Chapter 2 basic structure concepts
- 1. Chapter 2 Basic Structure Concepts January2015 1
- 2. 2.0 : BASIC STRUCTURE CONCEPT 2.1 Forces 2.2 Equilibrium and Reactions 2.3 Moments 2.4 Stress and Strain 2.5 Elastic and plastic range 2.6 Primary Loads & Secondary Loads January2015 2
- 3. 2.1 Forces A force is that which tends to exert motion, tension or compression on an object. The loads acting on a structure have mass which is usually measured in kilograms(kg). The basic unit of force is the newton (N). The force exerted on a structure by a static load is dependent upon both its mass and the force of gravity. The force exerted by a body as a result of gravity can be described as its weight. F =W= m x g Therefore, the force exerted by a mass of 1 kg is: F = m x g = 1 x 9.81 =9.81 N @ 0.00981 kN 1 kN = 1000 N January2015 3
- 4. 2.1 Forces Scalars and Vectors Scalars Vectors Examples: mass, volume force, velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font, a line, an arrow or a “carrot” January2015 4
- 5. 2.1 Forces Application of Vector Addition There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket ? January2015 5
- 6. 2.1 Forces Vector Operations Scalar Multiplication and Division January2015 6
- 7. 2.1 Forces Vector Addition Using Either the Parallelogram or Triangle Parallelogram Law: Triangle method (always ‘tip to tail’): How do you subtract a vector? How can you add more than two concurrent vectors graphically ? January2015 7
- 8. 2.1 Forces Resolution of a vector “Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse. January2015 8
- 9. 2.1 Forces Each component of the vector is shown as a magnitude and a direction. Cartesian Vector Notation We ‘ resolve’ vectors into components using the x and y axes system The directions are based on the x and y axes. We use the “unit vectors” i and j to designate the x and y axes. January2015 9
- 10. 2.1 Forces For example, F = Fx i + Fy j or F' = F'x i + F'y j The x and y axes are always perpendicular to each other. Together, they can be directed at any inclination. January2015 10
- 11. 2.1 Forces Addition of Several Vectors Step 1 is to resolve each force into its components Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector. Step 3 is to find the magnitude and angle of the resultant vector. January2015 11
- 14. 2.1 Forces Example 1 The screw eye in Figure below is subjected to forces F1 and F2. Determine the magnitude and direction (measured from x- positive axis) of the resultant force. F2= 150 N F1= 100 N10o 15o January2015 14
- 15. 2.1 Forces Solution 10o 15o 15o + 90o + 10o = 115o R =√(1002 +1502 – (2x 100 x 150 x cos 115 ) = 212.55 N Sin /150 = sin 115/212.55 Sin = (sin 115/212.55) x 150 = 39.76o Direction = 15o + 39.76o =54.76o January2015 15
- 16. 2.1 Forces Example 2 Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x-axis Solution a) Resolve the forces in their x-y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. January2015 16
- 17. 2.1 Forces F1 = { 15 sin 40° i + 15 cos 40° j } kN = { 9.642 i + 11.49 j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN = { -24 i + 10 j } kN F3 = { 36 cos 30° i – 36 sin 30° j } kN = { 31.18 i – 18 j } kN January2015 17
- 18. 2.1 Forces Summing up all the i and j components respectively, we get, FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN = { 16.82 i + 3.49 j } kN FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN = tan-1(3.49/16.82) = 11.7° January2015 18
- 19. 2.1 Forces Example 3 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis Solution a) Resolve the forces in their x-y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. January2015 19
- 20. 2.1 Forces F1 = { (4/5) 850 i - (3/5) 850 j } N = { 680 i - 510 j } N F2 = { -625 sin(30°) i - 625 cos(30°) j } N = { -312.5 i - 541.3 j } N F3 = { -750 sin(45°) i + 750 cos(45°) j } N { -530.3 i + 530.3 j } N January2015 20
- 21. 2.1 Forces Summing up all the i and j components respectively, we get, FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N = { - 162.8 i - 521 j } N FR = ((162.8)2 + (521)2) ½ = 546 N = tan–1(521/162.8) = 72.64° or From Positive x axis = 180 + 72.64 = 253 ° January2015 21
- 22. TUTORIAL 1 (FORCES) Question 1 Two forces are applied to an eye bolt fastened to a beam. Determine the magnitude and direction of their resultant. 6 kN 4.5 kN 25o 50o January2015 22
- 23. Question 2 Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 60 kN and Q = 100 kN. Determine the magnitude and direction of their resultant. (Ans: R = 150 kN, 76o towards x axis positive) P Q 15o 30o January2015 24
- 24. Question 3 The cables AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N in AD. Determine the magnitude and direction of the resultant of the forces exerted by cables at A. (Ans: R = 575 N, 113o towards x axis positive) A B C D 2 m 1.5 m 2.5 m January2015 25
- 25. Question 4 Determine the resultant and direction from x-axis positive of the five forces shown in Figure below by the graphical method. (Ans: R = 32.5 kN, = 124o) y x 8 kN 9 kN 4 kN 60o 25 kN 3 kN 15o 30o 20o January2015 26
- 26. Question 5 Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force of the three acting on the ring A. Take F1 = 500 N and = 20o (Ans: R = 1.03 kN, = 87.9o) y x 600 N F1 400 N 30o 3 4 5 January2015 27
- 27. Question 6 Three forces are applied at the end of the boom O. Determine the magnitude and orientation of the resultant force. (Ans: FR= 485 N, = 37.7o) y x F3 = 200N F2= 250 N F1 = 400 N 45o 3 4 5 January2015 29
- 28. • Question 7 • Two cables are attached to the frame shown in Figure below. Using trigonometry, determine : (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical. (b) the corresponding magnitude of R. (Ans: 489 N, 738 N) A A 25o 35o 360 N P January2015 30
- 29. 2.2 Equilibriumand Reactions Any structure subjected to loads must be provided with supports to prevent it from moving. The forces generated on the structure by these supports are called reactions. If the structure is in equilibrium (i.e. not moving) then the net forces from the loads and reactions must be zero in all directions. Fx = 0 Fy = 0 M = 0 January2015 31
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- 31. Example 1 Determine the magnitudes of F1 and F2 so that particle P is in equilibrium. (Ans: F1 = 435 N, F2 = 171 N) P F1 F2 400 N 3 4 5 30o 60o January2015 33
- 32. Solution The two unknown magnitudes F1 and F2 can be obtained from the two scalar equations of equilibrium, ∑Fx=0 and ∑Fy = 0. To apply these equations, the x, y axes are established on the free body diagram and forces must be resolved into its x and y components. P F1 F2 400 N 3 4 5 30o 60o 400 cos 30o 400 sin 30o F2 cos 60o F2 sin 60o F1 (4/5) F1 (3/5) January2015 34
- 33. Solution ∑ Fx =0 ; -400 sin 30o + F1(4/5) –F2sin 60o=0 ∑ Fy =0; 400 cos 30o -F1(3/5) –F2cos 60o=0 (1) (2) Simplify : 0.8 F1 – 0.866 F2 = 200 (3) -0.6 F1 – 0.5 F2 = -346.61 (4) Solving Eqs. (3) & (4) by simultaneous equations to determine F1 & F2. Ans : F1 = 435.08 N F2 = 170.97 N January2015 35
- 34. Applicationof equilibriumconceptsin truss A truss is a structure composed of slender members joined together at their end points. The members commonly used in construction consist of wooden struts or metal bars The joint connections are usually formed by bolting or welding the ends of the members to a common plate, called a gusset plate, as shown in Fig below. January2015 36
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- 36. 38 Timber Roof Trusses Steel Bridge Trusses January2015
- 37. Example • Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. 6 kN 6 kN 6 kN 4 m 3 m 3 m 3 m 3 m A B C D E F G H January2015 39
- 38. 2.3 Moments APPLICATION What is the net effect of the two forces on the wheel? January2015 40
- 39. APPLICATION What is the effect of the 30 N force on the lug nut? January2015 41
- 40. The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis. This tendency for rotation caused by force is sometimes called a torque, but most often it is called the moment of a force or simply the moment The magnitude of a moment about a point is the value of the force multiplied by the perpendicular distance from the line of action of the force to the point M= T = F x d The unit is Nm or kNm The typical sign convention for moment is that counter-clockwise is considered positive. January2015 42
- 41. Example 1 From Figure below, determine the moment of the force about point O. If the moment is increased to 5 kNm, what is the maximum mass can be supported by the diving board? 1.5 m Mass = 250 kg January2015 43
- 42. Solution F = mg = 250 (9.81)= 2452.5 N = 2.45 kN 1.5 m Mo = ( 2.45 x 1.5) = 3.68 kNm If Mo = 5 kNm, then new load is Mo = F x d 5 = F x 1.5 F = 3.33 kN = 3333.33 N F = m x g 3333.33 = m x 9.81 m = 339.79 kg F January2015 44
- 43. Example 2 For each case illustrated in Figure below, determine the moment of the force about point O. 5O kN 2 m 0.75 m Solution Mo = - (50 x 0.75) = - 37.5 kNm January2015 45
- 44. Example 3 For each case illustrated in Figure below, determine the moment of the force about point O. 6O kN 3 m 45o Solution 1 sin 45o Mo = 60 x (1 sin 45o) = 42.43 kNm January2015 46
- 45. Example 4 A 400 N force is applied to the frame and = 20o. Find the moment of the force at A. Solution 400 sin 20o 400 cos 20o MA = (400 sin 20ox 3) + (400 cos 20o x 2) = 1162.18 Nm January2015 47
- 46. Figure above shows a bridge deck, of weight 500 kN, supporting a heavy vehicle weighing 300 kN. Find the value of the support reactions at A and B when the load is in the position shown. Ans: RA = 430 kN, RB = 370 kN Example 5 January2015 48
- 47. Solution 500 kN 5 m 300 kN A B 4 m ∑ MA =0 ; ∑ Fy =0 ; RB (10) – ( 300 x 4) – (500 x 5) = 0 RB = 370 kN RA + RB – 300 – 500 = 0 RA = 430 kN January2015 49
- 48. Example 6 Figure shows an anchorage for the cable of a ski- lift, The cable produce a total pull of 220 kN at the top of the anchorage. Determine the support reactions. Ans: RAH = 191 kN, RAV = -196 kN RB = 306 kN January2015 50
- 49. Solution A B 220 cos 30o 220 sin 30o RBRAV RAH ∑ MA =0 ; RB (2.7) – ( 220 sin 30o x 1.8) – (220 cos 30o x 3.3) = 0 RB = 306 kN ∑ Fx =0 ; -RAH + 220 cos 30o = 0 RAH = 191kN RAV + RB – 220 sin 30o = 0 RAV = - 196 kN ∑ Fy =0 ; January2015 51
- 50. 2.4 Stress and Strain Stress Concept of stress To obtain distribution of force acting over a sectioned area Assumptions of material: 1. It is continuous (uniform distribution of matter) 2. It is cohesive (all portions are connected together) Normal stress Intensity of force, or force per unit area, acting normal to ΔA Symbol used for normal stress, is σ (sigma) σz = lim ΔA →0 ΔFz ΔA January2015 52
- 51. Tensile stress: normal force “pulls” or “stretches” the area element ΔA Compressive stress: normal force “pushes” or “compresses” area element ΔA Shear stress Intensity of force, or force per unit area, acting tangent to ΔA Symbol used for normal stress is τ (tau) τzx = lim ΔA →0 ΔFx ΔA τzy = lim ΔA →0 ΔFy ΔA Units (SI system) Newtons per square meter (N/m2) or a pascal (1 Pa = 1 N/m2) kPa = 103 N/m2 (kilo-pascal) MPa = 106 N/m2 (mega-pascal) GPa = 109 N/m2 (giga-pascal) January2015 53
- 53. Example 1 The 80 kg lamp is supported by two rods AB and BC as shown in Figure below. If AB has a diameter of 10 mm and BC has a diameter of 8 mm, determine the average normal stress in each rod. A B C 60o 3 4 January2015 55
- 54. A B C 60o 3 4 F = mg = 80 x 9.81 = 784.8 N = 0.78 kN F1 F2 F1cos 60o F1sin 60o F2cos F2sin Solution January2015 56
- 55. Solution ∑ Fx =0 ; – F1cos 60o + F2(4/5) =0 ∑ Fy =0; F1sin 60o+ F2(3/5) – 0.78 =0 (1) (2) Simplify : - 0.5 F1 + 0.8 F2 = 0 (3) 0.866 F1 + 0.6 F2 = 0.78 (4) Solving Eqs. (3) & (4) by simultaneous equations to determine F1 & F2. 0.6 x Eqs (3) - 0.3 F1 + 0.48 F2 = 0 0.8 x Eqs (4) 0.69 F1 + 0.48 F2 = 0.624 - 0.99 F1 = - 0.624 F1 = 0.63 kN Subs F1 into Eqs (3) - - 0.5 F1 + 0.8 F2 = 0 - 0.5 x 0.63 + 0.8 F2 = 0 F2 = 0.315/0.8 = 0.39 kN January2015 57
- 56. Solution 1 = F1 / A1 = 0.63 /( x 0.0052) = 8021.41 kN/m2 2 = F2 / A2 = 0.39/( x 0.0042) = 7758.8 kN/m2 January2015 58
- 57. Shear Stress Shear stress is the stress component that act in the plane of the sectioned area. Consider a force F acting to the bar For rigid supports, and F is large enough, bar will deform and fail along the planes identified by AB and CD Free-body diagram indicates that shear force, V = F/2 be applied at both sections to ensure equilibrium January2015 59
- 58. Average shear stress over each section is: P Aavg = avg = average shear stress at section, assumed to be same at each pt on the section V = internal resultant shear force at section determined from equations of equilibrium A = area of section January2015 60
- 59. Example 1 The bar shown in Figure below has a square cross section for which the depth and thickness are 40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar’s cross-sectional area, determine average normal stress and average shear stress acting on the material along section planes a-a. January2015 61
- 60. Internal loading Based on free-body diagram, Resultant loading of axial force, P = 800 N Average normal stress = P/A = 800/(0.04)(0.04) = 500 kPa Average shear stress No shear stress exists on the section, since the shear force at the section is zero. January2015 62
- 61. Example 2 Three plates are held together by two cyclindrical rivets. If a direct pull of 5 kN is applied between one plate and the other two, estimate the diameter of the rivets. The shear stress in the rivets is not to exceed 40 N/mm2. 5 kN January2015 63
- 62. Solution • 4 sliding areas (Double shear) P A = 40 = 5000 4(2/4) = 6.3 mm January2015 64
- 63. Question 1 A 50 kN axial load is applied to a short wooden post which is supported by a square concrete footing resting on distributed soil. Determine (a) The maximum bearing stress on the concrete footing (b) The size of the footing for which the average bearing stress on the soil is 150 kPa. (Ans: B = 4MPa, b= 577 mm) Plan 125 mm 100 mm b b 50 kN January2015 65 TUTORIAL 2 (STRESS
- 64. Question 2 The column is subjected to an axial force of 8 kN at its top. If the cross sectional area has the dimensions shown in the figure, determine the average normal stress at section a-a. (Ans: = 1.74 MPa) Plan Front Elevation 160 mm 160 mm a a 10 mm 10mm 10 mm 8 kN 160mm January2015 66
- 65. Question 3 The 20 kg lamp is supported by two steel rods connected by a ring A. Determine which rod is subjected to the greater average normal stress and compute its value. (Ans: = 2.33 N/mm2) B A C 60o 45o 12 mm 10 mm January2015 67
- 66. Question 4 A square hole having 12 mm sides is to be punched out of a metal plate 1.6 mm thick. The shear stress required to cause fracture is 350 N/mm2. What force must be applied to punch die? What would be the compressive stress in the punch? (Ans: 26.88 kN, 0.19 kN/mm2) January2015 68
- 67. ALLOWABLE STRESS When designing a structural member or mechanical element, the stress in it must be restricted to safe level Choose an allowable load that is less than the load the member can fully support One method used is the factor of safety (F.S.) F.S. = Ffail Fallow If load applied is linearly related to stress developed within member, then F.S. can also be expressed as: F.S. = σfail σallow F.S. = fail allow In all the equations, F.S. is chosen to be greater than 1, to avoid potential for failure Specific values will depend on types of material used and its intended purpose January2015 69
- 68. Strain Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position Normal strain () is a measure of elongation or contraction of small line segment in the body Normal Strain () = Change in length = L Original length L Has no unit Shear strain () is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to each other. Shear Strain () = x L L x January2015 70
- 69. Conventional stress-strain diagram Figure shows the characteristic stress-strain diagram for steel, a commonly used material for structural members and mechanical elements January2015 71
- 70. Elastic behavior. A straight line Stress is proportional to strain, i.e., linearly elastic Upper stress limit, or proportional limit; σpl If load is removed upon reaching elastic limit, specimen will return to its original shape January2015 72
- 71. Yielding. Material deforms permanently; yielding; plastic deformation Yield stress, σY Once yield point reached, specimen continues to elongate (strain) without any increase in load Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in elastic limit Material is referred to as being perfectly plastic January2015 73
- 72. Strain hardening. Ultimate stress, σu While specimen is elongating, its x- sectional area will decrease Decrease in area is fairly uniform over entire gauge length January2015 74
- 73. Necking. At ultimate stress, x-sectional area begins to decrease in a localized region As a result, a constriction or “neck” tends to form in this region as specimen elongates further Specimen finally breaks at fracture stress, σf January2015 75
- 74. Hooke’s Law When a material is worked within its elastic limit, the extension is proportional to the force. Strain Stress Stress = Constant (E) Strain This constant is known as the modulus of elasticity or Young’s Modulus January2015 76
- 75. Example 1 A tie-bar in a steel structure is of rectangular section 30 mm x 50 mm. The extension measured in a 250 mm length of the tie bar when load is applied to the structure is 0.1 mm. Find :- i) The tensile stress in the bar ii) The tensile force iii) The factor of safety used Take E = 205 kN/mm2 and Ultimate stress = 460 N/mm2 January2015 77
- 76. Solution i) The tensile stress in the bar E = stress/strain Strain (ε) = ∆L/L = 0.1/250 = 4 x 10-4 E = stress/strain 205 = / 4 x 10-4 = 205 x 4 x 10-4 = 0.082 kN/mm2 = 82 N/mm2 ii) The tensile force = F/A 82 = F/(30 x 50) F = 82 x (30 x 50) = 123,000 = 123 kN iiii) The factor of safety used F.S. = Ultimate stress/ Tensile stress = 460 /82 = 5.6 January2015 78
- 77. Example 2 The ultimate stress for a steel is 450 N/mm2. What is the maximum load which a rod 50 mm diameter can carry with a factor of safety of 5? If the rod is 1.5 m long, determine the extension under this loading.( E = 200 kN/mm2.) Solution The factor of safety F.S. = Ultimate stress/ Tensile stress 5 = 450 / tensile stress Tensile stress = 450/5 = 90 N/mm2 = F/A 90 = F/(π r2) F = 90 x (π x 252) = 176,714.59 N = 176.71 kN January2015 79
- 78. Solution The tensile stress in the bar E = stress/strain 200 x 103 = 90/ ε ε = 90/ 200 x 103 = 4.5 x 10-4 Strain (ε) = ∆L/L 4.5 x 10-4 = ∆L/1500 ∆L = 0.675 mm January2015 80
- 79. Exercise 1 A flat steel tie-bar 4.5 m long, is found to be 2.4mm short. It is sprung into place by means of drafts driven into holes in the end of the bar. Determine: (a) the stress in the bar (b) the factor of safety if the material of the tie-bar has an ultimate stress of 450 N/mm2. Take E for the material as 205 kN/mm2. ( Ans: 109.3 N/mm2, 4.117) January2015 81 TUTORIAL 3 (STRESS, STRAIN & FACTOR OF SAFETY)
- 80. Exercise 2 A metal tube of outside diameter 75 mm and length 1.65 m is to carry a compressive load of 60 kN. If the allowable axial stress is 75 N/mm2 , calculate the inside diameter of the tube. If E of the material is 90 kN/mm2., by how much will the tube shorten under this load? Ans: 67.87 mm, 1.375 mm) January2015 82
- 81. 2.5 PrimaryLoads & Secondary Loads DEADLOADS Dead Loads are those loads which are considered to act permanently; they are "dead," stationary, and unable to be removed. The self-weight of the structural members normally provides the largest portion of the dead load of a building. This will clearly vary with the actual materials chosen. Permanent non-structural elements such as roofing, concrete, flooring, pipes, ducts, interior partition walls, Environmental Control Systems machinery, elevator machinery and all other construction systems within a building must also be included in the calculation of the total dead load. These loads are represented by the red arrow in the illustration. Primary Loads are divided into three broad categories according to the way in which they act upon the structure or structural element. These are DEAD LOADS, LIVE LOADS(IMPOSED LOADS) and WIND LOADS January2015 83
- 82. 2.5PrimaryLoads Unit weights of various building materials (kN/m3) Materials Unit Weight Aluminium 24 Bricks 22* Concrete 24 Concrete blocks (lightweight) 12* Concrete blocks (dense) 22* Glass fibre composite 18 Steel 70 Timber 6* * Subject to considerable variation January2015 84
- 83. 2.5 PrimaryLoads Unit weights of various sheet materials (kN/m2) The dead load of a floor or a roof is generally evaluated for one square meter of floor or roof area Sheet materials kN/m2 Acoustic ceiling tiles 0.1 Asphalt (19 mm) 0.45 Aluminium roof sheeting 0.04 Glass (single glazing) 0.1 Plaster (per face of wall) 0.3 Plasterboard 0.15 Rafters, battens and felt 0.14 Sand/cement screed (25 mm) 0.6 Slates 0.6 Steel roof sheeting 0.15 Timber floorboards 0.15 Vinyl Tiles 0.05 January2015 85
- 84. • LIVE LOADS Live Loads are not permanent and can change in magnitude. They include items found within a building such as furniture, pianos, safes, people, books, cars, computers, machinery, or stored materials, as well as environmental effects such as loads due to the sun, earth or weather. • WIND LOADS Wind and earthquakes loads are put into the special category of lateral live loads due to the severity of their action upon a building and their potential to cause failure. January2015 86
- 85. Secondary loads Structures can be subjected to secondary loads from temperature changes, shrinkage of members and settlement of supports. January2015 87
- 86. Example 1 Figure shows a precast concrete Beam which is 10.5 m long. a) Calculate the weight of the beam per unit length in kN/m b) Calculate the total weight of the beam January2015 88
- 87. Solution a) Cross sectional area of the beam = (0.6 x 0.25) – (0.4 x 0.15) = 0.09 m2 Unit weight of concrete = 24 kN/m3 Weight per unit length = 0.09 x 24 = 2.16 kN/m b) Total weight of the beam = 2.16 x 10.5 = 22.68 kN January2015 89
- 88. Example 2 The floor in a multi-storey office. Building consists of the following: • Vinyl tiles • 40 mm sand/cement screed • 125 mm reinforced concrete slab • Acoustic tile suspended ceiling Determine the dead load in kN/m2 January2015 90
- 89. Solution From the table given • Vinyl tiles = 0.05 • 40 mm sand/cement screed = 0.6 x (40/25) = 0.96 • 125 mm reinforced concrete = 0.125 x 24 = 3.00 slab • Acoustic tile suspended ceiling = 0.10 The dead load = 4.11 kN/m2 January2015 91
- 90. Exercise 1 Figure shows the outer wall of a multi-storey building which is supported on a beam at each floor level. The wall consist of a 1.2 m height of cavity wall supporting 1.3 m high double glazing. The cavity wall construction is 102.5 mm of brickwork, a 75 mm cavity and 100 mm of plastered lightweight concrete blockwork. Determine the dead load on one beam in kN/m of beam. (Ans: 4.77 kN/m) January2015 92 TUTORIAL 4 (LOADS)