53

Lecture Note - 11

Behavior of RC Members: Torsion Torsion means twisting. The types of torsion in structure are

1. Primary or equilibrium torsion: - Primary or equilibrium torsion is that which is required to maintain the basic static equilibrium in a statically determinant structure.

2. Secondary or compatibility torsion: - Secondary or compatibility torsion is that which is required to maintain compatibility condition in a statically indeterminate structure.

The torsional stiffness (KT) of a member is the torsional moment required to produce a unit angle of twist.

Thus, L

GCTKT ==θ

Where, Torsional stiffness TK

T Torsional moment θ Total angle of twist in length ‘L’.

( )µ+=12EG Elastic shear modulous.

E Modulous of elasticity. µ Poission’s ratio.

C Torsion constant or polar moment of inertia 3KDb=K St. Venant’s torsional constant, which varies with (D/b) ratio.

The value of K for various ratios of (D/b) is given in Table: 1

54

Table: 1 Values of K & α in torsion of rectangles

⎟⎠⎞

⎜⎝⎛ ==

DbTKDbC 2max

3 ,α

τ

Value of bD

K α

1.0 1.2 2.5 2.0 2.5 3.0 4.0 5.0 10.0 α

0.141 0.166 0.196 0.229 0.249 0.263 0.281 0.291 0.312 0.333

0.208 0.219 0.231 0.246 0.258 0.267 0.282 0.291 0.312 0.333

A more convenient expression for C has been derived by Timoshenko as

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −=

363.01

3DbDbC

The following approximation can be made when dealing with flanged beam & other section, which can be assumed to be rectangular. For analysis, T, L or I sections are divided into component rectangles & the C value is the combined value of the component rectangles. The division should be such that the value of C obtained for the whole section, i.e.

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −=

363.01

3DbDbC

Should be the largest possible. Bending & torsional stiffness of RC beam The magnitude of distribution of moments as torsion to adjoining members is small.

55

This is because:

( )µ+=12EG

If, 15.0=µ then 2.3EG =

And 0µ = then 2EG =

Which shows that G is very low as compared to E. Again, For a rectangle D=2b, the value of I is about three times that of C So, the ratio of stiffness in bending to stiffness in torsion for adjoining members can therefore be obtained as

9.6: =L

GCLEI (approx. 7)

Hence the beams are several times more stiff in bending than in torsion Torsional rigidity of RC members On the basis of laboratory tests BS8110, Clause 2.4.3 states that for structural analysis or design , the torsional rigidity may be calculated by assuming G=0.42 times the modulous of elasticity of concrete & C is equal to half of the St. Venant value calculated for the plain concrete section. Torsional stress in flanged section To determine shear stress of a section, the section is to be divided with largest possible rectangle as one of its component, then by elastic analysis, the torsion is given by:

∑= 3

3

nnn

nnnn bDK

bDKTT

By plastic analysis its value is given by

∑= 3

3

nn

nnn bD

bDTT

The value of the maximum torsional stress by elastic & plastic analysis can be found from the following equations:

DbT

2max ατ = (Elastic method)

⎟⎠⎞

⎜⎝⎛ −

=

3

22 bDb

Ttτ (Plastic analysis)

56

Principles of design for torsion by IS 456 The clause 41.1 of IS 456 provides two options for design of torsion in the statically indeterminate structure.

(i) If torsion can be eliminated by releasing redundant restrains & torsion is not considered in the analysis of the structure, the structure may be designed for zero torsion. The normal shear & bending reinforcement will take care of any cracking if occurs.

(ii) If torsional stiffness is considered while analysis of the structure the member should be designed for compatibility torsion.

Design philosophy: Torsional reinforcement is not calculated separately from that of bending & shear. Instead total longitudinal reinforcement is determined for a fictitious bending moment which is a function of actual bending moment & torsion. Similarly the web reinforcement is determined for a fictitious shear which is a function of actual shear & torsion. Calculation for torsional shear As per clause 41.3 of IS456

bT

VV uue 6.1+=

Where, Equivalent shear eV

uV

uT

Factored shear force.

Factored torsional moment.

b Least lateral dimension. Equivalent nominal shear stress should not exceed maximum shear stress as given in table 2 (Ref. Table 20 of IS456) Table 2 Maximum shear stress Concrete grade M15 M20 M25 M30 M35 M40 & above

maxcτ 2.5 2.8 3.1 3.5 3.7 4.0

57

Calculation for equivalent bending moment As per clause 41.4.2 of IS456 Equivalent BM, tue MMM +=1

Where,

uM BM at cross sectional area and

( )7.1

1 bDT

Mu

t

+=

D Overall depth. b Width

If numerical value of is greater than then longitudinal reinforcement should also

be provided on the flexural compression face as equivalent BM: tM uM

2eM

Where, ute MMM −=2

The amount being taken as acting in the opposite sense to the moment 2eM uM

Transverse reinforcement Two legged transverse reinforcement should be provided with cross sectional area given

by ( ) ( ) 111 5.287.087.0 dfV

dbfT

Ay

u

y

usv +=

1d

( )

Where,

vs Spacing of stirrups

yf Characteristic strength of stirrups steel.

uT Torsioanal moment

uV Shear force

1b Centre to centre distance between corner bar in the direction of the width.

Centre to centre distance between corner bar However,

y

vcesv f

bsA

87.0ττ −

⊄

58

Design steps according to IS 456 Step: 1 Determine design moments, shear and torsion , and . uM uV uT

Step: 2 Determine equivalent moment & longitudinal steels. Calculate

( )7.1

1 bDT

Mu

t

+=

(a) tue MMM +=1

Design tension steel for 1eM

(b) If , then ut MM >

ute MMM −=2

Design steel on compression face for reversal of moment 2eM=

Step: 3 Determine the equivalent shear given by the equation

bT

VV uue 6.1+= .

Step: 4 Find shear stress & check for maximum shear. Determine the shear stress as

bdVe

ve =τ

Step: 5 Calculate area of shear links from the formula ( )

y

cve

v

sv

fb

SA

87.0ττ −

=

( )

Step: 6 Check for interaction of shear & torsion given by the relation

( ) 111 5.287.087.0 dfV

dbfT

Ay

u

y

usv +=

) 4/11 yx +

Adopt the larger value of step 5 & 6 for shear steel. The spacing should not exceed 1x

or ( or 300 mm as per clause 26.5.1.7 of IS456.

59

Lecture Note - 12

Analysis and Design T-Beam in Torsion Example:- (Analysis of T beam in torsion) A T beam is as shown in Fig 1. if the section is subjected to a factored torsion of 150 kNm. Calculate the torsion carried by two main rectangular portion of the T beam, assuming

(a) Elastic theory (b) Plastic theory

120800

700

350Fig. 1

Solution:- (a) Proportioning of torsion by elastic theory

∑= 3

3111

1 KDbbDK

TT

91.2120350

==bD , Thus from table, 26.01 =K For flange,

Similarly, for web, 30.2350800

==bD , 24.02 =K

333

2223111 35080024.012035026.0 ××+××=+ bDKbDK =(1.57+82.32) 4810 mm×

kNmT 2.14789.83

32.821501 =

×=

60

kNmT 8.289.83

57.11502 =

×=

(The major part is carried by the web only) (b)-Calculation by plastic method

322

311

311

1 bDbDbDTT+

=

Now, =3 3 3 31 1 2 2 350 120 800 350D b D b+ = × + × ( )8 86.05 10 343 10× + × 81005. ×=349

Thus, ⎟⎠⎞

⎜⎝⎛=

05.3493431501T =147.4 kNm.

And 6.205.349

05.61502 =⎟⎠⎞

⎜⎝⎛=T kNm

Example:- (Design of beam in torsion by IS 456) The T beam given in Fig. 1 is subjected to the following factored loads. Bending moment of 215 kNm, shear of 150 kN, and torsion of 105 kNm. Assuming and

(N/mm

30=ckf

415=yf 2), design the reinforcements according to IS456. Cover to centre of

steel is 50 mm. Solution:- Step 1: Assume that the torsion is fully taken by web. [ ref. Clause 41.1.1 of IS456] Step 2: Equivalent bending moment

⎟⎠⎞

⎜⎝⎛ ++=+=

bDTMMMM utue 1

7.1 [ ref. Clause 41.4.2 of IS456]

9.2022153508001

7.1105215 +=⎟

⎠⎞

⎜⎝⎛ ++= =417.9 kNm.

Step 3: Calculation of longitudinal steel If exceeds , provide compression steel for tM uM ut MM − )

61

u

If tM M< design for only. eM

Thus, 12.2750350

109.4172

6

2 =××

=bdM e

,646.0=p ( ) 21696750350100646.0 mmAs =×= [ref. SP 16]

Hence, provide 6 Nos. 20 mm bar.

Thus, percentage provided %72.07503501001885

=××

=

Step 4: Determine equivalent shear

bTVVe 6.1==

Using kNm/m units, we obtain

kNVe 63048015035.0

1056.1150 =+=+=

Step 5: Find shear stress

23

/4.275035010630 mmN

bdVe

e =××

==τ

This is more than 58.0=cτ for 0.72% steel (assuming full extension of steel) and less

than [ref. Table 19 & 20 of IS456] 2max /5.3 mmN=τ

Hence, it is OK. Step 6: Design of stirrups As per Clause 41.4.3 of IS456: Two conditions should be satisfied: Condition 1

( )111 5.2

87.0d

Vdb

Ts

fA

v

ysv +=8 3105 10 150 10 685.7

250 700 2.5 700× ×

= + =× ×

N/mm

Condition 2 ( ) ( )b

sfA

cev

ysv ττ −=87.0

= ( ) mmN /63735058.04.2 =×−

Hence, the value of ( )0.87sv

v

A fs

y will be 685.7 N/mm =6.9 kN/cm

Step 7: Design of stirrups

62

Adopting 10 mm bar at 8 cm gives( )0.87

7.09sv yus

v

A fVd s

= =

As per Cl. 26.5.1.7, Spacing should not exceed: x1 = 250 or 5.2374

7002504

11 =+

=+ yx

or 300 mm. Step 8: Provision of longitudinal steel Longitudinal steel consists of 6 nos of 20 mm diameter at bottom & nominal hangers, 12 mm at top. As the depth of beam is more than 750 mm, provide side reinforcement 0.05% on both faces.

Therefore, ( ) 2140350800100

05.0 mmAs =×=

Hence, provide 2 nos of 10 mm bars. Thus, 2157mmAs =

Spacing = mm2503

750=

Design for Torsion by British Code Design principle as per BS8110 If a section is subjected to bending moment (M), shear force (V) & torsion (T), it is necessary to design the transverse & longitudinal steel. It should be separately designed for SF, BM & torsion

bdV

v =τ Calculate

As per clause 2.4.4 of BS8110 part 1 -1985, the torsional shear stress assuming a plastic distribution will be:

⎟⎠⎞

⎜⎝⎛ −

=

3

22 bDb

Ttτ

As per BS8110, without torsional reinforcement, the value of tτ should not exceed by

cktc f067.0=τ or 0.4 N/mm2

Also, even if the section is fully reinforced for shear and torsion, the maximum value of shear stress ( tv )ττ + should not exceed ( ) ckc f8.0max =τ or 5 N/mm2

63

The rules for design is given in the following table Table 3 Design for shear and torsion (BS 8110)

Torsion shear stress Bending shear stress

tct ττ < tct ττ >

Less than safe in concrete ( v cτ τ< )

Greater than safe value in concrete( v cτ τ> )

Nominal shear steel, no Designed torsion steel torsional steel Designed shear steel, no Designed shear and torsional steel torsional steel

The area of links for the torsion may be calculated from the relation of:

1 1

(0.87 )0.8

svy

v

A Tfs x

=y

( )111

1 yxff

sA

Ay

y

v

svs += The area of longitudinal bar for torsion to be provided as:

Maximum spacing allowed for links and longitudinal steels in BS8110

The spacing of links should not exceed or1x 21y or 20 mm to control cracking.

The distance between the longitudinal steels should not exceed 300 mm. Design steps for Torsion by BS:8110 Step: 1 Find the area of tension steel for M.

Step: 2 CalculatebdV

v =τ

t

.

Step: 3 Calculate τ due to torsion.

Step: 4 Design shear & torsion steel as per table 3

Step: 5 Calculate additional longitudinal steel by equation ( )111

1 yxff

sA

Ay

y

v

svs +=

Where,

1sA Total area of longitudinal steel

svA Area of two legged stirrups

1yf Yield stress of the longitudinal steel

yf Yield stress of links

11, yx Centre to centre distance of links

64

Place this area as rods around the periphery of the beam. Example :-(Design of beam in torsion by BS 8110) The T beam given in Fig. 2 is subjected to the following factored loads. Bending moment of 215 kNm, shear of 150 kN, and torsion of 105 kNm. Assuming and

(N/mm

30=ckf

415=yf 2), design the reinforcements according to BS8110. Cover to centre of

steel is 50 mm. Solution:- Step1: Calculation of steel Bending moment capacity assuming the NA at bottom of slab:

( )( ) kNmM 63412042.07501207003036.01 =×−×××= As 215 kNm is very much less 634 kNm, NA is inside the slab. Hence, calculating the steel necessary for 215 kNm , we get

55.0750750700

10215 6

2 =××

×=

bdMU

2819100

750700156.0 mmAs =××

=156.0=p ,

Therefore, provide 3-20φ (tor).

%36.0750350100942

=××Thus, percentage of steel on web area =

350

120800

700

Fig. 2

Step 2: Calculation of steel for shear Here, the shear force, V=150kNm

57.075035010150 3

=××

==bdV

vτ

cτ for 0.3% steel =0.40N/mm2

Thus, the steel for shear will be required.

65

Step 3: Calculation of steel for torsion Here, the torsion moment, T=105 kNm Assume that torsion is taken fully by the beam part

Therefore, 2

2

6

2/51.2

3350800350

101052

3

2 mmNbDb

Tt =

⎟⎠⎞

⎜⎝⎛ −

××=

⎟⎠⎞

⎜⎝⎛ −

=τ

Thus, 2/37.030067.0067.0 mmNfcktct ===> ττ

Hence, tτ requires designed reinforcement.

Step 4: Determine )87.0( yv

sv fsA

Determine )87.0( yv

sv fsA

separately for (a) shear and (b) torsion

(a) )87.0( yv

sv fsA

for shear = ( )bcv ττ −

( ) mmN /5.5935040.057.0 =−=

(b) )87.0( yv

sv fsA

for torsion 118.0 yx

T=

( ) 280103023501 =−×−=x( ) 730103028001 =−×−=y

mmNfsA

yv

sv /6507302808.0

10105)87.0(6

=××

×=Thus, for torsion:

So the total value = 59.5 + 650 = 709.5 (design value) Step 5: Design for shear Design of shear steel by SP16 (using Fe415 steel)

cmkNinfsA

dV

yv

svus /)87.0(= ;09.71000

105.709=

×=

From SP 16, Table 62: Use usVd

= 8.167. Thus provide 12φ (tor) @100 mm c/c.

Step 6: Design of shear steel by formula

96.180.1165.087.0650

87.05.59

=+=+=yyv

sv

ffsA

Choose 12 mm rods , we get

66

2226mmAsv = for two legs

Therefore, mmsv 11596.1

226==

Step 7: Check spacing for max specified

vs should not be greater than x1 =280 mm ;y1=730 mm and

1 1 280 730 252 3004 4

x y mm or mm+ += =

Hence provide 12φ (tor) @115 mm c/c. Step 8: Extra longitudinal steel for torsion at corners

( ) ( ) 211

1

1818730280180.1 mmyxff

sA

Ay

yv

v

svst =+×=+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

Use 4 bars of 25 mm diameter Provide one bar at each corner 21968mmAs =

67

Lecture Note -13

Design for Torsion in Channel Section Example:- (Torsion in channel section) A precast edge beam of a building is of a channel section shown in the Fig. below. It is of span 12 m and is restrained at the ends. Precast floor slabs are placed on the lower flange as shown. Assuming that the load from the slab is 30 kN per meter length, determine the twisting moment for which the edge beam should be designed. Also, determine the torsional steel for the reinforced concrete channel section assuming

. 415,25 == yck ff

w=250

200

1100

t=200

160

30 kN/m

560

Solution: Analysis: Step1: Calculation of shear center Here the value of b = 560-250/2 = 435 mm and h = 1100 + 200/2 + 200/2 = 1300 mm

Thus,

( )btwh

be

611

2

+= from centre of web

68

mm134

2004356130025012

435=

⎟⎠⎞

⎜⎝⎛

×××

==

=134-250/2=9 mm from outer edge Step 2: Position of center of gravity of channel Taking moment about the outer edge, we get

( ) (( ) ( )

)22005602501100

28022005601252501100××+×

×××+××=x = insidemm195

104991097095

3

3

=××

( )

Step 3: Loads on channel Factored DL=1.5(0.499×25)=18.75kN Factored load on flange =1.5×30=45kN Thus, the total load =63.75kN Step 4: Design torsional moment per unit length on channel

mkNm /83.31000

9195=

+ Due to DL of channel =18.75

( ) mkNm /86.181000

1602509=

++ Due to load from slab = 45

Total moment =22.69 kNm Step 5:Maximum torsion

kNmT 1362

1269.22=

×= Maximum torsion at the end,

Design Step1: Torsion in main rectangle (web)

( )3

1 3

T t hT

t h=∑

( ) 933 104.231500250 ×=×=ht web

flanges ( ) 933 100.53102002 ×=××=ht

Therefore, kNmT 1124.28

4.231361 =

×=

Step 2: Design shear ( )vτ

69

kNwlV 5.3822

1275.632

=×

==

23

/06.11450250

105.382 mmNbdV

v =××

==τ

Assume cτ for grade M25 concrete =0.40 (Ref: SP16, Table 61)

( ) 21.06 0.40 0.66 /v c N mmτ τ− = − =

Step 3: Torsional shear stress( tτ )

( )6

2

2

2 2 112 10 2.53 /250 250 1500 250 3

3

tT N mm

tt hτ × ×= = =

× × −⎛ ⎞−⎜ ⎟⎝ ⎠

mm2 Total shear stress = 1.06+2.53=3.59 N/

This is less than 2max 0.8 4 /ckf N mmτ = = (Ref: BS 8110)

335.0067.0min

== cktfτ

But according to IS 456, this value ( maxτ ) will be 3.1 N/mm2

Step 4: Design of shear by BS8110 Here,

mintt ττ > Hence, design for shear & torsion

Here also, cv ττ >

( )v

ysv

sfA 87.0

for shear = ( ) 0.66 250 165 /v cVb N mmd

τ τ− = = × = Therefore,

( )v

ysv

sfA 87.0

for torsion = 6

1 1

112 10 543.9 /0.8 0.8 180 1430

T N mmx y

×= =

× ×

[Note: x1 = 250-(2×35) = 180]; and y1 = 1500-70 = 1430]

( )1000

101659.543 ×+=

dV in kN/cm=7.08kN/cm(708N/mm) Thus,

Therefore, Providing 12 φ (Tor) @100 mm c/c(this gives dVus =8.16 which is greater than 7.08

kN/cm) (Ref table 62 of SP16) Step 5

70

Design shear by IS456

bT

VV uue 6.1+= =382.5 +

250.01126.1 × = 1099 kN

d = 1500-50 =1450 mm

Capacity for concrete = bdcτ = 1000

15002504.0 ×× =150 kN

1099 150 6.5 /145

sV kN cmd

−⎛ ⎞= =⎜ ⎟⎝ ⎠

Therefore, providing 12 φ (Tor) @120 mm c/c gives 8.6=d

Vs

Step 6: Longitudinal steel for torsion by BS8110 According to BS,

( )1 11

ysvst

v y

fAA x ys f

⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠

50.141587.09.543

=×

=v

sv

sA

stA =(1.5)(180+1430)=2415 mm2

Therefore providing 8-20 φ(Tor) two on each corner [Area = 2513 mm2] In addition, provide steel for bending also.

101275.63

10

22

max×

==wlM =918 kNm

2

6

2 )1450(25010918×

=bdM =1.75

p=0.532%

1001450250532.0 ××

=sA =1928 mm2

Using 7-20 φ (Tor) (2199 mm2) Use 11-20 φ (Tor) on tension side (3455 mm2) & 4-20 φ (Tor) on compression side (1256 mm2) Step 7: Longitudinal steel by IS456

71

( )7.1

1 bDT

MMu

ue

++=

=918 +( )

7.1250

15001112 += 918 + 461 =1379 kNm ( ut MM < )

6

2 2

1379 10250(1450)

eMbd

×= =2.62

p= 0.845% (Ref table 3 of SP16)

Thus, 100

1450250845.0 ××=sA =3063 mm2

Using 10-20 φ (Tor) on tension side (3141 mm2) As compression steel is not required. ut MM <

Step 8: Detail steel according to standard practice

41587.01045

87.0

3

××

=yf

mload

=125 mm2/m stA required to carry load =

Hence ( )

v

ysv

sfA 87.0

for shear & torsion =708.9 N/mm

196341587.010009.708

=××

=svA mm2/meter length

Total steel required =1963 + 125 =2088 mm2/meter Area provide by the stirrups (12 mm @ 100 mm c/c)= 11312× =2262 mm2/meter >required (2088 mm2) Hence, O K.

72

Lecture Note – 14

Design of Column Axially loaded compression members All compression members are to be designed for a minimum eccentricity of load in two principal directions. Clause 25.4 of IS456 specifies the following minimum eccentricity

for the design of columns: mine

30500minDle += , 20 mm. (which ever is more) =mine

Where, l The unsupported length of the column D The lateral dimension of the column in the direction under consideration.

After determining the eccentricity, the section should be designed for combined axial load and bending. However, as a simplification, when the value of the minimum eccentricity calculated as above is less than or equal to 0.05D, clause 39.3 of IS456 permits the design of short axially loaded compression members by the following equation:

scycckuz AfAfP 67.04.0 +=

cA

scA

Where,

uP The axial load (ultimate),

The area of concrete, and

The area of reinforcement.

The above equation can be written as

10067.0

1004.0 g

yg

gcku

pAf

pAAfP +⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

gA

Where, The gross cross-sectional area

p The percentage of reinforcement Dividing both sides by gA

We get,

( )ckyckyckg

u ffpfpfpfAP

4.067.0100

4.0100

67.0100

14.0 −+=+⎟⎠⎞

⎜⎝⎛ −=

73

If the cross section of the column is known g

u

AP

can be calculated & reinforcement

percentage can be read from charts 24 to 26 of SP16.In the upper section of these charts

g

u

AP

is plotted against . The combined use of the upper & lower section would

eliminate the need for design for any calculation. This is particularly useful as aid for deciding the sizes of columns at the preliminary design stage of multistoried buildings.

gA

Tension in column In case of purely axial tension in column, We have,

( )yu fpbDP 87.0100

=

( )yckck

u ff

pfP

87.0100

=

Charts 66 to 75 in SP16 are given for rectangular sections with reinforcement on the two sides & charts 76 to 85 are for reinforcement on four sides. It is to be noted that these charts are meant for strength calculations only; they do not take into account crack control which may be important for tension members. Compression members subject to biaxial bending Exact design of members subject to axial load & biaxial bending is extremely laborious. Therefore the code permits the design of such members by the following equation.

1 1

1.0nn

uyux

ux uy

MMM M

αα ⎛ ⎞⎛ ⎞+ ≤⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

uyux MM ,

11 , uyux MM

n

Where, The moments about x,y axes

The maximum uniaxial moment capacities with an axial load ,

bending about x,y axes respectively. uP

An exponent whose value depends upon where uzu PP /α

sycckuz AfAfP 75.045.0 +=

uzu PP / nα

1.0 2.0≤ 2.0 8.0≥

74

For intermediate value linear interpolation can be done. Chart 63 of SP16 can be used for evaluating uzP

Slender compression member

DlexWhen the slenderness ratio or

bley of a compression member exceed 12, it is

considered to be a slender compression member. Where,

eyex ll , The effective lengths with respect to the major axis & minor axis

respectively. When a compression member is slender with respect to the major axis an additional moment given by the following equation should be taken into account in the design axM

2

2000⎟⎠⎞

⎜⎝⎛=

DlDP

M exuax

Similarly, a column slender about minor axis an additional moment should be

considered. ayM

2

2000 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

blbP

M eyuay

exuax ePM =

The expression for the additional moments can be written in the form of eccentricities of load, as follows:-

Where, 2

2000⎟⎠⎞

⎜⎝⎛=

DlDe ex

ax

2

20001

⎟⎠⎞

⎜⎝⎛=⇒

Dl

De exax

Table 1 of SP16 (page 106) gives different values of slenderness ratio for a given value of

Deax or

beay

In accordance with clause 39.7.1.1 of the code, the additional moment may be reduced by the multiplying factor k given below:-

75

1≤−−

=buz

uuz

PPPP

k

Where, sycckuz AfAfP 75.045.0 +=

Which may be obtained from chart 63 & and Pb is the axial load corresponding to the condition of maximum compressive strain of 0.0035 in concrete and tensile strain of 0.002 in outermost layer of tension steel. Though this modification is optional according to the Code, it should always be taken advantage of, since the value of k could be substantially less than unity. The value of Pb will depend on arrangement of reinforcement and the cover ratio d’/D, in addition to the grades of concrete and steel. The values of the coefficients required for evaluating pb for various cases are given in Table 60. The values given in Table 60 are based on the same assumptions as for members with axial load and uniaxial bending. The expression for k can be written as follows :

11

1≤

−

−=

uz

b

uz

u

PP

PP

k

Chart 65 can be used for finding the ratio of k after calculating the ratios uz

uP

P , and

uz

bP

P

Design of column for biaxial bending by BS 8110 method For the design of symmetrically reinforced rectangular column under biaxial bending, results comparable with those obtained by the Bresler method can be obtained by the simplified design procedure recommended in clause 3.8.4.5 of BS8110. The principle of the method is to transform the biaxial bending case, which should withstand an increased moment about that axis according to the two conditions of the code. Let the column be subjected to (P, , ) xM yM

Where, P Axial force

xM Moment about x-axis

yM Moment about y-axis

76

Then it can be designed for uniaxial bending of (P, ) or (P, ) depending upon the

following conditions:

/xM /

yM

Conditions 1:

xyx M

bM

dM

,/≥When controls the design and the column is to be designed for P and

, where /xM

yxx MbdMM /

/ α+=

Conditions2:

yyx M

bM

dM

,/< /yMWhen controls the design and the column is to designed for P and ,

where / 'y y

bxM M M

dα= +

Here, d Effective depth with respect to major axis and total depth D /b Effective depth with respect to minor axis and total depth b

⎟⎟⎠

⎞

⎝

⎛+ 26

71bdfP

ck⎜⎜ α Coefficient=

77

Lecture Note – 15

Example of Column Design Example Slender Column (with biaxial bending) Determine the reinforcement required for a column which is restrained against sway, with the following data: Size of column 40 x 30 cm Concrete grade M 30 Characteristic strength of reinforcement 415 N/mm2

=exl

y

Effective length for bending parallel to larger dimension, 6 m.

Effective length for bending parallel to shorter dimension, l = 5.0 m

Unsupported length =7.0m Factored load =1500kN Factored moment in the direction of larger dimension = 40 kNm at top & 22.5 kNm at bottom Factored moment in the direction of shorter dimension = 30 kNm at top & 20 kNm at bottom Solution:- The column is bent in double curvature. Therefore reinforcement will be distributed equally on four sides.

0.1540

1001.6=

×=

Dlex >12

7.16301005

=×

=bley >12

Therefore the column is slender about both the axes. From Table I,

0.15=Dlex , 113.0=D

ex For,

For, 7.16=bley , 140.0=D

ex

Additional moments:

10040113.01500 ××== xuux ePM =67.8 kNm

78

1003014.01500 ××== yuuy ePM =63.0 kNm

The above moments will have to be reduced in accordance with 39.7.1.1 of the Code; but multiplication factors can be evaluated only if the reinforcement is known. For first trial, assuming p=3.0 (with reinforcement equally on all the four sides). From chart 63

5.22=g

uzA

P N/mm2

32

101012005.22 ××=∴ uzP =2700 kN

Calculation of :bp

4025.5

=Assuming 25 mm dia bars with 40 mm cover d'/D (about xx-axis) = 0.13

Chart or Table for d’/d=0.15 will be used

d’/D(about yy-axis) 3025.5

= =0.17

Chart or Table for d’/d=0.20 will be used

From the table 60

bDffpkk ckck⎟⎟⎠

⎞

⎝

⎛+ 21⎜⎜ Pb(about xx- axis)=

32

1010403030

303203.0196.0 ××××⎟⎠⎞

⎜⎝⎛ ×+=bxP

=779 kN

32

1010303040

303028.0184.0 ××××⎟⎠⎞

⎝×+⎜

⎛ =672 kN Pb(about yy- axis) =

779270015002700−−

=−−

=bxuz

uuzx PP

PPk =0.625

672270015002700−−

=−−

=byuz

uuzy PP

PPk =0.592

The additional moments calculated earlier, will now be multiplied by the above values of k.

625.08.67 ×=axM =42.4 kNm

592.00.63 ×=ayM =37.3 kNm

79

The additional moments due to slenderness effects should be added to the initial moments after modifying the initial moments as follows (see Note 1 under 39.7.1 of the Code) :

( )5.224.0406.0 ×−×=uxM =15.0 kNm

( )204.0306.0 ×−×=uyM =10.0 kNm

The above actual moments should be compared with those calculated from minimum eccentricity consideration (see 24.4 of the Code) and greater value is to be taken as the initial moment for adding the additional moments.

3040

500700

30500+=+=

Dlex =2.73 cm

3030

500700

30500+=+=

Dley =2.4 cm

Both are greater than 2.0cm (20 mm) yx ee ,

Moments due to eccentricity:

10073.21500×=uxM =41.0 kNm >15.0 kNm

1004.21500×=uyM =36.0 kNm > 10.0 kNm

:. Total moments for which the column is to be designed are: =uxM 41.0 + 42.4 = 83.4 kNm

=uyM 36.0 + 37.3 =73.3 kNm

The section is to be checked for biaxial bending.

2

3

10403030101500×××

×=

bDfP

ck

u =0.417

10.030

0.3==

ckfp

Referring to chart 45(d’/D=0.15)

104.02 =bDfM

ck

u

63

1 101040403030104.0 ×××××=uxM =149.8 kNm

Referring to chart 46(d’/D=0.20)

096.02 =bDf

M

ck

u

63

1 101030304030096.0 ×××××=uyM =103.7 kNm

80

56.08.1494.83

1

==ux

ux

MM

; 71.07.1033.73

1

==uy

uy

MM

56.027001500

==uz

u

PP

Referring to chart 64 , the maximum allowable value of 1ux

ux

MM

corresponding to the above

value of 1uy

uy

MM

and uz

u

PP

is 0.58 which is slightly higher than the actual value of 0.56. the

assumed reinforcement of 0.30 % is therefore satisfactory.

10040300.3100 ××== pbDAs =3600 mm2

81

Lecture Note – 16

Design of RC Members in Tensions Elastic Method Assumption

• No crack at working load. • Use the principle of elastic theory and modular ratio. • Sometimes lower working stresses are adopted depending on the exposure

condition. • Both steel & concrete are assumed to be elastic. • Value of modular ratio is taken as 15. • The whole section including the concrete cover to reinforcement is assumed to be

effective in direct tension. • The area of reinforcement is calculated assuming that the whole tension is taken

by steel only. Allowable stress in steel for direct tension(As per BS:5337:1976) Type of stresses

Exposure Permissible stress(N/ mm2)

Plain bars Deformed bar Flexural tension and shear Compression

A B C A to C

85 100 115 130 125 140

125 140

Class A:- Exposed to wetting and drying, such as underside of roof of liquid retaining structures(Allowed crack width 0.1 mm) Class B:-Exposed to continuous contact with water e.g. walls of liquid retaining structures. (Allowed crack width 0.2 mm) Class C:-Not so exposed , for instance members exposed only to outside air. (Allowed crack width 0.3 mm) According to BS5337 (British code) the minimum cover should be 40 mm for surfaces in contact with water.

82

Design procedure Step 1: Calculate area of steel required

ss f

TA =1 where, Given in table sf

T Tensile force Area of steel required. 1sA

Provide this area of steel in the direction of T. 1sA

Find the size & spacing of bars. Maximum spacing ≤ 300 mm. Step 2: Check concrete stress cf

Equivalent area of concrete , ( ) 11 sce AmAA −+=

Where, =Area of concrete cA

e

c ATf = =Concrete stress

Allowable tensile stress in concrete ≤cf

Allowable stresses in concrete in direct tension without cracking of concrete. Concrete Grade Permissible direct tension(N/ mm2)

30 25

1.44 1.31

As per IS456, Annex B.2.1.1 Allowable stresses in concrete in direct tension allowing cracking of concrete are as follows: Grade of Concrete 15 20 25 30 35 40 Tensile Stress(N/mm2)

2.0 2.8 3.2 3.6 4.0 4.4

83

Step 3:-

Area of secondary reinforcement (Ast2) The minimum secondary steel to be provided, which should be based on the

concrete area is 0.3% for deformed bars & 0.5% for plain bars,

Thus, ( 3.01002

cst

AA = ) for deformed bars

( 5.0100

cA= ) for plain bars

Step 4:- Check cover to reinforcement

The rule for minimum cover should be satisfied. In addition, clear distance between bars should not exceed 300 mm. to limit crack

width in tension members. The lap length recommended in tension members are given in tension members is

given as follows by British code. Plain bar (Allowable stress) Deformed bar (Allowable stress) Grade of

concrete 85(A) 115(B) 100(A) 130(B) 25 30

24d 22d

32d 29d

20d 18d

26d 24d

Design of RC Flexural Members in Tensions Permissible Concrete stress for strength

Compression Tension Grade of Concrete Direct Bending Average Local

Shear

30 25

8.37 6.95

11.0 9.15

1.0 0.9

1.49 1.36

0.87 0.77

Strength Calculation of Flexural Member (By elastic method) (BS: 5337)

c

s

mff

dx+

=1

where, x N.A. depth d Effective depth

84

M Moment of resistance

⎟⎠⎞

⎜⎝⎛ −⋅=

321 xdfbM c ;

⎟⎠⎞

⎜⎝⎛ −

=

3xdf

MAs

s

Shear stress, zb

VC⋅

= , where 3xdz −=

Permissible concrete tension in bending

( )bzv=τGrade Tension in bending (MPa) Shear

30 25

2.02 1.84

2.19 1.94

Example: (Only bending)

A column of 300×300 mm cross section is carrying a moment of 25 KNm. Using M30 grade of concrete and 16 mm diameter Fe250steel, find the steel area required. Assume, cover =40 mm & m=15.

Solution:

Assume exposure condition as class B =115 N/mmstf 2. =11.0 N/mmcbf 2.

2522

1640300 =−−=dctf =2.02 N/mm2. mm.

N.A. Calculation

c

s

mff

dx+

=1

11151151

252

×+

=148.5 mm. =

⎟⎠⎞

⎜⎝⎛ −

=

3xdf

MAs

s =625 10

148.5115 2523

×⎛ ⎞−⎜ ⎟⎝ ⎠

=1073 mm2.

Hence provide 6-16φ longitudinal bars. Thus, As = 26 164π

× = 1206 mm2.

Thus % of reinforcement provided, 100 100 1206 1.6%300 252

sApbd

×= = =

×

Example: (Axial Tension)

85

A RCC column of 400 ×400 mm is subjected to axial tension of 150KN.Find the area of steel required using Fe415 & M30 concrete. Use m=15. Solution: Assume exposure class as B.

1154130

10150 6

1 =×

==∴s

st fTA mm2

Use 6 nos. of 16 φ 1206 mm2

Check concrete stress ( ) 11 sce AmAA −+=

( ) 1768841206115400400 =−+×= mm2

85.0176884

10150 3

=×

==e

ct ATf MPa

MPa. Hence OK 44.1<Secondary Steel

2 0.3%stA = (min.)

0.3 400 400100

= × × =480 mm2