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Home , Beam (structure), Bending moment

How to find 1. CH28 p355

Bending moment is a applied to each side of the 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 )

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.

Bending-Moment Page 3 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 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 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) M = + (5*4) -12 = -8 kNm = 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: 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 -10.8 kN Area of dist SFD; -2.8*5+0.5*-8*5 =- 34 BM = -8.4-34 = -42.4 kNm -8.4kNm

-42.4 kNm

Bending-Moment Page 18 Q 13: Clamp Tuesday, 26 July 2011 8:57 PM

Bending-Moment Page 19