1st task:Frame structure

Design of reinforcement

• Design the reinforcement of the middle column in the 1st floor

• Geometric imperfections of column

• Slenderness of the column

• Reinforcement of the column

• We calculated moments on ideal model of frame structure, but real structures are not perfect

• Geometric imperfections cause additional bending moments

Geometric imperfections

ei

Model Real structure

Geometric imperfection

• Geometric imperfection

• Additional moment due to geom. imperfection

Geometric imperfections

2

00

le mhi

Basic value of imperfection,

1/200

Reduction factor for

height

2h

h

21,0

3h

.

Clear length of the column

10,5 1m

m

Reduction factor for number of

members

Number of columns in one frame, m = 3

Effective length of the column,

in our case l0 = 0,8h

imp Ed iM N e Normal force in given cross-section

• Calculate bending moments with the influence of geometric imperfections MEd,I in the head and foot of the column for both combinations

• Use table to calculate the values

• We will use these values to check the load-bearing capacity (in the interaction diagram)

Geometric imperfections

• Slenderness of the column

• Limiting slenderness

Slenderness of the column

0l

i

c

I

Ai

Radius of gyration

3

col col

1

12I b h

Moment of inertia

Dimensions of the cross-section of the column

Cross-sectional area of the column

lim

2075

ABC

n

Effect of creep, A = 0.7

Effect of reinforcement ratio, B = 1.1

Effect of bending moments

Relative normal force

Ed

c cd

Nn

A f

You can´t consider lim

to be more than 75

• Effect of bending moments

• Calculate lim for all combinations and use the worst case (lowest lim)

Slenderness of the column

01

02

1,7 m

m

C r

Mr

M

M01 and M02 are bending moments in the head and foot of

the column, |M02| > |M01|

Compare absolute values, but assign values WITH

the sign into the equation for C

Example of calculation of C factor

• BUT: If the bending moments are causedpredominantly by the imperfections, we shouldalways take C = 0.7

• This is the case of our COMB1 (there are NO moments from internal forces, just the momentscaused by imperfections) => Take C = 0.7

• If ≤ lim, the column is robust• If > lim, the column is slender

Slenderness of the column

• 1st method: Estimation with the presumption of uniformly distributed compression over the whole cross-section

• If the equation gives As,req,1 < 0, minimum

reinforcement 4ø 12 mm should be designed

Design of reinforcement

Ed c cds,req,1

s

0,8N A fA

s yd

s yd yd

400 MPa if 400 MPa

if 400 MPa

f

f f

Stress in reinforcement

• 2nd method: Chart for design of symmetricalreinforcement

• Try both combinations and use higher value ofAs,req

Design of reinforcement

Ed,I chartEd

2

col col cd col col cd

M N

b h f b h f

Relative bending moment Relative normal force Relative exploitation of concrete part of the cross-section

c cds,req,2

yd

A fA

f

Design of reinforcement

• Final design:

• Check detailing rules

Design of reinforcement

s,req s,req,1 s,req,2 s,prov s,reqmax ;A A A A A

Eds,prov s,min c

yd

max 0.1 ; 0.002N

A A Af

s,prov s,max c0,04A A A

For example Design: 4x Ø16 (As,prov = 804 mm2)

Check of column – interaction diagram

Check of column – interaction diagram

• Calculate main points of interaction diagram

• Connect them by lines (simplification)

• Calculate minimum bending moment M0

• Restrict axial resistance

• If your column is slender, increase bending moments by approximately 30 % (simplification)

• If COMB1 and COMB2 lay inside the curve, column is checked

• If not, we will adjust the design (DO NOT recalculate the ID)

• See the example on my website

• Maximum normal force resistance (pure compression)

• In our case, for all points As1 = As2 and zs1 = zs2

because we have symmetrical reinforcement

ID – point 0

ss1s1s2s2s1s1s2s2Rd,0

ss2ss1cdcolcols2s1cRd,0

zAzAzFzFM

AAfhbFFFN

See design of reinforcement

As1

As2

FOR ALL POINTS OF ID:

• To calculate normal force capacity – sum the internal forces

• To calculate bending moment capacity – sum the moments of these forces

• Whole cross-section is compressed (strain in tensile reinforcement is 0)

ID – point 1

2syd2scdcol2s2sccRd,1

yd2scdcol2ccRd,1

4.02

8.0

8.0

zfAdh

dfbzFzFM

fAdfbFFN

Factor expressing the difference between real and idealized stress distribution,

see 3rd HW

• Maximum bending moment resistance (stress in tensile reinforcement s1 = fyd; that means x = xbal,1)

ID – point 2

s

1syd1s2s2s2s1,balcd1,balcol1s1s2s2sccRd,2

1s2s2scd1,balcol1s2scRd,2

4.02

8.0

8.0

zfAzAxh

fxbzFzFzFM

fAAfxbFFFN yd

df

dxyd

1,bal1,bal700

700

?

• How to find s2 (stress in compressed reinforcement) ?

ID – point 2

2s2 cd

bal,1

1d

x

yd

s2 yd s2 yd

s

s2 s2 s

If

else

ff

E

E

Elastic modulus of steel, 210000 MPa

Distance from surface of the column to the centroid

of compressed reinforcement

Limit strain of concrete, cd = 0.0035

• Pure bending

ID – point 3

Rd,3 c s2 s1

Rd,3 c c s2 s2 s1 s1 col cd s2 s2 s2 s1 yd s1

0

0.8 0.42

N F F F

hM F z F z F z b xf x A z A f z

We have two unknowns: height of compressed part of concrete cross section (x) and stress in

compressed reinforcement (s2)

• To find the value of s2, we can derive quadratic equation:

• By solving this equation, we will receive 2 roots

• Only one of them will „make sense“ – we will use that one to calculate x:

ID – point 3

08.0 2cdcolyd1sscdscd2syd1s2s2s

2

2s dfbfAEEAfAA

cdcol

2s2syd1s

8.0 fb

AfAx

• Whole cross-section is in tension (strain in compressed reinforcement is 0)

ID – point 4

1syd1s1s1sRd,4

yd1s1sRd,4

zfAzFM

fAFN

• Pure tension

ID – point 5

yd2s2s1s1s2s2s1s1sRd,5

yd2s1s2s1sRd,5

fzAzAzFzFM

fAAFFN

• Always consider minimum eccentricity

• Minimum bending moment

• Restriction of ID – pure compression can never occur, minimum bending moment always has to be taken into account

ID – moment M0

col0 max ;20 mm

30

he

0 Rd,0 0M N e