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Unit - 3


Slope-Deflection Method


1. Slope Deflection Method

  • Slope=
  • Deflection=
  • Sign Convention
  1. Simply supported Beam

 

At support form

At the simply supported end

is always zero.

is always zero.

2.     Cantilever beam

 

1. =0 at the fixed end at force end

At Fixed end

2. Slope deflection equation: 

When a member of a structure is loaded, it deforms from its original position and internal forces are developed. The end moments can be expressed in terms of

(i)                Fixed end moments (FEM) of the member due to transverse loading on the member

(ii)              The slopes or rotations at the end

(iii)           The end deflections or joint translations. These expressions are called as Slope-Deflection equations.

There are two slope deflection equations for each member.

These slope deflection equations are derived from superposition of the end moments caused by various actions and displacement.

Similarly

Where,

= be the slope or rotation at joint A

= be the slope or rotation at joint B.

= be the end deflection or translation of joint A.

= be the end deflection or translation of joint B.

B/A = Relative translation of joint B with respect to joint A

AB Angle between the unreformed member and the line joining the deflected joints.

Here > Therefore B/A is positive

MAB = Moment at A of member AB

МВА = Moment at B of member BA

 

3. Fixed End Moments: Standard Cases Fixed End Moments

Standard Cases

1.

 

2.

 

3.

 

4.

5.

 

4. Derivation of slope Deflection Method

  • In beam, a continuous beam ABCD, consider AB part and take fixed end moments and rotation at support as given below 

 

 

As per the diagram

Slop deflection equations are

.

5.  Types of numerical for slope deflection method

1. The beam is without sink (is zero.)

2. Beam with sink (is given)

3. Analyze of the frame without sink

4. Analyze the frame with the sink.

6. Steps for analysis of the slope deflection method

1. Find unknown slope.

2. Find fixed end moment

3. Apply the slope deflection equation.

4. Use the joint equilibrium equation at the joint.

5. Find final moments.

6. Find relation by equation &Draw SFI

7. Draw BMD

 


Problems:

  1. Analyze the continuous beam shown in figure Draw also B.M. And S.F. Diagram

 

 

Step-1 Find fixed end moments

Step-II Slope deflection method

Step-III Apply equilibrium condition at joint

Step-IV   Find final moments.

Step-V Find reaction

S.F. At A Take a moment at A is

SFD & BMD

 

 

2.     Determine support moments & draw BMD for the beam shown by using S.D.

 

Step-I) Find Dki

Step II) Find fixed end moments

For BC Beam

Step-III) Apply the SD equation

AB=>

BC=>

 

Step- IV) Apply equilibrium condition joint B

-------   due to Hinge support

Put in equation (1) (2) (3) & (4)

Draw BMD Diagram

 

3.     Analyze the continuous beam ABCD by S.D. Method draw BMD

Modify Diagram

Step- I) Fixed end moment

 

Apply S.D. Equation

Joint equilibrium equation

Joint at B

BMD and SFD

 

4.     Analyze the continuous beam as shown in the figure by S.D. Method Draw BMD

Step 1) Find fixed end moment

 

KNM

S.D. Equation

Span AB

 

 

Joint equilibrium equation

Joint B

KNM

 

SFD

BMD

 

5.     Analyze the continuous beam by S.D. Equation

Find fixed end moments:

 

 

Apply S.D. Equation

AB=>

2)

BC=>

Joint equilibrium condition

BMD

 

 

6.     Analyze the beam by slope deflection method. Draw SFD and BMD

Solution:

Step 1) Dki = 2 

Step 2) Find Fixed End Moments

For AB

For BC

 

 

For CD

 

Apply S.D. Equation

Joint equilibrium condition

7.     Determine the final moments by slope deflection method take EI is constant.

Step 1) Dki = 2

Step 2) Find fixed end moments

 

 

Step 3) Apply S.D. Equation

For beam BC

--------(3)

 

 

-------(4)

OR

For member, BC apply modified equation

 

Step4) Joint equilibrium equation:


 

9.5 + 1.46 EI + 0.33EI-------(A)

 

0-------(B)

Equating equation, A & B get

 

 

OR

Apply this condition to the modified equation

Step 5) BMD

Problems based on sink

Sink means deflection which occurs due to load which is acting on the beam

Note: always take EI in terms of KNM2

  1. Analyze continuous beam ABCD having

Point B is sunk by 2.5mm also draw BMD

Step 1) Dki = 2  

Step 2) Fixed end moments

STEP 3) Apply S.D. Equation

AB=> ,l = 3m

 

BC=>

 

Step 4) Apply equilibrium condition

Joint B

Joint C

KNM

KNM

KNM

KNM

KNM

BMD

 

Key takeaway:

Steps:

1. Find unknown slope.

2. Find fixed end moment

3. Apply the slope deflection equation.

4. Use the joint equilibrium equation at the joint.

5. Find final moments.

6. Find relation by equation &Draw SFI

7. Draw BMD

 


Problems:

  1. Analyze the portal frame using slope deflection method.
C:\Users\DELL\Desktop\1.jpeg

 

Solution:

It is symmetrical non-sway frame

Step 1) To find degree of kinematic indeterminacy

Dki= 2

Unknown displacement

Known displacement:

Step 2) To find fixed end moments

 

Step 3) To write slope deflection equation

Member AB

Member BC

Member CD

Step 4) To apply condition of equilibrium

At joint B

Adding equation 1and 2

Due to symmetry

Step 5) To find end moments at joint

Step 6) Draw BMD and SFD

C:\Users\DELL\Desktop\2.jpeg
C:\Users\DELL\Desktop\3.jpeg

SFD

C:\Users\DELL\Desktop\3.jpeg

BMD

C:\Users\DELL\Desktop\3.jpeg

 


Problems:

  1. Analyze the portal frame as shown by SD method draw BMD

 

1. Non-sway

2.     Find fixed end moments

 

 

 

AB =>

Joint equilibrium condition

Joint at B

 

 

2. Analyze the frame by SD method by BMD

Step 1) Find Dki

Step 2) Find fixed end moments

 

   ----due to no loading

Step 3) S.D. Equation

AB =>

BD =>

Step 4) Equilibrium condition joint B

BMD

 

3. Analyze the frame shown in Figure support A is fixed B& C are winged. Draw BMD

Step 1) Non sway & Dki =

Step 2) Find a fixed end

 

Step 3) S.D. Equation

For AB Span l =6m, ,I =2I,

 

For BD Span l =4m,  , I =I,

 

 

For BC Span l =4m, ,I =1.5I,

 

 

Step4) Equilibrium condition joint B

+ + =0

-16.67+3.83++ ---------(A)

------(B)

--------(C)

Equating Eq A, B & C you will get

 

Final moment:

 

Problem on sway Frame

  1. Analyze the frame shown in fig slope deflection method

 

Step 1) Sway Frame: due to lateral loud frame in sway

 

Find Fixed Ends Moments:

M FAB =   =             = 7.5KKNM

MFBA =   =            = -22.5KNM

M FBC      = M FCB = = = 60 KNM.

Apply Slope Deflection equations

For AB: ƟA=0, ƟB =?

MAB = MFAB  (2ƟA+ ƟB-)

MAB = 7.5 +    ------------1)

MBA = -22.5 +   ---------------2)

For BC: =?

MBC = +60+ ()   -------3)

MCB = -60+(ƟB)    -------------4)

MDC = MFDC(2ƟD+ ƟC-)

MDC   = (ƟC-) ---------5)

MCD   = (2ƟC-)   --------6)

4) Apply joint equilibrium equation

MBA+MBC = 0

-22.5 + = 0   ---------------(A)

MCB+MCD = 0

-60+(ƟB) + (2ƟC-)= 0   ------------(B)

 

Shear equation:                        HA +HD -40= 0   --------------7)

 

For AB Column

 

- HA 4 +MAB + MBA+ 40 *1=0

4HA = MAB + MBA+ 40 *1

Put Equation 1&2

4HA =7.5 + -----(8)

 

 

This Fig

4HD = MDC+MCD

 

Put Equation 5 & 6

 

4HD =  (ƟC-)  + (2ƟC-)    ---------9)

Put equation 8 & 9 in Equation (7) Equation u will get equation (C)

Then Equating Equation (A, B, &C)

QB = -39.25/EI

QC = 19.25/EI

      = 110/EI

Final moments

MAB = 29.12 KNM

MBA = -20.5 KNM

MBC = 20.5 KNM

MCB = -60.51 KNM

MCD = 60.51KNM

MDC = +50.875 KNM

 

Key takeaways:

Steps:

  1. To find degree of kinematic indeterminacy
  2. To find fixed end moment
  3. To write slope deflection equation
  4. To find moments at end
  5. Draw BMD

 

References:

1. Structural Analysis: deodas Menon---Narosa Publishing House.

2. Structural Analysis: Thandavamoorthy---Oxford University Press.

3. Structural Analysis: A Matrix Approach by Pundit and Gupta, McGraw Hills.

4. Structural Analysis by Hibbler, Pearson Education.

5. Structural Analysis: M. M. Das, B. M. Das---PHI Learning Pvt Ltd. Delhi.

6. Fundamentals of Structural Analysis: 2nded---West---Wiley.

7. Theory of Structures: Vol. I & II by B. C. Punmia, Laxmi Publication.

8. Theory of Structures: Vol. I & II by Peru mull & Vaidyanathan, Laxmi Publication.

9. Fundamentals of Structural Analysis: K. M. Leet, Vang, Gilbert—McGraw Hills

10. Matrix Methods for structural engineering. By Gere, Weaver.

11. Introduction to the Finite element method, Dr. P.N. Godbole, New Age Publication, Delhi.

12. Finite element Analysis, S.S. Bhavikatti, New Age Publication, Delhi.

13. Basic Structural Analysis: Wilbur and Norris.

 



Unit - 3


Slope-Deflection Method


1. Slope Deflection Method

  • Slope=
  • Deflection=
  • Sign Convention
  1. Simply supported Beam

 

At support form

At the simply supported end

is always zero.

is always zero.

2.     Cantilever beam

 

1. =0 at the fixed end at force end

At Fixed end

2. Slope deflection equation: 

When a member of a structure is loaded, it deforms from its original position and internal forces are developed. The end moments can be expressed in terms of

(i)                Fixed end moments (FEM) of the member due to transverse loading on the member

(ii)              The slopes or rotations at the end

(iii)           The end deflections or joint translations. These expressions are called as Slope-Deflection equations.

There are two slope deflection equations for each member.

These slope deflection equations are derived from superposition of the end moments caused by various actions and displacement.

Similarly

Where,

= be the slope or rotation at joint A

= be the slope or rotation at joint B.

= be the end deflection or translation of joint A.

= be the end deflection or translation of joint B.

B/A = Relative translation of joint B with respect to joint A

AB Angle between the unreformed member and the line joining the deflected joints.

Here > Therefore B/A is positive

MAB = Moment at A of member AB

МВА = Moment at B of member BA

 

3. Fixed End Moments: Standard Cases Fixed End Moments

Standard Cases

1.

 

2.

 

3.

 

4.

5.

 

4. Derivation of slope Deflection Method

  • In beam, a continuous beam ABCD, consider AB part and take fixed end moments and rotation at support as given below 

 

 

As per the diagram

Slop deflection equations are

.

5.  Types of numerical for slope deflection method

1. The beam is without sink (is zero.)

2. Beam with sink (is given)

3. Analyze of the frame without sink

4. Analyze the frame with the sink.

6. Steps for analysis of the slope deflection method

1. Find unknown slope.

2. Find fixed end moment

3. Apply the slope deflection equation.

4. Use the joint equilibrium equation at the joint.

5. Find final moments.

6. Find relation by equation &Draw SFI

7. Draw BMD

 


Problems:

  1. Analyze the continuous beam shown in figure Draw also B.M. And S.F. Diagram

 

 

Step-1 Find fixed end moments

Step-II Slope deflection method

Step-III Apply equilibrium condition at joint

Step-IV   Find final moments.

Step-V Find reaction

S.F. At A Take a moment at A is

SFD & BMD

 

 

2.     Determine support moments & draw BMD for the beam shown by using S.D.

 

Step-I) Find Dki

Step II) Find fixed end moments

For BC Beam

Step-III) Apply the SD equation

AB=>

BC=>

 

Step- IV) Apply equilibrium condition joint B

-------   due to Hinge support

Put in equation (1) (2) (3) & (4)

Draw BMD Diagram

 

3.     Analyze the continuous beam ABCD by S.D. Method draw BMD

Modify Diagram

Step- I) Fixed end moment

 

Apply S.D. Equation

Joint equilibrium equation

Joint at B

BMD and SFD

 

4.     Analyze the continuous beam as shown in the figure by S.D. Method Draw BMD

Step 1) Find fixed end moment

 

KNM

S.D. Equation

Span AB

 

 

Joint equilibrium equation

Joint B

KNM

 

SFD

BMD

 

5.     Analyze the continuous beam by S.D. Equation

Find fixed end moments:

 

 

Apply S.D. Equation

AB=>

2)

BC=>

Joint equilibrium condition

BMD

 

 

6.     Analyze the beam by slope deflection method. Draw SFD and BMD

Solution:

Step 1) Dki = 2 

Step 2) Find Fixed End Moments

For AB

For BC

 

 

For CD

 

Apply S.D. Equation

Joint equilibrium condition

7.     Determine the final moments by slope deflection method take EI is constant.

Step 1) Dki = 2

Step 2) Find fixed end moments

 

 

Step 3) Apply S.D. Equation

For beam BC

--------(3)

 

 

-------(4)

OR

For member, BC apply modified equation

 

Step4) Joint equilibrium equation:


 

9.5 + 1.46 EI + 0.33EI-------(A)

 

0-------(B)

Equating equation, A & B get

 

 

OR

Apply this condition to the modified equation

Step 5) BMD

Problems based on sink

Sink means deflection which occurs due to load which is acting on the beam

Note: always take EI in terms of KNM2

  1. Analyze continuous beam ABCD having

Point B is sunk by 2.5mm also draw BMD

Step 1) Dki = 2  

Step 2) Fixed end moments

STEP 3) Apply S.D. Equation

AB=> ,l = 3m

 

BC=>

 

Step 4) Apply equilibrium condition

Joint B

Joint C

KNM

KNM

KNM

KNM

KNM

BMD

 

Key takeaway:

Steps:

1. Find unknown slope.

2. Find fixed end moment

3. Apply the slope deflection equation.

4. Use the joint equilibrium equation at the joint.

5. Find final moments.

6. Find relation by equation &Draw SFI

7. Draw BMD

 


Problems:

  1. Analyze the portal frame using slope deflection method.
C:\Users\DELL\Desktop\1.jpeg

 

Solution:

It is symmetrical non-sway frame

Step 1) To find degree of kinematic indeterminacy

Dki= 2

Unknown displacement

Known displacement:

Step 2) To find fixed end moments

 

Step 3) To write slope deflection equation

Member AB

Member BC

Member CD

Step 4) To apply condition of equilibrium

At joint B

Adding equation 1and 2

Due to symmetry

Step 5) To find end moments at joint

Step 6) Draw BMD and SFD

C:\Users\DELL\Desktop\2.jpeg
C:\Users\DELL\Desktop\3.jpeg

SFD

C:\Users\DELL\Desktop\3.jpeg

BMD

C:\Users\DELL\Desktop\3.jpeg

 


Problems:

  1. Analyze the portal frame as shown by SD method draw BMD

 

1. Non-sway

2.     Find fixed end moments

 

 

 

AB =>

Joint equilibrium condition

Joint at B

 

 

2. Analyze the frame by SD method by BMD

Step 1) Find Dki

Step 2) Find fixed end moments

 

   ----due to no loading

Step 3) S.D. Equation

AB =>

BD =>

Step 4) Equilibrium condition joint B

BMD

 

3. Analyze the frame shown in Figure support A is fixed B& C are winged. Draw BMD

Step 1) Non sway & Dki =

Step 2) Find a fixed end

 

Step 3) S.D. Equation

For AB Span l =6m, ,I =2I,

 

For BD Span l =4m,  , I =I,

 

 

For BC Span l =4m, ,I =1.5I,

 

 

Step4) Equilibrium condition joint B

+ + =0

-16.67+3.83++ ---------(A)

------(B)

--------(C)

Equating Eq A, B & C you will get

 

Final moment:

 

Problem on sway Frame

  1. Analyze the frame shown in fig slope deflection method

 

Step 1) Sway Frame: due to lateral loud frame in sway

 

Find Fixed Ends Moments:

M FAB =   =             = 7.5KKNM

MFBA =   =            = -22.5KNM

M FBC      = M FCB = = = 60 KNM.

Apply Slope Deflection equations

For AB: ƟA=0, ƟB =?

MAB = MFAB  (2ƟA+ ƟB-)

MAB = 7.5 +    ------------1)

MBA = -22.5 +   ---------------2)

For BC: =?

MBC = +60+ ()   -------3)

MCB = -60+(ƟB)    -------------4)

MDC = MFDC(2ƟD+ ƟC-)

MDC   = (ƟC-) ---------5)

MCD   = (2ƟC-)   --------6)

4) Apply joint equilibrium equation

MBA+MBC = 0

-22.5 + = 0   ---------------(A)

MCB+MCD = 0

-60+(ƟB) + (2ƟC-)= 0   ------------(B)

 

Shear equation:                        HA +HD -40= 0   --------------7)

 

For AB Column

 

- HA 4 +MAB + MBA+ 40 *1=0

4HA = MAB + MBA+ 40 *1

Put Equation 1&2

4HA =7.5 + -----(8)

 

 

This Fig

4HD = MDC+MCD

 

Put Equation 5 & 6

 

4HD =  (ƟC-)  + (2ƟC-)    ---------9)

Put equation 8 & 9 in Equation (7) Equation u will get equation (C)

Then Equating Equation (A, B, &C)

QB = -39.25/EI

QC = 19.25/EI

      = 110/EI

Final moments

MAB = 29.12 KNM

MBA = -20.5 KNM

MBC = 20.5 KNM

MCB = -60.51 KNM

MCD = 60.51KNM

MDC = +50.875 KNM

 

Key takeaways:

Steps:

  1. To find degree of kinematic indeterminacy
  2. To find fixed end moment
  3. To write slope deflection equation
  4. To find moments at end
  5. Draw BMD

 

References:

1. Structural Analysis: deodas Menon---Narosa Publishing House.

2. Structural Analysis: Thandavamoorthy---Oxford University Press.

3. Structural Analysis: A Matrix Approach by Pundit and Gupta, McGraw Hills.

4. Structural Analysis by Hibbler, Pearson Education.

5. Structural Analysis: M. M. Das, B. M. Das---PHI Learning Pvt Ltd. Delhi.

6. Fundamentals of Structural Analysis: 2nded---West---Wiley.

7. Theory of Structures: Vol. I & II by B. C. Punmia, Laxmi Publication.

8. Theory of Structures: Vol. I & II by Peru mull & Vaidyanathan, Laxmi Publication.

9. Fundamentals of Structural Analysis: K. M. Leet, Vang, Gilbert—McGraw Hills

10. Matrix Methods for structural engineering. By Gere, Weaver.

11. Introduction to the Finite element method, Dr. P.N. Godbole, New Age Publication, Delhi.

12. Finite element Analysis, S.S. Bhavikatti, New Age Publication, Delhi.

13. Basic Structural Analysis: Wilbur and Norris.

 


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