Unit - III Limit State Design

Objectives of Design:Every material of the total structure takes part effectively for form, function,aesthetics, strength as well as safety and durability.1. The structures should have an acceptable probability of performingsatisfactorily during their intended life:- There are uncertainties in the designprocess both in the estimation of the loads likely to be applied on the structureand in the strength of the material.2. The designed structure should sustain all loads and deform within limits forconstruction and use: - Adequate strengths and limited deformations are thetwo requirements of the designed structure. The structure should havesufficient strength and the deformations must be within prescribed limits dueto all loads during construction and use. However, has to ensure that thefailure of the structures in case of any over loading should give sufficient timefor the occupants to vacate. The structures, thus, should give sufficientwarning to the occupants and must not fail suddenly.3 The designed structures should be durable.4. The designed structures should adequately resist to the effects of misuseand fire.

Characteristic load: Characteristic load is that load which has a 95%probability of not being exceeded during the life of the structure. The loadsare predicted based on probabilistic / statistical approach, where it is assumedthat the variation of the loads acting on structures follows the normaldistribution. Characteristic load should be more than the average/mean load.Characteristic load = Average/mean load + K x(standard deviation for load)The value of K is assumed such that the actual load does not exceed thecharacteristic load during the life of the structure in 95 per cent of the cases.

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Limit state ofcollapse

It deals with strength andstability of the structureunder maximum design load

it state ofserviceability

It deals with deflection and crackingunder service loads, durability underworking environment, fire resistance etc.

There are two main limit states: (i) limit state of collapse and (ii) limit state ofserviceability.(i) Limit state of collapse deals with the strength and stability of structuressubjected to the maximum design loads out of the possible combinations ofseveral types of loads. Therefore, this limit state ensures that neither any partnor the whole structure should collapse or become unstable under anycombination of expected overloads.(ii) Limit state of serviceability deals with deflection and cracking ofstructures under service loads, durability under working environment duringtheir anticipated exposure conditions during service, stability of structures as awhole, fire resistance etc.All relevant limit states have to be considered in the design to ensure adequatedegree of safety and serviceability. The structure shall be designed on thebasis of the most critical limit state and shall be checked for other limit statesPartial safety factors:(i) Structures are subjected to overloading. Hence, structures should bedesigned with loads obtained by multiplying the characteristic loads withsuitable factors of safety. These factors of safety for loads are termed aspartial safety factors (yf) for loads.(Design load, Fd) = (Characteristic load F)x(Partial safety factor for load yf)yf for DL and LL should be taken as 1.5 for limit state of collapseand 1.0 for limit state of serviceability.(ii) The characteristic strengths of materials may differ from sample tosample. Accordingly, the design strength is calculated dividing thecharacteristic strength further by the partial safety factor for the material (ym).(Design strength of material, fd) = (Characteristic strength of material, f ) /

(Partial safety factor of the material, ym)ym for concrete and steel should be taken as 1.5 and 1.15, respectively

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Assumptions:a) Plane sections normal to the axis remain plane after bending.b) The maximum strain in concrete at the outermost compression fibre istaken as 0.0035 in bending.c) The stress- strain relationship of concrete may be assumed any shape insubstantial agreement with the results. An acceptable stress - strain curve isgiven in Fig. The compressive strength of concrete in the structure shall beassumed to be 0.67 fck. The partial safety factor ym = 1.5 .

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d) Tensile strength of the concrete is ignored.e) The stresses in the reinforcement are derived from representative stress-strain curve for the type of steel used. Typical curves for mild and HYSDsteels are given in Fig. For design purposes the partial safety factor ym, equalto 1.15 shall be applied.

mm

f) The maximum strain in the tension reinforcement in the section at failureshall not be less than:-> ( fs/1.15 Es) + 0.002

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The stress-strain relations ship in elasto-plastic zone of HYSD bars -> Tabl

O.Sfyd0.85fyd

0.9fyd

0.95fyd

0.975fyd

fyd

00.0001

0.0003

0.0007

0.0010.002

fy415

a288.70

306.74

324.78

342.83

351.85

360.87

c0.00144

0.00163

0.00192

0.00241

0.00276

0.00380

fySOO

a347.83

369.57

391.30

413.04

423.91

434.78

e0.00174

0.00195

0.00226

0.00277

0.00312

0.00417fee = 1000x0.446 fck [s - 250e2] if e < 0.002fsc in case of doubly reinforced sections: (Table 2)

Fsc-fy415

500

d'/d0.005

355424

0.1353412

0.15342

395

0.2329370

Working Stress Method• The Stresses in an element is obtained from the working loads and

compared with permissible stresses.• The method follows linear stress-strain behaviour of both the materials.• Modular ratio can be used to determine allowable stresses.• Material capabilities are under estimated to large extent. Factor of safety

are used in working stress method.• Ultimate load carrying capacity cannot be predicted accurately.• The main drawback of this method is that it results in an uneconomical

section.Limit State Method• The stresses are obtained from design loads and compared with design

strength.• In this method, it follows linear strain relationship but not linear stress

relationship (one of the major difference between the two methods ofdesign).

• The ultimate stresses of materials itself are used as allowable stresses.• The material capabilities are not under estimated as much as they are in

working stress method. Partial safety factors are used in limit statemethod.

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Linearity ofstrain

DesignPhilosophy

Design approach

Design

constraints

Stress- strainrelationship of

materials

Factor of safety

Major concerns

Stress

StrainCompatibility

Loads

Economical

Area of section

%Reinforcement

Tensile strengthof concrete

WSM LSMPlane sections remains plane after bending

It is based on elastictheory

It is based ondeterministic approach

Stresses in materials

are permitted to acertain extent

Stress-strain diagram is

linear and It followsHooke's Law.

Factor of safety is used

to arrive at permissiblestresses

It's major concern isonly safety

The stresses arecompared withpermissible stresses

Modular ration can be

used to find stress, m=

280/3acbc

Strength is estimated

underworking(Service) loads and

compared with

allowable stresses

Conservative

More

less

It is based on Elastic andplastic theoriesIt is based on probabilistic

approach

Strain in material islimited to a certain extent.

Stress-strain diagram isnon- linear. Reservestrength is also considerd

Partial safety factor forloads and materials are

used

Major concern is safety aswell as serviceability

The stresses arecompared with designstresses

Modular ratio is used onlyfor strain compatibility,m= Es/Ec

Strength is estimated

under factored (ultimate)loads and compared with

characteristic stresses

Economical

less

more

Tensile strength of concrete is ignored; Section isanalysed on cracked section basis

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a) Depth of Limiting NA, xu,limFrom strain diagramxu.lim 0.0035

0.0035+- -+0.002i.iSEs

For fy 250, xulim/d = 0.53For fy 415, xulim/d = 0.48For fy 500, xu lim/d = 0.46

b) Depth of NA, xuArea of stress block == 0.446 fckx (3xu/7)

+(2/3)0.446fckx(4xu/7)= 0.19114+0.1699= 0.362 fck xu

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5 0.446 fck

20 -*4 C

fy/1.15E+0.002>T

Comp force, C = area of stress block x bFrom equilibrium : C=T -> :. (0.362 fck xu) b = Ast x 0.87 ry

Ast X 0.87 fy [fy/U5 = 0.87fy] [Ga]n xuHence, — =d 0.36 x/cfc xb xdc) Lever arm: Taking moments from the top of stress block (diagram)

0.19114(3/14) + 0.1699[3/7 + (3/8)4/7]V = XU =y 0.19114 + 0.1699

Lever arm jd = d - 0.416 xud) Moment of resistance: asC = TMR; Mu = Cx jd = [0.362 fck xu b ] x [d - 0.42 xu]

ORMR, Mu = T x jd = [Ast x 0.87 fy] x [d - 0.42 xu ]

Mu = Ast X 0.87 fy X d (l - 0.42

* Ast x0.87/y xd (l - ***x fy' J \dMu i Ast x

0.87/yd = ^St

The under reinforced solution,

----- [Gb]

fy \ x b x d>

4.6 Mu bdfckbcl2

e) Percentage of steel for section with limiting (balanced section) caseas C = T; [0.362 fck xu b ] = [Ast x 0.87 fy]

Ast 0.36 fck xu

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0,0035 .

3xu/7

xu ( .I 4xu/7

0X532 + 0. 87 f<

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Strain DiagramDesign constants for section with limiting NA

Q.4467fck

rf^

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Stress Diaeram

a) Depth of limiting NA, xu lim /d = 0.0035 / (0.0055+0.87 fy/ E)b) Area of stress block = (17/21 bh) = 0.362 fck xu = 0.36 fck xuc) Distance of resultant comp. from extreme comp. fibre =

9(3) + 8(6 + 3) /xuxy 1 1 fMlfi vii

17 VlJ °'41

d) Lever arm, jd = d-0.416 xu lim =d [ 1- 0.42 xu lim/d]e) MR, Mu = (0.36 fck xu) b x d (1-0.42 xu lim/d) =

= 0.36 (xu lim/d)( 1-0.42 xu lim/d) fck x bd2 = Qbd2 — [Gc]f) Percentage of steel, p = 100 x (0.36 fck/0.87fy) x (xu/d)

Mild/plain

Fe250xu lim/d = xu max / d 0.5314IfQ=qxfck; q 0.1496

MR Coeff, Q% Steel, p

2.99331.77%

MR Coeff, Q% Steel, p

3.74162.21%

Mild Tensisteel

M20

M25

M30MR Coeff, Q% Steel, p

4.48992.65%

e HYSD/CTITOR Fe 41

0.47920.138

2.77450.96%

3.46821.2 %

4.16181.44%

) HYSD 5005 Fe 500

0.45617 0.1336

2.67260.76%

3.34070.95%

4.0089 •1.14%

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Question 1: Derive the Design coefficients (for NA, lever arm, MR andpercent steel) for a rectangular section with M25 and Fe 415.Solutjpni

xu.lim 0.0035a) Coefficient for depth of Limiting NA, k;0.0035 + fy

1.1 sEs

= 0.48+0.002

b) The resultant compressive force passes through the CG of Stress block @0.42 xu from the extreme compression fibreThe lever arm coefficient, j = 1 - 0.42 (xu max/d) = 1 - 0.42 x 0.48 = 0.798c) Coefficient for ultimate Moment of resistance, Q = k xj x 0.36 fck

= 0.48(U042x0.48)x0.36 fck = 0.138 fck = 3.45d) Coefficient for percentage of steel = percentage of steel, p

41.4(fck/fy)(xu max/d) = 41.4(25/415)(0.48) = 1.197%

Question 2: A RCC beam 230mmx 500mm is reinforced with 3 x 16mm diabars. Find moment of resist ance if the effective cover is 40mm. AssumingM20 concrete & Fe415Solution:a) Depth of Limiting NA, xu max = 0.48 d = 0.48x460 = 220.8 mmb) Depth of NA, xu; from C = T -» 0.36 fck xu x b = (Ast) (0.87fy)

xu 0.87x/yx^st 0.87X415X603 _ ~oro— = = = U.toiJOd 0.36x/ckxbXd 0.36X20x230X460

xu = 131.47 mm [it is not Over Reinforced]c) MR, Mu = C x jd = (0.36 fck xu) b( d - 0.42 xu)= 0.36 x20x!31.47x230[ 460 - 0.42 x!31.47 ] =88.13xl06Nmm=88.13 kNm

OR -»MR, Mu = T x jd = Ast x 0.87fy ( d - 0.42xu)= 603 x 0.87 x 415 [460-0.42 xl31.47] = 88.13xl06Nmm =88.13 kNm

Question 3: Determine the service load (udl) which can be carried by asimply supported beam of 5m span having cross section 230 x 450mm(effective), the beam is reinforced with 3 x 20 mm ty placed at an effectivecover of 50 mm. M20 and Fe 500.Solution:a) Depth of Limiting NA, xu max = 0.48 d = 0.48x450 = 216 mm

b) Depth of NA, xu =0.87x/yx,4st 0.87x415x942

= 205.38 mm < xu max0.36xfckxft 0.36x20x230

c) MR, Mu = T x jd = Ast x 0.87fy ( d - 0.42xu)= 942x0.87x415 [450-0.42x205.38]= 123.71 kNm

d) We know that, BM = MR; ^ = 123.718

.-. Ultimate (factored) load, wu = 39.587 kN/tn

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Service (working) load, w = wu/1.5 = 39.587/1.5 =26.39 kN /me) Superimposed Service load (Service load it can carry):Self weight of beam/m = (1 x 0.23 x 0.5) 25 = 2.89 kN/mHence it can carry, 26.39 -2.89 = 23.5 kN/mQuestion 4:Find the area of reinforcement required in a rectangular beam of

230 x 500, if it has to resist an ultimate moment of 125 kNm. Use M 20 andFe4I5Solution:Effective cover = Nominal cover +<J)t + <j>m/2 *

Assume ~»30 + 8 + 20/2 = 48 [moderate exposure]Effective depth = 500 - 48 = 452 mm

[Gb] -> Mu = Ast x 0.87 fy x d (\

125 x 106 = Ast x 0.87 x 415 x 4521

fck xb xd,Ast X 415

20 X230 X452,1.99596 x 10~*Ast2 - Ast + 765.9567 = 0 -> Ast =943.72 mm2

OR

Ast =fck - 1- 4.6 Mu

fck b d2

10

4l51- 1-

4.6 x 125 x 106

20 X 230 X 4522bd

= 944 mm2Question 5: A singly reinforced beam of 6 m effective span has to carry a udlof 24 kN/m (inclusive of self weight) under service conditions. The width ofbeam is 250 mm and reinforce on tension side only. Design the smallestsection, calculate the depth of section and reinforcement. Use M20 & Fe 415.Solution:a) Factored load, wu = 1.5(DL+LL) = 1.5(24) = 36 kN/mb) Maximum BM @ mid-span = 36 x 62/8 = 162 kNmc) Depth of Beam :

' MUWe know that, BM s MR -> Mu = Qbd2; .-. d = [Q=kxjxO,36fck]

Effective depth required, d = f—-—-— = 484.5mmV 2.76 x 250

Effective cover = 20 + 6 +20/2 = 36 mmOverall depth required = 520.5 = Say 525 mm

Hence provide D = 525 mm & /. Effective depth, d = 525-36 ~ 489 mmd) Area of steel (reinforcement)

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Ast =fck2/y

i- i- 4.6 Mu

fck b d2

4.6 x 162 xlO6

20 x 250 x 4892bd

= 1137.7 mm2 [0.0096x250x484.5 =1163]Try 20 mm<J> -» No of bars, n = Ast/ast = 1137.7/314 = 3.62 NosProvide 4 Nos of 20 mm<j) [Ast provided = 1256]

ORTry 3 Nos of 20 mm <fr -> remaining steel = 1137.7 - 942 = 195.7 mm2No of 12 mm<f> bars required = 195.7/113 = 1.73 say 2 Nos.'. Area of steel = 3 x 20<J. + 2 x 12<j> [Ast provided^ 1168]Question 6: Determine Moment of Resistances of a rectangular sectionreinforced with a steel of area 2600mm2 on the tension side. The width of thebeam is 300mm, effective depth 600mm. The grade of concrete is M20 &Fe250 grade steel is usedSolution:a) Depth of Limiting NA, xu max = 0.53 d = 0.53x600 = 318 mm

b) Depth of NA, xu =Q.87xfyXA$t 0.87X250X2600

= 261.8 mm <xumaxQ.36Xfckxb 0.36x20x300

c) MR, Mu = T x jd = Ast * 0.87fy ( d - 0.42xu)= 2600 x 0.87 x 250 [ 600 - 0.42 x261.8] = 277.12 kNm

[ OR 0.36 x 20 x 261.8 x 300 ( 600 - 0.42 x261.8 ) = 277.11 kNm ]Question 7: A singly reinforced beam 300mmx600mm is reinforced with 4bars of 20mm dia with an effective cover of 50mm. effective span is 4m.Assuming M20 concrete & Fe415 steel, find the value of central load P thatcan be carried by the beamSolution:a) Depth of Limiting NA, xu max = 0.48 d = 0.48x550 = 264 mmu\ f x t A 0.87xfyxAst 0.87x415x1256b) Depth of NA, xu = Q36x/cfex, = 0.36X2QX300 = 209.9 mm < xu max

c) MR, Mu = T x jd = Ast x 0.87fy ( d - 0.42xu)= 1256 x 0.87 x 415 [ 550 - 0.42 x209.9 ] = 209.44 kNm

[ OR 0.36 x 20 x 209.9 x 300 ( 550 - 0.42 x209.9 ) ]d) We know that, BM = MR;Self weight = (1 x 0.3x0.6) 25= 4.5 kN/m

/Wl w!2\W 4.5x42\5 _ + — = 209.44 -> 1.5 —- + n = 209.44

V 4 8 / V 4 8 J.'. The value of central load P that can be carried by the beam, W = 130.6 kN

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Doubly reinforced sections:Doubly reinforced sections are preferred, when the section dimensions are/islimited (restricted), Or incase of restrained beams where both Sagging andhogging moments occur in the beam

a) Depth of Limiting NA, xu,limFrom strain diagramxu.lim 0.0035

0.0035+- fy -+0.002l.lSEsFor fy 250, xulim/d = 0.53For fy 415, xulim/d = 0.48For fy 500, xu lim/d = 0.46

b) Depth of NA, xuC ^ TCc + Cs = T

Cc = 0.36 fck xu bCs = Asc x (fsc-fcc)T =Astx0.87fy

ofy/1.15E+0.002

0.446 fck

dcfcc =0.446 fck, if ESC > 0.002

else fcc = 1000x0.446 fck [e - 250e2]Note: fsc is always less than 0.87 fy in case of HYSD bars

It can be either found from strain diagram or the table 2[ Initially fcc may be approximated to 0.446 fck

& fsc may be approximated to 0.87 fy ]From equilibrium: C = T -» Cc + Cs = T

.-. (0.36 fck xu)b+Asc (fsc-fcc) = Ast x 0.87 fy d.. xu Ast x 0.87 fy -Asc (fsc-fcc}Hence, — = — —-—-

d 0.36 x/cfe x b x dc) Lever arms: for Cc ; jd 1 = d - 0.42 xu

&forCs; jd2 = d -d ?

If MR, Mu is to determined on the basis of Tension, jd = d- y.... - Cc(0.42xu)+Cs(d')Where, y = — —

J Cc+Csd) Moment of resistance: asC = TMR; Mu = I (Cx jd ) = Cc x jdl + Cs x jd2

= [0.362 fck xu h ] [d - 0.42 xu] + [Asc(fsc-fcc)][d-d']OR _

MR, Mu = T x jd - [Ast x 0.87 fy] x [d - y]

_Cs_

d -0.42xud - y

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Question 8: A rectangular beam 250 x 550 mm in M20 grade concrete isprovided with Fe 415 type reinforcement of 3 - 20 mm§ + 2 -16 mm<}) ontension side with an effective cover of 46 mm and 2 - 1 2 mm o oncompression side with an effective cover of 42 mm. Calculate the ultimatemoment capacity of the section.

Solution:1. Depth of limiting NA, xu max = 0.48 d = 0.48x(550-46)=241.92 mm2. Depth of NA, xuAst = 3x 314 +2x201 = 1344 mm2; Asc = 2x1 13 = 226 mm2.

As C=T -> Cc+Cs =T -> (b xu) 0.36 fck + Asc( fsc-fcc) = (Ast) ( 0.87fy);Ast X O.o/ x fy — Asc(jsc — fcc)

u~X~0.36

Assuming fsc = 0.87 fy and fcc = 0.446 fck

1344 x 0.87 x 415 - 226(0.87 x 415 - 0.446x20)

MR< 241.92

u 0 .36x250x20=225.372 mm

, Mu = I Cxjd = Cc x jdl + Cs x jd2= 0.36 fck xu b [d-0.42 xu] + Asc( fsc-fcc) [d-d']= 0.36 x 20 x 225.37 x 250 x (504-0.42x225.37)

+ 226 (0.87 x 415 - 0.446 x 20) x (504 - 42)= 405666(504-0.42x225.37+79581.38 )x(504-42) = 202.824 kNm

ORAst x 0.87 x /y[0.42 xu] + Asc(fsc - fcc}[d]

y = -

x,, =

Ast x 0.87 x fy + Asc(fsc - /cc)405666(0.42 x 225.37) + 79581.38 x 42

= 86.0198405666 + 2958138

Mu = T x jd = [Ast x 0.87 fy] x [d - y] = 1344 x 0.87x 415[504-86.0198]= 202.825 kNm

Question 9: Calculate the moment of resistance of a doubly reinforced RCbeam of rectangular section of size 300 x 500 mm reinforced with 4 - 25$bars on tension side and 3 - 16$ bars on compression side. Use M20 and Fe250. Assume an effective cover of 45 mm on both sides.Solution:1. Depth of l i m i t i n g NA, \ max = 0.53 d = 0.53x(500-45}=241.2 mm2. Depth of NA, xu

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Cc+Cs =T -> (b xu) 0.36 fck + Asc( fsc-fcc) = (Ast) ( 0.87fy);_ Ast X 0.87 x fy - Asc(fsc - fcc}

*u~ 0.36 xbxfck

Assuming fsc = 0.87 fy and fee = 0.446 fck

x,, =4x491 x 0.87 x 250 - 3x201 (0.87 x 250 - 0.446x20)

0.36 x 300 x 20= 139.54 mm < 241,2

MR Mu = £ Cxjd = Cc x jdl + Cs x jd2= 0.36 fck xu b [d-0.42 xu] + Asc( fsc-fcc) [d-d']= 0.36 x 20 x 139.54 x 300 x (455-0.42x139.54)

+ 603 (0.87 x 250 - 0.446 x 20) x (455 - 45) = 171.04 kNmDesign of Doubly reinforced beam

A doubly reinforced beam is designed as Singly reinforced balanced sectionPLUS Compression steel and additional tensile reinforcement to resist theremaining moment.

Question 10 : A doubly reinforced beam of size 250 x 600 mm depth isrequired to resist an ultimate moment of 310 kNm. Using M20 and mild steelreinforcement, calculate the amount of steel required. The effective covers are55 mm and 40 mm on tension and compression side respectivelySolution:a) Ultimate moment of resistance of Singly reinforced balanced section =Qbd2 = 2.76 x 250 x 5452 = 204.95 kNMb) Area of steel required for SR Balanced section, Astl = 0.0096 x250 x545= 1308mm2.

[OR Ast = l- i-415 202/> fckb

C) Compression reinforcement:Balance Moment , Mu bal = 310 - 204.95 = 105.05 kNmAssuming fsc = 0.87 fy and fee = 0.446 fckMu bal = Asc( fsc-fcc) (d-d')105.05 x 106 = Asc (0.87 x 415 - 0.446 x 20)(545 - 40) ; Asc = 590.75mm2

Try 20 mmfc No of bars = 590/201 = 2.9 say 3Nosd) Additional tensile reinforcementMu bal = Ast2 x 0.87 fy (d-d') ,105.05 x 106 = Ast2 (0.87 x 4 15)( 545-40) ; Ast2 =576 mm2e) Total tensile reinforcementAst = Astl + Ast2 = 1308 + 576 = 1884 mm2

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Try 2 5 mnn|) ; n= Ast/ast = 1884/491 =3.83 say4Nos.

Question 11 :A doubly reinforced beam 230 x 500 mm size required to resistan ultimate moment of 170 kNm and ultimate shear force of 80 kN. UsingM20 and Fe 415 steel, calculate the quantity of steel required and shearreinforcement. The effective cover to tension steel is 50mm and compressionsteel is 45 mm.Solution:a) Ultimate moment of resistance of Singly reinforced balanced section =Qbd2 = 2.76 x 230 x 4502 = 128.55 kNMb) Area of steel required for SR Balanced section, Astl = 0.0096 x230 x450= 993.6 mm2.

[OR Ast = l- i-415 20

c) Compression reinforcement:Balance Moment , Mu bal = 170 - 128.55 = 41.45 kNmAssuming fsc = 0.87 fy and fee = 0.446 fckMu bal = Asc( fsc-fcc) (d-d')

41.45 x 106 = Asc (0.87 x 415-0.446 x 20)(450 -45) ; Asc = 290.65mm2

Try 12 mm<t>; No of bars = 290/1 13 = 2.6 say 3Nosd) Additional tensile reinforcementMu bal = Ast2 x 0.87 fy (d-d')41.45 x 106 = Ast2 (0.87 x 415)(450-45) ; Ast2 =283.47 mm2.'. Total tensile steel, Ast = Astl + Ast2 = 993.6 + 283.47 = 1277 mm2Try 25 mm$ ; n= Ast/ast =1277 7491 = 2.6 say 3 Nos.e) Check for shearShear stress, tv = V/(bd) =80 x 103/(230 x 450) =0.773 MPaPercentage of tensile steelt, p = 100 Ast/bd = 100xl277/(230x450) = 1.23%Permissible shear stress, TC (Table 1 9 - IS 456)1% - - - 0.621.25--- 0.671 .23 -» 0.62 + (0.05/0.25)(0.23) = 0.666 Mpaiv> t c - Unsafe; Hence provide shear reinforcementf) Design of shear reinforcement:Design shear force, Vus = V - TC bd = 80x1 03 -0.666x230x450 = l l x l 0 3

Try 2 legged Vertical slirrups of 6 mm <jt , Asv = 2 x 28.3 = 58.6 mm24sv(0.87/»d 58.6(0.87 x 415)450

sv = - - -- = - - — r-= - = 865 mm clcVus 11 x 103

Max permissible spacing (smaller of the three)

7 | P a g e R V R K P r a s a d , K D K C o l l e g e o f E n g i n e e r i n g , N a g p u r

i) 0.75 d = 0.75(450) = 337.5ii) 300....... . . ,- Asv(O.B7fy) 58.6(0.87X415) „„_in) Minimum shear reinforcement, sv = —•—-*« = = = 229

0.46 0.4X230

/. Maximum permissible spacing is 229 mm c/cHence provide 6 mm<j> 2 legged vertical stirrups @ 200 c/c

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:LQ c.Singly reinforced sections

Question 12: Design a RC beam 350x700mm effective section, subjected to abending moment of 300kNm. Adopt M20concrete and Fe415 steel. Sketchreinforcement detailsSolution:a) Design (Factored) Moment, Mu = 1.5 x 300 - 450 kNmb) MR of a balanced section = Qbd2 = 2.76 x 350 x 7002 = 473kNmHence it is singly reinforced and under reinforced sectionc) Area of tensile reinforcement

Ast =fck

2fy1- 1-

4.6 Mubd =

10

4151-

4.6 x 450 x 106

20 x 350 x 7002fck b d2i

= 2186 mm2 [0.0096x350x700 =2352]Try 25 mm<t>, n = 2186 7491 =4.45; provide 5 - 25$

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Question 13: A reinforced concrete beam is supported on two walls 250mmthick, spaced at a clear distance of 6m. The beam carries a super-imposed loadof 10 KN/m. Design the beam using M20 concrete and HYSD bars of Fe 415grade.Solution :1. Assumption of dimensions: d * l/20k = 60007(0.75x20) = 400

Effective cover = 30 + 8 + 2072 = 48 mm -> say 50 mmD = 450 mm, b » 0.5D =225 -» say 230 mm

2. Effective span [22.2]c7c of supports = 6+0.25 = 6.25mClear span + d = Lc + d = 6 + 0.4 = 6.4 m (Smaller)Effective span = 6.25m

3. Loading: P""- .Selfwt = (lx0.23x0.45)25=2.5875kN7m ^^JSuperimposed load = 10 kN7mTotal, w= 12.5875 kN7mDesign load, wu = 1.5 x 12.5875 = 18.88 kN7m

4. Calculation of BM & SFMax BM , Mu = wu x 1278 = 18.88x 6.25278= 92.19 kNmMax SF, Vu = wu x 17 2 = 18.88 x 6.257 2 = 59 kN

5. Computation of effective depth, d =V(M7Qb) =V(92.19x10672.78x230)= 379 mm < 400 mm Safe

[MR = C x jd = 0.36 fck b xc x (d-0.42 xc) = 0.138 fck bd2]c • c Q.Sfck [, l^~ 4.6 Mu

6. Area of reinforcement, Ast = —7— 1- I I — - rfy L fckbd2

10415

1- 1-4.6^92.19xl06

20 X 230 x 4002230x400 = 774mm2

Try 16 mm $, No of bars = 7747201 = 4 bars7. Shear ReinforcementShear stress, TV = Vu7bd = 59,0007(230x400) = 0.64 N7mm2Percentage of steel, p = 100 AsLOxi = 100x774 7(230x400)= 0.84%

0.75%-0.561.00%-0.62

TC = 0.84% - 0.56 +(0.0670.25)x0.09 =0.58 N7mm2.

TV > TC - unsafe - provide shear reinforcement

8 | P a g e R V R K P r a s a d , K D K C o l l e g e o f E n g i n e e r i n g , N a g p u r

Design shear force, Vus = V - TC bd = 59,000 - 0.58(230x400) =5640 N

Spacing of 6mm <f>, 2 legged vertical stirrups, sv = 0.87fy Asv d/Vus= 0.87 x415 x (2 x28.27) x400/ 5640 = 1447 mm c/c

Maximum spacing [26.5.1.5](i) 0.75 d = 0.75(400) = 300(ii) 300(iii) Min. shear reinforcement 0.87fy Asv /0.4b

= 0.87x415x(2x28.27) /0.4x230 = 221Provide 6 mm(|> 2 legged Vertical stirrups @ 200 mm c/cProvide 2 nos of 6 mm <|> anchor bars to hold the stirrupsCHECK for 88. Check for bondDevelopment length, Id = <t>(0.87fy)/4Tbd = <t>(0.87x415)/(4xl.6xl.2)

= 47<[> = 47x16 = 752 mmLd/3 = 752/3 = 250 mmExtend the 250 mm in side the support [26.2.3.3 a]

L d < M l / V + LO[ The value of Ml/V in the above expression may be increased by 30% whenthe ends of the reinforcement are confined by a compressive reaction]Bent up 2 bars @ 0.15 1 from supportOnly 2 bars are available at support, Ml * M/2 =92.19/2 = 46.2 kNmLO = 12$ or d (greater) 12<f> = 192 or 400 ; LO = 400Ml/V + LO = 46.2x106/59xl03 + 400 =1183 » 752 Safe

9. Reinforcement Detailing

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9 | P a g e R V R K P r a s a d , K D K C o l l e g e o f E n g i n e e r i n g , N a g p u r

Winter 14Q(5a)[ 14M]:Q(5b)[ 14M]:Q(6a)[ 14M]:Q(6b)[ 14M] :

OR

Summer 15Q (5a) [ 5M ]: Compare WSM and LSMQ (5b) [ 9M ] : Derive Equation for limiting depth of neutral axis. Also sketchthe stress-strain diagram for beam in flexure. [ OR ]Q (6a) [ 7M ]: Determine the service load which can be carried by a simplysupported beam of 5m span having cross section 230 x 450 mm effective, thebeam is reinforced with 3 x 1 6 mm § placed at an effective cover of 40 mm.M20 and Fe 500.Q (6b) [ 7M ]: Design a singly R/F rectangular beam for a clear span of 5 msubjected to superimposed load of 12kN/m over the entire span. Use M20and Fe415.

Winter 15Q (5a) [ 6M ] : A RCC beam 230mmx 500mm is reinforced with 3 x 16mmdia bars. Find moment of resistance if the effective cover is 40mm. AssumingM20 concrete & Fe415Q (5b) [ 7M ]: A singly reinforced beam of 4.5m span carries a udl of 30kN/m inclusive of self weight. The width of beam is 230 mm and reinforce ontension side only. Design the smallest section, calculate the depth of sectionand reinforcement. Use M20 and Fe 415.

ORQ (6a) [ 6M ] : Calculate the moment of resistance of a doubly reinforced RCbeam of rectangular section of size 300 x 450 mm reinforced with 6 - 20(|>bars on tension side and 4 - 20<J> bars on compression side. Use M20 and Fe250. Assume an effective cover of 35 mm on both sides.Q (6b) [ 7M ] : A doubly reinforced beam of size 250 x 600 mm depth isrequired to resist an ultimate moment of 310 kNm. Using M20 and mild steelreinforcement, calculate the amount of steel required. The effective covers are55 mm and 40 mm on tension and compression side respectively

Summer 160 (5a) [ 6M ] : Explain : ^i) Stress strain relationship for concrete,ii) Stress strain relationship for steel in LSM.

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Q (5b) [ 7M ] : A rectangular beam is 20cm wide and 40cm deep up to thecentre of reinforcement. Find the area of reinforcement require if it has toresist a moment of 25 kN/m. Use M 20 concrete mix and Fe 415 steel. Alsogive check for sectors

Q (6a) [ 6M ]: Derive Equation for limiting Moment of resistances forbalanced, underreinforced and over reinforced section by LSM of singlyreinforced beamQ (6b) [ 7M ]: A rectangular beam has a width of 250 mm and effectivedepth of 500 mm. The beam is provided with tension steel of 5 -25 mm<t> andcompression steel of 2 - 25 mm <f>. The effective cover to the compressionsteel being 50 mm. Calculate the ultimate moment capacity of the section offck = 20 MPa and fy = 250 MPa.

Winter 16

Q (5a) [ 7M ] : Design a RC beam 350- 700mm effective section, subjectedto a bending moment of 300kNm. Adopt M20concrete and Fe415 steel.Sketch reinforcement detailsQ (5b) [ 6M ] : Determine Moment of Resistances of a rectangular sectionreinforced with a steel of area 2000mm2 on the tension side. The width of thebeam is 200mm, effective depth 600mm. The grade of concrete is M20 &Fe250 grade steel is used

OR

Q (6a) [ 7M ] : A singly reinforced beam 230mm • 600mm is reinforced with4 bars of 16mm dia with an effective cover of 50mm. effective span is 4m.Assuming M20 concrete & Fe415 steel, find the value of central load P thatcan be carried by the beamQ (6b) [ 6M ] : Derive Equation for limiting depth of neutral axis and momentof resistances for balanced, under reinforced and over reinforced section byusing LSM

Summer 17Q (5) [ 13M ]: Design a singly reinforced rectangular beam for an effectivespan of 5m subjected to a live load of 12 kN/m over the entire span. Calculatemain reinforcement and shear reinforcement. Give all necessary checks as perIS 456. Draw the sketch of reinforcement details.ORQ (6) f 13M ] : A doubly reinforced beam 250 x 600 mm size required toresist an ultimate moment of 310 kNm and ultimate shear force of 70 kN.Using M20 and I'c 415 steel, calculate the quanti ty of steel required and shear

1 0 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

reinforcement. The effective cover to tension steel is 55mm and compressionsteel is 40 mm.JLV.k.1 1J -TV/ 1H111.

Recap from Structural AnalysisSupport Reactions, SFD. BMD. 6, 8 & FEM

I ) Cantilever beam with concentrated load at free end

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2) Cantilever beam with udl over the entire span

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4) SS Beam with concentrated load at ccL from A5) SS Beam with udl over the entire span

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reinforced R.C. beam of Rectangular section of size 300 x 450 mm.reinforced with 6-20 mm dia bars on tension side and 4-20 mm diabars on compression side.Use M20 grade concrete and Fe 250 grade steel.Assume eff. cover of 35 mm on both sides. (Use L. S.M.)Q (6b) [ 7M ]: A doubly reinforced beam of size 250mmx600mmdeep is required to resist on ultimate moment of 310 kNm. Usingconcrete M 20 and mild steel reinforcement, Calculate the amount ofsteel required. The effective cover to tension steel is 55mm while thatfor compression steel is 40mm. (Use L.S. M.)

TABU A SAUCNT POINTS OH THE DC6KJN 6THC56 STRAIN OJHVC fOUCnitV WORKED BARS

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Winter 15Q (5a) [ 6M ]: A RCC Beam 230 X 500 mm is reinforced with3-16mm dia bars. Find the moment of Resistance, if effective cover is40 mm and effective span 3 m. Use M 20 concrete and Fe 415 steel.(Use L.S.M.)0 (5b) [ 7M ]: A simply supported beam of 4.5 m span carriesa udL of 30 kN/m inclusive of self wt. The width of beam is 230 mmand is reinforced on tension side only. Design the smallest section,calculate depth of section and reinforcement. Use M20 concrete and (

Fe250 steel. (Use L. S. M.)Q (6a) [ 6M ] : Calculate the moment of resistance of a doubly

TABLE E MAXIMUM PERCENTAGE c*TENSILE REINFORCEMENT ft^ FOR

SINGLY RHWOROU1 RKTAWOUl-AR

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12 [ P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

TABLE D LIMITING MOMtNT OFR1SISTASO- r A4.TUR W,^_/W. S mm* TOR

SINGLY REINFORCFO RICIANCULARSECTIONS

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TABLE F STRESS IN COMfUFSSKJNRtlSfORCSMtNT/U, H'nxo* IN DOUBLV

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Summer 15

( la)[7]Discuss the merits and demerits of Working Stress method(Ib) [6] Derive design constants for neutral axis, lever arm, constant formoment of Resistance and constant for percentage of steel in WSMOR(2a)[6]Explain under reinforce, over reinforced and balanced section in WSM(2b)[7] A singly reinforced concrete beam is of width 400 mm and effectivedepth 615 mm. It is reinforced with 8-20<J> mild steel bars. Assume M25 gradeconcrete, determine its moment of resistance according to WSM.(3a)[6] Explain the advantages of pre-stressed concrete over RCC(3b)[7])Explain with the help of neat sketches any two of the following pre-s tress ing systems(i) Freyssinet system(ii) Magnel-Blaton system(iii)Gifford Udall systemOR(4X13] Explain(i) Different types of losses in pre-stressed concrete beam(ii) Pre tensioning and post tensioning(iii) Application of Pre-stressed concrete.(5a)[5] Compare WSM and LSM

[5b)[9]Derive equations for limiting depth of neutral axis. Also sketch thestress and strain diagram for beam in flexureOr(6a)[7]Determine the service load which can be carried by Simply supportedbeam of 5.0m span having 230 x 450 (eff), the beam is reinforced with 3-16<J>.Eff cover is 40 mm. M20 and Fe 500 are used.

(6b)[7] Design a Singly reinforced rectangular beam for a clear span of 5.0msubjected to a superimposed load of 12 kN/m over the entire span. Use M20and Fe415

(7a)[6]Determine the MR of a T-beam from the following detailsbf = 1000mm, Df = 120 mm, Ast = 6 -25<J> , d = 600 ,„ bw = 300 mm, M20and Fe415

..(7b)[7] Design a RCC Column of Rectangular section having unsupportedlength of 3m subjected to an axial compressive laod of 1200 kN using M20and Fe415. One dimension is restricted to 400 mm.OR(8)[13] Design a rectangular pad footing for a column of size 300 x 500 mmwith compressive load of 1000 kN. Use M20 concrete with Fe 415 steel, thedensity of soil is 21kN/m3 and SBC 150 kN/m2. Give all the necessarychecks as per IS 456 with neat sketch.

(9a)[7] What is limit state of serviceability? How it is ensured for beams?(9b)[6]ExpIain in brief the various measures for deflection control as per IS456-2000OR(10)[13]Design a rectangular beam section of size 250 x 450 mm subjecred toa bending moment of 30 kNm, Shear force of 40 kN and torsional moment of20 kNm at working condition. Use M20 And Fe 4 1 5.

(1 1)[14] Design a cantilever slab projecting 1.5 m from rhe face of column,the slab carries live load of 1.5 kN/m2. Use M20 andf Fe 415. Sketchreinforcement details.OR(12)[14]Design a slab panel for a hall of size 4 * 5.5m supported on 230 mmthick brick wall all around, the slab carries superimposed load of 2.5 kN/m2with floor finish of 0.75 kN/m2. Use M20 Grade of Concrete and Fe 415 type

13 ( P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n a g p u r