CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 96 TRY TO SOLVE THIS PROBLEM!!! Determine the second moment of area @ moment of inertia about xx axis and yy axis for the figure given below. 1. 4. 2. 5. 3. 6.
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 97 4.2 Bending stress formulae The general bending stress or flexural formula is = = Where: − − = . = = . − ℎ ℎ . − − 4.3 Bending stress diagram Bending stress diagram is linear and always maximum at extreme fibres and zero at neutral axis. Bending stress diagram for symmetrical and unsymmetrical sections are as shown. Bending stress is irrespective of shape of c/s. 4.1.4 a) Symmetrical section For symmetrical sections compression and tensile stresses are same since = , =
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 98 For simply supported beam = 2 = = 1 = For cantilever beam = 2 = = 1 = 4.1.4 b) Unsymmetrical section For unsymmetrical section like T, L, unsymmetrical I, inverted T etc. we have to calculate position of � from base � =11+22+⋯ 1+2+⋯ and then calculate = 1 + 2 + ⋯ since ≠ , ≠ . To calculate : = = Example 5: A steel strip 50mm wide and 5mm thick is subjected to end couples of 25Nm. Calculate radius of curvature of the bent strip if = 2 × 105. Solution: Flexural formula: = = Given; = 25 = 25 × 103 =3 12 = 50 × 53 12 = 5204 = 2 × 105 = 2 × 105 �2 Substituting in above equation 25 × 103 520 = 2 × 105 = = .
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 99 Example 6: Determine the radius to which a bar 10mm diameter will be required to bend such that maximum stress produced is 120 �2 . Given = 200. Solution: Flexural formula = = Where = 120 �2 = 2 = 10 2 = 5 = 200 = 200 × 103 �2 Substituting in above equation 120 5 = 200 × 103 = = . .
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 100 Example 7: A cantilever beam of solid circular cross section and 2.5m long carries a concentrated load of 30kN at its free end. If the maximum bending stresses does not exceed 120MPa, calculate diameter of beam. Solution: For cantilever beam, Maximum bending moment, = 30 × 103 × 2500 = 75 × 106 = = ℎ , = 75 × 106 = 64 ()4 = 0.044 = = 120 �2 = 2 = 0.5 75 × 106 0.044 = 120 0.5 625000 = 0.083 =
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 101 Example 8: A cantilever beam of rectangular section supports a point load of 10kN at free end. Span of beam is 2m. Maximum bending stress is 100N/mm2 . Determine size of beam if depth = 2 breadth. Solution: = = ℎ = × = 10 × 103 × 2000 = 20 × 106 =3 12 =(2)3 12 = 0.664 = 2 = 2 2 = = 100 �2 20 × 106 0.664 = 100 3 = 303030 ∴ = . , = .
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 102 Example 9: A simply supported beam is of symmetrical I section with flange 90 x 10 and web70 x 10 with a span of 7m carrying u.d.l of 5kN/m over entire span and a central point load of 15kN. Draw bending stress distribution diagram. Solution: ℎ @ = = 1. To calculate bending moment of beam: 2(7) = 15(3.5) + 5(7)(3.5) = 175 7 = 25 = 1
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 103 2. Calculate =3 12 −3 12 = 90(90)3 12 − 80(70)3 12 = 3.18 × 1064 3. Since the section is symmetrical ∴ = = = 90 2 = 45 = 56.88 × 103 3.18 × 106 = 45 = 805 �2 4. Bending stress distribution diagram
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 104 Example 10: A cantilever beam of 2.5m span is imposed with u.d.l of 5kN/m over entire span. C/s of beam is T section with flange 75 x 15, web 100 x 20. Calculate bending stress at connection of flange and web and also maximum stress. Solution: ℎ @ = = 1. To calculate bending moment; = 2 8 = 5(2500)2 8 = 16.9 × 106
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 105 2. Calculate a) To determine neutral axis from base, � = ∑ ∑ Section Area , A (mm2 ) y from origin/base, (mm) Ay, (mm3 ) 1 75 × 15 = 1125 15 2 + 100 = 107.5 120,937.5 2 100 × 20 = 2000 100 2 = 50 100,000 � = 3125 � = 220.937 × 103 ℎ, = ∑ ∑ = 220.937 × 103 3125 = 70.7 ℎ ℎ, = 115 − 70.7 = 44.3 ℎ b) Since this is unsymmetrical section, ∴ = 1 + 2 1 =3 12 + 2 ℎ = ( − �)2 1 = 75 × 153 12 + 1125(107.5 − 70.7)2 = . ×
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 106 2 = 20 × 1003 12 + 2000(50 − 70.7)2 = . × ∴ = 1.545 × 1064 + 2.524 × 1064 = . × ∴ , = (16.9 × 106) 44.3 4.069 × 106 = 183.994 2 ⁄ ∴ , = (16.9 × 106) 70.7 4.069 × 106 = 293.642 2 ⁄ , = 293.642 N/mm2 N.A 70.7mm 44.3mm , = 183.994 N/ mm2
CHAPTER 4 – BENDING STRESS IN BEAMS Mechanics of Civil Engineering Structures 107 END OF CHAPTER PROBLEM Problem 1 A simply supported beam of solid circular section carrier two-point loads of 50kN each placed at distance of 400mm from support A and B, span of beam is 1200mm. Determine bending stress if diameter is 100mm. Ans: , = = . 1. A cantilever beam of section 250mm wide and 400mm deep carries concentrated load ‘w’ at its free end. Span of beam is 3.5m and maximum bending stress is 130Mpa. Calculate value of ‘w’. Ans: w = 247.55 x 103 N Problem 2 A cross section of beam is as shown on sketch, span of beam is 5m. Find maximum u.d.l the beam can carry. Maximum stress is 120Mpa. Ans: w = 1.92 KN/m over entire span
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 108 CHAPTER 5 Shear Stress of Beam GENERAL SUMMARY This topic covers the basic knowledge of the relationship between shear stress, shear strain, and modulus of rigidity. It also covers the calculation of shear stress in plate, bolt, or rivet and shear stress distribution in beams. 5.0 Introduction Shear stress is a kind of stress that acts parallel or tangential to the surface. The shear stress is denoted by (tau). Shearing stresses are commonly found in rivets, pins, and bolts. 5.1 Understand the basic knowledge of shear stress in bolt or rivet in the beam If the plates, which are connected by a rivet as shown in Figure 5.1, are subjected to tension forces, shear stresses will develop in the rivet. The shear force P in the shear plane is equal to tension force F. The average shear stress in the plane is ave= F/A. This joint is said to be in single shear. Figure 5.1 Figure 5.2 If the plates, which are connected by a rivet as shown in Figure 5.2, are subjected to tension forces, shear stresses will develop in the rivet. This joint is said to be in double shear. To determine the average shear stress in each shear plane, free-body diagrams of the rivet and the portion of the rivet located between the two planes are drawn. Observing that the shear P in each of the planes is P = F/2, the average shearing stress is ave = F/2A.
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 109 Table 1: The formula of single shear and double shear Parameter Formula Description Diagram Shear Stress (Single Shear) Shear Stress (Double Shear) = × F = Shear force A = Area of rivet/bolt/plate F = Shear force A = Area of rivet/bolt/plate
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 110 Figure 5.3: The shearing stress failure in rivets in single shear and double shear There is a different types of bolted connections. They can be categorized based on the type of loading: i) Tension member connection and splice • It subjects the bolts to forces that tend to shear the shank • The shearing stress failure in rivets (Figure 5.3) • The shearing stress failure in plates (Figure 5.4)
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 111 Figure 5.4: The shearing stress failure in rivets where all rivets in the same diameter for single shear
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 112 Figure 5.5: The shearing stress failure in the plate for single shear ii) Beam and simple connection • It subjects the bolts to forces that tend to shear the shank. iii) Hanger connection • The hanger connection puts the bolts in tension. Where n = no. of plane = × If it is double shear, then; = ( ×) ×
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 113 5.1.1 Shear stress, shear strain, modulus of rigidity in bolt or rivet, and shear stress in the beam Shear stress When a beam is subjected to transverse loading, normal and shearing stress results in the beam. The influence of the shearing stress in the beam does not disturb the influence of the bending stress. The shear stress applied on the vertical surface of the beam exerts the same stress on the horizontal surfaces of the beam. Generally, the beam subjected to the transverse load exerts the longitudinal shear stress in the beam. The shearing stress in the beam is defined as the stress that occurs due to the internal shearing of the beam that results from the shear force subjected to the beam. It is denoted by the symbol t and is expressed in the unit of psi or N/mm2 . The equation of shearing stress is: When the shear load is applied, the impact of the shearing stress throughout the rectangular crosssection of the beam occurs. It can be resolved by estimating the shearing stress at a particular height from the neutral axis. The distribution of shearing stress on the cross-section of the beam represents a parabolic curve where the maximum shearing stress occurs at the neutral axis of the beam. Figure 5.3 shows the schematic of the rectangular cross-section of the beam. Figure 5.5: The schematic diagram of the rectangular cross-section of the beam Where: Shear stress = Shear load / force = V Second moment of area = Q Moment of inertia = I Thickness = t.
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 114 Shear strain Shear strain is measured as the displacement of the surface that is in direct contact with the applied shear stress from its original position. Modulus of rigidity Modulus of rigidity or shear modulus is the rate of change of unit shear stress concerning unit shear strain for the condition of pure shear within the proportional limit. The modulus of rigidity formula is: Where: Modulus of rigidity = G Elastic modulus = E Poisson's ratio = v Where: Shear strain = Base = w Height = L
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 115 The Modulus of rigidity also can be obtained from: = Typical values of modulus of rigidity: Aluminum 6061-T6: 24 GPa, Structural Steel: 79.3 GPa. Table 1 shows the Modulus of Rigidity (Shear Modulus) for common materials. Table 1: Modulus of Rigidity or Shear Modulus for common materials Modulus of Rigidity (Shear Modulus) of Materials Material GPa 106 psi Aluminum 26 3.8 Aluminum alloys, various 25.5 - 26.9 3.7 - 3.9 Aluminum, 6061-T6 26 3.8 Aluminum, 2024-T4 28 4 Aluminum, 7075-T6 26.9 3.9 Beryllium copper (C17200) 50 7.2 Brass, 70-30 (C26000) 41 6 Yellow Brass (C27000) 39 5.6 Red Brass (C23000) 44 6.4 Aluminum Bronze (C95500) 42 6.1 Manganese Bronze (C86350) 37 5.4 Oxygen-Free Copper (C10100) 44 6.4 Glass Ceramic (Mica) 26 3.7 Inconel X750 79.3 11.5 Invar 56 8.1 Iron 82 11.9 Iron, cast 36-57 5.2-8.2 Iron, ductile 63-66 9.1-9.6 Iron, gray cast (Class 20) 27-39 3.9-5.7 Iron, malleable 64 9.3 Lead 5.6 0.8 Magnesium 17 2.4 Where: Shear stress = Modulus of rigidity = G Shear strain =
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 116 Magnesium alloy 17 2.5 Molybdenum 120 17.4 Monel 400 65.3 9.5 9.5 Nickel silver 55-18 (C77000) 47 6.8 Nickel 76 11 Phosphor bronze (C51000) 41 5.9 Steel, A36 79.3 11.5 304 Stainless steel 77 11.2 Steel, cast 78 11.3 Steel, cold-rolled 79 11.5 Steel, all others including high-carbon, heattreated 76-82 11.0- 11.9 Titanium Grade 2 (99.0 Ti) 45 6.5 Titanium Grade 5 (Ti-6Al-4V) 44 6.4 Tungsten 160 23.2 Zinc 43 6. Source: https://amesweb.info/Materials/Modulus-of-Rigidity-Formula.aspx 5.1.2 The effect of shear stress on a loaded beam The shear force, V, along the length of the beam, can be determined from the shear diagram. The shear force at any location along the beam can then be used to calculate the shear stress over the beam's cross-section at that location. The average shear stress over the cross section is given by: The shear stress varies over the height of the cross-section, as shown in Figure 5.6(a) below: Figure 5.6(a): The diagram of shear stress varies over the height of the cross-section
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 117 The shear stress is zero at the free surfaces (the top and bottom of the beam), and it is maximum at the centroid. The equation for shear stress at any point located a distance y1 from the centroid of the cross-section is given by: Where; V = the shear force acting at the location of the cross-section Ic = the centroid moment of inertia of the cross-section b = the width of the cross-section. These terms are all constants Q = the first moment of the area bounded by the point of interest and the extreme fiber of the cross-section: Shear stresses for several common cross sections are discussed in the sections below: Shear Stresses in Rectangular Sections The distribution of shear stress along the height of a cross-section is shown in Figure 5.6 (b) below: OR Figure 5.6 (b): The diagram of shear stress in a rectangular section
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 118 The first moment of the area at any given point y1 along the height of the cross-section is calculated by: The maximum value of Q occurs at the neutral axis of the beam (where y1 = 0): The shear stress at any given point y1 along the height of the cross-section is calculated by: where Ic = b·h3/12 is the centroidal moment of inertia of the cross-section. The maximum shear stress occurs at the neutral axis of the beam and is calculated by: OR where A = b·h is the area of the cross-section. Note that the maximum shear stress in the cross-section is 50% higher than the average stress V/A. Shear Stresses in Circular Sections A cross-section is shown in Figure 5.6(c) below:
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 119 Figure 5.6 (c): A circular cross-section The equations for shear stress in a beam were derived using the assumption that the shear stress along the width of the beam is constant. This assumption is valid at the centroid of a circular crosssection, although it is not valid anywhere else. Therefore, while the distribution of shear stress along the height of the cross-section cannot be readily determined, the maximum shear stress in the section (occurring at the centroid) can still be calculated. The maximum value of the first moment, Q, occurring at the centroid, is given by: The maximum shear stress is then calculated by: Where; b = 2r is the diameter (width) of the cross-section, Ic = πr4/4 is the centroidal moment of inertia, A = πr2 is the area of the cross-section. Shear Stresses in Circular Tube Sections A cross-section is shown in Figure 5.6(d) below:
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 120 Figure 5.6 (d): A circular tube cross-section The maximum value of the first moment, Q, occurring at the centroid, is given by: The maximum shear stress is then calculated by: Where; b = 2 (ro − ri) is the effective width of the cross-section, Ic = π (ro 4 − ri 4) / 4 is the centroidal moment of inertia, A = π (ro 2 − ri 2) is the area of the cross-section. Shear Stresses in I-Beams The distribution of shear stress along the web of an I-Beam is shown in Figure 5.6(e) below:
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 121 Figure 5.6(e): The distribution of shear stress along the web of an I-Beam The equations for shear stress in a beam were derived using the assumption that the shear stress along the width of the beam is constant. This assumption is valid over the web of an I-Beam, but it is invalid for the flanges (specifically where the web intersects the flanges). However, the web of an I-Beam takes the vast majority of the shear force (approximately 90% - 98%, according to Gere), and so it can be conservatively assumed that the web carries all of the shear force. The first moment of the area of the web of an I-Beam is given by: The shear stress along the web of the I-Beam is given by: Where; tw is the web thickness and Ic is the centroidal moment of inertia of the I-Beam
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 122 The maximum value of shear stress occurs at the neutral axis (y1 = 0), and the minimum value of shear stress in the web occurs at the outer fibers of the web where it intersects the flanges (y1 = ±hw/2): 5.2 Shear stress formula applying in bolt and rivet and beam There are four types of stresses occur at riveted joints. Therefore, failure is possible in four locations as follows: • Shearing stress failure in rivets • Tension stress failure in plate. • Bearing stress failure between plate and rivet. • Shearing stress failure in plate. Assumption: Shearing stress in rivets is equal and uniform. This assumption is approximately true because the shearing stresses are distributed in a non-uniform way across the area of the cut.
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 123 Figure 5.7: Railway Bridge – riveted joints Example 1: In Figure 1 shown below, two plates are riveted with one rivet. The allowable stresses are 120 MPa for shearing in the plate material and 60 MPa for shearing of the rivet. Determine: (a) The force P (b) The minimum thickness of each plate Figure 1
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 124 Solution: a) The shearing stress of the rivet is 60MPa. = 60 = 60 2 ⁄ 60 = = � 2 4 � = 314.16 ∴ = 60 × 314.16 = 18849.6 b) The minimum thickness of each plate; = 120 = 120 2 ⁄ 120 = = 18849.6 20() 20() = 18849.6 120 ∴ = � 18849.6 120 � 20 = 7.854 Example 2: For the diagram shown in Figure 2, calculate the safe axial tensile force, P if allowable = 102 . Then determine the shear stress in plate.
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 125 Figure 2 Solution: Given allowable = 102 Diameter of rivet = 22mm or Example 3: Based on Figure 3, If P = 400kN: a) Calculate the shearing stress in plate the size of the plate is 350mm x 600mm b) Find the smallest diameter of the bolt that can be used in the connector if the shear stress in the bolt is 300MPa. Figure 3
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 126 Solution: Example 4: The simply supported wood beam in Figure 4 is fabricated by gluing together three 160-mm by 80-mm plans as shown. Calculate the maximum shear stress in i) The glue ii) The wood b) Given that the shear stress in bolt is 300MPa Since this is double shear; So, the smallest diameter of bolt that can be used to allow the 300MPa shearing stress is 29.13mm Given force, P = 400kN a) Size of plate = 350mm x 600mm Since this is double shear; ∴ = . × −⁄ = . Figure 4
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 127 Example 5: The I-section beam in Figure 5 is used as a first-floor beam. If the vertical shear acting at a certain section of the beam is 50 kN, determine the following at that section: a) i) the minimum shear stress in the web; ii) the maximum shear stress in the web b) Sketch the shear stress distribution diagram of the beam. Figure 5 Solution: The IXX of beam; Since this is symmetrical I beam, so; The neutral axis lies on; i) the minimum shear stress in the web The minimum shear stress in the web occurs at the junction with the flange; 1 = (200 × 10) = 2000 2 = 2 × 10−3 2
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 128 = 11 ̅= (2000)(150 − 5) = 0.29 × 106 3 = 0.29 × 10−3 3 ii) the maximum shear stress in the web The maximum shear stress is located at the neutral axis, Hence, b) The shear stress distribution diagram
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 129 Example 6: A T-section beam has a top flange of (120mm x 20mm) and a web of (20mm x 100mm) as shown in Figure 6. The overall depth is 120mm. It is subjected to a shear force of 60kN. i) Calculate; a) The moment of inertia IXX b) The shear stress at the web c) The shear stress at the junction of web and flange d) The maximum shear stress (shear stress at the neutral axis) ii) Draw the shear stress distribution diagram. Figure 6 Solution: a) Calculate i) The moment of inertia, Ixx
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 130 ii) the shear stress at the flange iii) The shear stress at the junction of web and flange iv) The maximum shear stress (shear stress at the neutral axis)
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 131 = . × = c) Draw the shear stress distribution diagram.
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 132 END OF CHAPTER PROBLEM Problem 1 Determine the maximum and minimum shearing stress in the web of the wide flange section in Figure 1 if V = 100 KN. Figure 1 Problem 2 The T section shown in Figure 2 is the cross-section of a beam formed by joining two rectangular pieces of wood together. The beam is subjected to a maximum shearing force of 60KN. Show that the NA is 34 mm from the top and the INA = 10.57 × 106 mm4 . Using these values, determine the shearing stress a) at the neutral axis b) at the junction between the two pieces of wood. Figure 2
CHAPTER 5 – SHEAR STRESS OF BEAM Mechanics of Civil Engineering Structures 133 Problem 3 The lap joint shown in Figure 3 is fastened by four 19mm diameter rivets. Calculate the maximum safe load P that can be applied if the shearing stress in the rivets is limited to 100 kPa and the bearing stress in the plates is limited to 180 kPa. Assume the applied load is uniformly distributed among the four rivets. Figure 3
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 134 CHAPTER 6 Slope and Deflections of Beam Due to Symmetrical Bending GENERAL SUMMARY This topic introduces the students to slope and deflection of determinate beam – simply supported and cantilever beam due to point load, uniformly distributed load, and moment. It also covers the determination of slope and deflection by using the Macaulay Method and Moment Area Method. 6.0 Introduction Generally, the deformation of a beam is usually expressed in terms of its deflection from its original unloaded position. It is fully designed to sustain failure due to either bending or shear. The deflection is measured from the original neutral surface of the beam to the neutral surface of the deformed beam. The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam. Deflection of beam exceeding the allowable limit will cause damage or failure to adjacent member structure (example column, slab, etc). 6.1 Slope and deflection of the beam In a beam when subjected to a load, the initial straight longitudinal axis is deformed into a curve known as the deflection curve of the beam. Distinct curvature and a beam that is not functioning well will cause failure or damage to the structure as a whole. To ensure this from occurring the beam should be designed in such a way that when loaded, it will not deflect greater than the allowable limit as stated in the code in example: × Generally, a beam supports various types of loadings, hence parameters such as shear load, bending moment, slope, and deflection do not have specific linear functions for the entire beam.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 135 However, an expression may be deduced for the entire beam without dividing the beam into several sections using a non-linear function. Several examples of linear and non-linear functions may be seen from the shear force diagrams of a beam loaded as shown in Figure 6.1. Linear function Non-linear function Non-linear function Figure 6.1: Linear and non-linear beam Standard Signs and Units Table 6.1(a) gives standard signs for bending moment, slopes, and deflection that will be used to determine the slope and deflection of the beam. These signs are in line with the sign convention for shear force and bending moment.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 136 Table 6.1 (a): Standard sign convention for bending moment, slope, and deflection Effects Symbols Differential Coefficient Integration Unit Sign Procedure Positive Negative Bending moment M 2 2 M Nm KNm Slope = � Radian Deflection ℎ � . mm @ M From the above differential coefficient, the following standard values for maximum slopes and deflections of simple beams are obtained. (These assume that the beams iareuniform for example, EI is constant throughout the beam). Table 6.1(b) gives standard formula values for maximum slopes and deflections for simple beams due to concentrated load and uniformly distributed load.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 137 Table 6.1(b): Standard formula values for maximum slopes and deflections for simple beam due to concentrated load (P) and uniformly distributed load (u.d.l). MAXIMUM SLOPE AND DEFLECTION FOR SIMPLE BEAM Loading condition , () () Cantilever beam with concentrated load, W at end Cantilever beam with u.d.l, w throughout the span Simply supported beam with concentrated load, W at the centre Simply supported beam with u.d.l, w across the complete span Source: https://www.dce-darbhanga.org/wp-content/uploads/2020/04/file_5e87d775f21ec.pdf
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 138 6.2 Determination of slope and deflection by using the Macaulay Method 6.2.1 The Moment Equation of Macaulay Mr. W.H Macaulay devised Macaulay’s Method. Macaulay’s Method is very efficient for discontinuous loading conditions. Macaulay’s Method (the double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams and this method is very useful for the case of discontinuous and/or discrete loading condition. Macaulay’s method allows us to obtain the slope and deflection at a particular point in a beam. This information is crucial to desithe gn of beams and shafts to ensure they meet the safe design criteria. The slope and deflection calculation is started from this formula; = () Where; = the flexural rigidity (2 or 2 or 2 or 2 ) = the deflection or displacement ( or ) () = the internal bending moment, expressed as a function of , which is the distance along the beam Then, the integration of the formula will get the (slope, ) and another time to get (deflection or displacement). Let’s take a look below:
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 139 Notice that, integrating twice produced two constants C1 and C2. To resolve this, the boundary conditions will be applied to the formula. The boundary conditions are usually taken at the supports. The characteristics of support are really important to be memorized. Please refer to TOPIC 1 for the subtopic of types of supports!! :-). There are two constants C1 and C2, therefore two boundary conditions need to apply to solve both C1 and C2.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 140 Table 6.2 (a): The boundary conditions for different type of supports SUPPORT @ Roller ≠ = Pinned ≠ = Fixed = = Symmetry = If supports and loading are completely symmetrical = = Source: http://www.engineeringcorecourses.com/solidmechanics1/C9-deflection-of-beams-and-shafts/C9.1-integration-method/theory/
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 141 6.2.2 Macaulay’s method for different types of loading Macaulay brackets will only be activated when the distance moves past where the load is applied: The formula for different types of loading is shown in Table 6.2 (b). Table 6.2 (b): Macaulay’s formula for different types of loading Loading Diagram Macaulay’s formula Point load, P = < − >1 Uniformly distributed load, w = < − >2 2 Moment, M = < − >0 Triangular distributed load, m = m < − >3 6
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 142 Example 1: Determine the slope at support and maximum deflection, in a simply supported beam of length L carrying a concentrated load P at midspan. Solution: Step 1 Moment equation; 1() − � − 2 � = Step 2 Macaulay Equation; 2 2 = () = 1[] − � − 2 � = 0.5[]2 2 − � − 2� 2 2 + 1 = 0.5[]3 6 − � − 2� 3 6 + 1[] + 2 ………… (1) ………… (2)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 143 Step 3 Boundary condition; (Refer to Table 6.2(a), for the boundary condition of pinned and roller support) Therefore, at support 1 at support 2 = 0 ≠ 0 = 0 Substitute the condition of support 1 in equation (2) (0) = 0.5(0)3 6 − �0 − 2� 3 6 + 1(0) + 2 ∴ 2 = 0 Substitute the condition of support 2 in equation (2) also where 2 = 0 (0) = 0.5()3 6 − � − 2� 3 6 + 1() + 0 0 = 0.5()3 6 − � − 2� 3 6 + 1() ∴ 1 = − � 0.5()3 6 − � − 2� 3 6 � = − 0.3752 6 = ≠ 0 = 0
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 144 Step 4 Complete equation of Macaulay = 0.5[]2 2 − � − 2� 2 2 − 0.3752 6 = 0.5[]3 6 − � − 2� 3 6 − 0.3752 6 () Step 5 Slope at supports; Use equation ….. (3) at support 1 = 0 ≠ 0 = 0.5(0)2 2 − �0 − 2� 2 2 − 0.3752 6 ∴ @ = − 0.3752 6 at support 2 = ≠ 0 = 0.5()2 2 − � − 2� 2 2 − 0.3752 6 ∴ @ = 0.3752 6 Max deflection Since the beam is symmetry, hence the max deflection occurs at the centre of the span where; ………… (3) ………… (4)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 145 = 2 (substitute into equation …….(4)) = 0.5 � 2� 3 6 − � 2 − 2� 3 6 − 0.3752 6 � 2 � =3 96 −3 32 ∴ =3 − 33 96 = − 23 96 = −3 48 (negative sign shows that the deflection is downward) So, the maximum deflection occurs downward at midspan.