Reinforced Concrete 2012 lecture 11/1

Budapest University of Technology and Economics

Department of Mechanics, Materials and Structures English courses Reinforced Concrete Structures Code: BMEEPSTK601 Lecture no. 11:

REINFORCED CONCRETE SECTIONS SUBJECTED TO AXIAL AND ECCENTRIC COMPRESSION

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Content: Introduction

1. Suppositions 2. Failure modes 3. The MR-NR capacity diagram of concrete sections 4. The MR-NR capacity diagram of reinforced concrete sections 5. Safe and unsafe applied force combinations 6. Applied force combinations to be investigated 7. Linearization of the MR-NR capacity diagram 8. Axial and eccentric tension 9. Use of a series of dimensionless capacity diagrams 10. MR-NR capacity diagram of nonsymmetric reinforced concrete

sections 11. The (MRz, MRy, NR) capacity body 12. Check of rc. sections subjected to compression with double

eccentricity

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b

hN

Rd

e

Introduction The Eurocode 2 does not distinguish axial and eccentric compression, because due to imperfection, the theoretically axial compression force does always have some eccentricity.

h

e

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1. Suppositions -plane sections remain plane -bound connectionbetween concrete and steel is perfect -the mechanical behavior of concrete and steel can be modelled by the idealized σ-ε relationships indicated below:

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b

h

ε =2 %oc2ε =3.5%ocu

ε =3.5%ocu

2%o

h

h

7

7

3

32 %o

NRd

(a) (b) (c) (d) (e)

e

2. Failure modes Failure modes of the rc. section subjected to axial or eccentric compression can be given by the ε-diagrams given below The deformation of the steel at rupture of the concrete extreme fibre is

s

yd1ss E

f=εε p , that is steel is in the elastic state: 700

560

cs −

ξ=σ (N/mm2)

For practical design the use of MR-NR capacity curves is recommended!

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3. The MR-NR capacity diagram of concrete sections For concrete sections the MR-NR capacity curve has the character indicated below. Because of zero tensile strength of the concrete, at NR=0, the capacity moment MR=0.

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4. The MR-NR capacity diagram of reinforced concrete sections Point 1: axial compression ,

uydscdcRd NfAfAN =+=

0MRd =

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Point 2: eccentric compression (xc=xco) cdcoRd fbxN = = Nbal

syd1sco

balRd zfA)2

x

2

h(NM +−= =

= =+∆ sR MM MRd,max

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Point 3: flexure NRd=0 ssyd1sRd MzfAM =≈

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5. Safe and unsafe applied force combinations unsafe force combination cases of small eccent- ricities safe force comb. great eccentricities

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6. Applied force combinations to be investigated Any of the below force combinations can be dangerous, should be investigated: 1. ))M(N,M( max,Edmin,Edmax,Ed

2. ))M(N,M( max,Edmax,Edmax,Ed

3. ))N(M,N( max,Edmax,Edmax,Ed

4. ))N(M,N( min,Edmax,Edmin,Ed

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7. Linearization of the MR-NR capacity diagram

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8. Axial and eccentric tension Part of the capacity diagram below the MR axis show capacity of the rc section in eccentric and axial tension For axial tension: yd2s1st,Rd f)AA(N +=

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= 0.8d1 h

µ = 0.0, 0.1, 0.2, ...., 1.0

µ = 0

µ = 1.0

n

n

m

0.5

0.5

0.0

1.0

1.5

2.0

-1.0

-0.5

0.40.30.20.10.0

9. Use of a series of dimensionless capacity diagrams

cd

yds

bhf

fA=µ

cd2Ed

fbh

Mm =

cd

Ed

bhf

Nn =

s1sis A2AA =∑=

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10. MR-NR capacity diagram of nonsymmetric reinforced concrete sections

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11. The (MRz, MRy, NR) capacity body

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12. Check of rc. sections subjected to compression with double eccentricity

The given force combination is safe!

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The given force combination is unsafe!

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13. Numerical example

The column section given below is subjected to an eccentric compression force NEd= 1620 kN, e= 27,7 mm. Check the section by determining the simplified (linearized) MR-NR capacity diagram of it! Materials: concrete: C20/25-32/KK steel: B60.50 Concrete cover 20 mm link diameter: φk = 8 mm

fcd = 5,1

20 = 13,3 N/mm2 fyd =

15,1

500 = 435 N/mm2

ξco = 0,49

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Solution: d = 350 - 20 - 8 - 8 = 314 mm Nu

' = bhfcd + As· 435 = (3502 · 13,3 + 1608 · 435) ⋅10-3 = = 1633 + 700,3 = 2333,3 kN Nbal = bξcodfcd = 350 ⋅0,49 ⋅314 ⋅13,3 ⋅10-3 = 716,2 kN

∆M = Nbal (2

h -

2

dcoξ) =716,2·(

2

350 -

2

31449,0 ⋅)·10-3 = 70,2 kNm

Ms= As1fydzs = 603 ·435 ·278⋅10-6 = 72,9 kNm

MRd,max = Ms + ∆M = 143,1 kNm MEd = NEde = 1620 ·0,0277 = 44,87 kNm

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The point (MEd, NEd) is inside the diagram: the cross-section is safe! Numerically: MRd(NEd)

kNmNN

NNMNM

balu

EduRdEdRd 12,63

2,7163,2333

16203,23331,143)(

,

,

max, =−−⋅=

−−= >

> MEd=44,87 kNm OK, safe!