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MECH ENG 2002

Stress Analysis & Design

Dr Leok Lee

TABLE OF CONTENTS

Lecture 1: Tension, compression and shear .................................................... 1 Lecture 2: Basics of structural design ........................................................... 22 Lecture 3: Axially loaded members .............................................................. 46 Lecture 4: Torsion of circular bars ................................................................ 71 Lecture 5: Shear force and bending moments ............................................... 96 Lecture 6: Stresses in beams – Part 1 .......................................................... 112 Lecture 7: Stresses in beams – Part 2 .......................................................... 140 Lecture 8: Deflection of beams and columns .............................................. 159 Lecture 9: Analysis of stress and strain – Part 1 ......................................... 185 Lecture 10: Analysis of stress and strain – Part 2 ....................................... 212 Lecture 11: Pressure vessels and shaft design ............................................. 230

These lecture notes were edited by Dr Ka Lok ‘Leok’ Lee in Semester II, 2019, authored by Dr Aditya Khanna in Semester II, 2018, and adapted from previous notes written by Prof Andrei Kotousov in 2005. Notes are available for individual study in MECH ENG 2002, Stress Analysis and Design at the University of Adelaide.

Lecture 1: Tension, compression and shear Lecture outline • Concept of stress and strain, • Hooke’s law, and Poisson’s ratio, • Small strain analysis. Recommended reading Mechanics of Materials 9th SI Edition: Chapter 1, Sections 1.4, 1.7, and 1.8

1

Concept of stress and strain Tension, compression and shear are three elementary ways in which a structural member can be loaded. The concept of stress and strain can be illustrated using these simple cases. P P V

A

γ L

L

L

L

A

A 

P

V



P



  2

Definition of normal stress m P

P n

!"# 

For a prismatic bar, the normal stress is defined as the axial force divided by the crosssectional area, i.e.

m P

P

σ=

n

P A



 In Eq. (1.1), we use the initial  cross-sectional area of the bar  and refer to σ as the nominal or   engineering stress.

$%

If the bar is stretched, the stress is tensile. If the bar is compressed, the stress is compressive. Tensile stresses are positive & compressive stresses are negative. Units: Newton per square meter (N/m2) or Pascal (Pa). More commonly, the stress is expressed in mega-Pascal (MPa), or its equivalent unit (N/mm2). 3

Limitations Eq. (1.1) is only valid if the stress is uniformly distributed over the crosssection of the bar. To satisfy this condition, the axial force must act through the centroid of the cross-sectional area.

P

P σ=

P A

&'%

4

Limitations Eq. (1.1) is generally not valid near the ends of the bar. This is because axial forces are typically transmitted through pin or bolt connections, which produce a non-uniform stress state in the vicinity of the connection. However, (1.1) can be used with good accuracy at distances greater than roughly one diameter (or largest lateral dimension of the bar) from the ends.

b

d

P

P ()# *+

5

Definition of normal strain For a prismatic bar, the normal strain is defined as the change in length divided by the original length, i.e.

L

P

P

ε=

ΔL L − L = L L



L ,-

In Eq. (1.2), L is the length before the load is applied and L is the length after the load is applied.

Elongation of the bar produces tensile strains, which are positive. Contraction of the bar produces compressive strains, which are negative. Units: Because normal strain is a ratio of two lengths, it is a dimensionless quantity. Typical values for strains in most engineering structures do not exceed 0.001 or 0.1%. 6

Hooke’s law in tension (compression) The linear relation between stress and strain for a bar in simple tension or compression is expressed by the equation σ = Eε

&

σ is the normal stress, ε is the normal strain, and E is the modulus of elasticity or Young’s modulus. Eq. (1.3) is only applicable for the analysis of simple tension/compression of a bar (uniaxial stress). For more complicated stress states, the generalised version of Hooke’s law needs to be used (Lecture 10).

7

Poisson’s effect When a prismatic bar is loaded in tension, the axial elongation is accompanied by lateral contraction (that is reduction of cross-sectional area). The normal strains in the axial and lateral directions are: ε =

L − L , L

ε =

D − D , D

(

The lateral strain, ε is proportional to the axial strain ε at the same point if the material is linear elastic. The ratio of these strains is a property of the material known as Poisson’s ratio. L

D

P

P

D

L .)% 8

Poisson’s ratio Poisson’s ratio is a dimensionless quantity, usually denoted by Greek letter ν (nu) and defined as: ν=−

ε laternal strain , =− axial strain ε

,

The minus sign means that the lateral and axial strains normally have opposite signs. If the Poisson’s ratio for the material is known, the lateral strain can be obtained as ε = −νε .

.

For steels the Poisson’s ratio is ν ~ 0.3, for rubber-like materials ν ~ 0.5, for cork ν ~ 0.

9

Definition of shear stress Shear stresses act tangential to the cross-section of the material. They are considered in the design of bolts, pins, rivets, keys, welds and glued joints.

$%

F V

V

a c b !"#  



d   

/- 10

Definition of average shear stress The average shear stress on the cross-section can be defined as τ$% = V⁄ A

/

where A is the area of the cross-section. Units: Newton per square meter (N/m2) or Pascal (Pa). More commonly, the stress is expressed in mega-Pascal (MPa), or its equivalent unit (N/mm2).

F

F τ

%$V a

a c b

τ$% c

V = F⁄2 d

A

b

d

01## 11

Limitations More precise investigations of the shear stress distribution over the critical sections often reveals that much larger magnitudes of shear stresses occur in the structure than those predicted by Eq. (1.7). However, application of (1.7) is generally acceptable for many engineering calculations. For example, engineering codes allow its use when: •

Designing fasteners such as bolts, rivets and,

•

Obtaining the bonding strength of joints subjected to shear loading.

12

Equality of shear stresses on perpendicular planes Consider a small element of material shown below. The equilibrium of this element requires that: 1. Shear stresses on opposite (and parallel) faces of the element are equal in magnitude and opposite in direction, 2. Shear stresses on adjacent (and perpendicular) faces of an element are equal in magnitude (τ( = τ) ) and have directions such that both stresses point toward, or both point away from, the line of intersection of the faces.

τ1

F

τ2 τ

%$τ

%$2$* 13

Sign convention for shear stresses Unlike normal stresses, the sign of shear stresses depends upon the choice of coordinate system. A shear stress acting on a positive face of an element is positive if it acts in the positive direction of one of the coordinate axes and negative if it acts in the negative direction of an axis. A shear stress acting on a negative face of an element is positive if it acts in the negative direction of an axis and negative if it acts in a positive direction.

+# y

τ

+%

τ %

x #

3$% 14

Definition of shear strain Shear stresses acting on an element of material (Fig. 1-11a) are accompanied by shear strains. The shear stresses have no tendency to elongate or shorten the element. Instead, the shear stresses produce a change in the shape of the element (Fig. 1-11b). Because of this deformation, the angles between the side faces change. The angle , is a measure of the distortion, or change in shape of the element and is called the shear strain. Because shear strain is an angle, it is usually measured in radians. γ y y τ 2

π −γ 2

τ 



x

x

γ 2

)4* 15

Sign convention for shear strain The sign of shear strain is also dependant upon the choice of coordinate system. Shear strain in an element is positive when the angle between two positive faces (or two negative faces) is reduced. The strain is negative when the angle between two positive (or two negative) faces is increased. +# y

γ 2

+% ) 

π −γ 2 % x

γ 2

# $% 16

Hooke’s Law in Shear Consider an element of material in the form of a rectangular parallelepiped. The shear stress τ produces a change in the shape of the element (Fig. 1-11). Similar to the previously considered case, the linear relation between shear stress and strain for the rectangular parallelepiped is expressed by the equation: τ = Gγ,

0

τ is the shear stress, γ is the shear strain, and G is the shear modulus or modulus of rigidity, which can be related to the Young’s modulus according to G=

E 2 1+ν

2

17

Small strain analysis Most engineering design applications only allow for small deformations. Deformations that occur during operation of most structures and machines are hardly noticeable. Even if the deflection of a structural member such as a thin plate or slender rod may appear to be large, the material from which it is made may only be subjected to very small strains. In this course we will assume that the strains within the material are very small compared to 1, that is, ε and γ =

T= b T>= L>= T!> L!> T! a =− , ϕ>= = = GI# GI# GI# GI#

0!

Substituting the above expressions in the compatibility equation (E1.2) yields: T!a T= b − =0 GI# GI#

0

Together, Eq. (E1.4) and the equilibrium equation (E1.1) form a system of two linear equations with two unknowns, which can be solved to obtain T! = University of Adelaide

T8 b T8 ab T8 a , ϕ> = ϕ!> = , T= = LGI# L L

0(

90

Analysis of a statically indeterminate bar T8

T8

Example 2 A solid cylinder S is enclosed in a circular tube C. The assembly is subjected to external torque. The sum of torques in the cylinder and tube must be equal to the external torque, i.e. the equilibrium equation is

T>

T? + T@ − T8 = 0

TA

T8

0

The cylinder and tube must twist by the same amount , i.e. the compatibility equation is

  .)& &

ϕA = ϕ>

University of Adelaide

0

91

Example 2 (contd.) Using Eq. (4.11), the angle of twist of the cylinder and tube can be obtained as: ϕA =

T> L TAL , ϕ> = GAI#A G> I#>

0!

Substituting the above expressions in the compatibility equation (E2.2) yields: TAL T> L = GAI#A G> I#>

0

Together, Eq. (E2.4) and the equilibrium equation (E2.1) form a system of two linear equations with two unknowns, which can be solved to obtain TA = T8

G> I#> GAI#A T8 L , ϕA = ϕ> = , T> = T8 GAI#A + G> I#> GAI#A + G> I#> GAI#A + G> I#> 0(

University of Adelaide

92

Analysis of Stress and Strain in Pure Shear Cut from the element abcd a wedge-shaped (or triangular) stress element having one face oriented at an angle θ to the x axis. The stresses on this face can be obtained from equilibrium equations of the triangular element using a freebody diagram. y τ

τ 

 τ

τ 

 τ

x

τD

σD

θ

τ

σD A sec θ

90° − θ

τ



θ

τD A sec θ

τA tan θ 



/1&%% & University of Adelaide

93

Analysis of Stress and Strain in Pure Shear From equilibrium equations, the stresses acting on the inclined plane are: σD = 2τ sin θ cos θ = τ sin 2θ

-

τD = τ cos  θ − sin θ = τ cos 2θ

.

Maximum/minimum normal stresses exist on planes at 450 to the axis x (when 2θ = π⁄ 2 , 3π⁄2 …). y σ,. = −τ σ = τ ∗ y τ τ





0*# τ



 τ

x

45°  ∗

∗

τ

∗

x

−τ

)θ = 0 - θ = 45- % University of Adelaide

94

Analysis of Stress and Strain in Pure Shear The existence of maximum tensile stresses on planes at 45o to the x axis explains why bars in torsion made of brittle materials fracture along a 45o helical surface.

3&4(  University of Adelaide

95

Lecture 5: Shear forces and bending moments Lecture outline • Shear Forces and Bending Moments, • Relationships between Loads, Shear Forces, and Bending Moments, • Shear and Moment Diagrams. Recommended reading Mechanics of Materials 9th SI Edition: Chapter 4, All sections

96

Beams Members that are slender and support loads that are applied perpendicular to their longitudinal axis are called beams. Beams may be considered among the most important structural elements in engineering. 





 97

Shear Forces and Bending Moments Consider a cantilever beam AB. We cut through the beam at a cross section  located at distance x and isolate the left-hand part of the beam as a free body. The free body is held in equilibrium by the force P and by stresses that act over the cut cross section.

!"# 98

Shear Forces and Bending Moments From statics – the resultant of these stresses can be reduced to a shear force V and bending moment M. Theses quantities are known as stress resultants. They can be found from equilibrium equations in the vertical direction and also taking moments about the cut section: ↑ ∑ F  0

or

PV0

or

PV

$%

↺ ∑ M  0

or

M  Px  0

or

M  Px

$!%

& 99

Sign convention Sign conventions for stress resultants are called deformation sign conventions because they are based upon how the material is deformed, rather than by the direction of the stress resultant in space. In the case of a beam, a positive shear force acts clockwise against the material and a negative shear force acts counter-clockwise against the material. Also, a positive bending moment compresses the upper part of the beam and a negative bending moment compresses the lower part. *

P

V M

M

V '() 100

Sign convention Distributed loads and concentrated loads are positive when they act downward on the beam and negative when they act upward. A couple as a load on a beam is positive when it is counter-clockwise and negative when it is clockwise.

+

101

Relationships between Loads, Shear Forces, and Bending Moments (1) Balance of shear forces gives ↑  F  0

V  qdx  V  dV  0

$&%

or dV ⁄dx  q

$'%

,)- 102

Relationships between Loads, Shear Forces, and Bending Moments (2) Balance of bending moments gives ↺  M  0 or

 M  q dx ⁄2  V  dV dx  M  dM  0 dM⁄ dx  V

$%

$,%

,$%)- 103

Relationships between Loads, Shear Forces, and Bending Moments Combining Eqs. (5.4) and (5.6), we have d M⁄ dx   q

$.%

These equations are very useful when making a complete investigation of the shear and bending moments in a beam. The table on the next slide illustrates application of the differential relationships between loading, shear forces and bending moments to some common cases.

104

Relationships between Loads, Shear Forces, and Bending Moments

105

Relationships between Loads, Shear Forces, and Bending Moments a) The bending moment does not change as we pass through the point of application of a concentrated load. b) At a point of application of a concentrated load P, the rate of change dM/dx of the bending moment decreases abruptly by an amount equal to P. c) The shear force does not change at the point of application of a couple. d) If the shear force diagram changes sign from posi...