Course File On By Assistant Professor Civil Engineering...

Design of Reinforced Concrete Structures 1 Mrs.S.A.Manchalwar

Course File On

Design of Reinforced Concrete Structures

By

Mrs. S.A.Manchalwar

Assistant Professor

Civil Engineering Department

K. G. Reddy College of Engineering and Technology

2019-2020

HOD PRINCIPAL


: Design of Reinforced Concrete Structure.

: Mrs.S.A.Manchalwar

: Professor.

: R18/ CE502PC

: III/I

: Civil Engineering

: 2019-20

Subject Name

Faculty Name

Designation

Regulation /Course

Code Year / Semester

Department

Academic Year

COURSE FILE


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Mrs.S.A.Manchalwar

COURSE FILE CONTENTS

PART-1

S. No Topic Page No.

1 Vision, Mission, PEO’s, PO’s & PSO’s 6

2 Syllabus (University Copy) 10

3 Course Objectives, Course Outcomes and Topic Outcomes 11

4 Course Prerequisites 14

5 CO’s, PO’s Mapping 15

6 Course Information Sheet (CIS) 16

a). Course Description

b). Syllabus

c). Gaps in Syllabus

d). Topics beyond syllabus

e). Web Sources-References

f). Delivery / Instructional Methodologies

g). Assessment Methodologies-Direct

h). Assessment Methodologies –Indirect

i). Text books & Reference books

7 Micro Lesson Plan 20

8 Teaching Schedule 22

9 Unit wise Hand-written Notes 23

10 OHP/LCD SHEETS /CDS/DVDS/PPT (Soft/Hard copies) 23

11 University Previous Question papers 24

12 MID Exam Descriptive Question Papers with Key 24


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13 MID Exam Objective Question papers with Key 35

14 Assignment topics with materials 36

15 Tutorial topics and Questions 36

16 Unit wise-Question bank 37

1 Two marks question with answers 5 questions

2 Three marks question with answers 5 questions

3 Five marks question with answers 5 questions

4 Objective question with answers 10 questions

5 Fill in the blanks question with answers 10 questions

17 Beyond syllabus Topics with material

18 Result Analysis-Remedial/Corrective Action

19 Record of Tutorial Classes

20 Record of Remedial Classes

21 Record of guest lecturers conducted


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PART-2

S. No Topics

1 Attendance Register/Teacher Log Book

2 Time Table

3 Academic Calendar

4 Continuous Evaluation-marks (Test, Assignments etc)

5 Status Request internal Exams and Syllabus coverage

6 Teaching Diary/Daily Delivery Record

7 Continuous Evaluation – MID marks

8 Assignment Evaluation- marks /Grades

9 Special Descriptive Tests Marks

10 Sample students descriptive answer sheets

11 Sample students assignment sheets


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Mrs.S.A.Manchalwar

1. Vision

To give the world new age civil engineers who can transform the society with their creative vibe for

the sustainable development by instilling scientific temper with ethical human outlook.

Mission

To become pioneers in technical education and serve the community, government in the form of

consultancy and research activities.

To impart quality education with application-oriented teaching methodologies and state of art

facilities.


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Mrs.S.A.Manchalwar

Program Educational Objectives

PEO 1 Graduates will utilize the foundation in Engineering and Science to improve lives

and livelihoods through a successful career in civil Engineering or other fields.

PEO 2 Graduates will become effective collaborators and innovators, leading or

participating in efforts to address Social, Technical and Business challenges.

PEO 3 Graduates will engage in Life-Long Learning and professional development

through Self-Study, continuing education or graduate and professional studies in

engineering & Business.

Program Outcomes

PO1 Fundamental engineering analysis skills: An ability to apply knowledge of

computing, mathematical foundations, algorithmic principles, and civil

engineering theory in the modeling and design of to civil engineering problems.

PO2 Information retrieval skills: An ability to design and conduct experiments, as

well as to analyses and interpret data.

PO3 Creative skills: An ability to design, implement, and evaluate a system, process,

component, or program to meet desired needs, within realistic constraints such as

economic, environmental, social, political, health and safety, manufacturability,

and sustainability. Graduates have design the competence.

PO4 Teamwork: An ability to function effectively on multi-disciplinary teams.

PO5 Engineering problem solving skills: An ability to analyze a problem, and identify,

formulate and use the appropriate computing and engineering requirements for

obtaining its solution.

PO6 Professional integrity: An understanding of professional, ethical, legal, security

and social issues and responsibilities. Graduates must understand the principles of

ethical decision making and can interpret the ASCE Code of Ethics. Graduates will

understand the proper use of the work of others (e.g., plagiarism, copyrights, and

patents). Graduates will understand the special duty they owe to protect the public's

health, safety and welfare by their professional status as engineers in society.

PO7 Speaking / writing skills: An ability to communicate effectively, both in writing

and orally. Graduates can produce engineering reports using written, oral and

graphic methods of communication.


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PO8 Engineering impact assessment skills: The broad education necessary to analyze

the local and global impact of computing and engineering solutions on individuals,

organizations, and society.

PO9 Social awareness: Knowledge of contemporary issues. Students are aware of

emerging technologies and current professional issues.

PO10 Practical engineering analysis skills: An ability to use the techniques, skills, and

modern engineering tools necessary for engineering practice.

PO11 Software hardware interface: An ability to apply design and development

principles in the construction of software and hardware systems of varying

complexity.

PO12 Successful career and immediate employment: An ability to recognize the

importance of professional development by pursuing postgraduate studies or face

competitive examinations that offer challenging and rewarding careers in Civil

Engineering

PROGRAMME SPECIFIC OUTCOMES (PSO’s) :

PSO 1: Problem Solving Skills – Graduate will be able to apply design techniques and analyze the

structures.

PSO2: Professional Skills – Graduate will be able to design the sustainable structures using

advanced design techniques.

PSO3: Successful Career – Graduates will be able to become entrepreneur and to pursue career in

designing field.


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1. University Syllabus copy


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Mrs.S.A.Manchalwar

DESIGN OF REINFORCED CONCRETE STRUCTURES

ech. III Year I Sem. L T/P/D C

Course Code: CE502PC 4 1/0/0 4

Pre-Requisites: Structural Analysis I & II

Course Objectives: Structural elements are subjected to different loading to with stand the

structures, for external loading we need to design the structures for its safety and serviceability.

Course Outcomes: At the end of the course, the student will be able to:

Design RC Structural elements.

Design the Reinforced Concrete beams using limit state Design.

Design Reinforced Concrete slabs.

Design the Reinforced Concrete Columns and footings Design structures for

serviceability.

Design staircases, canopy

UNIT – I Concepts of RC. Design – Working Stress Method – Limit State method – Material Stress-Strain Curves – Safety factors – Characteristic values. Stress Block parameters – IS – 456 – 2000. Beams:

Limit state analysis and design of singly reinforced, doubly reinforced, T and L beam sections

UNIT – II Limit state analysis and design of section for shear and torsion – concept of bond, anchorage and development length, I.S. code provisions. Design examples in simply supported and continuous beams, detailing; Design of canopy.

UNIT – III Short and Long columns – under axial loads, uniaxial bending and biaxial bending – I S Code provisions.

UNIT – IV Footings: Different types of footings – Design of isolated, square, rectangular, circular footings and combined footings.

UNIT – V Design of one way slab, Two-way slabs and continuous slab Using I S Coefficients Limit state design for serviceability for deflection, cracking and codal provision. Design of dog-legged staircase.

TEXT BOOKS: 1. Reinforced concrete design by S. Unnikrishna Pillai & Devdas Menon, Tata McGraw Hill, New

Delhi.

2. Reinforced concrete design by N. Subrahmanian Oxford University Press.

3. Limit state designed of reinforced concrete – P. C. Varghese, Prentice Hall of India, New Delhi.

REFERENCES:

1. Design of Reinforced Concrete Structures by I. C. Syal and A. K. Goel, S. Chand & company.

2. Fundamentals of reinforced concrete by N.C. Sinha and S.K Roy, S. Chand publishers

3. Design of concrete structures – Arthus H. Nilson, David Darwin, and Chorles W. Dolar, Tata McGraw-Hill, 3rd Edition, 2005.


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Mrs.S.A.Manchalwar

3. Course Objectives, Course Outcomes & Topic Outcomes

A) Course Objectives:

Structural elements are subjected to different loading to with stand the structures, for

external loading we need to design the structures for its safety and serviceability.

B) Course Outcomes: At the end of the course, the student will be able to:

CO1 Design RC Structural elements.

CO2 Design the Reinforced Concrete beams using limit state Design.

CO3 Design Reinforced Concrete slabs.

CO4 Design the Reinforced Concrete Columns and footings.

CO5 Design structures for serviceability.

CO6 Design staircases, canopy.

C) Topic Outcomes

Lecture

no.

Topic to be covered Topic outcome

L1 UNIT-I

Concepts of RCC._ Design – Working Stress Method.

Demonstrate concept of RCC, Working Stress Method and material stress.

L2 & L3 Limit state method Stress-Strain Curves,

Safety factors – Characteristic values

Describe about stress-strain curves for

concrete and steel. Explain about factor of safety and safety factor with Characteristic values.

L4 Stress block parameters, - IS 456-2000. Illustrate stress block parameters as per IS 456-2000.

L5 Beams: Limit State analysis Analyze the section as per limit state method.

L6 & L7 Singly reinforced beam. Designing of singly reinforced beam.

L8 & L9 Doubly reinforced beam. Designing of doubly reinforced beam.

L10 & L11

T beam sections. Designing of T beams.

L 12 L beam sections. Designing of L beams.

L 13 & L14

Revision. Solve the previous year university questions papers.

L 15 Presentation of construction videos. Demonstrate construction of R.C.C elements.

L16 & L17

UNIT-II Shear, Torsion and Bond:

Explain the concept shear and torsion; Analyse shear and torsion induced in


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Mrs.S.A.Manchalwar

Limit state analysis and design for shear and torsion.

beams due to application of loading.

L18 & L19

Concept of bond, anchorage and development length, IS code provisions.

Explain the concept of bond, anchorage and development length and calculate

the bond, anchorage and development length as per IS code provision.

L20 & L21

Simply supported beams. Design and detailing Simply supported beams to resist shear and torsion.

L22 & L23

Continuous beams Design and detailing continuous beams to resist shear and torsion.

L24 & L25

Limit state design for serviceability of deflection, cracking.

Illustrate limit state of serviceability (i.e. deflection and cracking).

L 26 Design of canopy. Canopy design with effect of shear

and torsion. L27 Revision Demonstration about shear, torsion,

bond, anchorage and development.

L28 Group discussion. Group discussion for one minute to

explain about failure of beam sections

due to shear and torsion.

L29 UNIT-III Short columns and Long columns

Demonstration columns.

of short and long

L30 Axially load columns Design the axially loaded short column.

L31 & L32

Uni axially loaded columns Design the Uni axially loaded short column.

L33 &

L44

Bi-axially loaded columns Design the Bi-axially loaded short

column.

L35 & L36

Long columns or slender columns Design the Long columns (or) Slender columns.

L37 & L38

Revision Solve previous year university question papers.

L39 &

L40

Model presentation Demonstration of various types of

column failures.

L41 UNIT-IV

Footings: Different types of footings –

isolated footing.

Explain about footings and types of

footings. Illustrate isolated footing.

L42 Isolated square footing design Design of square isolated footing.

L43 & L44

Rectangular isolated footing. Design of rectangular isolated footing.

L45 & L46

Circular footings. Design of circular footing.

L46 & L47

Combined footings Design of combined footing

L48 & L49


L50 Demonstration of designing software Over view on designing software which

are using in structural design (STAAD PRO, ETABS, STRUDS & SCADDS.)

L51 & L 52

UNIT-V One –way slabs

Designing of one way slab using IS coefficients.

L53 & L54

Two-way slabs Designing of one way slab using IS coefficients.


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Mrs.S.A.Manchalwar

L55 & L56

Continuous slabs Designing of continuous slabs using IS coefficients.

L57 & L58

Cracking and codal provision Illustrate the phenomenon of cracking and calculation of crack width as per codal conditions.

L59 & L60

Design of dog-legged staircase Demonstrate about types of staircases and designing of dog-legged staircase.

L61 & L62



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Mrs.S.A.Manchalwar

4. COURSE PRE–REQUISITES

Pre-Requisites: Structural Analysis I & II.


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Mrs.S.A.Manchalwar

5. a) CO’S, PO’S MAPPING

CO PO 1

PO 2

PO 3

PO 4

PO 5

PO 6

PO 7

PO 8

PO 9

PO 10

PO11 PO12 PSO1 PSO 2

PSO 3

CO 1

3 1 - - 2 - 1 1 - - 2 3 1 - 2

CO 2

2 - 3 - 2 - - - - 1 - 2 3 3 -

CO 3

3 1 2 1 1 - 2 - 2 - - 3 3 - 2

CO 4

- - 1 - 2 - - - - 1 - - 3 1 -

CO 5

1 2 3 - - - - 3 - 2 - 3 3 - 1

CO

6

- 2 - 2 1 - 3 1 1 - 2 1 1 2 -

Notes: 1 - Low, 2 - Medium, 3 - High


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Mrs.S.A.Manchalwar

6. COURSE INFORMATION SHEETS (CIS)

a) Course Description:

Programme: B. Tech. (Civil Engineering.) Degree: B Tech

Course: Design of Reinforced Concrete Structures

Year: III Sem: I Credits:

Course Code: A70143

Regulation: R18

Course Type: Core

Course Area/Domain: Design Contact Hours: 4+1 (L+T)) Hours/Week.

Corresponding Lab Course Code (If Any):

Nil

Lab Course Name: Nil


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Mrs.S.A.Manchalwar

b) SYLLABUS:

Unit Details Hours

I

Concepts of RC. Design – Working Stress Method – Limit

State method – Material Stress-Strain Curves – Safety factors

– Characteristic values. Stress Block parameters – IS – 456 –

2000. Beams: Limit state analysis and design of singly

reinforced, doubly reinforced, T and L beam sections

15

II

Limit state analysis and design of section for shear and

torsion – concept of bond, anchorage and development

length, I.S. code provisions. Design examples in simply

supported and continuous beams, detailing; Design of

canopy.

13

III Columns: Short and Long columns – under axial loads, uniaxial

bending and biaxial bending – I S Code provisions. 12

IV

Footings: Different types of footings – Design of isolated,

square, rectangular, circular footings and combined

footings.

10

V

Slabs: Design of one way slab, Two-way slabs and continuous slab Using I S Coefficients Limit state design for serviceability for deflection, cracking and codal provision. Design of dog-legged staircase.

12

Total No. of classes 62

c) Gaps in the Syllabus -

S.No. Description Proposed Actions

1 NIL NIL

d) Topics Beyond Syllabus / Advanced Topics:

S.No. Description Proposed Actions

1 Design of circular staircase, design of arches and

Analysis and design of retaining wall design

Guest Lecture


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e) Web Source References:

S. No. Name of book/ website

a. https://www.youtube.com/watch?v=0z6gjjrSn0M

b. https://www.youtube.com/watch?v=91fCagGP5Is

c. https://www.youtube.com/watch?v=PDBfY4FM9qY

https://www.youtube.com/watch?v=v0753-IzyB8

f) Delivery / Instructional Methodologies:

CHALK & TALK STUD. ASSIGNMENT WEB RESOURCES

LCD/SMART BOARDS STUD. SEMINARS ☐ ADD-ON COURSES

g) Assessment Methodologies - Direct

ASSIGNMENTS STUD.

SEMINARS

TESTS/MODEL

EXAMS

UNIV.

EXAMINATION

☐ STUD. LAB

PRACTICES

☐ STUD. VIVA MINI/MAJOR

PROJECTS

☐ CERTIFICATIONS

☐ ADD-ON COURSES ☐ OTHERS

h) Assessment Methodologies - Indirect

ASSESSMENT OF COURSE OUTCOMES

(BY FEEDBACK, ONCE)

STUDENT FEEDBACK ON

FACULTY (TWICE)

☐ ASSESSMENT OF MINI/MAJOR PROJECTS BY

EXT. EXPERTS

☐ OTHERS

i ) TEXT / REFERENCE BOOKS:

T/R Book Title/Authors/Publication

T/R BOOK TITLE/AUTHORS/PUBLICATION

Text Book 1. Reinforced concrete design by S. Unnikrishna Pillai & Devdas Menon, Tata

McGraw Hill, New Delhi.

https://www.youtube.com/watch?v=0z6gjjrSn0M

https://www.youtube.com/watch?v=91fCagGP5Is

https://www.youtube.com/watch?v=PDBfY4FM9qY

https://www.youtube.com/watch?v=v0753-IzyB8


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Mrs.S.A.Manchalwar

2. Reinforced concrete design by N. Subrahmanian Oxford University Press.

3. Limit state designed of reinforced concrete – P. C. Varghese, Prentice Hall of India, New Delhi.

Reference

Book 1. Design of Reinforced Concrete Structures by I. C. Syal and A. K. Goel,

S. Chand & company.

2. Fundamentals of reinforced concrete by N.C. Sinha and S.K Roy, S. Chand

publishers.

3. Design of concrete structures – Arthus H. Nilson, David Darwin, and Chorles

W. Dolar, Tata McGraw-Hill, 3rd Edition, 2005.


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Mrs.S.A.Manchalwar

7. MICRO LESSON PLAN

S. No. Topic Planned date Actual date

L1 UNIT-I

Concepts of RCC._ Design – Working

Stress Method.

L2 & L3

Limit state method Stress-Strain Curves, Safety factors – Characteristic values

L4 Stress block parameters, - IS 456-2000.

L5 Beams: Limit State analysis

L6 & L7

Singly reinforced beam.

L8 & L9

Doubly reinforced beam.

L10 & L11

T beam sections.

L 12 L beam sections.

L 13 & L14

Revision.

L 15 Presentation of construction videos.

L16 &

L17 UNIT-II

Shear, Torsion and Bond:

Limit state analysis and design for shear

and torsion.

L18 & L19

Concept of bond, anchorage and development length, IS code provisions.

L20 & L21

Simply supported beams.

L22 & L23

Continuous beams

L24 & L25

Limit state design for serviceability of deflection, cracking.

L 26 Design of canopy.

L27 Revision

L28 Group discussion.

L29 UNIT-III Short columns and Long columns

L30 Axially load columns

L31 & L32

Uni axially loaded columns

L33 & L34

Bi-axially loaded columns

L35 & Long columns or slender columns


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Mrs.S.A.Manchalwar

L36

L37 & L38

Revision

L39 & L40

Model presentation

L41 UNIT-IV

Footings: Different types of footings –

isolated footing.

L42 Isolated square footing design

L43 & L44

Rectangular isolated footing.

L45 & L46

Circular footings.

L46 & L47

Combined footings

L48 & L49

Revision

L50 Demonstration of designing software

L51 & L 52

UNIT-V One –way slabs

L53 & L54

Two-way slabs

L55 & L56

Continuous slabs

L57 & L58

Cracking and codal provision

L59 & L60

Design of dog-legged staircase

L61 & L62

Revision


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Mrs.S.A.Manchalwar

8. Teaching Schedule

Subject REINFORCED CONCRETE STRUCTURES DESIGN AND DRAWING

Text Books (to be purchased by the Students)

Book 1 Reinforced concrete design by Devdas Menon Book 2 Reinforced Concrete design by N. Krishna Raju and R. N. Pranesh.

Reference Books

Book 3 Fundamentals of reinforced concrete design by M. L. Gambhir. Book 4 Design of RCC Structural Elements- S. S. Bhavikatti

Unit

Topic Chapters Nos No of

classes Book 1 Book 2 Book 3 Book 4

I

Concept of Rc design, Stress

strain curve, stress block

parameters.

2

1

1

2

6

Design of singly, doubly, T

and L beam sections

2

1

2

4

9

II

Shear and Torsion

2

2

-

2

6

Development length

1

1

1

1

4

Deflection and cracking - 1 1 1 3

III

Short and long column 1 2 1 1 5

Axial, uni axial and biaxial load 1 2 1 3 7

IV Footings 2 3 2 3 10

V

One way slab 1 2 - 1 4

Two way slab 1 2 1 - 4

Continuous slabs - 1 - 1 2

Staircase - 1 - 1 2

Contact classes for syllabus coverage 56

Tutorial classes 6

Total No. of classes 62


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9 UNIT WISE HAND -WRITTEN NOTES

10 OHP/LCD SHEETS /CDS/DVDS/PPT (Soft/Hard copies)


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11 UNIVERSITY PREVIOUS QUESTION PAPERS



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Mr.S.A.Manchalwar




12 MID I EXAM DESCRIPTIVE QUESTION PAPER WITH KEY

13 MID II EXAM DESCRIPTIVE QUESTION PAPER WITH KEY


14 ASSIGNMENT TOPICS WITH MATERIALS

Unit 1: Design concepts, Design of Beams

1) OBJECTIVE QUESTIONS 1. The maximum area of tension reinforcement in beams shall not exceed

A.0.15%

B.1.5%

C.4%

D.1%

Answer: Option C

2. As per I.S. 456 - 1978, the pH value of water shall be

A. less than 6

B. equal to 6

C. not less than 6

D. equal to 7

Answer: Option C

3. The minimum number of main steel bars provided in R.C.C.

A. rectangular columns is 4

B. circular columns is 6

C. octagonal columns is 8

D.all the above.

Answer: Option D

4. A T-beam behaves as a rectangular beam of a width equal to its flange if its neutral axis

A. remains within the flange

B. remains below the slab

C. coincides the geometrical centre of the beam

D. none of these.

Answer: Option A

5. The maximum shear stress (qmax) in a rectangular beam is

A. 1.25 times the average

B. 1.50 times the average

C. 1.75 times the average

D. 2.0 times the average

E. 2.5 times the average.

Answer: Option B

6. If the permissible compressive and tensile stresses in a singly reinforced beam are 50 kg/cm2 and

1400 kg/cm2 respectively and the modular ratio is 18, the percentage area At of the steel required

for an economic section, is

A. 0.496%

B. 0.596%

C. 0.696%

D. 0.796%


Answer: Option C

7. If the maximum shear stress at the end of a simply supported R.C.C. beam of 6 m effective span is

10 kg/cm2, the share stirrups are provided for a distance x from either end where x is

A. 50 cm

B. 100 cm

C. 150 cm

D. 200 cm

Answer: Option C

8. An R.C.C. beam of 25 cm width and 50 cm effective depth has a clear span of 6 metres and carries

a U.D.L. of 3000 kg/m inclusive of its self-weight. If the lever arm constant for the section is

0.865, the maximum intensity of shear stress, is

A. 8.3 kg/cm2

B. 7.6 kg/cm2

C. 21.5 kg/cm2

D. 11.4 kg/cm2

Answer: Option A

9. If the ratio of the span to the overall depth does not exceed 10, the stiffness of the beam will

ordinarily be satisfactory in case of a

A. simply supported beam

B. continuous beam

C. cantilever beam

D. none of these.

Answer: Option C

10. The advantage of reinforced concrete, is due to

A. monolithic character

B. fire-resisting and durability

C. economy because of less maintenance cost

D. moulding in any desired shape

E. All the above.

Answer: Option E

2) FILL IN THE BLANKS

1. The shear reinforcement in R.C.C. is provided to resist

ANS: diagonal tension.

2. The width of the rib of a T-beam, is generally kept between

ANS:

3. A pre stressed rectangular beam which carries two concentrated loads W at L/3 from either end,

is provided with a bent tendon with tension P such that central one-third portion of the tendon

remains parallel to the longitudinal axis, the maximum dip h is


Ans:

4. Dimensions of a beam need be changed if the shear stress is more than_

Ans: 20 kg/cm2

5. An R.C.C. beam not provided with shear reinforcement may develop cracks in its bottom inclined

roughly to the horizontal at

Ans : 45°

6. The minimum cube strength of concrete used for a prestressed member, is

Ans: 400 kg/cm2

7. The reinforced concrete beam which has width 25 cm, lever arm 40 cm, shear force 6t/cm2, safe

shear stress 5 kg/cm2 and B.M. 24 mt is

Ans : unsafe in shear

8. In a beam the local bond stress Sb, is equal to

Ans :

9. According to I.S. : 456 specifications, the safe diagonal tensile stress for M 150 grade concrete, is

Ans : 5 kg/cm2

10. For initial estimate for a beam design, the width is assumed

Ans : 1/30th of span

3) SHORT QUESTIONS

1. What is meant by limit state? Ans: Limit state design also known as load and resistance factor design refers to a design method

used in structural engineering. A limit state is a condition of a structure beyond which it no

longer fuls the relevant design criteria.

2. With the help of neat sketch derive the stress block parameters for limit state of flexure.

Ans:

https://en.wikipedia.org/wiki/Structural_engineering


3. Explain characteristic strength of materials and characteristic loads.

Ans: Characteristic strength of concrete is one of the important properties of concrete which

indeed unanimously by design engineer or any other person involved in the construction sector.

The compressive strength of concrete is given in terms of the characteristic compressive strength of

150 mm size cubes tested at 28 days as per Indian Standards (ACI standards use cylinder of

diameter 150 mm and height 300 mm). The characteristic strength is defined as the strength of the

concrete below which not more than 5% of the test results are expected to fall.

This concept assumes a normal distribution of the strengths of the samples of concrete.

Normal Distribution curve on test specimens for determining compressive strength

The above sketch shows an idealized distribution of the values of compressive strength for a certain

number of test specimens. The horizontal axis represents the values of compressive strength in

MPa. The vertical axis represents the number of test samples for a particular compressive strength.

This is also termed as frequency. The average of the values of compressive strength (mean

strength) from the graph is 40 MPa. The characteristic strength (fck) is the value in the x-axis

below which 5% of the total area under the curve falls. From the graph we can clearly say that 30

MPa is the characteristic strength of the given concrete mix.

4. What is Ultimate Limit State (ULS)?

Ans: ULS in concerned with the maximum load – carrying capacity of the structure within the

limits of strength of the materials used.

5. What is characteristic load?

Generally, load on any structural members cannot be determined accurately. For most

http://en.wikipedia.org/wiki/Concrete

http://en.wikipedia.org/wiki/Design_engineer

http://en.wikipedia.org/wiki/Cube

http://en.wikipedia.org/wiki/Normal_distribution

http://en.wikipedia.org/wiki/Cartesian_coordinate_system

http://en.wikipedia.org/wiki/Types_of_concrete


Structures, it is uneconomical to design using anticipated maximum load. Therefore, in normal

design practice, the load to be used is based on the characteristic load. Characteristic load is defined

as the minimum load that statistically will not exceed during the design life of the structure.

6. State four objectives of the design of reinforced concrete structure?

Properly designed reinforced concrete structures should:

Have acceptable probability of performing satisfactorily during their intended life,

Sustain all loads with limited deformations during construction and use, be durable,

Adequately resist the effects of misuse and fire.

7. What are the three methods of design of reinforced concrete structural elements?

The three methods are:

Limit state method,

Working stress method,

Method based on experimental approach

8. Write a short notes on (i) Characteristics strength of materials (ii) Characteristics of loads

Ans: Characteristics strength: The term characteristic strength means that value of the strength of

the material below which not more than 5 present of the test results are expected to fall

Characteristics of loads: The term characteristic load means that the value of load which has a

95% probability of not being exceeds during the life of the stricture

9. What do you understand by limit state of collapse?

Ans: The limit state of collapse of the structure or part of the structure could be assessed from

rupture of one or more critical section and from buckling due to elastic or plastic instability or

overturning or fatigue etc. The resistance to bending, shear torsion and axial loads at every sections

shall not be less than the appropriate value at that section produced by the probable most

unfavourable combination of loads on the structure using the structure using the appropriate partial

safety factors.

4) Long questions:

1. What are the assumptions made in limit state of collapse in flexure?

Plane sections normal to the axis remain plane after bending i.e. the distribution of strain across

any section is linear to its distance from the neutral axis

The maximum strain in concrete at the outermost compression fiber is taken as 0.0035 in

bending.

The relationship between the compressive stress distribution in concrete and strain concrete may

be assumed to be rectangle, trapezoid, parabola or any other shape which results in the

prediction of strength in sustained agreement with results of tests .The partial safety γm = 1.5

The tensile strength of concrete is ignored.

The maximum strain in the tension reinforcement in the section at failure shall not be less than

fy/1.15Es +0.002

Where fy = characteristic strength of steel Es= modulus of elasticity of steel

2. What do you understand by limit state of collapse?

Ans: The limit state of collapse of the structure or part of the structure could be assessed from

rupture of one or more critical section and from buckling due to elastic or plastic instability or

overturning or fatigue etc. The resistance to bending, shear torsion and axial loads at every


section shall not be less than the appropriate value at that section produced by the probable most

unfavorable combination of loads on the structure using the structure using the appropriate

partial safety factors.

3. How to select cross sectional dimensions for beams?

The effective and overall depth of the beam is estimated from span/depth rations to satisfy the

limit state of serviceability. Overall depth to width should be in the range of 1.5 to 2

The width of the section should accommodate the required number of bars with sufficient

spacing between them with minimum side covers of 20mm to the links.

The depth of the beam should be such that the percentage of steel required is around 75% of any

one layer

The minimum number of bars on tension face should be not less than two and not more than six

in any one layer.

In flanged beams, the depth of the slab is generally taken as 20% of the overall depth.

Common widths of beams are 150,200,230,300mm.Also the width of the beam should be equal

to or less the dimensions of the column supporting the beam.

4. Write short note on balanced sections, over reinforced section and under reinforced section.

Balanced section: Reinforced concrete sections in flexure reach the failure stage when the

compressive strain in concrete reaches a value of 0.0035.when the sections are reinforced in

such a way that the section steel reaches the yield of єy= (0.87fy)+Es+0.002 and simultaneously

the concrete strain is єc = 0.0035, the section is termed as balanced sections

Under reinforced sections: In under reinforced sections, the tension steel reaches yield strain at

loads lower than the load at which concrete reaches the failure strain. When the steel yields

earlier than concrete, there will be excessive deflections and cracking with a clear indication of

impending failure. Hence preferable to design beams as under reinforced since failure will take

place after yielding of steel with clear warning signals like excessive deflections and cracking

before the ultimate failure.

Over reinforced sections: Over reinforced sections are those in which concrete reaches the yield

strain earlier than that of steel. Over reinforced beams fail by compression failure of concrete

without much warning and with very few cracks and negligible deflections. Over reinforced

concrete beams are not preferred since they require large quantities of steel and they fail

suddenly with explosive failures without any warning.

5. What is the necessity of doubly reinforced sections?

Ans: Doubly reinforced concrete sections are required in beams of restricted depth due to head

room requirements. When the singly reinforced section is insufficient to resist the bending

moment on the section, additional tension and compression reinforcements are designed based

on steel beam theory.

The doubly reinforced section comprises of two parts outlined as

1. singly reinforced section with the restricted depth providing the limiting moment of

resistance (Mu lim) which is less than the design moment Mu

2. Based on steel beam theory, a steel beam with tension and compression reinforcement

providing balance moment given by (Mu- Mu lim)

6. How to select cross sectional dimensions for beams?


Ans: I. The effective and overall depth of the beam is estimated from span/depth rations to

satisfy the limit state of serviceability. Overall depth to width should be in the range of

1.5 to 2

II. The width of the section should accommodate the required number of bars with

sufficient spacing between them with minimum side covers of 20mm to the links.

III. The depth of the beam should be such that the percentage of steel required is around

75% of any one layer

IV. The minimum number of bars on tension face should be not less than two and not more

than six in any one layer.

V. In flanged beams, the depth of the slab is generally taken as 20% of the overall depth.

VI. Common widths of beams are 150,200,230,300mm.Also the width of the beam should

be equal to or less the dimensions of the column supporting the beam.

7. Briefly explain about partial safety factor and limiting neutral axis.

Ans: When assessing the strength of a structure or structural member for limit state of collapse, the

value of partial safety factor γm should be taken as1.5 for concrete and 1.15 for steel. A higher

value of partial safety factor for concrete has been adopted because there are greater chances of

variation in strength of concrete due to improper compaction, inadequate curing and mixing and

variations in the properties of ingredients.

Limiting neutral axis, Xu(max) which gets formed when the strain in concrete and strain in steel

reaches their maximum permissible values ie 0.0035 and

(0.87fy/Es) +0.002

8. Find the moment of resistance of a singly reinforced concrete beam of 200 mm width 400mm

effective depth, reinforced with 3-16 mm diameter bars of Fe 415 steel. Take M20 grade of

concrete.

Solution

Ast =3 x 4 (16) 2 = 603.19 mm2

%pt= 100 x (603.19/200x400) = 0.754%

Xu/d= 2.417 pt( fy/fck) = 2.417x(0.754/100) x (415/20) = 0.378

Now for Fe415 grade of steel (Xu max/d) = 0.479

Hence the beam is under-reinforced.

The moment of resistance is given by

Mu = 0.87 fy Ast d[1-(fy Ast) /(Fck bd)]

= 0.87*415*603.19*400[1-(415-603.19)/(20*200*400)]

= 73.48 KN-m

9. A reinforced concrete beam is supported on two walls 250mm thick, spaced at a clear distance of

6m. The beam carries a super-imposed load of 9.8 KN/m. design the beam using M20 concrete and

HYSD bars of Fe 415 grade.

Ans: The minimum depth of beam is based on limit state of serviceability requirements. As per IS

456:2000 cl no 23.2.1 (a), for simply supported beam l/d=20 and for balanced section,pt,lim =

0.96%


Also for Fe 415 steel, fs=0.58x415=240 N/mm2

As per IS 456:2000 cl no 23.2.1(c) and fig 4 of IS 456:2000 modification factor=1.

Therefore, l/d=20x1;

d= l / 20 = 6000 / 20 = 300

This is the minimum value of d. actual value of d, based on bending may be more than this.

Now from experience, assume d=l/15=400mm

Therefore, overall depth= effective depth+ clear cover + diameter of stirrup +0.5(diameter of main

reinforcement)

=400+25+8+0.5x20=443mm = 450mm

Assume b=250mm

Therefore, try a trial section of dimension 250x450.

Load Calculation

Self-weight of beam (DL)= 0.25x0.45x1x25=2.8125 KN/m

Super-imposed load (LL)= 9.8 KN/m

Therefore, total load, w = (DL+LL)=(2.8125+9.8)=12.6125 KN/m

Design load, Wu = 1.5*w = 18.9187 KN/m

Calculation of effective span

As per IS 456:2000, cl no 22.2 (a), the effective span of a simply supported beam is lesser of the

following two.

Clear span+ the effective depth of beam or slab

Or center to center distance between supports.

Clear span =6m

Effective depth of beam, d=450-25-8-0.5x20=407mm

Therefore, clear span + effective depth of beam=(6+0.407)m=6.407m

Centre to centre distance between support=(6+0.25/2+0.25/2)m=6.25m

Lesser of two=6.25m

Therefore, effective span =6.25m

Calculation of BM and SF

Maximum BM = wul2/8 = 18.91*6.252/8 = 92.37 KN-m

Maximum SF = wul/2 = 18.91*6.25 / 2 = 5.12 KN

Computation of effective depth, d

For M20 grade of concrete and Fe 415 grade of steel

Mu = 0.138 M f bd2





nswe Optio C

nswe ption

UNIT – II SHEAR, BOND & TORSION

1) OBJECTIVE QUESTIONS

1. As compared to uniaxial tension or compression, the strain energy stored in bending is

only

A B C D

A r: n

2. In a loaded beam, the point of contraflexture occurs at a section where

A. bending moment is minimum

B. bending moment is zero or changes sign

C. bending moment is maximum

D. shearing force is maximum

E. shearing force is minimum.

Answer: Option B

3. Pick up the correct statement from the following: A.A pile is a slender member which transfers the load through its lower end on a strong strata B. A pile is a slender member which transfers its load to the surrounding soil

C. A pile is a slender member which transfers its load by friction

D. A pile is a cylindrical body of concrete which transfers the load at a depth greater than its width.

Answer: Option B

4. The shear reinforcement in R.C.C. is provided to resist

A. Vertical shear B. horizontal shear C. diagonal compression D. diagonal tension.

Answer: Option D

5 If bending moment is M, shear force is F effecive depth is d, lever arm is la area of reinforcement is As

and sum of the circumferences of main reinforcement is 0, the bond stress based on working stress method,

is A. B. C . D

A r: O A.

6 The length of lap in tension reinforcement should not be less than the bar diameter x (actual tension /

four times the permissible average bond stress) if it is more than

A.18 bar diameters, B.24 bar diameters, C.30 bar diameters, D.36 bar diameters

Answer: Option C

7 The maximum ratio of span to depth of a cantilever slab, is

A.8, B.10, C.12, D 14

Answer: Option C

8 Pick up the incorrect statement from the following. The intensity of horizontal shear stress at the

elemental part of a beam section, is directly proportional to


The transverse reinforcements provided at righ

A. shear force, B. area of the section, C. distance of the C.G. of the area from its neutral axis

D. moment of the beam section about its neutral axis

Answer: Option D

9 If the maximum shear stress at the end of a simply supported R.C.C. beam of 16 m effective span is 10

kg/cm2, the length of the beam having nominal reinforcement, is

A. 4 cm, B. 6 m, C. 8 m, D. 10 m

Answer: Option C

10 If the diameter of longitudinal bars of a square column is 16 mm, the diameter of lateral ties should

not be less than

A. 4 mm, B. 5 mm, C. 6 mm, D. 8 mm, E. 10 mm

Answer: Option B

2) FILL IN THE BLANKS 1. Distribution of shear intensity over a rectangular section of a beam is --------------

Ans: a parabolic curve

2. The width of the flange of a T-beam should be less than-----------

Ans: least of the above.

3. If the average bending stress is 6 kg/cm2 for M 150 grade concrete, the length of embedment of a

bar of diameter d according to I.S. 456 specifications, is----------

Ans: 58 d

4. Dimensions of a beam need be changed if the shear stress is more than---------

Ans: 20 kg/cm2

5. An R.C.C. beam not provided with shear reinforcement may develop cracks in its bottom inclined

roughly to the horizontal at--------

Ans: 45°

6. In a beam the local bond stress Sb, is equal to--------

Ans:

7. t angles to the main reinforcement------

Ans: resist the shrinkage stress

8. The anchorage value of a hook is assumed sixteen times the diameter of the bar if the angle of the bend,

is----------

Ans: 45°

9. The length of the lap in a compression member is kept greater than bar diameter x (Permissible stress in

bar / Five times the bond stress) or-------

Ans: 24 bar diameters

10. If diameter of a reinforcement bar is d, the anchorge value of the hook is------

Ans: 16d

3) SHORT QUESTIONS

1. How do we place the vertical stirrups in a beam?

Ans:

Design of Reinforced Concrete Structures Mrs.S.A.Manchalwar 43

The stirrups in beams shall be taken around the outer-most tension and compression bars. In T and L-

beams, the stirrups will pass around longitudinal bars located close to the outer face of the flange. In the

rectangular beams, two holder bars of diameter 10 or 12 mm are provided if there is no particular need for

compression reinforcement

2. Define torsion?

the twisting or wrenching of a body by the exertion of forces tending to turn one end or part about a

longitudinal axis while the other is held fast or turned in the opposite direction; also : the state of being

twisted.

3. Why is bond stress more in compression bars than that in tension bars?

Ans: The intensity of adhesion force at the adhesion force at the junction of steel and concrete is called the

bond stress. The surface of contact of steel bar and concrete gets improved under compression whereas

under tension it leads to separation. So the bond stress is more than that in tension bars.

4. Define: Flexural bond & Anchorage bond.

Ans: i) Flexural bond:Flexural bond (bf ) is one which arises from the change in tensile force

carried by the bar, along its length, due to change in bending moment along the length of the member.

Evidently, flexural bond is critical at points where the shear is significant. Since this occurs at a particular

section, flexural bond stress is known as local bond stress.

(ii) Anchorage bond:

Anchorage bond (bd) is that which arises over the length of anchorage provided for a bar. It also arises

near the end or cut-off point of a reinforcing bar. The anchorage bond resists the pulling out of the bar if it

is in tension or „pushing in of the bar if it is in compression.

5. What do you understand by the term Anchorage?

Ans: With modern high bond bars the mechanism of reinforcement anchorage is due to (i) Adhesion of

concrete and steel,

(ii) Shear strength of concrete and

(iii) Interlocking of ribs with concrete.

Design of Reinforced Concrete Structures Mrs.S.A.Manchalwar 44

6. What is development length? (or) What do you understand by development length of bar?

Ans: The development length (Ld) is defined as the length of the bar required on either side of the

section under consideration, to develop the required stress in steel at that section through bond. As

per clause 26.2.1 the development length (Ld) is given by,

7. Differentiate shear failure and bending failure.

Ans: i) Shear failure:

Shear failure observed in reinforced concrete structures are diagonal tension failure, flexural shear

failure, shear compression failure and shear bond failure.

(ii) Bending failure:

Flexure or bending failure is commonly encountered in structural elements of reinforced cement

concrete. Example: Beams and Slabs, which are transversely loaded, Flexure usually occurs in

combination with transverse shear and sometimes with axial compression or shear

8. What do you mean by primary and secondary torsion?

Ans: 1) Primary torsion:

Primary torsion (or) Equilibrium torsion is induced by an eccentric loading and equilibrium conditions

alone suffice in determining the twisting moments.

(ii) Secondary Torsion:

Secondary torsion (or) Compatibility torsion is induced by the need for the member to undergo an angle of

twist to maintain deformation compatibility, and the resulting twisting moment depends on the torsional

stiffness of the member

9. Write down the effect of torsion in RC beams?

Ans: Generally, the R.C. structures subjected to torsion of varying magnitude in addition with flexure and

shear. Torsion moment occurs as a secondary effect in many structures.

Example: (i) Peripheral beams in each floor of a multi-storey building are subjected to significant torsion

loading in addition in addition to conventional flexure and shear. (ii) The beams supporting cantilever

canopy slabs are always subjected to torsion.

10. How to overcome torsion on beams?

Ans: When torsion is present, different methods to overcome torsion is done by proper design as per Indian

Standard Code. When torsion is prevent along with bending shear, is recommends the use of are equivalent

shear for which the shear steels are calculated. Again Indian Standards when torsion is present as

combined with bending, an equivalent bending moment is calculated and reinforcement for this equivalent

gn of Reinforced Concrete Structures 45 Dr. T. S. Rames

bending moment is provided as longitudinal steel

4) LONG QUESTIONS

1. Name and explain the three different failure modes of reinforced concrete beams under the

combined effects of bending moment and shear force.

Ans:

Bending in reinforced concrete beams is usually accompanied by shear, the exact analysis of which is very

complex. However, experimental studies confirmed the following three different modes of failure due to

possible combinations of shear force and bending moment at a given section.

Web shear (Fig. 6.13.1a)

(ii) Flexural tension shear (Fig. 6.13.1b)

(iii) Flexural compression shear (Fig. 6.13.1c)

Web shear causes cracks which progress along the dotted line shown in Fig. 6.13.1a. Steel yields in

flexural tension shear as shown in Fig. 6.13.1b, while concrete crushes in compression due to flexural

compression shear as shown in Fig. 6.13.1c. An in-depth presentation of the three types of failure modes is

beyond the scope here. Only the salient points needed for the routine design of beams in shear are

presented here

2. How do you determine the critical sections for shear in a beam?

Ans:

Desi h Babu, Professor

Design of Reinforced Concrete Structures Mrs.S.A.Manchalwar

46

The critical section for shear and are as follows:

For beams generally subjected to uniformly distributed loads or where the principal load is located

further than 2d from the face of the support, where d is the effective depth of the beam, the critical

sections depend on the conditions of supports mentioned in below.

(i) When the reaction in the direction of the applied shear introduces tension into the end region of

the member, the shear force is to be computed at the face of the support of the member at that

section.

(ii) When the reaction in the direction of the applied shear introduces compression into the end

region of the member (Figs. 6.13.3b and c), the shear force computed at a distance d from the face

of the support is to be used for the design of sections located at a distance less than d from the face

of the support. The enhanced shear strength of sections close to supports, however, may be

considered as discussed in the following section.

3. When and why do we consider enhanced shear strength of concrete?

Ans:

Figure 6.13.4 shows the shear failure of simply supported and cantilever beams without shear


47

reinforcement. The failure plane is normally inclined at an angle of 30o to the horizontal. However, in

some situations the angle of failure is more steep either due to the location of the failure section closed to a

support or for some other reasons. Under these situations, the shear force required to produce failure is

increased.

Such enhancement of shear strength near a support is taken into account by increasing the design shear

strength of concrete to (2dτc/av) provided that the design shear stress at the face of the support remains

less than the value of τcmax given in Table 6.2 (Table 20 of IS 456). In the above expression of the

enhanced shear strength

d = effective depth of the beam,

τc = design shear strength of concrete before the enhancement as given in Table 6.1 (Table 19 of IS 456),

av = horizontal distance of the section from the face of the support (Fig. 6.13.4).

Similar enhancement of shear strength is also to be considered for sections closed to point loads. It is

evident from the expression (2dτc /av) that when av is equal to 2d, the enhanced shear strength does not

come into picture. Further, to increase the effectivity, the tension reinforcement is recommended to be

extended on each side of the point where it is intersected by a possible failure plane for a distance at least

equal to the effective depth, or to be provided with an equivalent anchorage.

4. What is meant by “Design shear strength of concrete τc”?

Ans: Recent laboratory experiments confirmed that reinforced concrete in beams has shear strength even

without any shear reinforcement. This shear strength (τc) depends on the grade of concrete and the

percentage of tension steel in beams. On the other hand, the shear strength of reinforced concrete with the

reinforcement is restricted to some maximum value τcmax depending on the grade of concrete. These

minimum and maximum shear strengths of reinforced concrete (IS 456, cls. 40.2.1 and 40.2.3,

respectively) are given below:

Table 19 of IS 456 stipulates the design shear strength of concrete τc for different grades of concrete with a

wide range of percentages of positive tensile steel reinforcement. It is worth mentioning that the reinforced

concrete beams must be provided with the minimum shear reinforcement as per cl. 40.3 even when τv is

less than τc given in Table 6.1.

5. How do you know the maximum shear stress of concrete beams τcmax with shear reinforcement?

Ans: able 20 of IS 456 stipulates the maximum shear stress of reinforced concrete in beams τcmax as given

below in Table 6.2. Under no circumstances, the nominal shear stress in beams τv shall exceed τcmax

given in Table 6.2 for different grades of concrete.

Table 6.2 Maximum shear stress, τcmax in N/mm2



48

Ans: When τv is more than τc given in Table 6.1, shear reinforcement shall be provided in any of the three

following forms:

(a) Vertical stirrups,

(b) Bent-up bars along with stirrups, and

(c) Inclined stirrups.

In the case of bent-up bars, it is to be seen that the contribution towards shear resistance of bent-up bars

should not be more than fifty per cent of that of the total shear reinforcement.

The amount of shear reinforcement to be provided is determined to carry a shear force Vus equal to

Vus = Vu – τc b d

The strengths of shear reinforcement Vus for the three types of shear reinforcement are as follows:

(a) Vertical stirrups

(b) For inclined stirrups or a series of bars bent-up at different cross-sections:

(c) For single bar or single group of parallel bars, all bent-up at the same cross-section:

7. State the conditions to be satisfied for the curtailment of tension reinforcement when designing the shear

reinforcement.

Ans: Curtailment of tension reinforcement is done to provide the required reduced area of steel with the

reduction of the bending moment. However, shear force increases with the reduction of bending moment.

Therefore, it is necessary to satisfy any one of following three conditions while terminating the flexural

reinforcement in tension zone:

(i) The shear stress τv at the cut-off point should not exceed two-thirds of the permitted value which

includes the shear strength of the web reinforcement. Accordingly

(ii) For each of the terminated bars, additional stirrup area should be provided over a distance of three-

fourth of effective depth from the cut-off point. The additional stirrup area shall not be less than 0.4 b s/fy,

where b is the breadth of rectangular beams and is replaced by bw, the breadth of the web for flanged

beams, s = spacing of additional stirrups and fy is the characteristic strength of stirrup reinforcement in

N/mm2. The value of s shall not exceed d/(8 βb), where βb is the ratio of area of bars cut-off to the total

area of bars at that section, and d is the effective depth.


49

(iii) For bars of diameters 36 mm and smaller, the continuing bars provide double the area required for

flexure at the cut-off point. The shear stress should not exceed three-fourths that permitted. Accordingly,

8. How do we design the shear reinforcement close to the support of a beam?

Ans: As stipulated in cl. 40.5.2 of IS 456, the total area of the required shear reinforcement As is obtained

from: As = av b (τv – 2d τc /av)/0.87 fy and ≥ 0.4 av b/0.87 fy

For flanged beams, b will be replaced by bw, the breadth of the web of flanged beams.

This reinforcement should be provided within the middle three quarters of av, where av is less than d,

horizontal shear reinforcement will be effective than vertical.

Alternatively, one simplified method has been recommended in cl. 40.5.3 of IS 456 and the same is given

below.

The following method is for beams carrying generally uniform load or where the principal load is located

further than 2d from the face of support. The shear stress is calculated at a section a distance d from the

face of support. The value of τc is calculated in accordance with Table 6.1 and appropriate shear

reinforcement is provided at sections closer to the support. No further check for shear at such sections is

required.

Explain, with examples, the difference between equilibrium torsion and compatibility torsion.

Ans: Torsion when encountered in reinforced concrete members usually occurs in combination with

flexure shear. Torsion in its associated with metal form shafts is generally rarely encountered in

reinforced concrete.

The interactive behaviour of torsion with bending moment and flexural shear in reinforced concrete

beams is fairly complex, owing to the no homogeneous, nonlinear and composite nature of the

material and the presence of cracks. For convenience in design, codes prescribe highly

simplified design procedures, which reflect a judicious blend of theoretical considerations and

experimental results.

These design procedures and their bases are described in this chapter, following a brief review of

the general behaviour of reinforced concrete beams under torsion.

10. Under what situations do the following modes of cracking occur in reinforced concrete beams: (a)

flexural cracks, (b) diagonal tension cracks, (c) flexural-shear cracks and (d) splitting cracks?

Ans: flexural cracks: Flexural cracks on the sides of a beam start at the tension face and will extend, at

most, up to the neutral axis. Crack widths will be greatest at the tension face and will reduce with distance

from that face. In general, the cracks will be uniformly spaced along the most heavily loaded portion of the

beam, i.e. near the mid-span in sagging or over the supports in hogging.

Flexural cracks on the soffit of a slab will run at right-angles to the span, again roughly uniformly spaced

in the region of maximum moment. In beams and slabs in buildings that have been correctly designed,

average crack widths should not exceed 0.3 mm.


50

Diagonal tension cracks: Beams must have an adequate safety margin against other types of failure, some

of which may be more dangerous than flexural failure. Shear failure of reinforced concrete, more properly

called “diagonal tension failure” is one example. Flexural-shear cracks.


51


UNIT – IV

DESIGN OF COLUMNS

1. If the diameter of longitudinal bars of a square column is 16 mm, the diameter of lateral ties should

not be less than

A. 4 mm, B. 5 mm, C. 6 mm, D. 8 mm, E. 10 mm

Answer: Option B

2. In a combined footing for two columns carrying unequal loads, the maximum hogging bending

moment occurs at

A. less loaded column

B. more loaded column

C. a point equidistant from either column

D. a point of the maximum shear force

E. a point of zero shears force.

Answer: Option E

3. If the size of a column is reduced above the floor, the main bars of the columns, are

A. continued up

B. bent inward at the floor level

C. stopped just below the floor level and separate lap bars provided

D. all the above.

Answer: Option D

4. The minimum clear cover for R.C.C. columns shall be

A. greater of 40 mm or diameter

B. smaller of 40 mm or diameter

C. greater of 25 mm or diameter

D. Smaller of 25 mm or diameter

Answer: Option C

5. Lapped splices in tensile reinforcement are generally not used for bars of size larger than

A. 18 mm diameter

B. 24 mm diameter

C. 30 mm diameter

D. 36 mm diameter

E. 32 mm diameter

Answer: Option D

6. In a pre-stressed member it is advisable to use

A. low strength concrete only

B. high strength concrete only

C. low strength concrete but high tensile steel


52

D. high strength concrete and high tensile steel

E. high strength concrete but low tensile steel

Answer: Option D

7. The ratio of the breadth to effective depth of a beam is kept

A. 0.25

B. 0.50

C. 0.70

D. 0.75

Answer: Option B

8. Spacing of stirrups in a rectangular beam, is

A. kept constant throughout the length

B. decreased towards the centre of the beam

C. increased at the ends

D. increased at the centre of the beam.

Answer: Option D

9. Steel bars are generally connected together to get greater length than the standard length by

providing

A. straight bar splice

B. hooked splice

C. dowel splice

D. all the above

Answer: Option D

10. By over-reinforcing a beam, the moment of resistance can be increased not more than

A. 10%

B. 15%

C. 20%

D. 25%

Answer: Option D


1. Minimum spacing between horizontal parallel reinforcement of the same size should not be less

than-----------

Ans: One diameter

2. The maximum ratio of span to depth of a slab simply supported and spanning in one direction, is-

---------------

Ans: 30

3. The load stress of a section can be reduced by---------

53

Ans: replacing larger bars by greater number of small bars

4. For M 150 grade concrete (1:2:4) the moment of resistance factor is----------

Ans: 8.50

5. For a ribbed slab--------------

Ans: clear spacing between ribs shall not be greater than 4.5 cm

6. An R.C.C. column is treated as short column if its slenderness ratio is less than----

Ans: 50

7. If the width of the foundation for two equal columns is restricted, the shape of the footing

generally adopted, is-----------

Ans: rectangular

8. Columns may be made of plain concrete if their unsupported lengths do not exceed their least

lateral dimension

Ans: four times

9. For a number of columns constructed in a rcjw, the type of foundation provided, is

Ans: strip

10 As the percentage of steel increases

Ans: depth of neutral axis increases

3) Short questions

1. Explain briefly about classifications of columns with neat sketchs.

Ans: (i) Columns subjected to axial loads only (concentric), as shown in Fig.1.a.

(ii) Columns subjected to combined axial load and uniaxial bending, as shown in Fig 1.b

(iii) Columns subjected to combined axial load and bi-axial bending, as shown in Fig. 1.c.

Figure 1(a) axial loading (concentric) Figure (b) Axial loading with uniaxial bending


Mrs.S.A.Manchalwar

54

Figure 1(c) axial loading with biaxial bending

2. Write short notes on Transverse Reinforcement

Transverse reinforcing bars are provided in forms of circular rings, polygonal links (lateral ties) with

internal angles not exceeding 1350 or helical reinforcement. The transverse reinforcing bars are

provided to ensure that every longitudinal bar nearest to the compression face has effective lateral

support against buckling. Clause 26.5.3.2 stipulates the guidelines of the arrangement of transverse

reinforcement. The salient points are

(a) Transverse reinforcement shall only go round corner and alternate bars if the longitudinal bars are

not spaced more than 75 mm on either side

(b) Longitudinal bars spaced at a maximum distance of 48 times the diameter of the tie shall be tied by

single tie and additional open ties for in between longitudinal bars

(c) For longitudinal bars placed in more than one row (i) transverse reinforcement is provided for the

outer-most row in accordance with (a) above, and (ii) no bar of the inner row is closer to the nearest

compression face than three times the diameter of the largest bar in the inner row.

(d) For longitudinal bars arranged in a group such that they are not in contact and each group is

adequately tied as per (a), (b) or (c) above, as appropriate, the transverse reinforcement for the

compression member as a whole may be provided assuming that each group is a single longitudinal bar

for determining the pitch and diameter of the transverse reinforcement. The diameter of such transverse

reinforcement should not, however, exceed 20 mm

3. Write short note on Pitch and Diameter of Lateral Ties

(a) Pitch: The maximum pitch of transverse reinforcement shall be the least of the following:

(i) The least lateral dimension of the compression members;

(ii) Sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied

(iii) 300 mm.

(b) Diameter: The diameter of the polygonal links or lateral ties shall be not less than one-fourth of the

diameter of the largest longitudinal bar, and in no case less than 6 mm.

4. Write a short note on Shallow Foundation

Shallow foundations are used when the soil has sufficient strength within a short depth below the

ground level. They need sufficient plan area to transfer the heavy loads to the base soil. These heavy

loads are sustained by the reinforced concrete columns or walls (either of bricks or reinforced concrete)

of much less areas of cross-section due to high strength of bricks or reinforced concrete when compared

to that of soil. The strength of the soil, expressed as the safe bearing capacity of the soil is normally

supplied by the geotechnical experts to the structural engineer. Shallow foundations are also designated

as footings. The different types of shallow foundations or footings are discussed below.

(i) Plain concrete pedestal footings

(ii) Isolated footings


Mrs.S.A.Manchalwar

55

(iii) Combined footings

(iv) Strap footings

(v) Strip foundation or wall footings

(vi) Raft or mat foundation

4) Long questions

1. Design the reinforcement in a column of size 400 mm x 600 mm subjected to an axial load of 2000 kN

under service dead load and live load. The column has an unsupported length of 4.0 m and effectively

held in position and restrained against rotation in both ends. Use M 25 con Solution

Step 1: To check if the column is short or slender

Given l = 4000 mm, b = 400 mm and D = 600 mm. Table 28 of IS 456 = lex = ley = 0.65(l) = 2600 mm.

So, we have

lex/D = 2600/600 = 4.33 < 12

ley/b = 2600/400 = 6.5 < 12

Hence, it is a short column.

Step 2: Minimum eccentricity

ex min = Greater of (lex/500 + D/30) and 20 mm = 25.2 mm

ey min = Greater of (ley/500 + b/30) and 20 mm = 20 mm

0.05 D = 0.05(600) = 30 mm > 25.2 mm (= ex min)

0.05 b = 0.05(400) = 20 mm = 20 mm (= ey min)

Hence, the equation given in cl.39.3 of IS 456 (Eq.(1)) is applicable for the design here.

Step 3: Area of steel

Pu = 0.4 fck Ac + 0.67 fy Asc

3000(103) = 0.4(25){(400)(600) – Asc} + 0.67(415) Asc

which gives, Asc = 2238.39 mm2

Provide 6-20 mm diameter and 2-16 mm diameter rods giving 2287 mm2 (> 2238.39 mm2) and p =

0.953 per cent, which is more than minimum percentage of 0.8 and less than maximum percentage of

4.0. Hence, o.k.

Step 4: Lateral ties

The diameter of transverse reinforcement (lateral ties) is determined from cl.26.5.3.2 C-2 of IS 456 as

not less than (i) θ/4 and (ii) 6 mm. Here, θ = largest bar diameter used as longitudinal reinforcement =

20 mm. So, the diameter of bars used as lateral ties = 6 mm.

The pitch of lateral ties, as per cl.26.5.3.2 C-1 of IS 456, should be not more than the least of

(i) The least lateral dimension of the column = 400 mm

(ii) Sixteen times the smallest diameter of longitudinal reinforcement bar to be tied = 16(16) = 256 mm

(iii) 300 mm

Let us use p = pitch of lateral ties = 250 mm.

2. What are the Types of Foundation Structures? Explain deep foundation.

Ans: 1. Shallow Foundation


Mrs.S.A.Manchalwar

56

of Reinforced Concrete Structures 56 Dr. T. S. Ramesh Babu,

2. Deep foundations

Deep foundations

As mentioned earlier, the shallow foundations need more plan areas due to the low strength of soil

compared to that of masonry or reinforced concrete. However, shallow foundations are selected when

the soil has moderately good strength, except the raft foundation which is good in poor condition of soil

also. Raft foundations are under the category of shallow foundation as they have comparatively shallow

depth than that of deep foundation. It is worth mentioning that the depth of raft foundation is much

larger than those of other types of shallow foundations.

However, for poor condition of soil near to the surface, the bearing capacity is very less and foundation

needed in such situation is the pile foundation. Piles are, in fact, small diameter columns which are

driven or cast into the ground by suitable means. Precast piles are driven and cast-in-situ are cast. These

piles support the structure by the skin friction between the pile surface and the surrounding soil and end

bearing force, if such resistance is available to provide the bearing force. Accordingly, they are

designated as frictional and end bearing piles. They are normally provided in a group with a pile cap at

the top through which the loads of the superstructure are transferred to the piles.

Piles are very useful in marshy land where other types of foundation are impossible to construct. The

length of the pile which is driven into the ground depends on the availability of hard soil/rock or the

actual load test. Another advantage of the pile foundations is that they can resist uplift also in the same

manner as they take the compression forces just by the skin friction in the opposite direction.

However, driving of pile is not an easy job and needs equipment and specially trained persons or

agencies. Moreover, one has to select pile foundation in such a situation where the adjacent buildings

are not likely to be damaged due to the driving of piles. The choice of driven or bored piles, in this

regard, is critical.

Exhaustive designs of all types of foundations mentioned above are beyond the scope of this course.

Accordingly, this module is restricted to the design of some of the shallow footings, frequently used for

normal low rise buildings only.


Mrs.S.A.Manchalwar

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3. Write column design steps.

(a) Minimum nominal cover (cl. 26.4.2.2 of IS 456)

The minimum nominal cover for the footings should be more than that of other structural elements of

the superstructure as the footings are in direct contact with the soil. Clause 26.4.2.2 of IS 456

prescribes a minimum cover of 50 mm for footings. However, the actual cover may be even more

depending on the presence of harmful chemicals or minerals, water table etc.

(b) Thickness at the edge of footings (cls. 34.1.2 and 34.1.3 of IS 456)

The minimum thickness at the edge of reinforced and plain concrete footings shall be at least 150 mm

for footings on soils and at least 300 mm above the top of piles for footings on piles, as per the

stipulation in cl.34.1.2 of IS 456.

(c) Bending moments (cl. 34.2 of IS 456)

1. It may be necessary to compute the bending moment at several sections of the footing depending on

the type of footing, nature of loads and the distribution of pressure at the base of the footing. However,

bending moment at any section shall be determined taking all forces acting over the entire area on one

side of the section of the footing, which is obtained by passing a vertical plane at that section

extending across the footing (cl.34.2.3.1 of IS 456).

2. The critical section of maximum bending moment for the purpose of designing an isolated concrete

footing which supports a column, pedestal or wall shall be:

(i) at the face of the column, pedestal or wall for footing supporting a concrete column,

pedestal or reinforced concrete wall, and

(ii) halfway between the centre-line and the edge of the wall, for footing under masonry wall.

This is stipulated in cl.34.2.3.2 of IS 456.

The maximum moment at the critical section shall be determined as mentioned in 1 above.

For round or octagonal concrete column or pedestal, the face of the column or pedestal shall

be taken as the side of a square inscribed within the perimeter of the round or octagonal

column or pedestal (see cl.34.2.2 of IS 456 and Figs.11.28.13a and b).

(d) Shear force (cl. 31.6 and 34.2.4 of IS 456)

Footing slabs shall be checked in one-way or two-way shears depending on the nature of

bending. If the slab bends primarily in one-way, the footing slab shall be checked in one-way vertical

shear. On the other hand, when the bending is primarily two-way, the footing slab shall be checked in

two-way shear or punching shear. The respective critical sections and design shear strengths

4. Design a brick masonry column which carries a super imposed axial compressive load of 100 KN at

base of column. Take effective heights of column as 3.0m Design the footing if SBC of soil is 100

KN/m2.

Solution

The super imposed load at the base of column =100 KN

Using H@ grade of mortar (CM=1:2), permissible axial compressive load for a short column (SR<= 12)

from table 8 =0.96 N/mm2


Mrs.S.A.Manchalwar

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Area of section required = 100x1000/0.96 = 104166 mm2

Provide 345mmx345mm square section of column.

Area provided =119625 mm2

Therefore, SR = 3x1000/345 = 8.695

Therefore it is a short column.

Load on soil assuming 15% as self weight of footing =100x1.15= 115 KN

Net SBC of soil = 115/100 =1.15 m2

Provide square footing 1.2mx1.2m

Area provide = 1.44 m2 >1.15 m2

Design is safe.


DESIGN OF FOOTING

1. To ensure uniform pressure distribution, the thickness of the foundation, is

A. kept uniform throughout

B. increased gradually towards the edge

C. decreased gradually towards the edge

D. kept zero at the edge.

Answer: Option C

2. The toe projection of foundation slabs is taken

A. as one third of the base

B. as one sixth of overall height of the wall

C. equal to heel slab

D. below ground surface.

Answer: Option A

3. The weight of a foundation is assumed as

A. 5% of wall weight, B. 7% of wall weight, C. 10% of wall weight, D. 12% of wall weight

Answer: Option C

4. The horizontal portion of a step in a stairs case, is known as

A. rises, B. flight, C. winder, D. tread.

Answer: Option D

5. Top bars are extended to the projecting parts of the combined footing of two columns L distance

apart for a distance of

A.0.1 L from the outer edge of column


Mrs.S.A.Manchalwar

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B. 0.1 L from the centre edge of column

C. half the distance of projection

D. one-fourth the distance of projection.

Answer: Option B

6. A raft foundation is provided if its area exceeds the plan area of the building by

A. 10%, B. 20%, C. 30%, D.,40%, E. 50%

Answer: Option E

7. In a combined footing if shear stress does not exceed 5 kg/cm2, the nominal stirrups provided are

A. 6 legged, B. 8 legged, C. 10 legged, D. 12 legged

Answer: Option B

8. A foundation is called shallow if its depth, is

A. one-fourth of its width

B. half of its width

C. three-fourth of its width

D. equal to its width

E. all the above.

Answer: Option D

9. In testing a pile by load test, pile platform is loaded with one and half times the design load and a

maximum settlement is noted. The load is gradually removed and the consequent rebound is

measured. For a safe pile, the net settlement (i.e. total settlement minus rebound) per tonne of test

load should not exceed

A. 10 mm

B. 15 mm

C. 20 mm

D. 25 mm

E. 30 mm

Answer: Option D

10. On piles, the drop must be at least

A. 80 cm

B. 100 cm

C. 120 cm

D. 140 cm

E. 150 cm

Answer: Option C


1. The diameter of transverse reinforcement of columns should be equal to one-fourth of the

diameter of the main steel rods but not less than-----


Mrs.S.A.Manchalwar

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Ans: 7 mm

2. The diameter of the column head support a flat slab, is generally kept--------------

Ans: 0.25 times the span length

3. The steel generally used in R.C.C. work, is-------------

Ans: mild steel

5. According to I.S.: 456, 1978 the thickness of reinforced concrete footing on piles at its edges, is

kept less than-------------

Ans: 15 cm

6. The self-weight of the footing, is------------------

Ans: not considered for calculating the upward pressure on footing

7. A very comfortable type of stairs is--------------

Ans: open newel.

8. The assumption for the design of a pre stressed concrete member from---

Ans: A transverse plane section remains a plane after bending

9. The weight of reinforced concrete, is generally taken as-----------

Ans: 2400 kg/m3

10. The Young's modulus of elasticity of steel, is-------------

Ans: 275 KN/mm2

3) SHORT QUESTIONS

1. What are the types of staircases?

A. straight stairs (with or without intermediate landing)

B. quarter-turn stairs

C. dog-legged stairs

D. open well stairs [Fig. 2.11 (d)]

E. spiral stairs

F. helicoidally stairs

2. What are the structural Classifications?

Structurally, staircases may be classified largely into two categories, depending on the predominant

direction in which the slab component of the stair undergoes flexure:

1. Stair slab spanning transversely (stair width wise);

2. Stair slab spanning longitudinally (along the incline).

3. Write short note on generalized design steps of foundation design.


Mrs.S.A.Manchalwar

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Ans: Design of foundations with variable conditions and variable types of foundation structures will be

different, but there are steps that are typical to every design, including:

1. Calculate loads from structure, surcharge, active & passive pressures, etc.

2. Characterize soil – hire a firm to conduct soil tests and produce a report that includes soil

material properties

3. Determine footing location and depth – shallow footings are less expensive, but the variability of

the soil from the geotechnical report will drive choices

4. Evaluate soil bearing capacity – the factor of safety is considered here

5. Determine footing size – these calculations are based on working loads and the allowable soil

pressure

6. Calculate contact pressure and check stability

7. Estimate settlements

8. Design the footing structure – design for the material based on applicable structural design codes

which may use allowable stress design, LRFD or limit state design (concrete).

4) LONG QUESTIONS

1. What are the structural Classifications?

Ans: Structurally, staircases may be classified largely into two categories, depending on the

predominant direction in which the slab component of the stair undergoes flexure:

1. Stair slab spanning transversely (stair width wise);

2. Stair slab spanning longitudinally (along the incline).

Stair Slab Spanning Transversely

The slab component of the stair (whether comprising an isolated tread slab, a tread-riser unit or a waist

slab) is supported on its side(s) or cantilevers laterally from a central support. The slab supports

gravity loads by bending essentially in a transverse vertical plane, with the span along the width of the

stair.

In the case of the cantilevered slabs, it is economical to provide isolated treads (without risers).

However, the tread-riser type of arrangement and the waist slab type are also sometimes employed in

practice, as cantilevers. The spandrel beam is subjected to torsion (equilibrium torsion) in addition to

flexure and shear.

When the slab is supported at the two sides by means of ‗stringer beams„ or masonry walls, it may be

designed as simply supported, but reinforcement at the top should be provided near the supports to

resist the ‗negative„ moments that may arise on account of possible partial fixity.

Stair Slab Spanning Longitudinally

In this case, the supports to the stair slab are provided parallel to the riser at two or more locations,

causing the slab to bend longitudinally between the supports. It may be noted that longitudinal

bending can occur in configurations other than the straight stair configuration, such as quarter-turn

stairs, dog-legged stairs, open well stairs and helicoidally stairs.

The slab arrangement may either be the conventional waist slab type or the tread riser type. The slab

thickness depends on the effective span which should be taken as the centre-to-centre distance

between the beam/wall supports, according to the Code Cl. 33.1a, c. In certain situations, beam or wall

supports may not be available parallel to the riser at the landing. Instead, the flight is supported


Mrs.S.A.Manchalwar

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between the landings, which span transversely, parallel to the risers.

2. Design a (waist slab type) dog-legged staircase for an office building, given the following data:

• Height between floor = 3.2 m;

• Riser = 160 mm, tread = 270 mm;

• Width of flight = landing width = 1.25 m

• Live load = 5.0 kN/m2

• Finishes load = 0.6 kN/m2

Assume the stairs to be supported on 230 mm thick masonry walls at the outer edges of the landing,

parallel to the risers Use M 20 concrete and Fe 415 steel. Assume mild exposure conditions.

Ans: Given: R = 160 mm, T = 270 mm ⇒RT22 = 314 mm Effective span = c/c distance between supports = 5.16 m

Assume a waist slab thickness ≈l20 = 5160/20 = 258 →260 mm.

Assuming 20 mm clear cover (mild exposure) and 12 θ main bars, effective depth

d = 260 – 20 – 12/2 = 234 mm.

The slab thickness in the landing regions may be taken as 200 mm, as the bending moments are

relatively low here.

Loads on going [fig. below] on projected plan area:

(1) Self-weight of waist slab @ 25 × 0.26 × 314/270 = 7.56 kN/m2

(2) Self-weight of steps @ 25 × (0.5x0.16) = 2.00 kN/m2

(3) Finishes (given) = 0.60 kN/m2

(4) live load (given) = 5.00 kN/m2

Total =15.16 kN/m2

=Factored load = 15.16 × 1.5 = 22.74 kN/m2

• Loads on landing

(1) self-weight of slab @ 25 × 0.20 = 5.00 kN/m2

(2) finishes @ 0.6 kN/m2

(3) live loads @ 5.0 kN/m2

Total =10.60 kN/m2

=Factored load = 10.60 × 1.5 = 15.90 kN/m2

• Design Moment

Reaction R= (15.90x1.365)+(22.74x2.43)/2 = 49.33 kN/m

Maximum moment at midspan:

Mu = (49.33 × 2.58) – (15.90 × 1.365) × (2.58 – 1.365/2) – (22.74) × (2.58 – 1.365)2/2

= 69.30 kN m

Main reinforcement

R= Mu/bd2 = 1.265 MPa

Assuming fck = 20 MPa, fy = 415 MPa,


Mrs.S.A.Manchalwar

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Pt/100= Ast= 0.38x10-2 = 892mm2 / m

Required spacing of 12 θ bars = 127 mm

Required spacing of 16 θ bars = 225 mm

Provide 16 θ @ 220c/c

Distributors

2 (Ast )req = 0.0012bt = 312mm2 / m

spacing 10 θ bars = 251 mm

Provide 10 θ @ 250c/c as distributors

3. Design an isolated footing of uniform thickness of a RC column bearing a vertical load of

600 KN and having a base of size 500x500 mm. the safe bearing capacity of soil may be taken as 120

KN/m2. Use M20 concrete and Fe 415 steel.

Solution

1. Size of footing

W=600 KN;

Self-weight of footing @ 10% =60 KN

Total load =660 KN

Size of footing = 660/120 = 5.5 m2

Since square footing, B= 5.5 =2.345 m2

Provide a square footing = 2.4mx 2.4m


Mrs.S.A.Manchalwar

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Net upward pressure, p0= 600/(2.4x2.4) = 104.17 KN/m2

2. Design of section

The maximum BM acts at the face of column

M= po*(B/8)*(B-b2) = 112.8KN-m

Mu = 1.5M =169.2 KN-m

Therefore d = 160 mm; D = 160+60 = 220mm

3. Depth on the basis of one-way shear

For a one way shear, critical section is located at a distance d from the face of the column where shear

force V is given by

V = poB{0.5(B-b) -d} = 104.17x2.4{0.5x(2.4-0.5) -0.001d}

Vu = 1.5V

Tc = Vu/bd= [375012(0.95 0.001)]/2400d

From table B.5.2.1.1 of IS 456:2000 k=1.16 for D = 220mm.

Also for under-

Hence design shear stress = k tc =0.445 N/mm2

From which we get d= 246.7 250 mm

4. write types foundations.

A. Shallow Foundation Types

B. Deep Foundation Types

A. Shallow Foundation Types

a. Spread footing – A single column bears on a square or rectangular pad to distribute the load

over a bigger area. Wall footing – A continuous wall bears on a wide pad to distribute the load.

b. Eccentric footing – A spread or wall footing that also must resist a moment in addition to the

axial column load.

c. Combined footing – Multiple columns (typically two) bear on a rectangular or trapezoidal

shaped footing.

d. Unsymmetrical footing – A footing with a shape that does not evenly distribute bearing

pressure from column loads and moments. It typically involves a hole or a non-rectangular shape

influenced by a boundary or property line.

e. Strap footing – A combined footing consisting of two spread footings with a beam or strap

connecting the slabs. The purpose of this is to limit differential settlements.

f. Mat foundation – A slab that supports multiple columns. The mat can be stiffened with a grid

or grade beams. It is typically used when the soil capacity is very low.

B. Deep Foundation Types

a. Retaining Walls – A wall that retains soil or other materials, and must resist sliding and

overturning. Can have counterforts, buttresses or keys.

b. Basement Walls – A wall that encloses a basement space, typically next to a floor slab, and


Mrs.S.A.Manchalwar

65

that may be restrained at the top by a floor slab.

c. Piles – Next choice when spread footings or mats won‟t work, piles are used to distribute

loads by end bearing to strong soil or friction to low strength soils. Can be used to resist uplift, a

moment causing overturning, or to compact soils. Also useful when used in combination, to

control settlements of mats or slabs.

d. Drilled Piers – Soil is removed to the shape of the pier and concrete is added.

e. Caissons –Water and possibly wet soil is held back or excavated while the footing is

constructed or dropped into place.

5. Write notes on Loads and Stresses in foundation.

Ans: Bearing loads must be distributed to the soil materials, but because of their variability and the

stiffness of the footing pad, the resulting stress, or soil pressure, is not necessarily uniform. But we

assume it is for design because dealing with the complexity isn‟t worth the time or effort.

The increase in weight when replacing soil with concrete is called the overburden. Overburden may also

be the result of adding additional soil to the top of the excavation for a retaining wall. It is extra

uniformly distributed load that is considered by reducing the allowable soil pressure

(Instead of increasing the loads), resulting in a net allowable soil pressure, qnet:

In order to design the footing size, the actual stress P/A must be less than or equal to the allowable

pressure:


Mrs.S.A.Manchalwar

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UNIT – V

DESIGN OF SLABS


1. In a simply supported slab the minimum spacing of distribution reinforcement, should be four

times the effective thickness of the slab or

A. 20 cm, B. 30 cm, C. 40 cm, D. 50 cm, E. 60 cm

Answer: Option E

2. If the sides of a slab simply supported on edges and spanning in two directions are equal, the

maximum bending moment is multiplied by

A.0.2, B.0.3, C.0.4, D .0.5

Answer: Option D

3. The toe projection of foundation slabs is taken

A. as one third of the base

B. as one sixth of overall height of the wall

C. equal to heel slab

D. below ground surface.

Answer: Option A

4. Distribution reinforcement in a simply supported slab, is provided to distribute

A. load B. temperature stress C. shrinkage stress D. all the above.

Answer: Option D

5. The thickness of the topping of a ribbed slab, varies between

A. 3 cm to 5 cm, B. 5 cm to 8 cm, C. 8 cm to 10 cm , D. 12 cm to 15 cm,

Answer: Option B

6. The maximum ratio of span to depth of a cantilever slab, is

A. 8, B. 10, C. 12, D. 14, E. 16

Answer: Option C

7. If the maximum bending moment of a simply supported slab is M Kg.cm, the effective depth of the slab

is (where Q is M.R. factor)

A. B. C. D.

Answer: Option d

8. If depth of slab is 10 cm, width of web 30 cm, depth of web 50 cm, centre to centre distance of beams 3


Mrs.S.A.Manchalwar

67

m, effective span of beams 6 m, the effective flange width of the beam, is

A. 200 cm B. 300 cm, C. 150 cm, D. 100 cm

Answer: Option C.

9. A flat slab is supported

A. on beams

B. On columns

C. On beams and columns

D On columns monolithically built with slab

E. all the above

Answer: Option D

10. The live load to be considered for an accessible roof, is

A. Nil

B. 75 kg/m3

C. 150 kg/m2

D. 200 kg/cm2

Answer: Option C


1. A circular slab subjected to external loading, deflects to form a-----------

Ans: parabola

2. In case the factor of safety against sliding is less than 1.5, a portion of slab is constructed

downwards at the end of the heel slab, which is known as--------

Ans: a rib

3. The allowable tensile stress in mild steel stirrups, reinforced cement concrete, is----

Ans: 1400 kg/cm2

4. If the length of an intermediate span of a continuous slab is 5m, the length of the end span is

kept---------

Ans: 4.5 m

5. The floor slab of a building is supported on reinforced cement floor beams. The ratio of the end

and intermediate spans is kept-------

Ans: 0.9

6. In a slab, the pitch of the main reinforcement should not exceed its effective depth--

Ans: three times

7. For a continuous floor slab supported on beams, the ratio of end span length and intermediate

span length, is----------


Mrs.S.A.Manchalwar

68

Ans: 0.9

8. The minimum thickness of a flat slab is taken------------

Ans: L/36 for end panels without drops

9. According to I.S. : 456, slabs which span in two directions with corners held down, are assumed

to be divided in each direction into middle strips and edge strips such that the width of the

middle strip, is-----------------

Ans: three-fourth of the width of the slab

10. The maximum ratio of span to depth of a slab simply supported and spanning in two directions,

is-----------------

Ans: 35

3) SHORT QUESTIONS

1. Explain clearly the difference between one way and two way slabs.

Ans:

2. What are the classifications of slabs?

Ans : Slabs are classified based on many aspects

1) Based of shape: Square, rectangular, circular and polygonal in shape.

2) Based on type of support: Slab supported on walls, Slab supported on beams, Slab supported on

columns (Flat slabs).

3) Based on support or boundary condition: Simply supported, Cantilever slab, Overhanging slab,

Fixed or Continues slab.

4) Based on use: Roof slab, Floor slab, Foundation slab, Water tank slab.

5) Basis of cross section or sectional configuration: Ribbed slab /Grid slab, Solid slab,

Filler slab, folded plate

One way slab Two way Slab

It is supported by beams in only 2 sides. Two way slab is supported by beams in all four

sides.

The ratio of longer span panel to shorter

span panel is equal or greater than 2

The ratio of longer span panel to shorter span

panel is less than 2

Main reinforcement is provided in only one

direction for one way slabs.

Main reinforcement is provided in both the

direction for two way slabs.

nforcement is provided along the spanning direction to resist one way b

inforced Concrete Structures 69 Dr. T. S. Ramesh Babu, Pr

6) Basis of spanning directions:

One way slab – Spanning in one direction

Two way slab _ Spanning in two direction

3. What are the methods of Analysis?

The analysis of slabs is extremely complicated because of the influence of number of factors stated

above. Thus the exact (close form) solutions are not easily available. The various methods are:

a) Classical methods – Levy and Navies solutions(Plate analysis)

b) Yield line analysis – Used for ultimate /limit analysis

c) Numerical techniques – Finite element and Finite difference method.

d) Semi empirical – Prescribed by codes for practical design which uses coefficients.

4. What are general guidelines for design of slab?

a. Effective span of slab:

Effective span of slab shall be lesser of the two

1. l= clear span + d (effective depth )

2. l = Centre to centre distance between the support

b. Depth of slab:

The depth of slab depends on bending moment and deflection criterion. the trail depth can be

obtained using:

• Effective depth d= Span /((l/d)Basic x modification factor)

• For obtaining modification factor, the percentage of steel for slab can be assumed from 0.2 to 0.5%

• The effective depth d of two way slabs can also be assumed using cl.24.1,IS 456

provided short span is 3.5m and loading class is < 3.5KN/m2

c. Load on slab:

The load on slab comprises of Dead load, floor finish and live load. The loads are calculated per unit

area (load/m2).

Dead load = D x 25 kN/m2 (Where D is thickness of slab in m)

Floor finish (Assumed as)= 1 to 2 kN/m2

Live load (Assumed as) = 3 to 5 kN/m2 (depending on the occupancy of the building)

5. Write short note on behaviour of one way slab?

Ans: When a slab is supported only on two parallel apposite edges, it spans only in the direction

perpendicular to two supporting edges. Such a slab is called one way slab. Also, if the slab is

supported on all four edges and the ratio of longer span(ly) to shorter span (lx) i.e ly/lx > 2, practically

the slab spans across the shorter span. Such a slabs are also designed as one way slabs. In this case, the

main rei ending.

Design of Re ofessor

Design of Reinforced Concrete Structures Mrs S.A.Manchalwar 70

6. Write short note on behaviour of two way slab?

Ans : A rectangular slab supported on four edge supports, which bends in two orthogonal directions

and deflects in the form of dish or a saucer is called two way slabs. For a two way slab the ratio of

ly/lx shall be lesser than 2.0

Since, the slab rest freely on all sides, due to transverse load the corners tend to curl up and

lift up. The slab looses the contact over some region. This is known as lifting of corner. These slabs

are called two way simply supported slabs. If the slabs are cast monolithic with the beams, the corners

of the slab are restrained from lifting. These slabs are called restrained slabs. At corner, the rotation

occurs in both the direction and causes the corners to lift. If the corners of slab are restrained from

lifting, downward reaction results at corner & the end strips gets restrained against rotation. However,

when the ends are restrained and the rotation of central strip still occurs and causing rotation at corner

(slab is acting as unit) the end strip is subjected to torsion.

7. What are the types of Two way slab? Explain any one.

Ans: Two way slabs are classified into two types based on the support conditions:

a) Simply supported slab

b) Restrained slabs

a) Two way simply supported slabs

The bending moments Mx and My for a rectangular slabs simply supported on all four edges with

corners free to lift or the slabs do not having adequate provisions to prevent lifting of corners are

obtained using

Mx = ax W l2x

My = y W l2x

Where, x and y are coefficients given in Table 1 (Table 27,IS 456-2000)

W- Total load /unit area

lx & ly – lengths of shorter and longer span

8. Explain the share of loads by the supporting beams in one- and twoway slabs.

Ans:

Figures 8.18.4a and b explain the share of loads on beams supporting solid slabs along four edges

when vertical loads are uniformly distributed. It is evident from the figures that the share of loads

on beams in two perpendicular directions depends upon the aspect ratio ly /lx of the slab, lx being

the shorter span. For large values of ly, the triangular area is much less than the trapezoidal area

(Fig.8.18.4a). Hence, the share of loads on beams along shorter span will gradually reduce with


increasing ratio of ly /lx. In such cases, it may be said that the loads are primarily taken by beams

along longer span. The deflection profiles of the slab along both directions are also shown in the

figure. The deflection profile is found to be constant along the longer span except near the edges for

the slab panel of Fig.8.18.4a. These slabs are designated as one-way slabs as they span in one

direction (shorter one) only for a large part of the slab when ly /lx > 2. On the other hand, for

square slabs of ly /lx = 1 and rectangular slabs of ly /lx up to 2, the deflection profiles in the two

directions are parabolic (Fig.8.18.4b). Thus, they are spanning in two directions and these slabs

with ly /lx up to 2 are designated as two-way slabs, when supported on all edges. It would be noted

that an entirely one-way slab would need lack of support on short edges. Also, even for ly /lx < 2,

absence of supports in two parallel edges will render the slab one-way. In Fig. 8.18.4b, the

separating line at 45 degree is tentative serving purpose of design. Actually, this angle is a function

of ly /lx .

9. How to determine the design shear strength of concrete in slabs of different depths having the same

percentage of reinforcement?

Ans: Experimental tests confirmed that the shear strength of solid slabs up to a depth of 300 mm is

comparatively more than those of depth greater than 300 mm. Accordingly, cl.40.2.1.1 of IS 456

stipulates the values of a factor k to be multiplied with c τ given in Table 19 of IS 456 for different

overall depths of slab. Table 8.1 presents the values of k as a ready reference below: Table 8.1 Values

of the multiplying factor k

Thin slabs, therefore, have more shear strength than that of thicker slabs. It is the normal practice to choose

the depth of the slabs so that the concrete can resist the shear without any stirrups for slab subjected to


uniformly distributed loads. However, for deck slabs, culverts, bridges and fly over, shear reinforcement

should be provided as the loads are heavily concentrated in those slabs. Though, the selection of depth

should be made for normal floor and roof slabs to avoid stirrups, it is essential that the depth is checked for

the shear for these slabs taking due consideration of enhanced shear strength as discussed above depending

on the overall depth of the slabs.

10. State the maximum diameter of a bar to be used in slabs

Ans: Both for one and two-way slabs, the amount of minimum reinforcement in either direction shall not

be less than 0.15 and 0.12 per cents of the total crosssectional area for mild steel (Fe 250) and high

strength deformed bars (Fe 415 and Fe 500)/welded wire fabric, respectively.

3) LONG QUESTIONS

1. Design a simply supported one –way slab over a clear span of 3.5 m. It carries a live load of 4 kN/m2 and

floor finish of 1.5 kN/m2. The width of supporting wall is 230 mm. Adopt M-20 concrete & Fe-415 steel.

Ans: 1) Trail depth and effective span

Assume approximate depth d =L/26

3500/26 = 134 mm

Assume overall depth D=160 mm & clear cover 15mm for mild exposure

d = 160-15 (cover) -10/2 (dia of Bar/2) =140 mm

Effective span is lesser of the two

i. l =3.5 + 0.23 (width of support) = 3.73 m

ii. l= 3.5 + 0.14 (effective depth) =3.64 m

effective span = 3.64 m

2) Load on slab

i. Self weight of slab = 0.16 x 25 = 4.00

ii. Floor finish = 1.50

iii. Live load = 4.00

= 9.5 kN/m2

Ultimate load Wu = 9.5 x 1.5 = 14.25 kN/m2

3) Design bending moment and check for depth

Mu = Wul2/8= (14.25*3.642)/8= 23.06KN.M

Minimum depth required from BM consideration

d= 92.4>140

4) Area of Reinforcement

Area of steel is obtained using the following equation

Mu= 0.87*fy*ast*d[1-ast/bd*fy/fck]

= 504mm2

Provide 10mm @ 150 C/C

5) Check for shear

Design shear Vu= Wu*l/2= 14.25*(8.64/2)= 25.93KN


2. Design the one-way continuous slab of Fig.8.18.6 subjected to uniformly distributed imposed loads of 5

kN/m2 using M 20 and Fe 415. The load of floor finish is 1 kN/m2. The span dimensions shown in the

figure are effective spans. The width of beams at the support = 300 mm.

Ans: Step 1: Selection of preliminary depth of slab

The basic value of span to effective depth ratio for the slab having simple

support at the end and continuous at the intermediate is (20+26)/2 = 23 (cl.23.2.1

of IS 456).

Modification factor with assumed p = 0.5 and fs = 240 N/mm2

is obtained as 1.18 from Fig.4 of IS 456.

Therefore, the minimum effective depth = 3000/23(1.18) = 110.54 mm. Let us take the effective depth d

= 115 mm and with 25 mm cover, the total depth D = 140 mm.

Step 2: Design loads, bending moment and shear force

Dead loads of slab of 1 m width = 0.14(25) = 3.5 kN/m

Dead load of floor finish =1.0 kN/m

Factored dead load = 1.5(4.5) = 6.75 kN/m

Factored live load = 1.5(5.0) = 7.50 kN/m

Total factored load = 14.25 kN/m

Maximum moments and shear are determined from the coefficients given in Tables 12 and 13 of IS

456.

Maximum positive moment = 14.25(3)(3)/12 = 10.6875 kNm/m

Maximum negative moment = 14.25(3)(3)/10 = 12.825 kNm/m

Maximum shear Vu = 14.25(3)(0.4) = 17.1 kN

Step 3: Determination of effective and total depths of slab

From Eq.3.25 of sec. 3.5.6 of Lesson 5, we have

Mu,lim = R,lim bd2

where R,lim is 2.76 N/mm2 from Table 3.3 of sec. 3.5.6 of Lesson 5. So, d =

{12.825(106)/(2.76)(1000)}0.5 = 68.17 mm

Since, the computed depth is much less than that determined in Step 1, let

us keep D = 140 mm and d = 115 mm.

Step 4: Depth of slab for shear force

Table 19 of IS 456 gives c τ = 0.28 N/mm2 for the lowest percentage of steel in the slab. Further for the

total depth of 140 mm, let us use the coefficient k of cl. 40.2.1.1 of IS 456 as 1.3 to get c τ k τ c = =

1.3(0.28) = 0.364 N/mm2

Table 20 of IS 456 gives c max τ = 2.8 N/mm2.

For this problem τ = uv / bdV = 17.1/115 = 0.148 N/mm2. Since, c max v c τ <τ <τ , the effective depth

d = 115 mm is acceptable.


Step 5: Determination of areas of steel

From Eq.3.23 of sec. 3.5.5 of Lesson 5, we have

Mu = 0.87 fy Ast d {1 – (Ast)(fy)/(fck)(bd)}

(i) For the maximum negative bending moment

12825000 = 0.87(415)(Ast)(115){1 – (Ast)(415)/(1000)(115)(20)}

or - 5542.16 A 2 Ast st + 1711871.646 = 0

Solving the quadratic equation, we have the negative Ast = 328.34 mm2

(ii) For the maximum positive bending moment

10687500 = 0.87(415) Ast(115) {1 – (Ast)(415)/(1000)(115)(20)}

or - 5542.16 A 2 Ast st + 1426559.705 = 0

Solving the quadratic equation, we have the positive Ast = 270.615 mm2

Step 6: Distribution steel bars along longer span ly

Distribution steel area = Minimum steel area = 0.12(1000)(140)/100 = 168 mm2

Since, both positive and negative areas of steel are higher than the minimum area, we provide:

(a) For negative steel: 10 mm diameter bars @ 230 mm c/c for which Ast = 341 mm2

giving ps = 0.2965.

(b) For positive steel: 8 mm diameter bars @ 180 mm c/c for which Ast = 279 mm2 giving ps =

0.2426

(c) For distribution steel: Provide 8 mm diameter bars @ 250 mm c/c for which Ast (minimum) =

201 mm2

3. Design the slab panel 1 of Fig.8.19.7 subjected to factored live load of 8 kN/m2 in addition to its dead

load using M 20 and Fe 415. The load of floor finish is 1 kN/m2. The spans shown in figure are

effective spans. The corners of the slab are prevented from lifting

Ans: Step 1: Selection of preliminary depth of slab The span to depth ratio with Fe 415 is taken

from cl. 24.1, Note 2 of IS 456 as 0.8 (35 + 40) / 2 = 30. This gives the minimum effective depth d

= 4000/30 = 133.33 mm, say 135 mm. The total depth D is thus 160 mm.

Step 2: Design loads, bending moments and shear forces

Dead load of slab (1 m width) = 0.16(25) = 4.0 kN/m2

Dead load of floor finish (given) = 1.0 kN/m2

Factored dead load = 1.5(5) = 7.5 kN/m2

Factored live load (given) = 8.0 kN/m2

Total factored load = 15.5 kN/m2

Step 3: Maximum bending moments and Maximum shear force


Vu = w(lx/2) = 15.5 (4/2) = 31 kN/m

Step 4: Determination/checking of the effective depth and total depth of slab

Using the higher value of the maximum bending moments in x and y

directions from above Table

Mu,lim = R,lim bd2

or d = [(18.6)(106)/{2.76(103)}] 1/2 = 82.09 mm,

where 2.76 N/mm2

Since, this effective depth is less than 135 mm and D = 160 mm.

Step 5: Depth of slab for shear force

Table 20 of IS 456 gives τ c max = 2.8 N/mm2

The computed shear stress

τ v = Vu/bd = 31/135 = 0.229 N/mm2

Since, τv < τc < τ c max , the effective depth of the slab as 135 mm and

the total depth as 160 mm are safe.

Step 6: Determination of areas of steel

The minimum steel is determined from the stipulation of cl. 26.5.2.1 of IS 456 and is

As = (0.12/100)(1000)(160) = 192 mm2


m reinforcement

B. cover to the Reinforcement

C. maximum diameter

D. spacing of Reinforcement

4.

Wh

at

are

the

req

uir

me

nts

of

rein

for

ce

me

nt

in

sla

bs?

An

s:

A.

min

imu

A. minimum reinforcement: the minimum reinforcement in slabs should be 0.12% of the gross

cross sectional area of the slab if height strength steel is used and 0.15%, if mild steel is used. In

one way slabs, the distribution reinforcement should be as per this minimum requirement.

B. cover to the Reinforcement: the clear cover to the reinforcement in slabs shall be as per clause

26.4 of code for bars up to 12mm diameter in slabs, this cover may be taken as 15mm for mild steel

exposure condition.

C. maximum diameter: the maximum diameter of the bar to be used in slab shall not be more than

one eight of the overall thickness of the slab.

D. Spacing of Reinforcement: for reinforcing bars the spacing shall not be more than the least of


i) three times the effective depth of the member

ii) 300mm for distribution of steel, the spacing shall not be more than the least of 450mm

5. write detailing of Reinforcement of slab.

Ans: (i) Restrained slabs The maximum positive and negative moments per unit width of the slab

calculated by employing are applicable only to the respective middle strips. There shall be no

redistribution of these moments. The reinforcing bars so calculated from the maximum moments are to

be placed satisfying the following stipulations of IS 456

Bottom tension reinforcement bars of mid-span in the middle strip shall extent in the lower part of

the slab to within 0.25l of a continuous edge, or 0.15l of a discontinuous edge (cl. D-1.4 of IS 456).

Bars marked as B1, B2, B5 and B6.

Top tension reinforcement bars over the continuous edges of middle strip shall extend in the upper

part of the slab for a distance of 0.15l from the support, and at least fifty per cent of these bars shall

extend a distance of 0.3l (cl. D-1.5 of IS 456). Bars marked as T2, T3, T5 and T6 in a and b are

these bars.

To resist the negative moment at a discontinuous edge depending on the degree of fixity at the edge

of the slab, top tension reinforcement bars equal to fifty per cent of that provided at mid-span shall

extend 0.1l into the span (cl. D-1.6 of IS 456). Bars marked as T1 and T4 in Figs.8.19.5 a and b are

these bars.

The edge strip of each panel shall have reinforcing bars parallel to that edge satisfying the

requirement of minimum amount as specified in sec. 8.18.15d of Lesson 18 (cl. 26.5.2.1 of IS 456)

and the requirements for torsion, explained in Step 7 of sec. 8.19.6 (cls. D-1.7 to D-1.10 of IS 456).

The bottom and top bars of the edge strips are explained below.

Bottom bars B3 and B4 (Fig.8.19.5 a) are parallel to the edge along lx for the edge strip for span ly,

satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).

Bottom bars B7 and B8 (Fig.8.19.5 b) are parallel to the edge along ly for the edge strip for span lx,


Top bars T7 and T8 (Fig.8.19.5 a) are parallel to the edge along lx for the edge strip for span ly,


Top bars T9 and T10 (Fig.8.19.5 b) are parallel to the edge along ly for the edge strip for span lx,


6. write short note on Selection of diameters and spacings of reinforcing bars.

Ans: The advantages of using the tables of SP-16 are that the obtained values

satisfy the requirements of diameters of bars and spacings. However, they are

checked as ready reference here. Needless to mention that this step may be omitted in such a situation.

Maximum diameter allowed, as given in cl. 26.5.2.2 of IS 456, is 160/8 = 20 mm, which is more that

the diameters used here.

The maximum spacing of main bars, as given in cl. 26.3.3(1) of IS 456, is the lesser of 300 mm. This is

also satisfied for all the bars.

The maximum spacing of minimum steel (distribution bars) is the lesser of 5(135) and 450 mm. This is

also satisfied.


UNIT - I

DESIGN CONCEPTS, DESIGN OF BEAMS

Part - A (Short Answer Questions)

1. What are the three methods of design of reinforced concrete structural elements?

2. State four objectives of the design of reinforced concrete structure.

3. Which of the three methods of design of reinforced concrete structural design is the best?

4. What do you mean by characteristic load?

5. What are the different kinds of loads?

6. Explain the limit state method of design?

7. What is the basis of the analysis of structures to be designed?

8. How do the beams and slabs primarily carry the transverse loads?

9. Differentiate between one-way and two-way slabs.

10. When do we go for doubly reinforced beams?

11. Explain the limiting moment of resistance

12. What is a doubly reinforced beam?

13. What do you mean by neutral axis?

14. What are the demerits of Working stress method?

15. State two assumptions made in working stress method?

Part - B (Long Answer Questions)

1. What is meant by characteristic load, what are those values?

2. State and explain the significance of the six assumptions of design of flexural members employing limit state of

collapse.

3. Draw a cross-section of singly reinforced rectangular beam and show the strain and stress diagrams.

4. Explain the limiting moment of resistance and give the expression forthis value forFe 250 and Fe415 grade steel ?

5. Moment of resistance of under and over reinforced sections?

6. Determine the depth of neutral axis for abeam section 250mmwide and 400mm deep (effective) . The beam is


reinforced with 3 bars of 20mm diameter. Use fck= 20 N/mm2and fy= 415 N/mm2

7. Give the stress block parameters used in limit state method along with the stress diagram.

8. Name the four different cases of flanged beams.

9. A singly reinforced R.C.C BEAM 250 mm wide and 400 mm deep (effective) is reinforced with 4 bars of 16 mm

diameter. Find the depth of neutral axis, limiting depth of neutral axis and specify the type of beam. Use fck =20

N/mm2 and fy=415 N/mm2.

10. Enumerate the steps of design of douby reinforced beam.

11. What do you mean by neutral axis and leaver arm? Explain briefly with neat sketches.

12. Describe with sketches T-and L-beams and indicate their principal components.

13. Explain with figure balanced, under reinforced, over reinforced sections

14. With the help of neat sketch derive the stress block parameters for limit state of flexure.

Part – A (Short Answer Questions)

UNIT - II

SHEAR, BOND & TORSION


1. State modes of failures in R.C beams?

2. Define „development length‟?

3. In which form we can provide Shear reinforcement?

4. What is the expression for spacing of vertical stirrups in R.C. beams for shear?

5. Why cover to be provided in design of reinforced concrete structures

6. Minimum shear reinforcement as per IS: 456.

7. How many forms we can provide shear reinforcement?

8. Explain the term tensional stiffness of beam?

9. What do you understand by nominal shear stress?

10. Write the formula for uniform formulae for rectangular section? What are the stresses produced by

torsion?


1. Explain modes of failures in R.C beams?

2. Step by step design procedure for shear reinforcement.

3. What are the various remedial measures for control of cracking?

4. A simply supported reinforced concrete beam is 220 mm wide and 400 mm effective depth and is

reinforced with 5 bars of 18mm diameter as tensile steel. If the beam is subjected to a factored shear of

62.5 KN at the support. Find the nominal shear stress at the support and design the shear reinforcement.

Use M20 concrete and Fe 415 steel.

5. Singly reinforced concrete beam is 400x450 mm deep to the centre of tension reinforcement which

consists of 4 bars of 16mm diameter. If the safe stresses on concrete and steel are 7 N / mm2 and 230 N

/ mm2 respectively, find the moment of resistance of the section. Take M = 14.

6. How do you determine the critical sections for shear in a beam?

7. When and why do we consider enhanced shear strength of concrete?

8. What is meant by “Design shear strength of concrete τc”?



UNIT-III

DESIGN OF COLUMNS

Part – A (Short Answer Questions)

1. Explain about column and pedestal.

2. What are the functions of column reinforcement?

3. Define effective length of column?

4. What is slenderness ratio? Explain.

5. Why cover to be provided in design of reinforczc0020ed concrete structures

6. What is the difference between behaviour of a short and long column?

7. Explain unsupported length of column?

8. What is meant by slenderness ratio of column?

9. Explain braced and unbraced column?

10. What is the minimum eccentricity specified for design of column?


Part – B (Long Answer Questions)

11. Write the design procedure for slender columns for both braced and unbraced column.

12. How columns are classified on the basis of different criteria?

13. Outline the procedure for design of axially loaded reinforced concrete column?

14. Differentiate between

i) Unsupported length and effective length of a compression member.

ii) Braced and unraced column.

15. Check the column safety having column dimensions 400 X 600 mm reinforced with 6 no‟s of 20 mm

dia equally distributed on opposite sides subjected to ultimate load of 2000 kN and ultimate moment of

120 kN-m .Use M-20 and Fe-415

16. A short RC column 300 mm wide and 500 mm deep is reinforced with 6 no‟s of 20 mm dia equally

distributed on opposite sides. Determine the bending moment Mu about an axis bisecting depth when it

is also subjected to PU= 800KN. Assume M20 grade concrete and Fe415grade steel

17. Design a short column, square in section carry an axial load of 2000 KN. Using i) Mild steel M20 ii)

Fe250 and M20

18. What is slenderness ratio? Explain.

19. Differentiate between short and long column.

20. A short column, 600 mm × 600 mm in section, is subject to a factored axial load of 1500 kN.

Determine the minimum area of longitudinal steel to be provided, assuming M 20 concrete and Fe 415

steel.

UNIT-IV

DESIGN OF FOOTING, STAIR CASE



2. How the foundations are classified?

3. What are the different types of foundations?

4. Explain about combined footing?

5. Explain about need of tensile reinforcement in footings?

6. Show how the pressure distribution beneath footings?

7. Provision of dowel bars as per IS: 456-2000 code of practice?

8. Explain shear and bond in footings?

9. How you can find tensile reinforcement for footing?

10. What are the types of footings?

11. Explain about the following stair cases

(A) A stair case

(B) A dog legged stair


12. What are the different types of foundations explain with fig?

13. Explain design procedure for footing as per IS:456

14. Explain pressure distribution under footing explain with figure?

15. Explain about one-way and two-way shear in footings?


16. What are the situations in which combined footings are preferred to isolated footings?

17. What are the Indian standard code recommendations for design of footings as per IS: 456-2000?

18. Explain about one-way and two-way shear in footings?

19. Design a plain concrete footing for a column of 400 mm x 400 mm carrying an axial load of 400

KN under service loads. Assume safe bearing capacity of soil as 300 kN/m2 at a depth of 1 m

below the ground level. Use M 20 and Fe 415 for the design. Detail as per IS 456-2000.

20. Design a sloped footing for a square column of 400 mm x 400 mm with 16 longitudinal bars of 16

mm diameter carrying a service load of 1400 KN. Use M 20 and Fe 415 both for column and

footing slab. The safe bearing capacity of soil is 150 KN/m2. Detail as per IS 456-2000.

21. Design footing for a column of 300 mm x 300 mm carrying an axial load of 325 KN under service

loads. Assume safe bearing capacity of soil as 250 KN/m2 at a depth of 1 m below the ground

level. Use M 20 and Fe 415 for the design. Detail as per IS 456-2000.

UNIT-V

DESIGN OF SLABS


1. What is difference between one-way and two-way slabs


3. Classification according to the method of supports for slabs?

4. What are the methods of analysis of slabs?

5. Effective span for slabs as per IS: 456?

6. Minimum and maximum reinforcement requirement for slabs.

7. What are the considerations that govern thickness of one way and two way slabs?

8. Describe the code provisions for calculating the deflection due to creep?

9. Name the two types of two way slabs?

10. What are the major factors that affect the deflection?


11.Design procedure for two way slabs, without transverse reinforcement or corners not held

down?

12.Design a two-way slab simply supported on all the four edges and corners are restrained for a

room 6 m X 4 m clear in size. Apply the superimposed load of 300 kN/m2.use M-20 and Fe 415.

13.Give neat sketches for the reinforcement details for one way simply support and two way

continuous slabs.

14.Design an interior panel of 6 m X 5 m carrying a superimposed load of 300 kN/m2 use M-20

and Fe 415.

15.Design a slab panel having only short edge is continuous of 7 m X 4 m carrying a superimposed

load of 300 kN/m2 use M-20 and Fe 415.

16.Design a slab panel having only long edge is continuous of 5 m X 4 m carrying a superimposed


17. What are the major factors which influence the crack width in flexural members?


18. Design procedure for two way slab?

19.Design a slab panel having only long edge is continuous of 5 m X 4 m carrying a superimposed


20. What are the various remedial measures for control of cracking?

17 BEYOND SYLLABUS TOPICS WITH MATERIAL

18 RESULT ANALYSIS-REMEDIAL/CORRECTIVE ACTION

19 RECORD OF TUTORIAL CLASSES

20 RECORD OF REMEDIAL CLASSES

21 RECORD OF GUEST LECTURERS CONDUCTED

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