The Three Moment Equation

1) Review of the “Moments by Parts” approach:

Due to

connecting

moment --- _

M is

Known

M

8

2L +

B.M.D

_ Due to load

+

+

+

8

2L

8

2L

2) Concept: M0 M1 M2

L0,(EI)0 L1,(EI)1

Three consecutive moments

)

11

12

00

10(6

11

12

)

11

1

00

0(1

2

00

00 IE

r

IE

r

IE

LM

IE

L

IE

LM

IE

LM

Due to Loads Due to connecting moments

M0 M1 M1 M2

0 1 1 2 0 1 1 2

10 12 10 12

Conjugate beam method

M0 M2

10r 12r

Bending Moment are plotted from Tension Side M1

10r 12r

00

10

10IE

r

11

1212

IE

r OR

00

10

10IE

r

11

1212

LE

r

M0 M2

2

00 LM

2

12 LM

( )() 12121010 + +

)()( 12101210

2

01LM M1

2

11LM

Negative sign due

to opposite slopes

;0

?

0

10

M

r

010

00100 )3

2

2()

32( Lr

LLMLLM o 00

01

00

0010

010010

3636 IE

LM

IE

LMLMLMr

;0

?

2

12

M

r

112111112 )

3

2

2()

32( Lr

LLMLLM

11

11

11

1212

111212

3636 IE

LM

IE

LMLMLMr

)(6

)33

(6 11

12

00

10

11

12

11

1

00

0

1

00

00

1210IE

r

IE

r

IE

LM

IE

L

IE

LM

IE

LM

Multiply both sides by 6 )(6)(211

12

00

10

11

12

11

1

00

0

1

00

00

IE

r

IE

r

IE

LM

IE

L

IE

LM

IE

LM

-Note that a more compact form of the three moment equation can be written by using the

following notations:

00

0

0IE

LL ;

11

11

IE

LL ;

00

10

10IE

rr ;

11

1212

IE

rr

Thus, )1210

(612

)10

(1

200

rrLMLLMLM

4) Numerical example: 8t

EI = constant 3t/m M1

Plot B.M.D and S.F.D. 0 2

4m 2m 2m

Solution:

L0=4, L1=4, r10=8, r12=8 12t 8t

Apply three moment equation:

)88

(6)44

(2 1EIEIEIEI

M

..69616 11 mtMM

8 8 8

68

2

L

6t.m

B.M.D. 6t.m

12t 8t.m

8t

6t.m 6t.m

Due to Loads: 6 6 4 4

Due to connecting moments: 6/4 6/4 6/4 6/4

4.5t 13t 2.5t

Total:

5.5

4.5 +

S.F.D. +

- 2.5

7.5