CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 146 Example 2: Determine the maximum deflection in a simply supported beam of length 5m carrying a uniformly distributed load of intensity 20KN/m applied over its entire length. Solution: Step 1 Moment equation; = 50() − 20() � 2 � ∴ = 50() − 20 () 2 2 Step 2 Macaulay equation; 2 2 = () = 50[] − 20 [] 2 2 ………… (1) ………… (2)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 147 = 50[]2 2 − 20 [] 6 3 + 1 = 50[]3 6 − 20 [] 24 4 + 1[] + 2 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) = 50[0]3 6 − 20 [0] 24 4 + 1[0] + 2 ∴ 2 = 0 Substitute the condition of support 2 in equation (2) where 2 = 0 (0) = 50[5]3 6 − 20 [5] 24 4 + 1[5] + 0 (0) = 1041.67 − 520.83 + 51 0 = 1041.67 − 520.83 + 51 −51 = 1041.67 − 520.83 1 = − 520.84 5 = −104.17 = 5 ≠ 0 = 0
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 148 Step 4 Complete equation of Macaulay = 50[]2 2 − 20 [] 6 3 − 104.17 = 50[]3 6 − 20 [] 24 4 − 104.17[] Step 5 Since the beam is symmetry, hence the max deflection occurs at the centre of the span where; = 2.5 (Substitute into equation …… (4)) = 50[2.5]3 6 − 20 [2.5] 24 4 − 104.17[2.5] = 130.21 − 32.55 − 260.43 ∴ = 130.21 − 32.55 − 260.43 = − 162.77 So, the maximum deflection . occurs downward at midspan, 2.5m. ………… (3) ………… (4)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 149 Example 3: A simply supported beam carries a load moment as shown in figure. Determine the equation of moment and complete equation of Macaulay for slope and deflection. Solution: Step 1 Moment equation; & � = 0 0 = −10 − (7) 10 7 = − ∴ = −1.43 = 1.43 (↓) Therefore, = 1.43 (↑) ∴ = 1.43() − 10( − 5)0
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 150 Step 2 Macaulay equation; 2 2 = () = 1.43[] − 10[ − 5]0 = 1.43[]2 2 − 10[ − 5]1 + 1 = 1.43[]3 6 − 10[ − 5]2 2 + 1[] + 2 Step 3 Boundary condition; (Refer to Table 6.2(a), for the boundary condition of pinned and roller support) Therefore, at Support A at support B = 0 ≠ 0 = 0 Substitute the condition of support A in equation (2) (0) = 1.43[0]3 6 − 10[0 − 5]2 2 + 1[0] + 2 ∴ 2 = 0 Substitute the condition of support B in equation (2) where 2 = 0 (0) = 1.43[7]3 6 − 10[7 − 5]2 2 + 1[7] + 0 0 = 81.74 − 20 + 1[7] ………… (1) ………… (2) = 7 ≠ 0 = 0
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 151 0 = 61.74 + 1[7] ∴ 1 = − 61.74 7 = −8.82 Step 4 Complete equation of Macaulay = 1.43[]2 2 − 10[ − 5]1 − 8.82 = 1.43[]3 6 − 10[ − 5]2 2 + 1[] ………… (3) ………… (4)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 152 Example 4: A cantilever beam is applied with loads as shown in the figure. Determine the slope at the free end and deflection at point B. Take EI = 200 KNm2 . Solution: Step 1 Moment equation; = 20(1.5) + 10 = 40(↑) = 20(1.5) � 1.5 2 � + 10(3) = 52.5 ( ) = −52.5() + 40() − 20() � 2 � + 20( − 1.5) � − 1.5 2 � − 10( − 3) ∴ = −52.5()0 + 40() − 20 ()2 2 + 20 ( − 1.5) 2 2 − 10( − 3)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 153 Step 2 Macaulay equation; 2 2 = () = −52.5[]0 + 40[] − 20 []2 2 + 20 [ − 1.5] 2 2 − 10[ − 3] = −52.5[]1 + 40 []2 2 − 20 []3 6 + 20 [ − 1.5] 6 3 − 10[ − 3]2 2 + 1 = −52.5 [] 2 2 + 40 []3 6 − 20 []4 24 + 20 [ − 1.5] 24 4 − 10[ − 3]3 6 + 1[] + 2 Step 3 Boundary condition; (Refer to Table 6.2(a), for the boundary condition of fixed support) Therefore, at Support A at support A at free end = 0 = 0 = 0 Substitute the condition of support A in equation (1) (0) = −52.5[0] 1 + 40 [0]2 2 − 20 [0]3 6 + 20 [0 − 1.5] 6 3 − 10[0 − 3]2 2 + 1 ∴ 1 = 0 Substitute the condition of support A in equation (2) where 1 = 0 = 5 ≠ 0 = ………… (1) ………… (2)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 154 (0) = −52.5 [0] 2 2 + 40 [0]3 6 − 20 [0]4 24 + 20 [0 − 1.5] 24 4 − 10[0 − 3]3 6 − 0 + 2 ∴ 2 = 0 Step 4 Complete equation of Macaulay = −52.5[] 1 + 40 []2 2 − 20 []3 6 + 20 [ − 1.5] 6 3 − 10[ − 3]2 2 = −52.5 [] 2 2 + 40 []3 6 − 20 []4 24 + 20 [ − 1.5] 24 4 − 10[ − 3]3 6 Step 5 Slope at the free end = 5 Substitute into equation ………… (3) @() = −52.5[5]1 + 40 [5]2 2 − 20 [5]3 6 + 20 [5 − 1.5] 6 3 − 10[5 − 3]2 2 () = −262.5 + 500 − 416.67 + 142.92 − 20 ∴ = −56.25 = −56.25 200 = −0.28 = 16° 54′ Deflection at point B is where = 3 Substitute into equation ………… (4) = −52.5 [3] 2 2 + 40 [3]3 6 − 20 [3]4 24 + 20 [3 − 1.5] 24 4 − 10[3 − 3]3 6 = −236.25 + 180 − 67.5 + 4.22 − 0 = 180 − 67.5 + 4.22 ∴ = −119.53 = −119.53 200 = −0.59 ………… (3) ………… (4)
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 155 Try to solve these problems 1. Refer Figure 1 By using the Macaulay method; i) Write the moment equation ii) Write the Macaulay equation iii) Determine the slope at 2m from support A iv) Determine the deflection at mid span Figure 1 2. Refer Figure 2 By using the Macaulay method; i) Write the moment equation ii) Write the Macaulay equation iii) Determine the slope at 4m from support A iv) Determine the deflection at point C Figure 2
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 156 3. Refer Figure 3 By using the Macaulay method; i) Write the moment equation ii) Write the Macaulay equation iii) What is the value of EI if the slope at point C is 0.05rad? iv) Calculate for deflection at point D by using the EI determined from (iii). Figure 3 4. A cantilever beam with a rectangular cross-section, is loaded as shown in Figure 4. By using the Macaulay equation; i) Calculate the slope and deflection at free end, A, and point B. Take E = 200 MPa. Figure 4
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 157 6.3 Moment Area Method 6.3.1 Moment diagram as a method of determining the slopes and deflections in beams Figure 6.3.1: Moment Area Diagram Theorems of Area-Moment Method Theorem I The change in slope between the tangents drawn to the elastic curve at any two points A and B is equal to the product of 1 multiplied by the area of the moment diagram between these two points: = ( )
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 158 Theorem II The deviation of any point B relative to the tangent drawn to the elastic curve at any other point A, in a direction perpendicular to the original position of the beam, is equal to the product of multiplied by the moment of an area about B of that part of the moment diagram between points A and B. Rules of Sign 1. The deviation at any point is positive if the point lies above the tangent, and negative if the point is below the tangent. 2. Measured from the left tangent, if θ is counterclockwise, the change of slope is positive, negative if θ is clockwise. / = ( ) ∙ � / = ( ) ∙ � and
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 159 6.3.2 Slope and deflection of simply supported beam –Moment Area Method The deflection delta, at some point B of a simply supported beam, can be obtained by the following steps: 1. Compute / / = 1 () ∙ � 2. Compute / / = 1 () ∙ � 3. Solve by ratio and proportion (see figure above). + / =/
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 160 Common equations for calculating the area and centroid of different shapes may be seen in Table 6.3 Table 6.3: Areas and Centroids of Common Shapes 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 161 Example 1: By using the Moment Area Method, find the value of slope and deflection at points B and C for the simply supported beam shown in Figure 1 below. Figure 1 Solution: � = 0 0 = 200(3) + 400(7) − (8) (8) = 200(3) + 400(7) ∴ = 200(3) + 400(7) 8 = 425(↑) ℎ ; = 200 + 400 + 425 = 175(↑) Moment area diagram:
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 162 / = () ∙ � / = � 1 � �� 1 2 × 8 × 1400� × � 1 3 × 8� − � 1 2 × 5 × 1000� × � 1 3 × 5� − � 1 2 × 1 × 400� × � 1 3 × 1�� ∴ / = 107003 / = () ∙ � / = � 1 � �� 1 2 × 7 × 1225� × � 1 3 × 7� − � 1 2 × 4 × 800� × � 1 3 × 4�� ∴ / = 47225 6 3 / = () ∙ � 1 7 = 1400 8 ∴ 1 = 1400 8 × 7 = 1225 2 4 = −1000 5 ∴ 2 = −1000 5 × 4 = −800 3 = 1400 8 ∴ = 1400 8 × 3 = 525
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 163 / = 1 � 1 2 × 3 × 525� × � 1 3 × 3� ∴ / = 1575 2 3 By ratio and proportion; 3 =/ 8 =/ 8 × 3 = � 10700 � 8 × 3 ∴ = 4012.5 7 =/ 8 = � 10700 � 8 × 7 ∴ = 9362.5 To find slope at B and C; To find deflection at B and C; = − / = 4012.5 − 1575 2 ∴ = 3225 3 = − / = 9362.5 − 47225 6 ∴ =.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 164 = � = 10700 8 = 10700 8 = 1337.5 But, assuming a counterclockwise direction, the slope at A should be negative! = + � � = 1 () � = � 1 � � 1 2 × 3 × 525� = 787.5 = + � ∴ = − 1337.5 + 787.5 = −550 = + � � = � 1 � �� 1 2 × 7 × 1225� − � 1 2 × 4 × 800�� � = � 1 � [(4287.5) − (1600)] � = 2687.5 ∴ = −. + . =
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 165 Example 2: For a simply supported beam shown in Figure 2 below, determine the slope and deflection at point B. Use = 200 × 1032. Figure 2 Solution: � = 0 0 = 80(4)(2) − (6) (6) = 640 ∴ = 640 6 = 106.67(↑) ℎ ; = 80(4) − 106.67 = 213.33(↑) Moment area diagram; 1 4 = 1279.98 6 ∴ 1 = 1279.98 6 × 4 = 853.32 2 4 = −1440 6 ∴ 2 = −1440 6 × 4 = −960
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 166 / = ( ) ∙ � / = � 1 � ��1 2 × 6 × 1279.98� × � 1 3 × 6� − �1 3 × 6 × 1440� × � 1 4 × 6� + � 1 3 × 2 × 160� × � 1 4 × 2�� / = � 1 � (7679.88 − 4320 + 213.33) ∴ / = 3573.21 3 / = ( ) ∙ � / = 1 ��1 2 × 4 × 853.32� × � 1 3 × 4� − �1 3 × 4 × 960� × � 1 4 × 4�� / = 1 (2275.52 − 1280) ∴ / = 995.52 3 By ratio and proportion; 4 =/ 6 =/ 6 × 4 = � 3573.21 � 6 × 4 ∴ = 2382.14 To find slope at B; = � To find deflection at B; = − / = 2382.14 − 995.52 = 1386.62 3 ℎ = 200 × 1032. ∴ =. × = . × − = .
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 167 = 3573.21 6 ∴ =. But, assuming a counterclockwise direction, the slope at A should be negative! = + � = 595.54 + � ℎ � = 1 () � = 1 �� 1 2 × 4 × 853.32� − � 1 3 × 4 × 960�� ∴ � = 1 [(1706.64) − (1280)] = 426.64 , = −. + . = −. × = −. × −
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 168 Example 3: A simply supported beam as shown in Figure 3 is loaded with concentrated load and moment. By using the moment area method, find the value of deflection at point C. Use = 200 × 1032. Figure 3 Solution: � = 0 0 = 500(1)−200 − (4) (4) = 300 ∴ = 300 4 = 75(↑) ℎ ; = 500 − 75 = 425(↑) Moment area diagram; 1 = 1700 4 ∴ = 425 1 3 = 1700 4 ∴ 1 = 1700 4 × 3 = 1275 2 2 = −1500 3 ∴ 2 = −1500 3 × 2 = −1000
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 169 / = () ∙ � / = � 1 � �� 1 2 × 4 × 1700� × � 1 3 × 4� − � 1 3 × 3 × 1500� × � 1 3 × 3� − [1 × 200] × � 1 2 �� / = � 1 � (4533.33 − 1500 − 100) ∴ / =. / = () ∙ � / = � 1 � �� 1 2 × 3 × 1275� × � 1 3 × 3� − � 1 3 × 2 × 1000� × � 1 3 × 2�� / = � 1 � (1912.5 − 444.44) ∴ / =. By ratio and proportion; 3 =/ 4 =/ 4 × 3 = � 2933.33 � 4 × 3 ∴ = To find deflection at C; = − / = 2200 − 1468.06 = 731.94 3 ℎ = 200 × 1032. ∴ =. × = . × − = .
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 170 6.3.3 Slope and deflection of a cantilever beam Generally, the tangential deviation t is not equal to the beam deflection. In cantilever beams, however, the tangent drawn to the elastic curve at the wall is horizontal and coincidentally therefore with the neutral axis of the beam. The tangential deviation in this case is equal to the deflection of the beam as shown below. Figure 6.3 From Figure 6.3 above, the deflection at B denoted as is equal to the deviation of B from a tangent line through A denoted as / . This is because the tangent line through A lies with the neutral axis of the beam.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 171 Example 1: Figure 1 shows a cantilever beam loaded with two-point loads of 4KN and 2KN respectively. Calculate the deflection of the slope and deflection at the free end and point B. Use EI = 10GPa. Figure 1 Solution: � = 0 0 = −4(4)−2(2) + ∴ = 20 ↻ ∴ = 4 + 2 = 6(↑) Moment area diagram; 2 = −16 4 ∴ = −8
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 172 / = ( ) ∙ � / = � 1 � ��1 2 × 2 × (−4)� × �( 2 3 × 2) + 2� − �1 2 × 4 × 16� × � 2 3 × 4�� / = � 1 � (−13.33 − 85.33) ∴ / = −98.66 3 , −. / = ( ) ∙ � / = 1 ��1 2 × 2 × (−8)� × � 2 3 × 2�� / = 1 (−10.67) ∴ / = −10.67 3 By ratio and proportion; 2 =/ 4 =/ 4 × 2 = � −98.66 � 4 × 2 ∴ = −49.33 To find deflection at B; = − / = −49.33 − (−10.67) ∴ = −.
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 173 To find slope at free end; = 0 ( ) - since = � = ( ) = 1 EI��1 2 × 2 × (−4)� − �1 2 × 4 × 16�� ∴ = 1 EI[−4 − 32] = −36 EI To find slope at point B; = � ∴ � = () = � × × (−)� = −
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 174 Example 2: Calculate the deflection at the free end for a cantilever beam shown in Figure 2 below. Figure 2 Solution: � = 0 0 = −2 + 0.25(4) − ∴ = 1 ↺ ∴ = 0.25(↑) Moment area diagram; / = ( ) ∙ � / = � 1 � ��1 2 × 4 × (1)� × �( 1 3 × 4)� − [2 × 2] × � 1 2 × 2� + �(1 × 4) × 1 2 × 4�� 1 2 = 1 4 ∴ 1 = 0.5 ∴ 2 = −2
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 175 / = � 1 � (2.67 − 4 + 8) / = 6.67 , =. () To find slope at the free end; = � = () = � = 1 �( 1 2 × 4 × (1) − (2 × 2) + 4� ∴ = � =
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 176 Try to solve these problems 1. Find the maximum deflection for the cantilever beam loaded as shown in Figure 1 if the cross section is 50 mm wide by 150 mm high. Use E = 69 GPa. Figure 1 2. For the beam loaded as shown in Figure 2, determine the deflection 6 m from the wall. Use E = 1.5 × 106 Pa and I = 400 mm4 . Figure 2 3. For the cantilever beam shown in Figure 3, what will be the value of P to cause zero deflection at A? Figure 3
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 177 END OF CHAPTER PROBLEM Problem 1 The cantilever beam shown in Figure 1 has a rectangular cross-section 50 mm wide by h mm high. Find the height h if the maximum deflection is not to exceed 10 mm. Use E = 10 GPa. Prove that, the maximum deflection is the same as calculated by the Macaulay method. Figure 1 Problem 2 An overhanging beam shown in Figure 2 is loaded with a uniformly distributed load along the 10m spans. If the deflection at the centre of the beam is 0.005m, determine the reactions at A and B. Use Macaulay method and take EI = 10 GPa. Figure 2
CHAPTER 6 – SLOPE AND DEFLECTION OF DETERMINATE BEAM Mechanics of Civil Engineering Structures 178 Problem 3 By using the Macaulay method, calculate the slope and deflection at the free end for a cantilever beam shown in Figure 3 below. Figure 3
REFERENCES R.C. Hibbeler / K.S. Vijay Sekar (2013). Mechanics of Materials. (9th Edition) Pearson Education Inc. Ferdinand P. Beer, E Russell Johnston Jr., John T. De Wolf. (2015). Mechanics of Materials. (7th Editon). New York, NY: McGraw-Hill Education. James M. Gere, Barry J. Goodno (2013). Mechanics of Materials. (8th Edition). Stamford, Ct Cengage Learning. Keith M. Walker. (2014). Applied Mechanics for Engineering Technology. (8 th Edition). Pearson New International Edition, United State, Pearson. Pant M K. (2016). Fundamental of Structural Analysis. (7 th Edition). New Delhi: S. K. Kataria & Sons. R.C. Hibbeler. (2004). Statics And Mechanics Of Materials. (2 nd Edition). Upper Saddle River, NJ,Prentice Hall. W.Morgan, D.T. Williams. (1981). Mekanik Struktur. (Cetakan Pertama). Dewan Bahasa dan Pustaka MATHalino Engineering Mathematics. (2022). Mechanics and strength of materials. Drupal.org. Retrieved from https://mathalino.com/reviewer/mechanics-and-strength-of-materials/mechanicsand-strength-of-materials Lemonis, D. M. (2015-2022). CALC RESOURCE Calculation Tools & Engineering Resources. Retrieved 2022, from calcresource.com
BIOGRAPHY Mohammed Azmi bin Ladi is a lecturer at the Department of Civil Engineering, Polytechnic Kuching Sarawak, and Polytechnic Mukah. He received his Diploma in Civil Engineering from Polytechnic Kuching Sarawak (2002), Bachelor of Civil Engineering (Hons.) from MARA University of Technology Shah Alam, Selangor (2006), and Master in Engineering (Civil) from the University of Malaysia Sarawak (2016). He taught Mechanics of Civil Engineering Structures and Theory of Structures at Polytechnic for almost 16 years. Normah binti Jainudin is a lecturer at the Department of Civil Engineering, Polytechnic Mukah Sarawak (2007-2011) and Polytechnic Kuching Sarawak (2011 - current). She received her Diploma in Civil Engineering from Universiti Teknologi Mara (2002). In 2005, she received a Bachelor of Civil Engineering (Hons.) from Universiti Teknologi Mara Shah Alam, Selangor. She had her Master in Engineering (Civil) from the University of Malaysia Sarawak in 2017. She taught Mechanics of Civil Engineering Structures and Theory of Structures at Polytechnic for almost 13 years.