1

Moment Area Theorems:

When a beam is subjected to external loading, it under goes deformation. Then the intersection angle between tangents drawn at any two points on the elastic curve is given by the area of bending moment diagram divided by its flexural rigidity.

Theorem 1:

2

Moment Area Theorems:

The vertical distance between any point on the elastic curve and intersection of a vertical line through that point and tangent drawn at some other point on the elastic curve is given by the moment of area of bending moment diagram between two points taken about first point divided by flexural rigidity.

Theorem 2:

3

Fixed end moment due to a point load at the mid span:

(1) ----4

WL - MM

0EI

L2

MML

4WL

21

BA

BA

AB

)2(WL8

3M2M

EI

2L

L4

WL21

L32

LM21

L31

LM21

AA

BA

BA1

4

Both moments are negative and hence they produce hogging bending moment.

8

WL

8

WL

4

WLM

4

WLM

8

WL M

get we(2) and (1) From

BA

B

5

Stiffness coefficients

a) When far end is simply supported

EI

MomentBB

Babout B &A between BMD of area of '

EI3

ML

EI

L 32

L M 21

2

EI3

ML LBB

2

A I

A L

EI3 M

6

b) When far end is fixed

EI

LM2

1 - LM

2

1

EI

BMD of area

BA

A

1 -------- 2

M - B

EI

LM A

2--------- 2M M

0EI

32

LM21

3L

LM21

EI

Aabout B & A between BMD of Areathe of Moment AA

BA

BA

1

7

Substituting in (1)

AB

ABB

ABA

L

2EI M

EI2

LM - M2

EI2

LMM

ABA L

4EI M 2 M )2(From

8

Fixed end moments due to yielding of support.

M M Hence

0)MM(.ie

L2

MM

0EI

B and A between BMD of area

BA

BA

BA

AB

2

1

L 6

2 -

3

2

2

1

32

1

EI

Babout B andA b/n BMD of area ofMoment

EI

MM

EI

LLML

LM

BB

AB

AB

9

26

BM hogging Hence 2

6

6

2

26

2-

L

EIMMNow

L

EIM

EI

LM

MMSinceLEI

MM

AB

A

A

BAAA

Hence sagging BM

10

Fixed end moment for various types of loading

11

12

Assumptions made in slope deflection method:

1) All joints of the frame are rigid

2) Distortions due to axial loads, shear stresses being small are neglected.

3) When beams or frames are deflected the rigid joints are considered to rotate as a whole.

13

Sign conventions:

Moments: All the clockwise moments at the ends of members are taken as positive.

Rotations: Clockwise rotations of a tangent drawn on to an elastic curve at any joint is taken as positive.

Sinking of support: When right support sinks with respect to left support, the end moments will be anticlockwise and are taken as negative.

14

Development of Slope Deflection Equation

Span AB after deformation

Effect of loading

Effect of rotation at A

15

Effect of rotation at B

Effect of yielding of support B

LL

EIF

L

EI

L

EI

L

EIFMSimilarly

LL

EIF

L

EI

L

EI

L

EIFM

BABAABBABA

BAABBAABAB

32

2624

32

22

624Hence

2

16

Slope Deflection Equations

L

32

L

EI2FM

L

32

L

EI2FM

BABABA

BAABAB

17

18

Example: Analyze the propped cantilever shown by using slope

deflection method. Then draw Bending moment and shear force

diagram.

Solution:

12

wLF,

12

wLF

2

BA

2

AB

19

Slope deflection equations

)1(L

EI2

12

wL

2L

EI2FM

B

2

BAABAB

)2(L

EI4

12

wL

2L

EI2FM

B

2

ABBABA

20

Boundary condition at BMBA=0

0L

EI4

12

wLM B

2

BA 48

wLEI

3

B

Substituting in equations (1) and (2)

8

wL

48

wL

L

2

12

wLM

232

AB

048

wL

L

4

12

wLM

32

BA

21

Free body diagram

wL8

5RA

0V wL

8

5wLRwLR AB

wL8

3

2

LwL

8

wLLR

2

A 0MB

BR

22

0wXwL8

3SX

L8

3X

2max wL

128

9M

L8

3

L8

3

2wL128

9

23

Example: Analyze two span continuous beam ABC by slope deflection method. Then draw Bending moment & Shear force diagram. Take EI constant

24

Solution:

KNM44.446

24100

L

WabF

2

2

2

2

AB

KNM89.886

24100

L

bWaF

2

2

2

2

BA

KNM67.4112

520

12

wLF

22

BC

KNM67.4112

520

12

wLF

22

CB

25

Slope deflection equations

BAABAB 2L

EI2FM

B6

EI244.44

)1(EI3

144.44 B

ABBABA 2L

EI2FM

6

2EI289.88 B

)2(EI3

289.88 B

CBBCBC 2L

EI2FM

CB25

EI267.41

)3(EI5

2EI

5

467.41 CB

BCCBCB 2L

EI2FM

BC25

EI267.41

)4(EI5

2

5

EI467.41 BC

26

Boundary conditions

MBA+MBC=0

ii. MCB=0

CBBBCBA EI5

2EI

5

467.41EI

3

289.88MM

)5(0EI5

2EI

15

2222.47 CB

)6(0EI5

4EI

5

267.41M CBCB

83.20EI B 67.41EI C Solving

i. -MBA-MBC=0

Now

27

KNM38.5183.203

1 44.44 – MAB

KNM00.7583.203

2 88.89 MBA

KNM00.7567.415

283.20

5

441.67 – MBC

067.415

483.20

5

241.67 MCB

28

Free body diagram

Span BC:

Span AB:

75+ 2

5×5×20 = 5×R 0 = M BC

KN 65 = R B

100KN = 5×20 = R+R 0=V CBKN 35 = 65-100 = RC

51.38-75+4×100 = ×6R 0 = M BAKN 70.60 = RB

100KN = +RR0 =V BAKN 29.40=70.60-100 = R A

29

BM and SF diagram

30

Example: Analyze continuous beam ABCD by slope deflection method and then draw bending moment diagram. Take EI constant.

Solution:

M KN 44.44 - 6

24100

L

Wab F

2

2

2

2

AB

KNM 88.88 6

24100

L

bWa F

2

2

2

2

BA

31

KNM 41.67- 12

520

12

wLF

22

BC

KNM 41.67 12

520

12

wLF

22

CB

M KN 30 - 5.120FCD

32

Slope deflection equations:

1--------- EI3

144.442

L

EI2FM BBAABAB

2--------- EI3

289.882

L

EI2FM BABBABA

3-------- EI5

2EI

5

467.412

L

EI2FM CBCBBCBC

4-------- EI5

2EI

5

467.412

L

EI2FM BCBCCBCB

KNM 30MCD

33

Boundary conditions 0MM BCBA

0MM CDCB

CBBBCBA EI5

2EI

5

467.41EI

3

289.88MM,Now

5-------- 0EI5

2EI

15

2222.47 CB

30EI5

2EI

5

467.41MM,And BCCDCB

6EI5

4EI

5

267.11 CB

Solving

67.32EI B 75.1EI C

34

Substituting

KNM 00.6167.322

144.44MAB

KNM 11.6767.323

289.88MBA

KNM 11.6775.15

267.32

5

467.41MBC

KNM 00.3067.325

275.1

5

467.41MCB

KNM 30MCD

35

36

Example: Analyse the continuous beam ABCD shown in figure by slope deflection method. The support B sinks by 15mm.Take 4625 m10120Iandm/KN10200E

Solution:

KNM44.44L

WabF

2

2

AB

KNM89.88L

bWaF

2

2

BA

KNM67.418

wLF

2

BC

KNM67.418

wLF

2

CB

KNM305.120FCD

37

FEM due to yielding of support B

For span AB:

2baab L

EI6mm KNM6

1000

151012010

6

2006 652

2cbbc L

EI6mm KNM64.8

1000

151012010

5

2006 65

2

For span BC:

38

Slope deflection equation

1---------- EI3

144.50

6EI3

144.44-

L

EI62

L

EIFM

B

B2BAABAB

2------------ EI3

289.82

6EI3

288.89

L

EI6)2(

L

EI2FM

B

B2ABBABA

3--------- EI 5

2EI

5

403.33

64.82EI5

241.67-

L

EI6)2(

L

EI2FM

CB

CB2CBBCBC

39

4--------- EI 5

2EI

5

431.50

64.82EI5

241.67

L

EI6)2(

L

EI2FM

BC

BC2BCCBCB

5--------- KNM 30MCD

Boundary conditions

0MM

0MM

CDCB

BCBA

0EI5

4EI

5

231.20MM

0EI5

2EI

15

2286.49MM

CBCDCB

CBBCBA

Now

Solving 35.31EI B 71.9EI C

40

Final moments

KNM 89.6035.313

144.50MAB

KNM 99.6135.313

289.82MBA

KNM 99.6171.95

235.31

5

403.33MBC

KNM 00.3035.315

271.9

5

431.50MCB

KNM 30MCD

41