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## PowerPoint Slideshow about ' Properties of Sections' - reagan-lucas

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### Material Properties

### Material Properties

### Material Properties

Properties of Sections

- Centre of gravity or Centroid
- Moment of Inertia
- Section Modulus
- Shear Stress
- Bending Stress

Centre of gravity

- 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 [email protected] 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

Example 1:

- 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

Centroid

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

Moment of inertia, I

- 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

Rectangle at one edge

Iuu= bd3/3

Ivv= db3/3

Triangle

Ixx= bd3/36

Inn= db3/6

Moment of Inertia of common shapesv

b

d

d

x

x

u

u

n

n

v

b

Izz = Ixx + AH2

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 axesx

x

H

z

z

Example 2:

- Calculate the moment of inertia of the following structural section

H= 212mm

12mm

400mm

24mm

200mm

Solution

- 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

Section Modulus, Z

- 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

Section Modulus, Z

- Z = 1/y
- f = M/Z = My/I
- Safe allowable bending moment, Mmax = f.Z

where

- f = bending stress
- y = distance from centroid

Example 3

- 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 sectionsy1

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.

Solution

- 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

Elastic Shear Stress Distribution

- 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

Bending Stress Distribution

- f=σ= bending stress = My/I

Concrete

Concrete

- Concrete compressive strength: fcu
- C30,C35,C40,C45 and C50
- Where the number represent compressive strength in N/mm2

Modulus Elasticity, E

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

Poisson ratio υc

- Refer to Clause 2.4.2.4 BS 8110
- The value = 0.2

Steel Reinforcement Strength: fy

- 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)

Steel

Modulus Elasticity, E

- 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

Section classification

- 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

Aspect ratio

- ε=(275/py)0.5

Types of sections

- I sections
- H sections
- Rectangular Hollow Sections (RHS)
- Circular Hollow Sections (CHS)
- Angles (L shape or C shapes)

Timber

Moisture content

- 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

Material Properties

- Elastic Modulus E = 4600 – 18000 N/mm2
- Poisson’s Ratio υ = 0.3

Grade of timber

- Timber can be graded by
- Visual Inspection
- Machine strength grading
- 3 grade only
- Select
- Standard
- Common

Less defect

Defects in timber

- 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

Reference to design

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

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