STUDY UNIT 8 Shear force and Bending Moment Synopsis This study unit develops the fundamental theory of shear forces and bending moments. It determines these mechanical phenomena via graphical method. Pre-requisites Mechanics I, Mathematics I and II Objectives To enable the learner to understand to draw shear force and bending moment diagrams. Learning Outcomes Estimated study period • • • • 4 hrs To draw free body diagrams of different beams To determine the reactions at different supports To draw shear force and bending moment diagrams and determine the maximum values and their locations To determine the points of contra-flexure on the beam Keywords Shear forces; Bending moment; Point of contraflexure; Free body diagram; Beams Prescribed TextBook(s) Drotsky JA. Strength of Materials for Technicians, Pearson Education South Africa (Pty) Ltd, Cape Town, 2011. ISBN: 9781775782421. Gere JM, Goodno BJ. Mechanics of Materials, Cengage Learning, Stamford, 2009. ISBN – 13: 9780495438076 Open Rubric 8.1. Introduction Beams: a member or bar that is subjected to lateral forces or moments that lie in a plane that contains the longitudinal axis. Figure 8.1 shows the four types of beams which are so classified according to their support conditions (Drotsky under Beams; Gere and Goodno under Types of Beams, Loads, and Reactions). W W RR RL M R Simply supported beam: supported at the ends by at least one roller and a pin. They are both replaced by vertical reactions. The roller ensures that there are no longitudinal forces induced in the beam by the lateral forces. d by a fixing moment M and a reaction P for static equilibrium. Cantilever: rigidly fixed at one end. The fixed end is replaced by a fixing moment M and a reaction R for static equilibrium. WL W MR ML RL RL WR RR RR Encastre’ or Built-in beam: both ends are rigidly fixed. This is statically indeterminate and will not be dealt with in this module. Overhanging beam: One or both of the ends of the beam extend beyond the supports. Figure 8.1: Beam types Loads: Beam loads are classified as: Point or concentrated loads, Uniformly distributed loads, and Uniformly varying loads. In this module we mainly deal with the first two types of loads [See Drotsky under Types of Loads for a diagram of the uniformly varying loads]. 8.2. Shear Force and Bending moment The lateral loads acting on the beams in Figure 8.1 will induce direct and shear stresses which may result in the deflection of the beams. The deflections and their attendant stresses are not uniform along the span of the beam and for cases of design one may want to know where their maximum values occur. Shear force and bending moment diagrams provide a means of determining such critical points on the beam graphically. Drotsky under Shear force and bending moment diagram shows the following ideas with the aid of diagrams. • • The shear force at any section of a beam is numerically equal to the algebraic sum of the lateral components of the forces to one side of the section, and The bending moment at any section of a beam is equal to the algebraic sum of the moment of the forces to one side of the section. Sign conventions: Shear forces are positive if an applied force tends to cause the left hand section to slide upwards relative to the right hand section, and the bending moment is positive if an applied load tends to cause the beam to sag or bend upwards [For clarity, see diagrams in Drotsky under Shear force and bending moment]. Calculating Shear Forces For Point Loads: You always use the method of sections when determining the shear force at a particular point on the beam. Assume that a beam is cut at that point, then for the section to be in static equilibrium, the forces must balance on both sides of the cutting line. Consider the beam in Figure 8.2. 5 kN 15 kN 2 kN/m C B A 0.75 E D 1 1 2 Step 1: Calculate the reactions at the supports First Condition of Static Equilibrium: F = 0 RA + RE − 15 − 5 − 2 2 = 0 R A = 24 − RE Second condition of static equilibrium: [8.1] M A =0 RE 3.75 − 15 0.75 − 5 2.75 − (2 2) 2.75 = 0 3.75RE = 36 RE = 36 = 9.6 kN 3.75 Using Eqn 8.1: R A = 24 − 9.6 = 14.4 kN The shear forces between the end of the 8.3. Point of Contra-Flexure of Inflexion The point on the beam where the bending moment changes sign (i.e. the bending moment at that point is zero). It can be located graphically from the properly scaled bending moment diagram. Alternatively, it can be calculated mathematically from the bending moment equation taken at that point. Illustrative example 8 Go through Examples 6.1and 6.2 in Drotsky. The given steps are the appropriate steps for solving these problems. A good solution must start with neat free body diagrams and then solving for the reactions. These are mandatory steps if a learner has to acquire the full score. Note that Example 2 is for a uniformly distributed load and the bending moment yields a parabolic curve. Whenever you have a uniformly distributed load over any section of the beam, you should know that the plot over that section becomes a curved line not linear as for point loads. Activity 8 Draw the free body diagrams and calculate the reactions for the following problems in Drotsky Self-assessment: Figure 6.30; Figure 6.34 and Figure 6.35. Practice Exercises 8 • Do the following problems in Drotsky under Self-assessment: Question 6.1 to 6.16. • Do the following problems in Gere and Goodno under Shear-Force and Bending-Moment Diagrams: 4.5-1 to 4.5-7; 4.5-9 to 4.5-17. [skip problem 4.5-8 please].