CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 45 Example 1: An overhanging beam is loaded as shown in Figure 1 below. Calculate: i) The reaction at supports ii) Draw the shear force and bending moment diagram Figure 1 Solution: From the load diagram; Segment AB;
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 46 Segment BC; Segment CD; To draw the Shear Diagram: i. In segment AB, the shear is uniformly distributed over the segment at a magnitude of -30 kN. ii. In segment BC, the shear is uniformly distributed at a magnitude of 26 kN. iii. In segment CD, the shear is uniformly distributed at a magnitude of -24 kN. To draw the Moment Diagram: 1. The equation MAB = -30x is linear. at x = 0, MAB = 0 and at x = 1 m, MAB = -30 kN·m. 2. MBC = 26x - 56 is also linear. at x = 1 m, MBC = -30 kN·m; at x = 4 m, MBC = 48 kN·m. when MBC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B. 3. MCD = -24x + 144 is again linear. at x = 4 m, MCD = 48 kN·m; at x = 6 m, MCD = 0
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 47
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 48 Example 2: A simply supported beam shown in Figure 2 is loaded with point load and moment at a particular point. A is pinned support while B is roller support. i) Calculate the reaction at A and B ii) Determine the shear force and bending moment at the centre of the beam iii) Draw the shear force diagram and bending moment diagram of the beam Figure 2 Solution: � = 0 2000(3) − 4800 − (12) = 0 6000 − 4800 = (12) ∴ = 1200 12 = 100(↑) ℎ, = 2000 − 100 = 1900(↑) Shear force at the centre, x = 6m � ↑ = � ↓ = 2000 + 1800 = 2000 + 1900 − 2000 =
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 49 Bending moment at the centre, x = 6m Shear force diagram and bending moment diagram � ↻ = � ↺ (6) − 2000(3) − = 0 1900(6) − 6000 = ∴ = 5400(↺)
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 50 Example 3: For the cantilever beam shown in Figure 3: i) Determine the support reactions at A ii) Draw the shear force and bending moment diagrams for the beam. iii) Determine the shear force and bending moment at 1.8m from support A. Figure 3 Solution: i) Since the support is fixed, therefore 3 unknowns , are reacted at the support � = 0 = 30(1.8) � 1.8 2 � + 50(1.8) + 40(3) + 200 + = 30(1.8) � 1.8 2 � + 50(1.8) + 40(3) + 200 + −458.6 = ℎ, = 458.6 (↺) ℎ � ↑= � ↓ = 30(1.8) + 50 + 40 ∴ = 144 (↑) ℎ ℎ ℎ , ℎ = 0
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 51 ii) Shear force diagram and bending moment diagram iii) The shear force and bending moment at 1.8m from support A: From the SFD and BMD; ∴ 1.8 = 90 ∴ 1.8 = 248
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT Mechanics of Civil Engineering Structures 52 END OF CHAPTER PROBLEM Problem 1 Sketch and name four (4) types of beam Problem 2 List three (3) types of load. Problem 3 Name the support and load in the empty box with the correct answer.
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT DCC20053 – Mechanics of Civil Engineering Structures 54 Problem 4 Solve the reactions for each case below: a. b. c.
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT DCC20053 – Mechanics of Civil Engineering Structures 55 d. e. f. g.
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT DCC20053 – Mechanics of Civil Engineering Structures 56 h. i. j. k. h.
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT DCC20053 – Mechanics of Civil Engineering Structures 57 Problem 5 Draw the shear force and bending moment diagram for a beam below: a. b. c. c.
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT DCC20053 – Mechanics of Civil Engineering Structures 58 d. . e. f.
CHAPTER 2 – EQUILIBRIUM FORCES, SHEAR FORCE AND BENDING MOMENT DCC20053 – Mechanics of Civil Engineering Structures 59 g. h. i.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 60 CHAPTER 3 Direct Stress GENERAL SUMMARY This topic introduces the relationship between stress and strain, Hooke’s law, and elastic modulus. The experiments cover the tensile test for steel bar specimens and also the determination of Young’s modulus of a steel beam. 3.0 Introduction 3.1 The relationship between direct stress and direct strain Figure 3.1: The figure above shows how force acts on a material. What do you think occurs in material? How much force is needed to compress or pull the material? Will the material get longer or shorter?
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 61 3.1.1 Effect of axial force on direct stress and direct strain When a load is applied to an elastic body it will deform but the particles of material will try to resist this deformation and the state will reach when resisting force and applied force are the same as an effect the further deformation is stopped. This internal resistance produced is termed stress and the resistance per unit area is called as the intensity of stress. In short, stress is the resistance per unit area. If the stress is acting normal in the section it is called normal stress and if it is tangential then it is called shear stress. Axial force and bending moment produce normal stress and shear force and torsional moment produce shear stress. When there is axial pull, then it is tensile stress, and when it is axial push it is compressive stress. σ.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 62 3.1.2 Definition of direct stress and strain Direct stress, σ When a force is applied to an elastic body, the body deforms. How the body deforms depends upon the type of force applied to it. A compression force makes the body shorter (Figure 3.1.2a). A tensile force makes the body longer (Figure 3.1.2b) Figure 3.1.2 (a) Figure 3.1.2 (b) Tensile and compressive forces are called Direct Forces. Stress is the force per unit area upon which it acts. Unit (N/m2) OR Pa (Pascals) The symbol σ is called SIGMA
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 63 Direct Strain, ɛ In each case, a force F produces a deformation x. In engineering, we usually change this force into stress and the deformation into strain and we define these as follows: Strain is the deformation per unit of the original length, *Strain has no units since it is a ratio of length to length. *The symbol ɛ is called EPSILON 3.1.3 Calculating the cross-sectional area, direct stress, direct strain, and deformation of materials on the prismatic bar and sectional bar For a simple homogenous bar (prismatic bar) with a constant cross-section and a constant applied load, the total deflection of the bar can be determined in terms of P, L, A, and E. Starting with the one-dimensional Hooke's Law, and substituting P/A for stress and δ/L for strain gives, = * Where E is Modulus Young (will be discussed in next subtopic) =
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 64 So, for the prismatic bar, this can be rearranged to give: = If there are a series of bars (cross sectional bar or composite bar), then the deflection of each section can be determined and then all deflections summed. Composite bar is when a member is made up of two different material. This can be written as: Example 1: A metal wire is 2.5mm diameter and 2m long. A force of 12N is applied to it and it stretches 0.3mm. Assume the material is elastic. Determine the following; i. The stress in wire, σ ii. The strain in wire, ɛ Solution: =4 4 = × 2.54 4 = 4.909 2 ∴ = = . = . � :
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 65 ∴ = =. = . Example 2; A prismatic bar with dimension 20cm x 40cm is applied with tensile force of 20kN. Determine the stress in bar. Solution : A = 20 × 40 = 800cm2 , = ∴ = = . � Example 3: A steel rod with cross-sectional area of 175mm2 is compressed with 370N. Calculate the stress in rod. If length of the rod is 2m, calculate the deformation if strain is limited to 0.0005. Solution: , = ∴ = = . � , =
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 66 ∴ = = . × = . = Example 4: A hollow rod of 600mm length, having an outer diameter of 30mm and inner diameter of 20mm respectively. It is stretched with 50kN and the elongation is 0.2mm. Determine the direct stress and strain in rod. Solution: Given; Outer diameter, d1 = 30mm Inner diameter, d2 = 20mm Length, L= 600mm Force, F = 50kN , = 0.2mm , = =(1)4 4 −(2)4 4 ∴ = . ∴ = = . = . � ∴ = =. = . × −
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 67 Example 5: A steel compound bar as shown in sketch is imposed with compressive load of 25N. Calculate; i. Stress for each section ii. The total deformation Take E = 210GN/m2 (Fill in the box with correct answer) Solution: Given; = 25 = 210/2 1 = 12 2 = 17 1 = 150 2 = 200 i. = = = = 0.22N/mm 2 = = = = 0.11N/mm 2 ii. = + Σ = � 1 1 + 2 2 �
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 68 Σ = � + � ∴ Σ = ____________________ 3.2 Hooke’s Law 3.2.1 Stress-strain Diagram Figure 3.2: Stress-strain diagram Suppose that a metal specimen be placed in tension-compression-testing machine. As the axial load is gradually increased in increments, the total elongation over the gauge length is measured at each increment of the load and this is continued until failure of the specimen takes place. Knowing the original cross-sectional area and length of the specimen, the normal stress σ and the strain ε can be obtained.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 69 The graph of these quantities with the stress σ along the y-axis and the strain ε along the x-axis is called the stress-strain diagram. The stress-strain diagram differs in form for various materials. The diagram shown below is that for a medium-carbon structural steel. From the origin O to the point called proportional limit, the stress-strain curve is a straight line. This linear relation between elongation and the axial force causing was first noticed by Sir Robert Hooke in 1678 and is called Hooke's Law that within the proportional limit, the stress is directly proportional to strain or or The constant of proportionality k is called the Modulus of Elasticity E or Young's Modulus and is equal to the slope of the stress-strain diagram from O to P. Then 3.2.2 Characteristics of material Metallic engineering materials are classified as either ductile or brittle materials. A ductile material is one having relatively large tensile strains up to the point of rupture like structural steel and aluminium, whereas brittle materials has a relatively small strain up to the point of rupture like cast iron and concrete. An arbitrary strain of 0.05 mm/mm is frequently taken as the dividing line between these two classes. The behaviour of materials can be broadly classified into two categories; brittle and ductile. Steel and aluminium usually fall in the class of ductile materials. Glass and cast iron fall in the class of brittle materials. The two categories can be distinguished by comparing the stressstrain curves. The material response for ductile and brittle materials are exhibited by both qualitative and quantitative differences in their respective stress-strain curves. Ductile materials will withstand large strains before the specimen ruptures; brittle materials fracture at much lower strains. The yielding region for ductile materials often takes up the majority of the stress-strain curve, whereas for brittle materials it is nearly non-existent. Brittle materials often have relatively large Young's moduli and ultimate stresses in comparison to ductile materials. *The energy absorbed (per unit volume) in the tensile test is simply the area under the stress strain curve. =
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 70 These differences are a major consideration for design. Ductile materials exhibit large strains and yielding before they fail. On the contrary, brittle materials fail suddenly and without much warning. Thus, ductile materials such as steel are a natural choice for structural members in buildings as we desire considerable warning to be provided before a building fails. The energy absorbed (per unit volume) in the tensile test is simply the area under the stress strain curve. Clearly, by comparing the curves in Figure 8, we observe that ductile materials are capable of absorbing much larger quantities of energy before failure. Finally, it should be emphasized that not all materials can be easily classified as either ductile or brittle. Material response also depends on the operating environment; many ductile materials become brittle as the temperature is decreased. With advances in metallurgy and composite technology, other materials are advanced combinations of ductile and brittle constituents. Elastic Limit The elastic limit is the limit beyond which the material will no longer go back to its original shape when the load is removed, or it is the maximum stress that may e developed such that there is no permanent or residual deformation when the load is entirely removed. Elastic and Plastic Ranges The region in stress-strain diagram from O to P is called the elastic range. The region from P to R is called the plastic range. Yield Point Yield point is the point at which the material will have an appreciable elongation or yielding without any increase in load. Ultimate Strength The maximum ordinate in the stress-strain diagram is the ultimate strength or tensile strength.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 71 Rupture Strength Rupture strength is the strength of the material at rupture. This is also known as the breaking strength. Modulus of Resilience Modulus of resilience is the work done on a unit volume of material as the force is gradually increased from O to P, in N·m/m3 . This may be calculated as the area under the stress-strain curve from the origin O to up to the elastic limit E (the shaded area in the figure). The resilience of the material is its ability to absorb energy without creating a permanent distortion. Modulus of Toughness Modulus of toughness is the work done on a unit volume of material as the force is gradually increased from O to R, in N·m/m3 . This may be calculated as the area under the entire stress-strain curve (from O to R). The toughness of a material is its ability to absorb energy without causing it to break. Working Stress, Allowable Stress, and Factor of Safety Working stress is defined as the actual stress of a material under a given loading. The maximum safe stress that a material can carry is termed as the allowable stress. The allowable stress should be limited to values not exceeding the proportional limit. However, since proportional limit is difficult to determine accurately, the allowable tress is taken as either the yield point or ultimate strength divided by a factor of safety. The ratio of this strength (ultimate or yield strength) to allowable strength is called the factor of safety. Normal Stresses Stress is defined as the strength of a material per unit area or unit strength. It is the force on a member divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa. where P is the applied normal load in Newton and A is the area in mm2 . The maximum stress in tension or compression occurs over a section normal to the load.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 72 Normal stress is either tensile stress or compressive stress. Members subject to pure tension (or tensile force) is under tensile stress, while compression members (members subject to compressive force) are under compressive stress. Compressive force will tend to shorten the member. Tension force on the other hand will tend to lengthen the member.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 73 END OF CHAPTER PROBLEM Problem 1 The following data (Table 1) were recorded during the tensile test of a 14-mm diameter mild steel rod. The gage length was 50 mm. Load (N) Elongation (mm) Load (N) Elongation (mm) 0 0 46200 1.25 6310 0.010 52400 2.50 12600 0.020 58500 4.50 18800 0.030 68000 7.50 25100 0.040 59000 12.5 31300 0.050 67500 15.5 37900 0.060 65000 20.0 40100 0.163 65500 Fracture 41600 0.433 Table 1: Data of tensile test Plot the stress-strain diagram and determine the following mechanical properties: i. proportional limits ii. modulus of elasticity iii. yield point iv. ultimate strength v. rupture strength
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 74 Problem 2 Refer to Figure 2, a hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400KN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2 . Figure 2 Problem 3 A rod is composed of an aluminium section rigidly attached between steel and bronze sections, as shown in Figure 3. Axial loads are applied at the positions indicated. If P = 3000 lb and the crosssectional area of the rod is 0.5 in2 , determine the stress in each section. Figure 3
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 75 Problem 4 An aluminium rod is rigidly attached between a steel rod and a bronze rod as shown in Figure 4. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa. Figure 4 Problem 5 A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end (Figure 5). It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E = 200 × 103 MN/m2 , find the total elongation of the rod. Figure 5
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 76 Problem 6 A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi and the total elongation is not to exceed 0.20 in. Assume E = 29 × 106 psi. Problem 7 A bronze bar is fastened between a steel bar and an aluminium bar as shown in Figure 7. Axial loads are applied at the positions indicated. Find the largest value of P that will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in the steel, 120 MPa in the bronze, and 80 MPa in the aluminium. Assume that the assembly is suitably braced to prevent buckling. Use Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa. Figure 7 Problem 8 A steel rod having cross-sectional area of 175mm2 experiences a compressive force of 370N. Calculate stress in the rod. Problem 9 A bar of 300mm length with an area of 176m2 is stretched by 0.8mm due to axial pull of 15kN. Calculate modulus of elasticity.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 77 Problem 10 What is the meaning of Tensile Stress and Compressive Stress? Problem 11 Define direct stress, direct strain and modulus of elasticity. Problem 12 A metal bar which is part of a frame is 50 mm diameter and 300 mm long. It has a tensile force acting on it of 40 KN which tends to stretch it. The modulus of elasticity is 205 GPa. Calculate the stress and strain in the bar and the amount it stretches. Problem 13 A metal wire is 2.5 mmdiameter and 2 mlong. A force of 12 N is applied to it and it stretches 0.3 mm. Assume the material is elastic. Determine the following: a) The stress in the wire σ. b) b) The strain in the wire ε. Problem 14 A steel bar is 10 mm diameter and 2 m long. It is stretched with a force of 20kN and extends by 0.2 mm. Calculate the stress and strain.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 78 Problem 15 A rod is 0.5 m long and 5 mm diameter. It is stretched 0.06 mm by a force of 3 kN. Calculate the stress and strain. Problem 16 A steel tensile test specimen has a cross sectional area of 100 mm2 and a gauge length of 50 mm, the gradient of the elastic section is 410 x 103N/mm. Determine the modulus of elasticity. Problem 17 A Steel column is 3 mlong and 0.4 m diameter. It carries a load of 50 MN. Given that the modulus of elasticity is 200 GPa, calculate the compressive stress and strain and determine how much the column is compressed. Problem 18 A bar is 500 mm long and is stretched to 505 mm with a force of 50 kN. The bar is 10 mm diameter. Calculate the stress and strain. The material has remained within the elastic limit. Determine the modulus of elasticity.
CHAPTER 3 – DIRECT STRESS Mechanics of Civil Engineering Structures 79 Problem 19 A steel bar is stressed to 280 MPa. The modulus of elasticity is 205 GPa. The bar is 80 mm diameter and 240 mm long. Determine the following: i. The strain. ii. The force. Problem 20 A circular metal column is to support a load of 500 Tonne and it must not compress more than 0.1 mm. The modulus of elasticity is 210 GPa. the column is 2 m long. Calculate the cross sectional area and the diameter. *Note 1 Tonne is 1000 kg
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 80 CHAPTER 4 Bending Stress in Beams GENERAL SUMMARY This topic introduces the determination of the neutral axis and second moment of area for a section and covers the calculation of the maximum value of bending stress and drawing the distribution. 4.0 Introduction of bending stress Bending stress is the normal stress that an object encounters when it is subjected to a large load at a particular point that causes the object to bend and become fatigued. Bending stress occurs when operating industrial equipment and in concrete and metallic structures when they are subjected to a tensile load. 4.1 Understand the basic knowledge of bending stress in beams. When a beam is imposed with the load it is likely to bend or deform when subjected to bending moment or shear force. The process of deformation or bending stops when every section reacts with resistance i.e. stresses. The stresses which resist bending moment are called bending stresses and stresses resisting shear force are called shearing stresses. Sources: https://www.thevitruviusproject.com/blog/formula-for-bending-stress-shear-stress-and-normal-stress-in-beam-calculations/ Figure 4.1: Bending beam
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 81 When a member is being loaded similar to that in Figure 4.1 bending stress (or flexure stress) will result. Bending stress is a more specific type of normal stress. When a beam experiences load like that shown in Figure 4.1 the top fibers of the beam undergo normal compressive stress. The stress at the horizontal plane of the neutral is zero. The bottom fibers of the beam undergo normal tensile stress. It can be concluded therefore that the value of the bending stress will vary linearly with distance from the neutral axis. 4.1.1 The effect of bending moment in a beam Figure 4.1.1: The effect of bending moment in a beam To understand the bending stress in an arbitrary loaded beam, consider a small element cut from the beam as shown in the diagram at the left. The beam type or actual loads does not affect the derivation of the bending strain equation. (Recall the basic definition of normal strain in Chapter 3!!!)
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 82 When a beam is bent due to transverse or virtual loading its pattern of bending depends on type of support. For a simply supported beam and cantilever beam the pattern of bending is shown in sketch these patterns are as sagging or hogging pattern.
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 83 4.1.2 Centroid and second moment of area of a section. 4.1.2.1 Centre of gravity A point through which the whole weight of the body acts is called center of gravity represented by C.G. 4.1.2.1 (a) A point through which the total area of a plane figure is assumed to be concentrated is called as centroid of that area. Centre of gravity of plane figures by method of moments. The figure shows a plane figure of total area A which center of gravity is to be calculated. Let us divide this area into small parts a1, a2, a3 etc.: ∴ = 1 + 2 + 3 + ⋯ Let 1 = distance of C.G of area 1 from axis . 2, 3, 4 = . 1, 2, 3, 4 … Take moments of all these areas about point 0. REMEMBER!!! Moment = Force x distance = + + + ⋯ Let C.G be the point of centre of gravity of the total area A. Let distance of this C.G from be equal to ̅. Thus, moment of C.G @ = Total area x distance = ̅ The moment of all small areas about the axis must be equal to the moment of total area about the same axis. Thus; 1 1 + 22 + 3 3 + ⋯ ̅=1 1 + 22 + 3 3 + ⋯ ̅=̅ ℎ � = 1 + 2 + 3 + ⋯ Similarly, when we take moments about axis : � =1 1 + 22 + 3 3 + ⋯ 1 + 2 + 3 + ⋯ � =�
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 84 4.1.2.1 (b) Centroid of some common figures: TYPE SHAPE �� AREA Rectangle � = � = = × Triangle � = � = � = � = = 1 2 × × ℎℎ = 1 2 × × ℎℎ
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 85 TYPE SHAPE �� AREA Circle � = � = = 2 Semi-circle � = � = =2 2 Quarter circle � = � = =2 4
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 86 4.1.2.1 (c) Procedure for solving problems of centroid @ C.G: Example 1: Determine the position of centroid gravity for the given sketch below. Calculate � and � for the given sketch below; Solution: Calculate � and � for the given sketch. 1. Divide the sketch into section:
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 87 2. Tabulate a table given below: SECTION AREA, A (mm2) � () � () � () � () 1 75 × 12.5 = 937.5 1 = 75 2 = . 937.5 × . = , . 1 = 12.5 2 + 87.5 + 12.5 = . 937.5 × . = , . 2 87.5 × 12.5 = 1,093.75 2 = 12.5 2 + 62.5 = . 1,093.75 × . = , . 2 = 87.5 2 + 12.5 = . 1,093.75 × . = , . 3 100 × 12.5 = 1,250 3 = 100 2 + 62.5 = . 1,250 × . = , 3 = 12.5 2 = . 1,250 × . = , .
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 88 TOTAL � = , . ∑ ̅ = , . �� = , . ∴ � = ∑ ̅ ∑ = 250,976.56 3,281.25 = . @ (, ) ∴ � = ∑ � ∑ = 168,945.32 3,281.25 = . @ (, ) 3. Determine the C.G in figure:
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 89 TRY TO SOLVE THIS PROBLEM!!! Determine the position of C.G for diagram shown below: 1. 4. 2. 5. 3. 6.
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 90 4.1.3 Second moment of area @ Moment of inertia As shown in sketch, consider lamina of area A and x and y be the distance of C.G of area A from axis oy and ox respectively. Moment of area about oy = area x perpendicular distance of C.G from oy. First moment of area = Ax If this equation is multiplied by perpendicular distance then = (Ax)x = Ax2 this is called as second moment of area or also moment of inertia about axis oy. Instead of area, if we consider mass (m) of the body then this is called as second moment of mass or moment of inertia. Thus, product of area and square of the distance of the C.G of area from axis is called moment of inertia. Moment of inertia @ Second moment of area @ Ixx@ Iyy = AR2 Unit; mm2 (mm2 ) = mm4 / cm4 / m4 4.1.4 Moment of inertia for standard cases Name Shape ( − ) ( − ) Rectangle = = Hollow rectangle = − = −
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 91 Symmetrical I section = − ℎ = ( − ) = − ℎ = ( − ) Circle For solid circle : = = For hollow circle : = = [− ] Example 2: Calculate moment of inertia about xx and yy axis and also polar moment of inertia for a diagram shown in Figure 1 below. Figure 1 Solution: The sketch is symmetrical at xx and yy axis. = () − () 1. =3 12 = 100 × 1003 12 = 8.33 × 1064
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 92 = = 2. = 64 (4) = 64 (504) = 0.30 × 1064 3. = = 8.33 × 106 − 0.30 × 106 = 8.03 × 1064 4. = + = 2(8.03 × 106) = . ×
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 93 Example 3: Find the polar M.I of a rectangular section 60 x 80mm and also calculate minimum radius of gyration. Solution: 1. To calculate =3 12 = 60(80)3 12 = . × 2. To calculate =3 12 = 80(60)3 12 = . × 3. To calculate polar moment of inertia, = + = 2.56 × 106 + 1.44 × 106 = × 4. Radius of gyration, R = 2 ∴ = � @ = � ( ℎ ) = �2.56 × 106 60 × 80 = 23.09 = �1.44×106 60×80 = 17.32 (So take = . )
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 94 Example 4: Calculate moment of inertia @ second moment of area for the figure below at its horizontal and vertical axis passing through C.G. Solution: 1. Calculate ̅ � Section Area (mm2 ) x from origin (mm) y from origin (mm) Ax (mm3 ) Ay (mm3 ) 1 320 × 10 = 3200 10 2 = 5 320 2 = 160 3200 × 5 = 16000 3200 × 160 = 512,000 2 200 × 10 = 2000 200 2 + 10 = 110 10 2 + 310 = 315 2000 × 115 = 220,000 2000 × 315 = 630,000 3 250 × 10 = 2500 250 2 + 10 = 135 10 2 = 5 2500 × 135 = 337,500 2500 × 5 = 12,500 TOTAL � = 7,700 � = 573.5 × 103 � = 1154.5 × 103 ∴ ̅= ∑ ∑ = 573.5 × 103 7,700 = . ℎ
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 95 � = ∑ ∑ = 1154.5 × 103 7,700 = . ℎ 2. Calculate second moment of area @ moment of inertia at xx axis and yy axis