UNIT-IV

SLOPE DEFLECTION METHOD

Continuous beams and rigid frames (with and without sway) – Symmetry and

antisymmetry – Simplification for hinged end – Support displacements

Introduction:

This method was first proposed by Prof. George A. Maney in 1915.

It is ideally suited to the analysis of continuous beams and rigid jointed frames.

Basic unknowns like slopes and deflections of joints are found out.

Moments at the ends of a member is first written down in terms of unknown slopes and

deflections of end joints.

Considering the joint equilibrium conditions, a set of equations are formed and solutions

of these simultaneous equations gives unknown slopes and deflections.

Then end moments of individual members are determined.

It involves solutions of simultaneous equations; a problem with more than three

unknowns is considered a difficult problem for hand calculations. Hence this method was

sidelined by moment distribution method with the help of computers; solutions for any

number of simultaneous equations can be obtained early.

The development of this method in the matrix form is “Stiffness Matrix Method” (it is

commonly used for the analysis of large structures with the help of computers.

Assumptions made in slope-deflection method

All joints are rigid.

The rotations of joints are treated as unknowns.

Between each pair of the supports the beam section is constant.

The joint in structure may rotate or deflect as a whole, but the angles between the

members meeting at that joint remain the same.

Distortions due to axial deformations are neglected.

Shear deformations are neglected.

Sign Conventions:

Moments:

Clockwise moments = (+)ive

Anti-clockwise moments = (-)ive

Rotations:

Clockwise rotations = (+)ive

Anti-clockwise rotations = (-)ive

Settlements:

Right side support is below left side support = (+)ive

Left side support is below right side support = (-)ive

Applications of Slope Deflection Equations:

Rigid jointed structures can be analyzed.

Continuous Beams

Frames without side sway (Non-Sway)

Frames with side sway (Sway)

The beam shown in Fig. is to be analyzed by slope-deflection method. What are the unknowns

and, to determine them, what are the conditions used?

Unknowns: ƟA, ƟB, ƟC

Equilibrium equations used: (i) MAB = 0 (ii) MBA + MBC = 0 (iii) MCB = 0

Write down the slope deflection equation for a fixed end support.

Write down the equilibrium equations for the frame shown in Fig.

Limitations of slope deflection method

It is not easy to account for varying member sections

It becomes very cumbersome when the unknown displacements are large in number.

Why slope-deflection method is called a ‘displacement method’?

In slope-deflection method, displacements (like slopes and displacements) are treated as

unknowns and hence the method is a „displacement method‟.

Degrees of freedom

In a structure, the numbers of independent joint displacements that the structure can

undergoes are known as degrees of freedom.

Write the fixed end moments for a beam carrying a central clockwise moment.

Problems:

1. Analyse the continuous beam given in figure by slope deflection method and draw the

B.M.D&S.F.D.

Step 1: Fixed end moments

MF

AB = -WL2/12 = -10×4

2/12 = -13.33 KNM

MF

BA = WL2/12 = 10×4

2/12 = -13.33 KNM

MF

BC = - Wab2/L

2 = -6×2×3

2/5

2 = -4.32 KNM

MF

CB = Wa2b/L

2 = 6×2

2×3/5

2 = 2.88 KNM

Step 2: Slope deflection equation

MAB = MF

AB+2EI/L (2θA+θB)

MAB = -13.33+EIθA + 0.5EIθB --------1

MBA = MF

BA+2EI/L (2θB+θA)

MBA = 13.33+0.5EIθA + EI θB --------2

MBC = MF

BC+2EI/L (2θB+θC)

MBC = -4.32+0.8EI θB --------3

MCB = 2.88+0.4EI θB --------4

Apply equilibrium conditions

MAB = 0

EI θA + 0.5EI θB = 13.33 --------5

MBA + MBC = 0

13.33 + 0.5EI θA +EI θB - 4.32 + 0.8EI θB = 0

0.5EI θA +1.8EI θB = -9.01 -------- 6

Solve eqn 5 & 6, we get

EIθA = 18.39

EIθB = -10.11

This values sub in eqn 1 to 4

MAB = 0 KNM

MBA = 12.67 KNM

MBC = -12.67 KNM

MCB = -1.16 KNM

Step 3: Find the Reactions

Span AB

RA = 16.83 KN

RB1 = 23.17 KN

RB2 = 6.312 KN

RC = -0.312 KN

𝑤 2. Analyze the frame given in figure by slope deflection method and draw the B.M.D & S.F.D.

Step 1: fixed end moments

MF

AB = -WL/8 = - 10×4/8 = -5 KNM

MF

BA = WL/8 = 10×4/8 = 5 KNM

MF

BC = -WL2/12 = - 10×2

2/12 = - 3.33 KNM

MF

CB = WL2/12 = 10×2

2/12 = 3.33 KNM

MF

CD = -WL/8 = - 10×4/8 = -5 KNM

MF

DC = WL/8 = 10×4/8 = 5 KNM

Step 2: Slope deflection equation

MAB=MFAB+2EI/L(2θA+θB)

MAB=-5 +0.5EIθB--------1

MBA = MF

BA+2EI/L (2θB+θA)

MBA = 5+EIθB --------2

MBC = MF

BC+2EI/L (2θB+θC)

MBC = -3.33+2EIθB + EIθC --------3

MCB = MF

CB+2EI/L (2θC+θB)

MCB = 3.33+2EIθC +EIθB --------4

MCD = MF

CD+2EI/L (2θC+θD)

MCD = -5+EIθC ---------5

MDC = MF

DC+2EI/L (2θD+θC)

MDC = 5+0.5EIθC ----------6

Apply equilibrium conditions

MBA+MBC = 0

5+EIθB-3.33+2EIθB +EIθC = 0 ----------7

MCB+MCD = 0

3.33+2EIθC +EIθB-5+EIθC = 0 ---------8

Solve eqn 7 & 8 we get

EIθB = -0.835

EIθC = 0.835

Sub this values eqn 1 to 6

MAB = -5.42 KNM

MBA = 4.17 KNM

MBC = -4.17 KNM

MCB = 4.17 KNM

MCD = -4.17 KNM

MDC = 5.42 KNM

Step 3: find reactions

Span AB:

RA = 5.31 KN

RB1 = 4.69 KN

Span BC:

RB2 = 10 KN

RC1 = 10 KN

Span CD:

RC2 = 4.69 KN

RD = 5.31 KN

3. Draw the SFD&BMD for th continuous beam shown in fig. Take E=2×105

N/mm2,I=3×10

6 mm

4 .The support B sinks by 30 mm. Using slope deflection method.

Step 1: fixed end moments

MF

AB = -WL2/12 = - 10×4

2/12 = -13.33 KNM

MF

BA = WL2/12 = 10×4

2/12 = -13.33 KNM

MF

BC = - Wab2/L

2 = -50×2×3

2/5

2 = -36 KNM

MF

CB = Wa2b/L

2 = 50×2

2×3/5

2 = 24 KNM

Step 2: Slope deflection equation

MAB = MF

AB + 2EI/L (2θA+θB) – 6EI∆/l2

MAB = EIθB - 20 -------1

MBA = MF

BA + 2EI/L (2θB+θA) – 6EI∆/l2

MBA = EIθB + 6.58 --------2

MBC = MF

BC + 2EI/L (2θB+θC) + 6EI∆/l2

MBC = 0.8EIθB -31.68 --------3

MCB = MF

CB + 2EI/L (2θC+θB) + 6EI∆/l2

MCB = 28.32 + 0.4EIθB --------4

Applying equilibrium conditions

MBA + MBC = 0

EIθB + 6.58 + 0.8EIθB - 31.68 = 0

EIθB = 13.94

This values sub in eqn 1 to 4

MAB = -13.03 KNM

MBA = 20.52 KNM

MBC = -20.52 KNM

MCB = 33.89KNM

Step 3: Find the reactions

Span AB

RA = 18.13 KN

RB1 = 21.87 KN

Span BC

RB2 = 27.33 KN

RC = 22.67 KN

4) Analyze continuous beam ABCD by slope deflection method and then draw bending

moment and SF diagram. Take EI constant.

Solution:

FEM 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

KNM 41.67- 12

520

12

wLF

22

BC

KNM 41.67 12

520

12

wLF

22

CB

M KN 30 - 5.120FCD

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

:

5-------- 0EI5

2EI

15

2222.47

EI5

2EI

5

467.41EI

3

289.88MM,Now

CB

CBBBCBA

6EI5

4EI

5

267.11

30EI5

2EI

5

467.41MM,And

CB

BCCDCB

clockwise B @ Rotation 75.1EI

iseanticlockw B @ Rotation 67.32EI

C

B

KNM 30M

KNM 00.3067.325

275.1

5

467.41M

KNM 11.6775.15

267.32

5

467.41M

KNM 11.6767.323

289.88M

KNM 00.6167.322

144.44M

CD

CB

BC

BA

AB

Reactions: Consider free body diagram of beam AB, BC and CD as shown

ABSpan

KN 31.32R100R

KN 69.67R

6111.6741006R

BA

B

B

BC Span

KN 42.57R520R

KN 58.42R

11.673052

5205R

BB

C

C

Maximum Bending Moments:

KNM 26.684

6

6111.676133.133Max

SpanBC: where SF=0, consider SF equation with C as reference

m 13.220

58.42x

0x2058.42SX

M KN 26.15302

13.22013.258.42M

2

max

5) Analyse the portal frame shown in figure and also drawn bending moment and shear force

diagram

Solution:

FEM

KNM 106.67 - 6

2480

6

4280

L

cdW

L

abWF

2

2

2

2

2

2

2

2

2

1BC

KNM 67.106L

dcW

L

bWaF

2

2

2

2

2

CB

Slope deflection equations:

1-------- EI2

10

4

EI202

L

EI2FM BBBAABAB

2------- EI024

EI202

L

EI2FM BBABBABA

3------ EI3

2EI

3

467.106)2(

6

I2E267.106

2L

EI2FM

CBCB

CBBCBC

4------ EI3

2EI

3

467.106)2(

6

I2E267.106

2L

EI2FM

BCBC

BCCBCB

5------- EI)02(4

EI20

2L

EI2FM

CC

DCCDCD

6------- EI2

1)0(

4

EI20

2L

EI2FM

CC

CDDCDC

Now (7)------- 0EI3

2EI

3

767.106MM CBBCBA

Clockwise 6445

3 03.960 EI

0EI3

45960.03-

subtracts

0EI3

14EI

3

434.213

0EI3

14EI

3

4969.746

B

B

CB

CB

Using equation (7)

ckwise Anticlo64643

767.106

2

3-

EI3

767.106

2

3EI BC

are moments Final

KNM -3264 2

1 M

KNM 64 M

KNM 64 643

2)64 (

3

467.106M

KNM 64 643

264

3

467.106M

KNM 64M

KNM 322

64M

DC

CD

CB

BC

BA

AB

Consider free body diagrams of beam and columns as shown

By symmetrical we can write

KNM 80R R

KNM 60R R

CD

BA

KN 24H

32644 H

0M

A

A

B

Apply

KN 24H

32644 H

0M

D

A

C

6 )Analyse the portal frame and then draw the bending moment diagram

Solution:

Assume sway to right.

Here 0,0,0,0 DBDA

FEMS:

KNM 75.938

3580

L

bWaF

KNM 25.568

3580

L

WabF

2

2

2

2

CB

2

2

2

2

BC

Slope deflection equations

2-------- EI8

3EI

4

302

4

2EI0

L

32

L

EI2FM

1-------- EI8

3EI

2

1

4

30

4

2EI0

L

32

L

EI2FM

BB

ABBABA

BB

BAABAB

6--------- EI8

3EI

2

1

4

30

4

2EI0

L

32

L

EI2FM

5--------- EI8

3EI

4

302

4

2EI0

L

32

L

EI2FM

4--------- EI4

1EI

2

175.932

8

2EI75.93

2L

EI2FM

3--------- EI4

1EI

2

125.562

8

2EI25.56

2L

EI2FM

CC

CDDCDC

CC

DCCDCD

BCBC

BCCBCB

CBCB

CBBCBC

0MMMM

04

MM

4

MM,e.i

condition Shear - -- 0PHH

0MM

conditionsintJo0MM

DCCDBAAB

DCCDBAAB

HDA

CDCB

BCBA

70EI8

3EI

4

1EI

2

325.56

0EI4

1EI

2

125.56EI

8

3EIMM,Now

CB

CBBBCBA

80EI8

3EI

2

3EI

4

175.93

0EI8

3EIEI

4

1EI

2

175.93MM,And

CB

CBCCDCB

90EI2

3EI

2

3EI

2

3

EI`8

3EI

2

1

EI8

3EIEI

8

3EIEI

8

3EI

2

1MMMM,And

CB

C

CBBDCCDBAAB

(8) & (7) in Substitute

EIEIEI (9) From CB

(7) Eqn

10-------0EI8

1EI

8

925.56

0EIEI8

3EI

4

1EI

2

325.56

CB

CBCB

)8(Eqn

11----------0EI8

9EI

8

175.93

0EIEI8

3EI

2

3EI

4

175.93

CB

CBCB

Solving equations (10) & (11) we get 25.41EI B

By Equation (10)

5.3775.7825.41EIEIEI

75.7825.418

925.568

EI8

925.568EI

CB

BC

Hence

5.37EI,75.78EI,25.41EI CB

KNM31.255.378

375.78

2

1M

KNM69.645.378

375.78M

KNM69.6475.414

175.78

2

175.93M

KNM31.5575.784

125.41

2

125.56M

KNM31.555.378

325.41M

KNM69.345.378

325.41

2

1M

DC

CD

CB

BC

BA

AB

Reactions

SpanBC:

17.51R80R

KN83.288

38069.6431.55R

BC

B

Column CD:

5.224

31.2569.64HD

7) Frame ABCD is subjected to a horizontal force of 20 KN at joint C as shown in figure.

Analyse and draw bending moment diagram.

Solution:

Frame is Symmetrical and unsymmetrical loaded hence there is a sway. Assume sway

towards right

FEM

0FFFFFF DCCDCBBCBAAB

Slope deflection equations are

2EI3

2EI

3

4

3

32

3

EI2

L

32

L

EI2FM

1--------- EI3

2EI

3

2

3

3

3

EI2

L

32

L

EI2FM

B

B

ABBABA

B

B

BAABAB

5EI3

2EI

3

4

3

32

3

2EI

L

32

L

EI2FM

4-------- EI5.0EI

24

EI2

2L

EI2FM

3-------- EI5.0EI

24

EI2

2L

EI2FM

C

C

DCCDCD

BC

BC

BCCBCB

CB

CB

CBBCBC

6--------- EI3

2EI

3

2

3

3

3

EI2

L

32

L

EI2FM

C

c

CDDCDC

060MMMM

0203

MM

3

MM,e.i

020HH.III

0MM.II

0MM.I

DCCDBAAB

DCCDBAAB

DA

CDCB

BCBA

70EI3

2EI5.0EI

3

7

EI5.0EIEI3

2EI

3

4MMNow

CB

CBBBCBA

80EI3

2EI

3

7EI5.0

EI3

2EI

3

4EI5.0EIMMand

CB

CBCCDCB

9060EI3

8EI2EI2

60EI3

2EI

3

2EI

3

2

EI3

4EI

3

2EI

3

4EI

3

2EI

3

260MMMMand

CB

C

CBBDCCDBAAB

77.34EI

,18.8EI

,18.8EI

C

B

KNM73.1777.343

218.8

3

2M

KNM27.1277.343

218.8

3

4M

KNM27.1218.818.85.0M

KNM27.1218.85.018.80M

KNM27.1277.343

218.8

3

4M

KNM73.1777.343

218.8

3

2M

DC

CD

CB

BC

BA

AB

Reactions: Consider the free body diagram of the members

Member AB:

KN103

27.1273.17HA

MemberBC:

downwardsR of direction indicates sign ve-KN135.6RR

KN135.64

27.1227.12R

BCB

C

Member CD:

righttoleftisH of direction the indicates sign ve- KN103

27.1273.17H DD