Worked Example Moment Distribution

Beam Moment Distribution: Worked Example

1.22

Beam Moment Distribution Worked Examples Question 1 Use the moment distribution method to draw a dimensioned sketch of the bending moment diagram for the following beam. The section is uniform.

40 kN 36 kN/m

A

B

C

D

4m

4m

2m

2m

Solution 1 Working in units of kN and m: FEMs: BC = 

wL2 36  4 2   48 12 12

CB =

wL2 36  4 2   48 12 12

CD = 

WL 40  4   20 8 8

DC =

WL 40  4   20 8 8

Distribution factors: Joint B

Joint C

BA K

1

I I  4 4

1

1 4

DF 1 4

 14 1 2

© M DATOO

BC I I  4 4

CB 1

1 4 1 4

 14 1 2

I I  4 4

CD 1

1 4 1 4

 14 1 2

I I  4 4 1 4

1 4

 14 1 2

Beam Moment Distribution: Worked Example

1.23

Distribution table: Joint

A

End

AB

B BA

C BC

 DF

1 2

-48

48

-20

24

-14

-14

-7

12

3.5

-6

-3

1.8

1.5

-0.9

-0.5

0.8

0.3

0.2

-0.4

-0.4

29

-29

41

-41

12 3.5

CO

1.8

Bal

1.5

CO

0.8

Bal Final EM

15

DC

 1 2

24

Bal

CD

1 2

FEM CO

CB

 1 2

Bal

D

20 -7

-6 -3 -0.9 -0.5 10

The maximum free moments, which are sagging (that is tension on the bottom of the member) are: BC =

wL2 36  4 2   72 8 8

CD = 20  2  40

(maximum at mid-span; parabolic in-between) (maximum at mid-span; linear in-between)

The final bending moments are: In AB:

at A = 15 (tension on bottom)

at B = 29 (tension on top)

In BC:

at B = 29 (tension on top)

at C = 41 (tension on top)

 29  41  at mid-span = 72     37 (tension on bottom)  2 

In CD:

at C = 41 (tension on top)

at D = 10 (tension on top)

 41  10  under point load = 40     15 (tension on bottom)  2 

© M DATOO

Beam Moment Distribution: Worked Example

1.24

Drawing the bending moment diagram (kN m) on the tension side: 41 29 10

15

15 37

Shear calculations: In AB: 15

29

RA

RB 4

R A  4  15  29  0

 R A  11.0

RB  0  (11.0)  11.0 In BC: 29

41 36

RB

RC 4

RB  4  41  36  4  2  29

 RB  69.0

RC  (36  4)  69.0  75.0

© M DATOO

Beam Moment Distribution: Worked Example

1.25

In CD: 40

41

10

RC

RD 4

RC  4  10  40  2  41

 RC  27.8

RD  40  27.8  12.2

Shear table: Span

AB

End

AB

Shears

-11.0

BC

Reaction -11

CD

BA BC

CB CD

DC

11.0 69.0

75.0 27.8

12.2

80

103

12

Shear force diagram (kN):

69 28

11

12

75

© M DATOO

Beam Moment Distribution: Worked Example

1.26

Question 2 Use the moment distribution method to draw a dimensioned sketch of the bending moment diagram for the following beam:

100 kN

80 kN

40 kN

30 kN/m

A

2I

B

2.5 m 2.5 m

3I

C

6m

1.25 m

4I 2.5 m

D 1.25 m

Solution 2 Working in units of kN and m: FEMs: AB = 

WL 100  5   62.5 8 8

BA =

WL 100  5   62.5 8 8

BC = 

wL2 30  6 2   90 12 12

CB =

wL2 30  6 2   90 12 12

CD = 

Wab 2 80  1.25  3.75 2 40  3.75  1.25 2     65.6 L2 52 52

Wa 2 b 80  1.25 2  3.75 40  3.75 2  1.25    46.9 DC = L2 52 52

Distribution factors: Joint B BA K

1

2I 2I  5 5

DF

BC 1

3I I  6 2

2 5 2 5

 4 9

© M DATOO

Joint C CB 1

3I I  6 2

1 2 1 2

2 5

 5 9

CD 3 4 I 3I   4 5 5 3 5

1 2 1 2

1 2

 5 11

3 5

1 2

 6 11

3 5

Beam Moment Distribution: Worked Example

1.27

Distribution table: Joint

A

End

AB

B BA

C BC



-62.5

CB

CD

4 9

5 9

5 11

6 11

62.5

-90

90

-65.6

46.9

-23.5

-46.9 0

Pin at D FEM*

-62.5

Bal CO

62.5

-90

90

-89.1

12.2

15.3

-0.4

-0.5

-0.2

7.7

0.1

-3.5

-4.2

94

-94

6.1

Bal

DC



DF FEM

D

0.1

CO

-1.8

Bal Final EM

0.8

1.0

76

-76

-56

0

The maximum free moments, which are sagging (that is tension on the bottom of the member) are: AB = 100 1.25  125 BC = RC 

wL2 30  6 2   135 8 8

(maximum at mid-span; linear in-between) (maximum at mid-span; parabolic in-between)

40  1.25  80  3.75  70 5

M 80  RC  1.25  70  1.25  88

RD 

80  1.25  40  3.75  50 5

M 40  RD  1.25  50  1.25  63

The final bending moments are: In AB:

at A = 56 (tension on top)

at B = 76 (tension on top)

 56  76  Under point load = 125     59 (tension on bottom)  2 

In BC:

at B = 76 (tension on top)

at C = 94 (tension on top)

 76  94  at mid-span = 135     50 (tension on bottom)  2 

© M DATOO

Beam Moment Distribution: Worked Example

In CD:

1.28

at C = 94 (tension on top)

at D = 0

 94  under 80 kN point load = 88    3.75   17 (tension on bottom)  5   94  under 40 kN point load = 63    1.25   39 (tension on bottom)  5 

Drawing the bending moment diagram (kN m) on the tension side: 94 76 56

17 59

© M DATOO

50

39

Beam Moment Distribution: Worked Example

1.29

Question 3 Use the moment distribution method to draw a dimensioned sketch of the bending moment diagram. Hence, draw a dimensioned sketch of the shear force diagram.

80 kN

160 kN

12 kN/m

A

24 kN/m

B

I

2m

1.5 m

C

I

1.5 m

D

2I

1m

E

3I

3m

6m

Solution 3 Working in units of kN and m:

M BA  12  2  1  24

Cantilever action at BA

 M BC  24

FEMs: BC = 

WL 80  3   30 8 8

CB =

WL 80  3   30 8 8

CD = 

Wab 2 160  1  3 2    90 L2 42

DC =

Wa 2 b 160  12  3   30 L2 42

DE = 

wL2 24  6 2   72 12 12

ED =

wL2 24  6 2   72 12 12

Distribution factors: Joint C CB K

3 I I   4 3 4

DF

1 4 1 4

 1 3

© M DATOO

Joint D CD

1

2I I  4 2

DC 1

2I I  4 2

1 2 1 2

1 4

 2 3

DE 1

3I I  6 2

1 2 1 2

1 2

 1 2

1 2 1 2

1 2

 1 2

1 2

Beam Moment Distribution: Worked Example

1.30

Distribution table: Joint

B

End

BC

C CB

D CD

DC

E DE

 DF



1 3

2 3

1 2

1 2

-90

30

-72

72 72

FEM

-30

30

MBA=-24

+6

+3

FEM*

-24

33

-90

30

-72

19

38

21

21

10.5

19

-7.0

-9.5

-4.8

-3.5

3.2

1.8

0.9

1.6

-0.3

-0.6

-0.8

-0.8

50

-50

60

-60

Bal CO Bal

-3.5

CO Bal

1.6

CO Bal Final EM

-24

Ed

10.5 -9.5 -4.8 1.7 0.9 79

The maximum free moments, which are sagging (that is tension on the bottom of the member) are: BC = 40  1.5  60

RC 

(maximum at mid-span; linear in-between)

160  3  120 4

M 160  RC  1  120  1  120

(maximum under point load; linear in-between)

wL2 24  6 2   108 DE = 8 8

(maximum at mid-span; parabolic in-between)

The final bending moments are: In AB:

at A = 0

at B = 24 (tension on top)

In BC:

at B = 24 (tension on top)

at C = 50 (tension on top)

 24  50  Under point load = 60     23 (tension on bottom)  2 

In CD: © M DATOO

at C = 50 (tension on top)

at D = 60 (tension on top)

Beam Moment Distribution: Worked Example

1.31

(60  50)   under point load = 120   50   1  68 (tension on bottom) 4  

In DE:

at D = 60 (tension on top)

at E = 79 (tension on top)

 60  79  at mid-span = 108     39 (tension on bottom)  2 

Drawing the bending moment diagram (kN m) on the tension side:

79 60 50

24

23 39

68

Shear calculations: In AB: 0

24 12

RA

RB 2

R A  2  24  12  2  1 RB  24  0  24 In BC: © M DATOO

 RA  0

Beam Moment Distribution: Worked Example

1.32

24

50

80

RB

RC 1.5

1.5

RB  3  50  80  1.5  24

 RB  31.3

RC  80  31.3  48.7

In CD: 50

60

160

RC

RD 1

3

RC  4  60  160  3  50

 RC  117.5

RD  160  117.5  42.5 In DE: 60

79 24

RE

RD 6

RD  6  79  24  6  3  60

 RD  69.0

RE  24  6  69  75.0

© M DATOO

Beam Moment Distribution: Worked Example

1.33

Shear table: Span

AB

End

AB

Shears

0

BC

CD

DE

BA BC

CB CD

DC DE

ED

24.0 31.3

48.7 117.5

42.5 69.0

75.0

55

166

112

Reaction 0

75

Shear force diagram (kN): 117 69 31 E B

A

D

C

24 49

43 75

© M DATOO