how to find bending moment bending moment 1. ch28...
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Bending Moment 1.
CH28 p355How to find Bending Moment
Calculate BM: M = Fr (Perpendicular to the force)
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
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Calculate BM: M = Fr (Perpendicular to the force)
In equilibrium, so MA = 0But 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;
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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.
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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)MR = 0 = (RL*10) - (9.3 * 2.2)9.3 * 2.2 = 20.46 RL= 20.46 /10 = 2.046 kN
RL+RR = 9.3RR = 9.3-2.046 = 7.254 kN
Take a slice thru the b force, to get BMoment. (Moment eqn for left side only)BMleft = (2.046 * 7.8) = +15.9588 kNm
We can check this for the right hand side.(Moment eqn for right side only)BMright = (7.254*2.2) = -15.9588 kNm
So at the cross section, there are two moments - equal and opposite.Hogging or sagging? Positive = sagging.
Simple example:Tuesday, 30 April 20136:45 PM
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Bending Moment 1.
CH28 p355
Shear force & Bending Moment
Positive Shear ForceUp on LHS
Shear Force is in all beams,but usually only seen as a problem in SHORT beams.Long beams fail by bending.
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Shear Force Diagram
Step 1. Reactions
(Working in kNm)0 = +(130*3) - (FR*8) FR = 90/8 = 11.25 kN
0 = FL -30 + 11.25FL = 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
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Shear Force Diagram > BMD
Step 1. Reactions
(Working in kNm)0 = +(130*3) - (FR*8) FR = 90/8 = 11.25 kN
0 = FL -30 + 11.25FL = 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
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Example 28.1 f. (P361)
Double Check
(LHS)M=1l *2 = 22 kNm(RHS)M = - (5x2) -(3x4) = -10-12 = -22 kNm + Positive BM
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Example 28.1 d
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 = SAGGINGNegative bending moment = HOGGING
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.
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M = + (5*4) -12 = 8 kNm
M = - (4*2) = -8 kNm
Example 28.1 h
The area of the SFD = height of the BMD
Positive Shear Force
Positive Bending Moment
The maximum bending moment = 12The maximum negative BM = -8
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.
Wall Reaction:M = -(4*8)+(9*6)-(5*2) = 12So wall reaction moment is -12 (CCW)
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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..
Example 28.1(j)Tuesday, 20 March 20126:11 PM
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Get reactions:Find Volume:V = 0.14*0.26*12 = 0.4368 m3
M = *V = 7150*0.4368 = 3123.12 kgReactions = 3123.12*9.81/2 = 15318.9 N = 15.3189 kNArea in SFD = BMBM = 0.5*15.3*6 = 45.9 kNm
Check by cutting in half and find Moment: = 3123.12 /2 = 1561.56 kg = 15.3189 kNMoment = +(15.3*3)- (15.3*6) = -45.9 kNmForget 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
WeightTuesday, 26 July 20117:27 PM
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Distributed Loads.
Bending Moment 5.
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Get reactions: (Moment eqn)ML = + (170) - (FR*10) = 0FR*10 = 170FR = 17 kN
Fy = 0FL = -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.
Applied MomentTuesday, 26 July 20118:14 PM
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Take moments at A:FBy*5.5 = 1240*9.81*3.3FBy = (1240*9.81*3.3)/5.5 = 7298.64 N
Fy = 0FAy = Fw-FBy = 1240*9.81-7298.64 = 4865.76 N
Q7: JibTuesday, 26 July 20118:35 PM
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FAy = Fw-FBy = 1240*9.81-7298.64 = 4865.76 N
Area in SFD = 4865.76*3.3 = 16057.008 Nm
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SFD:Down on LHS = negative-2.8kNDist Load: 1.6*5 = 8
= -2.8 -8 + RRY = 0RRY = 10.8 kNMR = -(2.8*8)-(8*2.5) = -42.4 kNm
BMD:
-2.8kN
Q11: CantileverTuesday, 26 July 20118:48 PM
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= -42.4 kNm
BMD:(All hogging = all negative)3*-2.8 = -8.4kNmArea of dist SFD;-2.8*5+0.5*-8*5 =- 34 BM = -8.4-34 = -42.4 kNm
-10.8 kN
-8.4kNm
-42.4 kNm
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Q 13: ClampTuesday, 26 July 20118:57 PM
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