centroidsmomentsofinertia-121222071213-phpapp02

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University of Manchester

School of Mechanical, Aerospace and Civil Engineering

Mechanics of Solids and StructuresDr D.A. Bond

Pariser Bldg. C/21

e-mail: [email protected]

Tel: 0161 200 8733

1

UNIVERSITY OFMANCHESTER

1st YEAR LECTURE NOTES

MECHANICS OF SOLIDS AND STRUCTURES

SEMESTER 2

11: CENTROIDS AND MOMENTS OF AREA 12: BEAM SUPPORTS AND EQUILIBRIUM

13: BEAM SHEAR FORCES & BENDING MOMENTS

14: BENDING THEORY


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Centroids and Moments of Area

2

11. CENTROIDS AND MOMENTS OF AREAS

11.1 Centroid and First Moment of Area

11.1.1 Definitions

The Centroid is the geometric centre of an area. Here the area can be said to be concentrated, analogous to

the centre of gravity of a body and its mass. In engineering use the areas that tend to be of interest are crosssectional areas. As the z axis shall be considered as being along the length of a structure the cross-sectional

area will be defined by the x and y axes.

An axis through the centroid is called the centroidal axis. The centroidal axes define axes along which the

first moment of area is zero.

The First Moment of Area is analogous to a moment created by a Force multiplied by a distance except this

is a moment created by an area multiplied by a distance. The formal definition for the first moment of area

with respect to the x axis (QX):

= ydAQX (11.1)Similarly for the first moment of area with respect to the y axis (QY) is:

= xdAQY (11.2)where x, y and dA are as defined as shown in Figure 11.1.

X

y

Y dA

x

x

C

Total Area = A

y

Figure 11.1

The X and Y subscripts are added to indicate the axes about which the moments of area are considered.

11.1.2 Co-ordinates of the Centroid

The centroid of the area A is defined as the point C of co-ordinates x and y which are related to the first

moments of area by:

AxxdAQ

AyydAQ

Y

X

==

==

(11.3)

An area with an axis of symmetry will find its first moment of area with respect to that axis is equal to zero

i.e. the centroid is located somewhere along that axis. Where an area has two axes of symmetry the centroid

is located at the intersection of these two axes


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3

11.1.3 Example: Centroid of a Triangle

Determine the location of the centroid of a triangle of base b and height h.

C

Y

Y^

X^

x

y

b

h

X

dx

dAx

Figure 11.2

Ans: x = 2b/3 and y = h/3

11.1.4 Centroid of a Composite Area

Where an area is of more complex shape a simple method of determining the location of the centroid may be

used which divides the complex shape into smaller simple geometric shapes for which the centroidal

locations may easily be determined. Consider Figure 11.3 which shows a complex shape of Area A madefrom three more simple rectangular shapes of Areas A1, A2 and A3. As the centroids of the rectangular shapes

are easily determined from symmetry the locations of their respective sub-area centroids are used to calculate

the location of the centroid of the composite shape.

X

Y

A2 C2

C3

C

C1

A3

A1

x

y

Figure 11.3

Ax

AxAxAx

xdAxdAxdA

xdAQ

332211

AAA

Y

321

=

++=

++=

=

The same method can be used to calculate the y-wise location of the centroid of the composite area.


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11.1.5 Example: Centroid of a L section

C

x

y

A1Y

A2

C1

C2t

X

b

h

t

(((( ))))(((( ))))hb

htbx

++++

++++====

2

2

(((( ))))

(((( ))))hb2

hbthy

2

++++

++++++++====

2

Figure 11.4

11.2 Second Moment of Area

11.2.1 Definitions

The second momentof the area about the x axis (IX) is defined as:

dAy=I2

X (11.4)

and the second momentof the area about the y axis (IY) similarly as:

dAx=I2

Y (11.5)

Some texts refer to second moment of Area as Moment of Inertia. This is not technically correct and SecondMoment of Area should be preferred.

11.2.2 Example: Rectangle (of dimensions b h)

Derive an expression for the second moments of Area for a rectangle with respect to its centroidal axes (Usethe symbol ^ to indicate centroidal axes and properties with respect to these axes).

The centroid is easily located by using intersecting axes of symmetry.

dy

C

b

h/2

yh/2

dA

X

Y

Ans:12

3bhI

X====

Figure 11.5


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The solution for second moment of area for a rectangle is frequently used as many composite shapes are

broken into rectangular sections to determine their composite second moments of area. The rule is often

recalled as:

The second moment of area of a rectangle about its horizontal centroidal axis is equal

to one-twelfth its base (b) multiplied by its height (h) cubed.

Similarly YI may be determined to be equal to b3

h/12.

11.2.3 Relationship to Polar Second Moment of Area

The Rectangular Second Moments of Area IX and IY are able to be related to the Polar Second Moment ofArea about the z axis (J) which was introduced in the section on Torsion.

( )

dAr

dAxy

dAxdAy=II

2

22

22

YX

=

+=

++

ZYX J=II + (11.6)

11.2.4 Radii of Gyration

The radius of gyration of an Area A with respect to an axis is defined as the length (or radius r) for which:

ArdArJ

ArdAxI

ArdAyI

2

Z

2

Z

2

Y

2

Y

2

X

2

X

==

==

==

(11.7)

As for the First Moment of Area, the X, Y and Z subscripts are added to indicate the axes about which the

second moments of area or radii of gyration are considered.11.2.5 Parallel Axis Theorem

If the axes system chosen are the centroidal axes the Second Moments of Area calculated are known as theSecond Moments of Area about the centroidal axes. Such axes are often annotated differently to other

axes indicating that they are centroidal axes. In this course the symbol ^ shall be used. If the second momentof area about another set of axes is required then the Parallel Axis Theorem may be used rather than havingto recalculate the Second Moments of Area.

Y dA

C

Total Area = A

d

X

y

X

y

Y

Figure 11.6


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Figure 11.6 shows an area with a centroid at C (with a centroidal x axis shown as X ) and for which theSecond Moment of Area with respect to the X axis is required. The X axis is a distance d away from the

centroidal axis.

( ) { }

{ }axescentroidalabout0dAyasdAddAy

dAddAyd2dAy

dyyasdAdy

dAy=I

22

22

2

2

X

=+=

++=

+=+=

2

XXAdI=I + (11.8)

This demonstrates that if the Second Moment of Area is known around an areas centroidal axis the SecondMoment of Area of that area about another axis distance d from the centroidal axis is simply the sum of the

Centroidal Second Moment of Area and the product Area d2.

This theorem applies to the Second Moments of Area IX and IY as well as to the Polar Second Moment of

Area provided the appropriate centroidal values are used.11.2.6 Example: Second moment of Area of a Rectangle about its base axis

For the rectangle shown; determine its second moment of area about its base and left edge axes (X and Y).

C

X

h/2

h/2

b

Y

X

Y

Figure 11.7

Ans:33

33hb

I,bh

IYX ========


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11.2.7 Second Moment of Area for a Composite Section

Consider the composite area shown in Figure 11.8. To determine the Second Moment of Area of such a

complex structure a similar approach to that used for calculating the centroids of complex areas is used.

Follow these steps:

i. Determine the Centroids of the sub-Areas

ii. Calculate the Second Moments of Area of the sub-Areas about their centroidal axesiii. Use the Parallel axis theorem to move sub-Area Second Moments of Area to axis of interest

iv. Sum the contributions of each sub-Area to the overall Second Moment of Area.

X

Y

A2

C2

C3

C1

A3

d1

A1

d2

d3

Figure 11.8

The validity of the above approach can be seen below for determining I Y of the area in Figure 11.8:

(((( )))) (((( )))) (((( ))))

(((( )))) (((( )))) (((( ))))

(((( )))) (((( )))) (((( )))) (((( )))) (((( )))) 022

222

33

321

321

321

321

2

33

2

22

2

11

22

333

2

3

22

222

2

2

2

111

2

1

2

33

2

22

2

11

222

2

========++++++++++++++++++++====

++++++++++++++++++++++++++++++++====

++++++++++++++++++++====

++++++++====

====

AA

AYAYAY

AAA

AAA

AAA

Y

dAxddAdxasdAIdAIdAI

dAddxxdAddxxdAddxx

dAdxdAdxdAdx

dAxdAxdAx

dAxI

This method is often well suited to a tabular layout or a spreadsheet.

11.3 Tabulated Centroids and Second Moments of Area

Many text books list the locations of standard area centroids and provide the Second Moment of Area aroundthese centroids. The departmental databook has such a table and will be allowed for use in exams

therefore students should become familiar with the use of this table.

11.4 Units

First Moment of Area has units of Length3.

Second Moment of Area has units of Length4.


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11.4.1 Example: A Regular I section

Derive an expression for the second moment of Area for a regular I beam with respect to its centroidal x axis.

a

b

t

t

Y

X

t

Y

X

Figure 11.9

The I section may be represented as being comprised of a rectangle of dimensions b(2t+a) from which two

smaller rectangles of dimensions (b-t)a have been taken out all of which have the same x-wise centroidalaxis. The total second moment of area is then simply the sum of all the contributions (with the missing areas

being subtracted).

(((( )))) (((( ))))

(((( )))) (((( ))))12

2

122

12

2

33

3

213

atbtab

atb.

tabI

X

++++====

++++====

This solution could also have been derived by considering the three rectangles separately and using parallelaxis theorem although there would have required significantly more work.


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11.4.2 An Unsymmetrical I section

Consider an unsymmetrical section shown below. The section is symmetrical about the vertical centroidal

(

) axis only. The y-wise position of centroid is to be found so that the second moment of area about its x-

wise centroidal axis can be determined. In examples such as this where the component is constructed from

"regular sub-areas" it is best to follow a tabular method as shown below.

2 6cm

10cm

1cm

1.5cm

1

X

1cm

5cmY

3

y1

y

y2

y3

X^

Figure 11.10

The section above is divided into three rectangular areas, (1), (2), (3). The bottom x-axis is used as datum.The tabular method of finding the centroid and the second moment of area are demonstrated in the following

Table.

Section Area (Ai) iy ( )iyA iy-yd=

Ad

2

iXI

2

iiiXdAI ++++

i (cm2) (cm) (cm

3) (cm) (cm

4) (cm

4) (cm

4)

1 10 0.5 5 3.208 102.913 0.833 103.746

2 6 4 24 -0.292 0.512 18 18.512

3 7.5 7.75 58.125 -4.042 122.533 1.406 123.939

Totals 23.5 87.125 246.197

4

2

iiiXX

i

i

cm197.246=

dAI=I

cm708.3=

A

)Ay(=y

+

Use first three columns to find y before proceeding to calculate d, Ad2

etc.


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Beam Supports and Equilibrium

10

12. BEAM SUPPORTS AND EQUILIBRIUM IN BENDING

12.1 Introduction

12.1.1 What is beam bending?

Tension, compression and shear are caused directly by forces. Twisting and bending are due to moments

(couples) caused by the forces.

VV

FF

FF

Tension

Compression

Shear

Torsion

Bending MM

T T

MM

T T

VV

FF

FF

Figure 12.1 Structural Deformations

Bending loads cause a straight bar (beam) to become bent (or curved). Any slender structural member onwhich the loading is not axial gets bent. Any structure or component that supports the applied forces

(externally applied or those due to self weight) by resisting to bending is called a beam.

12.1.2 Eraser Experiment

What is the basic effect of bending? Mark an eraser on the thickness face with a longitudinal line along thecentre and several equi-spaced transverse lines. Bend it. The centre line has become a curve.

Question:

What happens to the spacing of the transverse lines?

Bending causes compression on one side and extension on the other. By inference there is a section whichdoes not extend or shrink. This is called the Neutral Plane. On the eraser this will be the central longitudinalline. Consistent with extension and compression, bending must cause tensile (pulling) stresses on one side

and compressive (pushing) stresses on the other side of the neutral plane.

Bending is predominantly caused by forces (or components of forces) that are act perpendicular to the axis of

the beam or by moments acting around an axis perpendicular to the beam axis.


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free to move along the z-axis and rotate about the x axis). Imagine these supports as being similar to the

supporting wheel of a wheelbarrow.

z

y

RAySupport y-direction

Reaction Load

Beam

z

yA

Figure 12.3: Simple support reaction loads

12.3.3 Hinged or Pinned end support

Hinged or pinned supports provide similar support to simple supports with the addition of support in the

plane on which they are mounted i.e displacements in the axial direction are prevented (in Figure 12.4 the

beam is only free to rotate about the x axis). Imagine these supports as being the same as the connection at

the top of a grandfather clock pendulum.

z

y


Reaction Load

Beam

RAzSupport z-directionReaction Load

z

yA

Figure 12.4: Hinged/Pinned support reaction loads

12.3.4 Fixed or built-in end support

Fixed end supports (also called encastre) support moments in addition to lateral and axial forces. No axial,lateral or rotational movements are possible at a built in end (i.e in Figure 12.5 the end is not able to move ineither the y or z direction nor can it rotate about the x axis). Imagine these supports as being like the

connection of a balcony onto a building.

z

yMA

Support Reaction

Moment


Reaction Load

Beam

RAzSupport z-direction

Reaction Load

z

y

A

Figure 12.5: Fixed support reaction loads and moments

12.4 Distributed loads

To simplify the analysis of a distributed load it is usually easier to replace the distributed load with a point

load acting at an appropriate location. As the units of distributed loads are load per unit length the equivalent

point load may be determined by statics.


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L

w(z)

z

y

L

z

yWee

z

w(z)

z

y

Area under

w(z) = Aw

dz

dAw = w(z).dz

Figure 12.6: Replacing a distributed load with an equivalent point load

For the two cases to be equivalent the sum of the forces in the y and z directions have to be the same and thesum of the moments about any point have to be the same. Considering forces in the y direction first:

============ ewL

y WAdz)z(wF0

That is, the equivalent point force of a distributed load (We) is equal to the area under the w(z)

function (Aw).

Now consider moments about the origin:

e.WdAzdzz).z(wM e

L

w

L

o ============00

Note the similarity between this equation and Equation (11.3) for the first moment of area which allows the

previous expressions to be re-written as:

ze

e.Ae.W

AzdAz

we

w

L

w

====

====

====0

That is, the point along the beam at which the effective force (W e) must act is at the centroid of the area

(Aw) under the distributed load curve w(z).


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12.5 Equilibrium considerations for a beam

Consider a beam carrying loads as shown in the figure below. The right hand support at B is a simple support

and can only carry vertical forces. All the horizontal force components have to be supported by the left hand

(hinged) support, at A.

z

y

A

MW

C

w

l1

l2l3

l4L

B

c

Figure 12.7 A beam hinged at the left and simply supported on the right, loads as shown

Consider a point C where the left and right hand parts of the beam are to be separated into two free bodydiagrams. To maintain equilibrium in the separated sections additional forces and moments must be applied

at the new ends to keep both sections of the beam in the same geometry as when the beam was intact. These

forces and moments are known as the axial and shear forces and bending moments at position C. Theseforces and moments determine how a beam deforms under loading. To determine these forces and moments

the support reactions must first be obtained from the conditions of equilibrium of forces and moment for the

whole beam. Then the forces at the point C (shear force and bending moment) may be obtained from forceand moment equilibrium of the part of the beam to the left or to the right of the section. The left and right

hand parts and all the possible forces acting the two new ends are shown in the free body diagrams below.12.5.1 The support reactions

The support forces are obtained from the conditions of equilibrium of forces and moments on the wholebeam so DRAW a FBD of the beam and apply equilibrium conditions.

z

y

A

MW

w

B

RAy RBy

RAz

We

We is equivalent

point load to

distributed load w

Figure 12.8: FBD of entire beam used to calculate support loads and moments

Equilibrium conditions require:

============ 000 M,F,F yz First with the condition

MA = 0, we get the vertical support reaction at B.


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FC = Sum of all x-direction (axial) forces to one side of the point C is called the AXIAL FORCEacting on the vertical face of the beam at position C.

VC = Sum of all the y-direction (transverse/shear) forces to one side of the point C is called theSHEAR FORCE acting on the vertical face of the beam at position C.

z

y

M

C

w

l2-c

l3-c

l4-cL-c

B

RBy

VC

PC

MC

z

y

A

W

C

c

MCVC

RAy

RAz FC

c-l1

Figure 12.10: Left Hand and Right Hand FBDs for beam sectioned at C

Thus, in the above problem, MC, FC and VC can be found by considering the LH end FBD to be:

( ) ( )

( ) ( )

0RF

L

Mll

L2

w

L

WlllwWRV

WlL

Mcll

L

wc

L

cWlcllw)lc(Wc.RM

AzC

2

2

2

31

23AyC

1

2

2

2

31

231AyC

==

+++=+=

+++=+=

or by using the RH end FBD to be:

( )( )

( ) ( ) ( )

( ) ( ) ( )0F

L

M

llL2

w

L

Wl

llwRllwV

WlL

Mcll

L

wc

L

cWlcllwcLRMc

2

llllwM

C

2

2

2

3

1

23By23C

1

2

2

2

31

23By23

23C

=

+++=+=

+++=+

+=

Note that the numerical quantities of bending moment, axial force and shear forces must be the same in

magnitude and sense (sign) at a section irrespective of whether they are calculated considering the free body

diagrams of the beam to the left or to the right of the section.

The importance of MC, FC and VC are that the beams performance at position C is directly related to these

forces i.e., the stress, the deflection and the local rotation (angle) of the beam are all determined by thesemoments and forces.


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12.6.2 Example (easy): Beam forces at mid span for a cantilever beam

Determine the beam bending moment, shear and axial forces at mid span (C) for the beam and loading

shown in Figure 12.11.

z

y

A

3kN

C

2m

B

A

C

RAz

RAy

VC

2kN/m1m

MC

FC

y2kN/m

MA

Figure 12.11: A cantilever beam with concentrated and distributed loads

Answers: RAy = 5 kN, MA = 7 kN.m, FC= 0kN, MC= 3 kN.m, VC= -3 kN

12.6.3 Example (more difficult): Beam forces at mid span for a simply supported beam

Determine the beam bending moment, shear and axial forces at mid span (C) for the beam and loading

shown in Figure 12.12.

z

y

A

2kN

C

2m

B

1/3m

A

C

RAz

RAy

VC

2kN/m

1/3m

MC

FC

y2kN

1kN/m

Figure 12.12: A simply supported beam with concentrated and distributed loads

Answers: RAy = 59/27 kN, RBy = 31/27 kN, FC= 0kN, MC= 7/9 kN.m, VC= -2/27 kN


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12.7.2 Relationship between V and a Distributed Load

Consider the equilibrium of a small length of the beam between z and z+ z upon which a Distributed Load

(w) is applied. Assume that the Shear Force V varies along the length of the beam such that at z+

z the Shear

Force is V+ V. This element is shown in Figure 12.14.

V+ VV

z

w

z

z

yw+ w

Figure 12.14: FBD of a beam element to relate Q and a UDL

Equilibrium of forces in the y direction gives:

( )

zwzw2

1zwV

0zw2

1z.wVVVF

y

=

=++++=

In the limit when z approaches zero this reduces to:

====

====

wdzV

or

wdz

dV

(12.2)

Note that the second order quantity, the product of w and z, is neglected because it will be negligibly

small.

Therefore if a small element of a beam has no distributed load on it then w = 0 and the shear force in thesection must be constant. If w is non-uniform (i.e. w = w(z)), this analysis assumes that dz is so small that the

distributed load across dz may be considered uniform at a level defined by w(z). Thus this equations is valid

whether the distributed load is uniform or a function of z (i.e. w may equal w(z)).Equations (12.1) and (12.2) can be used to check the consistency of predicted shear force and bendingmoment variation along a beam.

Combining the previous two equations gives the key relationship:

2

2

dz

Md

dz

dVw ========

(12.3)


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12.7.3 Example: Simply supported beam with a Uniformly Distributed Load

Determine the shear force and bending moment at mid-span of the beam shown in Figure 12.15 using:

a. The Free Body Diagram approach of section 12.6.1, and

b. Equation (12.3)

y

A5kN/m

2m

B

Figure 12.15 Simply supported beam with a Uniformly Distributed Load

Answers: Vmid-span = 0kN, Mmid-span = -2.5kN/m


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Shear Forces and Bending Moments

21

13. SHEAR FORCE & BENDING MOMENT DIAGRAMS

13.1 Introduction

Bending moments cause normal tensile and compressive stresses simultaneously in different parts of a beamsection. Shear forces cause shear stresses that try to cut the beam. The magnitudes of bending moment and

shear forces generally vary from one section to another in a beam. As shown in the previous section, the twoquantities are dependent on each other. Graphs showing the variation of M and V along the length of the

beam are called Bending Moment (BM) and Shear Force (SF) diagrams. The BM and SF diagrams help toidentify the critical sections in beams where bending moments and shear forces are highest.

13.2 Examples of SF and BM diagrams

13.2.1 Example: Cantilever beam with a concentrated load

Consider beam AB fixed at A, carrying a concentrated load (-W) at any position C (distance z = c from A).

z

MA

RAy

RAz

z

y

A

Wc

L

B

C

Wy

c

Figure 13.1 Cantilever beam with a concentrated load and its FBD

To draw the SF and BM diagrams first calculate the support reactions by considering force andmoment equilibrium conditions. DRAW the FBD of the beam (see Figure 13.1).

WcMM

WRF

RF

AA

Ayy

Azz

====

====

====

0

Determine the shear force and bending moment relationships for each section of the beam - where

sub-length boundaries are defined by point loads, moments and the start/finish of DistributedLoads. Again use a FBD for each section (see Figure 13.2).

z

y

MA

RAy

W

z

y

MA

RAyV

z

M

V

M

z-c

c

z

Figure 13.2: Sub-lengths (A z C) and (Cz B) FBDs

In length (A

z

C):

(((( ))))

(((( )))) (((( ))))zcWWzWczRMzM

WRzV

AyA

Ay

============

========

In section (C

z

B):

(((( ))))

(((( )))) (((( )))) (((( )))) 0

0

====++++====++++====

========

x-cWWzWcz-cWzRMzM

-WRzV

AyA

Ay


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Beam Bending Theory

22

Plot the shear force (V) and Bending Moment (M) diagrams along the length of the beam.

z

VA

-W

BCz

M

A

Wc

BC

Figure 13.3: Shear Force and Bending Moment Diagrams for cantilever beam with concentrated load

Check consistency of shear force and bending moment expressions between sections and at locations

where concentrated loads/moments are applied or where UDLs start or finish.

Hints & Checks for BM and SF diagrams

Notice that the magnitudes of the shear force and bending moments at A (i.e. V(0) & M(0)) are equalto the magnitudes of the end reaction load and moment

Note that the Shear Force (V) is equal to the negative value of the end reaction load (V(0) = -RAy) as

by our definition for shear forces, RAy is acting in the ve direction. That is, in the +ve y direction but

on a face with a normal in the ve z direction.

Whenever there is a concentrated load or bending moment applied to a beam the corresponding Shear

Force and Bending Moment diagrams should show a step of the same magnitude. In this example the

steps at A due to the point load and moment RA and MA are from zero to -W and Wc respectively.

Notice that the shear force and bending moment at a free end are zero.

Note that the bending moment varies linearly from Wc to 0 over the distance of z = c in the region 0 z c; i.e., it has a constant gradient of -W as expected due to V being equal to -W throughout thatsection of the beam (see equation (12.1)).


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Beam Bending Theory

23

13.2.2 Example: Cantilever beam with a Uniformly Distributed Load.

Consider beam AB fixed at A, carrying a uniformly distributed load w positioned between C and D as shown

in Figure 13.4.

y

MA

RAy

RAz

y

z

A

ca

L

BC

z

D

w w

Figure 13.4 Cantilever beam with a UDL and its FBD

To draw the SF and BM diagrams first calculate the support reactions by considering force and

moment equilibrium conditions. DRAW the FBD of the beam (see Figure 13.4).

{{{{ }}}}

{{{{ }}}}centroidatactingforceeffectivebycreatedMomentc

a+wcMM

curveLoaddDistributeunderAreawcRF

RF

AA

Ayy

Azz

====

====

========

====

2

0

Determine shear force and bending moment relationships for each sub-length of the beam. In this

example sub-lengths are defined by the reaction loads/beam ends and the start and finish of the UDL.

Again use a FBD for each sub-length (see Figure 13.5).

z

y

MA

RAyV

z

M

x

y

MA

RAy

w

V

M

z-a

a

z

z

y

MA

RAy

w

V

M

z-a-c

a

z

z-a

Figure 13.5: FBDs for the sections (A z C), (Cz D) and (D z B) respectively.

In section (A z C):

(((( ))))

(((( ))))

====++++

++++========

========

22

cz-awcwcz

ca-wczRMzM

wcRzV

AyA

Ay

Again M(0) = MA as MA is a concentrated moment input to the beam at the left hand end. This is one

good check to see that derived expression for M is correct.


24/62


25/62

Beam Bending Theory

25

13.2.3 Example: Non-uniform distributed load on a cantilever

y

z

L

wo

Figure 13.7: Non-uniform distributed load on a cantilever

Derive an expression for the variation of shear force and bending moment in a cantilever beam loaded by a

non-uniform distributed load as shown in Figure 13.7. Try this using both the FBD approach and by usingequations (11.1) and (11.2).

Answers:

( ) ( )22o LzL2

wzV = and ( ) ( )323o L2zL3z

L6

wzM +=


26/62

Beam Bending Theory

26

13.2.4 Example: Simply supported beam AB of length L.

Consider a beam, pin supported at one end and simply supported at the other (this combination is often

referred to as simply supported). A concentrated load (-W) acts at distance c from A.

y

A

W

C

L

B

c

Figure 13.8: A simply supported beam with concentrated load

Determine support reactions using the FBD of the entire beam:

(((( ))))L

L-cW

L

WcW= W-RRF

L

Wc=RM

=RF

ByAyy

ByA

Azx

========

0

Derive expressions for Shear Force and Bending Moment in each section:

z

y

RAy

W

z

y

RAyV

z

M

V

M

z-c

c

Figure 13.9 Simply supported beam with a concentrated load FBDs

In section (A z C):

(((( ))))(((( ))))

(((( ))))(((( ))))

L

zL-cWzRzM

L

cLWRzV

Ay

Ay

========

========

In section (C z B):

(((( ))))

(((( )))) (((( ))))(((( ))))

(((( ))))(((( ))))

L

L-zWcz-cW

L

zL-cWz-cWzRzM

L

WcRWRzV

Ay

ByAy

====++++

====++++====

========++++====


z

V

A B

C

z

M

A BC

(((( ))))cLL

Wc-(((( ))))cL

L

W-

L

Wc

Figure 13.10: Shear force and Bending Moment diagrams for a simply supported beam with a concentrated load


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Beam Bending Theory

27

13.2.5 Example: A simply supported beam with a UDL.

y

A

w

C

L

B

a c

D

Figure 13.11 Simply supported beam with a UDL and the beam FBD

Determine support reactions using the FBD of the entire beam:

c

aLL

wc

RM

ca

L

wcRM

RF

AyB

ByA

Azz

++++====

++++====

====

2

2

0

Derive expressions for Shear Force and Bending Moment in each section:

z

y

RAyV

z

M

z

y

RAy

w

V

M

z-a

a

z

z

y

RByV

M

z

Figure 13.12: FBDs for beam sections (A z C), (Cz D) and (D z B) respectively.

In section (A z C):

(((( ))))

(((( )))) zc

aLL

wczM

caL

L

wczV

++++

====

++++

====

2

2

In section (C z D):

(((( )))) (((( ))))

(((( ))))(((( ))))

22

2

2azw

zc

aLL

wczM

azwc

aLL

wczV

++++

++++

====

++++

++++

====

In section (D z C):

(((( ))))

(((( )))) (((( ))))zLc

aL

wczM

ca

L

wczV

++++

====

++++====

2

2


z

V

A B

C

z

M

A BC

(((( ))))(((( ))))2

caLL

wc-++++

(((( ))))L

2cawc ++++

D

DE

Mmax

Figure 13.13: Shear force and Bending Moment diagrams for a simply supported beam with a UDL

How might Mmax be determined?


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Beam Bending Theory

28

13.3 Principle of Superposition

13.3.1 Theory

The support reactions and fixing moments as well as shear forces and bending moments (and all other

mechanical entities such as stresses and displacements) at a given section (or point) due to the individualloads can be calculated separately and summed up algebraically to obtain the total effect of all the loads

acting simultaneously. This is applicable to conservative (i.e. linearly elastic) systems only.

13.3.2 Example: Cantilever beam with a concentrated load and UDL

Consider a cantilever beam with a concentrated load (N) and a UDL (w) as shown in Figure 13.14. This can

be separated into two more simple to analyse cantilever loading cases: a single concentrated load and a singleUDL. The separate results for these two loading cases may be added together to obtain the results for the

complete beam (provided the beam remains in the linear elastic region of its stress-strain curve). This cangreatly simplify analysis as separate, simple expressions for SF and BM may be obtained for each of the

loading cases and then be added together to obtain the SF and BM expressions for the overall beam.

y

z

A

ed

L

BC E

w

D

cN

y

z

A

Concentrated Load Case (CL)

BC ED

cN

y

z

A

ed

BC E

w

D

UDL Case (UDL)

Figure 13.14: Example of Principle of Superposition

The results of sections 13.2.1 and 13.2.2 may be used to quickly write the SF and BM expressions for thecombined load case:

In section (A z C):

(((( ))))

(((( )))) (((( ))))

++++====++++====

++++====++++====

2

e

z-dweczNMMzM

weNVVzV

)UDL()CL(

)UDL()CL(

In section (C z D):

(((( ))))

(((( ))))

====++++====

====++++====

2

ez-dweMMzM

weVVzV

)UDL()CL(

)UDL()CL(


29/62

Beam Bending Theory

29

In section (D z E):

(((( )))) (((( ))))

(((( )))) (((( ))))22

edzw

MMzM

edzwVVzV

)UDL()CL(

)UDL()CL(

++++++++====++++====

++++++++====++++====

In section (E z B):(((( ))))

(((( )))) 0

0

====++++====

====++++====

)UDL()CL(

)UDL()CL(

MMzM

VVzV

Note that there are more sections in the overall beam than in either of the individual beams. Simply add theappropriate expressions from each separated load case to determine the overall expression for each section.

This method may also be used to determine the support loads and moments for the beam. In the exampleabove:

++++====++++====

====++++====

2

edweNcMMM

weNRRR

)UDL(A)CL(AA

)UDL(Ay)CL(AyAy

Check these results using FBDs for the complete beam.

13.4 Summary of Procedure for drawing SF and BM diagrams:

i. Find support reactions by considering force and moment equilibrium conditions.

ii. Determine shear force and bending moment expressions.

iii. Plot shear force (V) vs z and bending moment (M) vs z to appropriate scales.

iv. Check for consistency of sign convention and agreement of values at the supports and at the ends of

the beam and that equations (11.1) and (11.2) are valid along the length of the beam.

13.5 Macauley1 Notation

13.5.1 Introduction

Frequently it is beneficial to use a single expression for the shear force and bending moment distributionsalong a beam rather than the collection of sub-length expressions.

Consider a cantilever beam on which two concentrated forces and a UDL are acting.

Y

z

b

wW2W1

L

aw

W2W1

RAy

RAz

MA

A B

Figure 13.15: Beam for which Macauley expression is to be derived

1Macauley W.H. Note on the deflection of beams, Messenger of Mathematics, Vol 48 pp 129-130. 1919.


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Beam Bending Theory

30

There are three lengths of the beam in which the bending moment is different:

(((( ))))

(((( )))) (((( ))))

(((( ))))

2

0

2

21

1

bzw

bzWazWzRMMLzb

azWzRMMbza

zRMMaz

AyA

AyA

AyA

====

====

====

Clearly the expression for the length b z L contains the terms in the other two lengths. To reduce the

tedium of working with three separate equations and sub-lengths, the following notation (due to Macauley)may be used:

{{{{ }}}} {{{{ }}}}{{{{ }}}}

2

2

21

bzwbzWazWzRMM

AyA

====

In this version of the bending moment equation the terms within { } should be set to zero if the valuewithin these brackets become -ve.

In the general case:

{{{{ }}}}

(((( ))))


31/62

Beam Bending Theory

31

Integrating this term again to obtain an expression for bending moments (from Equation 12.3) gives:

{{{{ }}}} {{{{ }}}} {{{{ }}}} 21

2

1

1

2

2CbzWazWzRbz

wVdzM

Ay ++++========

Similar to the relationship between the shear force constant and point loads, the constant C2 represents aMacauley expression for all point moments along the length of the beam. Note: that anti-clockwise moments

applied to the beam are considered positive as they introduce a +ve step in the variation of M along the beam.

AMC ====2

Combining these last two expressions gives the full Macauley expression for bending moment along the

beam as:

{ } { }{ }

2

bzwbzWazWzRMM

2

21AyA

=

13.5.3 Example: Macauley Expression for a beam

Derive the Macauley expressions for shear force and bending moment for the beam and loading shown in

Figure 13.16.

Then use the expressions derived to determine the value and location of the maximum bending moment in

the beam.

y

A

w

4a

a a

B

Figure 13.16 Simply supported beam with a UDL

Answers: { } { } { } { }120 a2z4

wa15z

4

wa3az

2

wMa2z

4

wa15

4

wa3azwV +=+=


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33/62

Beam Bending Theory

33

14. BENDING THEORY

14.1 Introduction

Bending causes tensile and compressive stresses in different parts of the same cross-section of a beam. These

stresses vary from a maximum tension on one surface to a maximum compression on the other passing

through a point where the stress is zero (known as the neutral point). The maximum stresses areproportional to the bending moment at the cross-section. As the magnitude of the maximum stress dictatesthe load bearing capacity of the beam (i.e. for most engineering applications the stresses should be kept

below yield), it is important to find out how the stresses the bending moments are related. The relationshipbetween stresses and bending moments will be developed in this section. The analysis is restricted by the

assumptions stated in section 14.2. The assumptions can be relaxed and improved analysis can be made but

this is beyond the scope of the first year course and will be covered in future years.

14.2 Assumptions

The beam is made of linear-elastic material.

The cross section of the beam is symmetrical about the plane in which the forces and moments act (i.e.

the YZ plane).

A transverse section of the beam which is plane before bending remains plane after bending.

Young's Modulus is same in tension and compression.

The lateral surface stresses (in the y-direction) are negligible. The lateral stress within the beam and

the shear stresses between adjacent "layers" throughout the depth of the beam are ignored (until nextyear).

It is possible to do the analysis without these assumptions. But the algebra becomes very complicated.

14.3 The beam bending equation

14.3.1 Location of the neutral axis

Bending moment generally varies along the length of the beam. However it is reasonable to assume thatbending moment is constant over a very small (infinitesimally small) length of the beam. So the case of sucha small length of a beam subjected to a constant bending moment along its length (known as pure bending)

is analysed below.A small element of a beam is schematically shown in Figure 14.1. An initially straight beam element a'b'c'd'

is bent to a radius R at point z by bending moments M, to abcd. The layers above line e'f' (i.e. on the convex

side) lengthen and those below e'f' (i.e. on the concave side) shorten. The line e'f' (and ef) is therefore thelayer within the beam that neither lengthens nor shortens i.e. the NEUTRAL AXIS.

3The initial length of

e'f' is also equal to the arc length ofef, i.e:

Refdze'f' ============

Consider the line gh at a distance y from the neutral axis. The original length ofgh was the same as for allother layers within the element i.e dz. The new length ofgh may is related to its bend radius (R+y) and bend

angle ( ) so that the new length may be written as (R+y). Therefore as strain is the ratio of change in lengthto original length and stress (

) = strain (

) modulus of elasticity (E):

(((( ))))

(((( )))) (((( ))))R

y

R

y

R

RyR

dz

dzyR

'h'g

'h'ggh)y(

)y(utral axisy from nedistancestrain atdirectionzghlayerinStrain

z

z

========++++

====++++

====

====

====

and

3Note: A prime (') is used to indicate points in the undeformed condition.


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Beam Bending Theory

34

(((( ))))

R

Ey=E=)y(

)y(axisneutralfromydistancetastressdirectionz=ghlayerinStress

zz

z

z

y1

z

M

R

dz

M

y2

a'

b' c'

d'

b c

da

e' f'e f

g hy

y

Figure 14.1 Element of a beam subject to pure bending

Therefore as E and R are constants for a given position z and bending moment M the variation of stress througha beam is linear as shown in Figure 14.2. Note in Figure 14.2 positive stress is defined using the convention for

the right hand end of beam.

z

y1

z

M M

y2

z = Ey1/R

y

z

z = Ey2/R

Figure 14.2: Axial stress distribution at z


35/62

Beam Bending Theory

35

The effect of the axial stress at any point y on a small cross-sectional area dA is to create a small elemental load

(dF) on the elemental cross-section area dA (of thickness dyand widthb(y) - where b(y) is used to indicate thatb may vary with position y).

The stress and load are related bydF = dA = .b.dy.

x

y

y

dy

b(y)

dA

Figure 14.3: Cross-section of beam at z

The total axial load (F) on the cross-sectional face may then be related to the beam cross sectional dimensions

and radius of curvature by integrating across the face:

ydAR

Ebydy

R

E=dy

R

bEy=bdy.dF=F

y

y

y

y

y

y

y

y

2

1

2

1

2

1

2

1

==

As Figure 14.2 shows, the beam is not actually subjected to any axial load so the total axial load on the beam

must equal zero i.e., F= 0 and with E and R both non-zero the only way this relationship can equal zero is for y

dA to equal zero.

0QydAydA

0R&0E

0ydA

R

E=F

X

y

y

y

y

2

1

2

1

===

=

y dA is by definition, the first moment of area of the cross section (QX) and only equals zero if the axis fromwhich y is measured (i.e. the X axis) passes through the centroid of the cross-section.

Therefore the neutral axis of a simple beam must pass through the beam cross-section centroid. When

analysing beams the z axis is therefore located along the neutral axis.

14.3.2 The bending equation

Consider now the elemental moments (dM) caused by the elemental loads (dF) about the neutral axis; dM =

dF.y = .b.dy.y. The total applied moment (M) may then be found by integrating across the surface:

dAyR

Edyby

R

E=dy

R

Eyb=dy.y.b.=dMM

2

y

y

2

y

y

2y

y

y

y

2

1

2

1

2

1

2

1

==

as IX (the second moment of area about the neutral axis for the cross section) = y2

dA, the bending moment atpoint z may be related to the radius of curvature (R), the Youngs Modulus of the beam (E) and the second

moment of Area of the beam cross-section (IX):


36/62

Beam Bending Theory

36

R

EIM X====

The bending moment may then be combined with the expression for axial stress to provide the general bendingequation (also known as the Engineers bending equation or bending theory):

(((( ))))XI

MyREyy ========

(14.1)

This is one of the most important equations in Structural Engineering.

14.4 Maximum stresses in beams subjected to simple bending

The bending equation may be used to calculate the maximum tensile and compressive stresses in a beam.Maximum tensile stress occurs where My is maximum positive and maximum compressive stress is where

My is maximum negative, i.e. on the outer edges of the beam. For a positive bending moment his

corresponds to:

I

yM&

I

yM=

X

2)ecompressivmax.(z

X

1)tensilemax.(z =

Note that the magnitude of stress depends only on:

the moment (loads and their locations on the beam) and

geometry (cross section) of the beam

It does not depend on the material from which the beam is made. So, to determine the stresses in a beamsubjected to known bending moments, the only parameters needed are the cross section (shape) the beam, the

position of its centroid (to know the position of the neutral axis and to know y1 and y2) and the second momentof area IX.

EI (product of the modulus of elasticity of the material and the second moment of area of the cross section of thebeam) is called as the flexural stiffness of the beam. Compare this to the torsional rigidity (GJ) for a shaft

under torsion. Both indicate the resistance of the beam/shaft to bending/torsion deflections.

Note that the strain and hence the stress distribution through the thickness of the beam are linear and have zero

values at the neutral axis.

In summary, a positive bending moment M (as shown below) will generate tensile stresses along thelength of the beam (i.e in the z direction) in parts of the beam above the neutral axis (i.e where y is

positive) whereas compressive stresses will be generated in parts of the beam below the neutral axis(remember that the neutral axis is located at the centroid of the beam cross-section) according to the

Engineers theory of bending ( = My/I where y is the distance from the neutral axis).

z

Neutral Axis

z = My1/IX

z = My2/IX

y

y1

y2

z

x

M

compression

tension

Figure 14.4: Distribution of stress in a beam subject to bending


37/62

Beam Bending Theory

37

14.4.2 Example: Simply supported beam AB of length L.

Consider the beam of Figure 14.5, pin supported at one end and simply supported at the other of rectangular

cross-section h thick and b wide. A concentrated load W acts at distance c from A. Determine the maximum

axial tensile and compressive stresses in the beam.

y

A

W

C

L

B

cD

D

Section DD

Constant along length

h

b

Figure 14.5: A simply supported beam with concentrated load

The Macauley expression for beam bending moment (check with Example 13.2.4) is:

(((( )))) (((( )))) {{{{ }}}}czWL zL-cWzM ++++

====

The Bending Moment (M) diagrams is shown below with a maximum magnitude of

|M|(max) = -Wc(L-c)/L occurring at z = c.

z

M

A BC

(((( ))))cLL

Wc-

Figure 14.6 Bending Moment diagrams for a simply supported beam with a concentrated load

The second moment of area of the beam cross section about the neutral axis is equal to the second

moment of area about the centroidal x-axis (IX) = bh3/12

The maximum tensile stress (which will occur where M is a maximum) is found using the bending

equation (Equation (14.1))

Lbh

c)h-Wc(L6

12

bh

2

h.

L

c)-Wc(L-

I

yM

2

h33

X

z)tensile.(maxz =

==

=

similarly the maximum compressive stress may be found to be:

Lbh

c)h-Wc(Lhx.)compr.(maxx 3

6

2

====

====


38/62

Beam Bending Theory

38

14.5 Deflection of Beams

Whilst the beam bending equation is useful for calculating the maximum stress at a point along a beam, its

real strength is in how it helps to determine the deflection of a beam at any point along its length. The

deflection of a beam at any point on the beam shall be denoted by v. Note that v is defined as being positiveupwards in the same direction as the y axis.

Consider a beam deflected under a combination of loading (may be point loads, moments, DLs etc) as shownin Figure 14.7.

R

Y,v

z

s90-

=

dz

dds

d

d

dv

At any small section the beam can be considered

to be in pure bending so that a small length of

the neutral axis (ds) is at a radius R from thecentre of a circle.

defined as positive according to the right hand

screw rule (anti-clockwise) therefore:

ds = -Rd

as +ve s is in the same sense of direction as z

Deflected beam neutral axis

z

Figure 14.7: Derivation of curvature expressions

Here R is the radius of curvature of the neutral axis (NA) at point z. A new axis (s) is introduced which is

located on the neutral axis of the beam. Figure 14.7 shows that the angle between s and the z axis (the initial

location of the neutral axis) at any given value of z may be written as and can be seen to equal the angle that the radius of curvature makes with the vertical. The magnified portion of the beam in Figure 14.7 shows

a small length ds defined by a small angle in the radius of curvature d. The corresponding change in isdefined as d and is equal to d.

As ds = -R d = -R d the variation of along s (d/ds) may be written as:

==

R

1

ds

d

This is known as the curvature of the beam at z. The Greek letter (capital Kappa) is used to indicate this

parameter and has units of L

-1

. To determine the curvature in terms of the z and v co-ordinates of the beamsome mathematical manipulation is required.


39/62

Beam Bending Theory

39

2

2

2

1

dz

dv1

dz

vd

dz

d

dzdvtan

dz

dvtan

+

=

=

=

&

2

22

222

dz

dv1

dz

ds

dzdv1

dzds

dvdzds

+=

+=

+=

=

=

=

=

dz

dv+1

dz

vd

dzds

dzd

ds

dz.

dz

d

ds

d

23/2

2

2

This is the relationship between curvature and the deflection curve of the beams neutral axis.

Introduce the common nomenclature v' = dv/dz = slope of the neutral axis at point z and v'' = d2v/dz

2. For

beams with small deflections dv/dz is small and (dv/dz)2

is very small such that Equation (14.2) may be

simplified to:

''vdz

vd

R

1=

2

2

==(14.2)

Equations (14.1) and (14.2) may be combined to give:

''vdx

vd

EI

M

R

12

2

X

===

Which is more commonly written as:

M''vEIX = (14.3)M may be expressed in terms of applied loading and distances along the beam so that equation (14.3) may be

integrated to derive the slope and integrate again to find the deflection of the beam at any point z. Deriving

expressions for beam deflection using this approach is known as the beam deflection by integration

method. There are many other techniques that may be used for determining beam deflections but this is theonly one covered in this course.

14.6 Equation for the deflection curve

Consider the cantilever beam and loading shown in Figure 14.9

Y,v

z

b

w

W2W1

L

a

A B

Figure 14.8: Cantilever beam with loading

Section 13.5 showed that the Macauley expression for this beam and loading is:

{ } { }{ }

2

bzwbzWazWzRMM

2

21AyA

=


40/62

Beam Bending Theory

40

This expression for the bending moment may now be used to determine the vertical deflection along the

length of the beam.

Step 1: Integrate to bending moment expression to obtain v' and v:

{ } { }{ }

{ } { }{ }

{ } { }{ }

21

432313Ay2A

X

1

322212Ay

AX

2

21AyAX

czc24

bzwbz

6

Waz

6

Wz

6

Rz

2

MvEI

c6

bzwbz

2

Waz

2

Wz

2

RzM'vEI

2

bzwbzWazWzRMM''vEI

++

++++=

+

++++=

++++==

where c1 and c2 are constants to be obtained using boundary conditions.

Step 2: Solve for the integration constants using boundary conditions (bcs)

The boundary conditions are points (z values) at which the slope (v') or deflection (v) of the beam arealready known.

This commonly occurs at support points (v = 0) and points of inflection (v' = 0) such as fully supported ends,centre of symmetrically loaded beams etc. In the example under consideration the slope and deflection at the

fixed end are zero. By substituting these boundary conditions into the expressions for v' and v above thevalue of the constants may be obtained.

0c0v0zat&0c0'v0zat 21 ======

With the constants obtained the final beam deflection equation may be written as:

{ } { }{ }

++++=

24

bzwbz

6

Waz

6

Wz

6

Rz

2

M

EI

1v

432313Ay2A

X

Similarly the expression for the beam slope is:

{ } { }{ }

++++=

6

bzwbz

2

Waz

2

Wz

2

RzM

EI

1'v

322212Ay

A

X


41/62

Beam Bending Theory

41

14.7 Examples of Beam deflection by Integration method

14.7.1 Example: Cantilever beam with an end load (-W).

z

Y, v W

L

Figure 14.9: Cantilever beam with an end load

Macauley expression for bending moment is:

(((( ))))LzWzRMM AyA ========

Using equation (14.3) and integrating gives:

( )

( )

( ) 213

X

1

2

X

X

czcLz6

WvEI

cLz2

W'vEI

LzW''vEI

++=

+=

=

Identify and apply the boundary conditions:

600

200

3

2

2

1

WLcvzat

WLc'vzat

============

============

Therefore deflection and slope at the end of the beam (z = L) may be determined to be:

( ) ( )

( ) ( )X

3323

X

X

222

X

EI3

WL

6

WLL

2

WLLL

6

W

EI

1Lv

EI2

WL

2

WLLL

2

W

EI

1L'v

=

+=

=

=


42/62

Beam Bending Theory

42

14.7.2 Example: Cantilever with uniformly distributed load (-w).

Y, v

z

L

w

Figure 14.10: Cantilever with uniformly distributed load

Determine the deflection and slope at the end of the beam (z = L).

Answers:

( )

( )X

44343

X

X

3332

X

EI8

wL

48

wLL

8

wL

24

wL

2

LL

6

wL

EI

1Lv

EI6

wL

8

wL

6

wL

2

LL

2

wL

EI

1L'v

=

+

=

=

=


43/62


44/62

Beam Bending Theory

44

14.7.4 Example: Simply supported beam of with a UDL (-w).

v w

L

Figure 14.12: Simply supported beam of with a UDL

Determine the slope at either ends of the beam and the maximum deflection of the beam.

Answers:

( )

( )

( )X

4343

X

2L

max

X

2332

X

X

3332

X

EI384

wL5

2

L

24

wL

2

L

24

w

2

L

12

wL

EI

1vv

EI24

WL

24

wLL

6

wL

4

wL

EI

1L'v

EI24

wL

24

wL0

6

w0

4

wL

EI

1L'v

=

==

=

=

=

=


45/62

Beam Bending Theory

45

14.7.5 Example: Simply supported beam load (-W) acting at a distance a from the right end.

Y, v W

L

a

BA

C

Figure 14.13: Simply supported beam load (-W) acting at a distance a from the right end

Macauley expression for bending moment is:

{{{{ }}}}(((( ))))

{{{{ }}}}LzL

aLWz

L

WaLzRzRM

ByAy ++++

========


( ){ }

( ){ }

( ){ }

21

33

X

1

22

X

X

czcLzL6

aLWz

L6

WavEI

cLzL2

aLWz

L2

Wa'vEI

LzL

aLWz

L

Wa''vEI

+++

+=

++

+=

+

+=

Identify and apply the boundary conditions.

60

000

1

2

WaLcvLzat

cvzat

============

============

Maximum magnitude of deflection may occur at point in length AB or BC. In AB the maximum deflection

will occur at a value of z where dv/dz = 0 i.e. where v' = 0.

3

Lz:issolutionvalidonlybut

3

Lz

06

WaLz

L2

Wa'vEI 2

X

==

=+=

The deflections at this value of z and at z = L+a may now be evaluated to see which is the maximum.

( )

( ) ( )( )

{ } )aaL2L(

L6

)aL(Wa)aL(

6

WaLLaL

L6

aLWaL

L6

WaaLvEI

39

WaL

3

L

6

WaL

3

L

L6

WavEI

2333

X

23

3

LX

+

=++++

++=+

=

+

=


46/62

Beam Bending Theory

46

14.7.6 Example: Cantilever beam of supporting an end moment

z

Y, v

M

L

Figure 14.14 Cantilever beam of supporting an end moment

Determine the slope at the free end of the beam and the maximum deflection of the beam.

Answers:

( ) ( )

( )X

22

X

XX

EI2

MLL

2

M

EI

1Lv

EI

MLML

EI

1L'v

=

=

==


47/62


48/62


49/62

Beam Bending Theory

49

14.8.3 Example: A simply supported beam with multiple DLs over the beam

In more complex loading cases superposition is used to help derive the Macauley expression for bending

moment and the problem is solved by the integration method. For example, to determine the deflection of thebeam shown in Figure 14.17 separate the loading into two open ended DLs:

Y, v

wo

L/2L/2

z

Y, v

2wo

L/2L/2

z

Y, v

2wo

L/2L/2

z

DL1 load case DL2 load case

Figure 14.17 A simply supported beam with multiple DLs over the beam

The Macauley expression for bending moment is obtained by combining the appropriate expressions for theseparate load cases:

( ) ( )

3

0300

3

0

DLAy

30

DLAy

21

2

Lz

L3

w2z

L3

wz

4

Lw

2

Lz

L3

w2zRz

L3

wzR

DLtodueonContributiDLtodueonContributiM

21

+=

+

+=

+=

Using equation (14.3) and integrating and utilising two of the three boundary conditions that exist exist (v =0 at both ends and v = 0 at L/2) the slope and deflection may be found to be:

( ) ( )

( )X

4

0

2L

max

X

3

0

EI120

Lwvv

EI192

Lw5L'v0'v

==

==


50/62

Beam Bending Theory

50

14.9 Statically Indeterminate Beams

14.9.1 First degree static indeterminacy

One of the most important uses of the calculation of beam deflections is in the solution of staticallyindeterminate beams. Consider a beam with one end fully supported and the other pinned with a point loadapplied as shown in Figure 14.18.

W

L

a b

RBy

RBz

RAy

MB

zY, v

W

L

a

A BC

b

Figure 14.18: Beam with first degree static indeterminacy

The only reaction force or moment that may be determined by use of statics (i.e. Fz, Fy and M equal to zero)

is RBz = 0. That leaves three unknowns (RAy, RBy and MB) and two equations (from Fy or M being equal tozero).

=

=+

=

WbLRMM

WRRF

0RF

y

yy

z

ABB

BAy

Bz

This is referred to as being statically indeterminate to the first degree. If there were two more unknowns thanstatic equations available to solve them then the problem would be statically indeterminate to the seconddegree. As with all preceding statically indeterminate problems the method to solve them is to consider thegeometric compatibility (in this case the deflections and slope) of the structure.

The general approach is determine the value of one of the reaction forces using the knowledge that the deflection

of the beam at the support is equal to zero.The Macauley expression for bending moment is:

}az{WzRM Ay +=


{ }

{ }21

33Ay

X

1

22Ay

X

czcaz6

Wz

6

RvEI

caz2

Wz

2

R'vEI

++=

+=

This provides another two equations and two more unknowns bringing the total unknowns to five and thetotal number of equations to four. The number of equations may be increased by using the deflection and

slope expressions at different points on the beam (each new boundary condition produces a new equation).Thus the process is simply a matter of identifying and applying the boundary conditions to eliminateunknowns:


51/62

Beam Bending Theory

51

{ }

{ } 0LcaL6

WL

6

R0vLzat

0caL2

WL

2

R0'vLzat

0c0v0zat

1

33Ay

1

22Ay

2

=+==

=+==

===

c1 may be eliminated from the last two expressions by multiplying the second by L and then subtracting the

second from the first to get:

{ }

{ }

( ) ( )( )

( ) 2Ay

23Ay

1

33Ay

1

23Ay

L

a1

L2

aL2WR

0aLL3aL6

WL

3

R

secondfromfirstsubtract

0LcaL6

WL

6

R

0LcaL2

WLL

2

R

+=

=

=+

=+

The problem is now reduced to two unknowns and two equations (the statics equations Fy and MB) so RByand MB may be determined, the bending moment and shear force diagrams drawn, the beam axial stresses

calculated or the deflections and slopes at any point found.

14.9.2 Second degree static indeterminacy

Figure 14.19 shows an example of a beam with second order indeterminacy. The same method as used for

solving the first order indeterminate beam above is used to solve this problem. The extra equation required tosolve the problem comes from the extra boundary condition of v' = 0 at z = 0 thus giving four boundarycondition equations and three static equilibrium equations (total 7) to solve for the five supportloads/moments and the two constants of integration.

W

L

a b

RBy

RBz

RAy

MB

z

W

L

Frictionless

Surface

BA

MA

Figure 14.19 Beam with second degree static indeterminacy


52/62

Beam Bending Theory

52

14.9.3 Solving Statically Indeterminate Beams using Superposition

Y, v w

2L/3

L

Y, vw

L

Y, vRBy

2L/3

L

A CB

z z z

Figure 14.20: Solving Statically Indeterminate Beams using Superposition

Superposition may also be used to solve statically indeterminate problems. Consider the staticallyindeterminate beam shown in Figure 14.20 which may be considered as being equivalent to the two

simplified loading cases. Note that as there are one too many reaction loads to solve the problem staticallyRBy is considered as a separate load and is solved for using geometric compatibility.

Section 14.7.4 showed that the equation for the deflected beam under the UDL is:

( )zLzLz2EI24

wv 343

X

)UDL( =

This can be used to determine the deflection at B due to the UDL:

( )X

43

43

X

3L2

EI

wL01132.0

3

L2L

3

L2

3

L2L2

EI24

wv:UDL =

=

The deflection of the beam at B due to the point load may be found from Section 15 to be:

( )X

3

By

X

22

By

X

22

By

3L2

EI

LR01646.0

LEI3

3

L

3

L2R

LEI3

baRv:LoadintPo =

==

as the deflection at B must be zero the sum of these two deflections must equal zero. Thus R By may be

calculated to be:

( )

wL688.0R

EI

wL01132.0

EI

LR01646.00v

By

X

4

X

3

By

3L2

=

==

Now with RBy known the problem is no longer statically indeterminate and may be solved as for a standardstatically determinate beam.


53/62

Exam Formulae Sheet

53

15. EXAM FORMULAE SHEET

Stress and Strain

A

P=

A

Q=

oL

=

TL

o

TT ==

Properties of Elastic Materials

E= G=

allongitudin

transverse

=

EEE

EEE

EEE

yxzz

zxy

y

zyxx

=

=

=

)1(2

EG

+=

( )213 =

EK

Torsion of Circular Cross-section Shafts

LGr ====

GJTL=

JTr==== )RR(

2J

4

i

4

o = TP =

Centroids and Area Moments

AyydAQX == AxxdAQY ==

dAy=I2

X dAx=I2

Y ZYX J=II +

2

XXAdI=I + (parallel axis theorem)

Beam Relations

2

2

dz

Md

dz

dV

w ==

( )X

zI

My

R

Eyy ==

M''vEIX =


54/62

Exam Formulae Sheet

54

Table of Beam Deflection Equations

Beam and Loading Equation of deflection curve Maximum deflectionand location

ZL

Y,v Wa b

( ) ( ) { }[ ]32

X

azza3zEI6Wzv = ( ) ( )

X

2

maxEI6

aL3WaLv =

Z

w

L

Y,v

( ) ( )22X

2

zLz4L6EI24

wzzv += ( )

X

4

maxEI8

wLLv =

Z

M

L

Y,v

( )X

2

EI2

Mzzv = ( )

X

2

maxEI2

MLLv =

Z

W

L

Y,va b

a b

( ) ( ) { }[ ]3222X

azLzbLbzLEI6

Wzv +=

( ) ( )X

23

22

3bL

maxLEI39

bLWbv

22 =

Z

w

L

Y,v

( ) ( )323X

zLz2LEI24

wzzv += ( )

X

4

2L

maxEI384

wL5v =

Z

L

Y,vM

( ) ( )zLz

LEI6

MLzv 23

X

2

=

( ) X

2

3

L

max EI39

MLv =

Transformations of Stress and Strain (Mohrs Circle)

y

x

xy

xy

xx

y

y

1

1

2

2

p

max

ave

min

Pt B (y , xy)

Pt A (x ,

2pxy

2

yx

+2

-R

2

+=

2xy

yx

2

yx

ave

=


55/62

Tutorial 3: Moments of Area and Beam Bending

55

TUTORIAL 3: MOMENTS OF AREA & BEAM BENDING


3.1

Determine the location (horizontal & vertical distance) of the centroid from each of the reference corners for

the following cross-sections.

Ref

100

40 30

20

100

20

25

45

100

Ref

80

Ref

20 20

80

All dimensions in mm

Answers: (a. 52.143mm, 70mm b. 50.978mm, 62.935mm c. 60mm, 50.66mm)

3.2Calculate the second moment of area about the centroidal axes for the cross-sections shown in the Question

3.1.

Answers: (a. 17.033106mm4, 3.601106mm4 b. 13.565106mm4, 9.365106mm4 c. 3.598106mm4,

3.467106mm

4)

3.3

Find the position of the centroid and the second moment of area about the centroid for the beam section

shown in Question 3.16.

Answers (on axis of symmetry and 6.43mm above base, 1.44cm4)


56/62


56

Bending Moment and Shear Force

3.4

Find the support reactions and maximum bending moment and draw the SF and BM diagrams for a beam, of

length 2m, simply supported at its ends and carrying two equal loads of 10kN applied at distances 0.5m fromthe supports.

Answers: (-10kN, -10kN, 5kNm)

z

V(kN)

2.01.5

z

M(kNm)

B

0.50.5

-10

10

1.5

5

3.5

For the beam loaded by the distributed load w shown in below, calculate the reactions at A and B, derive anexpression for the shear force variation along the beam and determine the magnitude and location of the

maximum magnitude of bending moment in the beam.

A B

L

Lzsinww o====

Answers: ( ( )2

2

o2L

max

ooByAy

LwMM,

L

zcos

LwV,

LwRR

==

=

== )

3.6

Calculate the support reactions and maximum bending moment magnitude and draw the SF and BMdiagrams for the beam AB shown below. Treat the bracket CE as rigid (i.e. it transfers the applied load from

E to C with an accompanying moment).

A C

E

BD

5m 5m 5m

3T

Answers: (T, 2T, 20T)

z

V

1510 z

M

155

5-T

2T

10

-5T

-20T


57/62


57

3.7

A simply supported beam AB of length L carries a distributed load whose intensity varies from zero at its left

end A to wo per unit length at the right end B. Determine where the maximum bending moment occurs

(distance from A) and an expression for the maximum bending moment.

Answers: ( 393

2

0Lw

,

L

)

3.8

Calculate the applied load at C and reaction loads at B (parallel and perpendicular to the beam axis) for thebeam AB which is pinned at B and supported by a strut CD. The member CD is pin-ended. Then draw the

SF, BM and axial force diagrams for beam AB.

y

D

C

30 B

1.5m

A10kN

1.5m

60

z

Answers: (10kN, 53kN, 5kN)

z

V(kN)

5

31.5 z

M(kN.m)

31.5

z

P(kN)

-5 3

3

-5 -7.5

3.9

Draw the shear force and bending moment diagrams for the beam and loading shown below. Determine the

location and magnitude of the maximum bending moment.

3m

A B

20kN/m

6m

C

Answers: (4m, 160kNm)

z

V(kN)

96 z

M(kNm)

94

4

-80

406

-120

-160


58/62


58

Macauley Expressions

3.10

Derive the Macauley expressions for shear force and bending moment in terms of z, W and L for the beam

and loading shown below:

2W

L

A B

Ww = W/L

L L L

Answers (

{ } { } { } { }

{ } { } { } { }22

1010

L3z

L2

WL3z

4

W13L2z

L2

WLz

4

W3WzM

L3zL

WL3z

4

W13L2z

L

WL3z

4

W3WV

+=

+=

)

3.11

The overhanging beam ACDB shown below is simply supported at B and pinned at A. Write the Macauleyexpression for the variation of bending moment along the beam using superposition to assist in thedevelopment of the expression

w = 12kN/m

1.2m

C DA B

1.2m1.2m

Answers ({ } { }

{ } kNm4.2z122

4.2z

2

2.1zz4.2M

122

+= )

3.12

Write the Macauley expression for the variation of bending moment along the beam shown below. Assume

bar CE is rigid.

A C

E

B

0.4m 0.4m

D

0.4m

14kN14kN

Answer ( { } { } kNm8.0z288.0z6.5z15z32M 102 +++= )


59/62


59

Axial Stress due to Bending

3.13

A beam rests on supports 3m apart and carries a uniformly distributed load of 10kN. The beam is of

rectangular cross-section and is 75mm deep. How wide should it be if the skin stress must not exceed60MNm

-2?

Answer (66.7mm)

3.14

A light alloy I-beam of 100mm overall depth has flanges of width 50mm and depth l0mm. The thickness of

the web is 5mm. If the maximum stress must not exceed 120MNm-2

, find the maximum moments that can becarried about the X and Y axes shown.

Web

Flange

Flange Answers (5.39kNm, 1.004kNm)

3.15

What is the allowable bending moment about the X axis passing through the centroid for the T-sectionshown if the maximum allowable stress is 150MNm

-2?

100mm

10mm

10mm

90mm

Answers (3.78kNm)

3.16

A beam of length 1m and cross-section as shown is subjected to a pure bending moment. A measuring device

records a vertical deflection at mid-span of 5mm. What is the strain at the top surface of the beam?

1m

P

MM

60mm

10mm

10mm

10mm

Answer (542.8)

3.17 (Combination of bending, torsion and Mohrs circle good exam question!)

A 150mm shaft is subjected to a torque of 14kNm and simultaneously to a bending moment of 11kNm. Findthe maximum principal stress and the maximum shear stress in the shaft.

Answers: (434MPa, -10.2MPa, 25.9)3.18 (Combination of bending, torsion, axial stress and Mohrs circle really good exam question!)

An element on the periphery of a propeller shaft is subjected simultaneously to a torsional shear of 28MPa, abending stress of 7MPa caused by its own weight and a compressive stress of 3.5MPa due to propeller thrust.Use Mohrs circle to determine the principal stresses and the planes on which they act.

Answers: (23.2MPa, -33.7MPa, 38.67)


60/62


60

Beam Deflections

3.17

A brass bar, 900mm long, is simply supported at its ends and carries a concentrated load of 90 N at a point

200mm from the right hand end. If IX = 0.22 cm4, calculate the deflection at midspan. (use E = 96 GPa).

Answer (4.03mm)

3.18

Find the magnitude and position of the maximum deflection for the beam shown below, given that EI=

700kNm2.

1kN

0.5m

A B

2kN

1.5m 1.5m 1m

5kN4kN

0.5m

Answer (0.0315m)

3.19

Find the required flexural rigidity (EI) of the beam shown below so that the end deflection of the beam doesnot exceed 20mm.

20kN3m

5m

Answer (18MPa)

3.20

For the steel beam shown in figure below, find the second moment of area of section if the maximumdeflection must not exceed 5mm. Take E = 200Gpa:

0.5m

10kN

1m

A B

10kN0.5m

1m 1m

Answer (3.4210-6

m4)


61/62


61

3.21

A horizontal cantilever of uniform section has an effective length of 2.5m and carries a load of 50KN at the

tip. As an experiment, this 50kN load is replaced by two equal loads one at the tip and the other 1.5m from

the fixed end. If the maximum deflection is the same as in the first case, find (a) the magnitude of the equal

loads and (b) the ratio of the maximum bending stresses produced in the two cases.

Answers (34.9kN, 1.117)3.22

The beam shown has a steel prop at B, which for all intents and purposes keeps the deflection of the beam at

that point zero. Design a suitable prop, so that the deflection at the prop is < 1% of the end deflection withouta prop. (use Esteel =207 GPa)

20kN

3m 2m

Prop ofcross-sectional

area A

h

100mm

50mm

Cross-section of beam

Answer (L/A


62/62


Statically Indeterminate Beams

3.24

For the beam shown below, find the reaction at the roller support and the deflection at the end of the beam

Wa

L

Answers ( ( )a4L3EI12

Wa,

L2

a31WR

X

2

+=

+= )

3.25

Assume one end of the beam shown below is mounted in a frictionless slot. Write the general expression for

the bending moment at any cross-section along the length of the beam shown below using Macauleynotation. Determine the deflection at B and determine the forces and moments exerted at A and C. Sketch the

shear force and bending moment diagrams.

100kN

3m 2m

A B C

Answer (deflection = 0.29mm)

3.26

Assume one end of the beam shown below is mounted in a frictionless slot. What degree of indeterminacydoes the have? Find the maximum deflection of the beam.

50kN/m

1m

40mm

50mm

Cross-section of beam

R10mm

Typ.

Answers (2nd, 6.3mm)

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