Unit-9

UNIT 9 WORKING STRESS METHOD

Working Stress Method

Structure 9.1

Introduction Objectives

9.2

Basic Assumptions and Permissible Stresses 9.2.1 Basic Assumptions 9.2.2 Permissible Stresses

9.3

Development of Flexural Equations and Types of Singly Reinforced Rectangular Beam Section 9.3.1 Development of Flexural Equations 9.3.2 Types of Singly Reinforced Rectangular Beam Section 9.3.3 Shear 9.3.4 Torsion

9.4

Design of Column 9.4.1 Design of Short Columns with Axial Load and Lateral Ties 9.4.2 Design of Short Columns with Axial Load and Helical Reinforcement 9.4.3 Design of Long Columns 9.4.4 Eccentrically Loaded Columns

9.5

Summary

9.6

Answers to SAQs

9.1 INTRODUCTION Working Stress Method of Design for structures and their elements is the same as for Limit state Method except for safety considerations. To be more specific, Codal provision for durability, fire resistance, detailing of reinforcement, serviceability, stability, etc remain the same as for Limit Stress Method. However, the design loads are characteristic loads and the stresses developed due to loads shall be within the corresponding permissible stresses (i.e. characteristic strength divided by Factor of Safety).

Objectives After studying this unit, you should be able to

•

explain the basics of working stress method,

•

design the beams of rectangular cross sections, and

•

understand the design of columns.

9.2 BASIC ASSUMPTIONS AND PERMISSIBLE STRESSES 9.2.1 Basic Assumptions Basic assumptions for design applicable to flexural and compression members are as follows : (a)

Plane section before bending remains plane after bending.

(b)

The tensile stress of concrete is neglected unless otherwise mentioned.

175

Theory of Structures-II

* This is not the same as by

m

=

E S E C

formula

because it partially takes into consideration long-term effect such as creep.

(c)

The strain-stress relation for concrete as well as for steel reinforcement is linear.

(d)

The modular ratio ⎜⎜ i.e.

⎛ ⎝

stress in reinforcement ⎞

⎟⎟ ⎠

stress in concrete

m* =

280 3σ cbc

.

9.2.2 Permissible Stresses Permissible stresses are defined as characteristic strength divided by factor of safety. The factor of safety is not unique values either for concrete or for steel; therefore, the permissible stresses are given in tabular form as under. Table 9.1 : Permissible Direct Tensile Stress Grade of Concrete Tensile Stress 2 (N/mm )

σ

t

in Tension Members

M 10

M 15

M 20

M 25

M 30

M 35

M 40

M 45

M 50

1.2

2.0

2.8

3.2

3.6

4.0

4.4

4.8

5.2

Table 9.2 : Permissible Stress in Concrete Grade of Concrete

Permissible Stress in Compression

Permissible Stress in Bond (Average) for Plain Bars in Tension

Bending

Direct

(2)

(3)

(4)

σ cbc

σcc

τ hd

M 10

3.0

2.5

M 15

5.0

4.0

0.6

M 20

7.0

5.0

0.8

M 25

8.5

6.0

0.9

M 30

10.0

8.0

1.0

M 35

11.5

9.0

1.1

M 40

13.0

10.0

1.2

M 45

14.5

11.0

1.3

M 50

16.0

12.0

1.4

(1)

Note : The bond stress given in Column 4 shall be increased by 25 percent for bars in compression.

Table 9.3 : Permissible Stresses in Steel Reinforcement Permissible Stress in N/mm2 Sl. No.

Type of Stress in Steel Reinforcement

(1)

(2)

(i)

Mild Steel Bars Conforming to Grade 1 of IS 432 (Part 1) (3)

Tension ( σ st or σ sv ) : (a) Up to and including 20 mm (b) Over 20 mm

140 130

(ii)

Compression in column bars ( σ sc )

130

(iii)

Compression in bars in a beam or slab when the compressive resistance of the concrete is taken into account.

(iv)

(4)

Half the guaranteed yield stress subject to a maximum of 190 130

High Yield Strength Deformed Bars conforming to (Grade Fe 415) (5)

230 230 190

The calculated compressive stress in the surrounding concrete multiplied by 1.5 times the modular ratio or

Compression in bars in a beam or slab when the compressive resistance of the concrete is not taken into account : (a) Up to and including 20 mm (b) Over 20 mm

Medium Tensile Steel Conforming to IS 1786 (Part 1)

σ sc

whichever is lower

Half the guaranteed yield stress subject to a maximum of 190 140 130

190 190

Note : For high yield strength bars of grade Fe 500 the permissible stress in direct tension and flexural tension shall be 0.55 f y. The permissible stresses for shear and compression reinforcement shall be as for Grade Fe 415.

176


Table 9.4 : Permissible Shear Stresses in Concrete 100

A s

Permissible Shear Stress in Concrete,

, N/mm2

c

bd

Grade of Concrete

M 15

M 20

M 25

M 30

M 40

M 40 and above

(1)

(2)

(3)

(4)

(5)

(6)

(7)

≤ 0.15

0.18

0.18

0.19

0.20

0.20

0.20

0.25

0.22

0.22

0.23

0.23

0.23

0.23

0.50

0.29

0.30

0.31

0.31

0.31

0.32

0.75

0.34

0.35

0.36

0.37

0.37

0.38

1.00

0.37

0.39

0.40

0.41

0.42

0.42

1.25

0.40

0.42

0.44

0.45

0.45

0.46

1.50

0.42

0.45

0.46

0.48

0.49

0.49

1.75

0.44

0.47

0.49

0.50

0.52

0.52

2.00

0.44

0.49

0.51

0.53

0.54

0.55

2.25

0.44

0.51

0.53

0.55

0.56

0.57

2.50

0.44

0.51

0.55

0.57

0.58

0.60

2.75

0.44

0.51

0.56

0.58

0.60

0.62

3.00 and above

0.44

0.51

0.57

0.60

0.62

0.63

Note : As is that area of longitudinal tensile reinforcement which continues at least one effective depth beyond the section being considered except at supports where the full area of tension reinforcement may be used provided the detailing conforms to Cl 26.2.2 and Cl 26.2.3.

Table 9.5 : Maximum Shear Stress,

c,

2 max N/mm in

Concrete

Concrete Grade

M 15

M 20

M 25

M 30

M 35

M 40 and above

2 max N/mm

1.6

1.8

1.9

2.2

2.3

2.5

c,

SAQ 1

(a)

How working stress method of design differs from limit state method?

(b)

What are the basic assumptions applicable to flexural and compression members?

(c)

Why modular ratio ( m) is not calculated as by the formula m =

E s E c

?

177


9.3 DEVELOPMENT OF FLEXURAL EQUATIONS AND TYPES OF SINGLY REINFORCED RECTANGULAR BEAM SECTION 9.3.1 Development of Flexural Equations Under the action of transverse loads on a beam strains, normal stresses and internal forces developed on a cross section are as shown in Figure 9.1.

Figure 9.1 : Explaining Bending Mechanics of a Rectangular Beam

Since applied moment on the section is in equilibrium with the resisting moment developed, three basic equations namely (a)

Strain compatibility equation

(b)

Force equilibrium equation and

(c)

Moment equilibrium equation is developed as described below.

Strain Compatibility

From basic assumption (i) plane section remains plane means that the strain variation across the cross section is linear. The strain is linearly proportional to stress (i.e . f cbc= E ε cbc ) as per basic assumption (c). The stress equation may be developed as follows :

ε cbc kd

=

ε ct (d − kd )

f cbc

or, or, or,

f ct

E kd

=

f cbc

=

kd f ct =

E (d − kd ) f ct

(d − kd )

f st m

= f cbc

(1 − k ) k

Form Force Equilibrium

Applied force = Internal force or,

C – T = 0

or,

C = T

or, 178

1 2

f cbc bkd = f st Ast


From Moment Equilibrium

Applied bending moment = Resisting BM M applied = M R = C la = T la

or,

M R

⎛ ⎝

=

1

=

1

=

1

f cbc b k d ⎜ d −

2

2

2

⎛ ⎝

f cbc k ⎜1 −

kd ⎞

⎛ kd ⎞ ⎟ = f st Ast ⎜ d − ⎟ 3 ⎠ 3 ⎠ ⎝

k ⎞ 2 ⎟bd 3 ⎠

f cbc k j b d 2

k ⎞ = f st Ast ⎛ ⎜1 − ⎟d ⎝ 3 ⎠

⎛ ⎝

= f st Ast j d where

j = ⎜1 −

k ⎞

⎟

3 ⎠

= Rbd 2 = f st Ast jd where

R =

1 2

f cbc kj

9.3.2 Types of Singly Reinforced Rectangular Beam Section Depending on percentage of tensile reinforcement a section may be put in one of the three categories : (a)

a balanced section

(b)

an under-reinforced section

(c)

an over-reinforced section

Balanced Section

A balanced section (Figure 9.2) is one, in which under the applied bending moment, the permissible compressive stress in extreme fibre ( σ cbc ) and permissible tensile stress ( σ st ) in tensile reinforcement reaches simultaneously.

Figure 9.2 : A Balanced Section

The coefficient for neutral axis depth, compressive force, tensile force and coefficient for lever arm are denoted with suffix B as k Bd , C B, T B and j B, B

respectively. Hence, M RB =

1 2

B

B

σ cbc k B j B bd 2 = σ st AstB j B d

= R B bd 2 = σ st AstB j B d

179


Under-reinforced Section

An under reinforced section is one, in which due to applied bending moment, the tensile stress in steel reaches permissible value where as the compressive stress in outer face remains below permissible limit (Figure 9.3). This happens when the steel area provided is less than the balanced steel area.

Figure 9.3 : An Under-reinforced Section

According to mechanics of structure, neutral axis depth is always the centroidal axis depth. It is determined by equating moment of area of concrete in compression and equivalent area of steel in terms of concrete about natural axis, i.e. b k d

kd

2

=

m Ast ( d − kd ) . The resulting value

(Figure 9.3(d)) of kd < k Bd . B

Over-reinforced Section

A section in which the moment resisting capacity ( M R) is limited by the compressive stress on outer face of concrete reaching its limiting value (σ cbc ) . The stress in steel at this stage ( f st ) is less than the permissible stress (σ st ) indicating that steel provided is more than required to resist M R (Figure 9.4).

Figure 9.4 : An Over-reinforced Section

180

The neutral axis depth (kd ) is determined as usual by equating moment of area of concrete in compression and equivalent area of tensile steel in terms of concrete about neutral axis.

kd

Mathematically, b . k . d . Hence,

M R

=

2

1 2


= m Ast (d − kd )

kd ⎞ 1 σcbc .b . k d ⎛ ⎜ d − ⎟ = σ cbc k j bd 2 3 ⎠ 2 ⎝

To Determine Neutral Axis Depth for Balanced Section

From stress diagram

σcbc k B d

=

σ st / m σ st = ( d − k B d ) md (1 − k B )

or,

mσ cbc

or,

k B (σ st + mσ cbc )

or,

k B

=

− k B mσcbc = k B σ st = m σ cbc

mσ cbc

σ st + mσ cbc

1

= 1+

σ st mσ cbc

Hence, neutral axis depth for balanced section

⎛ ⎞ ⎜ ⎟ 1 ⎟ d k B d = ⎜ ⎜ σ st ⎟ ⎜ 1 + mσ ⎟ cbc ⎠ ⎝ To Determine Percentage of Tensile Reinforcement for Balanced Section

Equating the compressive force and tensile force, C = T 1 2

σ cbc b k B d = σ st AstB

ptB % =

AstB bd

× 100, or AstB =

( ptB %)

× bd

100

Substituting AstB in the above equation 1 2

or,

σ cbc b k B d = ptB %

=

( ptB %) × bd

σ st 100 50 k B σ cbc

σ st

Three types of problems may be encountered. These are : (a)

To determine moments of resistance ( M R) for a given cross section.

(b)

To determine maximum compressive stress in concrete ( f cbc) and tensile stress in steel ( f st ) for given cross section and M applied.

(c)

To determine M RB and AstB for a given concrete cross-section.

Example 9.1

Determine moment of resistance of the cross section shown in Figure 9.5 for f ck = 20 N/mm2 and f y = 415 N/mm2. 181


Figure 9.5 : Cross-section

Solution

Equating moment of area about neutral axis. bx 2

= m Ast (d – x)

2

(Here, m =

280

=

3σ cbc

x 2

280 3× 7

≈ 13 )

= 13 × 4 × 314 × (500 − x)

or,

250

or,

125 x2 +16328 x – 8164000 = 0

or,

x + 130.624 x – 65312 = 0

or,

2

2

x =

− 130.624 ±

(130.624 2

+ 4 × 65312)

2

k Bd =

1

σ st 1+ mσ cbc

d =

1

⎛ 1 + 230 ⎞ ⎜ ⎟ ⎝ 13 × 7 ⎠

= 198.46 mm

× 500 = 0.283 × 500 = 141.5 < 198.46

Hence, the section is over reinforced Thus,

M R

1 x ⎞ = σcbc b x ⎛ ⎜ d − ⎟ 2 ⎝ 3 ⎠ 1 198.46 ⎞ = × 7 × 250 × 198.46 × ⎛ ⎜ 500 − ⎟ × 10 −6 2 3 ⎠ ⎝

= 75.339 kN-m. Example 9.2

Determine tensile reinforcement and moment of resistance for the concrete cross section shown in Figure 9.6 for the following design parameters :

σcbc = 8.5 N/mm2 ; σ st = 230 N/mm2 and m = 11

182



Solution

As concrete cross section only is given, Ast and M R may be determined for balanced section :

∴

1

k B

=

j B

=1−

M RB

σ st 1+ mσ cbc k B

3

=

=1−

1

⎛ 1 + 230 ⎞ ⎜ ⎟ ⎝ 11 × 8.5 ⎠

0.289 3

= 0.289

= 0.904

=

1 2

cbc

=

1

× 8.5 × 250 × 0.289 × 500 × 0.904 × 500 × 10 − 6

σ

2

b k B djB d

= 69.396 kN-m. Equating

C = T 1 2

or,

σ cbc . b k B d = σ st AstB

Ast B

=

8.5 × 250 × 0.289 × 500 2 × 230

= 667.53 mm 2 .

Example 9.3

Evaluate stresses in concrete and tensile reinforcement for the section shown in Figure 9.7 to resist M R = 45 kNm for the following design parameters :

σ cbc = 7 N/mm2 ; σ st = 140 N/mm2 and m = 13.


Solution

Equating moment of areas about n.a. bx 2

2

= m Ast (d − x) x 2

or,

300

or,

150 x 2

or,

2

= 13 × 4 × 314 × (500 − x) = 8164000 − 16328 x

2

x + 108.85 – 54426.67 = 0

183

or,

⎛ ⎜ − 108.85 ± ⎝ x =

or,

x =185.133 mm


1

M =

2

or,

f cbc

+ 4 × 54426.67) ⎞⎟ ⎠

2

⎛ ⎝

f cbc bx ⎜ d −

45 × 10 6

or,

(108.852

x ⎞

⎟

3 ⎠

1 185.133 ⎞ ⎛ = × f cbc × 300 × 185.133 × ⎜ 500 − ⎟ 2 3 ⎠ ⎝ 2 × 45 × 10 6

=

⎛ ⎝

300 × 185.133 × ⎜ 500 −

185.133 ⎞ 3

⎟ ⎠

= 3.697 N/mm 2

⎛ ⎝

M = f st Ast ⎜ d −

x ⎞

⎟

3 ⎠

45 × 10 6

f st =

⎛ ⎝

4 × 314 × ⎜ 500 −

185.133 ⎞ 3

⎟ ⎠

= 81.75 N/mm2.

9.3.3 Shear The formula for calculating nominal shear stress ( τ v ) remains the same as in limit state method except that V u and M u are replaced by V and M , i.e. τ v =

V bd

for

rectangular cross section. The permissible shear stresses ( τc ) for different grades of concrete and percentage of tensile reinforcement are given in Table 9.4. Similarly maximum shear stresses τc, max in concrete are given in Table 9.6. Table 9.6 : Maximum Shear Stress, Concrete Grade c,

2 max N/mm

c,

2 max N/mm in

Concrete

M15

M20

M25

M30

M35

M40 and above

1.6

1.8

1.9

2.2

2.3

2.5

Example 9.4

Design vertical stirrups for a rectangular beam of cross section shown in Figure 9.8 for a shear force of 70 kN. Use M 20 concrete and Fe 250 steel.


184


Solution

τv =

V bd

=

70 × 103 300 × 400

= 0.583 N/mm2

τc,max for M 20 = 1.8 N/mm2 Percentage of tensile reinforcement pt % =

∴

100 Ast bd

=

100 × 4 × 314 300 × 400

% = 1.05%

From Table 9.4

τc = 0.39 +

(0.42 − 0.39) (1.25 − 1.0)

(1.05 − 1.0)

= 0.4 N/mm2

τ c,max > τv > τc ; therefore shear reinforcement is required for V s = V − τc bd or,

V s

= (70 − 0.4 × 300 × 400 × 10−3 ) = 22.00 kN

If transverse reinforcement of φ 6 is provided Asv

∴

sv

=

π = 2 × × 62 = 56 mm2 4

σ s v Asv d 140 × 56 × 400 = = 142 c/c V s 22 × 103

Hence, provided

6-2 legged vertical stirrups @ 140 mm c/c.

9.3.4 Torsion Codal provisions for design of torsion is the same as those for limit state method except that design forces V u, M u and design value of stresses are replaced by V , M and permissible stresses. Example 9.5

Determine the reinforcements required for a rectangular section (Figure 9.9) for the following data : Size of beam

= 300 × 600; Grade of concrete

Grade of steel

= Fe 415;

Torsional moment T = 20 kN-m;

Shear force

= M 25 V = 60 kN

Bending moment M = 60 kN-m


185


Solution

Assuming φ 25 for main reinforcement and φ 8 for shear reinforcement. d = 600 – 20 – 8 – 12.5 = 559.5 mm V e

⎛ T ⎞ = V + 1.6 ⎜ ⎟ ⎝ b ⎠ 20 ⎞ = 60 + 1.6 × ⎛ ⎜ ⎟ = 166.67 kN ⎝ 0.3 ⎠

Equivalent shear stress V e

τve =

bd

=

166.67 × 103

= 0.993 N /mm2

300 × 559.5

τ c,max = 1.9 N/mm2 Assuming value of pt = 0.5%, τc = 0.31 N/mm2 Equivalent BM,

⎛ ⎛ 0.6 ⎞ ⎞ ⎛ 1 + D ⎞ ⎜ ⎜1 + ⎟⎟ ⎜ ⎟ 0.3 ⎠ ⎟ ⎝ ⎜ b ⎟ = 60 + 20 × = M + M t = M + T ⎜ ⎜ 1.7 ⎟ ⎜ 1.7 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

M el

= 95.294 kN-m 280

m=

3σ cbc

280 3 × 8.5

1

k B

=

j B

= ⎛ ⎜1 − ⎝

M B

=

σ st 1+ mσ cbc

=

= 10.98 ≈ 11 1

⎛ 1 + 230 ⎞ ⎜ ⎟ ⎝ 11× 8.5 ⎠

= 0.289

k B ⎞

⎛ 0.289 ⎞ = 0.904 ⎟ = ⎜1 − ⎟ 3 ⎠ ⎝ 3 ⎠

1

= σ cbc b k B d j B d 2

=

1 2

× 8.5 × 300 × 0.289 × 559.5 × 0.906 × 559.5 × 10 − 6

= 104.274 kN-m > 95.294 kN-m Hence, the section will be under-reinforced M R

= σ st Ast j B d

or,

95.294 × 106 = 230 × Ast × 0.904 × 559.5

or,

Ast = 819.16 mm

Hence, provided 3

2

20.

Provision of Shear Reinforcement 100 Ast bd

186

=

100 × 3 × 314 300 × 559.5

= 0.561%

∴

τc = 0.31 +

(0.36 − 0.31) (0.75 − 0.5)


(0.561 − 0.5)

= 0.322 N/mm 2 < 0.993 N/mm 2 Hence, shear reinforcement will be provided

∴

Asv

=

or,

sv

=

T sv b1d 1σ sv

+

Vsv

2.5d 1σ sv

=

⎛ T V ⎞ ⎜⎜ ⎟ + σ sv ⎝ b1d 1 2.5d 1 ⎠⎟ sv

2 × 50 × 230

⎛ 20 × 106 60 × 103 ⎞ ⎜ ⎟ ⎜ 219 × 519 + 2.5 × 519 ⎟ ⎝ ⎠

= 103.51, i.e. 100 mm c/c. Maximum spacing = 0.75 d = 0.75 × 559.5 = 419.625 < 450 Asv

>

(τ vc

− τ c ) b s v (0.993 − 0.322) × 300 × 100 = 230 σ sv = 87.52

Hence, provided 2 legged

mm 2

< 100

mm 2

8 @ stirrups 100 mm c/c.

Bond and Anchorage

The Codal provisions for bond and anchorages remain the same as for design by limit state method except that the permissible bond stresses are as given in Table 9.7. Table 9.7 : Permissible Bond Stress for Plain Bars in Tension Grade of Concrete

M 15

M 20

M 25

M 30

M 35

M 40

M 45

M 50

Permissible Bond Stress ( bd N/mm2)

0.6

0.8

0.9

1.0

1.1

1.2

1.3

1.4

SAQ 2 (a)

How moment of resistance ( M R) is developed? 1

Prove that M R = f ck k j bd 2 = f st Ast jd . 2

(b)

Explain the difference between under-reinforced and over-reinforced sections.

(c)

How the equations for calculating τv and V s differ from those by limit state method?

(d)

Is there any basic difference in approach for calculating the V e and M e values by the two methods of design?

187


9.4 DESIGN OF COLUMNS 9.4.1 Design of Short Columns with Axial Load and Lateral Ties

*The minimum eccentricity criteria mentioned in Limit State Method may deemed to have been incorporated in the equation for P.

The permissible load*. P = σ cc Ac

+ σ sc Asc

where σ cc = Permissible stress in concrete in direct compression, Ac = Cross sectional area of concrete excluding any finishing material and

reinforcing steel,

σ sc = Permissible compressive stress for column bars, and Asc = Cross-sectional area of the longitudinal steel.

9.4.2 Design of Short Columns with Axial Load and Helical Reinforcement The Codal provisions for the design is the same as those for limit state method.

9.4.3 Design of Long Columns The permissible stresses both for concrete and steel are reduced by a reduction coefficient. C r = 1.25 −

lef

48b

where lef = Effective length of column, b = Least lateral dimension of column. For column with helical reinforcement b is diameters of the core.

For more exact evaluation of permissible load, the reduction coefficient, C r = 1.25 −

lef

160 imin

where imin = least radius of gyration. Example 9.6

A column has cross sectional area of 300 × 300 and is reinforced with 4 φ 25. Determine permissible load if effectively held in position at both ends, but not restrained against rotation having (a)

a length of 3 m, and

(b)

a length of 5 m. Use M 25 concrete and Fe 415 steel.

Solution

Permissible load assuming the column to be of short one P = σ cc Ac

+ σ sc Asc

= σ cc ( Ag − Asc ) + σ sc Asc = 6 × (300 × 300 − 4 × 491) + 190 × 4 × 491× 10−3 188

= 528.589 kN.

(a)


Slenderness Ratio (SR) Unsupported length, l = 3 m For the given end conditions lef =1.0 × l = 3 m

∴

SR =

lef b

=

3000 300

= 10 < 12

Hence, column is short one and = P′ = 528.589 kN (b)

Unsupported length = 5 m for the given end conditions lef =1.0 × l = 5 m SR =

lef b

=

5000 300

= 16.67 > 12

∴ Reduction coefficient, C r = 1.25 – ∴

P

lef

48b

= 1.25 −

5000 48 × 300

= 0.903

= C r P = 0.903 × 528.589 = 477.206 kN

9.4.4 Eccentrically Loaded Columns* Let the column shown in Figure 9.10 be applied with an axial compression P and a bending moment about x- x axis, M x. The stresses in extreme fibre due to P and M x will be f t ,b

=

P Ae

±

M x I ex

×

* Design based on uncracked section and for uniaxial BM

D

2

Figure 9.10 : Section Subjected to Axial Load and Uniaxial Bending

The cross section and the loads applied are such that the stresses in top fibre ( f t) and that in bottom fibre ( f b) are both compressive so that the concrete does not crack. The column will be considered safe if the following conditions are satisfied.

σ cc,cal σ cbc,cal + ≤1 σ cc σ cbc where σ cc, cal = Calculated direct stress in concrete,

σcbc,cal = Calculated bending compressive stress in concrete, σ cc = permissible axial compressive stress in concrete, and σ cbc = permissible bending compressive stress in concrete. The resultant tension in concrete is not grater than 25% of the resultant compression for unaxial bending or does not exceed three-fourth of the7 day modulus of rupture of concrete.

189


Ae = Equivalent area of concrete = Ac + mc Asc = Ac + 1.5 m Asc

and

I ex

= Equivalent Moment of Inertia =

bD3

3

+ (1.5 m − 1) ∑

2 Asci y sci

where Asci and ysci are area of steel reinforcement and their distances from n.a., respectively. Example 9.7

Check whether the column section shown in Figure 9.11 is safe. The design parameters are P = 300 kN; M x = 20 kN-m; f y = 415; Effective cover = 60.


Solution m= Ae

280 3σ cbc

12

1 12

bD

3

+ (1.5 m − 1) Asc

⎛ D − d ' ⎞ ⎜ ⎟ ⎝ 2 ⎠

2

× 250 × 4003 + (1.5 × 13 − 1) × 6 × 314 × (200 − 60) 2 = 2016471733 mm4 σ cc,cal =

P Ae

σ cbc, cal =

∴

= 13

3× 7

= ( Ac + 1.5 m Asc ) = (250 × 400 − 6 × 314) + 1.5 × 13 × 6 × 314 = 134854 mm 2 I ex =

1

280

=

=

300 × 10 3 134854

M x D

2

I ex

Resultant stresses =

P Ae

±

=

20

= 2.225 N/mm 2 × 10 6 × 200

2016471733

= 1.984 N/m 2

M x D I ex 2

= 2.225 ± 2.081 = 4.209 N/mm 2 and 0.241 N/mm 2 Since both are +ve, the section is uncracked. Check for Safety

σcc,cal σcbc,cal 1.984 + = σcc σcbc 7 190

+

2.225 5

= 0.728 < 1

Hence, O.K.


SAQ 3 (a)

Write down the formula for evaluation of permissible axial load ‘P’ for a short column with helical reinforcement.

(b)

How the permissible axial load for a long column is evaluated.

9.5 SUMMARY In working stress method of design, the design loads are the actual or characteristic loads and the design stresses are characteristic stresses divided by factor of safety. Due to above mentioned basic differences, the approach for evaluation of design load for a given cross section differs from that of limit state method. The other criteria for design such as serviceability, fire resistance, durability, stability, etc. remain the same as those in limit state method.

9.6 ANSWERS TO SAQs SAQ 1

(a)

Refer Section 9.1.

(b)

Refer Section 9.2.

(c)

Refer Section 9.2.

SAQ 2

(a)

Refer Section 9.3.1.

(b)


(c)


(d)


SAQ 3

(a)

Refer Section 9.4.

(b)

Refer Section 9.4.

191

Unit-9

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