properties of sections ert 348 controlled environmental design 1 biosystem engineering

Properties of Sections

ERT 348

Controlled Environmental Design 1

Biosystem Engineering

ERT 348 Controlled Environment Design 1

Siti Kamariah Binti Md Sa’at

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:1. Divide the body to several parts

2. Determine the area@ volume of each part

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

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



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



80mm

60mm 120mm 60mm

1 2 3



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.

a

axx

a

ayy



x = 1944000/14400 = 135 mm

80mm

Area a (mm2) y (mm) ∑ay

1 60x80=4800 30 144000

2 20x120=2400mm2 120 288000

3 60x120=7200mm2 210 1512000

Total 14400 mm2 1944000mm3

60mm 120mm 60mm

1 2 3

120m

m

Example 1:



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.

12

bdI

3

xx

12

dbI

3

yy

x x

b

d

y

y

Unit :

mm4 or cm4



Moment of Inertia of common shapes

Rectangle at one edge

Iuu= bd3/3

Ivv= db3/3

Triangle Ixx= bd3/36

Inn= db3/6

xx

v

v

d

b

b

d

uu nn



Moment of Inertia of common shapes

Ixx= Iyy = πd4/64 Ixx = (BD3-bd3)/12 Iyy = (DB3-db3)/12

x x

y

y

B

D

b

dx x

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

Hz z

x x



Example 2: Calculate the moment of inertia of the

following structural section

400mm

200mm

24mm

12mmH= 212mm



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

c

IZ xx

xx



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



If non-symmetrical sections

Mrc

= 1/y1 x compression stress

Mrt

= 1/y2 x tension stress

Safe bending moment

= f x least Z

x x

y2

y1



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 x

137

235

120

200

24

48

300



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.



Shear Stress



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

200mm

50mm

x xy y



Solution:

4633

10x33.3312

200x50

12mm

bdI

Ib

yA

26

33

75

33

2

N/mm197.050x 33.33x10

0x109.375x13x10τ

mm109.375x10yA

mm5.872/2575

mm 1250 50x25A 75mm,y

0 0,A then 100mm,y

y

1

2

3

4

5

y=50mm

y=25mm

y=0mm



Answer:



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

f=σ= bending stress = My/I



200mm

50mm

x x

M= 2.0 kNm

UFO lawat Menara KL?

Material Properties

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)

Material Properties

Steel



Design strength, py



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)



Types of sections

Material Properties

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



Group of timber We have Group

A B C D

Lower strength



Defects by nature



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



Typical sawing pattern



Reference to design MS 544 : Pt.1-Pt.11 : 2001 - Code of Practice

for Structural Use of Timber

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properties of sections ert 348 controlled environmental design 1 biosystem engineering

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