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# Properties of Sections - PowerPoint PPT Presentation

Properties of Sections. ERT 348 Controlled Environmental Design 1 Biosystem Engineering. Properties of Sections. Centre of gravity or Centroid Moment of Inertia Section Modulus Shear Stress Bending Stress. Centre of gravity.

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### Properties of Sections

ERT 348

Controlled Environmental Design 1

Biosystem Engineering

• Centre of gravity or Centroid

• Moment of Inertia

• Section Modulus

• Shear Stress

• Bending Stress

• A point which the resultant attraction of the earth eg. the weight of the object

• To determine the position of centre of gravity, the following method applies:

• Divide the body to several parts

• Determine the area@ volume of each part

• Assume the area @ volume of each part at its centre of gravity

• Take moment at convenient axis to determine centre of gravity of whole body

120mm

60mm

80mm

1

3

2

Example 1:

• Density=8000 kg/m3

• Thick=10 mm

• Determine the position of centre of gravity

x mm

W1= 0.08m x 0.06m x 0.01m x 8000kg/m3 x 10 N/kg

= 3.84 N

W2= 0.02m x 0.12m x 0.01m x 8000kg/m3 x 10 N/kg

= 1.92 N

W3= 0.12m x 0.06m x 0.01m x 8000kg/m3 x 10 N/kg

= 5.76 N

• Resultant, R = 3.84 +1.92 + 5.76 N

= 11.52 N

• Rx = 3.84 (30) + 1.92 (60+60) + 5.76 (120 + 60+30)

• x = 1555/11.52 = 135 mm

• Centre of gravity of an area also called as centroid.

120mm

60mm

Example 1:

80mm

120mm

1

3

2

• x = 1944000/14400 = 135 mm

• Or called second moment of area, I

• Measures the efficiency of that shape its resistance to bending

• Moment of inertia about the x-x axis and y-y axis.

y

d

Unit :

mm4 or cm4

x

x

b

y

Iuu= bd3/3

Ivv= db3/3

Triangle

Ixx= bd3/36

Inn= db3/6

Moment of Inertia of common shapes

v

b

d

d

x

x

u

u

n

n

v

b

Ixx= Iyy = πd4/64

Ixx = (BD3-bd3)/12

Iyy = (DB3-db3)/12

Moment of Inertia of common shapes

y

B

y

d

x

D

x

x

x

b

y

y

Example: b=150mm;d=100mm; H=50mm

Ixx= (150 x 1003)/12

= 12.5 x 106 mm4

Izz = Ixx + AH2

= 12.5 x 106 + 15000(502)

= 50 x 106 mm4

Principle of parallel axes

x

x

H

z

z

• Calculate the moment of inertia of the following structural section

H= 212mm

12mm

400mm

24mm

200mm

• Ixx of web = (12 x4003)/12= 64 x 106 mm4

• Ixx of flange = (200x243)/12= 0.23 x 106 mm4

• Ixx from principle axes xx = 0.23 x106 + AH2

AH2 = 200 x 24 x 2122 = 215.7 x 106 mm4

Ixx from x-x axis = 216 x 106 mm4

• Total Ixx = (64 + 2 x 216) x106 =496 x 106 mm4

• Second moment of area divide by distance from axis

• Where c = distance from axis x-x to the top of bottom of Z.

• Unit in mm3

• Example for rectangle shape:

• Ixx = bd3/12, c = d/2, Zxx= Ixx/c = bd2/6

• Z = 1/y

• f = M/Z = My/I

• Safe allowable bending moment, Mmax = f.Z

where

• f = bending stress

• y = distance from centroid

• A timber beam of rectangular cross section is 150mm wide and 300mm deep. The maximum allowable bending in tension and compression must not exceed 6 N/mm2. What maximum bending moment in N.mm?

• Z= bd2/6 = 150 x 3002/6 = 2.25 x 106 mm3

• Mmax= f.Z = 6 x 2.25x106 = 13.5x106 Nmm2

Mrc

= 1/y1 x compression stress

Mrt

= 1/y2 x tension stress

Safe bending moment

= f x least Z

If non-symmetrical sections

y1

x

x

y2

24

235

300

x

x

137

48

200

Example 4

• Figure shows an old type cast iron joist (in mm) with a tension flange of 9600 mm2, a compression flange 2880 mm2 and a web of 7200 mm2.

• The safe stress in compression is 5 N/mm2 and in tension 2.5 N/mm2. What is the safe bending moment for the section? What safe uniform load will the beam carry on a 4.8m span.

• x =∑ay/∑a

= (2880x360 + 7200x198 + 9600x24)

(2880 + 7200 + 9600)

= 137 mm

• Total Ixx = (120x243/12 + 2880x2232)+ (24x3003/12 + 7200x612) + (200x483/12 + 960x1132) = 348.57 x 106 mm4

• Mrt = 2.5 x 348.57 x 106 = 6.36x106 Nmm

• Mcr= 5.0 x 348.57 x 106 = 7.42x106 Nmm

• WL/8 = 6.36x106 Nmm

• W= 6.36 x 8/4.8 = 10.6 kN

• The shear forces induced in a beam by an applied load system generate shear stresses in both the horizontal and vertical directions.

• At any point in an elastic body, the shear stresses in two mutually perpendicular directions are equal to each other in magnitude.

x

x

200mm

Example 5: Elastic Shear Stress

• The rectangular beam shown in Figure is subject to a vertical shear force of 3.0 kN.

• Determine the shear stress distribution throughout the depth of the section.

A

y

y

3

y=50mm

4

y=25mm

1

5

y=0mm

2

• Average τ = V/A = 3 x 103/(50x200)

= 0.3 N/mm2

• Maximum = 1.5 V/A = 1.5x0.3 = 0.45 N/mm2

• f=σ= bending stress = My/I

x

x

200mm

M= 2.0 kNm

### Material Properties

Concrete

• Concrete compressive strength: fcu

• C30,C35,C40,C45 and C50

• Where the number represent compressive strength in N/mm2

Es(Modulus Elasticity of steel reinforcement) = 200 kN/mm2

• Refer to Clause 2.4.2.4 BS 8110

• The value = 0.2

• Refer to Clause 3.1.7.4 BS 8110 (Table 3.1)

• fy= 250 N/mm2 for hot rolled mild steel (MS)

• fy= 460 N/mm2 for hot rolled or cold worked high yield steel (HYS)

### Material Properties

Steel

• Modulus elasticity, E = 205 000 N/mm2

Poisson ratio υc

• The value = 0.3

Shear Modulus,G

• G=E/[2(1+υ)] = 78.85 x103 N/mm2

• Refer to Table 11 & Table 12 BS5950-1:2000 Clause 3.5

• Class 1 - Plastic Sections

• Class 2 - Compact Sections

• Class 3 - Semi-compact Sections

• Class 4 - Slender Sections

• ε=(275/py)0.5

• I sections

• H sections

• Rectangular Hollow Sections (RHS)

• Circular Hollow Sections (CHS)

• Angles (L shape or C shapes)

### Material Properties

Timber

• m1=mass before drying

• m2=mass after drying

• Unit in %

• The strength of timber is based on its moisture content.

• In MS 544, the moisture content – 19%

• >19% - moisture

• <19% - dry

• Elastic Modulus E = 4600 – 18000 N/mm2

• Poisson’s Ratio υ = 0.3

• Timber can be graded by

• Visual Inspection

• Select

• Standard

• Common

Less defect

• We have Group

• A

• B

• C

• D

Lower strength

• In addition to the defects indicated in Figure 7.1 there are a number of naturally occurring

• defects in timber. The most common and familiar of such defects is a knot

• MS 544 : Pt.1-Pt.11 : 2001 - Code of Practice for Structural Use of Timber