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



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.



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



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



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



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.

Edited by Foxit ReaderCopyright(C) by Foxit Software Company,2005-2008For Evaluation Only.



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




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.





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.





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




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)




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.




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



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.




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



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



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.



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)



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)



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]




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




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]



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.



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.


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



EMT 2407: 2.3 Torsion of Hollow Thin-walled Non-circular Members 21

Figure 2.3: Equivalent dimensions for other sections


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.




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)




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



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



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



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



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




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.




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




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




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]




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]




(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 ]



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.



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)




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.




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).




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)




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)



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




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)




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.




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



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.



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)




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




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



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




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




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




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




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




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

]




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

]




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



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.



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



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



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)




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.




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




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




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%



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



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




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)]




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]




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]



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



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.




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.




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)



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)



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



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)



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.



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.




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.




(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




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]



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



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




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




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




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




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



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



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



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



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)



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)




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




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 ]




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

solids andstructures mehanics iv notes

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