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 MA = 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.
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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)
MR = 0 = (RL*10) - (9.3 * 2.2)
9.3 * 2.2 = 20.46
RL= 20.46 /10 = 2.046 kN
RL+RR = 9.3
RR = 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.
Bending-Moment Page 4
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 5
Shear Force Diagram
Step 1. Reactions
(Working in kNm)
0 = +(130*3) - (FR*8)
FR = 90/8 = 11.25 kN
0 = FL -30 + 11.25
FL = 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
Bending-Moment Page 6
Shear Force Diagram > BMD
Step 1. Reactions
(Working in kNm)
0 = +(130*3) - (FR*8)
FR = 90/8 = 11.25 kN
0 = FL -30 + 11.25
FL = 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
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..
Bending-Moment Page 11
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)
ML = + (170) - (FR*10) = 0
FR*10 = 170
FR = 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 + RRY = 0
RRY = 10.8 kN
MR = -(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
-10.8 kN
-8.4kNm
-42.4 kNm
Bending-Moment Page 18
Q 13: Clamp
Tuesday, 26 July 2011
8:57 PM
Bending-Moment Page 19