1. Properties of Sections ERT 348 Controlled Environmental Design 1 Biosystem Engineering

2. 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

3. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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

4. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Example 1:  Density=8000 kg/m 3  Thick=10 mm  Determine the position of centre of gravity x mm W 1 = 0.08m x 0.06m x 0.01m x 8000kg/m 3 x 10 N/kg = 3.84 N W 2 = 0.02m x 0.12m x 0.01m x 8000kg/m 3 x 10 N/kg = 1.92 N W 3 = 0.12m x 0.06m x 0.01m x 8000kg/m 3 x 10 N/kg = 5.76 N 80mm 60mm120mm 60mm 1 2 3

5. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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

6. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Centroid  Centre of gravity of an area also called as centroid.

7. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at  x = 1944000/14400 = 135 mm 80mm Areaa (mm 2 )y (mm)∑ay 160x80=480030144000 220x120=2400mm 2 120288000 360x120=7200mm 2 2101512000 Total14400 mm 2 1944000mm 3 60mm120mm 60mm 1 2 3 120mm Example 1:

8. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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. xx b d y y Unit : mm 4 or cm 4

9. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Moment of Inertia of common shapes  Rectangle at one edge  I uu = bd 3 /3  I vv = db 3 /3  Triangle  I xx = bd 3 /36  I nn = db 3 /6 xx v v d b b d u u nn

10. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Moment of Inertia of common shapes  Ixx= Iyy = πd 4 /64  Ixx = (BD 3 -bd 3 )/12  Iyy = (DB 3 -db 3 )/12 xx y y B D b d xx y y

11. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Principle of parallel axes  Izz = Ixx + AH 2  Example: b=150mm;d=100mm; H=50mm  Ixx= (150 x 100 3 )/12 = 12.5 x 10 6 mm4  Izz = Ixx + AH 2 = 12.5 x 10 6 + 15000(50 2 ) = 50 x 106 mm 4 H zz xx

12. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Example 2:  Calculate the moment of inertia of the following structural section 400mm 200mm 24mm 12mm H= 212mm

13. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Solution  I xx of web = (12 x400 3 )/12= 64 x 10 6 mm 4  I xx of flange = (200x24 3 )/12= 0.23 x 10 6 mm 4  I xx from principle axes xx = 0.23 x10 6 + AH 2 AH 2 = 200 x 24 x 212 2 = 215.7 x 10 6 mm 4 I xx from x-x axis = 216 x 10 6 mm 4  Total I xx = (64 + 2 x 216) x10 6 =496 x 10 6 mm 4

14. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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 mm 3  Example for rectangle shape:  I xx = bd 3 /12, c = d/2, Z xx = I xx /c = bd 2 /6

15. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Section Modulus, Z  Z = 1/y  f = M/Z = My/I  Safe allowable bending moment, M max = f.Z where f = bending stress y = distance from centroid

16. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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/mm 2. What maximum bending moment in N.mm?  Z= bd 2 /6 = 150 x 300 2 /6 = 2.25 x 10 6 mm 3  M max = f.Z = 6 x 2.25x10 6 = 13.5x10 6 Nmm 2

17. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at If non-symmetrical sections M rc = 1/y 1 x compression stress M rt = 1/y 2 x tension stress Safe bending moment = f x least Z xx y2y2 y1y1

18. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Example 4  Figure shows an old type cast iron joist (in mm) with a tension flange of 9600 mm 2, a compression flange 2880 mm 2 and a web of 7200 mm 2.  The safe stress in compression is 5 N/mm 2 and in tension 2.5 N/mm 2. What is the safe bending moment for the section? What safe uniform load will the beam carry on a 4.8m span. xx 137 235 120 200 24 48 300

19. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Solution  x =∑ay/∑a = (2880x360 + 7200x198 + 9600x24) (2880 + 7200 + 9600) = 137 mm  Total I xx = (120x24 3 /12 + 2880x223 2 )+ (24x300 3 /12 + 7200x61 2 ) + (200x48 3 /12 + 960x113 2 ) = 348.57 x 10 6 mm 4  Mrt = 2.5 x 348.57 x 10 6 = 6.36x10 6 Nmm  Mcr= 5.0 x 348.57 x 10 6 = 7.42x10 6 Nmm  WL/8 = 6.36x10 6 Nmm  W= 6.36 x 8/4.8 = 10.6 kN

20. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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.

21. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Shear Stress

22. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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 xx y y

23. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Solution: 1 2 3 4 5 y=50mm y=25mm y=0mm

24. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Answer:

25. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Answer:  Average τ = V/A = 3 x 10 3 /(50x200) = 0.3 N/mm 2  Maximum = 1.5 V/A = 1.5x0.3 = 0.45 N/mm 2

26. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Bending Stress Distribution  f=σ= bending stress = My/I

27. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 200mm 50mm xx M= 2.0 kNm

28. UFO lawat Menara KL?

29. Material Properties Concrete

30. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Concrete  Concrete compressive strength: f cu  C30,C35,C40,C45 and C50  Where the number represent compressive strength in N/mm 2

31. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Modulus Elasticity, E E s (Modulus Elasticity of steel reinforcement) = 200 kN/mm 2

32. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Poisson ratio υ c  Refer to Clause 2.4.2.4 BS 8110  The value = 0.2

33. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Steel Reinforcement Strength: f y  Refer to Clause 3.1.7.4 BS 8110 (Table 3.1)  f y = 250 N/mm 2 for hot rolled mild steel (MS)  f y = 460 N/mm 2 for hot rolled or cold worked high yield steel (HYS)

34. Material Properties Steel

35. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Design strength, p y

36. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Modulus Elasticity, E  Modulus elasticity, E = 205 000 N/mm2 Poisson ratio υ c  The value = 0.3 Shear Modulus,G  G=E/[2(1+υ)] = 78.85 x10 3 N/mm 2

37. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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

38. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Aspect ratio  ε=(275/p y ) 0.5

39. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Types of sections  I sections  H sections  Rectangular Hollow Sections (RHS)  Circular Hollow Sections (CHS)  Angles (L shape or C shapes)

40. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Types of sections

41. Material Properties Timber

42. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Moisture content m 1 =mass before drying m 2 =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

43. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Material Properties  Elastic Modulus E = 4600 – 18000 N/mm 2  Poisson’s Ratio υ = 0.3

44. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Grade of timber  Timber can be graded by Visual Inspection Machine strength grading  3 grade only Select Standard Common Less defect

45. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Group of timber  We have Group A B C D Lower strength

46. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Defects by nature

47. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at 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

48. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Typical sawing pattern

49. ERT 348 Controlled Environment Design 1 Siti Kamariah Binti Md Sa’at Reference to design  MS 544 : Pt.1-Pt.11 : 2001 - Code of Practice for Structural Use of Timber

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