mech of solids

SESG 1001 Mechanics of Solids

Lecture 1 and 2 Introduction to Solid Mechanics

Block 1: Statically Determinate Structures

0 Introduction This module will provide the basic techniques required to analyze and design load-bearing structures. In general, when designing structures we are concerned with ensuring their “structural integrity”. This comprises three aspects of structural performance:

1) The structure should not deform excessively under load 2) The structure should not fail (break) 3) The structure should have sufficient life

Solid mechanics covers the analysis techniques to determine how structures transmit load (force) and how they deform under load. There is a close link to Materials (SESG1003), which covers aspects of elastic deformation, failure and factors that control life (fatigue, creep). Throughout SESG 1001 we will be invoking three basic principles (sometimes called the “three great principles”):

1) Equilibrium of forces 2) Compatibility of displacements 3) Constitutive behavior of materials

We will see how these apply to several classes of structure during the module. Finally, we will be constructing models for how structures behave. These are idealizations (simplifications) of the real structure that enable us to analyze it. A key skill for an engineer is to be able to develop models for physical situations that allow him/her to design and analyze engineering systems. For the moment we will rely on the experience and wisdom of those who have gone before us to examine existing models. In due course we will start developing our own models.

1.0 Statically Determinate Structures (See BCA, Ch1) These are structures for which the force distribution can be determined solely by applying equations of static equilibrium. We will examine three examples during SESG1001: 2-D trusses, bars of varying thickness and pressure vessels.

1.1 Trusses A truss is a structural configuration consisting of bars connected at joints. Generally these are 3-D, but we will only consider 2-D cases in SESG1001. This will start us thinking about models for structures and how we can analyze internal forces in structures.

1.2 Idealized Planar, Pin-Jointed Truss First define an idealized planar truss – this is the model

1. All bars are straight 2. Joints between bars are frictionless pins 3. Bars are massless and perfectly rigid (for loading analysis) 4. All loads and reactions are applied at the joints. 5. Loads in members are co-linear (axial), i.e. aligned with long axis of bar. Axial forces in bars also arise from (2).

1.3 Analysis of forces in pin-jointed trusses At this stage we are interested in determining the internal forces carried by the bars of the truss. In due course we will look at deformations and other factors related to determining the structural integrity of trusses Step 1. Determination of External Reactions Draw Free Body Diagram and apply equations of equilibrium EXAMPLE 1: A simple truss (3 bar truss). Note idealizations at support points: pins and/or pins on rollers:

Draw free body diagram (FBD)

Fbar = Fpin (by equilibrium)

Apply planar equations of equilibrium: Equilibrium of forces in x1, x2 directions, equilibrium of moments about x3 axis (do not worry about axis symbols for now: x1=x, x2=y, x3=z).

! F1 = 0 +" # " $ HA + HB = 0 (1)

! F2 = 0 " + # VB + 200N = 0 (2)

!M3 = 0 A+ ⇒ 10 •200 ! 5HB = 0 (3)

HB = 400N

From (1)

HA = !400N

Is this statically determinate? – so far yes, we have been able to evaluate all reactions using equilibrium equations. Three reactions apply to the three degrees of freedom (translation in x1 and x2 and rotation about x3) give three equilibrium equations that we can write down.

VB = !200N

Redrawing FBD

Step 2. Analysis of forces in bars (Method of Joints)

Now we want to evaluate the forces in the bars of the truss. This requires us to isolate part of the truss, replacing the cut bars by equivalent (equipollent) forces: Draw FBD for joint C, replace bars by equivalent forces (FCB and FCA)

! = tan"1 510

= 26.6!

sin! =5125

=15, cos! =

10125

=25

Apply equilibrium

!F1+" # " = 0$ %FCB cos& %FCA = 0

!F2 ' + = 0$FCB sin& + 200 = 0FCB = %200 ( 5 = %447N

NOTE: there is no opportunity to apply equilibrium of moments, all forces meet at C.

200 N

θ

FCB

FCA

Substitute in !F1

! "("200 5) 25"FCA = 0

⇒ ⇐

FCB is positive i.e. tensile, pulls out of the joint, FCB is negative, i.e. compression, pushes into the joint. Redraw joint:

Now consider joint, B:

Apply equilibrium

! F2 " + #$200 $ FCB %Sin& $ FBA = 0

# FBA = $200 $ ($200 5 %15) = 0

200 N

θ 400 N

FBA FBC

200 N 200√5 N

400 N

FCA = +400N

So what is the purpose of FBA?

As a check, look at joint A

In equilibrium – we have been consistent!

Finally, draw structure

If bar BA were not present a mechanism would result: this is a dynamic problem, not statically determinate

0 400 N FCA = 400

N

Consistent tensile forces

0

FBA = 0

This approach worked quite well for a truss with only a few bars, or if we want the load in every bar, then we work progressively through the truss. - but may not work for more complicated trusses - or if we only want to isolate a few bars - use method of sections Option 2: Method of Sections Isolate a section (part) of the truss of interest - Draw FBD, determine reactions - "cut" truss into sections (through bars) - Replace cut bars by coincident tensile forces - Apply planar equilibrium equations (positive tensile, negative compressive) All as for method of joints, but: - Can now use equilibrium of moments - Can now have 3 unknown bar forces in a section EXAMPLE 2

Problem: to find force in bar ED

B

A

D

C

E

45°

2L

2P2L

L

First draw FBD

B

A

D

C

E

45°

2L 2L

2P

~ ~

VA VC

HC

We are interested in bar ED. So take an appropriate cut.

x2

x1

L

A

2L

E DFED

FAB

FEB

45°

B

We want to find FED - try to find one equation with only one unknown

→ Take moments about B (draw in B)

!M B : +2!P ! !FED = 0

FED = !2P

Suppose we had wanted FEB. Still use same section:

Closed surface

Must include reactions crossing surface

P

! F2 = 0 " + # FEBSin45! + P = 0

= 12

! "

# $

FEP = + 2P

Suppose we had wanted FAB

→ Take moments about E

!M =0 E :

!PL + LFAB = 0FAB = P

Redraw FBD. Is this correct?

Is this correct? - Think about cutting members ED - shortens AB – lengthens NOTE: In both methods of joints, method of sections, we solve for the internal forces by isolating part of the structure. Two Tips: • Reduce computation by intelligent choice of method and section to analyze 1 equation, 1 unknown • Check, check and double check as you go!

1.4 Analysis of Truss Deformations For statically determinate trusses we have been able to use static equilibrium to obtain the bar forces. In order to calculate deformation (i.e. displacements of nodes due to loading), we need to invoke the other two great principles: constitutive behavior and compatibility. Constitutive Behavior of Bars in a Truss Will later show that the overall change in length of bar is (for a range of materials, loads)

! =FLAE

+ "#TL

Symbol Meaning Dimensions Si units

A = Cross-section area L2[ ] m2

L = Length of bar [L] m

F = Applied force [ML/T2] N/

Due to a change in temperature

Due to force

E = Young’s Modulus [ML/T2.L2] N/M2

α = Coefficient of thermal expansion

LL.![ ] M

MK

ΔT = Temperature difference ![ ] K

this is the same as Hooke’s law for a spring: F = K! for ΔT=0

! =FLAE

and rearranging gives AEL

" #

$ % ! = F

[NOTE: a real spring has added geometry]

Compatibility For Trusses For truss like structures: • bars extend or contract axially • bars rotate about pin joints • bars remain attached at joints i.e., deflections and rotations must be compatible Consider three bar truss from lecture 1, assume no temperature change:

Force (N) Length (m) FLAE

µm( )

AB 0 5 0

AC +400 10 571

BC -447 125 -714

"k" for a bar

A = 1!10"4m2

E = 70GPa

AE = 7 !106N

Note; the bar extensions are very small compared to bar lengths Consider what this implies about deformations of 3 bar truss.

we can enlarge the region of interest: assume deflections are small, circular arcs become straight lines at right angles to extension/contraction of bars: displacement diagram

Can extend to other joints by considering relative displacements. TIP: Do this on graph paper - measure deflections rather than calculate. 1.4 Concluding Remarks We have taken a quick look at how we can use equilibrium, constitutive behavior to examine the internal forces and deformations of a particular class of statically determinate structure; trusses. Most structures are not made up of discrete bars, instead they consist of continuous elements, e.g. aircraft skins, ship hulls, pressure vessel walls. We need to have analysis tools that allow us to deal with such cases. At the heart of these tools is an understanding of how forces and displacements can be understood in continuous structures. This leads to the ideas of stress and strain.

Lectures 4 and 5 Stress

Block 2: Continuum Mechanics: Stress, Strain and Elasticity We want to look more closely at how structures transmit loads. In general, structures are not made up of discrete, 1-D bars, but continuous, complex, 3-D shapes. If we want to understand how they transmit load we need new tools. To this end we will introduce the concept of stress (and later strain). See BCA, Chapters 2 and 11. Definition Stress is the measure of the intensity (per area) of force acting at a point. Use coordinate system

x, y, z! x1, x2, x3 Example:

For each square, increment of force and area given by:

!F1 = F1 nm

!A = A nm

The stress is the intensity, force per unit area, so let m and n go to infinity.

!1 = n"#m"#

lim $F1$A

units forcelength2!

" #

$

% & : N/m2: Pascals (Pa)

where

!1 is the stress at a point on the

x1 - face - Magnitude

!1 - Direction

i1 Consider a more general case:

Force F acting on x1face ( i1 is normal to plane of face).

Force, F, is a vector quantity

Resolve force into three components

F = F1i1 + F2 i2 + F3i3

Then take the limit as the force on the face is carried by a smaller set of areas.

!A" 0lim !F

!A= 0 (but stress vector has other components)

! = !11i1 + !12 i2 + !13i3 Where !1 is the stress vector and !1n are the components acting on the x1 normal face, in the x1, x2 and x3 directions. We obtain similar results if we looked at the x2 and x3 faces.

! 2 = ! 21i2 +! 22 i2 +! 23i3 on i2 face! 3 = ! 31i1 +! 32 i2 +! 33i3 on i3 face

We can write this more succinctly as:

! m = !mn in

!mn is the stress tensor,

! m is the stress vector. This method of representing a set of vector equations is an example of indicial, or tensor, notation:

Tensor (Indicial) Notation

This is a simple way to write formulae with several terms. Example:

xi = FijY j By Convention Latin subscripts

m,n, i, j, p,q, etc. Take values 1, 2, 3, (3-D) Greek Subscripts

!,",# Take values 1 or 2 (2-D)

!F1!A1

!F2!A1

!F3!A1

Subscripts or indices

Three Rules for Tensor Manipulation 1. A subscript occurring twice is a repeated (or dummy) index and is summed over 1, 2, (and)

3. Implicit summation

For

xi = FijY j = FijY jj=1

3!

i.e:

xi = Fi1Y1 + Fi2Y2 + Fi3Y3 2. A subscript occurring once in a term is called a free index, can take on the range 1, 2, (3) but

is not summed. It represents separate equations.

x1 = F11Y1 + F12Y2 + F13Y3x2 = F21Y1 + F22Y2 + F23Y3x3 = F31Y1 + F32Y2 + F33Y3

3. No index can appear in a term more than twice.

Stress Tensor and Stress Types

!mn is the stress tensor Consider a differential cubic element:

!mn

Acts in xn direction Stress acts on face with normal in the xm-direction

Types of Stress We can identify two distinct types of stress: Normal (Extensional) - acts normal to face !11,! 22, !33 - acts to extend element Shear Stress - acts parallel to face plane !12,!13,! 21,!23,! 31,!32 - acts to shear element i.e. 9 components of stress, but it turns out that not all of these are independent:

Symmetry of Stress Tensor Consider moment equilibrium of differential element:

Take moments about C +

2! 23"x 3"x112"x2

# $ %

& ' ( ) 2! 32"x2"x1

12"x 3

# $ %

& ' ( = 0

simplifies to:

! 23 = ! 32 Thus, in general !mn = ! nm Stress tensor is symmetric. Six independent components of the stress tensor.

!11!22!33

!12!23!31

===

!21!32!13

"

#

$ $

%

& ' '

What happens if stress varies with position? Stress Equations of Equilibrium Consider equilibrium in x1direction; let stress vary.

x3

x2

x1

!11

!21

!31

! 11 +"!11"x1

#x1

! 21 +"! 21"x2

#x2

! 31 +"! 31"x3

#x3

#x3

#x2

#x1

f1

Force equilibrium ! F1 = 0 +

!11 + "!11"x1

dx1# $ % &

' ( dx2dx3( ) )!11 dx2dx3( )

+ !21 + "! 21dx2

dx2# $ % &

' ( dx1dx2( ) )!21 dx1dx3( )

+ !31 +"! 31"x3

dx3# $ % &

' ( dx1dx2( ) ) !31( ) dx1dx2( )

+ f1dx1dx2dx3 = 0

Canceling out terms and dividing through by the common: dx1, dx2, dx3 . Gives:

!"11!x1

+!"21!x2

+!"31!x3

+ f1 = 0 x1 direction

Similarly, !"12

!x1+!" 22

!x2+!" 32

!x3+ f2 = 0 x2 direction

dx1 dx2 dx3 common to all terms

fn is a body force - acts over volume - gravity, electromagnetic forces, acceleration: it is a force per unit volume

!"13!x1

+!" 23!x2

+!"33!x3

+ f3 = 0 x3 direction

These are the three equations of stress equilibrium. In tensor form:

!"mn!xm

+ fn + 0

Stress Transformations Just as we need to resolve forces with respect to structural axes - need to resolve stress. Identify directions which "see" maximum stress Two Dimensional Case: Consider a unit square (of unit depth dx3)

What is the stress acting on a plane at an arbitrary angle ! ? Consider equilibrium of a triangular, wedge shaped element:

• Align ˜ x 1 perpendicular to cut face, ˜ x 2 parallel to cut face

• The element is of uniform depth dx3 • Must still be in equilibrium of forces:

F = 0! Σ(Stress component x length x depth) note: dx2 = d˜ x 2 cos!

dx1 = d˜ x 2 sin! Equation of equilibrium in ˜ x 1 direction: ˜ ! 11dx3d˜ x 2 " !11dx3 #˜ x 2 cos$[ ]cos$ " !22dx3 #˜ x 2 sin$[ ]sin$ " !12dx3 #˜ x 2 sin$[ ]cos$ " !21dx3 #˜ x 2 cos$[ ]sin$ = 0

cancelling out !˜ x 2 and combining ˜ ! 11 = cos2 " !11 + sin2 " ! 22 + 2 cos" sin" !12 This can be done for ˜ ! 12 and ˜ ! 22 , e.g.

˜ ! 12 = "!11 cos# sin# +! 22 sin# cos# + !12 cos2# " sin2 #( ) note one can always take equilibrium of an element Tensor form Transformation requires direction cosines (see below) Stresses are second order tensors (2 subscripts) and require two direction cosines for transformation. Thus:

˜ ! mn = ! ˜ m p! ˜ n q! pq

• 2nd order tensor requires 2 direction cosines • transforms stress from xn system to ˜ x m system

one angle for area, one for component of force

In 2-D ˜ ! "# = ! ˜ " $! ˜ # %!$%

Direction Cosines

Note: two angles: one for area; one for resulting force

area Resolved component

Define direction cosine as:

!m˜ n = cos xm ˜ x n[ ] such that

xm = !m˜ n x ˜ n

or ˜ x n = ! ˜ n m xm

where we mean

x1 = !1˜ 1 ̃ x ̃ 1 + !1˜ 2 ̃ x ̃ 2 + !1˜ 3 ̃ x ̃ 3

x2 = !2˜ n ̃ x ̃ n x3 = !3˜ n ̃ x ̃ n

We could use the same idea for forces:

˜ F n = ! ˜ n mFm

First order tensor is a vector (one subscript) requires one direction cosine for transformation. Angles measured positive counter-clockwise. ! ˜ m n = !n ˜ m Since cosine is an even function cos !( ) = cos "!( ) But ! ˜ m n ! ! ˜ n m - different angles 2-D Example

˜ y 1 !˜ 1 1 = cos! !˜ 1 2 = cos 90 "!( ) = sin! !˜ 1 3 = cos90 = 0

˜ y 2 !˜ 2 1 = cos "(90 +!)( ) !˜ 2 2 = cos! !˜ 2 3 = cos90 = 0

˜ y 3 !˜ 3 1 = cos90 = 0 !˜ 3 2 = cos 90 = 0 !˜ 3 3 = cos! = 1

y3

Angle between

Tensor equations

y2y1

One further way to transform stress is to use "Mohr's Circle" graphical method for rotating 2nd order tensors in 2-D coordinate system

Mohr's Circle for Stress Mohr's circle is a geometric representation of the 2-D transformation of stresses. Construction: Given the state of stress shown below, with the following definition (by Mohr) of positive and negative shear: "Positive shear would cause a clockwise rotation of the infinitessimal element about the element center." Thus: !21 (below) is plotted positive !12 (below) is plotted negative:

Following these rules for the sign of stress do the following:

1. Plot !11, -!12 as point A

2. Plot !22 , !21 as point B

3. Connect A and B. 4. Draw a circle with a diameter given by the line AB about the point where the line AB crosses the horizontal axis (denote this as C).

To read off the stresses for a rotated system: 1. Note that the vertical axis is the shear axis; the horizontal axis is the extensional stress axis. 2. Positive rotations are measured anti-clockwise as referenced to the original system and thus to the line AB.

3. Rotate line AB about point C by the angle 2θ where θ is the angle between the un-rotated and rotated system.

4. The points D and E where the rotated line intersects the circle are used to read off the stresses in the rotated system. The vertical location of D is ! ˜ " 1 2; the horizontal location of D is ˜ ! 1 1. The vertical location of E is ˜ ! 2 1, the horizontal location of E is ˜ ! 2 2 (recall Mohr's definition).

We can also immediately see the following: 1. The principal stresses, ! I and ! II are defined by the points G and F (along the horizontal axis ˜ ! 1 2 = 0 ). The rotation angle to the principal axis is θp which is half the angle from the line AB to

the horizontal line FG. 2. The maximum shear stress is defined by the points H and H' which are the endpoints of the vertical line. The line is orthogonal to the principal stress line and thus the maximum shear stress acts along a plane 45° (= 90°/2) from the principal stress system.

Shear

B

A

Mohr's Circle (Principal Stresses)

C Ext

2!p

H

H'

FG

SESG1001 Lectures 7 & 8: Strain

We have examined stress, the continuum generalization of forces, now let's look at the

continuum generalization of deformations:

Definition of Strain

Strain is the deformation of the continuum at a point. Or, the relative deformation of an

infinitessimal element.

Two ways that bodies deform: By elongation and shear:

Elongation (extension, tensile strain)

Tensile strain can be thought of as the change in length relative to the original length

tensile strain =Ldeformed ! Lundeformed

Lundeformed

but body can also deform in shear – angles change

Shear

This produces an angle change in the body (with no rotation for pure shear)

Consider infinitesimal element.

Undeformed:

Deformed :

Consider change in angle BPA = φ

φ = undeformed - deformed

NOTE: By convention positive strain is a reduction in angle (consistent with positive

shear stresses)

Δ =

!2" !2" #

$ % &

' ( )

* + ,

- . / = # radians

NOTE: Strain is a non-dimensional quantity

(Just as for stress, strain varies with position and direction. Use tensor formulation to

describe)

Formally Define !mn= strain tensor

There is a general definition of strain, but usually we are concerned with small strains:

change in length < 10% (0.1), change in angles < 5% (0.05)

Good for range of use of most engineering materials, most structural applications.

Allows us to neglect higher order terms and leads to:

Strain - Displacement Relations (for small strains)

Conceptually, want to separate out rigid body translations from

deformations

!11 = relative elongation in x1 direction

!22 = relative elongation in x2 direction.

!33 = relative elongation to x3 direction.

Consider infinitesimal element, side length dx1 undergoing displacements and deformations in the x1 direction defined by

u1 is a field variable = u1 x1,x2, x3( ) , i.e. displacements vary with position

let u1 be the displacement of the left-hand side of the infinitesimal element.

And

u1 + !u1!x1

dx1"

# $

%

& ' is the displacement of right-hand side

Recall:

!11 " !1 = ! deformed # ! undeformed! undeformed

!11 =dx1 + u1 + $u

$x1dx1 # u1

%

& '

(

) * # dx1

dx1

= $u1$x1

Similarly, ε22 = elongation in x2

!22 = "u2"x2

and

!33 = "u3"u3

Now for shear strain Need to be careful which angles we choose to define shear strain

!12 =12angle change =

12

/ 0 12" #

$ % etc.

= 12

= 12

!2" !2" #12

$ % &

' ( )

$ % &

' ( )

( undeformed - deformed )

Rigid body translation

! undeformed

!22 = ! deformed " ! undeformed! undeformed

=dx2 + u2 + #u2

#x2dx2 " u2

$

% &

'

( ) " dx2

dx2

e.g:

!12 + !21 = total angle change in x1 " x2 plane!13 + !31 = total angle change in x1 " x3 plane!23 + !32 = total angle change in x2 " x3 plane

NOTE: The strain tensor is defined such that there seem to be two parts in each angle change. But the stress tensor is symmetric and we would like the strain tensor also to be symmetric. Thus:

!12 = !21 =12angle change in x1 " x2 plane( )



Consider diagonal of parallelogram undergoing pure shear:

Angular charge of the x1 edge in the x2 direction

Formal Definitions

!12 = "1 + "2

Make small angle approximation:

tan! = ! Hence:

Angles change by same amount.

!1 =u1 + "u1

"x2dx2 # u1

dx2

!2 =u2 + "u2

"x1dx1 # u2

dx1

!12 = 12

"u2"x1

+ "u1"x2

#

$ %

&

' ( = !21

and similarly:

!23 = 12

"u2"x3

+ "u3"x2

#

$ %

&

' ( = !32

!31 = 12

"u3"x1

+ "u1"x3

#

$ %

&

' ( = !13

or

!ij = 12

"ui"x j

+"u j"xi

#

$ % %

&

' ( ( The strain tensor for small strains

Strain displacement relation 6 independent strain components (We have dealt with elongation and shear strains but remember, we were concerned to eliminate translation and rotational displacements) (- translation accounted for by subtracting u1, u2 etc. from strain equations) Rotation

!12

!12 = 12

"2# "2#

u2 + $u2$x1

dx1 # u2%

& '

(

) *

dx1+

u1 + $u1$x2

dx2 # u1%

& '

(

) *

dx2

+

,

- - - -

.

/

0 0 0 0

1

2 3 3

4 3 3

5

6 3 3

7 3 3

%

&

' ' ' ' '

(

)

* * * * *

!1 =u2 +

"u2"x1

dx1 # u2

dx1="u2"x1

as before

or !2 =" u1 +

#u1#x2

dx2 " u1$ % &

' ( )

dx2="#u1#x2

But may also have shear deformation of the cube !1 "!2 Define average rotation (of diagonal)

! =12

"u2"x1

#"u1"x2

$ % &

' ( )

note (-) sign

Transformation of Strain Identical to transformation of stress

˜ ! mn = ! ˜ m p! ˜ n q!pq

Mohr's Circle in 2-D Principal strains/directions -no associated shear strains

!1 =!2

Negative because it acts to increase enclosed angle - consistent with shear strain

Alternative notations for stress and strain Engineering Notation (see Benham, Crawford and Armstrong)

x1! xx2 ! yx3! z

Tensor Engineering!11 ! x!22 ! y!33 !z!12 ! xy!13 ! xz!23 ! yz

In addition τ (tau) is often used for shear stress

! xy = " xy! xz = " xz! yz = " yz

For stress this is just the substitution of different symbols, however for strain there is a fundamental difference from tensor notation. Engineering shear strain = total angle change Tensor shear strain = 1/2 (angle change)

Tensor Engineering

!11 " ! x

!22 " !y

!33 " ! z

!12 "12!xy or

12# xy

!13 "12! xz or

12# xz

!23 "12! yz or

12# yz

Matrix Notation

!mn =!11 !12 !13!21 !22 !23!31 !32 !33

"

#

$ $

%

&

' '

!mn =!11!12 !13!21 !22 !23!31 !32 !33

"

#

$ $ $ $

%

&

' ' ' '

- principal values are eigenvalues - principal directions are eigenvectors Next we will link stress and strain by constitutive behavior and to do this we will need to look at material properties.

Total angle change

SESG1001 Lectures 9 & 10: Elastic Stress-Strain Relations

These lectures will complete our quick journey through continuum mechanics, to provide

you with a continuum version of a constitutive law - at least for linear elastic materials

! pq = E ?"mn

Increasingly, materials are designed along with the structure, you need insight into what

contributes to material properties. These lectures will link to topics you will learn about

in materials lectures and form a foundation for material selection in design.

Engineering Elastic Properties of Materials In order to understand how we link stress and strain we need to understand that there are

two points of view to this matter. There is the experimental point of view that some

properties (behaviors) are easier to measure than others, and there is the mathematical

point of view that some representations of physical phenomena are mathematically easier

to handle than others. In the present case, engineering elastic constants are derived from

an experimental point of view, whereas the stress and strain tensors, are mathematically

useful. We need to resolve these two points of view.

Young’s modulus and Poisson’s ratio From materials lectures and the materials and solid mechanics labs you will become

familiar with at least two elastic constants.

If we apply a uniaxial tensile stress

! L to a constant cross-section rod of material, we

will obtain a biaxial state of strain, consisting of an axial tensile strain

!L and a

transverse strain

!T . The axial strain will be tensile for a tensile applied stress, and the

Elasticity Where does it come from?

transverse strain will usually be compressive. We can measure the strains using

resistance strain gauges.

For many materials, over some range of applied stress, the applied stress and the resulting

strains will follow a linear relationship. This observation is the basis for the definition of

the engineering elastic constants. The Young’s modulus, E, is defined as the constant of

proportionality between a uniaxial applied stress and the resulting axial strain, i.e:

! L = E"L

Note. This only applies for a uniaxial applied stress, and the component of strain in the

direction of the applied stress.

We can also define the Poisson’s ratio,

! , as the ratio of the transverse strain to the axial

strain. Since for the vast majority of materials the transverse strain is compressive for a

tensile applied stress, the Poisson’s ratio is defined as the negative of this ratio, to give a

positive quantity. I.e:

! = "#T#L

Typically, for metals, ν = ~ 0.3-0.4.

A similar process, of performing experiments in which a well-defined stress state is

applied and the resulting strain state is measured leads us to define other elastic constants,

we will concern ourselves with only one:

The Shear Modulus Application of a state of pure shear, leads to a shear strain:

Note angles are exaggerated in the figure.

An applied shear stress leads to an applied shear strain. The shear strain,

! , is defined in

engineering notation, and therefore equals the total change in angle:

! = " .

Consistent with the definition of the Young’s modulus, the Shear modulus. G, is defined

as:

! = G"

Again, note, that this relationship only holds if a pure shear is applied to a specimen.

For reasons that will become apparent shortly, The Young’s modulus, Shear Modulus and

Poisson’s ratio are linked. For most materials Poission’s ratio’s are approximately 0.33,

and for these materials

G ! 38E . However, values of the Young’s modulus can vary

widely.

The Young’s Moduli of Engineering Materials

A useful way of representing the range of Young’s moduli is to plot them for all the

classes of material against some other material property. Since we are often interested in

lightweight structures, we will choose the density of the materials for this purpose.

From: Material Selection in Mechanical Design, M.F Ashby, Pergamon Press, Oxford,

1992

Note, how the different classes of material tend to cluster: Metals, Polymer, Glasses and

ceramics have similar ranges of density and modulus. There are also some materials that

have quite wide ranges of moduli (and densities), while others are relatively narrowly

banded. Finally note how wide an overall range of moduli is represented, from 0.01 GPA

for foams to 1000 GPA for diamond. The range of densities is somewhat less, but still

spans more than two orders of magnitude. See SESG1003 for more details.

Generalized Hooke's Law We have met the engineering elastic constants, Young's moduli, Shear Moduli and

Poisson's ratio's, and understand that many structural materials behave elastically over

some range of stress and strain.

Now we want to add a mathematical formalism to this physical basis, i.e. our 3rd great

principle, that of constitutive behavior.

A couple of problems we would like to be able to solve: a) what are the strains due to

stresses at an arbitrary angle to the loading direction:

b) We would also like to be able to deal with any state of multiaxial stress and convert to

the resulting strains, or vice versa. To do this we need to revisit tensor stress and strain.

i.e. we want the elastic property that links the stress tensor to the strain tensor:

σmn=Emnpqεpq

Where Emnpq is the 4th order (i.e. 4 subscripts) ELASTICITY (or STIFFNESS) tensor.

e.g:

.

!11 = E1111"11 + E1112"12 + E1113"13 ( p =1,sum on q)

+E1121"21 + E1122"22 + E1123"23 p = 2,sum on q( )

+E1131"31 + E1132"32 + E1133"33 p = 3,sum on q( )

4th order tensor has 81 components, m,n,p,q =1, 2 and 3 therefore 34 = 81 terms

But fortunately there are symmetries which reduce this to only (!) 21 independent

components of the elasticity tensor. In matrix form, this can be written as:

!11

!22

!33

!23

!13

!12

"

#

$ $ $

%

$ $ $

&

'

$ $ $

(

$ $ $

=

E1111 E1122 E1133 2E1123 2E1113 2E1112

E1122 E2222 E2233 2E2223 2E2213 2E2212

E1133 E2233 E3333 2E3323 2E3313 2E3312

E1123 E2223 E3323 2E2323 2E1323 2E1223

E1113 E2213 E3313 2E1323 2E1313 2E1213

E1112 E2212 E3312 2E1223 2E1213 2E1212

)

*

+ + + + + + +

,

-

.

.

.

.

.

.

.

/11

/22

/33

/23

/13

/12

"

#

$ $ $

%

$ $ $

&

'

$ $ $

(

$ $ $

Since it links strains to stresses, Emnpq is also termed the "stiffness tensor". By inverting

the matrix we could also obtain the tensor linking stresses to strains, i.e. εmn=Smnpqσpq

Where Smnpq is the "Compliance" tensor (Why S is for compliance and E is used for

stiffness is unclear to me!)

Even with the simplifications, 21 independent terms seems rather too many to have to

deal with!. Let's go back to the engineering elastic constants and see if we can see how to

simplify this list further.

We know that there are several different classes of material. Most metals and ceramics

are isotropic, that is they have the same properties in any direction that you measure. By

contrast, fiber-reinforced composites may have different properties in different directions,

i.e. they are anisotropic, generalized elasticity provides the framework to analyze these.

Elasticity of Isotropic Materials

In SESG1001 we will only look at isotropic materials. However, we do want to be able

to consider the case in which an element of a structure is loaded by all possible

components of stress and we want to know the resulting strains. Also let's ignore thermal

expansion strains for the time being. We have six components of stress producing six

components of strain, therefore we need a six by six matrix to link them: ! x

! y

!z" zy" zx

" xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

) x

) y

) z

* zy

* zx

* xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

We also know that for small strains, and elastic materials the contributions of the separate

components of stress will superimpose. So let's consider the case of only σx applied and

all the other components of strain are zero:

! x

! y

!z" zy" zx

" xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

) x

0

0

0

0

0

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

Taking each component in turn:

What does εx equal?. This just reduces to a 1-D tensile test, so !x =" xE

And εy , εz are given by the Poisson contractions, so: !y = ! z = "#! x = "#$ xE

This allows us to fill in the first line of the matrix, and also by noticing that we could

have equally well applied σy or σz, and obtained similar relationships, we can fill in all of

the top left hand quadrant of the matrix:

! x

! y

!z

" zy

" zx

" xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

1E

)*E

)*E

_ _ _)*E

1E

)*E

_ _ _)*E

)*E

1E

_ _ _

_ _ _ _ _ _

_ _ _ _ _ _

_ _ _ _ _ _

#

$

% % % % % % % %

&

'

( ( ( ( ( ( ( (

+ x

+ y

+ z

, zy

, zx

, xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

If instead of applying an extensional stress we applied a shear stress, we know that the

shear stress and shear strain are linked by the shear modulus, so:

! x

! y

!z

" zy

" zx

" xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

1E

)*E

)*E

_ _ _)*E

1E

)*E

_ _ _)*E

)*E

1E

_ _ _

_ _ _ 1G

0 0

_ _ _ 0 1G

0

_ _ _ 0 0 1G

#

$

% % % % % % % % %

&

'

( ( ( ( ( ( ( ( (

+ x

+ y

+z

, zy

, zx

, xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

Finally we note that for isotropic materials the application of an extensional stress does

not result in a shear strain or vice versa, so the top right and bottom left quadrants are

populated by zeros:

! x

! y

!z

" zy

" zx

" xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

1E

)*E

)*E

0 0 0)*E

1E

)*E

0 0 0)*E

)*E

1E

0 0 0

0 0 0 1G

0 0

0 0 0 0 1G

0

0 0 0 0 0 1G

#

$

% % % % % % % % %

&

'

( ( ( ( ( ( ( ( (

+ x

+ y

+z

, zy

, zx

, xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

We have three separate elastic constants required, i.e: E, ν and G. However if we go

back to our knowledge of stress and strain transformation we can reduce this further.

Remember that the application of a shear stress can be thought of as a shear stress

resulting in a shear strain:

! ="G

or a biaxial stress state, of a combined tension and compression, at 45 degrees to the axis

of pure shear, i.e:

In terms of the Mohr's circles of stress and strain these appear as (note the factor of two

for the representation of shear strain on the Mohr's circle):

For the biaxial tensile and compressive stress the resulting (principal) strain is given by:

!I =" IE

#$" IIE

but for the case of pure shear: !I =12" (From Mohr's Circle -remember the

factor of two between tensor and engineering shear strain)

and: ! = " II = #" I

$% = 2 1E

+&E

' (

) * ! + G =

E2 1 +&( )

So we actually only have two independent elastic constants for an isotropic material.

Note that this only applies for isotropic materials

If we want to go in the reverse direction (i.e. have known strains and want to calculate

stresses) we need to invert the matrix of elastic constants. Note, this situation may arise

because we can experimentally measure strains using strain gauges. The inverse matrix

is usually expressed in terms of groupings of the elastic constant, known as Lamé's

constants, µ& ! , where:

! II = "! I

µ =!

2 1+ "( ) =G

# ="E

1 +"( ) 1 $ 2"( )

Thus:

! x

! y

! z

" xy

" xz

"yz

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

) + 2µ ) ) 0 0 0

) ) + 2µ ) 0 0 0

) ) ) + 2µ 0 0 0

0 0 0 µ 0 0

0 0 0 0 µ 0

0 0 0 0 0 µ

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

* x

* y

*z

+ xy

+ xz

+ yz

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

We can also include the effect of thermal expansion:

! x

! y

!z

" zy

" zx

" xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

=

1)

*+)

*+)

0 0 0

* +)

1)

* +)

0 0 0

*+)

*+)

1)

0 0 0

0 0 0 1G

0 0

0 0 0 0 1G

0

0 0 0 0 0 1G

#

$

% % % % % % % % %

&

'

( ( ( ( ( ( ( ( (

, x

, y

, z

- zy

- zx

- xy

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

+

.

.

.

0

0

0

#

$

% % % % % % %

&

'

( ( ( ( ( ( (

/T

We are now equipped to analyze stresses and strains in simple continuous structures

SESG1001 Lecture 11 1-D Structures In the rest of the course we are going to utilize the principles we have learnt so far (i.e the 3 great principles and their embodiment in the equations of elasticity) in order to be able to analyze simple structural members. These structural members are: Rods, Beams, Shafts and Columns. The key feature of all these structural members is that one dimension is longer than the others (i.e. they are one dimensional). There is a basic logical set of steps that we will follow for each in turn: 1) We will make general modeling assumptions for the particular class of structural member In general these will be on: a) Geometry b) Loading/Stress State c) Deformation/Strain State 2) We will make problem-specific modeling assumptions on the boundary conditions that apply (such as pins, clamps, rollers that we encountered with truss) a.) On stresses b.) On displacements 3) We will apply an appropriate solution method: 1.) Exact/analytical (SESG1001) 2.) Numerical (2 nd , 3rd and 4th year classes) - approximate. Such as energy

methods (finite elements - use computers) Let us see how this works:

At certain locations in the structure

Rods The first 1-D structure that we will analyzes is that of a rod (or bar), such as we encountered when we analyzed trusses. We are interested in analyzing for the stresses and deflections in a rod. First start with a working definition - from which we will derive our modeling assumptions: "A rod (or bar) is a structural member which is long and slender and is capable of carrying load along its axis via elongation" Modeling assumptions a.) Geometry

L = length (x1 dimension) b = width (x2 dimension) h = thickness (x3 dimension) assumption: L much greater than b, h (slender member) (think about the implications of this - what does it imply about the magnitudes of stresses and strains) b.) Loading - loaded in x1 direction only Results in a number of assumptions on the boundary conditions

Similarly on the x3 face - no force is applied

!31 = 0 = !13

!32 = 0 = !23

!33 = 0

on x1 face - take section perpendicular to x1

and c.) deformation Rod cross-section deforms uniformly (is this assumption justified? - yes, there are no shear stresses)

!21 = 0!22 = 0!23 = 0

!12 = 0!13 = 0

!11dA = PA"

!11dx2dx3"" = P

#!11 =Pbh

= PA

So much for modeling assumptions, Now let's apply governing equations and solve.

1. Equilibrium

!"mn!xn

+ fn = 0

only !11 is non-zero

!"11!x1

+ f1 = 0

!"11!x1

= 0#"11 = constant = P A

Constitutive Laws stress - strain equations:

!11 = 1E

" # $

% & ' (11

!22 =)*E

" # $

% & ' (11

!33 =)*E

" # $

% & ' (11

Now apply strain – displacement relations:

f1 = body force =0

Compliance Form

!mn = 12

"um"xn

+ "un"xm

#

$ %

&

' (

!11 = "u1"x1

, !22 = "u2"x2

, !33 = "u3"x3

Hence:

!11E

="u1"x1

# PAE = "u1

"x1

integration gives:

u1 =Px1AE

+ g(x2 , x3)

Apply B. C.

u1 = 0 @

x1 = 0 ! g(x2,x3) = 0

! u1 = Px1AE

similarly

u2 = !"PAE

x2

u3 = !"PAE

x3

check:

!12 = 12

"u1"x2

+ "u2"x1

#

$ %

&

' ( = 0 √

Further assumptions (Closer inspection reveals that our solutions are not exact.) 1) Cross section changes shape slightly. A is not a constant.

If we solved the equations of elasticity simultaneously, we would account for this. Solving them sequentially is acceptable so long as deformations are small. (δA is second order.)

2) At attachment point boundary conditions are different from those elsewhere on

the rod.

We deal with this by invoking St. Venant's principle: "Remote from the boundary conditions internal stresses and deformations will be insensitive to the exact form of the boundary condition." And the boundary condition can be replaced by a statically equivalent condition (equipollent) without loss of accuracy. How far is remote?

This is the importance of the "long slender" wording of the rod definition. This should have been all fairly obvious. Next time we will start an equivalent process for beams - which require a little more thought.

u1 = 01, u2 = 01, u3 = 0

SESG1001 Lectures 12 and 13 Beam Theory: Bending Moment and Shear Force

Distributions

Reading: Benham, Crawford and Armstrong, 1.10, Ch 6 and Ch 7 A beam is a structural member which is long and slender and is capable of carrying bending loads. i.e loads applied transverse to its long axis. Beams are very common structural members and are to be found across all areas of engineering. Wings, masts, gas turbine blades even whole ships, aircraft or spacecraft can be treated as beams to varying degrees. 1. We will continue, as for 1-D rods to make a set of modeling assumptions: a.) Geometry, slender member, L>>b, h

At this stage, we will assume arbitrary, symmetric cross-sections, i.e.:

b) Loading:

Similar to rods (traction free surfaces) but applied loads can be in the z direction c) Deformation

We will talk about this later 2) Boundary conditions As for rods, trusses

Pinned, simply supported, point load Cantilever, built in support, distributed load Analysis approach: Draw free body diagram, apply equilibrium to determine reactions. 3.) Governing equations Equations of elasticity (stress, strain, constitutive behaviour) 4.) Solution Method

Exact (exactly solve governing differential equations) Approximate (use numerical solution)

Before moving through this process we need to understand how beams transmit load. Internal Forces and Moments Apply methods of sections to beam (also change coordinate system – to x, y, z – consistent with BCA). Method exactly as for trusses. Cut structure at location where we wish to find internal forces, apply equilibrium, obtain forces. In the case of a beam, the structure is continuous, rather than consisting of discrete bars, so we will find that the internal forces (and moments) are, in general, a continuous function of position.

Internal forces

Opposite

directions on two faces - equilibrium

where M = Bending movement Beam S = Shear force Bending "beam bar" F = Axial force bar, rod Also drawn as:

Examples of calculating shear force and bending moment distribution along a beam. Example 1. A cantilever beam:

Free Body Diagram (note moment reaction at root)

Equilibrium:

HA = 0, VA = P, MA = !PL Take cut at point X ,distance x from left hand end (root). 0< x < L. Replace the effect of the (discarded) right hand side of beam by an equivalent set of forces and moments (M, S, F) which vary as a function of position, x.

Apply equilibrium

Fx = 0! "+

F(x) = 0

Fz = 0 # +! P $ S(x) = 0 S( x) = PMX! = 0 PL + M(x) $ Px = 0 M(x) = $P(L $ x)

+

Draw "sketches" - bending moment, shear force, loading diagrams

Shear Force Diagram

Bending moment diagram:

NOTE: At boundaries values go to reactions (moment at root, applied load at tip). These representations are useful because they provide us with a visual indication of where the internal forces on the beam are highest, which will play a role in determining where failure might occur and how we should design the internal structure of the beam (put more material where the forces are higher).

Example 2: A simply supported beam

Free body diagram:

take cut at 0< x< L/2

Equilibrium →F(x) = 0

! P2" S(x) = 0 # S(x) = P

2

Mx = 0 #" P2x + M (x) = 0# M (x) = Px

2

take cut at

L2

< x < L

Apply equilibrium (moments about X):

!F+ = 0 : P2

" # P # S(x) = 0$ S(x) = # P2

M = 0 :" # P2x + P x # L

2% & '

( ) * + M (x) = 0

M (x) = ! P2(x ! L)

= P2(L ! x)

+

0 < x < L

Draw Loading, Shear Force and Bending Moment Diagrams

Observations • Shear is constant between point loads • Bending moment varies linearly between discrete loads. • Discontinuities occur in S and in slope of M at point of application of concentrated

loads.

• Change in shear equals amount of concentrated loads. • Values of S & M (and F) go to values of reactions at boundaries

Distributed loads e.g. gravity, pressure, inertial loading. Can be uniform or varying with position.

q(x) = q 0 q(x) = q0 1!xL

" #

$ % &= q0@ x = 0, = 0@ x = L

[ q!] = [force/length]

We deal with distributed loads in essentially the same way as for point loads.

Example: Uniform distributed load, q (per unit length), applied to a simply supported

beam.

Free body diagram:

Apply method of sections to obtain bending moments and shear forces:

Apply equilibrium:

! + qL2" # Sx # q

0

x$ dx = 0

% qL2

# qx = S(x)

Mx! " qLx2

+Mx + q0

x# xdx = 0

" qL2

+Mx = qx2

2= 0$Mx =

qLx2

" qx2

2

+

Plot:

Observations

• Shear load varies linearly due to uniform distributed load. • Moment varies quadratically (parabolically) over region where distributed load

applied

• This suggests a relationship between M & S & P

General Relation Between q, S, M

Consider a beam under some arbitrary variable, distributed loading q(x):

Consider an infinitesimal element, length, dx, allow F, S, M to vary across element:

Now use equilibrium, replace partial derivatives by regular derivatives (F, S, M varying

only in x).

Fx! = 0"+: # F + F +

dFdx

dx = 0 dFdx

= 0

M + !M!x

dx

S + !S!xdx

F + !F!x

dx

Fz! = 0"+ S # S # dSdxdx + q(x)dx = 0

dSdx

= q(x)

M0! " / M + / M + dMdx

dx " S dx2" (S + dS

dxdx) dx

2= 0

note: q(x) has no net moment about O. dMdx

dx ! Sdx +12dSdx(dx)2 = 0

but (dx)2 is a higher order (small) term

" dMdx

= S

Summarizing:

dFdx

= 0 (unless a bar)

dSdx

= q

dMdx

= S (and d2Mdx2

= q)

This is a useful check, and also useful to automate process

Now we will move on to analyze how these internal forces and moments correspond to

the internal stress distribution and deformation of the beam.

+

Engineering Beam Theory Reading: Benham, Crawford and Armstrong Ch 6 and Ch 7

We have looked at the statics of a beam and we have seen that loads are transmitted by

internal forces: shear forces and bending moments.

Now look at how these forces imply stresses, strains, and deflections.

Recall model assumptions: slenderness

Geometry

L >> h,b

Loads

Leads to assumptions on stresses

Load in x - z plane →

! yy = ! xy = ! yz = 0

Also L >> h,b implies ! xx ,! xz >> ! zz

Consider moment equilibrium of a cross section of a beam loaded by some distributed

stress σzz, which is reacted by axial stresses on the cross section, σxx

General, symmetric, cross section

MX = 0 = ! zzL "! xx# h$! zz! xx

% hL% 0

i.e. geometry of beam implies σxx >> σzz

Assumptions on Deformations: The key to simple beam theory is the Bernoulli - Euler hypotheses (1750)

"Plane sections remain plane and perpendicular to the mid-plane after deformation."

It turns out that this is not really an assumption at all but a geometric necessity, at least

for the case of pure bending.

To see what the implications are of this, consider a beam element, which undergoes

transverse (bending)

deformation.

Obtain deflection in x-direction, displacement u of point k to K, defined by rotation

! .

Note, key assumption, "plane sections remain plane"

w= deflection of midplane/midline (function of x only)

axial displacement, u, of an arbitrary point on the cross-section arises from rotation of

cross, sections

u= -z sin θ

Note, negative sign here due to use of consistent definitions of positive directions for w, x

and dw/dx.

If deformations/angles are small sin! "!

! " dwdx

# u = $z dwdx

(1)

Hence obtain deformation field

u x,y,z( ) = !z dwdx

v x,y,z( ) = 0 nothing happening in y direction

w x,y,z( ) = w x( )

Hence we can obtain distributions of strain. (compatible with deformation). In absence of

deformations in transverse directions partial derivatives can be replaced by regular

derivatives, i.e.

! _! _

" d _d _

. Hence.

Deflection out of original plane i.e., cross sections remain rigid (2)

!xx = "u"x

= # d2wdx2

(3)

!yy = "v"y

= 0, !zz = "w"z

= 0

$ xy = 2!12 = "u"y

+ "v"x

%

& '

(

) * = 0, $ yz = "v

"z+ "w"y

%

& '

(

) * = 0

$ xz = "u"z

+ "w"x

% & '

( ) * = # dw

dx+ dwdx

% & '

( ) * = 0

If no shear - consistent with B-E assumption of plane sections remaining plane.

for constant bending moment: dMdx

= S = 0 (will revisit for S ! 0)

Next use stress - strain (for isotropic material)

! xx =" xx

E

! yy = #$ " xx

E

! zz = #$ " xx

E

% xy =& xyG

% yz =& yzG

% xz =& xzG

The inconsistency on the shear stress/strain arises from the plane/sections remain plane

assumption. Does not strictly apply when there is varying bending moment (and hence

non-zero shear force). However, displacements due to

! xz are very small compared to

those due to

!xx , and therefore negligible.

Note inconsistency -

! xz = 0, " xz # 0 (shear forces are non zero)

(4)

Finally apply equilibrium:

!"mn!xm

+ fm = 0

!" xx!x

+!# xy!y

+!# zx!z

= 0$!" xx!x

+!# zx!z

= 0

!" xy!x

+!# yy

!y+!" zy!z

= 0 $ 0 = 0

!" xz!x

+!" yz!y +

!# zz!z

= 0 $!" xz!z

= 0

Summarizing:

u x,y,z( ) = !z dwdx

(1)

w x,y,z( ) = w x( ) (2)

!xx = "z d2wdx2

(3)

! xx =" xxE (4)

!" xx!x

+!# zx!z

= 0 (5)

5 equations, 5 unknowns: w, u, ! xx , " xx , # xz

Solutions: Stresses & deflections

First consider relationship between stresses and internal force and moment resultants for

beams (axial force, F, shear force, S, and bending moment M). Consider rectangular

cross section (will generalize for arbitrary cross-section shortly)

(5)

Resultant forces and moments are related to the stresses via considerations of

equipollence:

F =!h2

h2" #xxbdz (6)

S = ! "xz!h2

h2# bdz (7)

M = ! "xxbzdz!h2

h 2# (8)

Now we combine and substitute between the equations derived above.

combine (3) and (4) to obtain:

! xx = E" xx = #Ezd2wdx2

Now substitute this in (7)

F = !Ed2wdx2

zbdz = !Ed2wdx2

z2

2b

"

# $ $

%

& ' ' !h

2

h2(

! h 2

h 2

= 0

Which is correct, since there is no axial force in the case of pure beam bending.

Similarly in (8)

M = Ed2wdx2

z 2bdz!h2

+h2"

.

We can write this more succinctly as:

M = EId2wdx 2

Where

I = z2bdz!h2

h2"

is the second moment of area (similar to moment of inertia in dynamics)

Recall that ! xx = "Ezd2wdx2

= "EzMEI

More general definition of I :

See Benham, Crawford and Armstrong Appendix A

For general, symmetric, cross-section

curvature

this is the moment curvature relation - positive moment results in

positive curvature (upward). The quantity “EI” defines the

stiffness of beam – sometimes called “flexural rigidity” – note

that it depends on both the shape and the material property, E.

!xx = "MzI

This is the moment - stress relationship

Implies a linear variation of stress, Maximum stress at edges (top and bottom) of beam

Iyy = z2A! dA =

"h 2

h 2! z2

"y(z)

+y(z)! dydz Iyy is second moment of area about y axis.

Note that there is also a moment of inertia about the z axis – Izz. In general we will be

concerned about the possibility of bending about multiple axes, and the bending of non-

symmetric cross-section. However, we will not consider these cases in SESG1001, see

2nd year structures courses.

Note, for particular case of a rectangular section I = 112bh3

We need to define centroid: centre of area (analogous to centre of mass). Position

calculated by taking moments of area about arbitrary axis. Hence z position given by:

zCentroid =

zdAA!

dAA!

In bending centroid is unstrained – also known as the “neutral axis” or “neutral plane”

Can also use parallel axis theorem to calculate second moments of area

Itotal = Ii! + Ai! di2

[L4 ]

No net moment about centroid

Second moment of area about centroid of area

area Distance of centroid from total section centroid

Useful to simplify problems, e.g. for I beam cross-section (structurally efficient form,

effectively used as spars in many aircraft).

is equivalent to:

Changing area distribution by moving material from the web to the flanges increases

section efficiency measured as I A - resistance to bending for a given amount of material

used.

I total= Iy1y1 + A1d12 + Iy2y2 + Iy3y3 + A3d3

2

112bh3 b3

Shear Stresses in Engineering Beam Theory

Note for information – not to be lectured Reading: Benham Crawford and Armstrong 6.14

Returning to the derivations of simple beam theory, the one issue remaining is to

calculate the shear stresses in the beam. We would like to obtain an expression for:

!zx z( ) .

Recall:

Shear stresses linked to axial (bending) stresses via:

!" xx!x

+!# zx!z

= 0$!# zx!z

= %!" xx!x (5)

Also, shear force (for rectangular cross-section, width b) linked to shear stress via:

S = ! " xz! h2

h2# bdz

(7)

Multiply both sides of (5) by b and integrate from z to h/2 (we want to know !zx z( )

and we know that !zx ±h2

" # $

% & ' = 0 )

b!" zx!zz

h2# dz = $

!% xx!xz

h2# bdz

b! xz (z)[ ]zh2 = "

"dMdx

# $ %

& ' (

z

h2)

zbIdz

recall dMdx

= S

b ! xz h2( ) " ! xz z( )( ) = SzbIyy

#

$ % %

&

' ( (

z

h2) dz

Note that at

z = h2

(top surface) !zx = 0

Also define

Q = zbdzz

h2! first moment of area above the center

Hence:

! xz (z) = "SQIyyb shear stress -force relation

For a rectangular section:

Parabolic, maximum at z = 0 . The centerline. Minimum = 0 at top and bottom

"ligaments"

!xx ="MzIyy

Q = zbdz = z2

2b

!

" # #

$

% & & z

h2'

z

h 2

= b2h2

4( z2

!

" # #

$

% & &

For non-rectangular sections this can be extended further if b is allowed to vary as a

function of z. I.e. can write: ! xz (z) = "SQ

Iyyb z( )

But need to beware cases in which we have thin walled open or closed sections where

shear is not confined to ! xz . Will also have ! xy (on flanges). E.g.

or

We will defer discussion of these cases until 2nd year courses. Next: Look at examples of

applying Engineering Beam Theory to examples of beams.

Lectures 17 and 18 Examples of Application of Beam

Theory Reading: Benham, Crawford and Armstrong Chs 6 and 7

Example: Calculate deflected shape and distribution of stresses in beam. Refer back to

example problem in lecture 13 & 14.

Example: Uniform distributed load, q (per unit length), applied to simply supported

beam.

Free Body diagram:

Apply method of sections to obtain bending moments and shear forces:

S(x) = qL2

! qx :

M (x) = qLx2

! qx2

2:

Let beam have rectangular cross-section, h thick, b wide. Material, Young’s modulus E:

Deflected shape of the beam: obtain via moment curvature relationship:

EI d2wdx2

= M ! d2wdx2

= M (x)EI

Integrate to obtain deflected shape w(x)

w x( ) = M (x)EI

!! dx dx

and for EI constant over length of the beam:

! w x( ) = 1EI

M (x)""

For our case:

M (x) = !qx2(x ! L)

dwdx

= ! 1EI

qx2

" (x ! L)dx

= !qEI

x3

6! Lx

2

4

#

$ % %

&

' ( (

)

* + +

,

- . . + C1

So w(x) = 1EI

(q2

"x3

6! Lx

2

4

#

$ % %

&

' ( (

+ C1)dx

w(x) = !q2EI

x4

24! Lx

3

12+ C1x + C2

"

# $ $

%

& ' '

C1 and C2 are calculated from values of w(x) and w’(x) at boundary conditions.

Two constants of integration, need two boundary conditions to solve.

x = 0, w = 0 ! C2 = 0x = L, w = 0

0 = !L4

24+ L4

12+ C1L = 0" C1 = !L3

24

Hence:

w(x) = ! q48EI

x4 ! 2Lx3 + L3x[ ]

Maximum deflection occurs where:

dw(x)dx

= 0

dwdx

! 4x3 " 6Lx2 + L3 = 0 satisified at x =L2

w = ! 5384

qL4

EIyy

Find Stresses:

!xx = "MzIyy

, I =112

bh3 , substitute for M(x)

Max tensile (bending) stress at

x = L2, Mmax = qL2

8, z = ±h 2

! xxmax = 3qL2

4bh2

Shear stresses (Not lectured on in 2009):

!xz ="SQIb

occurs where max shear force

S = qL2

@ x=0, x=L

for rectangular cross-section

Q(z) = ! z z

h2" bd ! z

= b ! z 2

! z

#

$ % %

&

' ( ( z

h2

= b2

h2

4) z2

#

$ % %

&

' ( (

! xzmax = bh2

8qL212bh3

1b

= 34qLbh

Note that this is much smaller than

! xxmax if L/h is large (i.e slender beam).

Application of Simple Beam Theory: Discontinuous Loading and

Statically Indeterminate Structures

See Benham, Crawford and Armstrong Ch 8

Example

Free body diagram

Shear forces:

Bending moments:

Note in passing M(x) and hence !xx =

"MzI

= constant for L # x # 2L - useful for

mechanical testing of materials – “four point bending”

Could solve for deflections by writing separate equations for: M(x) for 0 < x < L, L < x < 2L, 2L < x < 3L and integrating each equation separately and matching slope and displacements at interfaces between sections. This is a tedious process. Or we can use:.

Macaulay’s Method (BCA 7.3) Relies on the knowledge that the integral of the moment (i.e. the slope, and deflection) are continuous, smooth functions. Beams do not develop “kinks”. This is a particular example of a class of mathematical functions known as “singularity functions” which are discontinuous functions, defined by their integrals being continuously defined. Apply method of sections to the right of all the applied loads

Write down equilibrium equation for section, leaving quantities such as {x-L} grouped within curly brackets { } Mx=x = 0 M + P{x ! 2L} + P{x ! L} ! Px = 0

{x-a} is treated as a single variable, only takes a value for x > a otherwise { } = 0 Proceed as before - integrate moment-curvature relationship twice, applying boundary conditions

M = Px ! P(x ! 2L) ! P(x ! L) = EId2wdx2

EI dwdx

= Px2

2! P2

x ! 2L{ }2 ! P2

x ! L{ }2 + A

EIw = Px3

6! P6

x ! 2L{ }3 ! p6x ! L{ }3 + Ax + B

apply boundary conditions: w = 0 @ x = 0

w = 0 @ x = 3L

for x = 0 only Px3

6 is evaluated = 0, therefore B = 0

for x = 3L! 27PL3

6" PL3

6" P68L3( ) + Ax= 0

! A = "PL2

w(x) =1EI

Px3

6!P6

x! 2L{ }3 ! P6

x ! L{ }3 ! PL2 x"

# $ $

%

& ' ' (

Provides complete description of deflected shape of beam in one equation. Can also use

for distributed loads.

Statically Indeterminate Beams (not lectured in 2009) Reading: Benham, Crawford and Armstrong Ch 8.

(Two principal ways to approach.)

1.) By simultaneous application of three great principles

i.e., Set up: equilibrium

Moment - curvature

Geometric Constraints

2.) Superposition

Decompose problem into several statically determinate structures.

Replace constraints by applied loads

Solve for loads required to achieve geometric constraints.

Example: cantilever beam under uniform distributed load, with pin at right hand end.

Solve simultaneously (reactions remain as unknowns in moment-curvature equations)

Equivalent to:

(1)

+ (2)

From “standard solutions” or calculation obtain expressions for deflections in case (1)

and case (2):

RB = ?

!1 = " q!4

8EI,

!2 = + RBL3

3EI

!1 + !2 = 0

" qL4

8EI= RBL

3

3EI" RB = 3qL

8

This is a powerful technique for solving beam problems – statically indeterminate, or

otherwise complex loading.

Note. For superposition to work we must be superimposing the loads and boundary

conditions on the same beam. Only change one variable in each loading case.

L19 Shafts: Torsion of Circular Shafts Reading: Benham, Crawford and Armstrong, Ch 4

A shaft is a structural member which is long and slender and subject to a torque (moment) acting about its long axis. We will only consider circular cross-section shafts in SESG1001. These have direct relevance to circular cross-section shafts such as drive shafts for engines of all types. However, the basic principles are more general and will provide you with a basis for understanding how structures with arbitrary cross-sections carry torsional moments. Torsional stiffness, and the shear stresses that arise from torsional loading are important for the design of aerodynamic surfaces such as wings, and also play a key role in the design of ship hulls, automotive chassis and even hip and knee replacement prosthetics. Modelling assumptions

(a) Geometry (as for beam). Long slender, L >> r (b,h)

Note: For the time being we will work in tensor notation since this is all about shear stresses and tensor notation will make the analysis more straightforward. Remember we can choose the system of notation, coordinates to make life easy for ourselves!

(b) Loading Torque about x1 axis, T (units of Force x length). We may also want to consider the possibility of distributed torques (Force x length/unit length) (distributed aerodynamic moment along a wing, torques due to individual stages of a gas turbine)

No axial loads (forces) applied to boundaries (on curved surfaces with radial normal, or on x1 face)

!11 = !22 = !33 = 0 (c) Deformation

-Cross sections rotate as rigid bodies through twist angle φ, varies with x1 (cf. beams – plane sections remain plane and perpendicular)

- No bending or extensional deformations in x1 direction Cross-section:

!

u2 = "r#(x1)sin$

u3 = +r#(x1)cos$

where r = x22

+ x32

sin$ =x3

rcos$ =

x2

r

Positive x2, x3 quadrant

Torque, T right hand rule positive

!

so u2 = " x22

+ x32#

$ %

& ' ( )(x1)

x3

x22

+ x32

= ")(x1)x3

!

u3 = x22

+ x32"

# $

% & ' ((x1)

x2

x22

+ x32

= ((x1)x2

So

!

u1 = 0 1( )

u2 = "#(x1)x3 2( )

u3 = #(x1)x2 3( )

Governing Equations

!11 =du1

dx1

= 0

!22 =du2

dx2

= 0

!33 =du3

dx3

= 0

!12 =1

2

"u1"x2

+"u2"x1

#

$ % &

' ( = )

1

2x3d*

dx14( )

!13 =1

2

"u1"x3

+"u3"x1

#

$ % &

' ( =

1

2x2d*

dx15( )

!23 =1

2

"u2"x3

+"u3"x2

#

$ % &

' ( =

1

2)*(x1) + *(x1)( ) = 0

Consistent with assumption of no axial, or radial stresses, strains

Next, apply constitutive laws, i.e. stress-strain relations (assume isotropic),

remember we are working in tensor notation:

!

"mn =#mn

2G, therefore need “2G”

!23 =

"232G

= 0

!12 ="12

2G6( )

!13 ="13

2G7( )

Net moment due to shear stresses must equal resultant torque on section:

!

equipollent torque,T = x2"13 # x3"12( )$$ dx2dx3 (8)

Apply equilibrium:

!

"#mn

"xm+ fn = 0

i.e.:

!

"#11

"x1+"#21

"x2+"# 31

"x3= 0

!

"#12

"x1+"#22

"x2+"# 32

"x3= 0

!

"#13

"x1+"#23

"x2+"# 33

"x3= 0

Retaining non-zero terms we obtain

!

"#21

"x2+"# 31

"x3= 0 9( )

"#12

"x1= 0 10( )

"#13

"x1= 0 11( )

Solution

Go back to stress-displacement relationships (4,5, 6 and 7)

!12 = 2G"12 = 2G #1

2x3

d$

dx1

%

& ' (

) *

!12 = "Gx3d#

dx1

Similarly:

!13 = 2G13"13 = 2G1

2x2

d#

dx1

$

% & '

( ) = Gx2

d#

dx1

Substitute into equation for resultant moment (8)

T = x2!13 " x3!12( )## dx2dx3

= x22 d$

dx1

+ x32 d$

dx1

%

& ' (

) * ## dx2dx3

T =Gd!

dx1

x22+ x3

2( )dA""

define J = x2

2+ x3

2! "

# $ %% dA

=!R4

2 for circular cross-section

Polar 2nd moment of area

Hence T =GJ d!

dx1

(note, we can compare T =GJ d!dx1

for shafts with M = EI!2w

!x2 for beams)

Hence, relate stress to torque

!12 = "Tx3

J

!13 =Tx2

J

Express total shear as shear stress resultant, τ

!

" res = #122

+#132

=T

Jx32

+ x22

=Tr

J

compare

!

" =Tr

Jwith # xx =

Mz

I

$

% &

'

( )

Model works well for circular cross-sections and cylindrical tubes J = J1 ! J2

Torque-twist relation

Does not work for open sections:

We can approximate for other sections, e.g. square cross-section, J = 0.141a4 .

For a full treatment of torsion of slender members see 2nd year structures classes.

L20 Buckling of Simple Columns Benham Crawford and Armstrong: Chapter 10

Elastic instabilities, of which buckling is the most important example, are a key limitation on structural integrity. The key feature of an elastic instability is the transition from a stable mode of deformation with increasing applied load to an unstable one, resulting in collapse (loss of load carrying capability) and possibly failure of the structure. Examples of elastic collapse are the buckling of bars in a truss under compressive load, the failure of columns under compressive load, the failure of the webs of “I” beams in shear, the failure of skin panels or hull plates in shear and many others. The only particular case we will consider in SESG1001 is the failure of a bar or column loaded in axial compression, however, as for the other slender members we have considered, the basic ideas will apply to more complex structures. A structure is in stable equilibrium if, for all possible (small)displacements/deformations, a restoring forces arise. Now apply this idea to a continuous structural member. Model of a column (a) Geometry – identical to a beam, long, straight, slender, symmetric cross-section etc.

(b) Loading – axial compressive axial forces

(c) Deflection At low loads - same as a rod, axial (x) stresses, axial deformation only At higher loads - buckling deflection (transverse) governed by bending relations

i.e.,

d2wdx2

= MEI

L<< b , h

deflected shape:

Overall free body diagram:

We can take a “cut” (as before)

By equilibrium

Fx = 0!+

" P + F(x) = 0# F x( ) = $PF2 = 0" %+ S(x) = 0MA = 0 M (x) $ F(x)w(x) = 0"

M (x) + Pw(x) = 0

Substitute for curvature, obtain second order differential equation

EI d2wdx2

+ Pw = 0

Solution to equation given by:

w = e!x Write as:

d2wdx2

+ PEIw = 0

!2e!x + PEIe!x = 0

!2 = "PEI

#! PEI

± i i = "1

Can express as -complete homogeneous solution:

W = Asin PEIx + Bcos P

EIx + Cx + D

+

Stabilizing (restoring)

Tends to destabilize - compressive load As w increases

For the simply supported case:

x = 0 w = 0

M = EI d2wdx2

= 0! d2wdx2

= 0

x = L w = 0

M = EI d2wdx2

= 0! d2wdx2

= 0

d2wdx2

= ! PEI

Asin PEIx ! P

EIBcos P

EIx

So using B.C.’s

w x = 0( ) = 0! B + C = 0

d2wdx2

x = 0( ) = 0! B = 0

w x = L( ) = 0! +Asin PEIL + DL = 0 A, B, C, D = 0

d2wdx2

x = L( ) = 0! +Asin PEIL = 0

So: A sin

PEIL

!

" #

$

% & = 0

Satisfied if A=0, but that is not very interesting!

B = 0 C = 0

So:

sin PEIL = 0! P

EIL = n"

Thus buckling occurs if:

P = n2! 2EIL2

deformed shape

w = Asin n!xL

n = 1 n =2 n=3 lowest value is the critical buckling load:

Pcrit = ! 2EIL2

Other boundary conditions provide other values for A, B, C, & D, solution always of the form:

w = Asin PEI

x + Bcos PEIx + Cx + D

and buckling load always has the form:

Pcr =c!2EIL2

integer

Note: A is not defined

only achieved if 1st mode is restrained

Simply Supported c=1

clamped-free c=1/4

clamped-clamped c=4

mech of solids

Documents