DEPARTMENT OF CIVIL ENGINEERING

GEETHANJALI COLLEGE OF ENGINEERING & TECHNOLOGY

CHEERYAL (V), KEESARA (M), R.R. DIST. - 501 301

(Affiliated to JNTUH, Approved by AICTE, NEW DELHI, ACCREDITED BY NBA)

www.geethanjaliinstitutions.com

2015-2016

STRUCTURAL ANALYSIS-I

COURSE FILE

(Subject Code: A40115)

II Year B.TECH. (CIVIL ENGINEERING) II Semester

Prepared by T.Sandeep,B.Ravi Chand Asst.professor

GEETHANJALI COLLEGE OF ENGINEERING & TECHNOLOGY

CHEERYAL (V), KEESARA (M), R.R. DIST. 501 301

DEPARTMENT OF CIVIL ENGINEERING

(Name of the Subject /Lab Course): STRUCTURAL ANALYSIS-I

(JNTUH CODE:A40115) Programme: UG

Branch: CIVIL ENGINEERING Version No: 01

Year: II Updated on:

Semester: II No. of pages:

Classification status (Unrestricted/Restricted)

Distribution List:

Prepared by:

1) Name : T.Sandeep 1) Name: B.Ravi Chand

2) Sign. : 2) Sign :

3) Design.: Asst.Professor 3) Design: Asst.Professor

4) Date : 4) Date :

Verified by: * For Q.C Only.

1) Name : 1) Name:

2) Sign : 2) Sign :

3) Design : 3) Design. :

4) Date : 4) Date :

Approved by: (HOD)

1) Name :

2) Sign : 4) Date:

INDEX

Content Page

1. Introduction & Pre-requisites

2. Syllabus

3. Vision of the Department

4. Mission of the Department.

5. Program Educational Objects

6.Program outcomes

7. Course objectives and outcomes

8. Course outcomes

9. Instructional Learning

10. Course mapping with PEOs and POs

11. Class Time Table

12. Individual Time Table

13a. Unit wise Summary

13b. Micro Plan with dates and closure report

14. Detailed notes

15. University Question papers of previous years

16. Question Bank

17. Assignment topics

18. Unit wise Quiz Questions

19. Tutorial problems

20. Known gaps if any

21. References, Journals, websites and E-links

22. Quality Control Sheets

23. Student List

24. Group-Wise students list for discussion topics

1. Introduction to the subject

The structural analysis is based on engineering mechanics, mechanics of solids, laboratory research, model and

prototype testing, experience and engineering judgment. The basic methods of structural analysis are flexibility

and stiffness methods. The flexibility method is also called force method and compatibility method. The

stiffness method is also called displacement method and equilibrium method. These methods are applicable to

all type of structures; however, here only skeletal systems or framed structures will be discussed. The examples

of such structures are beams, arches, cables, plane trusses, space trusses, plane frames, plane grids and space

frames.

Pre-requisites

1.Engineering Mechanics

2. Strength of Materials-I

2. Syllabus

Sl.No Unit

No Topic

1 1

ANALYSIS OF PERFECT FRAMES: Types of frames- perfect, imperfect and

redundant pin jointed frames.. Analysis of determinate pin jointed frames using

method of joints, method of sections and tension coefficient method for vertical

loads horizontal and inclined loads.

2 2

ENERGY THEOREMS: Introduction-Strain energy in linear elastic system,

expression of strain energy due to axial load, bending moment and shear forces -

Castigliano‘s first theorem-Deflections of simple beams and pin jointed trusses.

THREE HINGED ARCHES: Introduction, types of arches- comparison

between three hinged and two hinged arches. Linear arch. Eddy‘s theorem

analysis of three hinged arches. normal thrust and radial shear in an arch

geometrical properties of parabolic and circular arch. Three hinged circular arch

at different levels. Absolute maximum bending moment diagram for a three

hinged arch.

3 3 PROPPED CANTILEVERS: Analysis of propped cantilevers-shear force and

Bending moment diagrams-Deflection of propped cantilevers.

FIXED BEAMS – Introduction to statically indeterminate beams with U.D.load

central point load, eccentric point load. Number of point loads, uniformly

varying load, couple and combination of loads shear force and Bending moment

diagrams-Deflection of fixed beams effect of sinking of support, effect of

rotation of a support.

4 4

CONTINUOUS BEAMS : Introduction-Clapeyron‘s theorem of three moments-

Analysis of continuous beams with constant moment of inertia with one or both

ends fixed-continuous beams with overhang, continuous beams with different

moment of inertia for different spans-Effects of sinking of supports-shear force

and Bending moment diagrams. Derivation of slope deflection equation,

application to continuous beams with and without settlement of supports.

Analysis of continuous beams with and without settlement of supports using

moment distribution method. Shear force and bending moment diagrams, Elastic

curve.

5 5

MOVING LOADS : Introduction maximum SF and BM at a given section and

absolute maximum S.F. and B.M due to single concentrated load U.D load longer

than the span, U.D load shorter than the span, two point loads with fixed distance

between them and several point loads-Equivalent uniformly distributed load-

Focal length.

INFLUENCE LINES: Definition of influence line for SF, Influence line for

BM- load position for maximum SF at a section- Load position for maximum

BM at a section single point load, U.D.load longer than the span, U.D.load

shorter than the span- Influence lines for forces in members of Pratt and Warren

trusses.

TEXT BOOKS:

1. Analysis of Structures-Vol I & Vol II by V.N. Vazirani & M.M.Ratwani, Khanna Publications, New Delhi.

2. Analysis of Structures by T.S. Thandavamoorthy, Oxford University Press, New Delhi

3. Structural Analysis by S S Bhavikatti – Vikas Publishing House.

REFERENCES:

1. Mechanics of Structures by S.B.Junnarkar, Charotar Publishing House, Anand, Gujrat

2. Theory of Structures by Pandit & Gupta; Tata Mc.Graw – Hill Publishing Co.Ltd., New Delhi.

3. Strength of Materials and Mechanics of Structures- by B.C.Punmia, Khanna Publications, New Delhi.

Websites:-

1. http://jntuhupdates.net/jntuh-b-tech-2-2-semester-r13-syllabus-book/

2. NPTEL Resources

Journals:-

1. International Journal of Strctural Engineering

2. International Journal of Advances in structural engineering

3. Vision of the Department: To develop a world class program with excellence in teaching, learning and research that would lead to growth,

innovation and recognition

4. Mission of the Department: The mission of the Civil Engineering Program is to benefit the society at large by providing technical education to

interested and capable students. These technocrats should be able to apply basic and contemporary science, engineering

and research skills to identify problems in the industry and academia and be able to develop practical solutions to them

5. Program Educational Objectives-PEOs:

The Civil Engineering Department is dedicated to graduating Civil engineers who:

1. Practice Civil engineering in the general stems of fluid systems, civil systems and design, and

materials and manufacturing in industry and government settings.

2. Apply their engineering knowledge, critical thinking and problem solving skills in professional

engineering practice or in non-engineering fields, such as law, medicine or business.

3. Continue their intellectual development, through, for example, graduate education or professional

development courses.

4. Pursue advanced education, research and development, and other creative efforts in science and

technology.

5. Conduct them in a responsible, professional and ethical manner.

6. Participate as leaders in activities that support service to and economic development of the region,

state and nation.

6. Program Outcomes (PO)

Graduates of the Civil Engineering Programme will be able to:

1. Apply the knowledge of mathematics, science, engineering fundamentals, and Civil Engineering principles to the

solution of complex problems in Civil Engineering.

2. Identify, formulate, research literature, and analyse complex Civil Engineering problems reaching substantiated

conclusions using first principles of mathematics and engineering sciences.

3. Design solutions for complex Civil Engineering problems and design system components or processes that meet

the specified needs with appropriate consideration for the public health and safety, and the cultural, societal, and

environmental considerations.

4. Use research-based knowledge and research methods including design of experiments, analysis and interpretation

of data, and synthesis of the information to provide valid conclusions related to Civil Engineering problems.

5. Create, select, and apply appropriate techniques, resources, and modern engineering tools such as CAD, FEM and

GIS including prediction and modelling to complex Civil Engineering activities with an understanding of the

limitations.

6. Apply reasoning informed by the contextual knowledge to assess societal, health, safety, legal and cultural issues

and the consequent responsibilities relevant to the professional Civil Engineering practice.

7. Understand the impact of the professional Civil Engineering solutions in societal and environmental contexts, and

demonstrate the knowledge of, and need for sustainable development.

8. Apply ethical principles and commit to professional ethics and responsibilities and norms of the Civil Engineering

practice.

9. Function effectively as an individual, and as a member or leader in diverse teams, and in multidisciplinary

settings.

10. Communicate effectively on complex Civil Engineering activities with the engineering community and with

society at large, such as, being able to comprehend and write effective reports and design documentation, make

effective presentations, and give and receive clear instructions.

11. Demonstrate knowledge and understanding of the engineering and management principles and apply these to

one’s own work, as a member and leader in a team, to manage Civil Engineering projects and in multidisciplinary

environments.

12. Recognise the need for, and have the preparation and ability to engage in independent and life-long learning in the

broadest context of technological change.

7. Course objectives

Students who successfully complete this course will have demonstrated ability to:

Course Objectives:

1. Ability to apply knowledge of mathematics and engineering in calculating slope, deflection, bending moment and shear force using various methods of approach.

2. Ability to identify, formulate and solve problems in structural analysis. Ability to analyse structural system and interpret ate data.

3. Ability to use the techniques, skills to formulate and solve engineering problem. Ability to communicate effectively in design of structural elements.

4. Ability to engage in life-long learning with the advances in structural problems.

7. Course Outcomes

1. Understands what different types of displacement methods are.

2. Understands how to solve different deflection related problems in beams, arches, cables.

3. Understands how to control the deflections and displacements under different loading conditions.

4. Understands the concept of influence lines.

5. Understands how to predict different mitigation problems by drawing shear force and bending moments.

9. Instructional learning

A mixture of lectures, tutorial exercises, and case studies are used to deliver the various topics. Some of these

topics are covered in a problem-based format to enhance learning objectives. Others will be covered through

directed study in order to enhance the students’ ability of “learning to learn.” Some case studies are used to

integrate these topics and thereby demonstrate to students how the various techniques are inter-related and how

they can be applied to real problems in an industry.

10. Course mapping with PEO’s and PO’s

PEO/PO Program Outcomes

1 2 3 4 5 6 7 8 9 10 11 12

Program A X X X X X X X X

Educational

Objectives

(PEO)

B X X X X X X X X X

C X X X X X X X

D X X X X X X X

E X X X X X X

F X X

11. Class Timetable

DEPARTMENT OF CIVIL ENGINEERING

Ref: TLE/2014-2015/23.12.2014/SADM /CT -1004

PROGRAMME: B.TECH. (CIVIL ENGINEERING)

SEMESTER: II Year II- SEMESTER

NOTE: “*” Represents Tutorial Classes.

Time Table Coordinator HOD PRINCIPAL

12. Individual Time Table

Name of the faculty: Load = 17 Rev: w.e.f.:

Section- II A and II B

Name of the faculty: T.Sandeep Load = 17 ; w.e.f.: 29/06/15

Time 9.30-

10.20

10.20-

11.10

11.10-

12.00

12.00-

12.50

12.50-

1.30

1.30-

2.20 2.20-3.10 3.10-4.00

Period 1 2 3 4

LU

NC

H

5 6 7

Monday SOM HHM BMC P&S S.A

Tuesday EVS SOM S.A BMC P&S HHM

Wednesday P&S BMC SA HHM SOM LIBRARY

Thursday EVS SA HHM P&S

BMC EVS

Friday LAB P&S

CRT EVS

Saturday EVS LAB

MENTOR SEMINAR

Time 9.30-10.20 10.20-

11.10 11.10-12.00

12.00-

12.50 12.50-1.30 1.30-2.20

2.20-

3.10 3.10-4.00

Period 1 2 3 4

LU

N

CH

5 6 7

Monday S.A

13. Unit wise Summary

Unit

no

Total

periods Topic

Reg/additio

nal

Lcd/ohp/

bb Remark

1

12 Analysis of perfect frames: types of frames Regular BB

2 Perfect, imperfect and redundant pin jointed frames Regular BB

3 Analysis of determinate pin jointed frames using

method of joints Regular BB

3 Method of sections Regular BB

2 Tension coefficient method for vertical loads horizontal

loads Regular BB

2 Tension coefficient method for inclined loads. Regular BB

2

20 Energy theorems: introduction Regular BB

2 Strain energy in linear elastic system Regular BB

4 Strain energy due to axial load, bending moment and

shear forces Regular BB

2 Castigliano‘s first theorem Regular BB

4 Deflections of simple beams and pin jointed trusses Regular BB

1 Three hinged arches: introduction, types of arches Regular BB

1 Comparison between three hinged and two hinged

arches. Linear arch. Regular BB

3 Normal thrust and radial shear in an arch geometrical

properties of parabolic and circular arch Regular BB

3 Three hinged circular arch at different levels Regular BB

3

Absolute maximum bending moment diagram for a

three hinged arch.

Regular BB

Tuesday S.A

Wednesday SA

Thursday SA

Friday

Saturday

3

15 Propped cantilevers: analysis of propped cantilevers Regular BB

3 Shear force and bending moment diagrams- Regular BB

3 Deflection of propped cantilevers. Regular BB

3 Fixed beams – introduction to statically indeterminate

beams Regular BB

3

U.d.load central point load, eccentric point load.

Number of point loads, uniformly varying load, couple

and combination of loads shear force and bending

moment diagrams-

Regular BB

3 Deflection of fixed beams effect of sinking of

support, effect of rotation of a support. Regular BB

4

18 Continuous beams : introduction-clapeyron‘s theorem

of three moments Regular BB

3

Analysis of continuous beams with constant moment

of inertia with one or both ends fixed-continuous

beams

Regular BB

3 Overhang, continuous beams with different moment

of inertia for different spans Regular BB

3 Effects of sinking of supports-shear force and bending

moment diagrams

Regular BB

3

Derivation of slope deflection equation, application to

continuous beams with and without settlement of

supports

Regular BB

3

Analysis of continuous beams with and without

settlement of supports using moment distribution

method.

Regular BB

3 Shear force and bending moment diagrams, elastic

curve. Regular BB

5

15 Moving loads : introduction maximum sf and bm at a

given section and absolute maximum s.f. and b.m Regular BB

3

Maximum sf and bm at a given section and absolute

maximum s.f. and b.m due to single concentrated load

u.d load longer than the span, u.d load shorter than

the span

Regular BB

3

Two point loads with fixed distance between them

and several point loads-equivalent uniformly

distributed load-focal length.

Regular BB

3

Influence lines: definition of influence line for sf,

influence line for bm- load position for maximum sf at

a section-

Regular BB

3

Load position for maximum bm at a section single

point load, u.d.load longer than the span, u.d.load

shorter than the span

Regular BB

3 Influence lines for forces in members of pratt and

warren trusses. Regular BB

Micro Plan with dates and closure report

Unit

no

Total

periods

Date Topic Reg/additional Lcd/ohp/bb Remark

1

12 Analysis of perfect frames: types

of frames Regular BB

2 Perfect, imperfect and redundant

pin jointed frames Regular BB

3 Analysis of determinate pin jointed

frames using method of joints Regular BB

3 Method of sections Regular BB

2 Tension coefficient method for

vertical loads horizontal loads Regular BB

2 Tension coefficient method for

inclined loads. Regular BB

2

20 Energy theorems: introduction Regular BB

2 Strain energy in linear elastic

system Regular BB

4 Strain energy due to axial load,

bending moment and shear forces Regular BB

2 Castigliano‘s first theorem Regular BB

4 Deflections of simple beams and

pin jointed trusses Regular BB

1 Three hinged arches:

introduction, types of arches Regular BB

1

Comparison between three hinged

and two hinged arches. Linear

arch.

Regular BB

3

Normal thrust and radial shear in

an arch geometrical properties of

parabolic and circular arch

Regular BB

3 Three hinged circular arch at

different levels Regular BB

3

Absolute maximum bending

moment diagram for a three

hinged arch.

Regular BB

3

15 Propped cantilevers: analysis of

propped cantilevers Regular BB

3 Shear force and bending moment

diagrams- Regular BB

3 Deflection of propped

cantilevers. Regular BB

3 Fixed beams – introduction to

statically indeterminate beams Regular BB

3

U.d.load central point load,

eccentric point load. Number of

point loads, uniformly varying

load, couple and combination of

loads shear force and bending

moment diagrams-

Regular BB

3

Deflection of fixed beams effect

of sinking of support, effect of

rotation of a support.

Regular BB

4 18

Continuous beams : introduction-

clapeyron‘s theorem of three

moments

Regular BB

3

Analysis of continuous beams with

constant moment of inertia with

one or both ends fixed-continuous

beams

Regular BB

3

Overhang, continuous beams with

different moment of inertia for

different spans

Regular BB

3 Effects of sinking of supports-

shear force and bending moment

diagrams

Regular BB

3

Derivation of slope deflection

equation, application to

continuous beams with and

without settlement of supports

Regular BB

3

Analysis of continuous beams with

and without settlement of

supports using moment

distribution method.

Regular BB

3 Shear force and bending moment

diagrams, elastic curve. Regular BB

5

15

Moving loads : introduction

maximum sf and bm at a given

section and absolute maximum

s.f. and b.m

Regular BB

3

Maximum sf and bm at a given

section and absolute maximum

s.f. and b.m due to single

concentrated load u.d load longer

than the span, u.d load shorter

than the span

Regular BB

3

Two point loads with fixed

distance between them and

several point loads-equivalent

uniformly distributed load-focal

length.

Regular BB

3 Influence lines: definition of Regular BB

influence line for sf, influence line

for bm- load position for

maximum sf at a section-

3

Load position for maximum bm at

a section single point load,

u.d.load longer than the span,

u.d.load shorter than the span

Regular BB

3

Influence lines for forces in

members of pratt and warren

trusses.

Regular BB

GUIDELINES:

Distribution of periods:

No. of classes required to cover JNTUH syllabus : 80

No. of classes required to cover Additional topics : Nil

No. of classes required to cover Assignment tests (for every 2 units 1 test) : 4

No. of classes required to cover tutorials : 2

No. of classes required to cover Mid tests : 2

No of classes required to solve University Question papers : 2

-------

Total periods 80

14.Detailed Notes

Unit-1

ANALYSIS OF PIN-JOINTED PLANE FRAMES

1. Explain about different types of frames and analysis of frames?

Ans:

The Different types of frames are:

(i) Perfect frame

(ii) Imperfect frame Imperfect frame may be deficient or a redundant frame.

Perfect Frame: The frame which is composed of such members, which are just sufficient to keep the frame in

equilibrium, when the frame is supporting an external load, is known as perfect frame.

Example:

(a) (b) (c) Suppose we add a set of two members and a joint again, we get a perfect frame as shown in the above

fig©.

Hence for a perfect frame, the number of joints and number of members are given by,

n = 2j – 3

Where, n = Number of members, and

j = Number of joints.

Imperfect frame: A frame in which number of members and number of joints are not given by,

n = 2j – 3 is known, an imperfect frame. This means that number of members in an imperfect frame will be either more

or less than (2j – 3).

(i) If the number of members in a frame are less than (2j – 3), then the frame is known as

deficient frame.

(ii) If the number of members in a frame are more than (2j – 3), then the frame is known as redundant frame.

The assumptions made in finding out the forces in a frame are:

(i) The frame is a perfect frame

(ii) The frame carries load at the joints

(iii) All the members are pin-joined.

Reactions of supports of a frame:

(i) On roller support or

(ii) On a hinged

Analysis of a frame consists of:

(i) Determinations of the reactions at the supports and

(ii) Determinations of the forces in the members of the frame.

A frame is analysed by the following methods:

(i) Method of joints,

(ii) Method of sections, and

(iii) Graphical method. ==========================================================================

2. Find the forces in the members AB, AC and BC of the truss as shown in the below figure?

Ans:

Now let us consider the equilibrium of the various joints. Joint B:

Resolving the forces acting on the joint B, vertically

Joint C:

3. A truss of span 7.5m carries a point load of 1 KN at joint D as shown in the below figure. Find the

reactions and forces in the members of the truss?

Ans: Moments about A,

&

Now let us consider the equilibrium of the various joints. Joint A:

Joint B:

Joint D:

Let the direction of F3 is assumed as shown in the below figure.

4. Determine the forces in the truss shown in the below figure. Which is subjected to inclined loads?

Ans:

Joint A:

Joint C:

Joint E:

Joint F:

Joint B:

Joint G:

MEMBER FORCE IN THE MEMBER NATURE OF FORCE

5. Find the forces in the members AB and AC of the truss as shown in the below figure

using method of section?

Ans:

6. A truss of span 9m is loaded as shown in the below figure. Find the reactions and foreces

in the members marked 1, 2 and 3.

Ans:

Now draw a section line (1-1), cutting the members 1, 2 and 3 in which forces are to be

determined.

Consider the equilibrium of the left part of the truss because it is smaller than the right part.

Moments about D:

Moments about G:

Moments about C:

SLOPE DEFLECTION METHOD

Example: Analyze the propped cantilever shown by using slope defection method. Then draw

Bending moment and shear force diagram.

Solution: End A is fixed hence A

End B is Hinged hence

=0

B ≠0

Assume both ends are fixed and therefore fixed end moments are

FAB wL2

,

12

FBA

wL2

12

The Slope deflection equations for final moment at each end are

MAB FAB 2EI 2

L A

2

B

wL

2EI

(1)

12 L B

M F

2EI 2

BA BA

wL2

L B A

4EI

(2)

12 L B

In the above equations there is only one unknown B . To solve we

have boundary condition at B;

Since B is simply supported, the BM at B is zero

ie. MBA=0.

From equation (2) M

wL2

4EI 0

BA 12 L

B

3

EIB

wL

48

- ve sign indicates the rotation is anticlockwise

L

Substituting the value of EIB in equation (1) and (2) we have end moments

2 3 2

M wL

2

wL wL

- ve sign

indicates moment is anticlockwise

AB 12

2

48 8

3

M wL

4

wL 0

BA 12

48

MBA has to be zero, because it is hinged.

Now consider the free body diagram of the beam and find reactions using equations of

equilibrium.

MB 0

RA L MAB

2

wL L

2

wL wL

L

5 wL

RA

8 2 8

5

wL

8

L

V 0

RA RB wL

RB wL RA

3

wL

8

wL 5

wL

8

Problem can be treated as

The bending moment diagram for the given problem is as below

L

2

The max BM occurs where SF=0. Consider SF equation at a distance of x from right

support

3

SX

8

wL wX 0

X 3

L

8

Hence the max BM occurs at 3 L from

8 2

support B

Mmax MX 3

wL 3

L

8 8

w 3

2 8

9

128

wL2

And point of contra flexure occurs where BM=0, Consider BM equation at a distance of

x from right support.

M 3

wLX w X

0

X 8 2

X 3

L

4

For shear force diagram, consider SF equation from B

S 3

wL wX X 8

SX 0 SB

SX L SA

3

wL 8

5

wL 8

Example: Analyze two span continuous beam ABC by slope deflection method. Then draw

Bending moment & Shear force diagram. Take EI constant

Solution: Fixed end moments are:

FAB Wab

L2

2

100 4 2

62

2

44.44KNM

F Wa b

100 4

2 88.89KNM

BA

FBC

FCB

L2

wL2

12

wL2

12

20 5

12

20 52

12

62

2

41.67KNM

41.67KNM

Since A is fixed A 0 , B 0, C 0,

Slope deflection equations are:

MAB FAB 2EI 2

L A

B

44.44 2EI

6 B

44.44 1

EI (1)

M F

2 2

3

2EI 2

B

BA BA L

B A

88.89 2EI 2B

6

88.89 2

EI

(2)

3 B

M F

2EI 2

BC BC L

B C

41.67 2EI 2

5 B C

41.67 4

EI 2

EI (3)

MCB

FCB

5

2EI 2

L C

B 5

C

B

41.67 2EI 2

5 C

B

41.67 4EI

2

EI (4)

5 C

5 B

In all the above four equations there are only two unknown

accordingly the boundary conditions are

i -MBA-MBC=0

MBA+MBC=0

ii MCB=0 since C is end simply support.

B and C . And

Now

M M

88.89 2

EI 41.67 4

EI 2

EI

BA BC

3 B

5 B

5 C

47.22 22

EI 2

EI 0 (5)

15 B

5 C

M 41.67 2

EI 4

EI 0 (6)

CB 5

B 5

C

Solving simultaneous equations 5 & 6 we get

EI B = – 20.83 Rotation anticlockwise.

EI C = – 41.67 Rotation anticlockwise.

1

2

Substituting in the slope definition equations

MAB = – 44.44 + 20.83 51.38 KNM

3

MBA = + 88.89 + 20.83 75.00 KNM

3

BC

CB

M = – 41.67+ 4 20.83

5

M = + 41.67+ 2 20.83

5

2 41.67 75.00 KNM

5

4 41.67 0

5

Reactions: Consider the free body diagram of the beam.

Find reactions using equations of equilibrium.

Span AB: ΣMA = 0 RB×6 = 100×4+75-51.38

RB = 70.60 KN

ΣV = 0 RA+RB = 100KN

RA = 100-70.60=29.40 KN

5

Span BC: ΣMC = 0 RB×5 = 20×5×

2

R

B = 65 KN +75

ΣV=0 RB+RC = 20×5 = 100KN RC =

100-65 = 35 KN

Using these data BM and SF diagram can be drawn.

Max BM:

Span AB: Max BM in span AB occurs under point load and can be found geometrically

Mmax=113.33-51.38 - 75 51.38

6

4 46.20 KNM

Span BC:Max BM in span BC occurs where shear force is zero or changes its sign.

Hence consider SF equation w.r.t C

Sx = 35-20x = 0

Max BM occurs at 1.75m from C

x 35 =1.75m

20

1.752

Mmax = 35 × 1.75 – 20 = 30.625 KNM

2

2

2

Example: Analyze continuous beam ABCD by slope deflection method and then draw bending

moment diagram. Take EI constant.

Solution:

A 0, B 0, C 0

FEMS

FAB Wab

L2

2

100 4 2

62

2

- 44.44

KN M

F Wa b

100 4 2

88.88 KNM

BA L2 62

2 2

FBC

FCB

wL

12 wL2

12

20 5

12

20 5

12

- 41.67

41.67

KNM

KNM

FCD 20 1.5 - 30 KN M

Slope deflection equations:

2

M F

2EI 2 44.44

1 EI - - - - - - - -- 1

AB AB L

A B 3

B

M F

2EI 2 88.89

2 EI - - - - - - - -- 2

BA BA L

B A 3

B

M F

2EI 2 41.67

4 EI

2 EI - - - - - - - - 3

BC BC L

B C 5

B 5

C

M F

2EI 2 41.67

4 EI

2 EI - - - - - - - - 4

CB CB L

C B 5

C 5

B

MCD 30 KNM

In the above equations we have two unknown rotations the

boundary conditions are:

MBA MBC 0

MCB MCD 0

B and C , accordingly

Now, M M 88.89

2 EI 41.67

4 EI

2 EI

BA BC

3 B

5 B

5 C

47.22 22

EI 2

EI 0 - - - - - - - - 5

15 B

5 C

And, M M 41.67

4 EI

2 EI 30

CB CD

5 C

5 B

11.67 2

EI 4

EI 6

5 B

5 C

Solving (5) and (6) we get

EIB 32.67 Rotation @ B anticlockwise

EIC 1.75 Rotation @ B clockwise

Substituting value of EIB and EIC in slope deflection equations we have

MAB

M

44.44 1 32.67 61.00

2

88.89 2 32.67 67.11

KNM

KNM BA

3

M 41.67 4 32.67

2 1.75 67.11

KNM

BC 5 5

M 41.67 4 1.75

2 32.67 30.00 KNM

CB

MCD 30

5 5

KNM

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

Span AB

RB 6 100 4 67.11 61

RB 67.69 KN

RA 100 RB 32.31 KN

Span BC

R 5 20 5 5 30 67.11

C 2

RC 42.58 KN

RB 20 5 RB 57.42 KN

Maximum Bending Moments:

Span AB: Occurs under point load

Max 133.33 61 67.11 61

4 68.26

KNM

6

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

SX 42.58 20x 0

x 42.58

2.13 m

20 2

Mmax 42.58 2.13 20 2.13

2

30 15.26 KN M

2

Example: Analyse the continuous beam ABCD shown in figure by slope deflection method. The

support B sinks by 15mm.

Take E 200 105

KN / m2

and I 120 106

m4

Solution:

In this problem A =0, B 0, C 0, =15mm

FEMs:

FAB

Wab

L2

44.44 KNM

FBA

Wa b

L2

2

88.89 KNM

FBC

wL

8

41.67 KNM

FCB wL2

8

41.67 KNM

FEM due to yield of support B

2

For span AB:

mab

mba

6EI

L2

6 200

105 120 10

6

62

15

1000

6 KNM

For span BC:

mbc

mcb

6EI

L2

6 200

105 120 106

52

15

1000

8.64KNM

Slope deflection equation

MAB FAB 2EI

(2

L A

B 3

) L

FAB EI 2

L A

B

6EI

L2

- 44.44 1

EI 6

3 B

50.44 1

EI

- - - - - - - -- 1

M F

3 B

2EI

(2 )

6EI

BA BA L

B A L2

88.89 2

EI 6

3 B

82.89 2

EI

- - - - - - - -- 2

M F

3 B

2EI

(2 )

6EI

BC BC L

B C L2

- 41.67 2

EI2 8.64

5 B C

33.03 4

EI 2

EI - - - - - - - -- 3

5 B

5 C

MCB FCB 2EI

(2

L C

B

) 6EI

L2

41.67 2

EI2

5 C

B 8.64

50.31 4

EI 2

EI - - - - - - - -- 4

MCD 30

5 C

5 B

KNM - - - - - - - -- 5

M

There are only two unknown rotations

conditions are

B and C . Accordingly the boundary

MBA MBC 0

MCB MCD 0

Now, BA

MBC 49.86 22

EI

15 B

2

EI 0

5 C

MCB MCD 20.31 2

EI

5 B

4

EI 0

5 C

Solving these equations we get

EIB 31.35

EIC 9.71

Anticlockwise

Anticlockwise

Substituting these values in slope deflections we get the final moments:

MAB

MBA

50.44 1 31.35 60.89

3

82.89 2 31.35 61.99

3

KNM

KNM

M 33.03 4 31.35

2 9.71 61.99 KNM

BC 5 5

M 50.31 4 9.71

2 31.35 30.00 KNM

CB

MCD 30

5 5

KNM

Consider the free body diagram of continuous beam for finding reactions

Reactions:

Span AB:

RB × 6 = 100 x 4 + 61.99 – 60.89

RB = 66.85

RA = 100 – RB

=33.15 KN

Span BC:

RB × 5 = 20 x 5 x

RB = 56.40 KN RC =

20 x 5 - RB

=43.60 KN

5 + 61.99 – 30

2

2

2

Example: Three span continuous beam ABCD is fixed at A and continuous over B, C and D. The

beam subjected to loads as shown. Analyse the beam by slope deflection method and draw

bending moment and shear force diagram.

Solution:

Since end A is fixed A 0, B 0, c 0, D 0

FEMs:

F Wl

60 4

- 30

KNM AB

8 8

F Wl

60 4

30 KNM

BA 8 8

FBC

FCB

M 12.5 KNM 4

M 12.5 KNM 4

F wl2

10 4

- 13.3 3 KNM

CD 12

wl2

F

12

10 4

13.33 KNM

DC 12 12

Slope deflection equations:

MAB FAB 2EI

L 2A B

- 30 2EI 0

4 B

- 30 0.5EIB - - - - - - - - 1

MBA FBA 2EI

L 2B A

30 2EI 2

4 B

0

30 EIB - - - - - - - -- 2

MBC FBC 2EI

L 2B C

12.5 2EI 2

4 B

C

12.5 EIB 0.5EIC

2EI

- - - - - - - -- 3

MCB FCB

L

2C B

12.5 2EI 2

4 C

B

12.5 EIC 0.5 EIB

2EI

- - - - - - - -- 4

MCD FCD

L

2C D

- 13.33 2EI 2

4 C

D

13.33 EI C 0.5EI D - - - - - - - - - - 5

MDC FDC 2EI

L 2D C

13.33 2EI 2

4 D

C

13.33 0.5EI C EI D - - - - - - - - - - 6

In the above Equations there are three unknowns, EI B ,EIC &

accordingly the boundary conditions are:

EID ,

i MBA MBC 0

ii MCB MCD 0

iii MDC 0

( hinged)

Now

MBA MBC 0

30 EIB 12.5 EIB 0.5EIC 0

2EIB 0.5EIC 42.5 0

7

MCB MBC 0

12.5 EIC 0.5EIB 13.33 EIC 0.5EID 0 0.5EIB

2EIC 0.5EID 0.83 0

MDC 0

13.33 0.5EIC EID 0

8

9

By solving (7), (8) & (9), we get

EIB 24.04

EIC 11.15

EID 18.90

By substituting the values of B, c and D in respective equations we get

MAB 30 0.5 24.04 42.02 MBA

30 24.04 5.96 KNM

KNM

MBC 12.5 - 24.04 0.5 11.15 - 5.96 MCB

12.5 11.15 0.5 24.04 11.63

KNM

KNM

MCD 13.33 11.15 0.5 18.90 11.63 KNM MDC

13.33 0.511.15 18.90 0 KNM

Reactions: Consider the free body diagram of beam.

Beam AB:

R 60 2 5.96 42.02

20.985 KN

B 4

RA 60 RB 30.015 KN

Beam BC:

R 11.63 50 5.96

13.92 KN

C 4

RB RC 13.92 KN RB is downward

Beam CD:

R 10 4 2 11.63

17.09 KN

D 4

RC 10 4 RD 22.91KN

2

Example: Analyse the continuous beam shown using slope deflection method. Then draw

bending moment and shear force diagram.

Solution: In this problem A 0, end A is fixed

FEMs:

F wl

10 8

- 53.33

KNM AB

12 wl2

F

12

53.33

KNM

BA 12

F Wl

30 6

- 22.50 KNM

BC

FCD

8

WL

8

8

22.50

KNM

Slope deflection equations:

MAB FAB 2EI

L 2A B

- 53.33 2E 3I

0

8 B

- 53.33 3

4

EIB - - - - - - - - 1

MBA FBA 2EI L

2

2B A

53.33 2E 3I

2

8 B

53.33 3

EI

0

- - - - - - - - 2

2 B

MBC FBC 2EI

L 2B C

- 22.5 2E2I

2

6 B

C

- 22.5 4

EI 2

EI - - - - - - - - 3

3 B

3 C

MCB FCB 2EI

L 2C B

22.5 2E2I

2

6 C

B

22.5 4

EI 2

3 C

3

EIB - - - - - - - - 4

In the above equation there are two unknown boundary

conditions are:

B and C , accordingly the

i MBA MBC 24 0

ii MCB 0 Now, M M 24 53.33

3 EI 22.5

4 EI

2 EI 24

BA BC

2 B

3 B

3 C

54.83 17

EI 2

EI 0 5

and M

22.5 4

EI

6 B

3 C

2

EI 0

CB 3

C 3

B

2 EI 11.25

1 EI - - - - - - - - - -- (6)

3 C

3 B

Substituting in eqn. (5)

54.83 17

EI 11.25 1

EI 0

6 B

3 B

44.58 15

EI 0

6 B

EIB

44.58 6 17.432 15 rotation anticlockwise

2

from equation (6)

EI 3 11.25

1 17.432

C

8.159

3

rotation

anticlockwise

Substituting EIB 17.432 we

get Final Moments:

and EIC 8.159 in the slope deflection equation

MAB

M

53.33 3 - 17.432 -66.40

4

53.33 3 17.432 27.18

KNM

KNM

BA 2

M 22.5 4 17.432

2 8.159 51.18 KNM

BC 3 3

M 22.5 4 8.159

2 (17.432) 0.00

CB 3 3

Reactions: Consider free body diagram of beams as shown

Span AB:

R 27.18 66.40 10 8 4

35.13 KN

B 8

RA 10 8 RB 44.87 KN

Span BC:

R 51.18 30 3

23.53 KN

B 6

RC 30 RB 6.47 KN

Max BM

Span AB: Max BM occurs where SF=0, consider SF equation with A as origin

Sx 44.87 - 10x 0

x 4.487 m 2

M max 44.87 4.487 10 4.487

2 64 36.67 KNM

Span BC: Max BM occurs under point load

BC Mmax 45 51.18

19.41KN M

2

Example: Analyse the beam shown in figure. End support C is subjected to an anticlockwise moment of

12 KNM.

Solution: In this problem A 0, end is fixed

FEMs:

F wl

20 4

26.67

KNM BC

FCB

12

wl2

12

12

26.67

KNM

Slope deflection equations:

M F 2EI

AB AB L

2A B

0 2E2I

0 4

B

EIB

MBA FBA

2EI

L

2B A

- - - - - - - -- 1

0 2E2I

2

4 B

2EIB

2EI

0

- - - - - - - -- 2

MBC FBC

L

2B C

2

2

- 26.67 2E 1.5I

2

4 B

C

- 26.67 3

EI 3

EI - - - - - - - -- 3

2 B

4 C

2EI

MCB FCB

L

2C B

26.67 2E 1.5I

2

4 C

B

26.67 3

EI 3

2 C

4

EIB - - - - - - - -- 4

In the above equation there are two unknowns boundary

conditions are

MBA MBC 0

MCB 12 0

B and C , accordingly the

Now, M M 2EI 26.67 3

EI 3

EI

BA BC B

2 B

4 C

7

EI 3

EI 26.67 0 - - - - - - - -- (5)

2 B

4 C

and, M 12 26.67

3 EI

3 EI 12

CB 2

C 4

B

38.67 3

EI 3

EI 0 - - - - - - - -- (6)

From (5) and (6)

7 EI

3

EI

4 B

2 C

26.67 0

2 B

4 C

3 EI

3 EI 19.33 0

8 B

4 C

25 EI

46 0

From (6)

8

EIB

B

46 8

25

14.72

EI 2 38.67

3 14.72

C 3 4

33.14 - ve sign indicates

rotation anticlockwise

Substituting EIB and EIC is slope deflection equations

MAB EIB 14.72 KNM

MBA 2EIB 2(14.72) 29.42 KNM

M 26.67 3

(14.72) 3 33.14 29.44 KNM

BC 2 4

M 26.67 3

(33.14) 3

(14.72) 12 KNM

CB 2 4

Reaction: Consider free body diagrams of beam

Span AB:

R 14.72 29.44

11.04 KN

B 4

Span BC: RA RB 11.04 KN

R 29.44 12 20 4 2

50.36 KN

B 4

RC 20 4 RB 29.64 KN

I. MOMENT DISTRIBUTION METHOD: This method of analyzing beams and frames was developed by Hardy Cross in 1930.

oment distribution method is basically a displacement method of analysis. But this method sid E

steps the calculation of the displacement and instead makes it possible to apply a series of

converging corrections that allow direct calculation of the end moments.

This method of consists of solving slope deflection equations by successive

approximation that may be carried out to any desired degree of accuracy. Essentially, the method

begins by assuming each joint of a structure is fixed. Then by unlocking and locking each joint in

succession, the internal moments at the joints are distributed and balanced until the joints have

rotated to their final or nearly final positions. This method of analysis is both repetitive and easy

to apply. Before explaining the moment distribution method certain definitions and concepts

must be understood.

Sign convention: In the moment distribution table clockwise moments will be treated

+ve and anti clockwise moments will be treated –ve. But for drawing BMD moments causing

concavity upwards (sagging) will be treated +ve and moments causing convexity upwards

(hogging) will be treated –ve.

Fixed end moments: The moments at the fixed joints of loaded member are called fixed end

moment. FEM for few standards cases are given below:

Member stiffness factor:

a) Consider a beam fixed at one end and hinged at other as shown in figure 3 subjected

to a clockwise couple M at end B. The deflected shape is shown by dotted line

BM at any section xx at a distance x from ‘B’ is given by

d2 y

EI

dx 2

= RBxM

w w

a b

wL/8 L/2 L/2 L

wL/8 wab²/L²

W W wa²b/L²

w /unit

length 2wL/

9

L/3 L/3 L/3 L wL²/12

w /unit

length

2wL/

9

wL²/12

w /unit

length

w

L²/20

L wL²/3

0

5wL²/96 L

5wL²/96

w/unit length L

11

wL²/192

L/2 L/2 5wL²/192 6EI /L²

6EI /L²

3EI/L²

Fig. 3

dy R x 2

Integrating EI B - Mx + C1

dx 2

Using condition x = l dy 0

dx

R l2

C1 = Ml - B

2

dy R x2 R l2

EI B

dx 2

Mx Ml

B

……….(1)

2

R x 3

Mx 2 R l2

Integrating again EI y = B Ml

6 2

B

2

x + C2

Using condition at x = 0 y = 0 C2 =

0

R x3 Mx2 R l2

EI y = B

6

Ml

2

B

x …………… (2)

2

Using at x = l y = 0 in the equation (2)

3M

RB =

2l

Substituting in equation (1)

EI dy

3M

x2 Mx Ml

………….. (3)

dx 4l 4

Substituting x = 0 in the equation (3)

Ml 4EI

EI B =

4

M =

B

l

The term in parenthesis

4EI

K L

For far end fixed …………………. (4)

is referred to as stiffness factor at B and can be defined as moment M required to rotate end B of

beam B = 1 radian.

b) Consider freely supported beam as shown in figure 4 subjected to a clockwise couple

M at B

By using MB = 0

M

RA =

l

()

M

And using V = 0 RB =

l

( )

2

l

BM at a section xx at distance x from ‘B’ is given by EI

d y

M x M

dx 2 l

Integrating EI

dy M x2

Mx C1

dx l 2

M x 3

Mx 2

Integrating again EI y =

l 6

C1 x C 2

2

At x = 0 y = 0 C2= 0

Ml

At x = l y = 0 C1 =

3 2

EI dy

M x Mx Ml

dx l 2 3

Substituting x = 0 in above equation

Ml

EI B =

3

3EI

M =

B

l

The term in parenthesis

K

3EI For far end hinged ………………(5)

is termed as stiffness factor at B when far end A is hinged.

Joint stiffness factor:

If several members are connected to a joint, then by the principle of superposition the total

stiffness factor at the joint is the sum of the member stiffness factors at the joint i.e., kT = k

Eg. For joint ‘0’ shown in fig 5

KT = K0A + KOB + KOC + KOD

Fig. 5

Distribution factors: If a moment ‘M’ is applied to a rigid joint ‘o’, as shown in figure 5, the

connecting members will each supply a portion of the resisting moment necessary to satisfy

moment equilibrium at the joint. Distribution factor is that fraction which when multiplied with

applied moment ‘M’ gives resisting moment supplied by the members.

To obtain its value imagine the joint is rigid joint connected to different members. If

applied moment M cause the joint to rotate an amount ‘’, Then each member rotates by same

amount.

From equilibrium requirement

M = M1 + M2 + M3 + ………………….

= K1 + K2 + K3 +……………..

= K

M

DF1 = 1

M

K1

= K1

K K

In general DF = K ………………. (6)

K

Member relative stiffness factor: In majority of the cases continuous beams and frames will be

made from the same material so that their modulus of electricity E will be same for all members.

It will be easier to determine member stiffness factor by removing term 4E & 3E from

equation (4) and (5) then will be called as relative stiffness factor.

Kr = I for far end fixed

l

Kr = 3 I for far end hinged,

4 l

M

Carry over factors: Consider the beam shown in figure

We have shown that M =

4EI

A

l

& RB = 3M .

2l

d2 y

BM at A = EI = 3M x

dx 2 l

= + M

2

at x l 2 x1

+ve BM of M at A indicates clockwise moment of

2

M at A. In other words the moment

2

‘M’ at the pin induces a moment of M at the fixed end. The carry over factor represents

2

the fraction of M that is carried over from hinge to fixed end. Hence the carry over factor

for the case of far end fixed is + same

direction.

1 . The plus sign indicates both moments are in the

2

Moment distribution method for beams:

Procedure for analysis:

(i) Fixed end moments for each loaded span are determined assuming both ends

fixed.

(ii) The stiffness factors for each span at the joint should be calculated. Using these

values the distribution factors can be determined from equation DF = for a

fixed end = 0 and DF = 1 for an end pin or roller support.

K . DF

K

(iii) Moment distribution process: Assume that all joints at which the moments in the

connecting spans must be determined are initially locked.

Then determine the moment that is needed to put each joint in equilibrium.

Release or unlock the joints and distribute the counterbalancing moments into

connecting span at each joint using distribution factors.

Carry these moments in each span over to its other end by multiplying each moment by

carry over factor.

By repeating this cycle of locking and unlocking the joints, it will be found that the

moment corrections will diminish since the beam tends to achieve its final deflected shape. When

a small enough value for correction is obtained the process of cycling should be stopped with

carry over only to the end supports. Each column of FEMs, distributed moments and carry over

moment should then be added to get the final moments at the joints.

Then superimpose support moment diagram over free BMD (BMD of primary structure)

final BMD for the beam is obtained.

The above process is illustrated in following examples

Ex: 1 Analyse the beam shown in figure 6 (a) by moment distribution method and draw the BMD.

Assume EI is constant

Fig. 6 (a)

Solution:

(i) FEM calculation

MFAB = MFBA = 0 20 x 12

2

MFBC =

12

MFCB = + 240 kNm

240 kNm

MFCB = - 2508

8

= - 250 kNm

FFDC = + 250 kNm

(ii) Calculation of distribution factors:

Jt. Member Relative

stiffness (K) K

DF = K

K

B BA

BC

I/12

I/12

I/6 0.5

0.5

C CB

CD

I/12

I/8

5I/24 0.4

0.6

(iii) The moment distribution is carried out in table below

Jt A B C D

Member AB BA BC CB CD

D.F 0 0.5 0.5 0.4 0.6 0

FEM 0 0 -240 +240 -250 +250

Balance +120 +120 4 6

C.O 60 2 60 3

Balance -1 -1 -24 -36

C.O -0.5 -12 -0.5 -18

Balance +6 +6 0.2 0.3

C.O 3 0.1 3 0.15

Balance -0.05 -0.05 -1.2 -1.8

C.O -0.03 -0.6 -0.03 -0.9

Balance +0.3 +0.3 0.01 0.02

C.O 0.15 0.01

Final

moments

62.62 125.25 -125.25 281.48 -281.48 234.26

After writing FEMs we can see that there is a unbalancing moment of –240 KNm at B & -

10 KNM at Jt.C. Hence in the next step balancing moment of +240 KNM & +10 KNM are applied at

B & C Simultaneously and distributed in the connecting members after multiply with D.F. In the

next step distributed moments are carried over to the far ends. This process is continued until the

resulting moments are diminished an appropriate amount. The final moments are obtained by

summing up all the moment values in each column. Drawing of BMD is shown below in figure 6

(b).

Fig.6(b)

Fig. 7 a

A

B C

10 kN

D 4m

3I 4m

10I

4m 1m 3m 2I

1m E

Ex 2: Analyse the continuous beam shown in fig 7 (a) by moment distribution method and draw

BMD & SFD

3 kN/m 25 kN 16 kN

Fig. 7 (a)

Solution:

FEM: MFAB = -

3x82

3x42

12

25 x 8

4 kNm;

MFBA 4 kNm

3x 82

25 x 8

MFBC = -

12 8

41 kNm MFAB= +

12 8

41kNm

MFDC = 16 x12 x 3

42

= + 3 kNm MDE = -10 x 1 = 10kNm

MFCD = 16 1 32

42

= -9 kNm

DF:

Jt. Member Relative

stiffness (K) K

DF = K

K

B BA 3 x

3I 0.56I

1.81I 0.31

4 4 BC 10I/8 = 1.25I 0.69

C CB 10I/8 = 1.25I 1.63I 0.77 CD 3

x 2I

= 0.38I 0.23

4 4

Note: Since support ‘A’ is simply supported end the relative stiffness value of 3 I has

4 l

been taken and also since ‘D’ can be considered as simply supported with a definite

moment relative stiffness of CD has also been calculated using the formula 3 I .

4 l

Moment distribution table:

Jt A B C D

Member AB BA BC CB CD DC DE

D.F 1 0.31 0.69 0.77 0.23 1 01

FEM -4 4 -41 +41 -9 3 -10

Release of +4 +7

joint A and 2 3.5

adjusting

moment at

‘D’

Initial 0 6 -41 41 -5.5 +10 -10

moments

Balance 10.9 24.1 -27.3 -8.2

C.O -13.7 12.1

Balance 4.2 9.5 -9.3 -2.8

C.O -4.7 4.8

Balance 1.5 3.2 -3.7 -1.1

C.O -1.9 1.6

Balance 0.6 1.3 -1.2 -0.4

C.O -0.6 0.7

Balance 0.2 0.4 -0.5 -0.2

C.O -0.3 0.2

Balance 0.09 0.21 -0.15 -0.05

Final

moments

0 23.49 -23.49 18.25 -18.25 10 -10

FBD of various spans is shown in fig. 7 (b) and 7 (c) and BMD, SFD have been shown in fig. 7 (d)

Fig. 7 (c)

Fig. 7 (b)

Fig. 7 (d)

Ex 3: Analyse the continuos beam as shown in figure 8 (a) by moment distribution method

and draw the B.M. diagrams

Fig. 8 (a)

Support B sinks by 10mm

E = 2 x 105 N/mm², I = 1.2 x 10-4 m4

Solution: Fixed End Moments:

MFAB = FEM due to load

+ FEM due to sinking 2

wl 6EI

12 l2

20 x 62

=

12

6 x 2 x105 x1.2 x10

4 x10

12 x10

6000 2 x10

6

= 60 – 40

MFAB = -100 kNm

MFBA = FEM due to load + FEM due to sinking

= + 60 40

MFBA = +20 kNm

MFBC = FEM due to loading

+ FEM due to sinking

Wab2

=

l2

=

6

EI l2

50 x 3 x 22

=

52

6 x 2 x105 1.2 x10

4 x10

12 x10

50002 x 10

6

= 24 + 57.6

MFBA = + 33.6 kNm

MFCB = + Wa 2 b

l2

2

6EI

l2

= 50 x 3

52

x 2 57.6

MFCB = 93.6kNm

MFCD = due to load only ( C & D are at some level)

MFCD=

wl2

12

20 x 42

12

26.67kNm

MFDC = + 26.67 kNm

Jt. Member Relative

stiffness (K) K

DF = K

K

B BA I/6 0.46 BC I/5 0.36I 0.54

C CB I/5 0.51 CD 3

x I

= 0.19I 0.39I 0.49

4 4

Jt A B C D

Member AB BA BC CB CD DC

D.F 0.46 0.54 0.51 0.49

FEM -100 +20 +33.6 +93.6 -26.67 +26.67

Release jt. -26.67

‘D’

CO -13.34

Initial -100 +20 +33.6 +93.6 -40.01 0

moments

Balance -24.66 -28.94 -27.33 -26.26

C.O -12.33 -13.67 --14.47

Balance +6.29 +7.38 +7.38 +7.09

C.O +3.15 +3.69 +3.69

Balance -1.7 -1.99 -1.88 -1.81

C.O -0.85 -0.94 -1

Balance +0.43 +0.51 +0.51 +0.49

C.O +0.22 +0.26 +0.26

Balance -0.12 -0.14 -0.13 -0.13

C.O -0.06

Final

moments

-109.87 +0.24 -0.24 +60.63 -60.63

Bending moment diagram is shown in fig. 8 (b)

109.87

60.63

0.24

20x6² / 8 = 90

KNM

50x3x2/5 = 60 KNM

Fig. 8 (b)

20x4²/8 = 40KNM

BMD

15. University Question papers of previous year

MODEL QUESTION PAPER

Time: 3 hours Max. Marks: 75

All questions carry equal marks

PART-A

I. Answer all the following (5x1=5)

1. Define structural Analysis?

2. What are the equations of Equilibrium?

3. Define carry over moment?

4. Write the equation of a perfect frame?

5. Define ILD?

II. Answer all the following (10x2=20)

1. Write down the difference between Method of joints and method of sections?

2. What is the difference between 3 hinged arch and 2 hinged arch?

3. What do you mean by rib shortening of arches and write down the formulae for deflection of

Arch due to rib shortening?

4. Write down the expression for strain Energy due to GVL on beam?

5. Write down the expression for deflection of propped cantilever beam carrying UDL on its

entire span?

6. Write down the Step by step procedure determining the bending moment for fixed beam?

7. What is the effect of sinking of a continuous beam and write down the expression for sinking

support?

8. Write the expression for slope deflection method?

9. Write the expression for maximum reaction for rolling loads spaced at equal distance?

10. Write the expression for absolute bending moment for rolling loads spaced equally?

PART-B

III. Answer the following questions (5x10=50)

1. Fig shows an inclined truss loaded as shown in fig. determine the

forces in the members of the truss by the method of joints.

2

OR

2. Using the method of tension coefficients Analyze the cantilever plane truss shown in fig. and find

the member forces.

3 .a. Find the deflection at the free end of a cantilever carrying a

concentrated load at the free end using strain energy method.

b. A simply supported beam carries a point load P eccentrically on the

span. Find the deflection under the load sing strain energy method.

OR

4. A symmetrical three hinged arch has a span of 20 meters and rise to the central hinge of 5 m. It

carries a vertical load of 10 kN at 4 m from the left support. Calculate the reactions at the supports and

bending moment at the load point.

5. A beam AB of uniform section and 6 m span is built at the ends. A u.d.l of 30 kN/m runs over left

half of the span and there is an additional concentrated load of 40 kN at right quarter. Determine the

fixed end moments at the ends and the reaction. Draw BMD and SFD.

OR

6. a) What is a propped cantilever? b) Determine the deflection at point ‘C’ in a propped cantilever shown in Figure?

3

90 kN

A

B

C

1 m 2 m

7. Analyze the continuous beam shown in figure by Clapeyron’s theorem of three moments. Draw

BMD and SFD.

OR

8..A continuous beam ABC covers two consecutive spans AB and BC of length 6 m and 8 m, carrying

loads of 10 kN/m and 15 kN/m respectively. If the ends A and B are simply supported, find the

support reactions at A, B and C. Use slope deflection method. Draw the shear force and bending

moment diagram. Draw elastic curve.

9. Two loads of 200 kN and 250 kN spaced at 5 meters apart crosses a girder of 25 meters span from

left to right with 200 Kn leading. Construct the maximum shearing force and bending moment

diagrams stating the absolute maximum values.

OR

10. A train of wheel loads crosses a span of 30 meters shown in figure. Calculate the maximum

positive and negative shear at midspan and absolute maximum bending moment anywhere in the span.

4

16. Question Bank

UNIT-I

1.Determine the axial force in the members of the frame the cross sectional area of the bars

AB and AC is 2a and that of other member is a.

2.Determine the forces in all the members of the redundant pin jointed frame shown in

Figure. The area of the cross section of the diagonals is twice that the other members.

3.A truss ABCD has both its ends A and D are provided with hinged supports and carries

two loads of 35kN and 60kN at B and C as shown in gure 4. Treat BC as redundant.

Calculate the forces in all the members. All the members have the same cross sectional area and are made of same material.

5

4. Determine the force in the member AB of the pin jointed frame work shown in

figure . All the members have the same area of the cross section and are of the same material.

UNIT-II

1. a) What is a propped cantilever?

b) Determine the prop reaction of the propped cantilever shown in fig.(1).? Also

Draw B.M.D

6

2. Determine the prop reaction RB the propped cantilever shown in fig . And also

Draw the S.F and B.M diagrams

3. The fixed beam AB of length 10m carries point loads of 150 and 160 KN at a

distance of 4 m and 6m from left end A. find the end moments and the reactions at

supports. Also draw SFD and BMD.

4. The propped cantilever shown in fig (3) .find the value of reaction. Also draw SFD

and BMD.

Unit-IV

1. Analyse the continuous beam shown in fig.(6) using moment distribution method.

Draw SFD and BMD.

7

2. Analyse the continuous beam shown in fig.(7) using moment distribution method.

Draw BM diagram.

3. Analyse the continuous beam shown in fig.(8) using moment distribution method.

Draw BM diagram.

4. Analyse the continuous beam shown in fig.(9) using moment distribution method.

Draw BM diagram.

Continuous beams

1. Derive the Clapeyron’s equation of three moments.

8

2. A Continuous beam ABC covers two consecutive span AB and BC of lengths 4m and 6m, carrying

uniformly distributed loads of 6KN/m and 10KN/m respectively. If the ends A and B are simply

supported, find the moments at A,B And C. Draw also B.M And S.F Diagrams.

3. A Continuous beam ABCD, Simply Supported a t A,B,C And D is Loaded ss shown In Fig.(10)

Find the moments over the beam and draw B.M And S.F Diagrams.

4. A Continuous beam ABC of uniform section, with span AB and BC as 6m each, is fixed at A and C

and supported at B as shown in fig.(11) Find the support moments and the reactions. Draw S.F And

B.M Diagrams.

UNIT-V

1.a) Define absolute maximum shear force.

b)Two point loads of 150kN and 300kN with 4m space between them rolls across the girder of span

20m. Calculate the equivalent UDL. 2.a) Define equivalent UDL.

b)Construct the influence line for bending moment at section of 2.5m from left support of a simple

beam of span of 6m. Determine the maximum bending moment when a UDL of 10 kN/m longer than

the span moves across the beam. 3.a) Define influence lines. b)Determine absolute maximum left and right reactions for a simple beam 15m span with a series of loads shown in Figure 6.

9

50 kN

100 kN 70 kN 200 kN

3 m 2m 4m

(1)

(2) (3) (4)

4. a)Define absolute maximum bending moment. b)Two concentrated loads of 75kN and 150kN separated by a distance of 3.5m between them rolls across a beam of 12m from left to right with 75kN load leading the train. Calculate the maximum negative shear force and maximum bending moment at a mid – span of the beam.

17. Assignment topics 1. Method of joints,sections,Tension cofficent

2. Propped cantilever analysis

3.Fixed beams and contionous beams analysis

4.Moment distribution method

5.Influence line diagram

18. Objective questions Unit-1

Multiple choice questions

1.The number of independent equations to be satisfied for static equilibrium of a plane structure is

a) 1 b) 2 c) 3 d) 6

2.Degree of static indeterminacy of a rigid-jointed plane frame having 15 members, 3 reaction components and

14 joints is

a) 2 b) 3 c) 6 d) 8

3. Independent displacement components at each joint of a rigid-jointed plane frame are

a) three linear movements

b) two linear movements and one rotation

c) one linear movement and two rotations

d) three rotations

4.If in a pin-jointed plane frame (m + r) > 2j, then the frame is…………… where m is number of members, r is

reaction components and j is number of joints

10

a) stable and statically determinate

b) stable and statically indeterminate

c) unstable

d) none of the above

5.A pin-jointed plane frame is unstable if

a) (m + r) < 2j

b) m + r = 2j

c) (m + r) > 2j

d) none of the above

6.where m is number of members, r is reaction components and j is number of joints

If in a rigid-jointed space frame, (6m + r) < 6j, then the frame is

a) unstable

b) stable and statically determinate

c) stable and statically indeterminate

d) none of the above

7. The deflection at any point of a perfect frame can be obtained by applying a unit load at the joint in

a) vertical direction

b) horizontal direction

c) inclined direction

d) the direction in which the deflection is required

8. which of the flowing is a perfect frame

a. triangle b. rectangle c. square d. trapiozdal

9. The degree of static indeterminacy of a rigid-jointed space frame is

a) m + r - 2j b) m + r - 3j

c) 3m + r - 3j d) 6m + r - 6j

where m, r and j have their usual meanings

10.The degree of static indeterminacy of a pin-jointed space frame is given by

11

a) m + r - 2j b) m + r - 3j

c) 3m + r - 3j d) m + r + 3j

where m is number of unknown member forces, r is unknown reaction components and j is number of joints

fill in the blanks

1. Triangle is a _____________________

2. Perfect frame can be analysed by _______ reactions

3. Equliburim condition means ____________________

4. Perfect frames can be solved by ________________________ methods

5. In a truss bottom chord members undergo__________________

6. In a truss top chord members under go _____________________

7. If a trusss force along x direction is _____- along Y direction is __________

8. Static determinacy means __________________________

9. Indeterminacy in a truss is given by ________________________

10. In Method of section moment is taken along an ______________

Unit-2

1.The principle of virtual work can be applied to elastic system by considering the virtual work of

a) internal forces only b) external forces only

c) internal as well as external forces d) none of the above

2. Castigliano's first theorem is applicable

a) for statically determinate structures only

b) when the system behaves elastically

c) only when principle of superposition is valid

d) none of the above

3.Principle of superposition is applicable when

a) deflections are linear functions of applied forces

b) material obeys Hooke's law

c) the action of applied forces will be affected by small deformations of the structure

d) none of the above

4. The Castigliano's second theorem can be used to compute deflections

a) in statically determinate structures only

b) for any type of structure

c) at the point under the load only

12

d) for beams and frames only

5.The principle of virtual work can be applied to elastic system by considering the virtual work of

a) internal forces only b) external forces only

c) internal as well as external forces d) none of the above

6. Normal trust=

a.H cos θ – V sin θ

b. H cos θ + V sin θ

c. H cos θ * V sin θ

d. H cos θ /V sin θ

7.Radial Shear=

a.H cos θ – V sin θ

b. H cos θ + V sin θ

c. H cos θ * V sin θ

d. H cos θ /V sin θ.

8.In a 3 hinged arch Y=

a. y = (4h/L2 ) (x (L – x))

b. y = (4h/L2 ) /(x (L – x))

c. y = (4h/L2 ) +(x (L – x))

d. y = (4h/L2 ) (x (L +x))

9. In a 3 hinged arch tan θ.=

a. y = (4h/L2 ) (L –2 x))

b. y = (4h/L2 )- (L –2 x))

c. y = (4h/L2 ) /(L –2 x))

d. y = (4h/L2 )+ (L –2 x))

10.strain energy in a beam is given by

a.PL/AE

13

B.PL/2AE

C.PL/3AE

D.PL/4AE

FILL IN THE BLANKS

1.An arch is a _________________________

2. Depending upon number of hinges the arches are classified in to __________

3. A 3 hinged is a ______________ determinate

4. A three hinged parabolic arch of 20 m span and 4 m central rise carries a point load of 40 kN at 4 m

horizontally from left support the vertical reactions are ______________

5. for a 3 hinged arch the ordinate y w.r.t X at any point of arch given by_______________________

6.Three-hinged arch is statically determinate structure and its reactions / internal forces are evaluated by

_________________________________

7. In a 3 Hinged arch the hinge at top of its rise is called as ___________________

8.A 3 hinged arch carrying UDl on its entire span than its horizontal trust is given by _____________

9.A 3 hinged arch carrying point at a distance a from its left end then horizontal trust is given by

______________

10. A 3 hinged arch carrying udl on its half of its span then horizontal trust is given by __________

1. A built-in beam AB of span L is loaded with a gradually varying load from zero at A to W per unit

length at B. The fixing moment A will be [ ]

A.WL2

/8 B. WL2

/12 C. WL2

/20 D. WL2

/30

2. When a continuous beam is fixed at the left end, then an imaginary span is taken to the left of the beam.

The support moment at the imaginary support is [ ]

A. Negligible B. Considerable C. Zero D. Infinity

3. If a continuous beam is fixed at its both ends, then the imaginary support is [ ]

A. not taken B. taken on left side only

C. taken on both the ends D. taken on right side only

4. The analysis of a structure by Slope Deflection method is ___________ method of analysis[ ]

A. Force method B. Displacement Method

C. Statically In-determinate D. Statically determinate

5. If the end B of a fixed beam AB settles by an amount of ‘Δ’ then the vertical reaction at the fixed end B

is [ ]

A. 4EIΔ /L2

B. 2EIΔ /L2

C. 6EIΔ /L2

D. 12EIΔ /L3

6. A fixed beam of span L is carrying a u.d.l of W kNper unit length. The Maximum deflection at the

center of the span is _______

7. The analysis of a beam by Three moment equation method is ___________ method of analysis

14

8. A 2 span continuous beam ABC is simply supported at the ends A and C and is carrying a u.d.l of W

kN/m over the entire 2 spans. If the length of the each span is L/2, the vertical reaction at the

support B is _________

9. If the end B of a fixed beam AB rotates by an amount of ‘α’ then the fixed end moment at the end B is

_________

10. If the end B of a fixed beam AB rotates by an amount of ‘α’ then the vertical reaction at the end A is

_________

.

Unit-IV

1. The analysis of a structure by Slope Deflection method is ___________ method of analysis [ ]

A. Force method B. Displacement Method

C. Statically In-determinate D. Statically determinate

2. If the end B of a fixed beam AB settles by an amount of ‘Δ’ then the vertical reaction at the fixed end B

is [ ]

A. 4EIΔ /L2

B. 2EIΔ /L2

C. 6EIΔ /L2

D. 12EIΔ /L3

3. Relative stiffness for a beam when the far end is fixed is [ ]

A.3EI/L B. 4EI/4L C. 3EI/4L D. 4EI/3L

4.The slope deflection equation is give the relationship between [ ]

a) slope and deflection only b)B.M and rotation only

c) B.M and vertical deflection only d)B.M rotation and deflections

5.For the fixed beam as shown in below figure, what is the fixed end moment at A for the given loading?

[ ]

a) 22WabCosLθ b) 22WbaCosLθ c) 22WabSinLθ d) 22WbaSinLθ

6. If the end B of a fixed beam AB rotates by an amount of ‘α’ then the fixed end moment at the end B is

_________

7. If the end B of a fixed beam AB rotates by an amount of ‘α’ then the vertical reaction at the end A is

_________

8. If the end B of a fixed beam AB settles by an amount of ‘Δ’ then the vertical reaction at the fixed end A

is _______

9.The number of simultaneous equations to be solved in the slope deflection method is equal

to=_______

15

10.A two span continuous beam ABC with end ‘A’ fixed and end ‘c’ hinged is having AB=4m, BC=6m. IAB

: IBC

=1:2, it is subjected to u.d.l of 10 KN/m over entire right span. Then the moment at ‘C’ is

=______

1. Relative stiffness for a beam when the far end is fixed is [ ]

A.3EI/L B. 4EI/4L C. 3EI/4L D. 4EI/3L

2. For a continuous beam ABCD, if the distribution factors in the members BA and BC are 0.4 and 0.6 and

if a moment of 25 kN-m acts at joint B. Then the moment in member BC is [ ]

A. 60 B. 50 C. 25 D. 15

3. For a continuous beam ABCD, if the distribution factors in the member CB is 6/13 then the distribution

factor in the member CD is [ ]

A.6/13 B. 7/13 C. 13/6 D. 13/7

4. The strain energy stored in a member due to axial load P is given as [ ]

A. P2

L/2AE B. 2AE / P2

L C. P2

L/AE D. AE / P2

L

5. When a continuous beam is fixed at the left end, then an imaginary span is taken to the left of the beam.

The support moment at the imaginary support is [ ]

A. Negligible B. Considerable C. Zero D. Infinity

6. If k is relative stiffness of a member and Σk is total stiffness of a joint. Then the distribution factor in

any member is given by _________

7. A built-in beam AB of span L is loaded with a gradually varying load from zero at A to w per unit

length at B. The fixing moment B will be __________

8. If U is the total energy stored in a structure and if P is the load acting on the structure, Then the

deflection under the load is given as________

9. The strain Energy method of structural analysis is a __________ method of analysis

10. The strain energy stored in a cantilever beam of span L meter and is subjected to a point load P at the

free end is _________

1. A cantilever of span L is fixed at A and propped at the other end B, If it is carrying a u.d.l of W kN/m,

then the prop reaction will be [ ]

A.3WL/8 B.5WL/8 C.3WL/16 D.5WL/16

2. The deflection at the center of a propped cantilever of span l carrying a u.d.l of W per unit length is

A. WL4

/48EI B. WL4

/96EI C. WL4

/128EI D. WL4

/192EI [ ]

3. A simply supported beam of span L is carrying a u.d.l of W per unit length. If the beam is propped at the

mid point, then the prop reaction is equal to [ ]

A.3WL/8 B.5WL/8 C.3WL2

/16 D.5WL2

/16

4. The degree of static indeterminacy for a Fixed beam is [ ]

A. 1 B. 2 C. 3 D. 4

5. The fixed end moments for a fixed beam carrying u.d.l over the span is [ ]

6. A uniform beam of span ‘l’ is rigidly fixed at both supports. If carries a u.d.l ‘w’ per unit length. The

bending moment at mid span is =_____________.

7. The elastic strain energy stored in a rectangular cantilever beam of length L subjected to a bending

moment M applied at the end is =________________.

8. In an intermediate structure, when there is no lack of fit, the partial derivative of strain energy with

respect to any of the redundant =______________.

16

9. If the strain energy absorbed in a cantilever beam in bending under its own weight is ‘K’ times greater

than the strain energy absorbed in an identical simply supported beam in bending under its own

weight, then the magnitude of ‘K’=______________.

10. Strain energy in linear elastic system (U) due to axial loading =_____________________.

20. Known gaps ,if any --NONE--

21. References, Journals, websites and E-links

TEXT BOOKS:

1. Analysis of Structures-Vol I & Vol II by V.N. Vazirani & M.M.Ratwani, Khanna Publications, New Delhi.

2. Analysis of Structures by T.S. Thandavamoorthy, Oxford University Press, New Delhi

3. Structural Analysis by S S Bhavikatti – Vikas Publishing House.

REFERENCES:

1. Mechanics of Structures by S.B.Junnarkar, Charotar Publishing House, Anand, Gujrat

2. Theory of Structures by Pandit & Gupta; Tata Mc.Graw – Hill Publishing Co.Ltd., New Delhi.

3. Strength of Materials and Mechanics of Structures- by B.C.Punmia, Khanna Publications, New Delhi.

Journals

1. The CSIR-Structural Engineering Research Centre, "Journal of Structural Engineering"

2. International Journal of Structural Engineering

3. The International Journal of Advanced Structural Engineering (IJASE)

Websites

1.www.sefindia.org

2. http://elearning.vtu.ac.in/CV42.html

22. Quality Control Sheets

EVALUATION SCHEME:

PARTICULAR WEIGHTAGE MARKS

End Examinations 75% 75

Two Sessionals 20% 20

Assignment 5% 5

TEACHER'S ASSESSMENT(TA)* WEIGHTAGE MARKS

*TA will be based on the Assignments given, Unit test Performances and Attendance in the class for a

particular student.

17

23. Student List II-A Section

S.No Roll No Student Name

1 14R11A0102 ATHIREK SINGH JADHAV

2 14R11A0103 BODAPATI ARVIND RAJ

3 14R11A0104 BODHASU MADHU

4 14R11A0105 BOLAGANTI YASHWANTH TEJA

5 14R11A0106 CHADA SHIVASAI REDDY

6 14R11A0107 D SATISH KUMAR

7 14R11A0108 E TEJASRI

8 14R11A0109 G DARSHAN

9 14R11A0110 GALIPELLI SRIKANTH

10 14R11A0111 GATTU MANASA

11 14R11A0112 GEEDI SRINIVAS

12 14R11A0113 GUNTUPALLY MANOJ KUMAR

13 14R11A0114 K ANJALI

14 14R11A0115 KASULA HIMA BINDU

15 14R11A0116 KASTHURI VINAY KUMAR

16 14R11A0117 KOPPULA KEERTHIKA

17 14R11A0118 KRISHNA VAMSHI TIPPARAJU

18 14R11A0119 MADDULA MANORAMA REDDY

19 14R11A0120 MALINENI VENKATA DILIP

20 14R11A0121 MANDA KUMIDINI

21 14R11A0122 MINNIKANTI NAGASAI GANESH BABU

22 14R11A0123 MOHD ABDUL WALI KHAN

23 14R11A0124 MOTUPALLI VENTAKA KIRAN

24 14R11A0125 MUDDETI HARI

25 14R11A0126 MUSHKE VAMSHIDAR REDDY

26 14R11A0127 NAGUNOORI PRANAY KUMAR

27 14R11A0128 NALLA UDHAY KUMAR REDDY

28 14R11A0129 P GAYATHRI

29 14R11A0130 PADALA SRIKANTH

30 14R11A0131 PASUPULATI SWETHA

31 14R11A0132 POLISETTY VINEEL BHARGAV

32 14R11A0133 PUNYAPU VENKATA SHRAVANI

33 14R11A0134 R DIVYA

34 14R11A0136 RAVULA VAMSHI

35 14R11A0138 S BARATH KUMAR

36 14R11A0139 S PRASHANTH REDDY

37 14R11A0140 S SAI RAGHAV

38 14R11A0141 SHAIK SHAMEERA

39 14R11A0142 SREEGAADHI SAICHARAN

40 14R11A0143 SRIRAM SURYA

41 14R11A0144 SUNKARI SHIVA

42 14R11A0145 VANAMALA SURENDER NIKITHA

43 14R11A0146 YADAVALLI PAVAN KUMAR

II-B-section

S. No Roll No Student Name

1 14R11A0149 A. SRAVAN KUMAR

2 14R11A0150 B MAHENDRA VARDHAN

3 14R11A0151

B. VIJAY

4 14R11A0152

B. KIRAN KUMAR

5 14R11A0153

B. SUNIL NAIK

18

6 14R11A0154

D. VENU CHARY

7 14R11A0155

D. VASANTHA KUMAR

8 14R11A0157

G. NIKHIL

9 14R11A0158

G. SANDEEP KUMAR

10 14R11A0159

G. CHARAN KUMAR

11 14R11A0160

J. HARISH KUMAR

12 14R11A0161

K.J. NANDEESHWAR

13 14R11A0162

K. SANTHOSH KUMAR

14 14R11A0163

K BHARATH KUMAR

15 14R11A0164

K ABHILASH

16 14R11A0165

K SAI KRISHNA

17 14R11A0168

MOHD. ABBAS

18 14R11A0169

M SRINIVAS

19 14R11A0170

N SANTHOSH

20 14R11A0172

OSA NITHISH

21 14R11A0173

P INDRA TEJA

22 14R11A0174

P NAVEEN KUMAR

23 14R11A0175

P BHARATH NARSIMHA REDDY

24 14R11A0176

P SURENDER

25 14R11A0177

R VIHARI PRAKASH

26 14R11A0178

S BHANU KISHORE

27 14R11A0179

SHAILESH KUMAR SINGH

28 14R11A0180

SYED OMER ASHRAF

29 14R11A0181

V SAI SHARATH

30 14R11A0182

Y VENKATA MOHAN REDDY

24. Group-Wise students list for discussion topics

II-A Section

S. No Group No Roll No Student Name

1 1 14R11A0102 ATHIREK SINGH JADHAV

2 1 14R11A0103 BODAPATI ARVIND RAJ

3 1 14R11A0104 BODHASU MADHU

4 1 14R11A0105 BOLAGANTI YASHWANTH TEJA

5 1 14R11A0106 CHADA SHIVASAI REDDY

6 1 14R11A0107 D SATISH KUMAR

19

7 2 14R11A0108 E TEJASRI

8 2 14R11A0109 G DARSHAN

9 2 14R11A0110 GALIPELLI SRIKANTH

10 2 14R11A0111 GATTU MANASA

11 2 14R11A0112 GEEDI SRINIVAS

12 2 14R11A0113 GUNTUPALLY MANOJ KUMAR

13 3 14R11A0114 K ANJALI

14 3 14R11A0115 KASULA HIMA BINDU

15 3 14R11A0116 KASTHURI VINAY KUMAR

16 3 14R11A0117 KOPPULA KEERTHIKA

17 3 14R11A0118 KRISHNA VAMSHI TIPPARAJU

18 3 14R11A0119 MADDULA MANORAMA REDDY

19 4 14R11A0120 MALINENI VENKATA DILIP

20 4 14R11A0121 MANDA KUMIDINI

21 4 14R11A0122 MINNIKANTI NAGASAI GANESH BABU

22 4 14R11A0123 MOHD ABDUL WALI KHAN

23 4 14R11A0124 MOTUPALLI VENTAKA KIRAN

24 4 14R11A0125 MUDDETI HARI

25 5 14R11A0126 MUSHKE VAMSHIDAR REDDY

26 5 14R11A0127 NAGUNOORI PRANAY KUMAR

27 5 14R11A0128 NALLA UDHAY KUMAR REDDY

28 5 14R11A0129 P GAYATHRI

29 5 14R11A0130 PADALA SRIKANTH

30 5 14R11A0131 PASUPULATI SWETHA

31 6 14R11A0132 POLISETTY VINEEL BHARGAV

32 6 14R11A0133 PUNYAPU VENKATA SHRAVANI

33 6 14R11A0134 R DIVYA

34 6 14R11A0136 RAVULA VAMSHI

35 6 14R11A0138 S BARATH KUMAR

36 7 14R11A0139 S PRASHANTH REDDY

37 7 14R11A0140 S SAI RAGHAV

38 7 14R11A0141 SHAIK SHAMEERA

39 7 14R11A0142 SREEGAADHI SAICHARAN

40 7 14R11A0143 SRIRAM SURYA

41 8 14R11A0144 SUNKARI SHIVA

42 8 14R11A0145 VANAMALA SURENDER NIKITHA

43 8 14R11A0146 YADAVALLI PAVAN KUMAR

II-B Section

S. No Group No Roll No Student Name

1 1 14R11A0149 A. SRAVAN KUMAR

2 1 14R11A0150 B MAHENDRA VARDHAN

3 1 14R11A0151

B. VIJAY

4 1 14R11A0152

B. KIRAN KUMAR

5 1 14R11A0153

B. SUNIL NAIK

6 1 14R11A0154

D. VENU CHARY

7 2 14R11A0155

D. VASANTHA KUMAR

8 2 14R11A0157

G. NIKHIL

9 2 14R11A0158

G. SANDEEP KUMAR

10 2 14R11A0159

G. CHARAN KUMAR

20

11 2 14R11A0160

J. HARISH KUMAR

12 2 14R11A0161

K.J. NANDEESHWAR

13 3 14R11A0162

K. SANTHOSH KUMAR

14 3 14R11A0163

K BHARATH KUMAR

15 3 14R11A0164

K ABHILASH

16 3 14R11A0165

K SAI KRISHNA

17 3 14R11A0168

MOHD. ABBAS

18 3 14R11A0169

M SRINIVAS

19 4 14R11A0170

N SANTHOSH

20 4 14R11A0172

OSA NITHISH

21 4 14R11A0173

P INDRA TEJA

22 4 14R11A0174

P NAVEEN KUMAR

23 4 14R11A0175

P BHARATH NARSIMHA REDDY

24 4 14R11A0176

P SURENDER

25 5 14R11A0177

R VIHARI PRAKASH

26 5 14R11A0178

S BHANU KISHORE

27 5 14R11A0179

SHAILESH KUMAR SINGH

28 5 14R11A0180

SYED OMER ASHRAF

29 5 14R11A0181

V SAI SHARATH

30 5 14R11A0182

Y VENKATA MOHAN REDDY