Hyd Civil 3 1 Drcs Set 1

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1/16

S.1Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad

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S e t - 1S o l u t i o n s

Code No: 09A50102/R09

III B.Tech. I Semester Examinations

May/June - 2012

DESIGN OF REINFORCED CONCRETE STRUCTURES

( Civil Engineering )

Time: 3 Hours Max. Marks: 75

Answer any FIVE Questions

All Questions carry equal marks

- - -

1. (a) What are the code recommendations for limit state design? (Unit-I, Topic No. 1.1)

(b) Draw the characteristic and design stress curves for concrete and steel (Fe415

). (Unit-I, Topic No. 1.1)

(c) Calculate stress block parameters of concrete. [15] (Unit-I, Topic No. 1.2)

2. Determine the moment of resistance of a T-beam section with the details given below.

bw= 250 mm, d = 550 mm, b

f= 1200 mm, D

f= 100 mm, A

st= 2946 mm2, f

ck= 25 MPa, f

y= 415 MPa. [15]

(Unit-II, Topic No. 2.2)

3. Design the reinforcement in a rectangular beam of section 300 mm 600 mm subjected to an ultimate twisting moment

of 50 kN-m combined with an ultimate bending moment of 120 kN-m. Assume M25

concrete and Fe415

steel. [15]

(Unit-III, Topic No. 3.1)

4. Design a slab for a two-room masonry building with internal dimension of each room as 4.00 m 5.00 m with one long

edge continuous. The masonry wall are of 300 mm thick. Assume a live load of 4.0 kN/m2and a finish load of 1.5 kN/

m2. Assume that the slab corners are not free to lift up. Consider M20

concrete and Fe415

steel. [15]

(Unit-IV, Topic No. 4.1)

5. Design an isolated footing for a column with an axial force of 2000 kN under working load. The size of the column is

300 mm 600 mm. Consider S.B.C of soil as 250 kN/m2. Consider M20

Concrete and Fe415

steel. Assume mild exposure

condition. [15] (Unit-V, Topic No. 5.1)

6. A corner column 300 mm 450 mm located in the multi storey of a system of braced frames, is subjected to factored

loads Pu

= 1600 kN, Mux

= 175 kNm and Muy

= 100 kN-m. The unsupported length of the column is 3.0 m. Design the

reinforcement in the column, assuming M30

concrete and Fe415

steel. [15] (Unit-VI, Topic No. 6.1)

7. (a) Distinguish between short and long term deflection and what are the limits prescribed by the code IS:456.

(Unit-VII, Topic No. 7.1)

(b) Explain effective and cracked moment of inertia. [7+8] (Unit-VII, Topic No. 7.1)

8. Design a suitable dog-legged stair in a public building, to be located in a staircase room 6 m long, 3.0 m wide and

the floor height is 3.5 m. Use M20

concrete and Fe415

steel. [15] (Unit-VIII, Topic No. 8.1)


2/16

S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012


SOLUTIONS TO MAY/JUNE-2012, SET-1, QP

Q1. (a) What are the code recommendationsfor limit state design?

Answer : May/June-12, Set-1, Q1(a)

Indian standard code has prescribed the following

recommendations for limit state design,

(i) Characteristic strength and loads (consider-

ations)

(ii) Partial safety factors for loads and materials

(iii) Design stress-strain curve for concrete

(iv) Design stress-strain curve for reinforcing steel.

(i) Characteristic Strength and Load Considerations

A characteristics load can be defined as the load which

has 95% probability of not exceeding its defined value dur-

ing the entire life of the structure.

In case, if the statistical data for calculating charac-

teristic strength or load is not available, the design charac-

teristic load values for dead, live and wind loads are recom-

mended to follow the code from IS - 875 : 1987 whereas the

seismic load follows IS : 1893 : 2002 standard.

(ii) Partial Safety Factor For Loads and Materials

In material, the partial factor of safety is employed so

as to reduce the characteristic strength value and obtain the

design strength values.

In case of loads, the partial safety factor is used to

increase the characteristic load and obtain the design load.

The partial safety factor commonly used in limit state method

are,

c= Partial safety factor for concrete = 1.5

s= Partial safety factor for steel = 1.15

f= Partial safety factor for loads = 1.5

Where, c=

dc

ck

f

f,

s=

ds

y

f

f,

f=

df

f

Where,fde

,fds

are design characteristic strengths.

fd= Design load.

(iii) Design Stress-strain Curve for Concrete

The curves shown in figure (1) adopted by IS-456 :

2000 for concrete under flexural compression. The maximum

stresses in characteristic and design curves are limited to

0.67fck

and 0.45fck.

0 0.001 0.002 0.003

0.45 fck

0.003

0.67 fck

Idealized curve

Characteristic curve

Design curve for concrete

Stress

fck

Strain

0 0.001 0.002 0.003

0.45 fck

0.003

0.67 fck

Idealized curve


Design curve for concrete

Stress

fck

Strain

Maximum characteristic stress,fc max

=5.1

ckf = 0.67fck

Maximum design stress, fd max

=5.15.1

ckf= 0.44f

ck

Where, 1.5 is a partial safety factor.

Figure (1): Stress Strain Curve of Concrete

(iv) Design Stress-strain Curve for Reinforcing Steel

0.0000

500.25 f

y

1000.50 f

y

150

2000.75 f

y

250 fy

0.87fy


Design curve m

fy

m= 1.15

Designyiel

dstrain

Es

0.87fy

y =

0.002 0.004 0.006 0.008

Strain

Stress

0.0000

500.25 f

y

1000.50 f

y

150

2000.75 f

y

250 fy

0.87fy


Design curve m

fy

m= 1.15

Designyiel

dstrain

Es

0.87fy

y =

0.002 0.004 0.006 0.008

Strain

Stress

0.0000

500.25 f

y

1000.50 f

y

150

2000.75 f

y

250 fy

0.87fy


Design curve m

fy

m= 1.15

Designyiel

dstrain

Es

0.87fy

y =

0.002 0.004 0.006 0.008

Strain

Stress

Figure (2): Stress Strain Curve for Mild Steel

The curve was given by IS - 456 : 2000. The charac-teristic strength is considered equal to the yield stress, while

the design strength is obtained by using a partial safety

factor of 1.15. As mild steel comprises of a well defined yield

point, it is linearly elastic to design stress 0.87fy

and hence

the design stress is considered to be constant.

The high yield steel bars do not have a well defined

yield point and hence it is assumed that the linear elastic

behaviour changes to linear inelastic behaviour at a strength

of 0.8fyin the characteristic curve. The design curve is as-

sumed to transform from elastic to inelastic state at a stress

level of 0.8fyd

.


3/16



Where,fyd

= Design yield strength.

0.001 0.002 0.003 0.00S0.004

0.00070.00030.0001

500

400

300

200

100

0

Fe500

Fe415

0.001 0.002 0.003 0.00S0.004

0.00070.00030.0001

500

400

300

200

100

0

Fe500

Fe415

Figure (3): Stress-Strain Curve for Steel (Fe415

and Fe500

)

(b) Draw the characteristic and designstress curves for Concrete and Steel(Fe

415).

Answer : May/June-12, Set-1, Q1(b)

Stress-Strain Curve for Concrete

For answer refer May/June-12, Set-1, Q1(a), Topic:

Figure (1).

Stress-strain Curve for Steel

For answer refer May/June-12, Set-1, Q1(a), Topic:

Figure (2).

(c) Calculate stress block parameters ofconcrete.

Answer : May/June-12, Set-1, Q1(c)

For answer refer Unit-I, Q8.

Q2. Determine the moment of resistance of aT-beam section with the details given below,

bw= 250 mm, d = 550 mm, b

f= 1200 mm,

Df= 100 mm, A

st= 2946 mm2, f

ck= 25 MPa,

fy= 415 MPa.

Answer : May/June-12, Set-1, Q2 M[15]

Given that,

Width of web, bw= 250 mm

Width of flange, bf= 1200 mm

Depth of flange,Df= 100 mm

Depth of web, d= 550 mm

Area of steel,Ast= 2946 mm2

M25

grade,fck

= 25 MPa = 25 N/mm2

Fe415

steel,fy= 415 MPa = 415 N/mm2

bf= 120 mm

d=

550

mm

450mm

250 mm

Df= 100 mm

Ast

= 2946 mm2

(1 MPa = 1 N/mm2)

bf= 120 mm

d=

550

mm

450mm

250 mm

Df= 100 mm

Ast

= 2946 mm2

(1 MPa = 1 N/mm2)

Figure

Let the neutral axis falls in the flange (i.e.,x4is less

than depth of flange Df)

ux

d=

dbf

Af

fck

sty

36.0

87.0

ux

d=

55012002536.0

294641587.0

ux

d= 0.179

xu

= 0.179 d

xu

= 0.179 550

xu = 98.450 mm xu(Hence safe)

The beam is under-reinforced

The moment of resistance,MR

MR

= 0.87fy

Ast

d

dbffA

fck

yst1

Moment of resistance,MR

= 0.87 415 2946 550

550120025

41529461

MR

= 541.662 106N-mm

MR

= 541.662 kN-mm

Therefore, the moment of resistance of T-beam,

MR

= 541.662 kN-m.


4/16



Q3. Design the reinforcement in a rectangularbeam of section 300 mm 600 mm subjectedto an ultimate twisting moment of 50 kN-m

combined with an ultimate bending momentof 120 kN-m. Assume M

25concrete and Fe

415

Steel.


Given that,

Width of the beam, b= 300 mm

Overall depth,D= 600 mm

Ultimate twisting moment, Tu= 50 kN-m

Ultimate bending moment,Mu= 120 kN-m

For M25

concrete,fck

= 25 N/mm2

For Fe415

concrete,fy= 415 N/mm2

Provide an effective cover of 40 mm all around the

beam

Effective depth, d = 600 40 = 560 mm

Design of Longitudinal Reinforcement

Equivalent bending moment due to torsion,

Mt= 1.7

uT

+b

D1

=7.1

50

+

300

6001

Mt= 88.235 kN-m

Total bending moment for design,Meq

Meq

=MtM

u

= 88.235 120

Meq

= 208.235 kNm or 31.765 kN-m

Meq1

= 208.235 kNm () orMeq2

= 31.765 kN-m ()

The momentMeq1

= 208.235 kNm is consider for pro-

viding flexural steel at top, while the momentMeq2

= 31.765

kNm is considered for providing flexural steel at bottom.

At Top

R1

=1

2

eqM

bd=

2

6

560300

10235.208

R1

= 2.213 N/mm2

100

tP=

bd

Ast=

y

ck

f

f

.2

ckf

R1598.411

100

tP=

4152

25

25

213.26.411 = 0.006929510

100

tP ~ 0.007

Area of steel required,

(Ast1

)req

=100

tP bd

= 0.007 300 560

= 1176 mm2

For a 20 mm bars, we get

Number of bars, n =1st r eqA

A=

u

220

1176

= 3.743

n ~ 4

Provide4 bars of 20 mm at top.

At Bottom

R2

= 22

bd

Meq= 2

6

560300

10765.31

R2

= 0.338 N/mm2

100

tP=

bd

Ast=

y

ck

f

f

.2

ckf

R26.411

100

tP=

4152

25

25

338.06.411

100

tP= 0.0009516600 ~ 0.001

(Ast2

)req

=100

tP= bd

= 0.001 560 300 = 168 mm2

Minimum steel =stA

bd=

0.85

yf


5/16



(on (Ast)

min=

415

85.0 300 560)

Ast,min

= 344.096 mm2> (Ast2

)req

Provide minimum area of steel.

(Ast2

)req

= 344.096 mm2

For a 12 mm bars, we get,

Number of bars =2( )st reqA

A=

4/)12(

096.3442

= 3.042 4

Provide 4 bars of 12 mm

At Side

Since overall depthD> 450 mm, torsional reinforce-

ment is provided at sides

(Ast3

)req

= 0.001 bD

= 0.001 300 600

(Ast3

)req

= 180 mm2

Provide minimum area of steel (Ast3

)req

=Amin

.

Provide 4 bars of 12 mm

40 mm

40 mm 4 bars of 12 mm

4 bars of 12 mm

4 bars of 20 mm

600mm

300 mm

40 mm

40 mm 4 bars of 12 mm

4 bars of 12 mm

4 bars of 20 mm

600mm

300 mm

Figure: Reinforcement in a Rectangular Beam

Q4. Design a slab for a two-room masonry build-ing with internal dimension of each room as4.00 m 5.00 m with one long edge continu-ous. The masonry wall are of 300 mm thick.Assume a live load of 4.0 kN/m2and a finishload of 1.5 kN/m2. Assume that the slab cor-ners are not free to lift up. Consider M

20con-

crete and Fe415

steel.

May/June-12, Set-1, Q4 M[15]

Answer :

Given that,

Internal dimension of slab = 4.00 m 5.00 mThickness of walls = 300 mm

Live load on slab = 4.0 kN/m2

Floor finish load = 1.5 kN/m2

For M20

concrete,fck

= 20 kN/mm2

For Fe415

steel,fy= 415 N/mm2

One long edge is continuous.

Design Constants

For M20

concrete,

, m a xuxd

=41587.01100

700+

= 0.479

Ru= 0.36f

ck.

, m a xux

d

d

x max,.416.01 4

= 0.36 20 0.479 (1 0.416 0.479)

Ru= 2.762

Loading Calculations

Consider the span/depth ratio as the average of sim-

ply supported slab and continuous slab. As high strength

bars of steel 415 are used multiply the value with a factor of0.8.

Span

Depth, x

l

d=

35 400.8

2

+ [Qfrom figure (1)]

d

4300= 30

d=30

4300

d= 143.333 mm

Assume a nominal cover of 20 mm and using 8 mm bars

Overall depth,D= 143.333 + 20 + 4 = 167.333 mm

Take overall depth,D= 170 mm

Effective depth, d = 170 20 4 = 146 mm

Dead load of slab = 0.17 1 1 25 = 4.25 kN/m2

Live load on slab = 4 kN/m2

Floor finish load = 1.5 N/m2

Total load = 9.75 kN/m2


6/16



Factored load, wu

= 1.5 9.75 = 14.625 kN/m2.

Effective length, lex

= 4300 + 146 = 4446 mm

ley

= 5300 + 146 = 5446 mm

r =ey

ex

l

l=

4446

5446= 1.225 1.225 < 2

The slab is a two way slab.

Moment in shorter span,Mux

= x. w

u. l

x

2

Moment in longer span,Muy

= y. w

u. l

x

2

Where, x,

yare obtained (from table 26 of IS 456 :

2000)

For one long edge continuous, we get,

x +ve

at mid span = 0.054

x veat continuous edge = 0.072y + ve

at mid span = 0.043

y ve

at continuous edge = 0

Muxve

= xve

wu l

x

2

= 0.072 14.625 4.4462

Muxve

= 20.815 kN-m

Negative moment at shorter span, M

ux ve= 20.815 kN-m

Mux + ve

= x + ve

wu l

x

2

= 0.054 14.625 4.4462

Mux + ve

= 15.611 kN-m

Positive moment at shorter span,

Mux +ve

= 15.611 kN-m

Mux + ve

= y + ve

14.625 4.662

= 0.043 14.625 4.4462

Muy + ve

= 12.431 kN-m

Positive moment at longer span,

Muy +ve

= 12.431 kN-m

Width of Strips

(i) Short Span

Width of middle strip =43 l

ey=

43 5.446 = 4.085 m

Width of edge strip =2

1(5.446 4.085) = 0.681 m

(ii) Long Span

Width of middle strip =4

3. l

ex=

4

3 4.446 = 3.335 m

Width of edge strip =2

1(4.446 3.335) = 0.556 m

Effective Depth Calculations

Effective depth from moment consideration is given

by,

d=bR

M

u

veux

.

=

1000762.2

10815.206

d= 86.811 mm

Therefore, provide depth according to deflection

considerationD= 170 mm

Effective depth in short span,dx= 170 20 4 = 146 mm

Effective depth in long span, dy= 146 8 = 138 mm

Area of Steel

Short Span

For an under reinforced section, area of steel at con-

tinuous edge.

vexstA 1 =y

ck

f

f5.0 2

4.61 1

.

ux ve

ck x

M

f bd

bd

=415

205.0

6

2

4.6 20.815 101 1

20 1000 146

1000 146

vexst

A1

= 420.159 mm2.

Minimum steel,

1 minst

A =100

12.0 1000 146

= 175.2 mm2 < vexstA 1

Provide 8 mm bars at spacing, s =1st x ve

A

A

1000

=159.420

4

82

1000

= 119.634 ~ 115 mm

Provide 8 mm bars at spacing of 115 mm c/c.

vexstA +2 =y

ck

f

f5.0 2

4.61 1 ux ve

ck x

M

f bd

+

bdx

=415

205.0

6

2

4.6 15.611 101 1

20 1000 146

1000 146


7/16



vexstA +2 = 309.951 mm2 >Amin

Provide 8 mm bars at spacing of,

S=951.309

4

82

1000

S= 162.172 mm 160 mm. Provide 8 mm bars at 160 mm c/cLong Span

Ast,y+ve

=y

ck

f

f5.0

+

2

6.411

yck

veuy

bdf

Mbd

y

veystA +, =415

205.0

26

138100020

10431.126.4

11

1000 138

veystA +, = 259.765 mm2>A

stmin.

Provide 8 mm bars at spacing,

S=765.259

4

82

1000 = 193.504

S~ 190 mm.

Provide 8 mm bars at 190 mm c/c.

Check for ShearShort Span

Shear force at edge of longer span

Vx

= wu.l

ex.

r

r

+2

= 14.625 4.446 225.12

225.1

+ V

x= 24.699 kN

Shear stress, vx

=xV

bdx=

324.699 10

1000 146

vx= 0.169 N/mm2Allowable shear stress in concrete,

bdx

A prvexst )(100 1 =

1461000

10001154

8100

2

=43709.115

146000= 0.299

From IS - 456 : 2000, table no: 19 ;c= 0.384 N/mm2>

vx

Hence safe.

Long Span

Shear force at edges of shorter span

Vy=y3wu. lex= 3

1

14.625 4.446

Vy= 21.674 kN

Shear stress, vy

=.

yV

b dy

=1381000

10674.213

vy

= 0.157 N/mm2

Allowable shear stress in concrete,

bdy

A prvesty )(100 + =1381000

1000190

4

8

100

2

= 0.192

From IS- 456 : 2000, table no : 19 ; c= 0.314 N/mm2>

vy

Hence safe.

Design of Torsional Reinforcement

Since the ends of the slabs are not allowed to lift up,

torsional reinforcement is required size of mesh,

lm

=5

exl=

5

446.4= 0.889 m from the centre of sup-

port.

or lm

= 0.889 +2

300.0= 1.039 1m from edge

Area of torsional steel,

Ast1,t

=4

3. vexstA 1

=4

3 420.159

Ast,t

= 315.119 mm2 Provide 8 mm bars at spacing of,

S=119.315

4

82

1000 = 159.513 mm

S 155 mm Provide 8 mm bars at 155 mm c/cNote

1. The area of steel vexstA 1 (with obtained spacing) at

continuous edge and at middle strip of length 4.085

m provide vexstA 2

2. Similarly, the area of steel vexstA +1 is provided only at

middle strip of length 3.335 m. The edge strip lengthof 0.556 m is provided with an area of steel equal to(1.2 D), whereD= Overall depth of slab.


8/16


9/16



M20

grade concrete,fck

= 20 N/mm2

Fe415

steel,fy= 415 N/mm2

d

xu max = 0.48

(For Fe415

d

xu max = 0.48from IS-456 : 2000 clause 38.1,

Pg.No : 70)

Ru

= 0.36fck

d

xu max,

d

xu max,416.01

Ru

= 0.36 20 0.48 (1 0.416 0.48)

Ru

= 2.766

(i) Size of Footing

Assume, weight of footing, w' = 10% of P

=100

10 2000

= 200 kN

Area of footing,A ='P w

SBC

+

=250

2002000 +

Area,A = 8.8 m2(L b)

Let,b

L=

600

300= 0.5 = 1/2

1/2LL= 8.8 L2 = 17.6

4.195mL = 4.2 m

B = 1/2 l

B = 0.5 4.2

2.1mB =

Provide footing of size,L B = 4.2 m 2.1 m

Net Upward Pressure Intensity (NUPI)

Pu

=BL

P

Pu=

1.22.4

2000

Pu= 226.757 kN/m2

(ii) Design of Section

On The Basis of Bending

Bending moment on the face AC of column,

M1

=8

UBP(la)2

=8

1.2757.226 (4.2 0.6)2

= 771.427 kNm

0.9 m

0.3 m

0.9 m

2.1 m

1.8 m 1.8 m0.6 m

A

4.2 m

A

D C

B

B

D C

0.9 m

0.3 m

0.9 m

2.1 m

1.8 m 1.8 m0.6 m

A

4.2 m

A

D C

B

B

D C

Figure (1)

M1

= 771.427 106 N/mm

Bending moment on the faceAB,

M2

=8U LP (B b)2

=8

2.4757.226 (2.1 0.3)2

M2

= 385.714 kNm

M2

= 385.714 106Nmm

Ultimate Moments

Mu1

= 1.5 771.427 106

Mu1

= 1157.141 106Nmm

Mu2

= 1.5 385.714 106

Mu2

= 578.571 106Nmm

Depth of Footing (d)

d =1u

u

M

R b(M

u1>M

u2)

d =300766.2

10141.11576

d = 1180.882

d 1181 mm


10/16



Nominal cover = 40 mm

Assuming 12 mm main bars

Overall depth,D= 1181 + 40 + 12/2 D= 1227 mm

Provide over all depth of D 1230 mm

Effective depth, d= 1230 40 6

d= 1184 mm

This depth is provided near the face of the column.

As the bending moment and shear force changes abruptly

near the edge.

Provide edge thickness of 0.2 d

= 0.2 1184

= 236.8 mmProvide over all depth of 285 mm

Effective depth of edge, de= 285 40 6 = 239 mm >

236.8 mm

Hence safe.

Checks

(i) Check for One-Way Shear

Critical section for one-way shear acts at a distance

of (d = 1184 mm) from the face of the column.

Shear force at critical section,

V= PuB{ 1/2 (L a) d }

= 226.757 2.1 { 1/2 (4.2 0.6) 1.184 m}

V= 293.333 kN

Vu= 1.5 V= 1.5 293.333

Vu= 440 kN

Effective Depth of Footing at Critical Section, d'

d'= de+

a

dd e

(a' d)

= 239 + 1800

2391184 (1800 1184)

d'= 562.4 mm

Top width of Footing at Critical Section, b'

b'= a +a

aL

d

b'= 600 +

1800

6004200 1184

b'= 2968 mm

Assumeux

d= 0.4 (For under reinforced section)

xu

= 0.4 562.4

xu= 224.960 mm

Width at neutral axis,( )

( )n u

e

L bb b x

d d

= +

nb = 2968 +(4200 2968)

(562.4 239)

224.960

nb = 3824.990

nb ~ 3825 mm Nominal shear stress,

v

v=

u

n

Vb d

v=

4.5623825

104403

v= 0.205 N/mm2

Assuming,P = 0.7% (For an under reinforced section)

c= k permissible stress

k = 1 (QD> 300 mm)

Permissible stress for m20

concrete ~ 0.35 N/mm2

(From IS 456 : 2000 Table no:23 Pg . No: 84)

c = 1 0.35

c= 0.35 N/mm2>

v

Hence safe.

(ii) Check for Two-Way Shear

Perimeter of (.PQRS)critical section of two-way shear,

P = 2 [(a+ d) + (b+d)]

= 2 [a+ b+ 2d ]

= 2 [600 + 300 + 2 1184]

P = 6536 mm

A

D C

B

d/2d/2

d/2

d/2

P Q

SR

D C

A B

1.273 m 0.527 m 0.527 m 1.1273 m0.6 m

4.2 m

A

D C

B

d/2d/2

d/2

d/2

P Q

SR

D C

A B

1.273 m 0.527 m 0.527 m 1.1273 m0.6 m

4.2 m

Figure (2)


11/16



Area of section PQRS,

A1

= (a+ d) (b+ d)

= (0.6 + 1.184) (0.3 + 1.184)

A1

= 2.647 mm2

Punching shear, Vp

Vp

=Pu [AA

1]

Vp

= 226.757 [(4.2 2.1) 2.647]

Vp

= 1399.771

Vp

~ 1400 kN

Shear stress, vp

= 1.5 .

pV

Pd

= 11846536

1014005.13

vp

= 0.271 N/mm2.

Allowable shear stress,

c= 0.25 ckf = 0.25 20

c= 1.118 N/mm2

Ks= (0.5 +

c) =

+

600

3005.0 = 1

Permissible stress = Ks

c

= 1 1.118

= 1.118 N/mm2> vp

Hence safe.

Reinforcement Design

As the section is an under reinforced section (dpr

> dreq

) the area of steel is given by,

Ast1

= 0.5 y

ck

f

f

2

16.411bdf

M

ck

ubd

= 0.5 415

20

2

6

118430020

10141.11576.411 300 1184

Ast1

= 3372.751 mm2

Provide 16 mm bars

Number of bars, n=A

Ast1=

4/)16(

751.33722

= 16.775 17 no.s

Therefore, provide 18 bars of 16 mm , which should be spaced uniformly along the length of 4.2 m(iii) Along Width

Ast2

= 0.5 y

ck

f

f

2

2

..

6.411

daf

M

ck

u a d


12/16


13/16



Q6. A corner column 300 mm 450 mm locatedin the multi storey of a system of bracedframes, is subjected to factored loads P

u=

1600 kN, Mux = 175 kNm and Muy = 100 kN-m.The unsupported length of the column is 3.0m. Design the reinforcement in the column,assuming M

30concrete and Fe

415steel.


Given that,

Size of column, bD= 300 450 mm

Factored load, Pu= 1600 kN

Factored moment in x-direction,Mux

= 175 kNm

Factored moment in y-direction,Muy

= 100 kNm

Unsupported length, l = 3.0 m = 3000 mmM

30concrete f

ck= 30 N/mm2

Fe415

steel fy= 415 N/mm2

Slenderness ratios,

exl

b= k

x.

b

l= k

x.

300

3000= 10 . Kx

eyl

D= k

y.

D

l= k

x.

450

3000= 6.67 Kx

For braced columns, kx, k

y< 1

exl

b,

eyl

D< 12

Therefore, the column will be designed as a short

column.

Check for Minimum Eccentricities

Eccentricities of loads,

ex

=1600

101753

= 109.375 mm

ey = 1600

101003

= 62.50 mm

Minimum eccentricities,

minxe 1 = 500

l+

30

b=

500

3000+

30

300= 16 mm

minye 1 = 500

3000+

30

450= 21 mm

Since, the applied eccentricities are greater than the

minimum eccentricities, modification of moments is not re-

quired.

Equivalent moment,

Mu

= 1.15 2 2

ux uyM M+

Mu

= 1.15 22 100175 +

Mu

= 231.79 kNm ~ 232 kNm

Assume effective cover of 60 mm,

Dd =

450

60= 0.133 0.15

Pu=

Dbf

P

ck

u

..=

45030030

1016003

= 0.395 0.4

Mu = 2.. Dbf

Mck

u = 2

6

4503003010232

= 0.127 0.13

From sp:16 design charts; for Pu= 0.4,M

u= 0.13 and

Dd =0.15

ckf

p= 0.125

p = 0.125 30 = 3.75%

Asc,req

= 3.75 300 100

450= 5062.5 mm2

Provide 12 bars of 24 mm .

Asc1,pr

= 12 4

242

= 5428.672 mm2 5429 mm2

Check for Moment Capacity of Section

. .

u

ck

P

f b D= 0.4

ppr

=1st pr

A

b D 100

ppr

=450300

1005429

= 4.021%.

ck

pr

f

p=

30

021.4= 0.134

Actual d ' for a clear cover of 40 mm

d' = 40 + 8 +2

24= 60 mm

Dd ~ 0.15


14/16



Referring to sp : 16 charts for Dd = 0.15,

ckfp

=

0.134 andpu

= 0.4

2

1

.. Dbf

M

ck

ux = 0.14

2

1

.. Dbf

M

ck

uy = 0.14

Mux1

= 0.14 30 300 4502= 255.15 106 Nmm >Mux

Muy1

= 0.14 30 300 4502= 255.15 106 Nmm >Muy

Hence safe.

Check for Safety Under bi-axial bending

n

ux

ux

M

M

1+

n

uy

uy

M

M

1< 1.0

Puz

= 0.45 .fck

Ag+ (0.75f

y 0.45f

ck) .A

sc

= 0.45 30 300 450 + (0.75 415 0.45 30)

5429

Puz

= 3438.985 kN ~ 3439 kN

u

uz

p

P=

3439

1600= 0.465 =x

Therefore, 0.2


15/16



Where,

Mcr

= Cracking moment of inertia =c

crg

Y

fI .

M= Maximum moment developed due to service

load

Z = Lever arm = d x/3

d = Effective depth

x = Depth of neutral axis

bw

= Width of web

bf

= Width of flange

m = Modular ratio =c

s

E

E

d' = Effective cover to compression reinforcement

Asc= Area of steel in compression reinforcement

Ast

= Area of steel in Tension reinforcement

Ig

= Gross moment of inertia of section =12

3bD

D = Overall depth of rectangular section

fcr

= Modulus of rupture = 0.7 ckf

yc= Distance of extreme compression fibre from

N-A

The effective moment of inertia Ieff calculated byequation (1) should be greater than cracked moment of

inertia (Icr) and less than the gross moment of inertia (I

g),

i.e. geffcr III


16/16



Weight of slab (on slope), w1

w' = Thickness Density

w' = 0.21 25

w' = 5.25 kN/m2

Dead weight horizontally (w1)

w1= w'

2 2R T

T

+

w1= 5.25

2 20.15 0.25

0.25

+

w1= 6.122 kN/m2

Dead weight of steps (w2)

w2

=b

R2

Density

=12

15.0

25

w2

= 1.875 kN/m Total death weight,

wd= 6.122 + 1.875

= 7.997 kN/m

wd 8 kN/m (Dead load per meter run, b= 1m)

Live load, wl= 3 kN/m2

Finishing load, wfl= 0.6 kN/m2

Total load, w = 8 + 3 + 0.6 w= 11.6 kN/m

Moment,M

M= 1.58

2effIW

M= 1.5 8

1.46.112

M= 36.562 KN/m

Design of Waist Slab

M20

concrete,fck

= 20 N/mm2

Fe415

Steel , fy= 415 N/mm2

For M20

concrete and Fe415

steel,

d

xu max,= 0.48

Ru= 0.36f

ck

d

xu max

,max1 0.416

ux

d

Ru

= 0.36 20 0.48 (1 0.416 0.48)

Ru

= 2.766

Depth or thickness of waist slab, (d)

d =bR

M

u

d =

636.562 10

2.766 1000

d = 114.971 ~ 115 mm

Provide overall depth of 150 mm with 10 mm barsand 30 mm nominal covers.

Effective depth,

d= 150 30 (10/2)

d= 115 mm

Reinforcement Design

As the provided depth is equal to the required depth,

the section is under reinforced.

Area of steel =y

ck

f

f

.2

2

6.411

bdf

M

ck

b d

=4152

20

2

6

115100020

10562.366.411

1000 115

Ast

= 1098.903 1099 mm2Provide 10 mm bars at spacing, S

S =st

A

A

1000

S =1099

4

102

1000

S = 71.465 mm

S 70 mm c/c Provide 10 mm bars at 70 mm c/c.Distribution Reinforcement

Provide a minimum reinforcement of 0.12% of cross

section.

(Ast)

d=

100

12.0 150 1000

(Ast)

d= 180 mm2

Provide 8 mm bars at spacing of

S=st

A

A

1000

S =180

4

82

1000

S = 279.253 mm

S 275 mm c/c Provide 8 mm bars at 275 mm c/c.

Documents

Hyd Civil 3 1 Drcs Set 1