Sd i-module2- rajesh sir

Dept. of CE, GCE Kannur Dr.RajeshKN

Design of Beams andOne-Way Slabs

for Flexure, Shear and Torsion

Dr. Rajesh K. N.Assistant Professor in Civil Engineering

Govt. College of Engineering, Kannur

Design of Concrete Structures


2

(Analysis and design in Module II, III and IV should be based on Limit State Method. Reinforcement detailing shall conform to SP34)

MODULE II (14 hours)

Design of beams with rectangular singly & doubly reinforced sections and flanged sections for flexure at ultimate limit state as per IS 456.

Design of one way slabs.

Shear, torsion and bond: Analysis and design with and without shear reinforcement at ultimate limit state as per IS 456,

deflection control, development length, splicing, curtailment, code requirements.

Reinforcement detailing.


Basic Design Concepts• The design problem is somewhat the reverse of the analysis problem

• A complete design of a beam involves considerations of:

• safety under the ultimate limit states in flexure, shear, torsion and bond,

• serviceability limit states of deflection, crack-width, durability etc.


Requirements of flexural reinforcement

Concrete Cover

Clear cover is the distance measured from the exposed concrete surface (without plaster and other finishes) to the nearest surface of the reinforcing bar (Ref. Cl. 26.4 & Tables 16, 16A).

Exposure Condition

Minimum Grade

Nominal Cover (mm)

Allowance permitted

Mild M 20 20 Can be reduced by 5mm for main rebars less than 12mm dia

Moderate M 25 30Severe M 30 45 } Can be reduced by 5mm if concrete

} grade is M35 or higherVery severe M 35 50

Extreme M 40 75

Nominal cover requirements based on exposure conditions


5

Spacing of Reinforcing Bars (Cl. 26.3, Page 45)

The Code specifies minimum and maximum limits for the spacing between parallel reinforcing bars in a layer.

The minimum limits are necessary to ensure that the concrete can be placed easily in between and around the bars during the placement of fresh concrete.

The maximum limits are specified for bars in tension for the purpose of controlling crack-widths and improving bond.


6

Minimum and Maximum Areas of Flexural Reinforcement

Minimum Flexural Reinforcement in Beams (Cl. 26.5.1.1)

y

st

fbdA 0.85)( min =

In the case of flanged beams, the width of the web bw should be considered in lieu of b.

• Minimum Flexural Reinforcement in Slabs

The minimum reinforcement (Ast)min in either direction in slabs is (Cl. 26.5.2)

min

0.15 %( )

0.12 %of for Fe 250

of for Fe 415 g

stg

AA

A⎧

= ⎨⎩

Maximum Flexural Reinforcement in Beams (Cl. 26.5.1 b & 26.5.1.2)

Tension reinforcement (Ast) as well as compression reinforcement (Asc) in beams shall be limited to 0.04 bD


7

Requirements for DEFLECTION Control

•The cross-sectional sizes of flexural members is mostly selected from deflections control criteria under service loads.

•The Code (Cl. 23.2a) specifies a limit of span / 250 to the final deflection due to all loads (including long-term effects of temperature, creep and shrinkage).

•The explicit computation of maximum deflection can be rather laborious

•Code recommends certain span/effective depth (l / d) ratios which are expected to satisfy the requirements of deflection control (Cl. 23.2.1)


8

Code Recommendations for Span/Effective Depth Ratios (Cl. 23.2.1)

ctbasic kkdldl ××= )()( max

( )⎪⎩

⎪⎨

⎧=

spans continuousfor 26spans supportedsimply for 20

spans cantileverfor 7 / basicdl

The modification factors kt (which varies with pt and fst) and kc (which varies with pc) are as given in Fig. 4 and Fig. 5 of the Code

where


9

Guidelines for selection of member sizes

Selection of flexural member sizes depends on:

serviceability criteria (control of deflections and crack-widths)

requirements related to the placement of reinforcement

other structural, economic and architectural considerations


10

•In general, the recommended ratio of overall depth (D) to width (b) in rectangular beam sections is in the range of 1.5 to 2.

•Often, architectural considerations dictate the sizes of beams. In the case of building frames, the width of beams should, in general, be less than or equal to the lateral dimension of the columns into which they frame

•In design practice, the overall depths of beams are often fixed in relation to their spans. Span to overall depth ratios of 10 to 16 are generally found to be economical in the case of simply supported and continuous beams.

•From practical considerations, it is desirable to limit the number of different beam sizes in the same structure to a few standard modular sizes, as this will greatly convenience the construction of formwork, and permit reusability of forms

General Guidelines for Beam Sizes


11

•The design problem is to determine the cross-sectional dimensions of a beam, and the area of tension steel Ast required to resist a known factored moment Mu.

•Accordingly, the design equation for flexure is:

DESIGN OF SINGLY REINFORCED RECTANGULAR SECTIONS

uRu MM ≤ max,uu xx ≤with

( )0.87 0.416uR y st uM f A d x= −We know, x xu u≤ ,maxfor

2

0.870.4160.87 10.362

0.87 1

y stu y st

ck

y stu sty

ck

f AM f A d

d f b

f AM Afbd bd f bd

⎡ ⎤= −⎢ ⎥

⎣ ⎦⎡ ⎤

= −⎢ ⎥⎣ ⎦

Rearranging,


⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=≡

1001

10087.02

t

ck

yty

u pffpfR

bdMThus lim,tt pp ≤for

,lim 2

000

lim

.1498 for Fe 250 .1389 for Fe 415

.1338 for Fe 500

cku

ck

ck

fM

R fbd

f

⎧⎪= = ⎨⎪⎩

0 870 362

,lim,max ..

y stu

ck

f Axd f bd

=

•The minimum value of d corresponding to the limiting case pt = pt,lim is obtained from


13

Fixing Dimensions of Rectangular Section

• Slabs:• b is taken as 1000 mm and d is governed by the limiting l/dratios for deflection control

• a trial value of d may be assumed as approximately l/25 for simply supported spans, l/32 for continuous spans and l/8 for cantilevers

•Beams:•The value of b may be suitably fixed and the value of dcorresponding to any R ≤ Rlim is given by:

RbMd u=

where Mu is the factored moment

2cover)clear ( tie

φ−φ−−= Dd


14

Determining Area of Tension Steel

For Mu = MuR and xu < xu,max, ( )( )

0.87 0.416u

st reqdy u

MAf d x

=−

where xu is obtained by solving ( )0 362 0 416. .u ck u uM f bx d x= −

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−⎟⎠

⎞⎜⎝

⎛=dx

dx

bdfM uucku 416.01362.0 2In terms of xu/d,

Alternative: Use of Design Aids

• The relationship between and pt are available in the form of charts or tables for various combinations of fy and fck in SP : 16

2uR M bd≡

• The designed section should be checked for strength (MuR>Mu)and deflection control (l/d ratio), and suitably redesigned if found inadequate


15

A rectangular reinforced concrete beam, under ‘moderate’ exposure condition, is simply supported on two masonry walls 300 mm thick and 7m apart (centre-to-centre). The beam has to carry, in addition to its own weight, a distributed live load of 20 kN/m and a dead load of 15 kN/m. Design the beam section for maximum moment at midspan. Assume Fe 415 steel and M25 concrete.

Design Problem 1

Take clear cover as 30 mm. (T 16 for moderate exposure condition )

Assume a trial cross-section b = 350 mm, and D = 700 mm (span/10).Let d = D – 50 = 650 mm.

Effective span (Cl. 22.2 of Code)∴7.0

(7.0 0.3) 0.65 7.35 m (distance between supports)

(clear span )l

d⎧

= ⎨ − + = +⎩

Taking the lesser value, 7.0 ml =

1. Trial section and Effective span

A rectangular reinforced concrete beam, under ‘moderate’ exposure condition, is simply supported on two masonry walls 300 mm thick and 7m apart (centre-to-centre). The beam has to carry, in addition to its own weight, a distributed live load of 20 kN/m and a dead load of 15 kN/m. Design the beam section for maximum moment at midspan. Assume Fe 415 steel and M25 concrete.

Design Problem 1


16

Self-weight 25 0.35 0.73 kN m m mDLwΔ = × × = 6.125 kN/m

15 0 6 125 21 125 kN m. . .DLw∴ = + = =LLw 20.0 kN/m (given)

Factored load (as per Code): ∴

)( 5.1 LLDLu www += =1.5 (21.125 + 20.0) = 61.69 kN/m

Factored Moment (maximum at midspan)2 261 69 7 0

8 8. .u

uw lM ×

= =

= 377.836kNm.

2. Loads and moment


17

3. Fixing up b, d and D

= 0.1389×25 = 3.472 MPa

With b = 350 mm, for a singly reinforced section, the minimum required effective depth d is:

bRM

d u

limmin =

6377 84 103 472 350

..

×=

×= 557.57 mm.

( )0 362 0 416,lim ,max ,max. .u ck u uM f bx d x= −

For Fe 415 steel, Mu,lim = 0.1389 fck bd2

,2im

limu lM

Rbd

=For M 25 concrete,

Adopt D = 650 mm. Assuming 25φ bars, 8φ stirrups and clear cover of 30 mm, d = 650 – 30 – 8 – 25/ 2 = 600 mm

20 362 1 0 416,max ,max. .u uck

x xf bd

d d⎛ ⎞

= −⎜ ⎟⎝ ⎠

Dept. of CE, GCE Kannur Dr.RajeshKN18

4. Determining (Ast)reqd

Alternatively, using ‘design aids’

For = 3.00 MPa,

for M 25 concrete and Fe 415 steel,

6

2 2377 84 10350 600

.uMR

bd×

= =×

[ ] 0.870.87 0.416 0.87 0.416

0.362

0.87

y stu y st u y st

ck

y stu y st

ck

f AM f A d x f A d

f b

f AM f A d

f b

⎡ ⎤= − = −⎢ ⎥

⎣ ⎦⎡ ⎤

= −⎢ ⎥⎣ ⎦

Solving this quadratic equation 2090 2mm( )st reqdA =

2090 2mm( )st reqdA =pt = 0.997 [Table 3, Page 49, SP: 16],

6 415377.84 10 0.87 415 60025 350

stst

AA ⎡ ⎤× = × × −⎢ ⎥×⎣ ⎦


5. DetailingProvide 5 # 25 ϕ bars

Ast, provided = 2454.37 mm2

650600

350

5 – 25φ

30+8+25/2

6. Design Checks

6.(a) Check for strength in flexure

Actual d = 650 – 30 – 8 – 25/2 = 600 mm

MuR = 428.5kNm > Mu = 377.84 kNm — Hence, safe.

0.87 y stuR y st

ck

f AM f A d

f b⎡ ⎤

= −⎢ ⎥⎣ ⎦

pt, provided 100 1.1687stAbd

= × =

415 2454.370.87 415 2454.37 60025 350×⎡ ⎤= × × −⎢ ⎥×⎣ ⎦


6.(b) Check for deflection control:

For =1.1687, and pt

⎟⎟⎠

⎞⎜⎜⎝

⎛=

)(

)(58.0providedst

reqstys A

Aff

20900 58 4152454 4

..

⎛ ⎞= × ×⎜ ⎟⎝ ⎠

= 204.96 MPa

kt = 1.1 (from Fig. 4 of Code),

and, as pc = 0 (singly reinforced beam), kc = 1 (from Fig. 5 of Code).

20 1 1 1max

.ld

⎛ ⎞ = × ×⎜ ⎟⎝ ⎠

= 22

7000600provided

ld

⎛ ⎞ =⎜ ⎟⎝ ⎠

= 11.6 <max

ld

⎛ ⎞⎜ ⎟⎝ ⎠

— Hence, OK.


1. Select a trial section and get effective span

Design Procedure summarised (Singly reinforced rectangular sections)

2. Find loads and factored moment

distance between supportsclear span

ld

⎧= ⎨ +⎩

Take the lesser value.

3. Fix up b, d and D

limmin

uMdR b

∴ =

E.g., For Fe 415 steel, Mu,lim = Rlim bd2 = 0.1389 fck bd2

Adopt a suitable D, thus get d assuming diameter of main bars, stirrups and clear cover.


4. Determining (Ast)reqd

Alternatively, use SP:16 to get pt for a given value of and for the given grades of steel and concrete.

2uM

bd

0.87 y stu y st

ck

f AM f A d

f b⎡ ⎤

= −⎢ ⎥⎣ ⎦

Solving this quadratic equation, get ( )st reqdA

5. Design Checks

5.(a) Check for strength in flexure

Find actual d provided and get

Ensure MuR > Mu . Else, redesign (e.g., by increasing d).

0.87 y stuR y st

ck

f AM f A d

f b⎡ ⎤

= −⎢ ⎥⎣ ⎦


5.(b) Check for deflection control:

For provided and ,t cp p

⎟⎟⎠

⎞⎜⎜⎝

⎛=

)(

)(58.0providedst

reqstys A

Aff

Get kt and kc from Fig. 4 & 5 of Code

max

ld

⎛ ⎞⎜ ⎟⎝ ⎠ maxprovided

l ld d

⎛ ⎞ ⎛ ⎞<⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and ensureGet Else, redesign.

6. Detailing


Design Procedure for Given Mu

• For a given rectangular section the Mu,lim for a singly reinforced section should be first determined

• If Mu,lim is greater than or equal to the factored moment Mu, the section should be designed as a singly reinforced section

• Otherwise (for Mu > Mu,lim), the section should be designed as a doubly reinforced section

Design of doubly reinforced rectangular sections


ΔMu is obtained from a couple comprising a (compressive) force (fsc –0.447fck)Asc and an equal and opposite (tensile) force (0.87fy ∆ Ast ), with a lever arm ( )d d′−

ΔAM M

f d dstu u

y=

−− ′

,

. ( )lim

0 87A

f Af fsc

y st

sc ck=

−

( . )( ).447

0 870

Δ


Design Formula M M M Mu uR u u= = +,lim Δ

• where Mu,lim is the limiting moment capacity of a singly reinforced ‘balanced’ section

• ΔMu is the additional moment capacity desired from the compression steel Asc and the corresponding additional tension steel

• ∆ Ast is the additional tension steel corresponding to the compression steel Asc

p p pt t lim t= +, ΔA A Ast st st= +,lim Δ

Ast,lim is the tension steel corresponding to Mu,lim

∆ Ast is the additional tension steel corresponding to ΔMu


Design the flexural reinforcement for the beam in Design Problem 1, if Mu = 520 kNm

Design Problem 2

Given b=350 mm and D = 650 mm.

Since the breadth and depth of the section are fixed here, it isnecessary to check whether the section is singly or doubly reinforced.

Effective span l = 7.0 m

Mu,lim = 0.1389 fck bd2 = 437.54 kNm < Mu

Hence the section has to be doubly reinforced.

For Fe 415 steel,

1. Singly reinforced/doubly reinforced?

Let d = D–50 =600 mm.


28

0 3620 87

,max,lim

..

ck ust

y

f bxA

f=

2. Determining Ast,lim

( )0 87 0 416,lim

,lim,max. .

ust

y u

MOR A

f d x=

− ×

22526 6 mm ,lim .stA =

3. Determining ΔAstAssuming 20 mm φ bars for compression steel,

mm 48≈′d (30 mm clear cover + 8 mm stirrup + φ/2 )

( ) ( )0 87lim,

.u u

st reqdy

M MA

f d d−

Δ =′−

= 413.75 mm2( )( )

6520 437 54 100 87 415 600 48

..

− ×=

× × −

( ), ,limst reqd st st reqdA A A= + Δ = 2526.6 + 413.75 = 2940.35 mm2

Provide 4 nos 32 mm ϕ [Ast,provided =3216.9mm2]

4. Determining Ast


Actual d (assuming 30 mm clear cover and 8 mm stirrups):d = 650-(30+8+32/2) = 596 mm < 600 mm assumed earlier

Revising the above calculations with d = 646 mm, assuming 20mm φcompression steel,

Mu,lim = 0.1389 fck bd2 = 431.7 kNm,

( )0 87lim,

, .u u

st reqdy

M MA

f d d−

Δ =′−

,st reqdA

= 446.18 mm2

= 2509.7 + 446.18 = 2955.96 mm2,

Hence actual (ΔAst) provided = 3216.9 – 2509.7 =707.2 mm2

( )( )

6520 431 7 100 87 596 48

.. yf

− ×=

−

Ast,provided = 3216.9mm2

( ) 00 87 0 416

2 25 9 7 mm,lim,lim

,max

.. .

ust

y u

MA

f d x= =

− ×

5. Revising steel areas based on the actual d provided


( )0 870 447,

..y st

sc reqdsc ck

f AA

f fΔ

=−

( )( )

0 87 415 707 2361 0 447 25

. ..

× ×=

− ×

= 729.9 mm2

Provide 4 nos 16 mm ϕ

[Asc = 804. 2 mm2 > 729.9 mm2 ].

6. Determining Asc

480 0035 1 0 0035 10 48 596

,max

. ..sc

u

dx

ε⎛ ⎞′ ⎛ ⎞= × − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠

= 0.002913

0 87 361 MPa.sc yf f≅ =From design stress-strain curve

350

650

30 clear

30 clear54

4 – 16 φ

4 – 32 φ

beam section

46

8 φ


7. Design check

Actual d provided: d = 596 mm 30 16 2 8 46d mm′ = + + =

460 0035 10 48 596

..scε ⎛ ⎞= −⎜ ⎟×⎝ ⎠

= 0.00294

( )0 870 447

lim,* ..

y st stsc

sc ck

f A AA

f f−

=−

( )0 87 415 3216 9 2509 7361 0 447 25

. . ..

× × −=

− ×= 729.9 mm2

*,sc provided scA A> Section is under-reinforced.

0 87 361 MPa.sc yf f≅ =From design stress-strain curve

Hence, OK.

Asc,provided = 804. 2 mm2

(Note: Else, if , increase Asc )*

,sc provided scA A<


8. Check for deflection control

3216 9 100 1 542350 596

. .tp = × =×

⎟⎟⎠

⎞⎜⎜⎝

⎛=

)(

)(58.0providedst

reqstys A

Aff

2955 960 58 4153216 9

...

⎛ ⎞= × ×⎜ ⎟⎝ ⎠

= 221.18 MPa


kc = 1.15 (from Fig. 5 of Code).

20 1 0 1 15max

. .ld

⎛ ⎞ = × ×⎜ ⎟⎝ ⎠

= 23

7000596provided

ld

⎛ ⎞ =⎜ ⎟⎝ ⎠

= 11.75 <max

ld

⎛ ⎞⎜ ⎟⎝ ⎠

— Hence, OK.

804 2 100 0 3855350 596

. .cp = × =×


1. For the given b and D, and effective span l check whether the section is singly or doubly reinforced.

E.g., For Fe 415 steel, Mu,lim = 0.1389 fck bd2

Design Procedure summarised (Doubly reinforced rectangular sections)

If Mu,lim < Mu the section has to be doubly reinforced.

0 3620 87

,max,lim

..

ck ust

y

f bxA

f=

2. Determine Ast,lim

( )0 87 0 416,lim

,lim,max. .

ust

y u

MOR A

f d x=

− ×

3. Determine ΔAst Assuming dia of compression steel,

( ) ( )0 87lim,

.u u

st reqdy

M MA

f d d−

Δ =′−


Get actual d and revise the calculations of Mu,lim = 0.1389 fck bd2

( )0 87lim,

, .u u

st reqdy

M MA

f d d−

Δ =′−( )0 87 0 416

,lim,lim

,max. .u

sty u

MA

f d x=

− ×

5. Revise steel areas based on the actual d provided

lim, , ,st reqd st st reqdA A A= + Δ

Hence actual lim, , ,st provided st provided stA A AΔ = −

( ), ,limst reqd st st reqdA A A= + Δ

Choose dia and number of bars. Get Ast,provided

4. Determine Ast


( )0 870 447,

..y st

sc reqdsc ck

f AA

f fΔ

=−

6. Determine Asc

0 0035 1,max

.scu

dx

ε⎛ ⎞′

= × −⎜ ⎟⎜ ⎟⎝ ⎠

scfFrom design stress-strain curve find corresponding to

Choose dia and number of bars. Get Asc,provided

7. Design check

Get actual d provided.

460 0035 10 48 596

..scε ⎛ ⎞= −⎜ ⎟×⎝ ⎠

( )0 870 447

lim,* ..

y st stsc

sc ck

f A AA

f f−

=−

scfFrom design stress-strain curve find corresponding to

If *,sc provided scA A> section is under-reinforced.

Else, if , increase Asc*

,sc provided scA A<



For provided and ,t cp p

⎟⎟⎠

⎞⎜⎜⎝

⎛=

)(

)(58.0providedst

reqstys A

Aff

Get kt and kc from Fig. 4 & 5 of Code

max

ld

⎛ ⎞⎜ ⎟⎝ ⎠ maxprovided

l ld d

⎛ ⎞ ⎛ ⎞<⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and ensureGet Else, redesign.

9. Detailing


ONE-WAY SLABS AND TWO-WAY SLABS

• When the slab is supported only on two opposite sides, the slab bends in one direction only. Hence, it is called a one-way slab.

• When the slab is supported on all four sides, and the length is less than twice the width, it bends in two directions. Hence, it is called a two-way slab.

one–way slab two–way slab

l < 2 b


b

l > 2 b

one–way slab

• If the length is greater than about twice the width, the bending along the longitudinal direction is negligible compared to the transverse (short-span) direction, and the resulting slab action is one-way.


• To resist secondary moments (arising as a result of the necessity to resist lateral deformation).

• To resist bending moments in the transverse direction generated locally when the slab is subject to concentrated loads.

• To resist secondary stresses introduced by shrinkage and temperature effects

Distribution steel


Design a one-way slab, with a clear span of 5.0m, simply supported on 230 mm thick masonry walls, and subjected to a live load of 3 kN/m2

and a surface finish load of 1 kN/m2, using Fe 415 steel and M25 concrete. Assume that the slab is subjected to mild exposure condition.

1. Trial depth and effective span

Assume an effective depth 500025

d ≈ = 200 mm

Problem 3

Let overall depth 200 40 240 mmD = + =

Effective span 5000 230 52305000 200 5200

mm (c/c distance) mm

l+ =⎧

= ⎨+ =⎩

Taking the lesser values (as per Code), l = 5.2 m.

Design of one-way slabs


Distributed load due to self-weight,

25 0 24 6 03 2 kN m m kN m. .DLwΔ = × =

6 0 1 0 7 0 2 kN m. . .DLw = + = 3 0 2 kN m.LLw =

Factored load )(5.1 LLDLu www += = 1.5 (7.0+3.0) =15.0 kN/m2

Factored Moment (maximum at midspan)

2

8u

uw lM = 50 7kNm.=

215 0 5 28

. .×=

2. Loads and moment (per metre width)


3. Determining Ast (main bars)

0.87 y stu y st

ck

f AM f A d

f b⎡ ⎤

= −⎢ ⎥⎣ ⎦

6 41550.7 10 0.87 415 20025 1000

stst

AA ⎡ ⎤× = × × −⎢ ⎥×⎣ ⎦

Alternatively, using ‘design aids’

For = 1.2675 MPa,

for M 25 concrete and Fe 415 steel,

6

2 2350 7 10

20010.uM

Rbd

×= =

×

Solving this quadratic equation 749 2mm( )st reqdA =

[Table 3, Page 49, SP: 16],

( )( ) ( )0 385 0 369

0 369 1 2675 1 251 3 1 25

. .. . .

. .tp−

= + × −−

0 375 1000 200 749 2100

2mm.( ) .st reqdA = × × =

0 375.=


Spacing of bars 1000

st

As

Aφ=

Provide 12 mm ϕ bars. 21000 12

4749

π⎛ ⎞× ×⎜ ⎟⎝ ⎠= 151mm=

3 3 200 600300 mm v

ds

= × =⎧≤ ⎨⎩

Maximum spacing (Cl.26.3.3 a)

Provide 10 mm ϕ bars at 150 mm c/c spacing as main reinforcement

4. Distribution bars (Cl. 26.5.2.1, Page 48)

0 0012 1000 240( ) .st distA = × × = 288 mm2/m.

Provide 8 mm ϕ distribution bars.

Spacing 1000

st

As

Aφ=

21000 84

288

π⎛ ⎞× ×⎜ ⎟⎝ ⎠= 174 53mm.=


5 5 200 1000450 mm v

ds

= × =⎧≤ ⎨⎩

Maximum spacing (Cl.26.3.3 b)

Provide 8 mm ϕ bars at 170 mm c/c spacing as distribution reinforcement

5. Detailing

• The complete detailing of the slab is given in Figure. This meets the Code requirements.

• Alternate bars of the main reinforcement are bent up (cranked) near the supports at a distance of 0.1 l from the support (Cl. D-1.6, Page 90)

• This is in order to resist any flexural tension that may arise due to partial fixity at the support


SECTION A – A

12 φ@ 300c/c distributors 8 φ

230

520 12 φ @ 150 c/c (alternate bars

bent up)

240 mm

clear cover 30 mm8 φ@ 170 c/c (distributors)

8 φ@ 170 c/c

12 φ @ 150 c/c520

240 mm THICK

M25Fe415

PLAN

(0.1L)

A A

230 2305000

Details of a one-way slab



490 87 1001000 205,

.t providedp = ×

×

⎟⎟⎠

⎞⎜⎜⎝

⎛=

)(

)(58.0providedst

reqstys A

Aff 478 20 58 415

490 87...

⎛ ⎞= × ×⎜ ⎟⎝ ⎠

= 234.487 MPa


20 1 7max

.ld

⎛ ⎞ = ×⎜ ⎟⎝ ⎠

= 34

4200205provided

ld

⎛ ⎞ =⎜ ⎟⎝ ⎠

= 20.4<max

ld

⎛ ⎞⎜ ⎟⎝ ⎠

— Hence, OK.

Providing a clear cover of 30 mm, d = 240–30–10/2 = 205 mm.

21000 104

160,st providedA

π⎛ ⎞× ×⎜ ⎟⎝ ⎠=

490 87 2mm.= 0 2395.=


47

Curtailment of flexural tension reinforcement • For economy, it is desirable to progressively curtail (‘cut-off’) the flexural tension reinforcement, according to the decrease in bending moment.

• Other factors to be considered in fixing actual bar cut-off points

• unexpected shifts in maximum moments, • development length requirements, • influence on shear strength• development of diagonal tension cracks due to the effects of discontinuity.

• Accordingly, Code (Cl. 26.2.3, Page 44) specifies a number of requirements


Theoretical Bar Cut-off Points• The ‘theoretical cut-off point’ for a bar is that point beyond which it is (theoretically) no longer needed to resist the design moment.• In a prismatic beam the required area of tension reinforcementvaries nearly linearly with the bending moment

• The stress at the end of a bar is zero; it builds up gradually along its length through bond with the surrounding concrete.

• In order to develop the full design stress (0.87fy) in the bar at a section, a minimum development length Ld is required on either side of the section. (Ref: Cl. 26.2.1 of Code)

4s

dbd

L φστ

= s

bd

φστ

==

=

Nominal diameter of the bar

Stress in bar at the section considered

Design bond stress given in Cl. 26.2.1.1

Development length requirements:


Code Requirements

1.The shear at the cut-off point does not exceed two-thirds of the shear resisting capacity of the section.

2.Excess stirrups are provided over a distance of 0.75d from the cut-off point having an area Asv and a spacing sv such that:

Ab sfsv

w v

y≥

0 4. s dv

b≤

8β

is the ratio of the area of bars cut off to the total area of the bars at the section.

β b

3. For 36 mm ϕ and smaller bars, 1. the continuing bars provide at least twice the area required for

flexure at the cut-off point and2. the shear does not exceed three-fourth of the shear resisting

capacity of the section.


• The extension of bars beyond the theoretical cut-off points should not be less than the effective depth or 12 times the bar diameter

• If the anchorage length beyond the zero moment location is denoted as Lo , then as per Code requirement,

( )M V L LuR u o d* + ≥

• where (MuR/Vu)* = 1.3MuR/Vu at a simple support with a confining reaction, and MuR/Vu otherwise.

• At a point of inflection, the anchorage length Lo is limited to the effective depth d or 12φ, whichever is greater.


Splicing of reinforcement (Cl. 26.2.5)


Critical sections for shear design

Design for SHEAR

• The maximum shear force usually occurs in a flexural member at the face of the support,

• When a support reaction introduces transverse compression in theend region of the member, critical section is located at a distance d(effective depth) from the face of the support (Cl. 22.6.2.1, Page 36)


1. stirrups perpendicular to the beam axis;

2. stirrups inclined (at 45° or more) to the beam axis; and

3. longitudinal bars bent-up (usually, not more than two at a time) at 45° to 60° to the beam axis, combined with stirrups.

Types of Shear Reinforcement


Design shear strength without shear reinforcement

uc cV bdτ= where is obtained from Table19, Page 73cτ

Design shear strength with shear reinforcement (Cl. 40.1, Page 72)

For beams of uniform depth, nominal shear stress at a section isdefined as:

uv

Vbd

τ =

where is the factored shear force at the sectionuV


Code Recommendations on shear reinforcement

If is less than (Table19, Page 73) minimum stirrup shear reinforcement has to be provided

vτ cτ

If is between suitable shear reinforcement has to be designed for shear (as per Cl. 40.4, Page 72)

vτ and ,maxc cτ τ

If exceeds (Table 20, Page 73) the section has to be suitably redesigned — by increasing the cross-sectional area and/or improving the grade of concrete

vτ τ c,max

Minimum stirrup reinforcement (Cl. 26.5.1.6, Page 48)

0 40 87

..

sv

v y

Abs f

≥

Minimum shear reinforcement in the form of stirrups should be provided so that:

0 5 .v cτ τ>for


Design of shear reinforcement (Cl. 40.4)

us u uc u cV V V V bdτ= − = −

Shear reinforcement has to be provided for a shear force of:

Area of shear reinforcement required is calculated from:

0 87. y svus

v

f A dV

s= for vertical stirrups

( )0 87.

sin cosy svus

v

f A dV

sα α= + for inclined stirrups or

bars bent up at different cross sections

0 87. sinus y svV f A α= for bars bent up at same cross section


• The Code limits the value of the spacing sv to 0.75d for ‘vertical’stirrups, and d for inclined stirrups with α = 45°

• In no case shall the spacing exceed 300 mm

Spacing of shear reinforcement (Cl. 26.5.1.5, Page 47)


Check the adequacy of the given section with regard to shear. Assume that a support reaction introduces transverse compression in the end region of the member. End supports are 300 mm wide. Take clear span = 6.7 m and

Problem 3

650600

350

3 – 25φ

30+8+25/2

uw (factored load) = 61.69 kN/m

Max. factored shear force (at d from support)

( )61 69 7 61 69 0 15 0 62

. . . .uV ×= − × +

169 64 kN.=

Nominal shear stress at the section is:

uv

Vbd

τ =3169 64 10

350 600. ×

=×

0 807MPa.= ( )3 1MPa,max .cτ< =

(Table 20, Page 73)


Ast, provided = 1472.62 mm2 100 0 7.sAbd

=

(Table 19, Page 73)

( )( ) ( )0 57 0 49

0 49 0 7 0 500 75 0 50. .

. . .. .cτ

−= + × −

−0 554. vτ= <

Since is between suitable shear reinforcement has to be designed (as per Cl. 40.4, Page 72)

vτ and ,maxc cτ τ


169 64 116 34 53 3 Nk. . .us u ucV V V= − = − =

0 87. y svus

v

f A dV

s=

0 75 0 75 600 450300 mm

. .v

ds

= × =⎧≤ ⎨⎩

Hence provide 300 mm spacing (least of the above).

Maximum spacing

To find shear reinforcement)Shear reinforcement has to be provided for a shear force of:

Find out spacing of 2–legged 8ϕ vertical stirrups from

20 87 415 2 8 6000 87 4 40953 3

mm..

.y sv

vus

f A ds

V

π× × × × ×

= = =

Assume 2–legged 8ϕ vertical stirrups are provided.


Total design shear strength 116 34 149 93. .= +

0 75 0 75 600 450300 mm

. .v

ds

= × =⎧≤ ⎨⎩

Spacing of stirrups (Cl. 26.5.1.5, Page 47)

Hence safe.

Hence shear reinforcement = 2–legged 8ϕ vertical stirrups at 300 mm spacing.

Maximum allowed spacing

( )266 27 233 27kN. .uV>= =

Design shear strength without shear reinforcement

uc cV bdτ= 0 554 350 600.= × × 3116 34 10 N.= ×

Design shear strength with a shear reinforcement of 2–legged 8ϕ stirrups

=100.53 mm222 84,sv providedA π

= × ×

0 87. svus

v

A dVs

=0 87 100 53 600

350. .× ×

= 149 93kN.=

Provide 2–legged 8ϕ vertical stirrups at 350 mm spacing. Alternate calculation (to find shear reinforcement)


Design for TORSION

Torsional reinforcement has to be suitably designed when the torsional shear stress exceeds the shear strength of the sectionτ t τ c

Where flexural shear Vu occurs in combination with torsional shear (as is commonly the case), the combined shear stress has to be considered.

The equivalent nominal shear stress (Cl. 41.3.1, Page 75)

1 6 .u uve

V T bbd

τ +=

If exceeds (Table 20, Page 73) the section has to be suitably redesigned — by increasing the cross-sectional area (especially width) and/or improving the grade of concrete

veτ τ c,max


If is less than minimum stirrup reinforcement has to be provided

veτ cτ

If is between suitable torsional reinforcement (both transverse and longitudinal) has to be designed for the combined effects of shear and torsion (as per Cl. 41.4)

veτ and ,maxc cτ τ


Code Provisions for Transverse Reinforcement

If Tu and Vu are the factored twisting moment and factored shear force at the section under consideration,

AT s

b d fV s

d fsvu v

y

u v

y= +

1 1 1087 2 5 087( . ) . ( . )

Asv = 2At is the total area of two legs of the stirrup;

sv is the centre-to-centre spacing of the stirrups;

b1 and d1 are the centre-to-centre distances between the corner bars along the width and depth respectively

The Code (Cl. 41.4.3) specifies a minimum limit to the total area of transverse reinforcement:

Abs

fsvve c v

y≥

−

( ).

τ τ0 87


• The torsional moment Tu is converted into an effective bending moment Mt defined as:

Code Provisions for Longitudinal Reinforcement

( )11 7.t u

D bM T

+=

• ‘Equivalent bending moments’

M M Me t u1 = + M M Me t u2 = −

• The longitudinal reinforcement area Ast is designed to resist the equivalent moment Me1, and this steel is to be located in the ‘flexural tension zone’.

• In addition, if Me2 > 0 (i.e., Mt > Mu), then a reinforcement area Ast’ is to be designed to resist this equivalent moment, and this steel is to be located in the ‘flexural compression zone’.


Check the adequacy of the given section with regard to shear andtorsion, assuming a torsional moment per metre run of 5kNm. Assume fixity at the ends of the beam against torsion and flexure. Take effective span = 7 m and

Problem 4

650600

350

5 – 25φ

30+8+25/2

uw (factored load) = 61.69 kN/m

Max. twisting moment (at support)

Max. bending moment (at support)

Max. shear force (at support)

Tt l

uu=2

2u

uw lV =

5 7 17 52

kNm.×= =

2

12u

uw lM =

261 69 712

. ×=

61 69 72

. ×=

251 9 kNm.=

215 915 kN.=

Stress resultants


The equivalent nominal shear stress

1 6 .u uve

V T bbd

τ +=

3 6215 915 10 1 6 17 5 10 350350 600

. . .× + × ×=

×1 409 MPa.=

(Table 19, Page 73)

( )( ) ( )0 70 0 64

0 64 1 168 1 01 25 1 0. .

. . .. .cτ

−= + × −

−

Ast, provided = 2454.37 mm2 100 1 168.sAbd

=

0 68. veτ= <

3 1MPa,max .cτ = (Table 20, Page 73) ,maxc veτ τ>

Stresses

Since is between suitable torsional reinforcement (both transverse and longitudinal) has to be designed for the combined effects of shear and torsion (as per Cl. 41.4)

veτ and ,maxc cτ τ


Check for adequacy of longitudinal reinforcement

( )11 7.t u

D bM T

+= ( )1 650 350

17 51 7

..

+= 29 4kNm.=

1e t uM M M= +

2 0e t uM M M= − <

29 4 251 9. .= + 281 3kNm.=

Ultimate resisting moment of section

Ast = 2454.37 mm2 (pt = 1.168) at bottom ‘flexural tension zone’

0.87 y stuR y st

ck

f AM f A d

f b⎡ ⎤

= −⎢ ⎥⎣ ⎦

415 2454.370.87 415 2454.37 60025 350×⎡ ⎤= × × −⎢ ⎥×⎣ ⎦

= 428.5kNm 281 3kNm.eM> =

Hence longitudinal reinforcement is sufficient.


Check for adequacy of transverse reinforcement

Area of 2–legged 8ϕ stirrups provided, Asv = 100.53 mm2

This should exceed the requirements given by Cl 41.4.3

AT s

b d fV s

d fsvu v

y

u v

y= +

1 1 1087 2 5 087( . ) . ( . )

sv = 150 mm (say), b1 = 350-2x50=250 mm, d1 = 650-50-44=556 mm

( ) ( )3 317 5 10 150 215 915 10 150

250 556 0 87 415 2 5 556 0 87 415. .

. . .svA × × × ×= +

× × × ×= 64.59 mm2

< Asv provided


( )0 87

.ve c v

svy

bsA

fτ τ−

=

Minimum limit to the total area of transverse reinforcement (Cl. 41.4.3):

( )1 409 0 68 350 1500 87 415

. .

.− × ×

=×

( ) ( )1

1 1

250 25 10 2834 283 582 5 4 216 38

300 mm . .v

xs x y

= + + =⎧⎪≤ + = + =⎨⎪⎩

sv provided = 150 mm < sv required. Hence safe.

= 106 mm2

> Asv provided

Spacing of torsion reinforcement (Cl. 26.5.1.7, Page 48)

Hence provide 2–legged 10 ϕ stirrups, Asv = 157 mm2


Summary

Design of beams with rectangular singly & doubly reinforced sections and flanged sections for flexure at ultimate limit state as per IS 456.

Design of one way slabs.

Shear, torsion and bond: Analysis and design with and without shear reinforcement at ultimate limit state as per IS 456,

deflection control, development length, splicing, curtailment, code requirements.

Reinforcement detailing.

Sd i-module2- rajesh sir

Engineering

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