CE5510 Advanced Structural CE5510 Advanced Structural Concrete Design

2 Design for Action Effects2. Design for Action Effects

Part 1 Bending & Axial LoadPart 1 Bending & Axial Load

Professor Tan Kiang HweeDepartment of Civil EngineeringNational University of SingaporeNational University of Singapore

Action Effects

Flexure (Bending)Compression/Tension

MyNMx

pShearTorsionT T

V

MyT

weightMxz

NV xy

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BENDING with or without AXIAL FORCE

Method of strain compatibility & force equilibriump y q

3 basic requirements: Compatibility Compatibility Strains (deformations) at various locations are related to one another

Material Laws Material Laws Stress-strain relations (aka constitutive relations)

Equilibrium Externally applied forces/moments = Internal resisting forces/moments

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Uni-axial Bending

w

ULSSLS

w

w steelyieldingSLS

1

23 -Inelastic,cracked

3

cracking

1Assumptions : Plane sections remain plane

1 - Elastic,uncracked

2 - Elastic,cracked

Plane sections remain plane Perfect bond between concrete

& reinft. Tensile strength of concrete is

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gneglected after cracking

Uniaxial bendingg

cElastic,

uncracked

C

Ts

n.a.

M

x

TsTc

M

Elastic,Cn.a.

>0.4fc

Inelastic,k dM

C

T

n.a.

crackedM T

cu

UltimateMn Cn.a.

T

Compatibility: ~ c Material laws: ~ Equilibrium: F = 0 ; M = M

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Serviceability Limit State

A /bd

x/3 fcc CC

b d

As = As/bd = As/bdn = Es/Ec fs

s

s

n.a.

M

C

T

dh

xAs

Ass c

Section Strain Stress

s T

Strain compatibility:

/ /( d)

Material laws:

f = E c/s = x/(x-d)c/s = x/(d-x)

fc = Ec cfs = Es sf E fs = Es s

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x/3 fccs

n a CC

b

d

d

xAs = As/bd

fs

fss

n.a.

MT

hAs

s = As/bdn = Es/Ec

Section Strain Stress

Stress resultants:C = bxfc/2C=fsAs=nfcAs(x-d)/x

Equilibrium:C + C = TC fs As nfcAs (x d )/x

T=fsAs=nfcAs(d-x)/xM = C(d - x/3) + C(d-d)

x/d = [(+)2n2 + 2(+d/d)n]1/2 - (+)n Tan K H, NUS

Ultimate limit state fcd = cc fck / c

If width of compression zone decreases in the

i i fcu (0/00) = 3,5 for fck 50 MPacu (0/00) = 2,6 + 35[(90-fck)/100]4 for fck 50 MPa

direction of extreme compression fibre, fcdshould be reduced by 10%.

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10%.

Example RC Box GirderpCalculate the ultimate moment of resistance of the box girder shown.

1200 120 mm50

g

f 30 MP

1000 mm2

fck = 30 MPafyk = 500 MPa

800

2502000 mm2

5000 mm2

120

505000 mm

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0.567fck= 0.0035x=200

cu160 C C100 mm

M

mmCs C2s

Mu s1s2

Ts1Ts2

s=0.0035x150/200 = 0.0263>y=(fy/m)/Es = (500/1.15)/(200,000) = 0.0217s1> y ; s2 > yC1 =0.567x30x1200x120x10-3 = 2449 kNC2 =0.567x30x240x40x 10-3 = 163 kNCs =1000x435x 10-3 = 435 kN

Mu= [2449x140 + 163x60 + 435x150 + 870x300 s

Ts1=2000x435x 10-3 = 870 kNTs2 =5000x435x 10-3 = 2175 kNC1+C2+Cs- Ts1- Ts2 = 2 0 OK

+ 435x150 + 870x300 + 2175x550] kNmm= 1875 kNmC1 C2 Cs Ts1 Ts2 2 0 OK

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Example Isolated L-beamExample Isolated L beam

A simply-supported beam with an inverted L-section isan inverted L-section issubjected to vertical loads as shown. The beam is free to deflect vertically and laterally

200 mm

0

m

m

Mdeflect vertically and laterally between its supports. Determine the designmoment of resistance of the

6

0

50 mm

M

L-section. The reinforcement consisted of four uniformly spaced steel bars of equal di t ith t t l f

300 mm 300 mm

50 mm

diameter, with a total area of 2250 mm2. Assume vertical loads to pass through shear centre of section

Material properties: fck = 40 MPa; fyk =500 MPa; and centre of section.

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yk

Es = 200 GPa.

T = Asfyk/s=978.8 kNC = T gives (450 g) (0.567fck)

= 9788000C x

300 mm150

g = 9788000

g = 191.8 mm

0

0

m

m

M

C x

6

0

50 mm

M

T x

Check s >yk(= 0.00217)

Mn= 978.8 x (550 191.8/3) =978.8 x 486.1 = 475.7 kNm300 mm 300 mm

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What if beam is prevented from deflecting laterally?y

xn.a.

My = 491.4 kNm

xMz = ?

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Effect of prestressing cuxd

p gfcd

0.8x

dpsM

dsA peP

As

Aps ApsfpsAsfs

ps = pe + ce + cu (dps - x)/xcecu(dps/x-1)

N = 0p p pwhere pe = fpe/Eps = P/ApsEps

ce = fce/Ec = P[1/Ac + eo2/I]/Ec = (d - x)/x

N 0M = M

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s cu (ds x)/x

Example PC Box GirderpCalculate the ultimate moment of resistance of the prestressed concrete box girder shown. The effective prestessin the tendons is 1100 MPa

Ac = 374,400 mm2I = 30.8 x 109 mm4 yt = 356 mm 1200 120 mm50

in the tendons is 1100 MPa.

yt 356 mm fck = 30 MPafyk = 500 MPaE = 200 GPa

120 mm50

1000 mm2

yt Es = 200 GPafpk = 1860 MPafp0.1k = 1580 MPaE 195 GP

800

2501000 mm2

120

ytc.g.c.

Eps = 195 GPafpe = 1100 MpaAssume bi-linear relations for steel

250

504600 mm2

relations for steel.

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0.567fck= 0.0035

300

cuC C1 240

1000 2

50

x = 300 mm

M

CsC2

s

ce + pec.g.c.mmmm

2

250

200

56 n.a.

Ms

TpTs

ce pe1000 mm2

4600 mm2250

50

Stress in concrete at c.g.s. due to P (= 1100 kN)fce = P/Ac + Peo2/I = 1100x103/374,400 + 2500x103x1442/(30.8x109)

= 4.62 MPa

Strain components in tendonspe = fe/Eps = 1100/195,000 = 0.00564pe fe/Eps 1100/195,000 0.00564Ecm = 22(fck + 8)0.3 = 22 x (30 + 8)0.3 = 32.8 GPace = fce/Ecm = 4.62/32,800 = 0.000141

00705015.1/1580/

008120300500 10 skpf

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00705.0000,195

15.1/158000812.0300

300500 1.0

ps

skpcucepeps E

f

0.567fck= 0.0035

300

cuC C1 240

1000 2

50

x = 300 mm

M

CsC2

s

ce + pec.g.c.mmmm

2

250

200

56 n.a.

Ms

TpTs

ce pe1000 mm2

4150 mm2250

50

Steel strains

00217043515.1/500/

)(0029205030000350' sykfcomp 00217.0000,200000,200

)(00292.0300

0035.0 ss Ecomp

00217.0/

00525.0300

3007500035.0

syks Ef

all steel have yielded.300 ss E

Tan K H, NUS

0.567fck= 0.0035

300

cuC C1 240

1000 2

50

x = 300 mm

M

CsC2

s

ce + pec.g.c.mmmm

2

250

200

56 n.a.

Ms

TpTs

ce pe1000 mm2

4150 mm225050

Forces kNC 244910120120030567.0 31

kNC 49010120240305670 32 Ult. moment capacity (taken about n.a.)

kNTp 137410)15.1/1580(10003

kNC 4901012024030567.02 kNCs 43510)15.1/500(1000

3 M = 2449x240 + 490x120+435x250 + 1374x200+ 2001x450 kNmm

kNTs 200110)15.1/500(46003

2 Check for Equilibrium

+ 2001x450 kNmm= 588 + 58.8 + 109 + 275 + 9001

1931 kNTan K H, NUS

kNF 1200113744354902449 = 1931 kNm

Exercise PC GirderCalculate the ultimate moment of resistance of the 350PC girder shown. 200

150

100

2H16Ac = 290,000 mm2y = 356 mm

500

yt = 356 mm fck = 40 MPafyk = 500 MPaE 200 GP

1502H20

12-12.9 strands

150

Es = 200 GPafpk = 1860 MPafp0.1k = 1580 MPa 2H20200

150 100100100150p

Eps = 195 GPafpe = 1100 MPaA = 100 mm2 /strand

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Aps 100 mm /strand

Effect of axial loadN

c2 =0.002ed

N

0.0035

x

cu2 = 0.0035

x0.0035

d

M

N x0.0020.0020.002

e = 0 increasingM = Ne

N = Nx = decreasing

Tan K H, NUSM = M

N-M (Column) Interaction Diagram

N

( ) g

No Compression failured

N

NbBalanced

failure

d

M

N

M = Ne

e

NoTensionfailure

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Examplep

A reinforced concrete short column, hexagonal in cross- N, gsection, contains six 25 mm diameter longitudinal bars and is subjected to an

7

5

m

m

5

0

m

m

N

eccentric compression load. Given that fck = 30 MPa; fyk = 500 MPa and Es = 200 GP l l t th

2

7

200 GPa, calculate the design ultimate load N that the column can carry at an eccentricity e = 275 mm

450 mmeccentricity, e = 275 mm

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0

m

N2

7

5

m

m

5

0

m

m

450 mm

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Biaxial Bendingg

M Mzz due towind load

MyMz

y

due to

wind load

My due togravity load

y

z

MM1tan

22zy MMM

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yC22

zy MMM

MyyCzC

y

z

MM1tan

z

nMz

j

ji

i TCC )(

j

cjjy CzzTM1

n CTM C>0 comp

j

cjjz CyyTM1

C>0 comp.T>0 tension

Tan K H, NUS

My/bh2fck

Increasingb g

As/AcfcuAs/4As/4M

b

As/4As/4

My

M

h

Mz

Mz/b2hfckTan K H, NUS

Example Biaxial bendingSo,

A concrete beam has a 250 mm square section and is reinforced by 4T28 steel bars, one bar being placed at 50 mm from each edge in

h f th ti E l t th fl l t th f theach corner of the section. Evaluate the flexural strength of the section if it is subjected to biaxial bending moments of equal magnitude about axes parallel to the edges. The material properties are: fyk = 500 MPa, fck = 30 MPa, and Es = 200 GPa.are: fyk 500 MPa, fck 30 MPa, and Es 200 GPa.

50 250

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fcd = 0.9x0.567fck= 0.51fckck

x=148 0.8x CsC

s

Ts1

Ts2

s1s2

Ans:

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Biaxial bending with axial loading

N Mey

z

NNMe zy

ez

NM

e yz y

22zyr eee

e.g. corner columns

z

y

y

z

MM

ee 11 tantan

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N x ez x er Cy

x

0 8

T

N jTCN0.8x

j

j

j

ijczy zTCzNeMC

Tjj

ijcyz yTCyNeMTj

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N-My-Mzi t ti N

eyinteraction diagram for rectangular

section N

Nez

er

Note:section

1 kk MMN

When Pn is small, My + Mz 1 (i t i ht li ) 1 zy MM (i.e. a straight line)When Pn is large, M 2 + M 2 1

Mz=Ney

Mnx2 + Mny2 1 (i.e. a circle)

My=Nez

Mz Ney

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Exercise (Exam 2010)Exercise (Exam 2010)A corner column in a multi-storey building is to be designed with a cross-sectional shape as

shown in Fig. Q-1. At ultimate limit state, the column is to carry bi-axial bending moments, M and M of equal magnitude about two orthogonal axes y and z besides an axial load NMy and Mz, of equal magnitude, about two orthogonal axes, y and z, besides an axial load N. Determine the maximum value of My or Mz, if no tension is allowed in any part of the section. Assume fck = 30 MPa; fyk = 500 MPa; Es = 200 GPa; c = 1.5; s = 1.15. The section has eight 25 bars placed around the periphery, with a cover of 50 mm to each bar.

y

200

200

axis ofsymmetry

All

z200 All

dimensions are in mm.

200 200

200

Tan K H, NUSFig. Q-1

Further Reading

Nilson A H Darwin D & Dolan C W Design ofNilson, A.H., Darwin, D. & Dolan, C.W., Design of Concrete Structures, 14e in SI Units, Ch 3.

Wight, J.K. & MacGregor, J.G., Reinforced Concrete: g gMechanics & Design, 5 ed., Ch 4.

Martin, L. & Purkiss, J., Concrete Design to EN 1992, , Ch 6Ch 6.

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