How to find Bending Moment Bending Moment 1. CH28 p355 Bending moment is a torque applied to each side of the beam if it was cut in two - anywhere along its length. The hinge applies a clockwise (+) moment (torque) to the RHS, and a counter-clockwise (-) moment to the LHS. Example Calculate BM: M = Fr (Perpendicular to the force) Bending-Moment Page 1
Calculate BM: M = Fr (Perpendicular to the force) In equilibrium, so M A = 0 But to find the Bending Moment, you must cut the beam in two. Bending moment is INTERNAL, moment is EXTERNAL. BM for RHS of beam: M = Fr = -(6*3) = -18 knm Repeat for LHS: M = Fr = (6*3) = 18 knm Procedure to find BM at any cross-section in the beam: Cut beam thru that section, then add moments for the right or left side only. Ignore + or - sign, and use the following definition for +/-; Positive = sagging Negative = hogging A good way to double-check is to do moments for BOTH sides and compare. In engineering, we are concerned with the MAXIMUM BM. How do we find it? Try the same problem at 1m from left end; Bending-Moment Page 2
BM for LHS of beam: M = Fr = (6*1) = 6 knm Repeat for RHS: M = Fr = +(12*2) -(6*5) = -6 knm The BMD helps us know the MAXIMUM, but also what the BM is an any location along the beam. Bending-Moment Page 3
Bending-Moment Page 4 Simple example: Tuesday, 30 April 2013 6:45 PM Q1: If distance a=7.8m and Force b=9.3kn, find the maximum bending moment. Sag=(+), Hogg=(-) Find Reactions: (Take moment at RH Support) M R = 0 = (R L *10) - (9.3 * 2.2) 9.3 * 2.2 = 20.46 R L = 20.46 /10 = 2.046 kn R L +R R = 9.3 R R = 9.3-2.046 = 7.254 kn Take a slice thru the b force, to get BMoment. (Moment eqn for left side only) BM left = (2.046 * 7.8) = +15.9588 knm We can check this for the right hand side. (Moment eqn for right side only) BM right = (7.254*2.2) = -15.9588 knm So at the cross section, there are two moments - equal and opposite. Hogging or sagging? Positive = sagging.
Bending-Moment Page 5 Shear force & Bending Moment Bending Moment 1. CH28 p355 Positive Shear Force Up on LHS Shear Force is in all beams, but usually only seen as a problem in SHORT beams. Long beams fail by bending.
Bending-Moment Page 6 Shear Force Diagram Step 1. Reactions (Working in knm) 0 = +(130*3) - (F R *8) F R = 90/8 = 11.25 kn 0 = F L -30 + 11.25 F L = 18.75 kn Step 2. SFD Cut anywhere; Add forces on LHS (or RHS - which ever is easier). Use positive S.F. = Upwards on LHS
Shear Force Diagram > BMD Step 1. Reactions (Working in knm) 0 = +(130*3) - (F R *8) F R = 90/8 = 11.25 kn 0 = F L -30 + 11.25 F L = 18.75 kn Step 2. SFD Cut anywhere; Add forces on LHS (or RHS - which ever is easier). Use positive S.F. = Upwards on LHS Step 3. BMD Change of AREA of SFD = Change of HEIGHT of BMD (a) BM zero both ends (free ends) (b) 3*18.75 = 56.25 (c) 5*-11.25 = -56.25 Bending-Moment Page 7
Example 28.1 f. (P361) Double Check (LHS) M=1l *2 = 22 knm (RHS) M = - (5x2) -(3x4) = -10-12 = -22 knm + Positive BM Bending-Moment Page 8
Example 28.1 d The SFD (Shear Force Diagram) tells you how much the beam wants to SLIDE apart. The BMD (Bending Moment Diagram) tells you how much the beam wants to BEND apart by rotation. Check the max BM: Take section thru centre: Moment LHS = +(3*1.75) -(3*0.75) = 3.0 Moment RHS = -(3*1.75) +(3*0.75) = -3.0 We cannot determine +/- from the moment equation because it depends which side we choose. Positive bending moment = SAGGING Negative bending moment = HOGGING Bending-Moment Page 9
Example 28.1 h The area of the SFD = height of the BMD Positive Shear Force Wall Reaction: M = -(4*8)+(9*6)-(5*2) = 12 So wall reaction moment is -12 (CCW) Positive Bending Moment The maximum bending moment = 12 The maximum negative BM = -8 M = - (4*2) = -8 knm M = + (5*4) -12 = 8 knm Remember! You can't determine + or - by looking at the RHS or LHS, since they will always be opposing each other. Just use sagging = positive bending. Bending-Moment Page 10
Bending-Moment Page 11 Example 28.1(j) Tuesday, 20 March 2012 6:11 PM SFD from LHS: Starts at -2 because down on LHS. BMD from LHS: Starts at 0 because free end on beam. Max at 2 places..
Weight Tuesday, 26 July 2011 7:27 PM Get reactions: Find Volume: V = 0.14*0.26*12 = 0.4368 m3 M = *V = 7150*0.4368 = 3123.12 kg Reactions = 3123.12*9.81/2 = 15318.9 N = 15.3189 kn Area in SFD = BM BM = 0.5*15.3*6 = 45.9 knm Check by cutting in half and find Moment: = 3123.12 /2 = 1561.56 kg = 15.3189 kn Moment = +(15.3*3)- (15.3*6) = -45.9 knm Forget the minus sign when you cut in half because you get a different sign depending on which side you take! Scrap minus sign and use HOGG/SAG. Should be... (Sag) = +45.9kNm Bending-Moment Page 12
Distributed Loads. Bending Moment 5. Bending-Moment Page 13
Applied Moment Tuesday, 26 July 2011 8:14 PM Get reactions: (Moment eqn) M L = + (170) - (F R *10) = 0 F R *10 = 170 F R = 17 kn Fy = 0 FL = -17 kn Area in SFD = 2.5*-17 = -42.5 knm Area in SFD = 7.5*-17 = -127.5 knm Difference: +127.5--42.5 = 170 (knm) which is the same as the applied moment. Bending-Moment Page 14
Q7: Jib Tuesday, 26 July 2011 8:35 PM Take moments at A: FBy*5.5 = 1240*9.81*3.3 FBy = (1240*9.81*3.3)/5.5 = 7298.64 N Fy = 0 FAy = Fw-FBy = 1240*9.81-7298.64 = 4865.76 N Bending-Moment Page 15
FAy = Fw-FBy = 1240*9.81-7298.64 = 4865.76 N Area in SFD = 4865.76*3.3 = 16057.008 Nm Bending-Moment Page 16
Q11: Cantilever Tuesday, 26 July 2011 8:48 PM SFD: Down on LHS = negative -2.8kN Dist Load: 1.6*5 = 8 = -2.8-8 + R RY = 0 R RY = 10.8 kn M R = -(2.8*8)-(8*2.5) = -42.4 knm -2.8kN BMD: Bending-Moment Page 17
= -42.4 knm BMD: (All hogging = all negative) 3*-2.8 = -8.4kNm Area of dist SFD; -2.8*5+0.5*-8*5 =- 34 BM = -8.4-34 = -42.4 knm -8.4kNm -10.8 kn -42.4 knm Bending-Moment Page 18
Bending-Moment Page 19 Q 13: Clamp Tuesday, 26 July 2011 8:57 PM