file tinh cot 3

7/28/2019 File Tinh Cot 3

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1 Level:

Name : Reference level : 0.00 (m)

2 Column: Column3 Number: 1

2.1 Material properties:

Concrete : B15 R b = 8.50 (MPa)Type : heavyweightCuring method : normalConcreting in layers h>1.5 (m) : noHigh humidity/hydration : no

Longitudinal reinforcement: : A-II R s = 280.00 (MPa)

Transversal reinforcement: : A-I R s = 225.00 (MPa)

2.2 Geometry:

2.2.1 Rectangular 22.0 x 45.0 (cm)2.2.2 Height: L = 4.70 (m)2.2.3 Slab thickness = 0.00 (m)2.2.4 Beam height = 0.60 (m)2.2.5 Cover = 4.0 (cm)

2.3 Calculation options:

Calculations according to : SNiP 2.03.01-84 Region seismic type : none Pre-design : no Slenderness taken into account : yes stirrups: : to slab

2.4 Loads:

Case Nature Group f N My(s) My(i) Mz(s) Mz(i)

(daN) (kN*m) (kN*m) (kN*m)(kN*m)

COMB1design 3 1.00 76705.82 -23.71 12.57 0.000.00

COMB2design 3 1.00 77398.29 -34.21 17.94 0.000.00

COMB3design 3 1.00 70138.53 -76.69 82.45 0.000.00


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COMB4design 3 1.00 67007.00 24.01 -56.10 0.000.00

COMB5design 3 1.00 85525.86 -31.92 16.92 0.000.00

COMB6design 3 1.00 78110.07 -69.33 74.54 0.00

0.00COMB7design 3 1.00 75291.69 21.30 -50.16 0.000.00

COMB8design 3 1.00 78802.54 -79.83 79.91 0.000.00

COMB9design 3 1.00 75984.16 10.79 -44.79 0.000.00

COMB10 design 3 1.00 86930.10 -77.54 78.890.00 0.00

COMB11 design 3 1.00 84111.72 13.08 -45.810.00 0.00

f - load factor

2.5 Calculation results:

2.5.1 ULS Analysis

Design combination: COMB10 (B)Internal forces:

NSd = 86930.10 (daN) MSdy = 78.89(kN*m) MSdz = 0.00 (kN*m)

Design forces: Lower nodeNSd = 86930.10 (daN) NSd*etotz =78.89 (kN*m) NSd*etoty= 8.69 (kN*m)

2.5.1.1 Eccentricity:

Eccentricity: ez (My/N) ey (Mz/N)static ee: 9.1 (cm) 0.0 (cm)Not intended ea: 1.5 (cm) 1.0 (cm)Initial e0: 9.1 (cm) 1.0 (cm)total etot: 9.1 (cm) 1.0 (cm)

2.5.1.2 Detailed analysis -Direction Y:

2.5.1.2.1 Critical force (38)

Ncrit

= 6.4 * Eb / lo

2 * [Ib / l* (0.11/(0.1+ e)+0.1) + Eb/Es * Js ] =1270694.62 (daN)

lo = 3.08 (m)


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Eb = 23000.00 (MPa)Ib = 167062.5 (cm4)Es = 210000.00 (MPa)Is = 5842.6 (cm4) l =2.00

= 1.00Nd/N = 1.00

e = max (eo/h, 0.5 - 0.01 * lo /h - 0.01 * Rb ) = 0.36eo = 9.1 (cm)h = 45.0 (cm)Rb = 7.65 (MPa)

2.5.1.2.2 Slenderness analysis

Non-sway structurelcol

(m) lo

(m) lim4.40 3.08 23.71 14.00 Slender column

2.5.1.2.3 Buckling analysis

M1 = 78.89 (kN*m) M2 = -77.54 (kN*m)Case: Cross-section at the column end (Lower node),Slenderness not taken into accountMsd = 78.89 (kN*m)ee = Msd/Nsd = 9.1 (cm)ea = max (lcol/600, hy/30, 1.0cm) = 1.5 (cm)

lcol = 4.40 (m)hy = 45.0 (cm)

eo = max(ee , ea) = 9.1 (cm) (31)etot = *eo = 9.1 (cm) (36)

=1 (Slenderness not taken into account)

2.5.1.3 Detailed analysis-Direction Z:

M1 = 0.00 (kN*m)M2 = 0.00 (kN*m)Case: Cross-section at the column end (Lower node),Slenderness not taken into account

Msd = 0.00 (kN*m)ee = Msd/Nsd = 0.0 (cm)ea = max (lcol/600, hz/30, 1.0cm) = 1.0 (cm)

lcol = 4.40 (m)hz = 22.0 (cm)

eo = max(ee , ea) = 1.0 (cm) (31)etot = *eo = 1.0 (cm) (36)

=1 (Slenderness not taken into account)


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2.5.2 Reinforcement:

Reinforcing bars used in the section 25.0 (mm)Total number of bars in the section = 6

Number of bars (b side) = 2Number of bars (h side) = 3Real (provided) area Asr = 29.45 (cm2)Ratio: = Asr/Ac = 2.97 %

2.6 Reinforcement:

Main bars (A-II): 6 25 l = 4.66 (m)

Transversal reinforcement: (A-I):

stirrups: 21 6 l = 1.11 (m)

pins 21 6 l = 1.11 (m)

3 Material survey:

Concrete volume = 0.41 (m3) Formwork = 5.49 (m2)

Steel A-II Total weight = 107.78 (kG) Density = 265.53 (kG/m3) Average diameter = 25.0 (mm) Reinforcement survey:

Diameter Length Weight Number Total weight(m) (kG) (No.) (kG)

25 4.66 17.96 6 107.78

Steel A-I Total weight = 5.19 (kG) Density = 12.80 (kG/m3) Average diameter = 6.0 (mm) Reinforcement survey:

Diameter Length Weight Number Total weight


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(m) (kG) (No.) (kG)6 1.11 0.25 21 5.19

file tinh cot 3

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