AZ024 Lecture 1 (rev[1] a)

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Prepared by Dr. Thomas Tong

October 2006 (rev.a)

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AZ024 Technology and Science 4

Lecture 1 Revisions: Structural Mechanics

1.0 IntroductionsIn this lecture, a revision to structural mechanics will be given. It includes:

(a) Force and vectors

(b) Resolution of forces

(c) Structure and Loads

(d) Newtons law of mechanics

(e) Moment

(f) Strength of Materials: Stress, Strain and Elasticity

2.0 Nature of Force Force are vectorsIf an objected is moved from one place to another place, a force must be applied. However, how the object

moves will depend on how hard and which direction the force is pull.

Two things about a force are important: the size of the force and the direction of the force. Quantities, which

consist of both size and direction, are called a vector quantity.

To indicate that the force is a vector quantity, we usually use a symbol

F .

Forces can also be represented graphically by arrow The direction and size fo the force can be shown by the

length of the arrow drawn to a chosen scale respectively.

For example, a force

F of 2N pointing in the southwest direction can be represented by the arrow in Figure

1.

Figure 1:- Force is represented graphically by arrows

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- Compressive force and Tensile force (Figure 2)

Figure 2 :-Axial compression and Axial tension

Adding forces

Surely it is known that scalar quantities* can be added algebraically, e.g. 2 + 2 = 4. But for vector quantities,

like forces, 2 and 2 is not always equal to 4!

*Remarks: scalar means that the quantity without direction implication, for example, weigh and length.

For example:

1. Forces of 2N and 2N acting in opposite directions cancel out to give no resultant force at all. (Figure 3)

Figure 3:- Adding of force

The resultant force is (-2N) + (2N) = 0N (Take right direction as positive)

2. What would be the resultant force of 5N opposed by 2N (Figure 4) The 2N force cancels out part of the

5-N force. How much is left?

Take right as positive, the resultant force is (+5N)+(-2N) = +3N

Figure 4:- Adding of force

When two forces are not in the same straight line, their sum can be found graphically. Two forces can be

added by drawing a parallelogram of the forces. InFigure 5, the resultant force of the two forces 1

F

and 2

F is represented by the diagonal of the parallelogram. This method is called the method of

parallelogram of forces.

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Figure 5:-Adding two forces by the method of parallelogram of forces

Example:

What is the resultant force as shown in Figure 6?

Figure 6

Solution:

The resultant force = (-3N) + (4N) = +1N

Remarks: we always take right direction as positive side.

3.0 Resolution of forcesSometimes it is useful to use parallelogram method of adding vectors. The parallelogram method is used to

resolve a force into two perpendicular components.

Figure 7:-Resolving a force into two perpendicular components

Remarks: the parallelogram method is commonly used in analysis of framing structure.

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Example:

Using the method of parallelogram of vectors, find the horizontal and vertical component of the resultant force

5kN in Figure 8?

Figure 8

Solution:

The force in the horizontal direction should be

)(87.168cos5 == kNFx

The force in the vertical direction should be

)(635.468sin5 == kNFy

Example:Replace the force F shown in Figure 9 by two forces, one vertical and one horizontal, which together will have

the same effect as F.

Figure 9

Solution:

The force in the horizontal direction should be )(3.1540cos20 == kNFx

The force in the vertical direction should be )(9.1240sin20 == kNFy

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3.1 Resultant of concurrent forces

The resultant force of a group of concurrent forces may be determined using the following equation

=

=

iyy

ixx

FR

FR

whereyx RR , are xand ycomponent of resultant force R

iyix FF , are xand ycomponent of force iF

iF is the ith force of the concurrent forces

n is the number of forces in the concurrent system

The magnitude and direction of the resultant is then determined using the following equations:

22

yx RRR +=

x

y

R

R1tan =

where Ris the magnitude of the resultant force and,

is the angle of the force measured from xaxis

Example:Calculate the resultant force of the force shown in the Figure 10.

Figure 10

Solution:

For ,2501 NF = For ,5002 NF =

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)(177)45sin(250

)(177)45cos(250

1

1

==

==

NF

NF

y

X

)(433)60sin(500

)(250)60cos(500

2

2

==

==

NF

NF

y

X

Let xcomponent pointing to right is positive and ycomponent pointing to up is positive.

NFR ixx 70250177 ===

NFR iyy 610433177 ===

Magnitude of resultant force NR 6156107022 =+=

The angle of the resultant to x axis is == 8370

610tan 1

Figure 11

Remarks: the angle of the resultant can be determined using above equation but the direction of the resultant

force should be judged by its component. In this example, xR is to the left and yR is to downward, then R is

therefore pointing to lower left direction.

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4.0 Structure and LoadDuring the development of a buildings program of space, the architect should obtain from the client written

data listing the intended occupancy of each floor area as well as its loads, anticipated load and possible future

renovations. Moreover, the architect should obtain from all mechanical, electrical written data listing the

weight and sizes and any adverse structural anomalies of every mechanical element to be installed in the

building. Finally, the architect should include in the plans a schedule of loads listing total services loads (for

example, live load, dead load).

The various types of loadings that can act on a building in Hong Kong are as follows:-

a. dead load

b. live load

c. wind load

d. water and earth load

4.1 Dead load

The dead loads that act on the structure arise from the weight of the structure itself and from the weight of all

nonstructural materials attached to it to form the building. These nonstructural items include the roofing,

permanent internal partition, cladding walls, finishing, etc. These nonstructural loads can be estimated with a

fair degree of accuracy, for they involve mainly material weight and thickness.

The dead load that arises from the weight of the structure itself is considerably more difficult to estimate,

however, because the size of the structural members can be accurately known only after the design has beencarried out. Both architect and engineer must take intuitive guesses regarding the sizes of the members and,

after the design has been completed quickly verify that their guesses were reasonable.

4.2 Live load

Live loads are the superimposed user or occupancy loads that the structure must carry to perform its function

satisfactorily- the reason was built in the first place. It includes all moving weights that a structure may carry:

occupants, furnishing, machinery (building services installation), movable partitions, rain, wind, lateral subsoil

and hydrostatic pressure, and any temporary loads applied during construction or maintenance.

The estimation of the live load can never be either fully accurate or complete, because even the user can

change with time, thereby altering the user or occupancy load during the life of the building.

The problem of estimating the live users or occupancy load is not really all that complex. However, the

Uniform Building Code and other codes provide a value for the load that should be used in different locations.

For instance, in a residential, high-rise building, the codes suggest a value of 1.92kN/m2

of plan area in the

living areas. As an other example, the load suggested by the codes for libraries is 2.87 kN/m2

in reading room.

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4.3 Wind Load

The actual manner in which the wind load affects the structure is very complex, for it depend on, among other

factors, the shape of the building, surface quality, its elevation above ground, the amount of shelter afforded

by surrounding buildings, and openings in walls. (Figure 12.)

The basic wind pressure sq may then be determined from the wind velocity V by using the formula,

200256.0 Vqs =

Figure 12:- Variable wind pressure around buildings

4.4 Water and Earth Load

Water and earth loads have to be accounted for in designing those structures that are below groundm such as

retaining walls, beasement walls, floor slabs, some foundations, among others. These load are really form oflive load for such structures.

If the structure is entirely below water, the hydrostatic pressure is 2.99kN/m2

per foot of depth. The pressure

distribution is linear and is shown in Figure. A load that is caused by earth pressure is not as serve as one

caused by water, because soil is generally capable of maintaining itself unaided on a small slope, which is the

angle of repose. This naturally depends on the types of the soil. Sand, for instance, has a lower angle of

repose than stiff clay. If however, water enters the clay, the material then be a greater tendency to flow and

hence exerts greater earth pressure. In the absence of water, soil pressure may be assumed 1.20-1.68 per

foot of depth. This is shown in Figure 13.

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Figure 13: Pressure Distribution from underground water and earth movement

4.5 Combination of the Load

For rational design, it is essential to tale combinations of loadings that can realistically occur together and to

design the structural element for the worst case of these realistic combinations. Regarding wind action, most

building codes permit, in working stress design, a 44 percent overstress, or increase in the permissible stress

that the material can take. In strength design in concrete, the ultimate load U(will be discussed in the next

section) for which the structural element has to be designed is given by

( )WLLLDLU 7.17.14.175.0 ++=

The factor 1.4 is a factor of safety to the dead load, and the factors 1.7 are the factors of safety for the live

load and the wind load. The factor 0.75 is in fact a method of reducing the total ultimate load for which thedesign is being carried out.

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5.0 Newtons law of mechanics

5.1 Newtons first Law

A body remains a rest or in motion with a constant velocity unless a force is applied on it.

Everybody has a tendency to maintain its state of rest or of uniform motion. The tendency is called

inertia.

Mass is a measure of inertia of an object, i.e. greater the mass of an object, then greater the inertia.

Force is something, which changes the state of rest or uniform motion of an object.

Daily example

The inertia of an object tends to keep it moving. The greater the inertia, the larger force is needed to stop it.

If a car stops suddenly, the passengers inside will tend to keep on moving (through the windscreen) because

of their inertia. So they need seat belts, which exert forces to stop them.

Similarly, the passengers in a car are thrown back if the car suddenly starts moving. The headrests fitted on

car seats can reduce neck injuries when the car is bumped from behind suddenly.

5.2 Newtons second law

The relation between force, mass and acceleration can be expressed as an equation,F= kma .(1)

where k is a proportional constant.

Equation (1) may be rewritten as:

=

)2

(

)(Re

)(

ms

onaccelerati

kgmasssult

NForce

amF

Both force and acceleration are vectors. This equation implies that the direction of ais in the direction of F.

This is Newtons second law of motion:

The acceleration of an object is directly proportional to and in the same direction as the resultant force acting

on it, and inversely proportional to the mass of the object.

Note:

Fis the resultant (or unbalanced) force on the mass m, it must be measured in Newtons.

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Example

When a force of 6N is applied to a block of mass 2kg, it moves along a horizontal table at constant velocity.

(Figure 14)

(a) What is the frictional force?

(b) What is the acceleration if the applied force is increased to 10N?

Figure 14

Solution:

(a) When the block is moving at constant velocity (no acceleration), there is no resultant force on it (Newtons

first law).

This means that the forces are balanced.

From Figure 7, the frictional force = 6N.

(b) When the applied force = 10N, the frictional force is still 6N

Resultant (unbalance) force = (10 6) N = 4N (Figure 15(a))

Figure 15(a) and (b)

Formula first: F= ma

From Figure 15(b), 4N = 2kg x a

222

4 == mskg

Na

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5.3 Newtons third law

Newton noticed that forces were always in pairs and that the two forces were always equal in size but

opposite in direction. The two forces is called action and reaction respectively.

Newtons third law of motion is:

For every force, there is an equal but opposite reaction force.

Note that while the action acts on one body, reaction must act on a different body.

Example:

Now consider a wooden block resting on a table (Figure 16). Write down the equation for the two pairs of

forces.

Figure 16

Solution

There are four forces here, in two pairs!

First pair (red)

The force of the block

on the table

(downwards)

=

The force of the table

on the block

(upwards)

These forces act where the block touches the table.

Second pair (blue)

The force of gravity

of the Earth pulling down =

The force of gravity

of the block pulling up

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on the block

(this weight in effect acts

at the centre of the block)

on the Earth

(this force in effect acts

at the centre of the Earth)

When the block is on the table, these four forces are equal. But if the block is allowed to fall to the floor, only

the second pair of forces exists and as the block moves downwards, the Earth moves slightly upwards!

6.0 MomentA force (the weight) acting at a distance away from the fixed point Pproduces a turning effect. You can feel

the turning effect on your hand when you are holding the rod, The turning effect of a force is called a moment

or a torque. (Figure 17)

Figure 17- The turning effect of a force

The moment of a force depends on the size of the force and the distance of the force from the turning centre

or pivot. It is defined by Fd=

Where = moment of a force, F= force, d= perpendicular distance (from the force to the pivot).

The distance used is always the shortest (perpendicular) distance (Figure 18). Moments are measured in

newton metres (often written N m).

Figure 18:- The distance used in defining moment is the shortest distance

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Example:

In Figure 19, what is the moment of the force, about the nut at point P?

Figure 19

Solution

Perpendicular distance from the force to P= 20cm = 0.20 m

Formula first: Fd=

Then numbers: = 10N x 0.20m

= 2Nm (turning clockwise)

Remarks: For clarity, the direction of the moment should always be stated.

Example:

A uniform rod of weight 100 N is supported by two pages Pand Qas shown in the following Figure 20. A force

of 30 N acts vertically downwards at R. Calculate the reactions at Pand Q.

(Hints: a weight 100N of rod is supported by point P and Q, that mean there is point load of 100N acting on the

centre of the rod)

Figure 20

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Solution:

Take moment about point P,

NR

R

Q

Q

100

)80()100(30)50(100

=

=+

By Newtons Third Law,

NR

RR

P

PQ

30

30100

=

+=+

7.0 Strength of materials

A structural system is not only effected by external conditions, but also by the properties and behavior of the

materials which comprise it. These also determine the nature of the system's reaction(s) to external forces.

The study of Strength of Materials is concerned specifically with the following issues:

1. the internal forces of a member caused by the external forces acting on that member or system.

2. the changes in dimension of a member caused by these forces.

3. the physical properties of the material in the member.

Statics is the study of the behavior of rigid bodies at rest as they are acted upon by external forces. Although

most of these bodies are not absolutely rigid, the assumption of rigidity is valid for the purpose of determining

the reactions of the system. Actually, every material will deform under a load. Even a concrete slab deforms

microscopicly when a person walks on it. Some deformations in a structure can be detrimental to the overall

system's performance, while others might only be an issue of comfort. The recognition of the relative

importance of these deformations will be an important part of the study of structures.

External loads on a structural system create resisting forces within all of the members that form the load path

from the point of the application of the load to the ground beneath the foundation. This internal resistance

exists within every member and joint included in the load path and are known simply as the internal forces

acting on a member.

Some of these forces have already been examined: the connection between a beam and its support, and the

connection of a two-force member to a three-force member in a pin-connected frame. The internal forces

within a beam were demonstrated by cutting a beam. These internal forces were required at the cut section to

put the beam back into equilibrium. The forces and moments that were examined were applied externally to

the end of that cut section; they were exactly equal to the internal forces and moments.

The distribution of the internal forces on the cross-sectional area of a member may or may not be uniformly

distributed; it is dependent on the loading condition, the type of member, and how it is supported.

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8.0 Stress and Strain relationship, Modulus of Elasticity

8.1 Definition of stress

The internal forces of each member of a structural system are distributed in a particular way across that

member's cross-sectional area. Stress is a measure of the intensity of this force on a single unit of area.

Common units of measure of stress are KN/m2, N/mm

2. The distribution of the stress may be constant across

a cross-sectional area, or it may be variable. It can vary due to the loading conditions, the material or

geometry of the structural member.

Two common types of stress

1. Normal to a section

2. Parallel to a section

Figure 21:- Different kinds of stress

8.2 Definition of strain

Strain, is the fractional change per unit of original dimension.

= (Change in dimension of body) / (original dimension of body)

The common types of strain is

(a) Normal strain change in length per unit length

(b) Shear strain distortion, measured by the angle of the distortion, .

Remarks: shear strain will be discussed in the advanced course

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Figure 22:- Strain

Example:

Determine the stress in sectionaand binFigure 23.

Figure 23

Solution

Let tensile stress be positive sign, and compressive stress be negative sign

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Figure 24:- Stress-strain curve for different kind of materials

Figure 25:- stress against strain for two grades of steels

An example plot of a test on two grades of steel is illustrated in Figure 25. If one begins at the origin and

follows the graph a number of points are indicated. Point A is known as the proportional limit. Up to this

point the relationship between stress and strain is exactly proportional. The number, which describes the

relationship between the two, is the Modulus of Elasticity.

Strain increases faster than stress at all points on the curve beyond point A. Up to this point, any steel

specimen that is loaded and unloaded would return to its original length. This is known as elastic behavior.

Point B is the point after which any continued stress results in permanent, or inelastic, deformation. Thus,

point B is known as the elastic limit. Since the stress resistance of the material decreases after the peak of

the curve, this is also known as the yield point.

The line between points C and D indicates the behavior of the steel specimen if it experienced continued

loading to stress indicated as point C. Notice that the dashed line is parallel to the elastic zone of the curve

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(between the origin and point A). When the specimen is unloaded the magnitude of the inelastic deformation

would be determined. If the same specimen was to be loaded again, the stress-strain plot would climb back

up the line from D to C and continue along the initial curve. Point E indicates the location of the value of the

ultimate stress. Note that this is quite different from the yield stress. The yield stress and ultimate stress are

the two values that are most often used to determine the allowable loads for building materials and should

never be confused.

A material is considered to have completely failed once it reaches the ultimate stress. The point of rupture, or

the actual tearing of the material, does not occur until point F. It is interesting to note the curve that indicates

the actual stress experienced by the specimen. This curve is different from the apparent stress since the

cross sectional area is actually decreasing. There is quite a bit to be learned from both the study of the ideal

and actual behavior of all building materials.

* within the elastic range, the ratio of stress to strain for materials is a constant. This constant is referred themodulus of elasticity, E.

To determine the constant, the following formula should be used.

E= Tensile stress / tensile strain =

=AL

lF

All

Fl

lllA

F

=

=

)( 11

where Fis the applied force

A is the cross-sectional area of the object

lis the original length of the object, 1l is length of object after a force is applied.

( llL = 1 ) is the change in the length of the object

Remarks:

Eis the slope of the stress-strain curve (within elastic range)

Emeasure the stiffness of the materials, for example: resistance to deformation due to stress. The higher the

E, the smaller the strain under a given stress.

Example:

Determine E from the following stress-strain curve (steel) in Figure 26.

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Figure 26

Solution:

When stress is2/150 mmN , corresponding strain is 0.001

22

/150000001.0

/150mmN

mmNE ===

Example:

Determine the deformation of the member in Figure 27.

Figure 27

Solution:

Cross section area , A = 150 x 150 = 22500mm2

Internal force in all sections between A and B, kNF 90=

Stress in sections between A and B,

2

2 /422500

90

mmNmm

kN

A

F

=

==

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Strain in sections between A and B,

4

2

2

1067.6/6000

/4 =

== xmmN

mmN

E

Deformation of the member = mmmmx 67.240001067.6 4 = (the member is shortened under the

compressive force kN90 .

Example:

Determine the deflection of the steel road shown in Figure 28 under the given load (2500mmA = ;

25 /103 mmNxE= )

Figure 28:-Steel rod under the force

Solution

(1) Firstly, calculate the internal force of rod in each section

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P 23

Internal Force ,

In portion C-D, kNP 201 = (Tensile Force)

In portion B-D, kNP 3020502 =+= (Compressive force)

In portion A-D, kNP 202050503 =++= (Tensile force)

(2) Calculation the change of dimensions

Change of dimension in portion A-B, mmEA

LFL ABABAB 053.0

103500

40010205

3

=

=

= (extension)

Change of dimension in portion B-C, mmEALFL BCBCBC 08.0

1035004001030

5

3=

=

= (shortening)

Change of dimension in portion C-D, mmEA

LFL CDCDCD 053.0

103500

40010205

3

=

=

= (extension)

(3) The deflection of the rod =0.053 + (-0.08) + 0.053 = 0.026mm (extension)

-END-

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AZ024 Lecture 1 (rev[1] a)