APPENDIX

DESIGN PHILOSOPHIES

Since the inception of the concept of reinforced concrete in last

twenties of the nineteenth century, the following design philosophies

have been evolved for design of R.C. structures.

i. Working stress method.

ii. Ultimate load method.

iii. Limit state method.

Working stress method: - This has been the traditional method used

for reinforced concrete design. It is assumed that concrete is elastic,

steel and concrete act together elastically, and the relationship

between loads and stresses is linear right up to the collapse of the

structure. The sections are design in accordance with the elastic

theory of bending assuming that both materials obey the Hooke‘s law.

The elastic theory assumes linear variations of strain and stress from

zero at the neutral axis to a maximum at the extreme fibre. Typical

stress strain distribution in rectangular sections is shown below.

d

b

At

D

SECTION

E

E

st

C

NEUTRAL

AXIS

STRAIN

Nd

0 cb

LEVER ARMjd

C

st0

STRESS

STRESS - STRAIN CURVE IN WORKING STRESS DESIGN

T

Where At = Area of tension steel; b = Width of the section

C = Total force of compression; T = total force of tension.

D = Overall depth of the section; d = effective depth

jd = lever arm; nd = depth of neutral axis

Єc = compressive strain concrete; Єst = tensile strain in steel.

σst = permissible tensile stress in steel

σcbc = permissible compressive stress in concrete

Ultimate load method:- In this method the working loads are

increased by suitable factors called load factors to obtain ultimate

loads. The structure is then designed to resist the ultimate loads. The

term safety factor has been used in the working stress method to

denote the ratio between the yield stress and permissible stress. The

term load factor has been used to denote the collapse or ultimate load

to the working load. Whitney‘s theory is based on the assumptions

that ultimate strains in concrete is 0.3%. Whitney replaced the actual

parabolic stress diagram by a rectangular stress diagram such that

the centre of gravity of both the diagrams lies at the same point and

areas are also equal. The major advantage of this method over the

working stress method is that total safety factor the structure thus

found is nearer to its actual value. The structure designed by Ultimate

load method requires less reinforcement than those designed by

working stress method.

a = depth of rectangular stress block,

Єsy = Yield strain in concrete

σcb = Ultimate compressive strength of concrete

Єcu =Ultimate strain in concrete

d

b

A t

D

SECTION

E

E

sy

cu

NEUTRAL

AXIS

STRAIN

Xm

0 cu

LEVER ARMz

C

y0

ACTUAL STRESS

STRESS - STRAIN CURVE IN ULTIMATE LOAD DESIGN

T

k0 cu

y0

WHITNEY'S STRESS

aC

Limit state method: - Limit state design has originated from Ultimate

load or plastic design. The object of design based on the limit state

concept is to achieve an acceptable probability that the structure will

not become unserviceable in its lifetime for the use for which it is

intended, i.e. it will not reach limit state.

Where σy = characteristic strength of steel.

Єs = strain in steel at failure; Es = modulus of elastic of steel

d

b

At

D

SECTION

E

E

sy

cu

NEUTRALAXIS

STRAIN

Xm

k0 ck

LEVER ARMz

C

y

STRESS

STRESS - STRAIN CURVE IN LIMIT STATE DESIGN

T

0

Modes of failure:

If the ratio of steel to concrete in a beam is such that the

maximum strain in the two materials reaches simultaneously, a

sudden failure would occur with less alarming deflection. Such a

beam is referred as a balanced reinforced beam. When the amount of

steel is kept less than that in the balanced condition, the neutral axis

moves upwards to satisfy the equilibrium condition that force of

compression is equal to the force in tension. In this process center of

gravity of compressive force also shifts upwards. Under increasing

bending moment, steel is strained beyond the yield point and the

maximum strain in concrete remains less than 0.35%.

If the beam is further loaded, the strain in the section increases.

Once steel has yielded, it does not take any additional stress for the

additional strain and the total force of tension remains constant.

However, compressive stresses in concrete do increase with the

additional strain. Thus, neutral axis and the center of gravity of

compressive forces further shift upward to maintain equilibrium. This

results in an increase in the moment of resistance of the beam. This

process of shift in the neutral axis continues until maximum strain in

the concrete reaches its ultimate value, that is 0.35% and the concrete

crushes. Such a beam is referred to as an under reinforced beam.

When the amount of steel is kept more than that in the balanced

condition, the neutral axis tend to move downward and strain in steel

remains in the elastic region .If the beam is further loaded, the strain

and the stress in steel keep on increasing and so the force of tension.

The additional increase in the concrete stress is much slower. Thus to

maintain the equilibrium of tension and compression forces, the area

of concrete resisting compression has to increase. In this process,

neutral axis further shifts downwards until the maximum strain in

the concrete reaches its ultimate value, which is 0.35% and concrete

crush the steel is still well with in the elastic limit. Such a beam is

referred as an over reinforced beam and failure as a compression

failure.

Moment of Resistance:

Consider a simply supported beam subject to bending under

factored load. For equilibrium, total force of compression must be

equal to total force of tension. The applied bending moment at

collapse, that is, factored bending moment is equal to the resisting

moment on the section provided by internal stress, this is called the

ultimate moment of resistance.

Total force of compression in concrete =0.36 σck b Xu.

Total force of tension in steel = 0.87 σy Ast.

Moment of resistance with respect to concrete

= compressive force * lever arm = 0.36σck b Xu (d-0.42 Xu)

Moment of resistance with respect to steel = tensile force * lever arm

= 0.87 σy Ast (d-0.42 Xu)

Design of Beam Using Ordinary Grade Concrete

Beam dimensions : 100 X 150 X 1200 mm.

Grade of concrete : M20

Grade of steel : Fe 415

Xu max = 0.48d. = 0.48 X (150-30) = 57.6 mm

For balanced section,

0.36 x fck x b x Xu max = 0.87 x Fy x Ast.

0.36 x 20 x 100 x 57.6 = 0.87 x 415 x Ast. Ast = 114.86 mm²

Providing 2-10 mm dia. tor steel bars:

Ast = 2 (π /4) 10² = 157.1mm²

0.36 x 20 x 100 x Xu = 0.87 x 415 x 157.1. Xu = 78.78 mm.

Moment of resistance = 0.87 x 415 x 157.1(120-0.42 x 78.78)

= 4.93 kN-m.

W/2 W/2

166.6 166.6

1000 100 100

W/2 W/2

(W/2) x 333.3 =4.93x10 6 W = 29.58 kN.

Shear force : V = W/2 =29.58/2 = 14.79 kN. Vu = 14.79 kN.

Normal shear stress τv = Vu /(b.d) =14.79x10 3 /(100x120)

= 1.23 N/mm²

Percentage of steel p = 100 x 157.1/(100 x120) = 1.31

τc = 0.68 N/mm ² (from IS456:2000) τv > τc

Design for shear reinforcement

Vc = 0.68 x 100 x 120 Vc = 8.16 kN.

Vus = Vu-Vc =14.79-8.16 = 6.63 kN.

Spacing for 6mm dia-2 legged stirrups

Vus = 0.87 x Fy x Ast x d / Sv

Sv = 0.87 x 250 x 2 x 28.27 x 175/(6.63x10 3) =222.58mm c/c.

Maximum spacing = 0.75d = 0.75x120 = 90 mm.

Design of Beam Using Standard Grade Concrete

Beam dimensions : 100 X 150 X 1200 mm.

Grade of concrete : M20

Grade of steel : Fe 415

Xu max = 0.48d. = 0.48 X (150-30) = 57.6 mm

For balanced section,

0.36 x fck x b x Xu max = 0.87 x Fy x Ast.

0.36 x 40 x 100 x 57.6 = 0.87 x 415 x Ast. Ast = 229.73 mm²

Providing 2-12 mm dia. tor steel bars:

Ast = 2 (π /4) 12² = 226.19 mm²

0.36 x 40 x 100 x Xu = 0.87 x 415 x 226.19 Xu = 56.71 mm.

Moment of resistance = 0.87 x 415 x 226.19(120-0.42 x 56.71)

= 7.85 kN-m.

(W/2) x 333.3 = 7.85x10 6 W = 47.1 kN.

Shear force : V = W/2 =47.1/2 = 23.55 kN. Vu = 23.55 kN.

Normal shear stress τv = Vu /(b.d) =33.82x10 3 /(100x120)

= 1.96 N/mm²

Percentage of steel p = 100 x 226.19/(100 x120) = 1.85

τc = 0.86 N/mm ² (from IS456:2000) τv > τc

Design for shear reinforcement

Vc = 0.86 x 100 x 120 Vc = 10.32 kN.

Vus = Vu-Vc =23.55 – 10.32 = 13.23 kN.

Spacing for 6mm dia-2legged stirrups

Vus = 0.87 x Fy x Ast x d / Sv

Sv = 0.87 x 250 x 2 x 28.27 x 125/(13.23x10 3) = 111.4 mm c/c.

Maximum spacing = 0.75d = 0.75x120 =90 mm.

Therefore a spacing of 75 mm c/c is provided for both the beams.

# MATLAB Program #

fid_a=fopen('Sunil.txt','w');

fck=input('Enter the value for fck:');

eu=input('Enter the value for eu:');

A=input('Enter the value for A:');

B=input('Enter the value for B:');

Ast=input('Enter the value for Ast:');

d = 120;

b = 100;

for x=0.1:0.1:1

e=x*eu;

fa=(1/e)*fck*A*(eu/(2*B))*log10(1+((B*e*e)/(eu*eu)));

n=0.1;

for n=0.1:0.01:(n<0.99)

es=((1-n)/n)*eu;

if((es>=0.000416)&&(es<=0.00054))

fst=52.04+(((69.39-52.04)/(0.00054-0.000416))*(es-0.000416));

elseif((es>0.00054)&&(es<=0.000666))

fst=69.39+(((86.74-69.39)/(0.000666-0.00054))*(es-0.00054));

elseif((es>0.000666)&&(es<=0.00075))

fst=86.74+(((104.09-86.74)/(0.00075-0.000666))*(es-0.000666));

elseif((es>0.00075)&&(es<=0.00108))

fst=104.09+(((121.44-104.09)/(0.00108-0.00075))*(es-0.00075));

elseif((es>0.00108)&&(es<=0.001208))

fst=121.44+(((138.79-121.44)/(0.001208-0.00108))*(es-0.00108));

elseif((es>0.001208)&&(es<=0.001416))

fst=138.79+(((156.141-138.79)/(0.001416-0.001208))*(es-0.001208));

elseif((es>0.001416)&&(es<=0.001583))

fst=156.141+(((173.48-156.141)/(0.001583-0.001416))*(es-0.001416));

elseif((es>0.001583)&&(es<=0.001708))

fst=173.48+(((208.188-173.48)/(0.001708-0.001583))*(es-0.001583));

elseif((es>0.001708)&&(es<=0.00175))

fst=208.188+ (((225.53-208.188)/(0.00175-0.001708))*(es-0.001708));

elseif((es>0.00175)&&(es<=0.00185))

fst=225.53+(((242.88-225.53)/(0.00185-0.00175))*(es-0.00175));

elseif((es>0.00185)&&(es<=0.00191))

fst=242.88+(((260.23-242.88)/(0.00191-0.00185))*(es-0.00185));

elseif((es>0.00191)&&(es<=0.00208))

fst=260.23+(((277.58-260.23)/(0.00208-0.00191))*(es-0.00191));

elseif((es>0.00208)&&(es<=0.00241))

fst=277.58+(((312.282-277.58)/(0.00241-0.00208))*(es-0.00208));

elseif((es>0.00241)&&(es<=0.0025))

fst=312.282+(((329.63-312.282)/(0.0025-0.00241))*(es-0.00241));

elseif((es>0.0025)&&(es<=0.00265))

fst=329.63+(((346.96-329.63)/(0.00265-0.0025))*(es-0.0025));

elseif((es>0.00265)&&(es<=0.00275))

fst=346.96+(((364.3-346.96)/(0.00275-0.00265))*(es-0.00265));

elseif((es>0.00275)&&(es<=0.002958))

fst=364.3+(((381.66-364.3)/(0.002958-0.00275))*(es-0.00275));

elseif((es>0.002958)&&(es<=0.003166))

fst=381.66+(((399.02-381.66)/(0.003166-0.002958))*(es-0.002958));

elseif((es>0.003166)&&(es<=0.003708))

fst=399.02+(((416.376-399.02)/(0.003708-0.003166))*(es-0.003166));

elseif((es>0.003708)&&(es<=0.00385))

fst=416.376+(((433.725-416.376)/(0.00385-0.003708))*(es-0.003708));

elseif((es>0.00385)&&(es<=0.00408))

fst=433.725+(((451.06-433.725)/(0.00408-0.00385))*(es-0.00385));

else((es>0.00408)&&(es<=0.0166))

fst=451.06+(((468.42-451.06)/(0.0166-0.00408))*(es-0.00408));

end

cc=fa*b*n*d;

ct=Ast*fst;

y=(ct/cc);

if(y>0.95 && y<1.05)

% fprintf(fid_a,'\n n=%2f',n);

m=((b*n*n*d*d*fck*A)/(e*e*eu))*((((eu*eu*e)/B)-

((eu*eu*eu)/((B^1.5)))*atan((e*(B^0.5))/eu)));

fprintf(fid_a,'\n %2f %2f ',n,m);

break

end

end

end

fclose(fid_a);

PLATES

cell concentration – Nil (control)

cell concentration - 104/ml

cell concentration – 105/ml

cell concentration – 106/ml

cell concentration – 107/ml

Plate 1.1.1 Hydrated Structure of Cement-sand Mortar – Magnified

SEM Micrographs

Plate 1.1.2 Compression Test set up of Cement-sand Mortar cubes

Plate 1.2.1 Compression Test set up of concrete cubes

Plate 1.2.2 Split Tension test set up

Plate 1.3.1 Test set up for stress-strain behaviour on computer

controlled UTM

Loading beam (Rolled Steel

Joist)

Wing table of UTM

Deflection guage Support Blocks

Test Beam

Distribution plate

Curvature meter Loading blocks

Test set-up of Simply Supported Beam

Head - UTM

Plate 1.4.2

150mm

100mm

2 No. 6mm dia MS bars

2 No. 10mm dia Tor bars

Cross Section

1000 mm

Longitudinal section

Plate 1.4.1 The reinforcement details of UR Group–A Beams

2-L 6mm dia @75mm c/c

Plate 1.4.3 Experimental test set up for beams, with curvature meters

Plate 1.4.4 Failed beam specimen

Plate 1.4.5 Failed beam specimen

Plate 1.4.6 Failed beam specimen

Plate 1.4.7 Failed beam specimen

Plate 1.4.8 Failed beam specimen

Plate 2.1.1 Cubes immersed in 5% concentrated sulphuric acid

Plate 2.1.2 Cubes after 28 days of curing under 5% Concentrated

sulphuric acid waters.

Plate 3.2.1

Plate 3.2.2 ACCELERATED CORROSION TEST SET UP

Plate 3.2.3 Accelerated corrosion of concrete set up