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Properties of Sections

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

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

  2. Properties of Sections • Centre of gravity or Centroid • Moment of Inertia • Section Modulus • Shear Stress • Bending Stress

  3. 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 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

  4. 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

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

  7. 60mm 120mm 60mm Example 1: 80mm 120mm 1 3 2 • x = 1944000/14400 = 135 mm

  8. 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

  9. 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

  10. 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

  11. 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

  12. Example 2: • Calculate the moment of inertia of the following structural section H= 212mm 12mm 400mm 24mm 200mm

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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.

  19. 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

  20. 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. Shear Stress

  22. 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

  23. Solution: 3 y=50mm 4 y=25mm 1 5 y=0mm 2

  24. Answer:

  25. 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

  26. Bending Stress Distribution • f=σ= bending stress = My/I

  27. 50mm x x 200mm M= 2.0 kNm

  28. UFO lawat Menara KL?

  29. Material Properties Concrete

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

  31. Modulus Elasticity, E Es(Modulus Elasticity of steel reinforcement) = 200 kN/mm2

  32. Poisson ratio υc • Refer to Clause 2.4.2.4 BS 8110 • The value = 0.2

  33. 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)

  34. Material Properties Steel

  35. Design strength, py

  36. 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

  37. 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. Aspect ratio • ε=(275/py)0.5

  39. Types of sections • I sections • H sections • Rectangular Hollow Sections (RHS) • Circular Hollow Sections (CHS) • Angles (L shape or C shapes)

  40. Types of sections

  41. Material Properties Timber

  42. 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

  43. Material Properties • Elastic Modulus E = 4600 – 18000 N/mm2 • Poisson’s Ratio υ = 0.3

  44. Grade of timber • Timber can be graded by • Visual Inspection • Machine strength grading • 3 grade only • Select • Standard • Common Less defect

  45. Group of timber • We have Group • A • B • C • D Lower strength

  46. Defects by nature

  47. 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. Typical sawing pattern

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

  50. Continue structure design...

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