1

Moment Area Theorems:

When a beam is subjected toexternal loading, it under goesdeformation. Then the intersectionangle between tangents drawn atany two points on the elastic curveis given by the area of bendingmoment diagram divided by itsflexural rigidity.

Theorem 1:

2

Moment Area Theorems:

The vertical distance between anypoint on the elastic curve andintersection of a vertical linethrough that point and tangentdrawn at some other point on theelastic curve is given by themoment of area of bendingmoment diagram between twopoints taken about first pointdivided 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

>=+∴

=

++

××=θ

)2(WL83

M2M

EI

2L

L4

WL21

L32

LM21

L31

LM21

AA

BA

BA1

>−−−−−=+

××+

××+

××=

4

Both moments are negative andhence they produce hoggingbending moment.

8WL

8WL

4WL

M 4

WLM

8WL

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 '=

EI3ML

EI

L 32

L M 21

2

=

×

××=

EI3ML

LBB 2

A I =θ=∴

A LEI3

M θ=∴

6

b) When far end is fixed

EI

LM21

- LM21

EI

BMD of area

BA

A

==θ

( ) ( )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

θ=∴

θ=+

θ=−

ABA L

4EI M 2 M )2(From θ==

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

−=

=+−

+−=

==θ

( )

+=

×+××−=

==−

2

1

L 6

2 -

32

21

321

EI

Babout B andA b/n BMD of area ofMoment

EI

MM

EI

LLML

LM

BB

AB

AB

δ

9

( )

δ

δ

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 smallare neglected.

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

13

Sign conventions:

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

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

Sinking of support: When right support sinks with respect toleft support, the end moments will be anticlockwise and aretaken 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

δθθδθθ

δθθδθθ

32

2624

32

22

624Hence

2

16

Slope Deflection Equations

δ−θ+θ+=

δ−θ+θ+=

L3

2LEI2

FM

L3

2LEI2

FM

BABABA

BAABAB

17

18

Example: Analyze the propped cantilever shown by using slope

deflection method. Then draw Bending moment and shear force

diagram.

Solution:

12wL

F,12wL

F2

BA

2

AB +=−=

19

Slope deflection equations

( )

)1(LEI2

12wL

2LEI2

FM

B

2

BAABAB

→θ+−=

θ+θ+=

( )

)2(LEI4

12wL

2LEI2

FM

B

2

ABBABA

→θ+=

θ+θ+=

20

Boundary condition at BMBA=0

0LEI4

12wL

M B

2

BA =θ+= 48wL

EI3

B −=θ∴

Substituting in equations (1) and (2)

8wL

48wL

L2

12wL

M232

AB −=

−+−=

048wL

L4

12wL

M32

BA =

−++=

21

Free body diagram

wL85

RA =0V =∑

wL85

wLRwLR AB −=−=

wL83=

2L

wL8

wLLR

2

A ×+=×0MB =∑

BR

22

0wXwL83

SX =+−=

L83

X =∴

2max wL

1289

M =∴L

83

L83

2wL128

9

23

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

24

Solution:

KNM44.446

24100L

WabF

2

2

2

2

AB −=××−=−=

KNM89.886

24100L

bWaF

2

2

2

2

BA +=××+=+=

KNM67.4112

52012wL

F22

BC −=×−=−=

KNM67.4112

52012wL

F22

CB =×=+=

25

Slope deflection equations

( )BAABAB 2LEI2

FM θ+θ+=

B6EI2

44.44 θ+−=

)1(EI31

44.44 B →−−θ+−=

( )ABBABA 2LEI2

FM θ+θ+=

62EI2

89.88 Bθ×++=

)2(EI32

89.88 B →−−−−θ+=

( )CBBCBC 2LEI2

FM θ+θ+=

( )CB25EI2

67.41 θ+θ+−=

)3(EI52

EI54

67.41 CB →θ+θ+−=

( )BCCBCB 2LEI2

FM θ+θ+=

( )BC25EI2

67.41 θ+θ++=

)4(EI52

5EI4

67.41 BC →θ+θ+=

26

Boundary conditions

MBA+MBC=0

ii. MCB=0

CBBBCBA EI52

EI54

67.41EI32

89.88MM θ+θ+−θ+=+

)5(0EI52

EI1522

22.47 CB >−−−−=θ+θ+=

)6(0EI54

EI52

67.41M CBCB >−−−−−=θ+θ+=

83.20EI B −=θ 67.41EI C −=θSolving

i. -MBA-MBC=0

Now

27

( ) KNM38.5183.2031

44.44 – MAB −=−+=

( ) KNM00.7583.2032

88.89 MBA +=−++=

( ) ( ) KNM00.7567.4152

83.2054

41.67 – MBC −=−+−+=

( ) ( ) 067.4154

83.2052

41.67 MCB =−+−++=

28

Free body diagram

Span BC:

Span AB:

75+ 2

5×5×20 = 5×R 0 = M BCΣ

KN 65 = R B∴100KN = 5×20 = R+R 0=V CBΣ

KN 35 = 65-100 = RC

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

100KN = +RR0 =V BAΣKN 29.40=70.60-100 = R A∴

29

BM and SF diagram

30

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

Solution:

M KN 44.44 - 6

24100

LWab

F2

2

2

2

AB =××−=−=

KNM 88.88 6

24100

LbWa

F2

2

2

2

BA +=××+=+=

31

KNM 41.67- 12

52012wL

F22

BC =×−=−=

KNM 41.67 12

52012wL

F22

CB +=×+=+=

M KN 30 - 5.120FCD =×−=

32

Slope deflection equations:

( ) ( )1--------- EI31

44.442LEI2

FM BBAABAB >θ+−=θ+θ+=

( ) ( )2--------- EI32

89.882LEI2

FM BABBABA >θ++=θ+θ+=

( ) ( )3-------- EI52

EI54

67.412LEI2

FM CBCBBCBC >θ+θ+−=θ+θ+=

( ) ( )4-------- EI52

EI54

67.412LEI2

FM BCBCCBCB >θ+θ++=θ+θ+=

KNM 30MCD −=

33

Boundary conditions 0MM BCBA =+

0MM CDCB =+

CBBBCBA EI52

EI54

67.41EI32

89.88MM,Now θ+θ+−θ+=+

( )5-------- 0EI52

EI1522

22.47 CB >=θ+θ+=

30EI52

EI54

67.41MM,And BCCDCB −θ+θ++=+

( )6EI54

EI52

67.11 CB >−−−−−−θ+θ+=Solving

67.32EI B −=θ 75.1EI C +=θ

34

Substituting( ) KNM 00.6167.32

21

44.44MAB −=−+−=

( ) KNM 11.6767.3232

89.88MBA +=−++=

( ) ( ) KNM 11.6775.152

67.3254

67.41MBC −=+=−+−=

( ) ( ) KNM 00.3067.3252

75.154

67.41MCB +=−+++=

KNM 30MCD −=

35

36

Example: Analyse the continuous beam ABCD shown in figureby 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:

δ−==2baab LEI6

mm KNM61000

151012010

62006 652

−=×××××−= −

δ+==2cbbc LEI6

mm KNM64.81000

151012010

52006 65

2+=×××××+= −

For span BC:

38

Slope deflection equation

( )

( )1---------- EI31

44.50

6EI31

44.44- LEI6

2LEI

FM

B

B2BAABAB

>θ+−=

−θ+=δ−θ+θ+=

( )2------------ EI32

89.82

6EI32

88.89 LEI6

)2(LEI2

FM

B

B2ABBABA

>θ++=

−θ++=δ−θ+θ+=

( )

( )3--------- EI 52

EI54

03.33

64.82EI52

41.67- LEI6

)2(LEI2

FM

CB

CB2CBBCBC

>θ+θ+−=

+θ+θ+=δ+θ+θ+=

39

( )

( )4--------- EI 52

EI54

31.50

64.82EI52

41.67 LEI6

)2(LEI2

FM

BC

BC2BCCBCB

>θ+θ++=

+θ+θ++=δ+θ+θ+=

( )5--------- KNM 30MCD >−=

Boundary conditions

0MM

0MM

CDCB

BCBA

=+=+

0EI54

EI52

31.20MM

0EI52

EI1522

86.49MM

CBCDCB

CBBCBA

=θ+θ+=+

=θ+θ+=+Now

Solving 35.31EI B −=θ 71.9EI C −=θ

40

Final moments

( ) KNM 89.6035.3131

44.50MAB −=−+−=

( ) KNM 99.6135.3132

89.82MBA +=−++=

( ) ( ) KNM 99.6171.952

35.3154

03.33MBC −=−+−+−=

( ) ( ) KNM 00.3035.3152

71.954

31.50MCB +=−+−++=

KNM 30MCD −=

41