10/3/2011 - PEOPLE AT UNIVERSITI TEKNOLOGI MALAYSIA · 2013. 9. 13. · 10/3/2011 8 Maximum limit,...

10/3/2011

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Section design

Rectangular, T or L section?

Singly reinforced section

Doubly reinforced section

First principle EC2 method

Singly reinforced section

Doubly reinforced section

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Reinforcements are only provided in the tension zone (As)

d

b

fyk/1,15 = 0.87fyk

0.85fck/1.50 = 0.567fck

x s =

0.8

x

N. A

Fst = 0.87fykAs

0.4

x

As

Fcc = 0.567fck.s.b = 0.454fck bx

z = d – 0.4x

Stress Section Action

N.A

For equilibrium;

Fcc = Fst

0.454fckbx = 0.87fykAs

x = (0.87fykAs) / (0.454fckb)

Taking moment either towards Fcc or Fst:

M@steel = Fcc z

= 0.454fckbx (d – 0.4x)

M@conc = Fst z

= 0.87fykAs (d – 0.4x)

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The equation shows that M increases with x and hence with As

To avoid concrete failure in compression due to over-reinforced

section, EC2 limit the value of x = 0.45d

When x = xbal = 0.45d, M represent the maximum ultimate moment

capacity of the section which is known as the ultimate moment of

resistance of singly reinforced section or balanced moment, Mbal

Take M@steel and replace x = 0.45d, therefore:

M = Mbal = [0.454 fckb(0.45d)] [d – 0.4(0.45d)]

= [0.454 fckb(0.45d)] [0.82d]

= 0.167fckbd2

= Kbalfckbd2 where Kbal = 0.167

Mbal = Ultimate moment of resistance for singly reinforced section

If M < Mbal, ONLY tension reinforcement

is required and this is known as Single

Reinforced Section

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A rectangular section beam is required to resist a design moment, M = 175

kNm. If fck = 25 N/mm2 and fyk = 500 N/mm

2, determine As and the required

number of steel bars.

d = 500 mm

b = 250 mm

Mbal = 0.167 fckbd2

= 0.167(25)(250)(500)2

= 261 kNm M = 175 kNm (Only tension reinforcement is required)

Determine N.A. depth, x:

M = 0.454fckbx (d – 0.4x)

175 x 106 = 0.454(25)(250x)(500 – 0.4x)

x = 138.7 mm or 1111 mm

x = 138.7 mm < 0.617d (308.5 mm)

Lever arm, z = d – 0.4x = 500 – 0.4(138.7) = 444.5 mm 0.95d = 475 mm

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As = M/0.87fykz

= 175 x 106 / 0.87(500)(444.5) = 905 mm2

Provide 3H20 (As = 943 mm2)

3H20

Determine the moment of resistance of the beam provided with 3H20

bars. Use fck = 25 N/mm2 and fyk = 500 N/mm

2.

d = 500 mm

b = 300 mm

3H20

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d=

50

0 m

m

b = 300 mm

x

N. A

Fst = 0.87fykAs

0.4

x

3H20 (943 mm2)

Fcc = 0.454fck bx

z = d – 0.4x

0.8

x

Determine x and from equilibrium;

Fcc = Fst

0.454fckbx = 0.87fykAs

x = 0.87(500)(943) / (0.454)(25)(300)

= 120.4 mm 0.617d = 309 mm

Steel has yielded as assumed (under reinforced section)

Check x/d = 120.4/500 = 0.241 0.45 OK

What happens if x/d > 0.45?

Moment of resistance, M

M = Fst.z = 0.87fykAs.z = 0.87(500)(943) [500 – (0.4)(120.4)] 10-6

= 185.3 kNm

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Maximum limit, x = 0.45d Mbal = 0.167fckbd2 (tension reinforcement)

If M > Mbal, COMPRESSION REINFORCEMENT IS REQUIRED to take the

remaining moment (M – Mbal)

d

b cc = 0.0035

x s =

0.8

x Fst = 0.87fyk As

0.4

x

As

Fcc = 0.454fck bx

z

Strain Section Stress and Action

As’ d’

st

sc Fsc = 0.87fyk As’

0.567fck

z1 = d – d’

N.A

For equilibrium;

Fst = Fcc + Fsc

0.87fykAs = 0.454fckbx + 0.87fykAs’ ------------ (1)

x = (0.87fykAs – 0.87fykAs’) / (0.454fckb)

Taking moment resistance towards the tension reinforcement:

M = Fcc.z + Fsc.z1

= 0.454fckbx (d – 0.4x) + 0.87fykAs’ (d – d’)

For design according to EC2, x = 0.45d

M = 0.454fckb(0.45d) [d – 0.4(0.45d)] + 0.87fykAs’ (d – d’)

= 0.167fckbd2 + 0.87fykAs’ (d – d’)

= Mbal + 0.87fykAs’ (d – d’)

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Required area for COMPRESSION reinforcement

As’ = (M – Mbal) / 0.87fyk (d – d’)

or

As’ = (K – Kbal)fckbd2 / 0.87fyk (d – d’)

From Eq. (1):

0.87fykAs = 0.454fckbx + 0.87fykAs’ ( z)

0.87fykAs. z = 0.454fckbx. z + 0.87fykAs’. z

When x = 0.45d z = d – 0.4x = 0.82d

0.87fykAs. z = 0.454fckb (0.45d).(0.82d) + 0.87fykAs’. z

0.87fykAs. z = 0.167fckbd2 + 0.87fykAs’. z

Required area for TENSION reinforcement

As = (0.167fckbd2/ 0.87fy.z) + As’

or

As = (Kbal fckbd2/ 0.87fyk.z) + As’

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The equation is derived by assuming the compression reinforcement has

yield at 0.87fyk.

sc / (x – d’) = 0.0035 / x Only for Concrete Class C50/60

(x – d’) / x = sc / 0.0035

d’/x = 1 – (sc / 0.0035)

To reach 0.87fyk;

sc = 0.87fyk / Es

d’/x = 1 – [0.87fyk / Es (0.0035)] --------------------- (2)

From Eq. (2) and for fyk = 500 N/mm2 and Es = 200 kN/mm

2

d’/x = 0.38

Therefore, for compression reinforcement to reach its yielding point;

d’/x 0.38 and fsc= 0.87fyk

If d’/x 0.38 fsc 0.87fyk

fsc = Es.sc where sc = 0.0035(x – d’) / x

= 200 103 (0.0035)(1 – d’/x)

= 700 (1 – d’/x)

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Therefore the revised As and As’ are:

Required area for COMPRESSION reinforcement

As’ = (M – Mbal) / fsc (d – d’)

or

As’ = (K – Kbal)fckbd2 / fsc(d – d’)

Required area for TENSION reinforcement

As = (0.167fckbd2/ 0.87fyk.z) + As’ (fsc/0.87fyk)

or

As = (Kbal fckbd2/ 0.87fyk.z) + As’ (fsc/0.87fyk)

A rectangular section beam is required to resist a design moment M = 300

kNm. If fck = 25 N/mm2 and fyk = 500 N/mm

2, determine As and the

required number of steel bars. Assumed d’ = 50 mm.

d = 500 mm

b = 250 mm

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Mbal = 0.167 fckbd2

= 0.167(25)(250)(500)2

= 261 kNm < M = 300 kNm

(Compression reinforcement is required)

Check whether the compression steel has yielded:

d’ = 50 mm

x = 0.45d = 225 mm

d’/x = 0.22 0.38 Therefore the compression steel will have yielded

Therefore, fsc = 0.87fyk

and z = d – 0.4x = 410 mm

Required area for compression reinforcement:

As’ = (M – Mbal) / 0.87fyk(d – d’)

= (300 – 261) 106 / 0.87(500)(500 – 50)

= 200 mm2

Required area of tension reinforcement:

As = (Mbal/ 0.87fyk.z) + As’

= 261 106 / 0.87(500)(410) + 200 = 1663 mm2

Provide 2H12 (As’ = 226 mm2) for Compression Reinforcement

Provide 6H20 (As = 1885 mm2) for Tension Reinforcement

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d = 500 mm

b = 250 mm

6H20

2H12

Determine the moment of resistance of the beam provided with 6H20 and

2H12 for tension and compression reinforcement, respectively. Use fck =

25 N/mm2 and fyk = 500 N/mm2. Take d’ = 50 mm.

d = 500 mm

b = 250 mm

6H20

2H12

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b = 250 mm

d=

50

0 m

m x

N. A

Fst = 0.87fykAs

0.4

x

6H20 (1885 mm2)

Fcc = 0.454fck bx

z

0.8

x

Fsc = 0.87fyk As’

z1 = d – d’

2H12 (226 mm2)

Determine x and from equilibrium;

Fst = Fcc + Fsc

0.87fykAs = 0.454fckbx + 0.87fykAs’

x = 0.87fyk(As – As’) / (0.454fckb)

= 0.87(500)(1885 – 226)/0.454(25)(250)

= 254 mm 0.617d = 309 mm

Tension steel has yielded as assumed

Check d’/x = 50/254 = 0.20 0.38 Compression steel has yielded

as assumed

Moment of resistance, M

M = Fcc.z + Fsc.z1 = 0.454fckbx (d – 0.4x) + 0.87fykAs’ (d – d’)

= 0.454(25)(250)(254)[500 – 0.4(254)] +

0.87(500)(226)(500 – 50)

= 332 106 Nmm

= 332 kNm

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or xbal ( - k1) d k2

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Depth of stress block, sbal = 0.8xbal where xbal = xu

Lever arm, zbal = d – 0.5sbal

Moment of resistance of concrete (in compression):

Mbal = Fcc zbal = 0.567fckbsbal zbal

and

Kbal = Mbal / fckbd2 = 0.567sbal zbal/d

2

Arranged Kbal with previous equation gives:

Kbal = 0.454( - k1)/k2 – 0.182 [( - k1)/k2]2

where k1 = 0.44

k2 = 1.25(0.6 + 0.0014/cu2) = 1.25 (only for concrete class C50/60)

Kbal = 0.454 ( - 0.44)/1.25 – 0.182 [( - 0.44)/1.25]2

= 0.363 ( - 0.44) – 0.116 ( - 0.44)2

where

= Moment at section after redistribution

Moment at section before redistribution

1.0

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Concrete action (in compression), Fcc = 0.567fckbs

Lever arm, z = d – s/2 or s = 2(d – z)

Moment, M = Fcc . z = 0.567fckbs . z

= [0.567 fckb . 2(d – z)] . z

= [1.134fckb (d – z)] . z

= 1.134fckbdz – 1.134fckbz2 ( fckbd

2)

M/fckbd2 = 1.134fckbdz/fckbd

2 – 1.134fckbz2/fckbd

2

K = 1.134(z/d) – 1.134(z/d)2

0 = (z/d)2 – (z/d) + K/1.134

Quadratic solution for z/d gives:

z = d [0.5 + (0.25 – K/1.134)]

Determine = Moment at section after redistribution

Moment at section before redistribution

Determine Kbal Kbal = 0.167 (if 1.0)

Kbal = 0.454( - k1)/k2 – 0.182 [( - k1)/k2]2 (if 1.0)

Compression reinforcement required

No Compression reinforcement not required

Yes K Kbal

Determine K = M/fckbd2

A B

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z = d [0.5 + (0.25 – K/1.134)]

Tension reinforcement As = M/0.87fyk.z

A

Tension reinforcement As = Kbalfckbd

2 + As’ fsc 0.87fykz 0.87fyk

z = d [0.5 + (0.25 – Kbal/1.134)]

x = (d – z) / 0.4

Compression reinforcement As’= (K – Kbal)fckbd

2

0.87fyk(d – d’)

Compression reinforcement As’= (K – Kbal)fckbd

2

fsc(d – d’)

where fsc = 700(1 – d’/x)

Check d’/x

If d’/x 0.38 If d’/x 0.38

B

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Determine the area of reinforcement required for a rectangular beam subjected to an ultimate design moment of 350 kNm after a 20%

reduction due to moment redistribution. Use fck = 30 N/mm2 and fyk =

500 N/mm2. Take d’ = 45 mm.

d = 600 mm

b = 225 mm

K = M/fckbd2 = 350 106 / 30 225 6002

= 0.144

Redistribution = 20%, therefore = 0.8

Kbal = 0.454( - k1)/k2 – 0.182 [( - k1)/k2]2 where k1 = 0.44 and k2 = 1.25

= 0.363 ( - k1) – 0.116 ( - k1)2

= 0.116

Since K Kbal, compression reinforcement is required

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z = d [0.5 + (0.25 – Kbal/1.134)]

= 0.88d = 0.88 600 = 530.8 mm

x = (d – z) / 0.4 = 172.9 mm

d’/x = 45/172.9 = 0.26 0.38 The compression steel will have yield

Therefore, use fsc= 0.87fyk

Area for compression steel:

As’ = (K – Kbal)fckbd2 / 0.87fyk(d – d’)

= 286 mm2

Provide 3H12 (As’ = 339 mm2)

Area for tension steel:

As = [Kbalfckbd2 / 0.87fyk .zbal] + As’

= 1503 mm2


d =

60

0 m

m

b = 225 mm

3H12

5H20

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L-Beam T-Beam

Slab

Actual width Actual width

Effective width Effective width

Beam Beam Beam

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beff = bw + beff,i b

where beff, i = 0.2bi +0.1lo 0.2lo and also beff, i bi

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(a) Neutral axis lies in the flange

(b) Neutral axis falls below the flange

(c) Neutral axis falls below the flange with M > Mbal

This occur when s = 0.8x < flanged depth (hf)

b

bw

As

d z = d – 0.4x

Fcc = 0.454fck bx hf s = 0.8x

0.567fck

Fst = 0.87fykAs

Section Stress and Action

x N.A

h

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Moment of resistance

Mf = Fcc.z

= (0.567fck.b.0.8x) . (d – 0.4x)

In this case, 0.8x = hf

M = Mf = (0.567fck.b.hf) . (d – hf/2)

where Mf is the flange ultimate moment of resistance

If M < Mf, N.A lies in the flange. Therefore, design similar to

Rectangular Section

A T-beam is required to resist a design moment M = 225 kNm. If fck = 25

N/mm2 and fyk = 500 N/mm2, determine the area of reinforcement and the

required number of steel bars.

d = 320 mm

b = 1150 mm

bw = 250 mm

hf = 100 mm

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Flange moment of resistance

Mf = (0.567fck.b.hf) . (d – hf/2)

= 0.567(25)(1150)(100)(320 – 100/2)

= 440.1 kNm M = 225 kNm

Neutral axis LIES in the flange

K = M/fckbd2

= 225 106 / (25)(1150)(320)2 = 0.076 Kbal

z = d {0.5 + (0.25 – K/1.134)}

= 0.93d 0.95d

As = M/0.87fyk.z

= 225 106 / 0.87(500)(0.93 320)

= 1738 mm2


4H25

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(c) Neutral axis falls below the flange with M > Muf

This occur when M Mf

b

bw

As

d

z2

hf s = 0.8x

0.567fck

Fst = 0.87fykAs


x

z1

2 2 1

N.A h

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Internal forces

Fcc1 = (0.567fck)(bw.0.8x) = 0.454fcubwx

Fcc2 = (0.567fck)(b – bw)hf

Fst = 0.87fykAs

Lever arm,

z1 = d – 0.4x and z2 = d – 0.5hf


M = Fcc1.z1 + Fcc2.z2

= (0.454fckbwx)(d – 0.4x) + (0.567fckhf)(b – bw)(d – 0.5hf)

Ultimate moment of resistance is when x = xbal = 0.45d

Mbal = 0.454fckbw 0.45d[d – 0.4(0.45d)] + (0.567fckhf)(b – bw)(d – 0.5hf)

= 0.167fckbwd2 + (0.567fckhf)(b – bw)(d – 0.5hf)

If M < Mbal, compression reinforcement is NOT

required. As can be determine by taking

moment towards Fcc2

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M = Fst.z2 – Fcc1. (z2 – z1)

= 0.87fykAs (d – 0.5hf) – 0.567fckbwx [(d – 0.5hf) – (d – 0.4x)]

As = M + 0.567fckbwx (0.4x – 0.5hf)

0.87fyk (d – 0.5hf)

Taking x = 0.45d as required in EC2:

As = M + 0.1fckbwd (0.36d – hf)

0.87fyk (d – 0.5hf)

Equation above only applicable for hf 0.36d

A T-beam is required to resist a design moment, M = 445 kNm. If fck = 25

N/mm2 and fyk = 500 N/mm2, determine As and the required number of steel

bars.

d = 320 mm

b = 1150 mm

bw = 250 mm

hf = 100 mm

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Mf = (0.567fck.b.hf) . (d – hf/2)

= 0.567(25)(1150)(100)(320 – 100/2)

= 440.1 kNm M = 445 kNm

Neutral axis FALLS BELOW the flange

Mbal = 0.167fckbwd2 + (0.567fckhf)(b – bw)(d – 0.5hf)

= 0.167(25)(250)(320)2

+ 0.567(25)(100)(1150 – 250)[320 – 0.5(100)]

= 451.3 kNm M = 445 kNm

Compression reinforcement NOT required

As = M + 0.1fckbwd (0.36d – hf)

0.87fyk (d – 0.5hf)

= 445 106 + 0.1(25)(250)(320)(0.36 320 – 100)

0.87(500)(320 – 0.5 100)

= 3815 mm2

Provide 5H25 + 5H20 (As = 4026 mm2)

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d = 320 mm

b = 1150 mm

bw = 250 mm

hf = 100 mm

5H25 + 5H20



(c) Neutral axis falls below the flange with M > Muf

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Compression reinforcement REQUIRED so that x 0.45d

b

bw

As

d

z2

hf s = 0.8x

0.567fck

Fst = 0.87fykAs


x

z1

2 2 1

As’

z3

N.A

Internal forces

Fcc1 = 0.454fckbwx and Fcc2 = (0.567fck)(b – bw)hf

Fst = 0.87fykAs and Fsc = 0.87fykAs’

Lever arm,

z1 = d – 0.4x, z2 = d – 0.5hf and z3 = d – d’


M = Fsc.z3 + Fcc1.z1 + Fcc2.z2

= (0.87fykAs’)(d – d’) + (0.454fckbwx)(d – 0.4x)

+ (0.567fckhf)(b – bw)(d – 0.5hf)

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When x = 0.45d

M = 0.87fykAs’ (d – d’) + Mbal

As’ = (M – Mbal)

0.87fyk (d – d’)

From equilibrium of forces

Fst = Fcc1 + Fcc2 + Fsc

0.87fykAs = 0.454fckbwx + (0.567fck)(b – bw)hf + 0.87fykAs’

When x = 0.45d

0.87fykAs = 0.167fckbwd2 + (0.567fck)(b – bw)hf + 0.87fykAs’

As = 0.167fckbwd + 0.567fckhf (b – bw) + 0.87fykAs’

0.87fyk

A T-beam is required to resist a design moment M = 500 kNm. If fck = 25

N/mm2 and fyk = 500 N/mm2, determine the required area of reinforcement

and number of steel bars.

d = 320 mm

b = 1150 mm

bw = 250 mm

hf = 100 mm d’ = 50 mm

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Mf = 440.1 kNm M = 500 kNm

Neutral axis FALLS BELOW the flange

Mbal = 451.3 kNm M = 500 kNm

Compression reinforcement is REQUIRED

As’ = (M – Mbal) / 0.87fyk(d – d’)

= (500 – 451.3) / 0.87(500)(320 – 50)

= 415 mm2


0.87fyk

= 0.167(25)(250)(320) + 0.567(25)(100)(1150 – 250) + 0.87(500)(415)

0.87(500)

= 4115 mm2

Provide 5H25 + 5H16 (As = 4153 mm2)

Provide 4H12 (As’ = 453 mm2)

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d = 320 mm

b = 1150 mm

bw = 250 mm

hf = 100 mm d’ = 50 mm

5H25 + 5H16

4H12

Start

M Mf Yes

No

N.A lies in the flange

N.A below the flange K = M/fckbd

2

z = d {0.5 + (0.25 – K/1.134)} As = M/0.87fyk.z

Calculate Mbal = ffckbd

2

Check Mbal with M Check hf 0.36d

A

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M Mbal

A

Yes No Compression reinforcement is NOT

required

Compression reinforcement

required


0.87fyk As’ = (M – Mbal) / 0.87fyk (d – d’)

As = M + 0.1fckbwd (0.36d – hf)

0.87fyk (d – 0.5hf)

End

10/3/2011 - PEOPLE AT UNIVERSITI TEKNOLOGI MALAYSIA · 2013. 9. 13. · 10/3/2011 8 Maximum limit,...

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Transcript of 10/3/2011 - PEOPLE AT UNIVERSITI TEKNOLOGI MALAYSIA · 2013. 9. 13. · 10/3/2011 8 Maximum limit,...