design and detailing of rc structures

DESIGN & DETAILING OF RC STRUCTURES 10CV321

Dr.G.S.Suresh, Prof, CE Dept., NIE, Mysore

Reference books• Dr. H.J. Shah, “Reinforced Concrete Vol-1”, 8th Edition,

2009.• Dr. H.J. Shah, “Reinforced Concrete Vol-2”, 6th Edition,

2012.• S. Unnikrishna Pillai and Devadas Menon, “Reinforced

Concrete Design”, 3rd Edition, 2009• P.C.Varghese, “Limit State Design of Reinforced

Concrete”, PHI, 2nd edition, 2009• Ashok K Jain, “Reinforced Concrete, Limit State

Design”, Nem Chand and Bros, 7th Edition, 2012• M.L.Gambhir, “Design of Reinforced Concrete

Structures”, PHI, 2008

Reference books Contd.

• Dr. B.C. Punmia et al, “Limit State Design of Reinforced concrete”, Laxmi, 2007.

• J.N. Bandyopadhyay, “Design of Concrete Structures”, PHI, 2008.

• N.Krishna Raju, “Structural Design and Drawing, Reinforced concrete and steel”, Universities Press, 1992

• M.R. Dheerencdra Babu, “Structural Engineering Drawing”, Falcon, 2011

• Bureau of Indian Standards, IS456-2000, IS875-1987, SP16, SP34

Evaluation Pattern for Theory • Test : 25 Marks: Design of beams and Design of slabs• Mid sem Exam: 25 Marks: Design of beams,Design of Stairs,

Design of Shallow foundations• Make-up Test : 25 Marks: Design of slabs, Design of shallow

foundations, Design of Retaining walls• Semester End Exam: 100 MarksPart A: Answer 2 out of 3 questions: Design of Beams, Design of slabs, Design of StairsPart B: Answer 2 out of 3 questions: Design of shallow foundations, Design of Retaining Walls,

Design of Water TanksPart C: Compulsory having short questions or fill in the

blanks covering the entire syllabus

Evaluation Pattern for Drawing• Detailing of Simply supported beam, Cantilever beams,

lintels and continuous beams- 2 sheets• Detailing of Cantilever slabs, One way slabs, Two way

slabs, Inclined slabs and filler slabs- 2 sheets• Detailing of Different types of stair cases- 2 sheet • Detailing of Columns, Isolated footings, Combined footings

- 3 sheets• Detailing of Cantilever and Counter fort retaining walls

2 sheet• Detailing of Rectangular and Circular water tanks resting

on ground- 2 sheetsEach sheet is valued for 10 marksSemester end test will be conducted for 50 marks and then reduced to 30 marks

Dr.G.S.Suresh 6

Introduction to Fundamentals of Limit

State of Design

LIMIT STATE METHOD

Limit State

Collapse Serviceability

Flexure, Compression, Shear, Torsion

Deflection, Cracking, Vibration

LIMIT STATE METHOD (Contd)

• Different theories for different limit states• WSM for serviceability limit state• Ultimate Load theory for limit state of

collapse• Stability analysis for overturning• Provides unified rational basis

PROBABILISTIC ANALYSIS AND DESIGN

• Safety margins are provided in design to safeguard against the risk of failure

• Loads and material property varies randomly.

• Probability of certainty is m/n, m no of certain events out of n events.

• Variations in strength of materials and in the loads are analysed using statstical techniques.

Partial Safety Factors Contd.


• The characteristic strength of a material as obtained from the statistical approach is the strength of that material below which not more than five per cent of the test results are expected to fall

• Such characteristic strengths may differ from sample to sample also.

Partial Safety Factors• Accordingly, the design strength is

calculated dividing the characteristic strength further by the partial safety factor for the material (m), where m

depends on the material and the limit state being considered.

•

Partial Safety Factors


• Clause 36.4.2 of IS 456 states that m for

concrete and steel should be taken as 1.5 and 1.15, respectively when assessing the strength of the structures or structural members employing limit state of collapse.

• However, when assessing the deflection, the material properties such as modulus of elasticity should be taken as those associated with the characteristic strength of the material.


• It is worth mentioning that partial safety factor for steel (1.15) is comparatively lower than that of concrete (1.5) because the steel for reinforcement is produced in steel plants and commercially available in specific diameters with expected better quality control than that of concrete.


• In case of concrete the characteristic strength is calculated on the basis of test results on 150 mm standard cubes.

• But the concrete in the structure has different sizes. To take the size effect into account, it is assumed that the concrete in the structure develops a strength of 0.67 times the characteristic strength of cubes.


• Accordingly, in the calculation of strength employing the limit state of collapse, the characteristic strength (fck) is first multiplied with 0.67 (size effect) and then divided by 1.5 (m

for concrete) to have 0.446 fck

as the maximum strength of concrete in the stress block.

Dr.G.S.Suresh 18

Materials for RCC- Properties of Concrete

Stress-Strain Curve for Concrete

Dr.G.S.Suresh 19


Design Stress-Strain Curve for Concrete

Dr.G.S.Suresh 20

• Tensile strength is about 10 to 20% of compressive strength

• Flexure test or split tensile strength is conducted


Dr.G.S.Suresh 21

• However, the following expression gives an estimation of flexural strength (fcr) of concrete from its characteristic compressive strength (cl. 6.2.2)


Dr.G.S.Suresh 22

• Steel in the form of circular bars (Re-bars) are used as reinforcement to take care of tensile stress

• Induces ductility• Steel is strong in both compression and tension• Plain bars were used earlier to 60s• Plain bars have poor bond strength• Deformed bars with ribs on surface are generally

used in reinforced concrete structures

Materials for RCC- Properties of Steel

Dr.G.S.Suresh 23

• High strength cold twisted (CTD)deformed (HYSD) bars were popular for last four decades

• Recently Thermo mechanically treated (TMT) bars have replaced CTD bars

• Yield strength is characteristic strength • Characteristic strength:

MS bars-250 MPa, CTD/TMT bars- 415MPa, 500 MPa, 550 MPa

• CTD bars have better elongation property


Dr.G.S.Suresh 24


Stress-Strain Curve for MS Bars

Dr.G.S.Suresh 25


Stress-Strain Curve for HYSD Bars

26Dr.G.S.Suresh

• Plane sections normal to axis remain plane after bending

• Maximum strain in concrete of compression zone at failure is 0.0035

• Tensile strength of concrete is ignored• Design curve for concrete and steel is given in

IS456-2000• To ensure ductility, maximum strain in tension

reinforcement shall not be less than 0.002+fy/(1.15xEs)

• Perfect bond between concrete exists between concrete and steel

AssumptionsLimit State Method of Design

27Dr.G.S.Suresh

Assumptions

Dr.G.S.Suresh 28


Dr.G.S.Suresh 29


Dr.G.S.Suresh 30

• Members like beam and slab are subjected to bending moment, shear and torsion

• These members deforms and cracks are developed

• A RC member should have perfect bond between concrete and steel

• Members are primarily designed for flexure and then checked for other forces

General aspects of Flexural Ultimate strength

Dr.G.S.Suresh 31

• Elastic theory used for homogeneous materials cannot be used for RC members

• Elastic design method (WSM) do not give a clear indication of their potential strengths

• Study of behaviour of members under ultimate load has been published by several investigators

• Most of the codes in the world adopted Ultimate load behaviour in 1950s

• IS456 introduced it in 1964• Ultimate theory was later replaced by Limit state

method in IS456 during the year 1978

General aspects of Flexural Ultimate strength

Dr.G.S.Suresh 32

Behaviour of RC Beam

Dr.G.S.Suresh 33

Behaviour of RC Beam

34Dr.G.S.Suresh

• Stress diagram across the depth in compression zone up to neutral axis is called stress block

• Hognested introduced the concept of stress block

• Parabolic variation across the depth was proposed

• Whitney’s equivalent rectangular stress block replaced the actual stress block.

• Most of the codes of other country adopt Whitney’s equivalent rectangular stress block

Stress block parameters for limit state of collapse

35Dr.G.S.Suresh


36Dr.G.S.Suresh


1

20.002

cu=0.0035

xu

x1 = 0.57xu

x2 = 0.43xu

0.45 fck

Cu = k1 k2 fck xub x

37Dr.G.S.Suresh

• The stress block given in IS456 has parabolic(1) and rectangular (2) portion as shown

• The strain in extreme concrete fibre is 0.0035 and strain at which the stress reaches ultimate value is 0.002

• From strain diagram and using similar triangle properties

• From above equation x1 =0.57 x u x2=0.43 x u


1u x0.002

x0.0035

38Dr.G.S.Suresh

• Area of stress block is A= A1 +A2

• A1= (2/3)*0.45*fck*xu*0.57*xu = 0.171 f ck x u

• A2= 0.45*fck*0.43*xu = 0.1935 xu

• A = A1+A2= 0.3645 fck x u• Depth of Neutral axis of stress block is

obtained by taking moment of area 1 and 2 about top extreme fiber


u

u

x43.0A

2x43.0

)

2u1

i

ii Ax 0.57

83( A

axa

x

39Dr.G.S.Suresh

• Comparing the above result with the stress block shown in figure, the stress block parameters can be expressed ask1= 0.45, k2=0.42, k3= 0.3645/0.45 = 0.81


40Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sections,

x

T u

1

20.002

cu=0.0035

xu

x1 = 0.57xu

x2 = 0.43xu

0.45 fck

Cu

b

d

A st

z

L

NA

41Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sections

• Strain and Stress distribution in a beam at mid span at ultimate load is shown

• For horizontal equilibrium the compressive force (Cu) should be equal to tensile force (Tu)

• External ultimate moment at collapse is equal to internal couple developed by the compressive (Cu) and tensile force (Tu)

• Internal couple is called “Ultimate Moment of Resistance”

42Dr.G.S.Suresh


• From stress block parameters Cu = 0.36 fck b xu

• All the tensile stress is carried by steel and this stress is fy/m = 0.87 fy

• If Ast is the area of steel then total tensile force Tu = 0.87 fy Ast and from equilibrium Cu = Tu ; 0.36 fck b xu = 0.87 fy Ast ;

b f 0.36 Af 0.87

xck

styu

43Dr.G.S.Suresh


• From equilibrium equation M =0, Mu = MuR

where Mu = Applied ultimate momentMuR= Ultimate Moment of

Resistance• MuR = Cu x z or Tu x z;

=0.36 fck b xu z or = 0.87 fy Ast z where z= lever arm = d-0.42 xu

44Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section• Maximum strain in the two

materials reach simultaneously and failure would be sudden

• Depth of neutral axis is maximum

• Maximum depth of neutral axis is obtained from strain diagram

cu=0.0035

su=0.002+(0.875fy/Es)

xumax

d

A st

NA

b

d-xumax

45Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section• Applying similar triangle

properties

dxE

0.87f0.002

x0.0035

umax

s

y

umax

cu=0.0035

su=0.002+(0.875fy/Es)

xumax

d

A st

NA

b

d-xumax

s

yumax

E0.87f

0.0055

0.0035x

(1) • Value of xumax depends on

properties of steel

46Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section• Value of xumax depends on

properties of steel• Clause 38.1 (pp 70) of

IS456-2000 gives value of xumax/d for different grade of steel

cu=0.0035

su=0.002+(0.875fy/Es)

xumax

d

A st

NA

b

d-xumax

fy xumax/d

250 0.53

415 0.48

500 0.46 Es = 200 GPa

47Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section• For equilibrium Cu = Tu

0.36 fck xumax b = 0.87fy A st max

• Ratio of area of steel to effective cross-section is called percentage of steel (pt). In this case it is maximum percentage of steel (ptmax) or limiting steel (ptlim)

y

ckumaxstmax

0.87ff 0.36

)d

x(

bdA

y

ckumxtmax f

f 0.414

dx

p (2) d

x0.414

ffp umx

ck

ytlim (2a)

48Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section

• Ultimate moment of resistance (Mulim) is the internal moment of Cu and Tu.

• Mulim = Cu x z or Mulim = Tu x z• Mulim = 0.36 fck xulim b (d-0.42 xulim)

)(

)(

dx

0.42-1 d

xf 0.36 Q Where,

bdQ d

x 0.42-1bd

dx

fck 0.36

ulimulimcklim

2lim

ulim2ulim

(3)

49Dr.G.S.Suresh


• Table D of SP16 (pp 10) gives the values of Qlim for different combination of fck and fy

Value of Qlim

fckin N/mm2

fy in N/mm2

250 415 50020 2.98 2.76 2.6625 3.73 3.45 3.3330 4.47 4.14 3.99

50Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section• Moment of resistance with respect to steel is expressed

as• Mulim = 0.87 fy Ast (d-0.42 xulim)

• From equation (2) • • Mulim = 0.87fy Ast (d-0.42 (2.42 (fy/fck) ptlim) d)

= 0.87fy Ast d(1- (fy/fck) ptlim)

ck

ytlimumx

ffp

2.42 d

x

)pff

-(1p ff

0.87 bdf

Mtlim

ck

ytlim

ck

y2

ck

ulim (4)

51Dr.G.S.Suresh


• Using equation (2) and (4) values of for

different grades of steel is given in Table C (pp10) of SP16

ck

ytlim2

ck

ulim

ff p

and bdf

M

fy in N/mm2 250 415 500

Mulim/ fck bd2 0.149 0.138 0.133Ptlim fy / fck 21.97 19.82 18.87

ptlim is in %

52Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Under Reinforced Section• Tensile strain in steel attains

its limiting value first and strain in compression fibre of concrete is less than limiting value

• Depth of neutral axis is obtained from horizontal equilibrium equation Cu = Tu

c<cu

su=0.002+(0.875fy/Es)

xu

d

A st

NA

b

d-xu

53Dr.G.S.Suresh

• For equilibrium Cu = Tu

0.36 fck xumax b = 0.87fy A st max

bdA

ff

2.42 x

bA

ff

2.42 bf 0.36

A0.87f x

st

ck

yu

st

ck

y

ck

styu

d (5)

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Under Reinforced Section

dx

dx

section this In ulimu

54Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Balanced Section• Moment of resistance is calculated considering ultimate

tensile strength of steel Mu = Tu z• Mu = 0.87 fy Ast (d-0.42 xu)

• From equation (5) • • Mu = 0.87fy Ast d(1-0.42 (2.42 (fy/fck) (Ast/bd)) )

= 0.87fy Ast d(1- (fy/fck) pt/100)

ck

ystu

ff /bd)(A

2.42 dx

2

100p

ff

- 100p

bdf 0.87

M t

ck

yt2

y

u

55Dr.G.S.Suresh


• Solving the quadratic equation for pt

(6)2 ck

u

y

ckt bdf

M 4.611

ff

50 p (

56Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Over Reinforced Section• Strain in extreme fibre in

compression zone reaches its ultimate value before the steel reaches its ultimate value.

• Member fails suddenly due to crushing of concrete

• This type of section is not recommended by code

• Depth of neutral axis is computed from equation (5)

cu

s < su

xu

d

A st

NA

b

d-xu

57Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Over Reinforced Section• Force in concrete prevails in calculating the moment of

resistance)(

dx

0.42-1bddx

fck 0.36M u2uu

dx

dx

section this In ulimu

(7)

58Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Comparison of Strain diagrams• Position of neutral axis in case of under reinforced

section is above the position of balanced section and in case of over reinforced section it is below the position of balanced section

59Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsModes of Failure- Comparison of Strain diagrams

60Dr.G.S.Suresh

Ultimate flexural strengthof singly reinforced rectangular sectionsBeam with very small amount of steel- Minimum Steel• Section is large and intensity of loading is small• If designed steel is provided, failure is brittle• To prevent brittle failure minimum steel is required

• Clause 26.5.1.1 a (pp47) of IS 456 stipulates minimum

steel as

61

Ultimate flexural strengthof doubly reinforced rectangular sections

d’

T u

1

2sc

cu=0.0035

xu

x1 = 0.57xu

x2 = 0.43xu

0.45 fck

Cuc

b

d

A st

z

L

NA

A sc0.002

Cus

62

Doubly reinforced rectangular sections

• Two alternatives when applied moment is > than Mulim

1. To increase the depth of section2. To provide compression reinforcement

• The beam with its limited depth, if reinforced on the tension side only, may not have enough moment of resistance, to resist the bending moment.

• Architectural or other consideration restricts the depth of section

63

Doubly reinforced rectangular sections• By increasing the quantity of steel in the tension zone, the

moment of resistance cannot be increased indefinitely. Usually, the moment of resistance can be increased by not more than 25% over the balanced moment of resistance, by making the beam over-reinforced on the tension side.

• Hence, in order to further increase the moment of resistance of a beam section of unlimited dimensions, a doubly reinforced beam is provided.

• The external live loads may alternate i.e. may occur on either face of the member.

64

doubly reinforced rectangular sections• Strain at the level of compression steel is

• The stress corresponding to strain is obtained from the curve. The stress for various ratio of d’/d given table F of SP16 (pp10) is

mcu x

d'0.0035 1

Grade of Steel

d’/d0.05 0.1 0.15 0.2

250 217 217 217 217415 355 353 352 329500 424 412 395 370

65

Ultimate flexural strengthof doubly reinforced rectangular sections

b

T u

0.45 fck

CucCus

Cuc

+

T u2

d’

xu

d

A

st

NA

A sc

b

z1

A st1

b

z2

A

st2

A sc

T

uc

=

Cus

Dr.G.S.Suresh 66

Doubly reinforced rectangular sections

• Cuc = 0.36 fck b xulim Tu1 = 0.87 fy Ast• Mulim = 0.36 fck b xulim (d-0.42 xulim) • ptlim = 0.414 (xulim/d) (fy/fck) and Ast1= ptlim bd/100• When Mu > Mulim then Mu2 = Mu-Mulim

• For equilibrium Cus =Tu2 , Cus = fsc Asc, Tu2=0.87 fy Ast2

• Ast2= fsc Asc/(0.87 fy)• fsc is obtained corresponding to sc from stress-strain diagram• From internal couple Mu2 = Cus*z2 or Tu2*z2, z2 = d-d’• Mu2 = fsc*Asc* (d-d’) or Asc = Mu2 / (fsc * (d-d’)

Dr.G.S.Suresh 67

1. Find neutral axis from equilibrium equation Cuc +Cus = Tu

exact value of xu can be

found by trial and error2. MR is found from Mulim = Cu *z1 + Cusc*z2

Procedure for Analysis

bf 0.36AfAf 0.87

xck

scscstyu

Dr.G.S.Suresh 68

1. Check for need for doubly reinforced beam, ie., Mu >Mulim

2. Find Ast1 = Astlim using equation (2)3. Mu2 = Mu-Mulim

4. Compute Asc = Mu2/(fsc (d-d’)) obtain fsc corresponding to sc obtained from table F of SP16

5. Additional tension required is obtained from equilibrium equation Ast2 = fsc Asc/(0.87fy)

6. Total tension steel Ast = Ast1 + Ast2

Procedure for design

Dr.G.S.Suresh 69

1. Table 45 to 56 (pp 81-92) of SP16 gives pt and pc for different value of Mu/(bd2)

Use of SP-16

70

• Reinforced concrete slabs used in floors, roofs and decks are mostly cast monolithic from the bottom of the beam to the top of the slab.

• Such rectangular beams having slab on top are different from others having either no slab (bracings of elevated tanks, lintels etc.) or having disconnected slabs as in some pre-cast systems (Figs. a, b and c).

• Due to monolithic casting, beams and a part of the slab act together. Under the action of positive bending moment, i.e., between the supports of a continuous beam, the slab, up to a certain width greater than the width of the beam, forms the top part of the beam.

Introduction to flanged sections

71


a) Bracing of elevated Water tank

b) Lintel without effective chajjas

c) Precast slab on rectangular beams

72

• Such beams having slab on top of the rectangular rib are designated as the flanged beams - either T or L type depending on whether the slab is on both sides or on one side of the beam. Over the supports of a continuous beam, the bending moment is negative and the slab, therefore, is in tension while a part of the rectangular beam (rib) is in compression. The continuous beam at support is thus equivalent to a rectangular beam (Fig d to i)


73


74

d) Key Plan

e) Section 1-1

f) Section 2-2


75

e) Section 1-1

f) Section 2-2g) Detail at 3 (L-Beam)


76

e) Section 1-1

f) Section 2-2h) Detail at 4 (T-Beam)


77

e) Section 1-1

f) Section 2-2i) Detail at 5 (Rectangular Beam)


78

ISOLATED T-BEAM FOR BRIDGE


79

• The actual width of the flange is the spacing of the beam, which is the same as the distance between the middle points of the adjacent spans of the slab, as shown in Fig. e. However, in a flanged beam, a part of the width less than the actual width, is effective to be considered as a part of the beam. This width of the slab is designated as the effective width of the flange.


80

Reinforcement


81

IS code requirements

The following requirements (cl. 23.1.1 of IS 456) are to be satisfied to ensure the combined action of the part of the slab and the rib (rectangular part of the beam). (a) The slab and the rectangular beam shall be cast integrally or they shall be effectively bonded in any other manner.(b) Slabs must be provided with the transverse reinforcement of at least 60 percent of the main reinforcement at the mid span of the slab if the main reinforcement of the slab is parallel to the transverse beam (Figs. a and b).


82



83


84

Ultimate flexural strength of flanged sections: T-beam

• Analysis of T-beam depends on the position of neutral axis.• The neutral axis of a flanged beam may be either in the

flange or in the web depending on the physical dimensions of the effective width of flange bf, effective width of web bw, thickness of flange Df and effective depth of flanged beam d .

• The flanged beam may be considered as a rectangular beam of width bf and effective depth d if the neutral axis is in the flange as the concrete in tension is ignored.

• However, if the neutral axis is in the web, the compression is taken by the flange and a part of the web.

85

Ultimate flexural strength of flanged sections:T-beam

Typical T-Beam Section

86


• All the assumptions made in sec. 3.4.2 of Rectangular sections are also applicable for the flanged beams.

• The compressive stress remains constant between the strains of 0.002 and 0.0035.

• It is important to find the depth h of the beam where the strain is 0.002

• If it is located in the web, the whole of flange will be under the constant stress level of 0.446 fck.

• The relation of Df and d to facilitate the determination of the depth h where the strain will be 0.002 similar triangle properties is applied to strain diagram.

87

Ultimate flexural strength of flanged sections: T-beam

• From the strain diagram

88


• Hence the three values of h are around 0.2 d for the three grades of steel.

• The maximum value of h may be Df, at the bottom of the flange where the strain will be 0.002, if Df /d = 0.2.

• The thickness of the flange may be considered small if Df /d does not exceed 0.2 and in that case, the position of the fibre of 0.002 strain will be in the web and the entire flange will be under a constant compressive stress of 0.446 fck .

• On the other hand, if Df is > 0.2 d, the position of the fibre of 0.002 strain will be in the flange. In that case, a part of the slab will have the constant stress of 0.446 fck where the strain will be more than 0.002.

89

Ultimate flexural strength of flanged sections-Tbeam

• Thus, in the balanced and over-reinforced flanged beams (when xu = xu,max), the ratio of Df /d is important to determine if the rectangular stress block is for the full depth of the flange (when Df /d does not exceed 0.2) of for a part of the flange (when Df /d > 0.2).

• Similarly, for the under-reinforced flanged beams, the ratio of Df /xu is considered in place of Df /d.

• If Df /xu does not exceed 0.43, the constant stress block is for the full depth of the flange.

• If Df /xu > 0.43, the constant stress block is for a part of the depth of the flange.

90

T-beam: Case-1, xu<Df

• Section is analysed similar to rectangular beam of size bf and xu

91

T-beam: Case-2, xu>Df and (Df /d) 0.2

• In this case the rectangular portion of the stress block is greater than that of flange thickness

1

2 2

1

92

T-beam: Case-2, xu>Df and (Df /d) 0.2

• C= force of portion 1 + 2 x force of portion 2 = 0.36 fck bw xulim + 0.45 fck (bf-bw) Df

T = 0.87 fy Ast

Mulim = Mu1+ Mu2

= 0.36 fck bw xulim (d-0.42 xulim) + 0.45 fck (bf-bw) Df (d-Df/2)

93

T-beam: Case-3, xu>Df and (Df /d) > 0.2

T-beam: Case-3, xu>Df and (Df /d) > 0.2 • In this case the depth of rectangular portion of stress block is

within flange and part of parabolic portion lies within flange• Equating the areas of actual stress block o equivalent stress

block, yf = 0.142 xu +0.67 Df

As per IS456-2000 yf = 0.15 xu + 0.65 Df• C= force of portion 1 + 2 x force of portion 2

= 0.36 fck bw xulim + 0.45 fck (bf-bw) yf

T = 0.87 fy Ast

Mulim = Mu1+ Mu2

= 0.36 fck bw xulim (d-0.42 xulim) + 0.45 fck (bf-bw) yf (d-yf/2)

T-beam: Case-4, xumax >xu>Df

• When Df/xu 0.43, use equation of case-2 other wise use equation of case-3. In both cases use xumax in place of xu

96

Ultimate shear strength of RC sections

• Shear force in a beam is due to change in BM along the length

• Shear and BM act together • Generally in design the strength is governed by flexure and

the section is checked for shear• Failure due to shear is sudden• Most of the codes recommends the equations based on

laboratory tests• Shear and flexural stress induce principal stress called as

diagonal tension

97


• Before cracking the RC beam acts similar to homogeneous material

• The flexural and shear stress at any point across the depth is computed as f=M y /I, q= F(Ay)/ Ib

• Flexural stress varies linearly and shear stress varies parabolically with max stress being at neutral axis

• These stresses acting on element is indicated in sketch

fcfc

q

q

q

q

ft

q

q

ft

Compression zone Neutral Axis Tension zone

98


• After the concrete cracks, variation of shear stress across the depth is complex.

• The nominal shear stress in beams of uniform depth shall be obtained by the following equation:

• Nominal shear strength v should not exceed the maximum shear stress cmax given in table 20 of IS 456. If it exceeds then the depth of the beam to be revised

• Shear reinforcement in the form of stirrups or cranked bars is provided to take care of diagonal tension developed

• The design shear strength of concrete c in beams without shear reinforcement is given in Table 19 of IS456-2000

99


100


101


• If v < c then nominal shear reinforcement given in clause 26.5.1.6 page 48 to be provided. Minimum shear reinforcement as per this clause is

• When v > c then design of shear reinforcement is required • The applied ultimate shear force is resisted by both concrete

and steel. If Vu is the applied shear force, Vcu is the shear resistance of concrete and Vsu is the shear resistance of steel then Vu = Vcu + Vsu

• Vcu = c bd• Computation for different form of shear reinforcement is given

in clause 40.4 of IS456 page 72 and 73

102


103

Ultimate shear strength of RC sections• Where more than one type of shear reinforcement is used

to reinforce the same portion of the beam, the total shear resistance shall be computed as the sum of the reistance for the various types separately

• Where bent-up bars are provided, their contribution towards shear resistance shall not be more than half that of total shear reinforcement


Procedure for design of shear in RCC1. The ultimate shear force acting at critical section (ie., d from

face of support) is calculated as Vu2. Calculate the nominal shear stress as v = Vu /bd, v vmax

3. From the flexural design result calculate the percentage of steel provided as pt = 100 Ast / bd and obtain the shear strength of concrete c from table 19 page 73 of IS 456-2000

4. The shear strength of concrete is calculated as Vcu = c bd5. If v < c then provide nominal shear reinforcement as specified

26.5.1.6 page 48

6. If v >c then design the shear reinforcement as in clause 40.4

Ultimate torsion strength of RC sections

• Moment about axis of the beam is torsion1. Corner lintels2. Curved beams3. Balcony beams supporting slabs4. L-Beams5. Spiral staircase6. Non rectangular 3D structures

• Classified as Equilibrium torsion and Compatibility torsion• Equilibrium torsion occurs in statically determinate structures• Compatibility torsion occurs in both determinate and

indeterminate structures

Ultimate torsion strength of RC sections• Elastic Torsion equation for circular section is

• Torsion induces shear and in turn induces principal tensile stress

• When the magnitude of principal tensile stress is more than modulus of rupture of concrete, diagonal cracks are induced

• These cracks spiral around the beam

Rf

LG

IT s

P


• The torsional resistance of plain concrete can be improved by providing suitable reinforcement

• Combination of longitudinal reinforcement and stirrups are provided

• Design equations are based on space truss analogy proposed by Lampert and Collins

• Case of pure torsion does not occur• Combination of Flexure, Shear and Torsion is

considered• For design equivalent shear force and bending

moment are computed

Ultimate torsion strength of RC sections• Equivalent bending moment is Me1 = Mu + Mt

Where, Mu = Ultimate external Bending MomentMt = Tu (1+D/b)/1.7D = Overall depthb = Width of section

Longitudinal reinforcement is designed for Me1

• When numerical value of Mt exceeds numerical value of Mu, longitudinal reinforcement is provided on compression face also for Me2 = Mt - Mu

• Equivalent shear is Ve= Vu + 1.6 (Tu/b)• Equivalent nominal shear ve = Ve /bd• ve should not exceed cmax given in table 20 page 73 of IS456-

2000


• If ve >c (of table 19 page 73 of Is456-2000) then transverse reinforcement is provided

• Two legged closed hoops enclosing the corner longitudinal bars shall have c/s area Asv given by

• The total transverse reinforcement shall be less than)f87.o(d5.2

sV)f87.0(db

sTA

y1

vu

y11

vusv

y

vcve

f87.0bs)(

Ultimate torsion strength of RC sections• The transverse reinforcement for torsion shall be rectangular

closed stirrups shall be closed stirrups placed perpendicular to axis of members.

• The spacing of stirrups shall not exceed the least of 1. x1

2. (x1+y1)/43. 300 mm

x1 and y1 are short and long dimensions of the stirrups• The longitudinal reinforcement shall be placed as close as

practicable to the corners of the cross section and there shall be at least one longitudinal bar in each corner of the ties

Concepts of Development Length and Anchorage• The bond between steel and concrete is very important and

essential so that they can act together without any slip in a loaded structure.

• With the perfect bond between them, the plane section of a beam remains plane even after bending.

• The length of a member required to develop the full bond is called the anchorage length.

• The bond is measured by bond stress. • The local bond stress varies along a member with the variation

of bending moment.

Concepts of Development Length and Anchorage• The average value throughout its anchorage length is

designated as the average bond stress. • Thus, a tensile member has to be anchored properly by

providing additional length on either side of the point of maximum tension, which is known as ‘Development length in tension’.

• Similarly, for compression members also, we have ‘Development length Ld

in compression’.

Concepts of Development Length and Anchorage• The deformed bars are known to be superior to the smooth mild

steel bars due to the presence of ribs. • In such a case, it is needed to check for the sufficient

development length Ld only rather than checking both for the local bond stress and development length as required for the smooth mild steel bars.

• Accordingly, IS 456, cl. 26.2 stipulates the requirements of proper anchorage of reinforcement in terms of development length Ld only employing design bond stress bd.

Concepts of Development Length and Anchorage

Design Bond Stress bd

• The design bond stress bd is defined as the shear force per unit

nominal surface area of reinforcing bar. The stress is acting on the interface between bars and surrounding concrete and along the direction parallel to the bars.

• This concept of design bond stress finally results in additional length of a bar of specified diameter to be provided beyond a given critical section.

Concepts of Development Length and AnchorageDesign Bond Stress bd

• Though, the overall bond failure may be avoided by this provision of additional development length Ld, slippage of a bar may not always result in overall failure of a beam.

• It is, thus, desirable to provide end anchorages also to maintain the integrity of the structure and thereby, to enable it carrying the loads.

• Clause 26.2 of IS 456 stipulates, “The calculated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or by a combination thereof.”

Concepts of Development Length and AnchorageDesign bond stress – values • The local bond stress varies along the length of the reinforcement

while the average bond stress gives the average value throughout its development length.

• The average bond stress has been designated as design bond stress bd

and the values are given in cl. 26.2.1.1. The same is

For deformed bars conforming to IS 1786, these values shall be increased by 60 per cent. For bars in compression, the values of bond stress in tension shall be increased by 25 per cent.

Development Length

Concepts of Development Length and Anchorage(a) A single bar • Figure (a) shows a simply supported beam subjected to

uniformly distributed load. Because of the maximum moment, the Ast

required is the maximum at x = L/2. For any section 1-1 at a distance x < L/2, some of the tensile bars can be curtailed.

• Let us then assume that section 1-1 is the theoretical cut-off point of one bar. However, it is necessary to extend the bar for a length Ld

as explained earlier. Let us derive the expression to determine Ld of this bar.

Concepts of Development Length and Anchorage• Figure (b) shows the free body diagram of the segment AB of

the bar. At B, the tensile force T trying to pull out the bar is of the value T = (π φ2 σs

/4)where φ is the nominal diameter of the bar and σs

is the tensile stress in bar at the section considered at design loads. • It is necessary to have the resistance force to be developed by

τbd for the length Ld to overcome the tensile force. The

resistance force = π φ (Ld) (bd). • Equating the two, • π φ (Ld) (bd) = (π φ2 σs /4)


• The above equation is given in cl. 26.2.1 of IS 456 to determine the development length of bars. Table 65 of SP16 gives value of Ld

Concepts of Development Length and Anchorage• The above equation is given in cl. 26.2.1 of IS 456 to determine

the development length of bars.Checking of Development Lengths of Bars in Tension The following are the stipulation of cl. 26.2.3.3 of IS 456. (i) At least one-third of the positive moment reinforcement in simple members and one-fourth of the positive moment reinforcement in continuous members shall be extended along the same face of the member into the support, to a length equal to Ld/3.

Concepts of Development Length and Anchorage(ii) Such reinforcements of (i) above shall also be anchored to develop its design stress in tension at the face of the support, when such member is part of the primary lateral load resisting system. (iii) The diameter of the positive moment reinforcement shall be limited to a diameter such that the Ld

computed for σs = fd

in Eq. for Ld does not exceed the following:

where M1 = moment of resistance of the section assuming all reinforcement at the section to be stressed to fd, fd

= 0.87 fy, V = shear force at the section due to design loads,

Concepts of Development Length and AnchorageLo = sum of the anchorage beyond the centre of the support and the equivalent anchorage value of any hook or mechanical anchorage at simple support. At a point of inflection, Lo

is limited to the effective depth of the member or 12φ, whichever is greater, and φ = diameter of bar. It has been further stipulated that M1/V in the above expression may be increased by 30 per cent when the ends of the reinforcement are confined by a compressive reaction.


Anchoring Reinforcing Bars The bars may be anchored in combination of providing development length to maintain the integrity of the structure. Such anchoring is discussed below under three sub-sections for bars in tension, compression and shear respectively, as stipulated in cl. 26.2.2 of IS 456.

Concepts of Development Length and AnchorageThe salient points are: • Deformed bars may not need end anchorages if the development length requirement is satisfied. • Hooks should normally be provided for plain bars in tension. • Standard hooks and bends should be as per IS 2502 or as given in Table 67 of SP-16, which are shown in Figs.• The anchorage value of standard bend shall be considered as 4 times the diameter of the bar for each 45o bend subject to a maximum value of 16 times the diameter of the bar. • The anchorage value of standard U-type hook shall be 16 times the diameter of the bar.

Concepts of Development Length and AnchorageBars in compression (cl. 26.2.2.2 of IS 456) Here, the salient points are: • The anchorage length of straight compression bars shall be

equal to its development length. • The development length shall include the projected length of

hooks, bends and straight lengths beyond bends, if provided.


The salient points are: • Inclined bars in tension zone will have the development length equal to that of bars in tension and this length shall be measured from the end of sloping or inclined portion of the bar. • Inclined bars in compression zone will have the development length equal to that of bars in tension and this length shall be measured from the mid-depth of the beam. • For stirrups, transverse ties and other secondary reinforcement, complete development length and anchorage are considered to be satisfied if prepared as shown in Figs.

Control of DeflectionThe deflection of beams and slabs would generally be, within permissible limits if the ratio of span to effective depth of themember does not exceed the values obtained in accordance with 22.2.1 of the Code. The following basic values of span to effectivedepth are given:

Simply supported------------------------------- 20Continuous--------------------------------------- 26Cantilever ---------------------------------------- 7Correction factors for tension steel, compression steel and flanged section are to be used on these values

Control of DeflectionChart 22 of SP16 gives l/d after correction for tension steel and compression steel.The values read from these Charts are directly applicable for simply supported members of rectangular cross section for spans up to 10 m. For simply supported or continuous spans larger than 10 m, the values should be further multiplied by the factor (10/span in meters). For continuous spans or cantilevers, the values read from the charts are to be modified in proportion to the basic values of span to effective depth ratio. 1.3 for continuous beam and 0.35 for cantilever.In the case of cantilevers which are longer than 10 m the Code recommends that the deflections should be calculated in order to ensure that they do not exceed permissible limits

Control of Deflection

Dr.G.S.Suresh

CE DEPT NIE, MYSORE

GOOD DAY