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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

Properties of Sections

ERT 348

Controlled Environmental Design 1

Biosystem Engineering

properties of sections1
Properties of Sections
  • Centre of gravity or Centroid
  • Moment of Inertia
  • Section Modulus
  • Shear Stress
  • Bending Stress
centre of gravity
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
example 1

60mm

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 11
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
Centroid
  • Centre of gravity of an area also called as centroid.
example 12

60mm

120mm

60mm

Example 1:

80mm

120mm

1

3

2

  • x = 1944000/14400 = 135 mm
moment of inertia i
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

moment of inertia of common shapes
Rectangle at one edge

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

moment of inertia of common shapes1
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

principle of parallel axes
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 axes

x

x

H

z

z

example 2
Example 2:
  • Calculate the moment of inertia of the following structural section

H= 212mm

12mm

400mm

24mm

200mm

solution
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
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 z1
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
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
if non symmetrical sections
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

example 4

120

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.
solution1
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
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.
example 5 elastic shear stress

50mm

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

solution2
Solution:

3

y=50mm

4

y=25mm

1

5

y=0mm

2

answer1
Answer:
  • Average τ = V/A = 3 x 103/(50x200)

= 0.3 N/mm2

  • Maximum = 1.5 V/A = 1.5x0.3 = 0.45 N/mm2
bending stress distribution
Bending Stress Distribution
  • f=σ= bending stress = My/I
slide27

50mm

x

x

200mm

M= 2.0 kNm

concrete
Concrete
  • Concrete compressive strength: fcu
  • C30,C35,C40,C45 and C50
  • Where the number represent compressive strength in N/mm2
modulus elasticity e
Modulus Elasticity, E

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

poisson ratio c
Poisson ratio υc
  • Refer to Clause 2.4.2.4 BS 8110
  • The value = 0.2
steel reinforcement strength f y
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)
modulus elasticity e1
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
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
Aspect ratio
  • ε=(275/py)0.5
types of sections
Types of sections
  • I sections
  • H sections
  • Rectangular Hollow Sections (RHS)
  • Circular Hollow Sections (CHS)
  • Angles (L shape or C shapes)
moisture content
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 properties3
Material Properties
  • Elastic Modulus E = 4600 – 18000 N/mm2
  • Poisson’s Ratio υ = 0.3
grade of timber
Grade of timber
  • Timber can be graded by
    • Visual Inspection
    • Machine strength grading
  • 3 grade only
    • Select
    • Standard
    • Common

Less defect

group of timber
Group of timber
  • We have Group
    • A
    • B
    • C
    • D

Lower strength

defects in timber
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
Reference to design
  • MS 544 : Pt.1-Pt.11 : 2001 - Code of Practice for Structural Use of Timber
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