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LECTURE 1

Young's modulus,

Also known as the tensile modulus or elastic modulus is a measure of

the stiffness of an elastic material and is a quantity used to characterize materials.

It is defined as the ratio of the stress along an axis over the strain along that axis in

the range of stress in which Hooke's law holds. In solid mechanics, the slope of

the stress-strain curve at any point is called the tangent modulus. The tangent

modulus of the initial, linear portion of a stress-strain curve is called Young's

modulus. It can be experimentally determined from the slope of a curve created

during tensile tests conducted on a sample of the material. In anisotropic materials,

Young's modulus may have different values depending on the direction of the

applied force with respect to the material's structure.

Elastic properties

For the description of the elastic properties of linear objects like wires, rods,

columns which are either stretched or compressed, a convenient parameter is the

ratio of the stress to the strain, a parameter called the Young's modulus of the

material. Young's modulus can be used to predict the elongation or compression of

an object as long as the stress is less than the yield strength of the material.

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Bulk Elastic Properties

The bulk elastic properties of a material determine how much it will compress

under a given amount of external pressure. The ratio of the change in pressure to

the fractional volume compression is called the bulk modulus of the material.

A representative value for the

bulk modulus for steel is

and that for water is

The reciprocal of the bulk

modulus is called the

compressibility of the substance.

The amount of compression of

solids and liquids is seen to be

very small.

Material Density

(kg/m3)

Young's

Modulus

109 N/m

2

Ultimate Strength

(or “Ultimate

Stress”) Su

106 N/m

2

Yield Strength

(or “Yield

Stress”) Sy

106 N/m

2

Steels 7860 200 400 250

Aluminum 2710 70 110 95

Glass 2190 65 50b ...

Concrete 2320 30 40b ...

Wood 525 13 50b ...

Bone 1900 9b 170

b ...

Polystyrene 1050 3 48 ...

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LECTURE 2

CENTROIDS AND MOMENT INERTIA

Centroids axis in all body and structure

The centroid of a two dimensional surface is a point that corresponds to the center of

gravity of a very thin homogeneous plate of the same area and shape.

Formula centroid gravity

Area centroid gravity

=

0 = r

= bh =0

=

h =

=

b

=

h

b

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Example 1 :

150mm

30mm

70mm

30mm

Find the centroid gravity??

Find the y-axis :

Y1 = (70 +

) = 85

Y2 = (

) = 35

Use formula

(150*30)(85) + (30*70) (35) = [(150*30) + (30*70)] YC

YC = 69.1MM

d1=15.9

d=34.1 85MM

35 69.1

A1

A2

Calculation must from bottom

AIYI + A2Y2 = (A1+A2) YC

A2

A1

YC

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MOMENT INERTIA (I)

Also known as the Second Moment of the Area is a term used to describe the capacity of a

cross-section to resist bending.

It is a mathematical property of a section concerned with a surface area and how that area

is distributed about the reference axis. The reference axis is usually a centroid axis

It is used to determine the state of stress in a section.

It is used to calculate the resistance to bending.

It can be used to determine the amount of deflection in a beam.

FORMULA

Second Moment of the Area

Ixx = Sum (A)(y2)

In which:

Ixx = the moment of inertia around the x axis

A = the area of the plane of the object

y = the distance between the centroid of the object and the x axis

Ix:

Iy:

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Example 2:

5 cm

10 cm 10 cm

5 cm

I xx for a

5 (10 ) ^ 3 / 12 = 416.6 cm4

I xx for b

10 ( 5)^ 3 / 12 = 104 cm4

a

b

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Transfer formula

• There are many built-up sections in which the component parts are not symmetrically

distributed about the centroid axis.

• To determine the moment of inertia of such a section is to find the moment of inertia of

the component parts about their own centroid axis and then apply the transfer formula.

• The transfer formula transfers the moment of inertia of a section or area from its own

centroid axis to another parallel axis. It is known from calculus to be:

FORMULA

Or ( expand from above)

[

] [

]

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Example 3 :

150

150

30

d1=15.9

d=34.1 85MM

35 69.1 70

30

Find the moment inertia.

Use formula : [

] [

]

: [

] [

]

= 4774546

A2

A1

YC

A1

A2

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LECTURE 3

BENDING STRES

Formula

Unit: N/mm^2

WHERE: Z= I/Y Y= center gravity to participate section M= bending moment

F=

=

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Example 4:

100mm

Where M=15KNm

250mm

Find the bending stress.

Y= d/2 = 250/2 = 125mm

I =

=

=130208333.3 mm^4

Use formula : Z= I/Y

Z= 130208333.3/125

=1041666.667mm^3

Then use formula stress :f= m/z

F= (15*10^6)/(1041666.67)

=14.40N/MM^2

CHANGE M TO

MM.

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BENDING FORMULA in Combined Stress

Z is termed as Modulus of section = I/y

I – moment inertia; y – distance from Neutral axis to stress level

M – Applied moment (from loads)

Combined Stress – bending and compression

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EXAMPLE 5 :

1000 mm

100mm xx

yy

Column / wall section

I xx = 1000 (100)^3 /12 = 83.33 e 6 mm4

Zxx = 83.33e6 / 50 = 1.66 e 6 mm3

Axial Load P = 50 kN

Applied Moment = 5 kNm

Total Stress on section = P/A +- M/Z

50e3 / 1e5 + 5 e 6 / 1.66 e6 = 0.5 + 3 = 5.3 N/mm2

Or 0.5 – 3 = -2.5 N/mm2 tensile stress in wall

Compressive stress 5.3N/mm2

+ M/Z =

Tensile stress -2.5N/mm2

Compressive(P)

0.5N/mm2 3N/mm2

POSITIVE – COMPRESION

NEGATIVE – TENSILE

STRESS

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SHEAR STRESSES

The formula for shear stress in beams is a as below

Where :

V - shear force at that section

Q – first moment of area

I – moment of inertia

b- breadth of web

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Example:

for rectangular section :

b- 50 mm ( breadth )

d- 100 mm ( depth ) 50mm

I = 50 (100) ^3 /12 = 4.16e6 mm4 x x

V = 5 kN

Q = Ay = 50 x 50 = 2500 x 50/2

= 62.5e3 mm3

T ( shear stress at XX ) = 5e3 . ( 62.5e3 ) /( 4.16e6). 50 = 1.5N/mm2

Average shear stress = V/A = 5000 / 50 x 100 = 1 N/mm2

Maximum shear stress in timber beam = 0.9 to 1.45 parallel to grain

So above beam may fail if it is made from timber

If concrete beam grade 20 then the allowable shear is 0.26 N/mm2 so this beam

may too

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KOMPOSIT SECTIONS

Sometimes sections of beams or columns can be made of 2 or more different materials such as

a sandwich section or compbined beam and concrete deck as shown below.

Both materials may be bolted or glued but they are supposed to act together / combined as a

whole unit.

In sandwich section the internal core takes the shear and the outer skin takes the bending .

Composite section

If the section is acting as a whole , then the strain across section is still linear .

Outer skin

Inner core

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Ec (compressive strain)

Neutral axis

Et( tensile strain)

Because a stronger materials such as steel has a higher E value compared to concrete , a higher

load is taken by the steel and the stress is not uniform as in a homogenous section.

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We have to convert the section into a “transformed section” so that we can analyze the stress

in the section.

Skin stress

Core stress

Stress in steel

In timber

Steel

Timber

E steel

E timber

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The Modula ratio = 203.4 / 12.1 = 16.8

So in a section shown below the steel plate can be converted into a timber plate by;

B

16.8 B

timber

Steel

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Example :

Determine the maximum bending stress to the timber and steel if the applied moment on the

above section is 5.65 kNm.

E ( steel) = 203.4 kN/mm2

E ( timber) = 12.1 kN/mm2

M ( Modula ratio ) = 203.4/12.1 = 16.8

Transformation B = 75 x 16.8 = 1260 mm

1260mm

6.25mm

137.5mm

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Stress at top of section = 8.15 N/mm2

Actual Steel Stress at this top section = 8.15 x 16.8 = 136.92 N/mm2

Maximum bending stress ( at timber section ) where y = 137.5 / 2 = 68.75mm

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If the sandwich composite section is made of outer skin cement board and inner core

lightweight concrete of density 1000 kg/m3 , find out the maximum bending stress in the

cement board and stress at the LWC section.

Timber

stress

steel

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