44 CHAPTER 3 SHEAR FORCE AND BENDING MOMENT 3.0 TYPES OF BEAMS There 3 types of beams considered in this topic a) Simply supported beam - It is a type of beam that has pinned support at one end and roller support at the other end. Depending on the load applied, it undergoes shearing and bending. It is the one of the simplest structural elements in existence. The following image illustrates a simply supported beam b) Overhanging beam - It is defined as a beam that has its one or both ends stretching out past its supports. It can have any number of supports. In other words, it is a beam when a cantilever portion is hanging out of a simply supported beam. c) Cantilever beam - is a rigid structural element, such as a beam or a plate, anchored at only one end to a (usually vertical) support from which it is protruding, this could also be a perpendicular connection to a flat vertical surface such as a wall.
45 3.1 TYPES OF LOADS Types of loads in engineering application are: a) Concentrated Load - Point load is that load which acts over a small distance. Because of concentration over small distance this load can may be considered as acting on a point. Point load is denoted by P and symbol of point load is arrow heading downward b) Uniformly distributed load - Distributed load is that acts over a considerable length, or you can say “over a length which is measurable. Distributed load is measured as per unit length. Uniformly distributed load is that whose magnitude remains uniform throughout the length. c) Coupled load - Coupled load is that in which two equal and opposite forces acts on the same span. The lines of action of both the forces are parallel to each other but opposite in directions. This type of loading creates a couple load.
46 3.2 REACTION FORCE A reaction force is a force that acts in the opposite direction to an action force. 3.3 SHEAR FORCE Shear force is the force in the beam acting perpendicular to its longitudinal (x) axis. For design purposes, the beam's ability to resist shear force is more important than its ability to resist an axial force. Axial force is the force in the beam acting parallel to the longitudinal axis. The upward force is positive, and downward force is negative. 3.4 BENDING MOMENT A bending moment is the reaction induced in a structural element when an external force or moment is applied to the element causing the element to bend. The most common or simplest structural element subjected to bending moments is the beam.
47 3.5 SHEAR FORCE AND BENDING MOMENT DIAGRAMS Shear and bending moment diagrams are analytical tools used in conjunction with beam analysis to help perform beam design by determining the value of shear force and bending moment at a given point of a beam. These diagrams can be used to easily determine the type, size, and material of a member in a beam so that a given set of loads can be supported without beam failure. Another application of shear and moment diagrams is that the deflection of a beam can be easily determined using either the moment area method or the conjugate beam method. 3.5.1 Example 3.0 A beam is loaded as shown in Figure 1, determine: a) the reactions force at A and E. b) draw shear force diagram c) draw bending moment diagram 20 kN 10 kN 50 kN 0.1 m 0.2 m 0.2 m 0.1 m A B C D E Figure 1
48 SOLUTION Free Body Diagram: From FBD, we will determine the reaction RA and RD a) The reactions force at A and E. Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RA+RE = 50-20+10 = 40 kN --------------------------------------- (1) Fixed moment at A, + ΣMA 50(0.1)-20(0.3)+10(0.5)-RE(0.6) = 0 0.6RE = 4 RE = 6.7 kN -------------------- (2) Substitute equation (2) into equation (1) RA = 40 - 6.7 = 33.3 kN b) Draw shear force diagram Shear Force VA = 33.3kN VB = 33.3 - 50 = -16.7 kN VC = -16.7 + 20 = 3.3 kN VD = 3.3 - 10 = -6.7 kN VE = -6.7 + 6.7 = 0 kN 50 kN 20 kN 10 kN 0.1 m 0.2 m 0.2 m 0.1 m A B C D E R A R E
49 Shear force diagram c) Draw bending moment diagram Bending moment diagram Bending moment using area of shear force diagram MA = 0 kNm MB = 33.3(0.1) = 3.33 kNm MC = 3.33 - 16.7 (0.2) = -0.01 kNm MD = -0.01 + 3.3(0.2) = 0.65 kNm ME = 0.65 - 6.7(0.1) = 0 kNm Bending moment using moment from point to point MA = 0kNm MB = 33.3(0.1) = 3.33kNm MC = 33.3(0.3) - 50 (0.2) = -0.01 kNm MD = 33.3(0.5) - 50(0.4) + 20(0.2) = 0.65 kNm ME = 33.3(0.6) - 50(0.5) + 20(0.3) - 10(0.1) = 0 kNm 50 kN 20 kN 10 kN 0.1 m 0.2 m 0.2 m 0.1 m A B C D E 6.7 kN 3.3 kN 33.3 kN 16.7 kN 50 kN 20 kN 10 kN 0.1 m 0.2 m 0.2 m 0.1 m A B C D E 3.33 kNm 0.01 kNm 0.65 kNm
50 3.5.2 Example 3.1 A simply supported beam is given loads and support as below. Determine: a) the reactions at each support b) the shear and bending moment from point A to F c) sketch the shear and bending moment diagram SOLUTION Free Body Diagram: From FBD, we will determine the reaction RA and RD a) The reactions each support. Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RB+RF = 10(7) + 30 + 30 = 130 kN ---------------------------- (1) Fixed moment at B, + ΣMB -10(3)(3/2) + 10(4)(4/2) + 30(2) + 30(7) - RF(9) = 0 9RF = 305 RF = 33.9 kN -------------- (2) Substitute equation (2) into equation (1) RB = 130 - 33.9 = 96.1 kN 10 kN/m 30 kN 30 kN 3 m 2 m 2 m 3 m 2 m A B C D E F 10 kN/m 30 kN 30 kN 3 m 2 m 2 m 3 m 2 m A B C D E F R B R E
51 b) Shear force and bending moment Shear force diagram Shear Force VA = 0 kN VB1 = -10(3) = -30kN VB2 = -10(3) + 96.1 = 66.1 kN VC1 = -10(5) + 96.1 = 46.1 kN VC2 = 46.1 - 30 = 16.1 kN VD = -10(7) +96.1 - 30 = -3.9 kN VE = -3.9 - 30 = -33.9 kN VF = -33.9 + 33.9 = 0 kN 10 kN/m 30 kN 30 kN 3m 2m 2m 3m 2m A B C D E F RB RF 30 kN 66.1 kN 46.1 kN 16.1 kN 3.9 kN 33.9 kN x 2-x x 16.1 = 2 - x 3.9 3.9x = 32.2 - 16.1x 20x = 32.2 x = 1.61 m
52 Bending moment diagram Bending moment using area of shear force diagram MA = 0 kNm MB = -1/2(30)(3) = -45 kNm MC = -45 + 1/2(66.1+46.1)(2) = 67.2 kNm MD = 67.2 + 1/2(1.61)(16.1) - 1/2(0.39)(3.9) = 79.4kNm ME = 79.4 - 3.9(3) = 67.7 kNm MF = 67.7 - 33.9(2) = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = -10(3)(3/2)= -45 kNm MC = -10(5)(5/2) + 96.1(2) = 67.2 kNm MD = -10(7)(7/2) +96.1(4) - 30(2) = 79.4 kNm ME = -10(7)(7/2+3) + 96.1(7) - 30(5) =67.7 kNm MF = -10(7)(7/2+5) + 96.1(9) - 30(7) - 30(2) = 0 kNm 10 kN/m 30 kN 30 kN 3m 2m 2m 3m 2m A B C D E F RF RB 45kNm 67.2 kNm 79.4 kNm 67.7 kNm
53 3.5.3 Example 3.2 The figure shows a simply supported beam is subjected with concentrated load and moment 4.8 kNm, calculate: i) The reaction forces ii) Draw the shear force and bending moment diagram SOLUTION Free Body Diagram: From FBD, determine the reaction RA and RD a) The reaction forces Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RB+RF = 2 kN ---------------------------- (1) Fixed moment at B, + ΣMA = 0 2(3) - 4.8 - RD(12) = 0 12RD = 1.2 RD = 0.1 kN -------------------------- (2) Substitute equation (2) into equation (1) RA = 2 - 0.1 = 1.9 kN 2 kN RA 3 m 6 m 3 m A B C RD D 4.8 kNm
54 b) Draw the shear force and bending moment diagram Shear Force VA = 1.9kN VB = 1.9 - 2 = -0.1 kN VC = -0.1 kN VD = -0.1 + 0.1 = 0 kN Bending moment using area of shear force diagram MA = 0 kNm MB = 1.9(3) = 5.7 kNm MC = 5.7 - 0.1(6) - 48 = 0.3 kNm MD = 0.3 - 0.1(3) = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = 1.9(3) = 5.7 kNm MC = 1.9(9) - 2(6) - 48 = 0.3 kNm MD = 1.9(12) - 2(9) - 48 = 0 kNm 2 kN RA 3 m 6 m 3 m A B C RD D 4.8 kNM 5.7 kNm 1.9 kN 0.1 kN 0.1 kN 0.3 kNm
55 3.6 POINT OF CONTRA-FLEXURE In a bending beam, a point is known as a point of contra-flexure if it is a location at which no bending occurs. In a bending moment diagram, it is the point at which the bending moment curve intersects with the zero line. In other words where the bending moment changes its sign from negative to positive or vice versa. 3.7 MAXIMUM BENDING MOMENT The maximum bending moment is known based on the bending moment diagram. Maximum bending moment position is obtained at the highest peak on a bending moment diagram either positive or negative position. Two types can be used in investigating maximum bending moment: a) Obtaining the area of the shear force from the beginning to the zero point. b) Determining the gradient of the bending moment equation from the original position to the peak of the bending moment diagram.
56 3.7.1 Example 3.3 Based on Figure below, determine: a) The reactions at A and D. b) Diagram of shear force and bending moment and label the values of shear force and bending moment on the diagram. c) Determine the value of contra-flexure point. SOLUTION Free Body Diagram: From FBD, we will determine the reaction RA and RD a) The reactions of each supports . Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RA+ RD = 100 +60 +15(4) = 220 kN ------------------- (1) Fixed moment at A, + ΣMA 60(7) + 15(4)(4/2 + 12) - RD(12) = 0 2RD = 1260 RD = 105 kN ------------------- (2) Substitute equation (2) into equation (1) RA = 220 - 105 = 115 kN b) Diagram of shear force and bending moment 100 kN 60 kN 15 kN/m 3 m 4 m 5 m 4 m A B C D E RD RA
57 Shear Force VA1 = 115 kN VA2 = 115 - 100 = 15 kN VB = 15 kN VC = 15 - 60 = -45 kN VD = -45 + 105 = 60 kN VE = 60 - 15(4) = 0 kN Bending moment using area of shear force diagram MA = 0 kNm MB = 15(3) = 45 kNm MC = 45 +15(4) = 105 kNm MD = 105 - 45(5) = -120 kNm ME = -120 + 1/2(60)(4) = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = 115(3) -100(3) = 45 kNm MC = 115(7) - 100(7) = 105 kNm MD = 115(12) -100(12) -60(5) = -120 kNm ME = 115(16) -100(16) -60(9) + 105(4) - 15(4)(4/2) = 0 kNm 100 kN 60 kN 15 kN/m 3 m 4 m 5 m 4 m A B C D E RD RA 115 kN 15 kN 15 kN 45 kN 60 kN 45 kN/m 105 kN/m 120 kN/m contraflexture point, K x y
58 c) The value of contra-flexure point Using area of shear force diagram also produce value of contra-flexure point MK = 15(3) + 15(4) - 45(y) = 0 45(y) = 105 y = 2.33 m Therefore, contra-flexure point, K = 3 + 4 + 2.33 = 9.33 mm 3.7.2 Example 3.3 A beam is subjected to loads as shown below. Calculate: a) The reactions at B and E b) The value of the shear force and bending moment at A, B, C, D and E c) Draw a diagram of shear force and bending moment d) Calculate the maximum bending moment and contra-flexure point of the beam SOLUTION Free Body Diagram: From FBD, we will determine the reaction RB and RE a) The reactions at B and E Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RB+RE = 25 + 25 + 50(2) + 10 = 160 kN ------------------------------------- (1) Fixed moment at E, + ΣMA = 0 -25(5) + RB(4) - 25(3) - 50(2)(2/2+1) = 0 4RB = 400 25 kN 25 kN 50 kN/m 1 m 1 m 2 m 1 m A B C D E 10 kN 25 kN 25 kN 50 kN/m 1 m 1 m 2 m 1 m A B C D E 10 kN RB RE
59 RB = 100 kN ---------------------------------- (2) Substitute equation (2) into equation (1) RE = 160 - 100 = 60 kN a) The value of the shear force and bending moment at A, B, C, D and E Shear Force VA = -25 kN VB = -25 + 100 = 75 kN VC = 75 -25 = 50 kN VD = 50 - 50(2) = -50 kN VE1 = -50 - 10 = -60 kN VE2 = -60 + 60 = 0 kN Bending moment using area of shear force diagram MA = 0 kNm MB = -25(1) = -25 kNm MC = -25 +75(1) = 50 kNm MD = 50+1/2(50)(1)-1/2(50)(1) = 50 kNm ME = 50 - 50(1) = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = -25(1) = -25 kNm MC = -25(2) + 100(1) = 50 kNm MD = -25(4)+100(3)-25(2)-50(2)(2/2) = 50 kNm ME = -25(5)+100(4)-25(3)-50(2)(2/2+1) = 0 kNm
60 c) A diagram of shear force and bending moment d) The maximum bending moment and contra-flexure point of the beam Using area of shear force diagram to produce maximum bending moments Mmax = -25(1)+75(1)+1/2(50)(1) = 75 kNm Using area of shear force diagram to produce contra-flexure point when MK = 0 MK = -25(1) +75(y) = 0 75(y) = 25 y = 0.33 Therefore, contra flexure point, x = 1 + 0.33 = 1.33 mm 50 kN/m 25 kN 25 kN 1m 1m 2m 1m A B C D E 10 kN RB RE 25 kN 75 kN 50 kN 50 kN 60 kN x Contraflexture point 25 kNm 50 kNm 50 kNm Maximum bending moment y
61 3.7.4 Example 3.4 An 8.3 kN/m distributed load was placed on the simply supported beam from point A to point C and concentrated load 11 kN at point B. a) Draw the free body diagram of the beam b) Determine the reaction force at the support of A and C c) Determine all the shear forces and bending moment d) Determine maximum bending moment and show its location on the graph e) Sketch the shear force and bending moment diagram SOLUTION a) The free body diagram of the beam b) The reaction force at the support of A and C Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RA + RC = 11 + 8.3(4.5) = 48.35 kN ------------------------------------- (1) Fixed moment at E, + ΣMA = 0 11(0.9) + 8.3(4.5)(4.5/2) - RC(4.5) = 0 4.5RC = 93.9375 RC = 20.875 kN ---------------------------------- (2) Substitute equation (2) into equation (1) RA = 48.35 - 20.875 = 27.475 kN c) The all the shear forces and bending moment Shear Force VA = 27.475 kN VB1 = 27.475 - 8.3(0.9) = 20.005 kN VB2 = 20.075 - 11 = 9.005 kN 8.3 kN/m 11 kN A B C 0.9 m 3.6 m 8.3 kN/m 11 kN A B C 0.9 m 3.6 m RA RC
62 VC1 = 9.005 - 8.3(3.6) = -20.875 kN VC2 = -20.875 + 20.875 = 0 kN Bending moment using area of shear force diagram MA = 0 kNm MB = 0.5(27.475 + 20.005)(0.9) = 21.366 kNm MC = 21.366 + 0.5(9.005)(1.085) - 0.5(20.875)(2.515) = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = 27.475(0.9) - 8.3(0.9)(0.9/2) = 21.366 kNm MC = -27.475(4.5) - 8.3(4.5)(4.5/2) - 11(3.6) = 0 kNm d) The maximum bending moment and show its location on the graph Using area of shear force diagram to produce maximum bending moments Mmax = 0.5(27.475 + 20.005)(0.9) + 0.5(9.005)(1.085) = 26.251 kNm 8.3 kN/m 11 kN A B C 0.9 m 3.6 m RA R Mmax 27.475 20.005 9.005 kN 20.875 21.366 kNm x y 3.6 - y y 9.005 = 3.6 - y 20.875 20.875y= 9.005(3.6 - y) 29.88y = 32.418 y = 1 085 m
63 3.7.5 Example 3.5 Figure below shows a cantilever beam is subjected with concentrated loads of 20 kN and 30 kN and uniformly distributed load of 10 kN/m respectively, determine: a) the reaction force and moment at D b) value of shear force and bending moment at every point c) sketch the shear force and bending moment diagram d) the maximum bending moment and contra flexure point SOLUTION a) The reaction force and moment at D Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ RD = 20 + 10(2.5 + 30 RD = 75 kN Fixed moment at D, + ΣMD = 0 -20(6.5) - 10(2.5)(2.5/2+2) - 30(2) + MD = 0 MD = 271.25 kNm 30 kN 10 kN/m A 20 kN 2.0 m 2.5 m 2.0 m B C D MD RD 10 kN/m 30 kN A 20 kN 2.0 m 2.5 m 2.0 m B C D
64 b) Value of shear force and bending moment at every point Shear Force VA = -20 kN VB = -20 kN VC1 = -20 - 10(2.5) = - 45 kN VC2 = - 45 - 30 = -75 kN VD = -75 + 75 = 0 kN Bending moment using area of shear force diagram MA = 0 kNm MB = -20(2) = -40 kNm MC = - 40 - 0.5(45+20)(2.5) = -121.25 kNm MD1 = -121.25 - 75(2) = -271.25 kNm MD2 = -271.25 + 271.25 = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = -20(2) = -40 kNm MC = -20(4.5) - 10(2.5)(2.5/2) = -121.25 kNm MD1 = -20(4.5) - 10(2.5)(2.5/2 + 2) - 30(2) = -271.25 kNm MD2 = -271.25 + 271.25 = 0 kNm c) Sketch the shear force and bending moment diagram MD RD 30 kN 10 kN/m A 20 kN 2.0 m 2.5 m 2.0 m B C D 20 kN 45 kN 75 kN 40 kNm 121.25 kNm 271.25 kNm
65 d) The maximum bending moment is 271. 25 kNm and there is no contra flexure point. 3.7.6 Example 3.6 Figure below shows a cantilever beam is subjected with concentrated load and uniformly distributed load, calculate: a) The reaction force and moment at D b) Draw the shear force and bending moment diagram c) The contra flexure point SOLUTION Free Body Diagram: From FBD, we will determine the reaction RD and momen of area, MD a) The reactions force and moment at D Based on the principle of equilibrium of forces ∑F↑ = ∑F↓ 50 + RD = 25(4) + 20 RD = 70 kN Fixed moment at D, + ΣMD 50(6) - 24(4) - 20(2)(4/2+2) + MD = 0 25 kN/m 20 kN 50 kN 2 m 2 m 2 m A B C RD D MD
66 MD = 400 + 80 - 300 MD = 180 kNm b) Draw the shear force and bending moment diagram Shear Force VA = 50 kN VB1 = 50 - 25(2) = 0 kN VB2 = 0 - 20 = -20 kN VC = -20 - 25(2) = -70 kN VD = -70 + 70 = 0 kN Bending moment using area of shear force diagram MA = 0 kNm MB = 0.5(50)(2) = 50 kNm MC = 50 - 0.5(20+70)(2) = -40 kNm MD1 = -40 - 70(2) = -180 kNm MD2 = -40 - 70(2) + 180 = 0 kNm Bending moment using moment from point to point MA = 0 kNm MB = 50(2) -25(2)(2/2)= 50 kNm MC = 50(4) - 25(4)(4/2) - 20(2) = -40 kNm MD1 = 50(6) -25(4)(4/2 + 2) - 20(4) = -180 kNm MD2 = 50(6) -25(4)(4/2 + 2) - 20(4) + 180 = 0 kNm
67 c) Base on bending moment diagram, contraflexure point is at point BC when MK = 0 MK = 50(x) - 25(x)(x/2) - 20(x - 2) = 0 50x - 12.5x2 - 20x + 40 = 0 -12.5x2 + 30x + 40 = 0 x = -30 ± � 302 - 4(-12.5)(40) 2(-12.5) = -30 ± 53.9 -25 x = -30 + 53.9 -25 = -0.96 m or x = -30 - 53.9 -25 = 3.36 m (accepted) Contraflexure point is 3.36 m 50 kN 20 kN 70 kN 20 kN 25 kN/m A B C MD D 2 m 2 m 2 m 50 kN RD Contraflexture point 50 kNm x 40 kNm 180 kNm
68 3.8 EXERCISES 1. Figure 3(1) shows a 6 m simply supported beam carries a uniformly distributed load of 1.5 kN/m over the entire span and a point load of 2 kN at 2 m from the right support. a) By referring the figure with the aid of free body diagram, express the value of reaction forces b) Calculate the shear force along the beam and sketch the shear force diagram c) Calculate the bending moment value along the beam and sketch the bending moment diagram (Ans: 5.17 kN, 5.83 kN) 2. A simple supported beam carrying a few loads as shown in Figure 3(2) below. a) Calculate the reaction force at point A and F b) Calculate the shear force along the beam and sketch shear force diagram c) Calculate the bending moment value along the beam and sketch bending moment diagram. (Ans: 18 kN, 18 kN) 3. A simply supported beam is loaded as shown in Figure 3(3) a) Sketch a free body diagram for the beam and calculate the reaction forces if P = 20 kN b) Calculate the shear force and bending moment 1.5 kN/m 2 kN A B C 4 m 2 m Figure 3(1 ) 5 kN/m 6 kN 2 kN 1 m 5 m A B C D E F 4 m 1 m 2 m 4 kN/m Figure 3(2) 4 kN/m P A B C 5 m 5 m Figure 3(3)
69 c) Analyse the bending moment maximum and its position from the left of the beam based on shear force and bending moment diagram (Ans: 15 kN, 25 kN, 75 kNm, 5m) 4. A simply supported beam carrying a few loads as shown in Figure 3(4) below. a) Calculate the reaction forces for the beam and sketch the free body diagram b) Calculate the shear force and bending moment for the beam c) Draw the shear force and bending moment diagram for the beam and determine the maximum bending moment and its position (Ans: 24.8 kN, 22.2 kN, 30.4 kNm, 8m) 5. A beam shown in Figure 3(5) is subjected to a uniformly distributed load of 3 kN/m from point A to point C and concentrated load of 8 kN at point D. a) Sketch the free body diagram of the beam b) Determine the shear force and bending moment at point ABCDE c) Sketch the shear force and bending moment diagram d) Analyse the value of contraflexure point from point A (Ans: 23.25 kN, 2.75 kN, 7.43m) 6. An overhanging beam is loaded as shown in Figure 3(6) below. a) Sketch the free body diagram of the beam and find the reaction force at point A and D. b) Determine shear force and bending moment at point ABCDE 2 kN/m 15 kN 10 kN A B C D 4 m 4 m 2 m 7 kN/m Figure 3(4) 3 kN/m 8 kN 3 m 3 m A B C D E 4 m 2 m Figure 3(5 ) 10 kN 25 kN/m A B C D E 1 m 1 m 2 m 1 m Figure 3(6) 15 kN
70 c) i) Sketch the shear force and bending moment diagram ii) Determine the value of maximum bending moment and its position (Ans: 25.625 kN, 74.375 kN, 36.258 kNm, 2.025 m) 7. A simply supported beam is subjected to a uniformly distributed load and a concentrated load as shown in Figure 3(7). a) Calculate the reaction forces at the supports b) Construct the shear force and bending moment diagram of the beam c) Identify and specify the magnitude of the maximum bending moment (Ans: 8 kN, 8 kN, 18 kNm) 8. An 8.3 kN/m distributed load as shown in Figure 3(8) is placed on a simply supported beam from point A to point C and a concentrated load of 11 kN at point B a) Draw a free body diagram of the beam b) Determine the reaction forces at the support of A and C c) Determine all the shear force and bending moment d) Determine the maximum bending moment and show its location on graph e) Sketch the shear force and bending moment diagram (Ans: 47.475 kN, 20.875 kN, 26.251 kNm, 1.985m) 3 kN/m 4 kN A B C 2 m 2 m Figure 3(7) 1 m 1 m D E 8.3 kN/m 11 kN A B C 3.6 m Figure 3(8) 0.9 m