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9C402.16 1
Department of Technical Education
Andhra Pradesh
Name : G.DamodaraRao
Designation : Lecturer
Branch : Civil Engineering
Institute : Govt Polytechnic Srikakulam
Year/Semester : IV Semester
Subject : Design of RC Structures
Subject Code : C 402
Topic : Analysis and Design of Rectangular Beams
Duration : 50 Minutes
Sub Topic : Stress block parameters, determination of
N.A and lever arm
Teaching Aids :PPT, Diagrams and Animations
Revised by : A. Punyavathi
9C402.16 2
Objectives
On completion of this period, you would be able to
• Stress block parameters
• Calculate the of depth of Neutral Axis
• Calculate lever arm
9C402.16 3
Recap
In previous class we have learnt about
• Introduction to R.C.C
• Concepts of working stress design and limit state
method of design
9C402.16 4
What is a beam?
• A Transversely loaded horizontal structural member
9C402.16 5
What are the different types of beams?
• Cantilevers
• Simply supported
• Over Hanging
• Fixed
• Continuous
9C402.16 6
What are the internal actions developed in a beam?
Internal actions
Bending stresses Shear stresses
Bending Compressive
stresses
Bending tensile
stresses
9C402.16 7
Why Reinforcement is provided in beams?
• To counteract the tensile weakness of concrete in
tension zone
9C402.16 8
Where do we provide Reinforcement ?
• In tension zone
9C402.16 9
How do we locate tension zone ?
• Neutral axis divides a beam cross section into two zones
• A compression zone
• A Tension zone
• In a simply supported beam tension zone will be
below neutral axis
• In a cantilever beam tension zone will be above
neutral axis
9C402.16 10
Singly Reinforced Beam
• If steel reinforcement is provided in tension zone only, it is
a singly reinforced beam
9C402.16 11
Reinforcement in Simply Supported Beam
Fig.1
9C402.16 12
Reinforcement in Cantilever Beam
Fig.2
9C402.16 13
Stress Block Parameters
Fig.6
9C402.16 14
• Breadth of beam = b
Effective depth
• Effective depth is the distance between the centroid of
tension reinforcement and extreme compression fiber
• Effective depth (d)=overall depth - effective cover
• Effective cover=(clear cover+Ø/2)
• D=Overall depth
• Ø=Diameter of bar
9C402.16 15
Depth of parabolic part of stress block
Depth of rectangular part
0.002
0.0035
4
7
u
u
x
x
 
  
 

4
7
3
7
u u
u
x x
x
 

9C402.16 16
Area of stress block
The distance of centroid of stress block from top fiber
=0.42 xu
3 2 4
0.446 0.446
7 3 7
0.36
u ck u ck
ck u
x f x f
f x
     

9C402.16 17
To find Neutral axis
• Equate force of compression to force of tension of the
stress block diagram
• Force of compression (C)
= Average stress x compression area
=0.36fckbxu
• Force of Tension (T)
= Design yield stress x area of steel
=0.87 f y A st
9C402.16 18
For equilibrium condition
• C = T
0.36fckbxu = 0.87 f y A st
x u = 0.87 f y A st / 0 .36fckb
9C402.16 19
Maximum Depth of Neutral Axis
• Compression failure is brittle failure and is sudden and is to
be avoided
• Maximum depth of Neutral axis is limited to ensure that
tensile steel will reach its maximum strain before concrete
9C402.16 20
Strain Diagram
0.87fy/Es +0.002
0.0035
N A
Xu max
Fig.1
9C402.16 21
• From the similar triangles of strain diagram
Xu max / (d - xu max) =.0035/(0.87fy / Es)+0.002
Solving the above we can solve for Xu max
(Xu max / d) = 0.0035 / [ (0.87 fy / Es)+0.0055 ]
• The value of Xu max depends on grade of steel only
Substituting the values of fy & Es the values of Xu max can
be obtained
9C402.16 22
• The values of Xu max for three grades of steel
fy
N/mm2
Xu max
250 0.53d
415 0.48d
500 0.46d
Table 1
9C402.16 23
Lever arm (z)
• The distance between the lines of action of compression
and tension forces
Z = d – 0.42 xu
9C402.16 24
• The resultant compressive force acts at the centroid of the
compression zone of stress block
• At a distance of 0.42xu from top of extreme compression fiber
9C402.16 25
• As the tensile resistance of concrete is ignored
• The resultant Tensile force acts at the centroid of the
steel reinforcement
• i.e. at distance ‘d’ from top of extreme compression fiber
9C402.16 26
Fig.2
9C402.16 27
Lever arm (z) = d - 0.42xu
Fig.3
9C402.16 28
Summary
In this class we have
• Stress block parameter
• Depth of neutral axis of singly reinforced beams
• Lever arm
9C402.16 29
Quiz
1. A Singly reinforced beam is one in which steel
will be provided in
A) Only Tension zone
B) Only compression zone
C) Both
D) None
9C402.16 30
Frequently Asked Questions
1. Draw the stress block diagram for rectangular section
2. Define lever arm N.A

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9C402.16.ppt

  • 1. 9C402.16 1 Department of Technical Education Andhra Pradesh Name : G.DamodaraRao Designation : Lecturer Branch : Civil Engineering Institute : Govt Polytechnic Srikakulam Year/Semester : IV Semester Subject : Design of RC Structures Subject Code : C 402 Topic : Analysis and Design of Rectangular Beams Duration : 50 Minutes Sub Topic : Stress block parameters, determination of N.A and lever arm Teaching Aids :PPT, Diagrams and Animations Revised by : A. Punyavathi
  • 2. 9C402.16 2 Objectives On completion of this period, you would be able to • Stress block parameters • Calculate the of depth of Neutral Axis • Calculate lever arm
  • 3. 9C402.16 3 Recap In previous class we have learnt about • Introduction to R.C.C • Concepts of working stress design and limit state method of design
  • 4. 9C402.16 4 What is a beam? • A Transversely loaded horizontal structural member
  • 5. 9C402.16 5 What are the different types of beams? • Cantilevers • Simply supported • Over Hanging • Fixed • Continuous
  • 6. 9C402.16 6 What are the internal actions developed in a beam? Internal actions Bending stresses Shear stresses Bending Compressive stresses Bending tensile stresses
  • 7. 9C402.16 7 Why Reinforcement is provided in beams? • To counteract the tensile weakness of concrete in tension zone
  • 8. 9C402.16 8 Where do we provide Reinforcement ? • In tension zone
  • 9. 9C402.16 9 How do we locate tension zone ? • Neutral axis divides a beam cross section into two zones • A compression zone • A Tension zone • In a simply supported beam tension zone will be below neutral axis • In a cantilever beam tension zone will be above neutral axis
  • 10. 9C402.16 10 Singly Reinforced Beam • If steel reinforcement is provided in tension zone only, it is a singly reinforced beam
  • 11. 9C402.16 11 Reinforcement in Simply Supported Beam Fig.1
  • 12. 9C402.16 12 Reinforcement in Cantilever Beam Fig.2
  • 13. 9C402.16 13 Stress Block Parameters Fig.6
  • 14. 9C402.16 14 • Breadth of beam = b Effective depth • Effective depth is the distance between the centroid of tension reinforcement and extreme compression fiber • Effective depth (d)=overall depth - effective cover • Effective cover=(clear cover+Ø/2) • D=Overall depth • Ø=Diameter of bar
  • 15. 9C402.16 15 Depth of parabolic part of stress block Depth of rectangular part 0.002 0.0035 4 7 u u x x         4 7 3 7 u u u x x x   
  • 16. 9C402.16 16 Area of stress block The distance of centroid of stress block from top fiber =0.42 xu 3 2 4 0.446 0.446 7 3 7 0.36 u ck u ck ck u x f x f f x       
  • 17. 9C402.16 17 To find Neutral axis • Equate force of compression to force of tension of the stress block diagram • Force of compression (C) = Average stress x compression area =0.36fckbxu • Force of Tension (T) = Design yield stress x area of steel =0.87 f y A st
  • 18. 9C402.16 18 For equilibrium condition • C = T 0.36fckbxu = 0.87 f y A st x u = 0.87 f y A st / 0 .36fckb
  • 19. 9C402.16 19 Maximum Depth of Neutral Axis • Compression failure is brittle failure and is sudden and is to be avoided • Maximum depth of Neutral axis is limited to ensure that tensile steel will reach its maximum strain before concrete
  • 20. 9C402.16 20 Strain Diagram 0.87fy/Es +0.002 0.0035 N A Xu max Fig.1
  • 21. 9C402.16 21 • From the similar triangles of strain diagram Xu max / (d - xu max) =.0035/(0.87fy / Es)+0.002 Solving the above we can solve for Xu max (Xu max / d) = 0.0035 / [ (0.87 fy / Es)+0.0055 ] • The value of Xu max depends on grade of steel only Substituting the values of fy & Es the values of Xu max can be obtained
  • 22. 9C402.16 22 • The values of Xu max for three grades of steel fy N/mm2 Xu max 250 0.53d 415 0.48d 500 0.46d Table 1
  • 23. 9C402.16 23 Lever arm (z) • The distance between the lines of action of compression and tension forces Z = d – 0.42 xu
  • 24. 9C402.16 24 • The resultant compressive force acts at the centroid of the compression zone of stress block • At a distance of 0.42xu from top of extreme compression fiber
  • 25. 9C402.16 25 • As the tensile resistance of concrete is ignored • The resultant Tensile force acts at the centroid of the steel reinforcement • i.e. at distance ‘d’ from top of extreme compression fiber
  • 27. 9C402.16 27 Lever arm (z) = d - 0.42xu Fig.3
  • 28. 9C402.16 28 Summary In this class we have • Stress block parameter • Depth of neutral axis of singly reinforced beams • Lever arm
  • 29. 9C402.16 29 Quiz 1. A Singly reinforced beam is one in which steel will be provided in A) Only Tension zone B) Only compression zone C) Both D) None
  • 30. 9C402.16 30 Frequently Asked Questions 1. Draw the stress block diagram for rectangular section 2. Define lever arm N.A