P R4
64D
w = P r4
64D − P R2
32Dr2 +
P R4
64D
= P
64D[r4 − 2R2r2 + R4] =
P
64D[r2 − R2]2
The stresses are given by:
σr = 12M r
h3 , M r = −D
d2w
dr2 +
ν
r
dw
dr
σr =
12M hh3
, M h = −D1
r
dw
dr + ν
d2w
dr2
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EMT 2407: 4.4 Tutorial 3 53
w = P
64D[r4 − 2R2r2 + R4] =
P
64D[r2 − R2]2
dwdr
= P 64D
[4r3 − 4R2r]
d2w
dr2 =
P
64D[12r2 − 4R2]
M r = −Dd2w
dr2 +
ν
r
dw
dr
=
P
16
R2(1 + ν ) − r2(3 + ν )
similarly, M h =
P
16
R2(1 + ν ) − r2(1 + 3ν )
at r = 0,
M r = M h =
P R2
16 (1 + ν )
σr = σh = 3
8
P R2
h2 (1 + ν )
at r = R,
M r = −P R2
8
M h = −νP R2
8
σr = −
3
4
P R2
h2
σh = −3
4ν
P R2
h2
4.4 Tutorial 3
1. Write down the expressions for the deflection and bending moments at the center of a circular
plate of radius R loaded and supported as shown in Fig. 4.8. q = force per unit area. Take
ν = 0.3
Figure 4.8: Question 1
[Ans:
w = qR4
64DM r = M h = 0.08125qR2
]
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EMT 2407: 4.4 Tutorial 3 54
2. A cast iron disk valve is a flat plate 300mm in diameter and is simply supported. The plate
is subjected to uniform pressure supplied by a head of 60m of water. Find the thickness of
the disk if the allowable stress is 14MN/m2. Determine the maximum deflection of the plate
at this pressure. For cast iron, E =100GN/m2, ν = 0.2
[Ans: h=33.7mm, wmax = 0.061m]
3. A circular steel plate whose diameter is 2.54m and whose thickness is 12.7mm, is fixed at
its edges and is subjected to a uniformly distributed pressure P. The tensile yield stress of
the steel is 207MN/m2. Determine the maximum pressure that produces initial yielding.
Determine the maximum deflection at this pressure. Take E =200GPa, ν = 0.29.
[P max = 16.78kPa, wmax = 75mm]
4. A circular opening in the flat end of a nuclear reactor pressure vessel is 254mm in diameter.A circular steel plate 2.54mm thick with a tensile yield stress of 241MN/m2 is used as a
cover for the opening. When the cover is inserted in the opening, its edges are clamped
securely. Determine the maximum internal pressure to which the vessel may be subjected
if a factor of safety of 3 for the cover is to be used. Assume that the maximum pressure in
the vessel is limited by the strength of the cover. Determine the maximum deflection of the
plate at this pressure. Take E =200GPa, ν = 0.29.
[Hint: Yielding first occurs at the fixed edge. P max = 42.84kPa, wmax = 0.584mm]
5. A mild steel plate (E =200GPa, ν =0.29, σy=315MN/m2) has a thickness h =10mm and
covers a circular opening having a diameter of 200mm. The plate is fixed at the edges and
is subjected to a uniform pressure P .
(a) Determine the magnitude of the yield pressure P y and the deflection wmax at the center
of the plate when this pressure is applied. [P y =4.2MPa, wmax=0.361mm]
(b) Determine the working pressure based on a factor of safety of f = 2.0. [P w=2.1]
6. A circular plate radius R and thickness h, having its edge clamped all round is loaded at
the centre by a concentrated load P. Find equations for
(a) the deflection
(b) radial stress
(c) circumferential stress.
[ans:
w = P
16πD
2r2 ln
r
R + R2 − r2
σr = 3P z
πh
(1 + ν ) ln R
r − 1
σr = 3P z
πh
(1 + ν ) ln
R
r − ν
]
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EMT 2407: 4.4 Tutorial 3 55
7. A pressure control system includes a thin steel disk which is to close an electrical circuit by
deflecting 1mm at the centre when the pressure attains a value of 3 MPa. Calculate the
required disk thickness if it has a radius of 0.03 m and is built-in at the edge. ν = 0.3 E =
200 GPa
[h =1.275mm]
8. A circular plate is simply supported round the outer boundary r = a. If the plate carries a
point load P at the centre, derive the deflected shape of the plate and expressions for the
radial and circumferential bending moments.
Ans:
w = P
16πD2r2 ln
r
a +
3 + ν
1 + ν (a2
−r2)
M r = P
4π(1 + ν ) ln
a
r
M h = P
4π
(1 + ν ) ln
a
r + 1 − ν
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56
Chapter 5
Bending of Curved Beams with Plane Loading
5.1 Introduction
In theory of bending, the bending equation,
M
I =
E
R =
σ
y
was derived by assuming the beam to be initially straight (besides other fundamental assumptions).
However machine members and structures subjected to bending are not always straight as in the
case of crane hooks, chain links, bridge members, building trusses eg in warehouses.
The problem of curved beams can be classified into two:
1. Initially curved beams where the depth of the cross-section is small in relation to the initial
radius of curvature of the beam (Rd
> 10). Such beams are called slender beams.
2. Beams where the depth of the cross-section is significantly large in relation to the initial
radius of curvature of the beam (Rd
< 10) - Deeply curved beams.
5.2 Stresses and Strains in Curved Beams - Winkler Bach Analysis
Consider a section of a curved beam OAB as shown in Figure 5.1.
(a) Curved beam before bending (b) Curved beam after bending
Figure 5.1: Nomenclature of a curved beam.
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EMT 2407: 5.2 Stresses and Strains in Curved Beams - Winkler Bach Analysis 57
5.2.1 Assumptions
i. Plane sections remain plane during bending.
ii. Material obeys Hooke’s law (elastic limit is not exceeded)
iii. Radial strain is negligible ⇒ y = y
iv. The fibres are free to expand or contract without any constraining effect from the adjacent
fibre.
For a fibre AB on the neutral axis,
AB = A
B
, R1θ = R2φ⇒ φ =
R1
R2θ (5.1)
Strain on fibre C’D’, a distance y from the neutral axis is given by:
εC D = C D − CD
CD =
(R2 + y)φ − (R1 + y)θ
(R1 + y)θ
= R2φ + yφ − R1θ − yθ
(R1 + y)θ
but R1θ = R2φ
∴ εC D = y(φ − θ)
(R1 + y)θ (5.2)
Substituting the expression for φ, equation 5.1 in equation 5.2, we obtain:
εC D =yR1
R2
θ − θ
(R1 + y)θ =
y(R1 − R2)
R2(R1 + y) (5.3)
5.2.2 Case 1 - Slender Beam
y<<R1 ⇒ y can be neglected in relation to R1, i.e., R1 + y ≈ R1. The strain is then given by:
εC D = y(R1 − R2)
R2R1= y
1
R2− 1
R1
(5.4)
From equation 5.4, we can deduce that:
1. The strain is directly proportional to the distance y from the neutral axis.
2. The stress and strain distribution is linear.
3. The neutral axis and the centroidal axis coincide.1
1Transverse forces across the section, in absence of direct forces = 0 ∴
σdA = 0 or
y 1
R2
− 1
R1
EdA = 0 ⇒
E 1
R2
− 1
R1
ydA = 0 ⇒
ydA = 0
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EMT 2407: 5.3 Position of the Neutral Axis for a Deeply Curved Beam 58
From Hooke’s Law, σ = Eε
∴ σ = Ey
1
R2− 1
R1⇒ σ
y = E
1
R2− 1
R1and from the simple theory of bending,
M
I =
σ
y =
E
R
We can use a modified bending theory to determine the stress distribution, i.e:
M
I =
σ
y = E
1
R2− 1
R1
(5.5)
If the beam was initially straight, R1 → ∞ and equation 5.5 reduces to the simple theory of
bending.
5.2.3 Case 2: Deeply Curved Beams
In this case, the distance y is not negligible when compared to R1.
The strain distribution is no longer directly proportional to y i.e, the stress and strain distribution
are non-linear.
The neutral axis no longer passes through the centroid.
(a) slender (linear) (b) deeply curved (non-linear-hyperbolic)
Figure 5.2: Strain distribution for curved beams.
5.3 Position of the Neutral Axis for a Deeply Curved Beam
For equilibrium of transverse forces in absence of any applied direct load, the net force = 0.
σdA = 0
but σ = Eε = Ey(R1 − R2)
R2(R1 + y) (Hooke’s law) (5.6)
∴
σdA =
E (R1 − R2)
R2
y
R1 + ydA = 0
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EMT 2407: 5.4 Bending Moment on the Cross-section 59
Since E=0 and R1−R2
R2
= 0 only when R2 = R1, i.e, no bending,
⇒ y
R1 + y
dA = 0 (5.7)
Which implies that the neutral axis does not coincide with the centroidal axis.
5.4 Bending Moment on the Cross-section
Consider the cross-section shown in Figure 5.3
Figure 5.3: Beam section
The bending moment on the cross-section is given by:
M =
σdA · y =
E (R1 − R2)
R2
y2
R1 + ydA
y2
R1 + ydA =
y(y + R1 − R1)
R1 + y dA =
y[(R1 + y) − R1]
R1 + y dA
=
ydA − R1
y
R1 + ydA (5.8)
From equation 5.7,
yR1+ydA = 0,
∴ M = E (R1 − R2)
R2
ydA (5.9)
From Figure 5.4, y = yc + e, where e is the eccentricity, therefore, ydA =
ycdA +
edA, but
ycdA = 0
∴ ydA = Ae
M = E (R1 − R2)
R2· Ae
rearranging, M
Ae =
E (R1 − R2)
R2(5.10)
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EMT 2407: 5.4 Bending Moment on the Cross-section 60
Figure 5.4: Section of a curved beam
From equation 5.6,
E (R1 − R2
R2=
σ
y(R1 + y)
∴ σ = My
Ae(R1 + y) (5.11)
which is the general bending equation for a deeply curved beam.
5.4.1 Terminologies used with Curved Beams
y is measured from the neutral axis and is considered positive when measured away from the
centre of curvature.
M is considered positive when it tends to decrease R (increase curvature).
From equation 5.11,
σi = −Mh1
Aea ; σo =
M h2
Aec (5.12)
which implies that:
• The stress is always greater at the inside radius.
• The neutral axis moves towards the centre of curvature.2
5.4.2 To determine R1
Rectangular section
We need to determine R1, since Rc can be easily determined. Recall that for no applied load,
yR1+y
dA = 0. Consider an element, a distance r from the center of curvature:
y = r − R1 ⇒ R1 + y = r dy = drdA = bdy = bdr
2
y2
R1+ydA = Ae, for Ae to be positive, it implies that e should be positive, i.e measured from the N.A away
from the centre of curvature.
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EMT 2407: 5.4 Bending Moment on the Cross-section 61
Figure 5.5: Rectangular section y
R1 + ydA =
r − R1
r dA =
dA −
R1
r dA = 0 (5.13)
= A − R1
dA
r = 0 (5.14)
⇒ A = R1
dA
r (5.15)
∴ R1 = A
dAr
(5.16)
For a rectangular beam,3 dA = bdr, and A = bd
∴
dA
r = b
ca
dr
r = b ln
c
a
⇒ R1 = bh
b ln ca
= h
ln ca
e = Rc − R1
I-section
The section can be treated as a composite section consisting of three rectangles.
For the top flange, dA1
r =
a+h1a
b1dr
r = b1 ln
a + h1
a
For the web, dA2
r =
a+h1+h2a+h1
b2dr
r = b2 ln
a + h1 + h2
a + h1
For the bottom flange, dA3
r =
ca+h1+h2
b3dr
r = b3 ln
c
a + h1 + h2
3To design a curved beam such that maximum stress=minimum stress, Mh2Aec
= Mh1Aea
⇒ h1h2
= ac
and R1 = 2aca+c
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EMT 2407: 5.4 Bending Moment on the Cross-section 62
Figure 5.6: I - section
For the whole section, dA
r = b1 ln
a + h1
a + b2 ln
a + h1 + h2
a + h1+ b3 ln
c
a + h1 + h2
and,
R1 = b1h1 + b2h2 + b3h3
b1 ln a+h1a
+ b2 ln a+h1+h2a+h1
+ b3 ln ca+h1+h2
e = Rc − R1
Example 5.4.1. A curved rectangular bar has a mean radius of curvature Rc = 100mm and a
cross-section of width b = 50mm and depth h = 25mm as shown in Figure 5.7 . Determine the
largest tensile and compressive stresses given that the bending moment in the bar is M = 500Nm
Figure 5.7: Example
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EMT 2407: 5.4 Bending Moment on the Cross-section 63
Solution
a = Rc − h2
= 100 − 12.5 = 87.5mm
c = Rc + h
2 = 100 + 12.5 = 112.5mm
R1 = h
ln ca
= 25
ln 112.587.5
= 99.477mm
e = Rc − R1 = 100 − 99.477 = 0.523mm
Using the bending equation for curved beams,
σ =
My
Ae(R1 + y)
Maximum tensile stress
σmax = Mh2
Ae(R1 + h2)
h2 = c − R1 = 112.5 − 99.477 = 13.023 A = bh = 25 × 50 = 1250 × 10−6m2
σmax = 500 × 13.023 × 10−3
1250 × 10−6 × 0.523 × 10−3 × (99.477 + 13.023) × 10−3
= 88.54 Mpa
Maximum compressive stress;
σmax = −Mh1
Ae(R1 − h1)
h1 = R1 − a = 99.477 − 87.5 = 11.977
σmax = −500 × 11.977 × 10−3
1250 × 10−6 × 0.523 × 10−3 × (99.477 − 11.977) × 10−3
= −104.688 Mpa
Example 5.4.2. Compare the percentage error 4 computed in calculating the maximum tensile and
compressive stresses if the initial curvature of the beam was neglected.
solution
For a straight beam,
σmax,min = ±M h2
I = ±Mh
2 · 12
bh3 =
6M
bh2
= ± 6 × 500
50 × 10−3 × (25 × 10−3)2 = ±96 MP a
% error, σt = 96
−88.54
88.54 × 100 = 8.4%
% error, σc = 104.688 − 96
104.688 × 100 = 8.3%
4Normally in design, % error between predicted value and measured value should not exceed 5%
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EMT 2407: 5.5 Combined Direct and Bending Stresses 64
5.5 Combined Direct and Bending Stresses
Most structural members are usually subjected to combined bending and direct stresses5. In such
cases, the total stress is computed by using the principle of superposition. For a curved beam,
σ = M y
Ae(R1 + y) ± P
A, M xx = P × R
Figure 5.8: Example of a curved beam subjected to combined loading
Example 5.5.1. A punch press is loaded as shown in Figure 5.9 .
(i) Determine the location of
(a) the centroidal axis
(b) the neutral axis
(ii) Determine the limiting value of P if the resultant stress at point A and B are 44N/mm 2 and
-31N/mm 2 respectively.
Solution
Table 5.1:Part Ai xi Aixi
1 300 × 70 = 21000 35 7350002 380 × 200 = 76000 70+190=260 19.76×106
3 340 × 100 = −34000 70+170=240 -8.16×106 63000 12.355×106
x̄ =
Aixi
Ai
= 12.355 × 106
63000 = 195.79mm
5
Bending moment may be induced in a number of ways:• due to externally applied moment
• due to longitudinal eccentric loading
• due to the member buckling
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EMT 2407: 5.5 Combined Direct and Bending Stresses 65
(a) Press (b) section AB
Figure 5.9: Punch press
The distance from the centre of curvature to the centroidal axis is given by:
Rc = a + x̄ = 195.79 + 300 = 495.79mm
The distance from the centre of curvature to the neutral axis is given by:
R = A
dArDivide the section into three rectangular areas.
Figure 5.10:
dA
r 1 =
370300
300dr
r =
300ln r
370300
= 62.916
dA
r
2 = 750
370
200dr
r
= 200ln r750
370 = 141.314
dA
r 3 = −
710370
100dr
r = −100ln r
370300
= −62.916 dA
r = 62.916 + 141.314 − 65.176 = 139.054
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EMT 2407: 5.6 Tutorial 4 66
R = A
dAr
= 63000
139.054 = 453.061mm
e = Rc − R = 495.061 − 453.061 = 42.729mm
h1 = R − a = 453.061 − 300 = 153.061mm
h2 = c − R = 750 − 453.061 = 296.939mm
The bending moment is given by
M = P (800 + Rc) = −P (800 + 495.79) = −1295.79P N mm
and is taken to be -ve since it tends to reduce curvature.
The total direct stress at any point is determined by superposition of the direct stress and bending
stress using the equation:
σmax,min = P
A ± My
Ae(R + y)
At point A,
σmax = P
A +
Mh1
Aea
44 = P
63000 + 1295.79
×153.061P
63000 × 42.729 × 300 = 2.615 × 10−
4P
⇒ P = 168.28kN
At point B,
σmin = P
A − Mh2
Aec
44 = P
63000 − 1295.79 × 153.061P
63000 × 42.729 × 750 = −1.747 × 10−4P
⇒ P = 177.44kN
The limiting value of P is 168.28kN.
5.6 Tutorial 4
1. Figure 5.11 shows a bracket in the form of a curved beam having a T cross-section. Determine
the magnitude of the maximum tensile stress and maximum compressive stress along section
AB if the bracket is subjected to a load of 150kN. All dimensions are in mm.
[Ans: σA = 98.75 MPa (tensile), σB = 102.29MPa (compressive)]
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EMT 2407: 5.6 Tutorial 4 67
Figure 5.11: Curved beam
2. A steel bar of diameter d=38 mm is bent into a curve of mean radius, R c=31.7mm. If thebar is subjected to a bending moment of 4.6Nm tending to increase curvature acts on the
bar, find the intensities of maximum tensile and compressive stresses.
Take R1 = 12
Rc +
R2c −
d2
2[Ans: 0.56MN/m2 (tensile), 1.6MN/m2 (compressive)]
3. Figure 5.12 shows a C-frame subjected to a load P. If the maximum stresses in compression
and tension are 63MPa and 120MPa respectively, determine the limiting value of P. All
dimensions are in mm.
Figure 5.12: Curved beam
[Ans: P = 121 N]
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EMT 2407: 5.6 Tutorial 4 68
4. Figure 5.13 shows a crane hook lifting a load of P = 150 kN. Determine the maximum
compressive and tensile stress in the critical section AB on the hook.
Figure 5.13: Crane hook
[Ans: σA= 42.5 MPa (compressive), σB=82.39MPa]
5. If the maximum stress both in tension and compression is not to exceed 120 MPa, determine
the maximum load that the hook can carry.
[Ans: 218kN]
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69
Chapter 6
Bending due to Thermal Stresses
6.1 Introduction
When beams are subjected to temperature gradient, they experience thermal strain as shown in
Figure 6.1.
Figure 6.1: Metal strip subjected to a temperature rise
δ T = α(T )L ⇒ ε = δ T
L = αT (6.1)
where alpha = coefficient of thermal expansion. Different materials have different coefficient of
expansion.
Figure 6.2: Un-bonded strips with α1 > α2
By fusing two strips of different materials together, a bimetallic strip is formed which bends undertemperature change due to the different expansion rates of the metals. The bimetallic strip is
used as a sensor in thermostats.
A thermostat is a device for regulating the temperature of a system so that the systems temper-
ature is maintained near a desired set point.
Table 6.1: Examples of αMaterial α/oC
Aluminium 2.4×10−
5
Iron (steel) 1.2×10−5
Copper 1.7×10−5
Brass 1.9×10−5
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EMT 2407: 6.2 Analysis 70
Figure 6.3: bonded strips with α1 > α2
Common sensors used in thermostats:
• Bimetallic strip
• Electronic thermistors (change in resistance with temp change)
• Electrical thermocouples.
Applications of thermostats:
• Clothing iron thermostat
• Toaster oven
• Air-conditioning unit
• Wall heater thermostat
• circuit breakers
• Cloth drier
6.2 Analysis
Consider a bimetallic strip of length L, with α1 > α2 shown in Figure 6.4.
Figure 6.4: Bimetallic strip
When the strip is heated through a T o C, it will bend since α1 > α2 and both strips will deform
together introducing compressive stress in strip 1 and tensile stress in strip 2.
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EMT 2407: 6.2 Analysis 71
Figure 6.5: Bimetallic strip
Metal 2 will exert a compressive force on 1 reducing its free expansion of α1LT and metal 1 will
exert tensile a force on metal 2 and further increasing its free expansion of α2LT
For longitudinal equilibrium,
P 1 − P 2 = 0 ⇒ P 1 = P 2 = P (6.2)
Since R1 >> h, R1 = R2 = R.
From theory of bending,M
I
= E
R ⇒M =
EI
R
(6.3)
At any section, P 1 and P 2 constitute a couple M = P h
But,
M = M 1 + M 2 (6.4)
= E 1I 1
R +
E 2I 2R
(6.5)
∴ P h = E 1I 1 + E 2I 2
R =
bh3
12 (E 1 + E 2) (6.6)
P = bh2
12R(E 1 + E 2) (6.7)
Direct strain due to bending on the outer fibres of strip 1 is:
ε1 = h
2R ∵
M
I =
σ
y =
E
R (6.8)
or σ
E = ε =
y
R =
h
2 ∵ y =
h
2R at outer fibres (6.9)
Direct strain due to bending on the outer fibres of strip 2 is:
ε2 = −h2R
(6.10)
The force P 1 = P introduces a compressive strain in strip 1 while P 2 = P introduces a tensile
strain in strip 2.
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EMT 2407: 6.2 Analysis 72
The resultant strain due to the direct strain and thermal strain is equal to the bending strain.
Resultant strain in strip 1 ε1 = α1T − P
bhE 1(6.11)
Resultant strain in strip 2 ε1 = α2T + P
bhE 1(6.12)
Subtracting equation 6.12 from 6.11,
ε1 − ε2 = h
2R − −h
2R
=
α1T − P
bhE 1
−
α2T + P
bhE 2
(6.13)
h
R = (α1 − α2)T − P
bh
1
E 1+
1
E 2
(6.14)
h
R
= (α1
−α2)T
− P
bhE 1 + E 2
E 1E 2 (6.15)
substituting the value of P from equation 6.7 in equation 6.15, we get:
h
R = (α1 − α2)T − bh2
12R(E 1 + E 2) · 1
bh
E 1 + E 2E 1E 2
(6.16)
= (α1 − α2)T − h
12R
E 21 + 2E 1E 2 + E 22E 1E 2
(6.17)
h
R
1 +
E 21 + 2E 1E 2 + E 22E 1E 2
= (α1 − α2)T (6.18)
h
R12E 1E 2 + E 21 + 2E 1E 2 + E 22
E 1E 2
= (α1 − α2)T (6.19)
or h
R =
12E 1E 2E 1 + E 2 + 14E 1E 2
(α1 − α2)T (6.20)
For most thermostats, E 1 = E 2 = E and h = d2
where d is the total depth of the bimetallic strip.
∴
1
R =
12E 2
16E 2(α1 − α2)T
h =
3
4h(α1 − α2)T (6.21)
or 1
R =
3
2d(α1 − α2)T (6.22)
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EMT 2407: 6.3 Stresses in Bimetallic Strips 73
6.3 Stresses in Bimetallic Strips
6.3.1 Bending Stress
The bending moment in each strip is given by:
M 1 = M 2 = M = EI
R = EI · 3
2d(α1 − α2)T (6.23)
whereI = b
12
d
2
3=
bd3
96 (6.24)
Maximum bending stress occurs at y = d4
,
σmax = M d
4
I = EI
·
3
2d(α1
−α2)T
·
d
4 ·
1
I (6.25)
= ±3
8E (α1 − α2)T (6.26)
Figure 6.6: Bending stress Distribution
6.3.2 Direct Stress
From equation 6.7,
P = bh2
12R(E 1 + E 2) =
bd2E
24R (6.27)
= bd2E
24 · 3
2d(α1 − α2)T = Ebd
16 (α1 − α2)T (6.28)
σd = ∓P
A = ∓Ebd
16 (α1 − α2)T · 1
bd2
= ∓1
8E (α1 − α2)T (6.29)
6.3.3 Combined Stress
The total normal stress is obtained by superposition of the direct and bending stresses.
At the upper surface,
σT = 38
E (α1 − α2)T − 18
E (α1 − α2)T (6.30)
= E
4 (α1 − α2)T (6.31)
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EMT 2407: 6.4 Types of Thermostats Bimetallic Strips 74
At the interphase,
σT = ±3
8E (α1 − α2)T ± 1
8E (α1 − α2)T (6.32)
= ±E 2
(α1 − α2)T (6.33)
At the lower surface,
σT = −3
8E (α1 − α2)T +
1
8E (α1 − α2)T (6.34)
= −E
4 (α1 − α2)T (6.35)
Figure 6.7: Normal stress Distribution
6.4 Types of Thermostats Bimetallic Strips
6.4.1 Simply Supported Beam Type
Figure 6.8: Simply supported strip
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EMT 2407: 6.4 Types of Thermostats Bimetallic Strips 75
From Figure 6.8,
R2 = (R − δ )2 + L
2 2
(6.36)
= R2 − 2Rδ + δ 2 + L2
4 (6.37)
⇒ 2Rδ − δ 2 = L2
4 (6.38)
since the deflection is small, δ 2 → 0 (6.39)
∴ 2Rδ = L2
4 (6.40)
or δ = L2
4 · 1
2R (6.41)
but 1
R = 3
2d (α1 − α2)T (6.42)
∴ δ = L2
4 · 3
4d(α1 − α2)T (6.43)
The term 34
(α1 − α2) = K s is known as the strip deflection constant and is used by manufacturers
in classifying bimetallic strips.
δ = L2
4dK sT (6.44)
Equation 6.44 is known as the free deflection of the strip.
Restraining Force
If a restraining force W was applied to restrain the strip from deflecting, from simple theory of
bending,
δ = W L3
48EI =
L2
4dK sT (6.45)
⇒ W = 48K sEI L2T
4dL3 =
12K sEI T
d but I =
bd3
12 (6.46)
Hence W = K sEbd2
L
T (6.47)
Manufacturers have also specified another constant: F s = E 4
which is known as the strip force
constant.
⇒ W = 4K sF sbd2
L T (6.48)
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EMT 2407: 6.5 Minimum Volume Concept 76
Figure 6.9: Cantilever strip
6.4.2 Cantilever Type
From Figure 6.9,
(R − δ )2 + L2 = R2 (6.49)
R2 − 2δR + δ 2 = R2 (6.50)
2δR = L2 ⇒ δ = L2
2R (6.51)
δ = L2
d
3
4(α1 − α2)T =
L2K sT
d (6.52)
Restraining Force
The maximum deflection for a cantilever beam is given by:
W L3
3EI
= L2
d
K sT
⇒W =
3EI K sT L2
L3
d
(6.53)
W = bd2
4LEK sT (6.54)
= E
4
bd2
L T (6.55)
= K sF sbd2
L T (6.56)
6.5 Minimum Volume Concept
For proper operation, the free deflection of the bimetallic strip must be applied through a temper-ature range rT where r < 1. For the rest of the temperature range (r − 1), a force W is exerted.
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EMT 2407: 6.5 Minimum Volume Concept 77
Figure 6.10: Restrained deflection
For a cantilever thermostat,
δ = L2
d K srT (6.57)
and W = K sF sbd2
L (1 − r)T (6.58)
Multiplying equation 6.57 and 6.58, we get:
W δ = K
−s2F s
L2
d
bd2
L
(1
−r)rT 2
= K 2sF sLbd(1 − r)rT 2
V olumeV = Lbd
∴ W δ = K 2sF sV (1 − r)rT 2
⇒ V = W δ
K − s2F sT 21
r − r2 =
C
r − r2
where C is a constant. For minimum volume,
dV
dr
= 0 = C d
dr
1
r − r2
0 = (r − r2)0 − 1(1 − 2r)
r − r2
= 2r − 1
r − r2 ⇒ 2r − 1 = 0
∴ r = 1
2
Therefore, one half of the temperature range is taken up for free deflection while the other half is
used to exert the force W.
Example 6.5.1. A cantilever bimetallic strip used in a commercial thermostat is to operate at a temperature range T = 100oC . The strip has to deflect 2mm and then exert a force of 7N.
Calculate the minimum volume of material of the bimetallic strip required. The strip has a strip
deflection constant K s = 14 × 10−6/oC and a strip force constant F s = 46GN/m 2.
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EMT 2407: 6.6 Tutorial 5 78
Solution
Let rT = T 1 = 50oC and (1 − r)T = T 2 = 50oC.
δ = L2
d K sT 1 (6.59)
Restraining force W = bd2
L K sF sT 2 (6.60)
Multiplying equation 6.59 and 6.60,
W δ = bdLK 2sF sT 1T 2
bdL = V = W δ
k2sF sT 1T 2
= 7 × 2 × 10−3
(14
×10−6)2
×46
×109
×502
= 621.12 × 10−9 m3 or 621.12 mm3
6.6 Tutorial 5
1. A bimetallic thermostat is 12.5mm wide by 25.0mm long. The thickness of each material is
1.0mm. It is clamped at one end and free at the other end. In operation, the thermostat is
to press a spring set at the free end.
Calculate the maximum temperature range of operation of the thermostat if the force on
the spring is not to exceed 110N and the maximum stress on the thermostat material isnot to exceed 138MN/m2. The Young’s modulus, E may be assumed to be the same for
both materials and = 185GN/m2. The difference between the coefficients of expansion of
the materials is 16×10−6/oC.
[Ans: 93.24oC]
2. Fig. 6.11 shows a bimetallic strip of a commercial thermostat in form of a cantilever beam
havering the following specifications:
• Strip Deflection Constant, K s =17 × 10−6/oC
• Strip Force Constant, F s =48 GN/m2
• Effective Length, L = 80mm
• Width, b = 6mm
• Total Thickness, d = 0.8mm
• Thickness of each metal, h = 0.4mm
The modulus of elasticity for each material is the same and α1 > α2. If contact is made
between the thermostat and the rigid stopper when the temperature is raised by 100 oC,
calculate:
(a) the distance, δ between the edge of the thermostat and the stopper.
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EMT 2407: 6.6 Tutorial 5 79
(b) the normal stress at the upper surface, interface and lower surface of the bimetallic
strip.
(c) the force that would be exerted on the stopper if the temperature is further increased
by 40oC. You may assume that:
1
R =
3
2d(α1 − α2)T
where R is the radius of curvature and T is the temperature change.
Figure 6.11:
[Ans: δ =13.6mm; σu=108.8MN/m2, σi = ±217.6 MN/m2, σl=-108.8MN/m2; W =1.567N ]
3. (a) A bi-metal element made of materials whose width, b, length, L, thickness of each
material, d2
and Young’s modulus, E are the same is used for making thermostats to be
heated through a temperature range, T . Show that the expression for the maximum
normal stress, σmax, on the element is given by:
σmax = AK sF sT
where K s is the strip deflection constant, F s is the strip force constant and A is a
numerical constant. Evaluate A [A = 83
]
(b) A particular bi-metal element has the following specifications: K s = 17 × 10−6/oC,
F s = 55GN/m2, b = 10mm, L = 60mm, and d = 0.8mm. If the element is used as
a simply supported beam, determine the maximum normal stress if the temperature
range of operation is 60oC. [Ans: 149.6MN/m2]
(c) If the element was restrained from deflecting, determine the restraining force, W .
[Ans: W =23.9N]
4. Fig. 6.12 shows a bimetallic strip used in a commercial thermostat in form of a simply
supported beam of length L with the following characteristics:
Strip force constant (F s) = 49 GN/m2
Strip deflection Constant (K s) = 14×10−6 /oCWidth (b) = 6mm
Depth (d) = 0.8mm
Effective length, L = 108mm
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EMT 2407: 6.6 Tutorial 5 80
Figure 6.12:
(a) Determine the temperature change required for the thermostat to make contact with
the spring located at δ = 3mm. [Ans: 58.8oC]
(b) If the bimetallic strip deflects the spring by 2mm, determine the total temperature
change subjected to the bimetallic strip. The spring has a stiffness constant of k=1kN/m.[Ans: 118.5oC]
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81
Chapter 7
Rotating Discs and Cylinders
7.1 Introduction
A rotating disc is a uniformly thin disc which on rotating at constant velocity, is subjected to
stresses induced by centrifugal forces.
Components modeled as uniform discs include:
- gas turbine rotors
- flywheel
- rotating shrink-fit assemblies e.g. shaft-hub assemblies, pulleys
For a thin disc, plane stress is assumed such that we only have
i. σh = Circumferential (hoop stress)
ii. σr = Radial stress
iii. σa = 0 (axial stress)
For a solid disc,
Figure 7.1: Solid disc
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EMT 2407: 7.2 Circumferential and Radial Strains 82
Consider the element of the disc, a distance r from the center. Assuming unit thickness, for
equilibrium of forces in the radial direction, we have:
2σh sin
dθ
2 · dr + σr · rdθ − σr +
dσr
dr dr
r + dr
dθ − F = 0 (7.1)
For small angles, sin dθ
2 ≈ dθ
2 (7.2)
∴ 2σh · dθ
2 dr + σrrdθ − σrrdθ − dσr
dr rdrdθ − σrdrdθ − dσr
dr dr2dθ − F = 0 (7.3)
ignoring higher powers of small quantities,
∴ σhdr − rdσrdr
dr − σrdr − F
dθ = 0 (7.4)
where, F is the centrifugal force on the element given by:
F = mω2r = ρdr · rdθ · rω2 = ρr2ω2drdθ
where ω is the angular speed of the disc and ρ is the density of the disc
Equation 7.4 becomes,
rdσrdr
+ σr − σh + ρr2ω2 = 0 (7.5)
or dσr
dr +
σr − σhr
+ ρrω2 = 0 (7.6)
7.2 Circumferential and Radial Strains
Figure 7.2: Deformed element
εh = New circumference - initial circumference
initial circumference
= 2π(r + u) − 2πr
2πr =
u
r
εr = New thickness - Initial thickness
Initial thickness
= u + du
drdr + dr − u) − dr
dr =
du
dr
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EMT 2407: 7.2 Circumferential and Radial Strains 83
From the general stress-strain relationships:
εh = σh
E − ν
σrE
= u
r (7.7)
εr = σrE
− ν σh
E = du
dr (7.8)
Differentiating equation 7.7 with respect to r, we get,
r dudr
− u
r2 =
1
E
dσhdr
− ν dσrdr
1
r
du
dr − u
r2 =
1
E
dσhdr
− ν dσrdr
du
dr =
u
r +
r
E dσhdr
− ν dσrdr
From equation 7.8, 1E
(σr − σh) = 1E
(σh − σr) + rE
dσhdr
− ν dσrdr
σr(1 + ν ) − σh(1 + ν ) = r
dσhdr
− ν dσrdr
(1 + ν )
σr − σhr
= dσh
dr − ν
dσrdr
or σr − σh
r =
1
1 + ν
dσhdr
− ν dσrdr
From equation 7.6,
dσrdr
+ 11 + ν
dσhdr
− ν dσrdr
+ ρrω2 = 0
(1 + ν )dσrdr
+ dσh
dr − ν
dσrdr
+ (1 + ν )ρrω2 = 0
dσrdr
(1 + ν − ν ) + dσh
dr + (1 + ν )ρrω2 = 0
dσrdr
+ dσh
dr + (1 + ν )ρrω2 = 0 (7.9)
Integrating equation 7.9,
σr + σh = −(1 + ν ) ρr2ω2
2 + 2A (7.10)
from equation 7.6 σh − σr − rdσrdr
= ρrω2 (7.11)
equation 7.10 − equation 7.11 (7.12)
2σr + rdσrdr
=
− 1 + ν
2 − 1
ρω2r2 + 2A (7.13)
2σr + rdσrdr
= 2A − 3 + ν
2 ρω2r2 (7.14)
Rewriting equation 7.14,
1
r
2σrr + r2
dσrdr
= 2A − 3 + ν
2 ρω2r2
or d
dr
σrr2
− 2Ar +
3 + ν
2 ρω2r3 = 0
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EMT 2407: 7.2 Circumferential and Radial Strains 84
Integrating,
σrr2 − 2Ar2
2 +
3 + ν
2 ρω2 r4
4 + B = 0 (7.15)
or σr = A − Br2
− 3 + ν 8
ρω2r2 (7.16)
From equation 7.10,
σh = 2A − 1 + ν
2 ρω2r2 − σr (7.17)
= 2A − 1 + ν
2 ρω2r2 − A +
B
r2 +
3 + ν
8 ρω2r2 (7.18)
= A + B
r2 − 4 + 4ν − 3 − ν
8 ρω2r2 (7.19)
= A + Br2
− 1 + 3ν 8
ρω2r2 (7.20)
7.2.1 Solid Disc with Unloaded Boundaries
(a) Rotating disc Disc (b) Stress distribution
Figure 7.3:
Apply boundary conditions,
When r = 0, both σr and σh → ∞. This is not practical and so we set B = 0
When r = R2, σr = 0
∴ 0 = A − 3 + ν
8 ρω2R2
2
⇒ A =
3 + ν
8 ρω2
R2
2
σr = 3 + ν
8 ρω2R2
2 − 3 + ν
8 ρω2r2 =
3 + ν
8 ρω2(R2
2 − r2)
σh = 3 + ν
8 ρω2R2
2 − 1 + 3ν
8 ρω2r2
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EMT 2407: 7.2 Circumferential and Radial Strains 85
at r = R2, σr = 0 and:
σh = 3 + ν − 1 − 3ν
8 ρω2R2
2 = 2 − 2ν
8 ρω2R2
2
= 1 − ν 4
ρω2R22
7.2.2 Maximum Speed for Initial Yielding
Tresca’s Criterion (Max shear stress criterion)
Tresca’s Yielding criterion states that:
σmax − σmin2
= σy
2 = τ max
The maximum stress occurs at r = 0 and is:
σr = σh = 3 + ν
8 ρω2R2
2
Minimum stress σz = 0
∴ σmax = σy3 + ν
8 ρω2
yR22 = σy
ωy = 1
R2 8σy
(3 + ν )ρ
von-Mises Criterion (Shear strain energy criterion)
(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2 = 2σ2y
σ1 = σ2 = 3 + ν
8 ρω2R2
2
σ3 = 0 ⇒ σ1 = σy
ωy = 1
R2 8σy
(3 + ν )ρ
which is the same value as for the Trescas criterion
7.2.3 Increase in Radius
εh = 1
E (σh − νσr) =
u
R2∵ r = R2
∴ u = R2
E
(σh
−νσr but atr = R2, σr = 0
⇒ u = R2
E σh =
R2
E
1 − ν
4 ρω2R2
2
u = ρω2R2
2
4
1 − ν
E
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EMT 2407: 7.2 Circumferential and Radial Strains 86
7.2.4 Change in Thickness
Let the original thickness be t. The change in thickness in the z-direction is ∆t
εz = σzE
− ν E
(σr + σh) but σz = 0
∴ εz = − ν
E (σr + σh) =
∆t
t
at r = 0, σr = σh
∆t
t = − ν
E
2
3 + ν
8 ρω2R2
2
∆t = − ν
E 2
3 + ν
4 ρω2R2
2t
when r = R2, σr = 0,
∆t
t = − ν
E
0 +
1 − ν
4 ρω2R2
2
= − ν
E
1 − ν
4 ρω2R2
2
∆t = − ν
E
1 − ν
4 ρω2R2
2
t
Figure 7.4: Change in thickness
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EMT 2407: 7.3 Disc with Central Hole and Unloaded Boundaries 87
7.3 Disc with Central Hole and Unloaded Boundaries
Boundary conditions:
at r = R1, σr = 0 and at r = R2, σr = 0
0 = A − B
R22
− 3 + ν
8 ρω2R2
2
0 = A − B
R21
− 3 + ν
8 ρω2R2
1
⇒ B = 3 + ν
8 ρω2R2
1R22, A =
3 + ν
8 ρω2(R2
1 + R22)
∴ σr = 3 + ν
8 ρω2
R21 + R2
2 − 2R21R2
2
r2 − r2
σh =
3 + ν
8 ρω2
R2
1 + R2
2 − 2R2
1R22
r2 − 1 + 3ν
3 + ν r2
Maximum σh occurs at r = R1
σhmax = 3 + ν
8 ρω2
R2
1 + R22 − 1 + 3ν
3 + ν r2
Maximum σr occurs at r =√
R1R2. Stresses are either zero or tensile
7.4 Disc Shrunk onto a Shaft
Applied in shaft-hub assemblies.
Use shrink-fit (interference fit) between components instead of mechanical fittings e.g. keys and
keyways, rivetting, bolts etc.
Same principle used for compound cylinders applies.
At the mating surface, we have radial compressive stresses set up by the shrink fit and on rotation,
radial tensile stresses are set-up. It is therefore necessary to ensure that the shrink-fit stress is
always greater than the rotational stresses to avoid the disc running free from the shaft.
Figure 7.5: Free body diagrams of shrink-fit components
Let
δ = shrinkage between the hub and the shaft.
= difference in radius between the mating surfaces
= uhub − ushaft
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EMT 2407: 7.4 Disc Shrunk onto a Shaft 88
when the assembly is stationary, ω = 0 and the contact pressure is P c. Therefore,
σr = A − B
r2
σh = A + Br2
We can determine the contact pressure by using a similar analysis to that of compound cylinders.
Applying the boundary conditions:
Considering the hub,
at r = R1, σr = −P c
at r = R2 = σr = 0
∴ −P c = A − B
R21
(7.21)
0 = A − B
R22
(7.22)
subtracting equation 7.22 form 7.21,we obtain:
−P c = −B 1
R21
− 1
R22
⇒ B =
R21R2
2
R22 − R2
1
Eliminating B,
−P cR21 = AR2
1 − B
subtracting 0 = AR22 − B
−P cR21 = −A(R2
2 − R21) ⇒ A =
P cR21
R22 − R2
1
The radial and circumferential strains become:
σr = P cR
21
R22 − R2
1
1 − R2
2
r2
σh = P cR
2
1R22 − R2
1
1 +
R2
2r2
Considering the shaft,
σr and σh → ∞ at r = 0 ⇒ B = 0
∴ σr = σh = = A = −P c everywhere on the shaft
For the hub,
εh =
uh
R1 =
1
E (σh − νσr) =
1
E
P cR21 + R2
2
R22 − R2
1 + νP c
uh = R1P c
E
P c
R21 + R2
2
R22 − R2
1
+ ν
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EMT 2407: 7.4 Disc Shrunk onto a Shaft 89
For the shaft:
εh = usR1
= 1
E (σh − νσr) =
1
E (−P c + νP c)
us = −P cR1
E (1 − ν )
δ = uh − us
= R1P c
E
P c
R21 + R2
2
R22 − R2
1
+ ν
+ P cR1
E (1 − ν )
= R1P c
E
P c
R21 + R2
2
R22 − R2
1
+ ν + 1 − ν
= R1P c
E P
c
R21 + R2
2 + R22 − R2
1
R22 − R21
= R1P c
E
P c
2R22
R22 − R2
1
⇒ P c = Eδ (R2
2 − R21)
2R1R22
Loosening speed
When rotating freely, σr = 0 at the interface.
• Hub can be considered as a disc with a central hole and unloaded boundaries
σh = 3 + ν
8 ρω2
R2
1 + R22 − 2R2
1R22
r2 − 1 + 3ν
3 + ν r2
at r = R1 σh = 3 + ν
8 ρω2
R2
1 + R22 − 1 + 3ν
3 + ν r2
uh = R1
E (σh − νσr) =
R1
E
3 + ν
8 ρω2
R21 + R2
2 − 1 + 3ν
3 + ν r2
• Shaft can be considered as a solid disc with unloaded boundaries.
At the loosening speed,
δ = uh − us
Example 7.4.1. A solid steel disc of small constant thickness has a steel ring of outer diameter
610 mm and same thickness shrunk onto it. The assembly has an interface diameter of 457mm. If
the interface pressure is reduced to zero at a rotational speed of 3000rpm, calculate the difference
in diameters of the mating surfaces of the disc and the ring before assembly and the interface
pressure. Take ν = 0.29 and E =207 GN/m 2
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EMT 2407: 7.5 Disc with Loaded Outer Boundary 90
7.5 Disc with Loaded Outer Boundary
Rotor discs normally have a large number of blades mounted on the outer boundary layer. These
will themselves have a centrifugal force component acting at the periphery of the disc. Given themass of each blade, its effective centre of mass and the number of blades, we can compute the
total force acting on the outer surface. Dividing this force by the area of the outer surface gives
us the required value of σr to be used as a boundary condition.
F c = N × ω2 × re
and,
at r = R2, σr = F c2πR2t
Example 7.5.1. A steel rotor disc of uniform thickness 50mm has an outer rim of diameter 750mm and a central hole of diameter 150mm. There are 200 blades each of mass 0.22kg at
an effective radius of 430mm pitched evenly around the periphery of the rotor. Determine the
rotational speed at which yielding first occurs according to the maximum shear stress criterion.
Yield stress in simple tension for the steel is 700MN/m 2, ν = 0.29, ρ = 7300kg/m 3 and E =
207GN/m 2
Solution
centrifugal force caused by each blade, = mω2re
= 0.22ω2 × 0.43 = 0.0946ω2
Total centrifugal force F = 200 × 0.0946ω2 = 18.92ω2
Radial stress on the outer surface,
σr|r=R2 =
F
2πR2t =
18.92ω2
2π × 0.375 × 0.05 = 160.598ω2
General expressions for radial and hoop stress are:
σr = A − B
r2 − 3 + ν
8 ρω2r2
σh = A − B
r2 +
1 + 3ν
8 ρω2r2
Boundary conditions:
at r = R1 = 0.075, σr = 0
at r = R1 = 0.375, σr = 160.598ω2
Applying Boundary conditions,
A − 177.78B = 16.89ω2 (7.23)
A − 7.11B = 582.77ω2 (7.24)
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EMT 2407: 7.6 Disc of Uniform Strength 91
Equation 7.24 - 7.23 gives,
170.67B = 565.88ω2
⇒ B = 3.316ω2
A = 606.35ω2
and the general equations for the radial and hoop stress become:
σr = 606.35ω2 − 3.316ω2
r2 − 3002.125ω2r2
σh = 606.35ω2 + 3.316ω2
r2 + 1706.375ω2r2
(σr)max occurs at dσr
dr = 0
dσrdr
= −r2(0) − 2r × 3.316ω2
(r2)2 − 2 × 3002.125ω2r = 0
6.632ω2
r3 = 6004.25ω2r
r4 = 6.632
6004.25 ⇒ r = 0.182
(σr)max = 606.35ω2 − 3.316ω2
0.1822 − 3002.125ω2 × 0.1822 = 406.78ω2
The maximum circumferential stress occurs at r = R1
σh = 606.35ω2 + 3.316ω2
0.0752 + 1706.375ω2 × 0.0752 = 1186.26ω2
at r = R1, σr = 0
σmax = 1186.26ω2
σmin = 0σmax − σmin
2 =
σy2
⇒ σmax=σy
1186.26ω2 = 700 × 106 ⇒ ω = 768.173ω2 rad/s
N = 60ω
2π = 7335.5 rpm
7.6 Disc of Uniform Strength
A disc of uniform strength is one in which the values of radial and circumferential stresses are
equal in magnitude for all values of the radius r. This means that the disc of uniform strength
must have varying thickness.
In practice, components such as rotor of a steam turbine which have constant strength throughout
are designed by varying their thickness. centrifugal force on the element is given by:
F = ρ · dr · t · rdθω2r = ρω2tdrdθr2 (7.25)
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EMT 2407: 7.7 Tutorial 6 92
Figure 7.6: Disc with varying thickness
Considering equilibrium of forces in the radial direction,
2σdθ
2 tdr + σrtdθ − σ(r + dr)dθ(t + dt) − F = 0
σtdrdθ + σrtdθ − σrtdθ − σtdrdθ − σrdtdθ − σdrdtdθ − ρω2tdrdθr2 = 0
Ignoring products of small quantities,
σrdt + ρtω2r2dr = 0
dt
t = −ρω2r
σ dr
integrating, ln t = −ρω2r2
2σ + A
Applying the boundary conditions,
at r = 0, t = to
∴ ln to = A
ln t = −ρω2r2
2σ + ln to
or ln t
to= −ρω2r2
2σ
t = e−ρω2r2
2σ
7.7 Tutorial 6
1. A steel ring has been shrunk onto the outside of a solid steel disc. The interface radius is
250mm and the outer radius of the assembly is 356mm. If the pressure between the ring
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EMT 2407: 7.7 Tutorial 6 93
and the disc is not to exceed 34.5 N/mm2 and the circumferential pressure must not exceed
207N/mm2;
(a) Determine the maximum speed at which the assembly can be rotated(b) Determine the stress at the centre of the disc at this maximum speed.
Take ν = 0.28 and ρ = 7570kg/m3
[Ans: 3323 rpm; -11MN/m2]
2. A steel rotor which is part of a turbine assembly has a uniform thickness of 80mm. The
outside diameter of the rotor is 800mm while the inside diameter is 360mm. 240 blades each
of mass 0.16kg are pitched evenly around the periphery of the disc at an effective radius
of 430mm. Calculate the required speed of revolution for the internal radius to change by
0.14mm. Take ν = 0.29, E = 200GN/m2 and ρ = 7560kg/m3
[Ans: 3376rpm]
3. A steel rotor disc which is part of a turbine assembly has a uniform thickness of 60mm.
The disc has an outer diameter of 700mm and a central hole of diameter 150mm. If there
are 300 blades each of mass 0.225kg pitched evenly around the periphery of the disc at an
effective radius of 370mm, determine the rotational speed at which the yielding of the disc
first occurs according to the maximum shear stress criterion. The yield stress in simpletension is 550MN/m2. Take ν = 0.3, E = 200GN/m2, ρ = 7470kg/m3
[Ans: 6870rpm]
4. A steel disc of uniform thickness is hollow and has an outer diameter of 660mm and an inside
diameter of 120mm. The disc is made to rotate at a speed of 600 rad/s and a radial stress
of 92MN/m2 is generated at the outer circumference due to blades attached to the disc. At
this speed, calculate:
(a) the change in inner diameter,
(b) the change in outer diameter.
Take ν = 0.3, E = 200GN/m2, ρ = 7740kg/m3
[u1=0.1334mm, u2=0.218mm]
5. A disc of uniform thickness having inner and outer diameters, 100mm and 400mm respec-
tively is rotating at 5000rpm about its axis. The density of the material of the disc is
7800kg/m3 and Poisson’s ratio is 0.28. Determine the maximum radial and circumferential
stresses, and maximum shear stress. [Ans: (σr)max=19.74MN/m2
, (σh)max=71.09MN/m2
,τ max=35.54MN/m2 ]
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EMT 2407: 7.7 Tutorial 6 94
6. A steam turbine rotor is to be designed so that the radial and circumferential stresses are
to be the same and constant throughout and equal to 90MN/m2, when running at 4000rpm.
If the axial thickness at the centre is 20mm, what is the thickness at a radius of 400mm?
Assume the density of the material of the rotor is 7800kg/m3.
[Ans: t= 5.93mm]
Assignment 3
Thermal stresses arise in the rotor of the turbine of a gas turbine engine, or steam turbine when
there are radial variations in temperature due to the action of cooling air which is applied to the
surfaces of the disc in order to prevent it from reaching excessive temperature. In these turbines,
the thermal effect is of interest because it tends to offset the radial and circumferential stressescaused by rotation.
If plane stress is assumed, equilibrium of an element of the disc requires that:
dσrdr
+ σr − σh
r + ρω2r = 0 (7.26)
For a linear thermo-elastic material, the stress-strain relationships are:
εr = du
dr =
σrE
− ν σhE
+ αT (7.27)
εh = ur
= σhE
− ν σrE
+ αT (7.28)
where T = temperature change in the disk which is a function of the radial position r, measured
from the centre of the disc and α is the linear coefficient of expansion.
(a) Show that the general expressions of radial and circumferential stress with thermal effects are:
σr = A − B
r2 − 3 + ν
8 ρω2r2 − αE
r2
T · rdr