three moment theorem2

Three moment theorem

Asst. Prof. Hang Thu Vu

[email protected]

Lecture outline

The purpose of carrying out structural analysis is to calculate the maximum values of actions (moment, shear force and axial force) for design member or check member capacity

Continuous beam with n spans which subjects to various loading configurations is one of most popular structural member for building

Continuous beam is indeterminate structure with the number of unknown reactions is more than the number of static equilibrium equations (3 equations).

Lecture outline A simple technique to solve continuous beam

problem is to use the “Theorem of Three bending moments”. First, the bending moment can be constructed straightforward by applying the theorem. Then, other unknowns can be found from the beam equations. For example, the shear force can be found by differentiating the bending moment

dx

dMV

Lecture outline

This lecture will detail the theorem and its applications Theorem of Three bending moments Applications

Example 1: 4 spans, patch load Example 2: 4 spans, 2-span loaded Example 3: 4 spans, loads on alternate spans Example 4: Fixed ends

Theorem of Three bending moments

Revision: Structural analysis

Free body diagram

Vertical reaction Ay, equivalent load P. At the cut of distance x from left end: shear force V, bending moment M

w (kN/m)

L (m)

M

V

Revision: Structural analysis The reaction at A is The equivalent point load at a distance x/2 from A is Take moment about the cut,

Force equilibrium for y direction,

Maximum shear force happens at end Maximum bending moment happens at middle

wLAy 2

1

wxP

22

1

022

1

2wxwLxM

xwxwLxM

wxwLV

wxwLV

2

1

02

1

wLV2

1max

8

2

max

wLM

Revision: Structural analysis

The area underneath the bending moment curve Integral

The distance from either support to centroid of A: x1=x2=L/2

1246

223222

333

0

23

0

2

0

wLwLwL

xwLxwdx

wxLwxMdxA

LLL

Point load In a similar manner, the bending moment M

and bending moment area A can be obtained for span subjected to point load

Point load

Distance from left end to centroid of bending moment area A is

bxaPaPL

Pbx

axL

Pbx

M

0

2

PabA

31

Lax

Three bending moment theorem The theorem applies to any two adjacent

spans in a continuous beam. For constant section, the theorem states

Where MA, MB, MC are the bending moment values at three subsequent supports A, B, C

22

22

11

11

2

2

2

2

1

1

1

1 62IL

xA

IL

xA

I

LM

I

L

I

LM

I

LM CB

A

For any load system

Solution procedure

For each patch formed by 2 adjacent spans of the continuous beam, write the theorem equation. Establish the system of n linear independent equations for n unknown bending moments at supports

Solve the system for support bending moments The BM of the continuous beam is constructed

individually for each span by summing the basic bending moment MLoad (due to applied load on the single span) and the linear function fitted through computed bending moments at 2 corresponding supports, M3BMtheorem

Applications

Example 1: UDL on 4 spans

To generate the bending moment diagrams, nominate values for UDL w and span length L. Assume w = -10 kN/m (negative sign to indicate downward force) and L = 10 m.

Example 1: UDL on 4 spans Apply the “Three bending moment theorem”

equation to spans (AB, BC), (BC, CD), (CD, DE). Substitute L=10, A1 = A2 = -833.33, MA = ME = 0 into 3 equations, we obtain

For each span , fit a linear function through the bending moment at its 2 ends.

14.107

43.71

14.107

500

500

500

410

141

014

D

C

B

D

C

B

M

M

M

M

M

M

Lx 0


Revise mathematics: linear function y(x) fitted through point 1 (x1, y1) and point 2 (x2, y2) satisfies

12

1

12

1

xx

xx

yy

yy

Example 1: UDL on 4 spans Sum the bending moment due to applied load

MLoad and the fitted function M3BMtheorem to obtain the bending moment M on each span

Plot the bending moment for the continuous beam from constructed bending moment M in the above table


UDL over 4 equal spans

-100

-50

0

50

100

150

0 5 10 15 20 25 30 35 40

Distance (m)

Be

nd

ing

mo

me

nt

M (

kN

.m)


Having L=10; spans (AB, BC): A1 = A2 = -833.33, spans (BC, CD): A1 = -833.33, A2 =0, spans (CD, DE): A1 = A2 = 0; MA = ME = 0.

Hence,


Fitting the linear function through the bending moment data at ends for each span

93.8

714.35

07.116

0

250

500

410

141

014

D

C

B

D

C

B

M

M

M

M

M

M


M = MLoad + M3BMtheorem

Example 2: UDL on 2 spansUDL on 2 adjacent spans

-100

-50

0

50

100

150

0 5 10 15 20 25 30 35 40

Distance (m)

Be

nd

ing

mo

me

nt

M (

kN

.m)

Example 3: Load on alternate spans

Having L=10; spans (AB, BC): A1 = -833.33, A2 =0; spans (BC, CD): A1 =0, A2 = -833.33; spans (CD, DE): A1 = -833.33, A2 =0; MA = ME = 0. Hence,


Fitting the linear function through the bending moment data at ends for each span

571.53

714.35

571.53

250

250

250

410

141

014

D

C

B

D

C

B

M

M

M

M

M

M

Example 3: Load on alternate spans M = MLoad + M3BMtheorem


UDL loaded on alternate spans

-120

-100

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30 35 40

Distance (m)

Be

nd

ing

mo

me

nt

M (

kN

.m)

Example 4: Continuous beam with fixed ends For a continuous beam of (n) spans, (n-1) equations

can be generated from the three moment theorem. The total of supports on the beam are (n+1).

If 2 end supports are pins, the number of unknown BMs at supports is (n-1). We can solve the system of (n-1) equations for these (n-1) BMs as detailed in previous examples

For continuous beam with one fixed end, the number of unknown BMs is (n). For continuous beam with two fixed ends, the number of unknown BMs is (n+1). We can not solve for these unknown BMs as the number of equations is smaller than the number of unknowns.

Example 4: Continuous beam with fixed ends Consider this example of 2-span continuous

beam ABC of const. cross section, built in fixed ends at A and C, supported at B

One equation for 3 unknowns MA, MB, MC

Example 4: Continuous beam with fixed ends Effect of a fixed end is similar to placing a

mirroring span at the wall end

B1 is mirror of B over A, B2 is mirror of B over C. Have MB1= MB, MB2= MB

Example 4: Continuous beam with fixed ends With , from the left hand side,

having AB

1A = -157.5, AAB = -157.5,

ABC = -125, ACB2 = -125

Consider spans (B1A, AB)

12

3wLA

)1.(5.1572

3

5.15.157

3

5.15.157633323

1

1

EqMM

MM

MMM

BA

BB

BAB

Example 4: Continuous beam with fixed ends Consider spans (AB, BC)

Consider spans (BC, CB2)

)2.(5.28217.133.4

5.3

75.1125

3

5.15.157633323

1

EqMMM

MMM

CBA

BAB

)3.(1.1072

5.3

75.1125

5.3

75.112565.35.35.325.3

2

2

EqMM

MM

MMM

CB

BB

BCB

Example 4: Continuous beam with fixed ends From Eqs. (1), (2), (3), the bending moments at A,

B, C are found

After the bending moments at all supports are known, the bending moment is recovered for each span in the usual manner. The BMD for the continuous beam can then be plotted.

84.31

47.43

01.57

C

B

A

M

M

M

Next lecture

Design of tension member Reading:

AS4100:1998, Section 7 and related parts in Section 6 and Section 9

three moment theorem2

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