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
Unit Sl.No Topic No
ANALYSIS OF PERFECT FRAMES: Types of frames- perfect, imperfect and redundant pin jointed frames.. Analysis of determinate pin jointed frames using 1 1 method of joints, method of sections and tension coefficient method for vertical loads horizontal and inclined loads. ENERGY THEOREMS: Introduction-Strain energy in linear elastic system, 2 2 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. PROPPED CANTILEVERS: Analysis of propped cantilevers-shear force and 3 3 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. CONTINUOUS BEAMS : Introduction-Clapeyron‘s theorem of three moments- 4 4 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. 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 5 5 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 B X X X X X X X X X Objectives (PEO) 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
9.30- 10.20- 11.10- 12.00- 12.50- 1.30- Time 2.20-3.10 3.10-4.00 10.20 11.10 12.00 12.50 1.30 2.20 Period 1 2 3 4 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 LUNCH SOM LIBRARY Thursday EVS SA HHM P&S BMC EVS
Friday LAB P&S CRT EVS
Saturday EVS LAB MENTOR SEMINAR
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
10.20- 12.00- 2.20- Time 9.30-10.20 11.10-12.00 12.50-1.30 1.30-2.20 3.10-4.00 11.10 12.50 3.10
Period 1 2 3 4 5 6 7
CH Monday LUN S.A
Tuesday S.A Wednesday SA Thursday SA Friday Saturday
13. Unit wise Summary
Unit Total Reg/additio Lcd/ohp/ Topic Remark no periods nal bb
12 Analysis of perfect frames: types of frames Regular BB
2 Perfect, imperfect and redundant pin jointed frames Regular BB
Analysis of determinate pin jointed frames using 3 Regular BB method of joints 1 3 Method of sections Regular BB
Tension coefficient method for vertical loads horizontal 2 Regular BB loads
2 Tension coefficient method for inclined loads. Regular BB
20 Energy theorems: introduction Regular BB
2 Strain energy in linear elastic system Regular BB
Strain energy due to axial load, bending moment and 4 Regular BB shear forces
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 2 Comparison between three hinged and two hinged 1 Regular BB arches. Linear arch.
Normal thrust and radial shear in an arch geometrical 3 Regular BB properties of parabolic and circular arch
3 Three hinged circular arch at different levels Regular BB
Absolute maximum bending moment diagram for a 3 three hinged arch. Regular BB
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
Fixed beams – introduction to statically indeterminate 3 Regular BB beams 3 U.d.load central point load, eccentric point load. Number of point loads, uniformly varying load, couple 3 Regular BB and combination of loads shear force and bending moment diagrams-
3 Deflection of fixed beams effect of sinking of Regular BB support, effect of rotation of a support. Continuous beams : introduction-clapeyron‘s theorem 18 Regular BB of three moments
Analysis of continuous beams with constant moment 3 of inertia with one or both ends fixed-continuous Regular BB beams
Overhang, continuous beams with different moment 3 Regular BB of inertia for different spans
3 Effects of sinking of supports-shear force and bending Regular BB 4 moment diagrams Derivation of slope deflection equation, application to 3 continuous beams with and without settlement of Regular BB supports
Analysis of continuous beams with and without 3 settlement of supports using moment distribution Regular BB method.
Shear force and bending moment diagrams, elastic 3 Regular BB curve.
Moving loads : introduction maximum sf and bm at a 15 Regular BB given section and absolute maximum s.f. and b.m
5 Maximum sf and bm at a given section and absolute maximum s.f. and b.m due to single concentrated load 3 Regular BB 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 3 Regular BB distributed load-focal length.
Influence lines: definition of influence line for sf, 3 influence line for bm- load position for maximum sf at Regular BB a section-
Load position for maximum bm at a section single 3 point load, u.d.load longer than the span, u.d.load Regular BB shorter than the span
Influence lines for forces in members of pratt and 3 Regular BB warren trusses.
Micro Plan with dates and closure report
Unit Total Topic Reg/additional Lcd/ohp/bb Remark no periods Date
Analysis of perfect frames: types 12 Regular BB of frames
Perfect, imperfect and redundant 2 Regular BB pin jointed frames
Analysis of determinate pin jointed 3 Regular BB 1 frames using method of joints 3 Method of sections Regular BB
Tension coefficient method for 2 Regular BB vertical loads horizontal loads
2 Tension coefficient method for Regular BB inclined loads. 20 Energy theorems: introduction Regular BB
Strain energy in linear elastic 2 Regular BB 2 system
Strain energy due to axial load, 4 Regular BB bending moment and shear forces
2 Castigliano‘s first theorem Regular BB
Deflections of simple beams and 4 Regular BB pin jointed trusses
Three hinged arches: 1 Regular BB introduction, types of arches
Comparison between three hinged 1 and two hinged arches. Linear Regular BB arch.
Normal thrust and radial shear in 3 an arch geometrical properties of Regular BB parabolic and circular arch
Three hinged circular arch at 3 Regular BB different levels
Absolute maximum bending moment diagram for a three 3 hinged arch. Regular BB
Propped cantilevers: analysis of 15 Regular BB propped cantilevers
Shear force and bending moment 3 Regular BB diagrams-
3 Deflection of propped Regular BB cantilevers. Fixed beams – introduction to 3 Regular BB statically indeterminate beams 3 U.d.load central point load, eccentric point load. Number of point loads, uniformly varying 3 Regular BB load, couple and combination of loads shear force and bending moment diagrams-
Deflection of fixed beams effect 3 of sinking of support, effect of Regular BB rotation of a support. Continuous beams : introduction- 4 18 clapeyron‘s theorem of three Regular BB moments
Analysis of continuous beams with constant moment of inertia with 3 Regular BB one or both ends fixed-continuous beams
Overhang, continuous beams with 3 different moment of inertia for Regular BB different spans
Effects of sinking of supports- 3 shear force and bending moment Regular BB diagrams
Derivation of slope deflection equation, application to 3 Regular BB continuous beams with and without settlement of supports
Analysis of continuous beams with and without settlement of 3 Regular BB supports using moment distribution method.
Shear force and bending moment 3 Regular BB diagrams, elastic curve.
Moving loads : introduction maximum sf and bm at a given 15 Regular BB section and absolute maximum s.f. and b.m
Maximum sf and bm at a given section and absolute maximum s.f. and b.m due to single 3 Regular BB concentrated load u.d load longer than the span, u.d load shorter 5 than the span
Two point loads with fixed distance between them and several point loads-equivalent 3 uniformly distributed load-focal Regular BB length.
3 Influence lines: definition of Regular BB
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, 3 Regular BB u.d.load longer than the span, u.d.load shorter than the span
Influence lines for forces in 3 members of pratt and warren Regular BB trusses.
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 =0
End B is Hinged hence B ≠0
Assume both ends are fixed and therefore fixed end moments are 2 2 wL wL FAB , FBA 12 12 The Slope deflection equations for final moment at each end are 2EI MAB FAB 2 B L A 2
wL 2EI (1) B 12 L 2EI M F 2
BA BA L B A
(2) 2 4EI wL B 12 L
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. 2 4EI wL 0 From equation (2) M
BA 12 L B 3
EIB wL - ve indicates the rotation is anticlockwise sign 48
Substituting the value of EIB in equation (1) and (2) we have end moments 2 3 2 wL wL M 2 wL - ve indicates moment is anticlockwise sign
AB 12 L 2 48 8 3 wL M 4L 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 L RA L MAB wL 2 2 L 5 wL wL wL 8 2 8 5 wL R A 8
V 0
RA RB wL
5 RB wL RA wL wL 3 8 wL 8
Problem can be treated as
The bending moment diagram for the given problem is as below
The max BM occurs where SF=0. Consider SF equation at a distance of x from right support
3 wL wX 0 SX 8 3 X L 8 3 Hence the max BM occurs at L from support B 8 2 3 3 w 3 Mmax MX wL L L 8 8 2 8
2 9 wL 128
And point of contra flexure occurs where BM=0, Consider BM equation at a distance of x from right support. 3 X M wLX w 02 X 8 2 3 X L 4
For shear force diagram, consider SF equation from B 3 S wL wX X 8 3 SX 0 SB wL 8 S L S 5 X A 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:
2 2 FAB Wab 100 4 2 L2 62 44.44KNM 2 2 2 Wa b 100 4 88.89KNM F BA L2 62
2 wL 2 20 5 41.67KNM 12 12 FBC 2 2 wL 20 5 F CB 12 41.67KNM 12
Since A is fixed A 0 , B 0, C 0,
Slope deflection equations are:
2EI M FAB AB 2 B L A 2EI 44.44
6 B 1 44.44 EI (1) M F
3 B 2EI 2
BA BA L B A
2EI 2B 88.89 6 2 88.89 EI (2)
3 B
2EI M F 2
BC BC L B C 2EI 41.67 2
5 B C 4 2 41.67 EI EI (3)
B C 5 5
MCB 2EI B FCB 2 L C 2EI 41.67 2 B 5 C 4EI 2 41.67 EI (4)
5 C 5 B In all the above four equations there are only two unknown B and C . And accordingly the boundary conditions are
i -MBA-MBC=0
MBA+MBC=0 ii MCB=0 since C is end simply support. 2 4 2 88.89 EI 41.67 EI EI Now M M BA BC 3 B 5 B 5 C 22 2 47.22 EI EI 0 (5) B C 15 5 2 4 M 41.67 EI 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.
Substituting in the slope definition equations
MAB = – 44.44 + 1 20.83 51.38 KNM 3
MBA = + 88.89 + 2 20.83 75.00 KNM 3
4 2 M = – 41.67+ 20.83 41.67 75.00 KNM BC 5 5 2 4 M = + 41.67+ 20.83 41.67 0 CB 5 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 4 46.20 KNM
6 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
35 Sx = 35-20x = 0 x =1.75m Max BM occurs at 1.75m from C 20
2 1.75 Mmax = 35 × 1.75 – 20 = 30.625 KNM 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
Wab 2 100 4 2 2 FAB FEMS L2 62 - 44.44 KN M 2 2
Wa b 100 4 2 88.88 KNM F BA 2 2 L 6 2 2
FBC 20 5 - 41.67 KNM wL 12 FCB 12 2 2 41.67 wL 20 5 KNM 12 12
FCD 20 1.5 - 30 KN M
Slope deflection equations:
2EI 1 M F 2 44.44 EI ------ 1
AB AB L A B 3 B
2EI 2 M F 2 88.89 EI ------ 2
BA BA L B A 3 B
M F 2EI 4 2 2 41.67 EI EI ------ 3
BC BC L B C 5 B 5 C
M F 2EI 4 2 2 41.67 EI EI ------ 4
CB CB L C B 5 C 5 B
MCD 30 KNM
In the above equations we have two unknown rotations the B and C , accordingly boundary conditions are:
MBA MBC 0
MCB MCD 0
2 4 2 Now, M M 88.89 EI 41.67 EI EI BA BC 3 B 5 B 5 C 22 2 47.22 EI EI 0 ------ 5 B C 15 5
And, M M 4 2 41.67 EI EI 30
CB CD 5 C 5 B 2 4 11.67 EI 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 1 44.44 32.67 61.00 KNM MAB 2 2 M 88.89 32.67 67.11 KNM BA 3 4 2 KNM M 41.67 32.67 1.75 67.11
BC 5 5 4 2 M 41.67 1.75 32.67 30.00 KNM CB 5 5 M 30 CD 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
5 R 5 20 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 67.11 61 Max 133.33 61 KNM 4 68.26 6 Span BC: where SF=0, consider SF equation with C as reference
SX 42.58 20x 0 42.58 x 2.13 m 20 2 2.13 KN M Mmax 42.58 2.13 20 30 15.26 2
Example: Analyse the continuous beam ABCD shown in figure by slope deflection method. The support B sinks by 15mm.
5 2 6 4 Take E 200 10 KN / m and I 120 10 m
Solution:
In this problem A =0, B 0, C 0, =15mm
FEMs: 2 Wab L2 44.44 KNM FAB
FBA 88.89 KNM 2 Wa b L2 2
FBC wL 41.67 KNM 8 2 wL FCB 41.67 KNM 8
FEM due to yield of support B
For span AB:
m ab mba 6EI
L2 6 200 105 120 106 15
62 6 KNM 1000
For span BC: m bc 6EI mcb
L2 6 200 105 120 106 15
52 8.64KNM 1000 Slope deflection equation
2EI M F 3 AB AB (2 B ) L L A EI FAB 2 B 6EI L A L2 1 - 44.44 EI 6
3 B 1 50.44 EI ------ 1
3 B 6EI 2EI ) M F (2
BA BA L B A L2 2 88.89 EI 6
3 B 2 82.89 EI ------ 2
3 B 6EI 2EI ) M F (2
BC BC L B C L2 2 - 41.67 EI2 8.64
5 B C 4 2 33.03 EI EI ------ 3
5 B 5 C
2EI M CB FCB (2 B 6EI ) L C L2 2 41.67 EI2 B 8.64 5 C 4 2 50.31 EI EI ------ 4
C B 5 5 ------ 5 KNM MCD 30
There are only two unknown rotations B and C . Accordingly the boundary conditions are
MBA MBC 0
MCB MCD 0
Now, 22 2 MBC 49.86 EI EI 0 M BA B 15 5 C 2 4 MCB MCD 20.31 EI EI 0
5 B 5 C Solving these equations we get EI 31.35 B Anticlockwise EI 9.71 C Anticlockwise
Substituting these values in slope deflections we get the final moments:
1 MAB 50.44 31.35 60.89 KNM 3 2 82.89 31.35 61.99 M BA 3 KNM 4 2 M 33.03 31.35 9.71 61.99 KNM BC 5 5 4 2 M 50.31 9.71 31.35 30.00 KNM CB 5 5 M 30 CD 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 5 + 61.99 – 30
2 RB = 56.40 KN RC =
20 x 5 - RB
=43.60 KN
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: Wl 60 4 F - 30 KNM AB 8 8
Wl 60 4 30 KNM F BA 8 8 M FBC 12.5 KNM 4 M 12.5 KNM 4 FCB
2 F 2 wl 10 4 - 13.3 3 KNM CD 12 12
2 13.33 KNM wl 2 10 4 F DC 12 12
Slope deflection equations:
2EI MAB FAB L 2 A B 2EI - 30 0
4 B
- 30 0.5EIB ------ 1
MBA FBA 2EI 2B A L 2EI 30 2 0 4 B
30 EIB ------ 2
MBC FBC 2EI 2B C L 2EI 12.5 2 C 4 B
12.5 EIB 0.5EIC ------ 3 2EI
MCB FCB 2C B L 2EI 12.5 2 B 4 C
12.5 EIC 0.5 EIB ------ 4 2EI
MCD FCD 2C D L 2EI - 13.33 2 D 4 C
13.33 EI C 0.5EI D ------ 5
MDC FDC 2EI 2D C L 2EI 13.33 2 C 4 D
13.33 0.5EI C EI D ------ 6
In the above Equations there are three unknowns, EI B ,EIC & EID , accordingly the boundary conditions are:
i MBA MBC 0
ii MCB MCD 0
iii MDC 0 ( hinged)
Now
M M 0 BA BC
30 EIB 12.5 EIB 0.5EIC 0 7 2EIB 0.5EIC 42.5 0
MCB MBC 0
12.5 EI 0.5EI 13.33 EI 0.5EI 0 0.5EI C B C D B 2EIC 0.5EID 0.83 0 8 MDC 0 13.33 0.5EIC EID 0
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
KNM 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
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: 60 2 5.96 42.02 R 20.985 KN B 4
RA 60 RB 30.015 KN
Beam BC:
11.63 50 5.96 R 13.92 KN
C 4
RB RC 13.92 KN RB is downward
Beam CD:
10 4 2 11.63 R 17.09 KN D 4
RC 10 4 RD 22.91KN
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:
wl 2 2 F 10 8 - 53.33 KNM AB 12 12 KNM 2 wl F 53.33 BA 12 - 22.50 KNM Wl 30 6 F BC 8 8
FCD KNM WL 22.50 8
Slope deflection equations:
MAB FAB 2EI 2 A B L 2E 3I - 53.33 0 8 B 3 - 53.33 EIB ------ 1 4
MBA FBA 2EI L
2B A 2E 3I 53.33 2 0 8 B 3 ------ 2 53.33 EI
2 B
MBC FBC 2EI 2B C L 2E2I - 22.5 2 C 6 B 4 2 - 22.5 EI EI ------ 3
3 B 3 C
MCB FCB 2EI 2C B L 2E2I 22.5 2 B 6 C 4 2 22.5 EI EIB ------ 4 3 C 3
In the above equation there are two unknown boundary B and C , accordingly the conditions are:
i MBA MBC 24 0
ii MCB 0 Now, M 3 4 2 M 24 53.33 EI 22.5 EI EI 24
BA BC 2 B 3 B 3 C 17 2 54.83 EI EI 0 5 6 B 3 C 4 22.5 EI 2 and M EI 0
CB C B 3 3 2 1 EI 11.25 EI ------ (6) 3 C 3 B Substituting in eqn. (5)
17 1 54.83 EI 11.25 EI 0
6 B 3 B 15 44.58 EI 0
6 B
44.58 6 EI B 17.432 15 rotation anticlockwise
from equation (6) 3 1 EI 11.25 17.432
C 2 3 8.159 anticlockwise rotation
Substituting EIB 17.432 we and EIC 8.159 in the slope deflection equation get Final Moments: M 3 AB 53.33 - 17.432 -66.40 KNM 4 3 53.33 17.432 27.18 M KNM
BA 2 4 2 M 22.5 17.432 8.159 51.18 KNM BC 3 3 4 2 M 22.5 8.159 (17.432) 0.00 CB 3 3
Reactions: Consider free body diagram of beams as shown
Span AB: 27.18 66.40 10 8 4 R 35.13 KN B 8
RA 10 8 RB 44.87 KN Span BC:
51.18 30 3 R 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 4.487 M max 44.87 4.487 10
2 64 36.67 KNM Span BC: Max BM occurs under point load 51.18 BC Mmax 45 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:
wl 2 F 20 4 26.67 KNM
2 BC 12 12 2 KNM wl 26.67 FCB 12
Slope deflection equations: 2EI M F
2 A B AB AB L 2E2I 0 0 4 B EI B ------ 1
MBA FBA 2EI 2 L B A 2E2I 0 2 0 4 B ------ 2 2EIB 2EI
MBC FBC 2B C L
2E 1.5I - 26.67 2 C 4 B 3 3 - 26.67 EI EI ------ 3
2 B 4 C 2EI
MCB FCB 2C B L 2E 1.5I 26.67 2 B 4 C 3 3 26.67 EI EIB ------ 4 2 C 4
In the above equation there are two unknowns boundary B and C , accordingly the conditions are
MBA MBC 0
MCB 12 0 Now, M 3 3 M 2EI 26.67 EI EI BA BC B 2 B 4 C 7 3 ------ (5) EI EI 26.67 0
2 B 4 C and, M 3 3 12 26.67 EI EI 12
CB 2 C 4 B 3 3 38.67 EI EI 0 ------ (6)
B C 4 2 From (5) and (6) 7 3 EI EI 26.67 0
B C 2 34 3 EI EI 19.33 0 8 B 4 C 25 EI 46 0
B 8
8 46 14.72 EI B 25 From (6)
2 3 EI 38.67 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 3 3 M 26.67 (14.72) 33.14 29.44 KNM BC 2 4 3 3 M 26.67 (33.14) (14.72) 12 KNM CB 2 4
Reaction: Consider free body diagrams of beam
Span AB:
14.72 29.44 R 11.04 KN B 4
RA RB 11.04 KN Span BC:
29.44 12 20 4 2 R 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:
w w a b L/2 L/2 L wL/8 wL/8 wab²/L² wa²b/L² W W w /unit
L length 2wL/ L/3 L/3 L/3 2wL/ wL²/12 wL²/12 9 w /unit 9 w /unit length length
w L L wL²/3 5wL²/96 5wL²/96 L²/20 0 w/unit length L
L/2 L/2 11 5wL²/192 6EI /L² wL²/192 6EI /L²
3EI/L²
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 = RBxM 2 dx
Fig. 3
2 dy R x B Integrating EI - Mx + C1 dx 2
Using condition x = l dy 0 dx 2 R l B C1 = Ml - 2 dy R x2 R l2
B EI B Mx Ml ……….(1) dx 2 2 3 2 2 R x Mx R l
B Integrating again EI y = Ml B x + C2 6 2 2
Using condition at x = 0 y = 0 C2 = 0 3 2 2 R x Mx R l
B EI y = B Ml x …………… (2) 6 2 2 Using at x = l y = 0 in the equation (2) 3M
RB = 2l Substituting in equation (1) Ml dy 3M Mx ………….. (3) EI x2 dx 4l 4 Substituting x = 0 in the equation (3) Ml 4EI EI B = M = B
4 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
d y M x M BM at a section xx at distance x from ‘B’ is given by EI 2 dx 2 l
2 dy M x Mx C1 Integrating EI dx l 2
3 2 M x Mx
Integrating again EI y = C1 x C 2 2 l 6
At x = 0 y = 0 C2= 0 Ml
At x = l y = 0 C1 = 3 2 Ml EI dy M x Mx dx l 2 3 Substituting x = 0 in above equation Ml
EI B = 3 3EI M = B
l
The term in parenthesis 3EI K For far end hinged ………………(5) l 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 K K 1 1 1 DF1 = = K K M K In general DF = ………………. (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.
I Kr = for far end fixed l
Kr = 3 I 4 for far end hinged, l
Carry over factors: Consider the beam shown in figure
4EI & R = 3M B . A We have shown that M = 2l l 2 d y 3M x = BM at A = EI dx 2 l M 2 at x l x1 M = + 2 M M +ve BM of at A indicates clockwise moment of at A. In other words the moment 2 2 M ‘M’ at the pin induces a moment of 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 1 for the case of far end fixed is + same . The plus sign indicates both moments are in the direction. 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
K values the distribution factors can be determined from equation DF = for a . DF fixed end = 0 and DF = 1 for an end pin or roller support. 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 2 20 x 12 MFBC =
12 240 kNm MFCB = + 240 kNm
MFCB = - 2508 = - 250 kNm
8
FFDC = + 250 kNm
(ii) Calculation of distribution factors:
K Relative DF = Jt. Member K stiffness (K) K B BA I/12 I/6 0.5 BC I/12 0.5 C CB I/12 5I/24 0.4 CD I/8 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
62.62 125.25 -125.25 281.48 -281.48 234.26 Final moments
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)
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
A 10 kN C B D 4m 4m 1m 3I 4m 3m 1m E 10I 2I Fig. 7 a Fig. 7 (a)
Solution:
FEM: MFAB = - 2 3x4 MFBA 4 kNm 2 3x8 2 4 kNm; 3x 8
12 25 x 8 25 x 8 MFBC = - 41 kNm MFAB= + 41kNm 12 8 12 8 2 MFDC = 16 x1 x 3 = + 3 kNm MDE = -10 x 1 = 10kNm
42
2 MFCD = 16 1 3 = -9 kNm 42
DF: K Relative DF = Jt. Member K stiffness (K) K B BA 3 3I 1.81I 0.31 x 0.56I 4 4 BC 10I/8 = 1.25I 0.69 C CB 10I/8 = 1.25I 1.63I 0.77 CD 3 2I 0.23 x = 0.38I 4 4
Note: Since support ‘A’ is simply supported end the relative stiffness value of 3 I 4 has 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 0 23.49 -23.49 18.25 -18.25 10 -10 moments
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 (b)
Fig. 7 (c)
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 5 -4 4 E = 2 x 10 N/mm², I = 1.2 x 10 m
Solution: Fixed End Moments: MFAB = FEM due to load
+ FEM due to sinking 2 wl 6EI = 2 12 l
2 5 4 12 20 x 6 6 x 2 x10 x1.2 x10 x10 x10 = 12 2 6 6000 x10
= 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 2 Wab = l2
6 I 2 E l 2 5 4 12 50 x 3 x 2 6 x 2 x10 1.2 x10 x10 x10 = 2 2 6 5 5000 x 10
= 24 + 57.6
MFBA = + 33.6 kNm
2 MFCB = + Wa b 6EI l2
l2
2
50 x 3 x 2 = 57.6 52
MFCB = 93.6kNm
MFCD = due to load only ( C & D are at some level)
2 2 wl 20 x 4 12 M = 26.67kNm FCD 12
MFDC = + 26.67 kNm
K Relative DF = Jt. Member K stiffness (K) K B BA I/6 0.46 BC I/5 0.36I 0.54 C CB I/5 0.51 CD 3 I 0.39I 0.49 x = 0.19I 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 -109.87 +0.24 -0.24 +60.63 -60.63 moments
Bending moment diagram is shown in fig. 8 (b)
109.87
60.63
0.24
20x6² / 8 = 90 50x3x2/5 = 60 KNM KNM 20x4²/8 = 40KNM
Fig. 8 (b)
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.
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?
2
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.
3
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.
4
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
5
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.
6
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.
7
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.
8
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 9
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
10
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 11
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
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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 [ ] 2 2 2 2 A.WL /8 B. WL /12 C. WL /20 D. WL /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 [ ] 2 2 2 3 A. 4EIΔ /L B. 2EIΔ /L C. 6EIΔ /L D. 12EIΔ /L
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
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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 [ ] 2 2 2 3 A. 4EIΔ /L B. 2EIΔ /L C. 6EIΔ /L D. 12EIΔ /L
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=______
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10.A two span continuous beam ABC with end ‘A’ fixed and end ‘c’ hinged is having AB=4m, BC=6m. I : I =1:2, it is subjected to u.d.l of 10 KN/m over entire right span. Then the moment at ‘C’ is AB BC =______
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 [ ] 2 2 2 2 A. P L/2AE B. 2AE / P L C. P L/AE D. AE / P 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 4 4 4 4 A. WL /48EI B. WL /96EI C. WL /128EI D. WL /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 [ ] 2 2 A.3WL/8 B.5WL/8 C.3WL /16 D.5WL /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 =______. 15
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.
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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 14R11A0151 3 B. VIJAY 14R11A0152 4 B. KIRAN KUMAR 14R11A0153 5 B. SUNIL NAIK
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14R11A0154 6 D. VENU CHARY 14R11A0155 7 D. VASANTHA KUMAR 14R11A0157 8 G. NIKHIL 14R11A0158 9 G. SANDEEP KUMAR 14R11A0159 10 G. CHARAN KUMAR 14R11A0160 11 J. HARISH KUMAR 14R11A0161 12 K.J. NANDEESHWAR 14R11A0162 13 K. SANTHOSH KUMAR 14R11A0163 14 K BHARATH KUMAR 14R11A0164 15 K ABHILASH 14R11A0165 16 K SAI KRISHNA 14R11A0168 17 MOHD. ABBAS 14R11A0169 18 M SRINIVAS 14R11A0170 19 N SANTHOSH 14R11A0172 20 OSA NITHISH 14R11A0173 21 P INDRA TEJA 14R11A0174 22 P NAVEEN KUMAR 14R11A0175 23 P BHARATH NARSIMHA REDDY 14R11A0176 24 P SURENDER 14R11A0177 25 R VIHARI PRAKASH 14R11A0178 26 S BHANU KISHORE 14R11A0179 27 SHAILESH KUMAR SINGH 14R11A0180 28 SYED OMER ASHRAF 14R11A0181 29 V SAI SHARATH 14R11A0182 30 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
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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 14R11A0151 3 1 B. VIJAY 14R11A0152 4 1 B. KIRAN KUMAR 14R11A0153 5 1 B. SUNIL NAIK 14R11A0154 6 1 D. VENU CHARY 14R11A0155 7 2 D. VASANTHA KUMAR 14R11A0157 8 2 G. NIKHIL 14R11A0158 9 2 G. SANDEEP KUMAR 14R11A0159 10 2 G. CHARAN KUMAR
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14R11A0160 11 2 J. HARISH KUMAR 14R11A0161 12 2 K.J. NANDEESHWAR 14R11A0162 13 3 K. SANTHOSH KUMAR 14R11A0163 14 3 K BHARATH KUMAR 14R11A0164 15 3 K ABHILASH 14R11A0165 16 3 K SAI KRISHNA 14R11A0168 17 3 MOHD. ABBAS 14R11A0169 18 3 M SRINIVAS 14R11A0170 19 4 N SANTHOSH 14R11A0172 20 4 OSA NITHISH 14R11A0173 21 4 P INDRA TEJA 14R11A0174 22 4 P NAVEEN KUMAR 14R11A0175 23 4 P BHARATH NARSIMHA REDDY 14R11A0176 24 4 P SURENDER 14R11A0177 25 5 R VIHARI PRAKASH 14R11A0178 26 5 S BHANU KISHORE 14R11A0179 27 5 SHAILESH KUMAR SINGH 14R11A0180 28 5 SYED OMER ASHRAF 14R11A0181 29 5 V SAI SHARATH 14R11A0182 30 5 Y VENKATA MOHAN REDDY
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