CM01 04 [režim kompatibility]) - cvut.cz
Transcript of CM01 04 [režim kompatibility]) - cvut.cz
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
• 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
• 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