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Table Of Contents Table Of Contents ....................................................................................................................................... 2
I. GENERAL INFORMATION & BASIC DATAS ................................................................................. 3
II. SPREAD FOOTING: FOUNDATION 1…134 ................................................................................... 4
III. SINGLE COLUMN FOR TANK HOLDER) .................................................................................. 10
IV. EXTERNAL COLUMN (COL 26, 176 and COL 206) ................................................................... 12
V.INTERNAL COLUMN (Ref: COL 65 AND COL 94 )...................................................................... 17
VI. LONGITUDINAL BEAM 142, 220 and 234 .................................................................................... 22
VII. TRANSVERSAL BEAM (BEAM 133, 212 and 226 ...................................................................... 33
VIII. SLAB: ............................................................................................................................................... 43
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I. GENERAL INFORMATION & BASIC DATAS
Client : ……………….
PC No: 3/03/04/1311
Uburezi village
Mbugangali Cell
Gisenyi Sector
Rubavu District
Engineer responsible :
Eng. Dieudonne
NTEZIYAREMYE
No A 298/EC/IER/2015
Autodesk robot structural analysis 2017
Geotechnic calculations according to : ACI
Concrete calculations according to : BS 8110
Relevant Building
Regulations and
Design Code
RESIDENTIAL BUILDING Intended use of the
building
Roof –Imposed 0.26 kN/m2
-Finishes 1.5 kN/m2
Floor –Imposed (0.48 )and partitions(1) 0.48+0.19 kN/m2
Stairs –Imposed 4 kN/m2
Loads: 1. self-weight; 147 241 ; PZ Negative 2. uniform load; PZ=-0.48(kN/m) 3. uniform load; PZ=-0.26(kN/m) 4. (FE) uniform; 147 241 ; PZ=-0.24(kN/m2) 5. self-weight; 147 241 ; PZ Negative 5. (FE) uniform; 147 241 ; PZ=-4.79(kN/m2) Combination/Component and Definition ULS/8; = (1+2+3+4+5)*1.20+6*1.60 SLS/9; = (1+2+3+4+5+6)*1.00
General loading
conditions
Severe(external )and Mild (internal) Exposure conditions
Variable according to the site conditions:
Average allowable bearing pressure = 234kN/ m2
Subsoil conditions
Reinforced Concrete footing to columns. Foundation type
Concrete: grade C 25 (with 20mm max. aggregates.
Mix ratio 400 kg/ m3
Reinforcement :-Characteristic strength:
ƒy = 460 N/mm
for stirrups ƒy = 250 N/mm2
Material data
Self weight of Reinforced concrete = 25kN/ m3
Self weight of masonry = 18.9kN/ m3
Other relevant
informations
For dead load : 1.4
For live load : 1.6
Partial safety factor
4
II. SPREAD FOOTING: FOUNDATION 1…134
1 Spread footing: Foundation1...134
1.1 Basic data
1.1.1 Assumptions
· Geotechnic calculations according to : ENV 1997-1:1994 · Concrete calculations according to : BS 8110 · Shape selection : without limits
1.1.2 Geometry:
A = 1.40 (m) a = 0.30 (m) B = 1.40 (m) b = 0.20 (m) h1 = 0.25 (m) ex = 0.00 (m)
h2 = 0.20 (m) ey = 0.00 (m)
h4 = 0.05 (m)
5
a' = 30.0 (cm) b' = 20.0 (cm) c1 = 5.0 (cm) c2 = 5.0 (cm)
1.1.3 Materials
· Concrete : C20; Characteristic strength = 20.00 MPa Unit weight = 2501.36 (kG/m3) · Longitudinal reinforcement : type T Characteristic strength = 460.00
MPa · Transversal reinforcement : type R Characteristic strength = 250.00
MPa · Additional reinforcement: : type T Characteristic strength = 460.00
MPa
1.1.4 Loads: Foundation loads: Case Nature Group N Fx Fy Mx My (kN) (kN) (kN) (kN*m) (kN*m) COMB1 design ---- 155.99 -2.86 0.67 -0.00 0.00 COMB2 design ---- 155.99 -2.86 0.67 -0.00 0.00 COMB1 design ---- 296.40 1.08 1.38 -0.00 -0.00 COMB2 design ---- 296.40 1.08 1.38 -0.00 -0.00 COMB1 design ---- 190.49 0.49 1.20 -0.00 -0.00 COMB2 design ---- 190.49 0.49 1.20 -0.00 -0.00 COMB1 design ---- 49.08 2.08 0.90 -0.00 -0.00 COMB2 design ---- 49.08 2.08 0.90 -0.00 -0.00 COMB1 design ---- 328.33 -4.74 -0.66 -0.00 0.00 COMB2 design ---- 328.33 -4.74 -0.66 -0.00 0.00 COMB1 design ---- 650.22 1.49 -0.66 -0.00 -0.00 COMB2 design ---- 650.22 1.49 -0.66 -0.00 -0.00 COMB1 design ---- 404.37 1.20 -0.96 -0.00 -0.00 COMB2 design ---- 404.37 1.20 -0.96 -0.00 -0.00 COMB1 design ---- 82.07 2.99 -1.55 -0.00 -0.00 COMB2 design ---- 82.07 2.99 -1.55 -0.00 -0.00 COMB1 design ---- 303.24 -4.51 -0.78 -0.00 0.00
Backfill loads: Case Nature Q1 (kN/m2)
1.1.5 Combination list 1/ ULS A : 1.00DL1 2/ ULS A : 0.95DL1 3/ ULS B : 1.35DL1 4/ ULS B : 1.00DL1 5/ ULS C : 1.00DL1 6/ ULS : COMB1 N=155.99 Fx=-2.86 Fy=0.67 7/ ULS : COMB1 N=296.40 Fx=1.08 Fy=1.38 8/ ULS : COMB1 N=190.49 Fx=0.49 Fy=1.20 9/ ULS : COMB1 N=49.08 Fx=2.08 Fy=0.90 10/ ULS : COMB1 N=328.33 Fx=-4.74 Fy=-0.66 11/ ULS : COMB1 N=650.22 Fx=1.49 Fy=-0.66
6
1.2 Geotechnical design
1.2.1 Assumptions
Soil cohesion reduction factor: 1.00
Sliding with soil pressure considered: none
Conditions without drainage
Partial factors for soil properties:
tan() c' cu qmax Case A 1.10 1.30 1.20 1.20 Case B 1.00 1.00 1.00 1.00 Case C 1.25 1.60 1.40 1.40 ALS 1.00 1.00 1.00 1.00
1.2.2 Soil:
Rubavu soil • Soil level: 0.00 (m) • Unit weight: 2243.38 (kG/m3) • Unit weight of solid: 2702.25 (kG/m3) • Internal friction angle: 42.0 (Deg) • Cohesion: 1.16 (MPa)
1.2.3 Limit states
Stress calculations Soil type under foundation: not layered Design combination ULS : COMB1 N=675.58 Fx=1.76 Fy=1.27 Load factors: 1.00 * Foundation weight 1.00 * Soil weight Calculation results: On the foundation level Weight of foundation and soil over it: Gr = 20.65 (kN) Design load: Nr = 696.23 (kN) Mx = -0.57 (kN*m) My = 0.79 (kN*m) Soil profile parameters: C = 1.16 (MPa) Stress in soil: 0.36 (MPa) Design soil pressure 1.16 (MPa) Safety factor: 3.2 > 1 Uplift Uplift in ULS Design combination ULS : COMB1 N=82.07 Fx=2.99 Fy=-1.55 Load factors: 0.95 * Foundation weight 0.95 * Soil weight Contact area: s = 0.01 slim = 0.17
Sliding Design combination ULS : COMB1 N=49.08 Fx=2.08 Fy=0.90
7
Load factors: 0.95 * Foundation weight 0.95 * Soil weight Weight of foundation and soil over it: Gr = 19.64 (kN) Design load: Nr = 68.72 (kN) Mx = -0.41 (kN*m) My = 0.94 (kN*m) Equivalent foundation dimensions: A_ = 1.40 (m) B_ = 1.40 (m) Sliding area: 1.96 (m2)
Foundation/soil friction coefficient: tg( = 0.82 Cohesion: C = 1.5 (MPa) Sliding force value F = 2.27 (kN) Value of force preventing foundation sliding: - On the foundation level: F(stab) = 0.00 (kN) Stabilility for sliding: 1.5 > 1 Average settlement Soil type under foundation: not layered Design combination SLS : COMB2 N=675.58 Fx=1.76 Fy=1.27 Load factors: 1.00 * Foundation weight 1.00 * Soil weight Weight of foundation and soil over it: Gr = 20.67 (kN) Average stress caused by design load: q = 0.36 (MPa) Thickness of the actively settling soil: z = 4.20 (m) Stress on the level z:
- Additional: zd = 0.02 (MPa)
- Caused by soil weight: z = 0.10 (MPa) Settlement: - Original s' = 0.2 (cm) - Secondary s'' = 0.0 (cm) - TOTAL S = 0.2 (cm) < Sadm = 5.1 (cm) Safety factor: 26.72 > 1 Settlement difference Design combination SLS : COMB2 N=341.61 Fx=-5.12 Fy=0.92 Load factors: 1.00 * Foundation weight 1.00 * Soil weight Settlement difference: S = 0.0 (cm) < Sadm = 5.1 (cm) Safety factor: 950.9 > 1 Rotation About OX axis Design combination ULS : COMB1 N=82.07 Fx=2.99 Fy=-1.55 Load factors: 0.95 * Foundation weight 0.95 * Soil weight Weight of foundation and soil over it: Gr = 19.64 (kN) Design load: Nr = 101.71 (kN) Mx = 0.70 (kN*m) My = 1.34 (kN*m) Stability moment: Mstab = 71.20 (kN*m)
Rotation moment: Mrenv = 0.70 (kN*m)
Stability for rotation: 102.3 > 1 About OY axis Design combination: ULS : COMB1 N=49.08 Fx=2.08 Fy=0.90
8
Load factors: 0.95 * Foundation weight 0.95 * Soil weight Weight of foundation and soil over it: Gr = 19.64 (kN) Design load: Nr = 68.72 (kN) Mx = -0.41 (kN*m) My = 0.94 (kN*m) Stability moment: Mstab = 48.10 (kN*m)
Rotation moment: Mrenv = 0.94 (kN*m)
Stability for rotation: 51.31 > 1
1.3 RC design
1.3.1 Assumptions
Exposure : mild
1.3.2 Analysis of punching and shear Shear Design combination ULS : COMB1 N=675.58 Fx=1.76 Fy=1.27 Load factors: 1.00 * Foundation weight 1.00 * Soil weight Design load: Nr = 696.25 (kN) Mx = -0.57 (kN*m) My = 0.79 (kN*m) Length of critical circumference: 1.40 (m) Shear force: 198.28 (kN) Section effective height heff = 0.19 (m) Shear area: A = 0.27 (m2)
Reinforcement ratio: = 0.77 % Shear stress: 0.75 (MPa) Admissible shear stress: 0.70 (MPa) Safety factor: 0.9344 > 1
1.3.3 Required reinforcement
Spread footing: bottom: SLS : COMB2 N=675.58 Fx=1.76 Fy=1.27 My = 73.26 (kN*m) Asx = 11.38 (cm2/m)
SLS : COMB2 N=675.58 Fx=1.76 Fy=1.27 Mx = 87.09 (kN*m) Asy = 14.55 (cm2/m)
As min = 3.25 (cm2/m)
top: SLS : COMB2 N=675.58 Fx=1.76 Fy=1.27 My = 73.26 (kN*m) Asx = 11.38 (cm2/m)
SLS : COMB2 N=675.58 Fx=1.76 Fy=1.27 Mx = 87.09 (kN*m) Asy = 14.55 (cm2/m)
9
As min = 3.25 (cm2/m)
Column pier: Longitudinal reinforcement A = 9.50 (cm2) A min. = 2.40 (cm2)
A = 2 * (Asx + Asy) Asx = 1.86 (cm2) Asy = 2.89 (cm2)
1.3.4 Provided reinforcement 2.3.1 Spread footing: Bottom: Along X axis: 15 T 16 l = 1.36 (m) e = 1*-0.65 + 14*0.09 Along Y axis: 15 T 16 l = 1.36 (m) e = 1*-0.65 + 14*0.09 Top: Along X axis: 15 T 16 l = 1.36 (m) e = 1*-0.65 + 14*0.09 Along Y axis: 15 T 16 l = 1.36 (m) e = 1*-0.65 + 14*0.09
2.3.2 Pier Longitudinal reinforcement Dowels Longitudinal reinforcement
8 T 16 l = 1.17 (m) e = 1*-0.06 + 1*0.00 + 2*0.06 + 1*0.00
2 Material survey:
Concrete volume = 0.50 (m3)
Formwork = 1.60 (m2)
Steel T
Total weight = 143.61 (kG)
Density = 286.07 (kG/m3)
Average diameter = 16.0 (mm)
Survey according to diameters:
Diameter Length Number: (m) 16 1.17 8 16 1.36 60
Steel R
Total weight = 1.09 (kG)
Density = 2.18 (kG/m3)
Average diameter = 8.0 (mm)
10
Survey according to diameters:
Diameter Length Number: (m) 8 0.92 3
III. SINGLE COLUMN FOR TANK HOLDER)
3.1 Column: Column4..8
3.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
3.2 Calculation results:
3.2.1 ULS Analysis Design combination: COMB1 (A) Internal forces: NSd = 82.07 (kN) MSdy = -9.25 (kN*m) MSdz = 4.80 (kN*m) Design forces: Upper node NSd = 82.07 (kN) NSd*etotz = -10.24 (kN*m) NSd*etoty= 4.80 (kN*m)
11
3.2.1.1 Eccentricity: Eccentricity: ez (My/N) ey (Mz/N) Static ee: -11.3 (cm) 5.8 (cm) II order eadd: -1.2 (cm) 1.2 (cm) Minimal emin: -1.0 (cm) 1.0 (cm) Total etot: -12.5 (cm) 5.8 (cm)
3.2.1.2 Detailed analysis-Direction Y:
3.2.1.2.1 Slenderness analysis Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 15.50 > 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Slender column
3.2.1.2.2 Buckling analysis
M2 = 0.00 (kN*m) M1 = -9.25 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness taken into account M = -9.25 (kN*m) emin = max (20mm ; 0.05 *hy) = -1.0 (cm) (3.8.2.4) hy = 20.0 (cm) Mmin = N*emin = -0.82 (kN*m)
au/2 = a*K*h/2 = -1.2 (cm) (32)
a = 0.12 (34) K = min((Nuz-N)/(Nuz-Nbal) ; 1) = 1.00 (33)
Nuz = 2/3*fcu/c*Ac+fy/s*Asc = 642.63 (kN) Ac = 0.04 (m2) Asc = 4.52 (cm2) Nbal = 165.82 (kN) h = 20.0 (cm) Madd/2 = au/2 * N = -0.99 (kN*m) (35) Md = max (Mmin;M+Madd/2) = -10.24 (kN*m)
3.2.1.3 Detailed analysis-Direction Z:
M2 = 4.80 (kN*m) M1 = 0.00 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = 4.80 (kN*m) emin = max (20mm ; 0.05 *hz) = 1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = 0.82 (kN*m) Md = max (Mmin;M) = 4.80 (kN*m)
3.2.2 Reinforcement:
Real (provided) area Asr = 4.52 (cm2)
Ratio: = 1.13 %
3.3 Reinforcement: Main bars:
4 T 12 l = 3.22 (m) Transversal reinforcement:
stirrups: 20 R 8 l = 0.61 (m)
12
IV. EXTERNAL COLUMN (COL 26, 176 and COL 206)
4 Column: Column26
4.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : fy = 460.00 (MPa)
Transversal reinforcement : fy = 250.00 (MPa)
4.2 Calculation results:
2.2.1 ULS Analysis Design combination: COMB1 (B) Internal forces: NSd = 582.60 (kN) MSdy = -3.86 (kN*m) MSdz = -2.84 (kN*m) Design forces: Lower node NSd = 582.60 (kN) NSd*etotz = -3.86 (kN*m) NSd*etoty= -5.83 (kN*m)
2.2.1.1 Eccentricity: Eccentricity: ez (My/N) ey (Mz/N) Static ee: -0.7 (cm) -0.5 (cm) II order eadd: 0.0 (cm) 0.0 (cm) Minimal emin: 0.0 (cm) -1.0 (cm) Total etot: -0.7 (cm) -1.0 (cm)
13
4.2.1.2 Detailed analysis-Direction Y:
4.2.1.2.1 Slenderness analysis Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 10.33 < 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Short column
4.2.1.2.2 Buckling analysis
M2 = 0.00 (kN*m) M1 = -3.86 (kN*m) Case: Cross-section at the column end (Lower node), Slenderness not taken into account M = -3.86 (kN*m) emin = 0 Mmin = N*emin = 0.00 (kN*m) Md = max (Mmin;M) = -3.86 (kN*m)
4.2.1.3 Detailed analysis-Direction Z:
M2 = 0.00 (kN*m) M1 = -2.84 (kN*m) Case: Cross-section at the column end (Lower node), Slenderness not taken into account M = -2.84 (kN*m) emin = max (20mm ; 0.05 *hz) = -1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = -5.83 (kN*m) Md = max (Mmin;M) = -5.83 (kN*m)
4.2.2 Reinforcement:
Real (provided) area Asr = 9.24 (cm2)
Ratio: = 1.54 %
4.3 Reinforcement: Main bars:
6 14 l = 3.22 (m) Transversal reinforcement:
stirrups: 19 8 l = 0.81 (m)
4.1 Column: Column176
4.1.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
4.1.2 Calculation results:
14
4.1.2.1 ULS Analysis
Design combination: COMB1 (B) Internal forces: NSd = 311.35 (kN) MSdy = -7.90 (kN*m) MSdz = -5.22 (kN*m) Design forces: Lower node NSd = 311.35 (kN) NSd*etotz = -7.90 (kN*m) NSd*etoty= -5.22 (kN*m)
4.1.2.1.1 Eccentricity:
Eccentricity: ez (My/N) ey (Mz/N) Static ee: -2.5 (cm) -1.7 (cm) II order eadd: 0.0 (cm) 0.0 (cm) Minimal emin: -1.5 (cm) -1.0 (cm) Total etot: -2.5 (cm) -1.7 (cm)
4.1.2.1.2 Detailed analysis-Direction Y:
4.1.2.1.2.1 Slenderness analysis
Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 10.33 < 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Short column
4.1.2.1.2.2 Buckling analysis
M2 = 7.76 (kN*m) M1 = -7.90 (kN*m) Case: Cross-section at the column end (Lower node), Slenderness not taken into account M = -7.90 (kN*m) emin = max (20mm ; 0.05 *hy) = -1.5 (cm) (3.8.2.4) hy = 30.0 (cm) Mmin = N*emin = -4.67 (kN*m) Md = max (Mmin;M) = -7.90 (kN*m)
4.1.2.1.3 Detailed analysis-Direction Z:
M2 = 5.12 (kN*m) M1 = -5.22 (kN*m) Case: Cross-section at the column end (Lower node), Slenderness not taken into account M = -5.22 (kN*m) emin = max (20mm ; 0.05 *hz) = -1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = -3.11 (kN*m) Md = max (Mmin;M) = -5.22 (kN*m)
4.1.2.2 Reinforcement: Real (provided) area Asr = 9.24 (cm2)
Ratio: = 1.54 %
4.1.3 Reinforcement: Main bars:
6 T 14 l = 3.07 (m)
15
Transversal reinforcement:
stirrups: 18 R 8 l = 0.81 (m)
4.2 Column: Column206
4.2.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
4.2.2 Calculation results:
4.2.2.1 ULS Analysis
Design combination: COMB1 (A) Internal forces: NSd = 40.62 (kN) MSdy = 4.28 (kN*m) MSdz = 3.84 (kN*m) Design forces: Upper node NSd = 40.62 (kN) NSd*etotz = 4.28 (kN*m) NSd*etoty= 3.84 (kN*m)
4.2.2.1.1 Eccentricity:
Eccentricity: ez (My/N) ey (Mz/N) Static ee: 10.5 (cm) 9.5 (cm) II order eadd: 0.0 (cm) 0.0 (cm) Minimal emin: 1.5 (cm) 1.0 (cm) Total etot: 10.5 (cm) 9.5 (cm)
4.2.2.1.2 Detailed analysis-Direction Y:
4.2.2.1.2.1 Slenderness analysis
Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 10.33 < 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Short column
4.2.2.1.2.2 Buckling analysis
M2 = 4.28 (kN*m) M1 = -1.17 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = 4.28 (kN*m) emin = max (20mm ; 0.05 *hy) = 1.5 (cm) (3.8.2.4) hy = 30.0 (cm) Mmin = N*emin = 0.61 (kN*m) Md = max (Mmin;M) = 4.28 (kN*m)
4.2.2.1.3 Detailed analysis-Direction Z:
16
M2 = 3.84 (kN*m) M1 = -1.23 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = 3.84 (kN*m) emin = max (20mm ; 0.05 *hz) = 1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = 0.41 (kN*m) Md = max (Mmin;M) = 3.84 (kN*m)
4.2.2.2 Reinforcement: Real (provided) area Asr = 9.24 (cm2)
Ratio: = 1.54 %
4.2.3 Reinforcement: Main bars:
6 T 14 l = 3.07 (m) Transversal reinforcement:
stirrups: 19 R 6 l = 0.79 (m)
17
V.INTERNAL COLUMN (Ref: COL 65 AND COL 94 )
5 Column: Column22 Number: 1
5.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
5.2 Calculation results:
5.2.1 ULS Analysis
Design combination: COMB1 (B) Internal forces: NSd = 461.63 (kN) MSdy = -1.32 (kN*m) MSdz = 2.54 (kN*m) Design forces: Lower node NSd = 461.63 (kN) NSd*etotz = -1.32 (kN*m) NSd*etoty= 4.62 (kN*m)
5.2.1.1 Eccentricity:
Eccentricity: ez (My/N) ey (Mz/N) Static ee: -0.3 (cm) 0.6 (cm) II order eadd: 0.0 (cm) 0.0 (cm) Minimal emin: 0.0 (cm) 1.0 (cm) Total etot: -0.3 (cm) 1.0 (cm)
5.2.1.2 Detailed analysis-Direction Y:
18
5.2.1.2.1 Slenderness analysis Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 10.33 < 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Short column
5.2.1.2.2 Buckling analysis
M2 = 0.00 (kN*m) M1 = -1.32 (kN*m) Case: Cross-section at the column end (Lower node), Slenderness not taken into account M = -1.32 (kN*m) emin = 0 Mmin = N*emin = 0.00 (kN*m) Md = max (Mmin;M) = -1.32 (kN*m)
5.2.1.3 Detailed analysis-Direction Z:
M2 = 2.54 (kN*m) M1 = 0.00 (kN*m) Case: Cross-section at the column end (Lower node), Slenderness not taken into account M = 2.54 (kN*m) emin = max (20mm ; 0.05 *hz) = 1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = 4.62 (kN*m) Md = max (Mmin;M) = 4.62 (kN*m)
5.2.2 Reinforcement:
Real (provided) area Asr = 9.24 (cm2)
Ratio: = 1.54 %
5.3 Reinforcement: Main bars:
6 T 14 l = 3.22 (m) Transversal reinforcement:
stirrups: 19 R 8 l = 0.81 (m)
5.1 Column: Column172 Number: 1
5.1.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
5.1.2 Calculation results:
5.1.2.1 ULS Analysis
19
Design combination: COMB1 (A) Internal forces: NSd = 251.17 (kN) MSdy = 2.87 (kN*m) MSdz = -5.30 (kN*m) Design forces: Upper node NSd = 251.17 (kN) NSd*etotz = 3.77 (kN*m) NSd*etoty= -5.30 (kN*m)
5.1.2.1.1 Eccentricity:
Eccentricity: ez (My/N) ey (Mz/N) Static ee: 1.1 (cm) -2.1 (cm) II order eadd: 0.0 (cm) 0.0 (cm) Minimal emin: 1.5 (cm) -1.0 (cm) Total etot: 1.5 (cm) -2.1 (cm)
5.1.2.1.2 Detailed analysis-Direction Y:
5.1.2.1.2.1 Slenderness analysis
Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 10.33 < 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Short column
5.1.2.1.2.2 Buckling analysis
M2 = 2.87 (kN*m) M1 = -2.84 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = 2.87 (kN*m) emin = max (20mm ; 0.05 *hy) = 1.5 (cm) (3.8.2.4) hy = 30.0 (cm) Mmin = N*emin = 3.77 (kN*m) Md = max (Mmin;M) = 3.77 (kN*m)
5.1.2.1.3 Detailed analysis-Direction Z:
M2 = 5.22 (kN*m) M1 = -5.30 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = -5.30 (kN*m) emin = max (20mm ; 0.05 *hz) = -1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = -2.51 (kN*m) Md = max (Mmin;M) = -5.30 (kN*m)
5.1.2.2 Reinforcement: Real (provided) area Asr = 9.24 (cm2)
Ratio: = 1.54 %
5.1.3 Reinforcement: Main bars:
6 T 14 l = 3.07 (m) Transversal reinforcement:
stirrups: 18 R 8 l = 0.81 (m)
20
5.2 Column: Column202 Number: 1
5.2.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
5.2.2 Calculation results:
5.2.2.1 ULS Analysis
Design combination: COMB1 (A) Internal forces: NSd = 38.99 (kN) MSdy = 0.52 (kN*m) MSdz = -2.63 (kN*m) Design forces: Upper node NSd = 38.99 (kN) NSd*etotz = 0.58 (kN*m) NSd*etoty= -2.63 (kN*m)
5.2.2.1.1 Eccentricity:
Eccentricity: ez (My/N) ey (Mz/N) Static ee: 1.3 (cm) -6.8 (cm) II order eadd: 0.0 (cm) 0.0 (cm) Minimal emin: 1.5 (cm) -1.0 (cm) Total etot: 1.5 (cm) -6.8 (cm)
5.2.2.1.2 Detailed analysis-Direction Y:
5.2.2.1.2.1 Slenderness analysis
Non-sway structure
lo (m) le (m)
3.10 1.00 3.10
ley/h = 10.33 < 15.00 (10-7) lez/b = 15.50 > 15.00 (10-7) Short column
5.2.2.1.2.2 Buckling analysis
M2 = 0.52 (kN*m) M1 = -0.59 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = 0.52 (kN*m) emin = max (20mm ; 0.05 *hy) = 1.5 (cm) (3.8.2.4) hy = 30.0 (cm) Mmin = N*emin = 0.58 (kN*m) Md = max (Mmin;M) = 0.58 (kN*m)
5.2.2.1.3 Detailed analysis-Direction Z:
M2 = 2.07 (kN*m) M1 = -2.63 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = -2.63 (kN*m)
21
emin = max (20mm ; 0.05 *hz) = -1.0 (cm) (3.8.2.4) hz = 20.0 (cm) Mmin = N*emin = -0.39 (kN*m) Md = max (Mmin;M) = -2.63 (kN*m)
5.2.2.2 Reinforcement: Real (provided) area Asr = 9.24 (cm2)
Ratio: = 1.54 %
5.2.3 Reinforcement: Main bars:
6 T 14 l = 3.07 (m) Transversal reinforcement:
stirrups: 19 R 8 l = 0.81 (m)
22
VI. LONGITUDINAL BEAM 142, 220 and 234
6 Beam: Beam142
6.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
Additional reinforcement: : T fy = 460.00 (MPa)
6.2 Calculation results:
6.2.1 Internal forces in ULS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 16.54 -0.00 -15.02 -25.72 33.94 -37.97 P2 8.05 -0.00 -18.56 -17.19 28.37 -28.96 P3 9.08 -0.00 -16.10 -18.55 27.55 -29.55 P4 8.58 -0.00 -15.87 -20.00 27.59 -31.76 P5 12.04 -0.00 -20.46 -18.42 32.40 -31.63 P6 4.94 -1.57 -8.49 -4.32 14.40 -11.35
23
6.2.2 Internal forces in SLS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 16.54 0.00 -15.02 -25.72 33.94 -37.97
P2 8.05 0.00 -18.56 -17.19 28.37 -28.96
P3 9.08 0.00 -16.10 -18.55 27.55 -29.55
P4 8.58 0.00 -15.87 -20.00 27.59 -31.76
P5 12.04 0.00 -20.46 -18.42 32.40 -31.63
P6 4.94 -1.57 -8.49 -4.32 14.40 -11.35
2. Required reinforcement area
0 5 10 15 2040
30
20
10
0
-10
-20
-30
-40
-50
-60
[m]
[kN*m]
Bending Moment ULS: M Mr Mc
0 5 10 15 20-80
-60
-40
-20
0
20
40
60
80
[m]
[kN]
Shear Force ULS: V Vr Vc(stirrups) Vc(total)
0 5 10 15 2020
15
10
5
0
-5
-10
-15
-20
-25
-30
[m]
[kN*m]
Bending Moment SLS: M Mr Md
0 5 10 15 20-40
-30
-20
-10
0
10
20
30
40
[m]
[kN]
Shear Force SLS: V Vr Vd
24
2. Deflection and cracking
6. Reinforcement: 2..1 P1 : Span from 0.30 to 4.70 (m)
Longitudinal reinforcement: bottom
2 T16 l = 4.13 from 0.03 to 4.16 2 T16 l = 2.63 from 3.48 to 6.12
assembling (top)
2 T16 l = 4.16 from 0.42 to 4.58
support (T)
2 T16 l = 1.42 from 0.03 to 1.36
Transversal reinforcement: main
stirrups 25 R8 l = 0.81 e = 1*0.04 + 24*0.18 (m)
.2 P2 : Span from 5.00 to 8.90 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.87 from 5.44 to 10.32
assembling (top)
2 T16 l = 3.44 from 5.44 to 8.88
0 5 10 15 2010
8
6
4
2
0
2
4
6
8
10
[m]
[cm2]
Reinforcement Area for Bending: Abt Abr Abmin Ades Aver_gross
0 5 10 15 206
4
2
0
2
4
6
[m]
[cm2/m]
Reinforcement Area for Shear: Ast Asr AsHang
0 5 10 15 202
1.5
1
0.5
0
-0.5
-1
-1.5
-2
[m]
[cm]
Deflections: ap(s-t) ap(l-t) at(s-t) a a adm
0 5 10 15 200.3
0.2
0.1
0
0.1
0.2
0.3
[m]
[mm]
Cracking: Cw Cw adm
25
support (T)
2 T16 l = 2.74 from 3.64 to 6.38 2 T16 l = 2.22 from 7.94 to 10.16
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.3 P3 : Span from 9.20 to 13.10 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.01 from 9.64 to 12.66
assembling (top)
2 T16 l = 3.86 from 9.22 to 13.08
support (T)
2 T16 l = 2.22 from 12.14 to 14.36
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.4 P4 : Span from 13.40 to 17.30 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.87 from 11.98 to 16.86
assembling (top)
2 T16 l = 3.86 from 13.42 to 17.28
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.5 P5 : Span from 17.60 to 21.90 (m) Longitudinal reinforcement: bottom
2 T16 l = 5.19 from 16.18 to 21.38
assembling (top)
2 T16 l = 4.10 from 17.70 to 21.80
support (T)
2 T16 l = 2.30 from 16.34 to 18.64
Transversal reinforcement: main
stirrups 24 R8 l = 0.81 e = 1*0.08 + 23*0.18 (m)
.6 P6 : Span from 22.20 to 24.10 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.67 from 20.70 to 24.37
support (T)
2 T16 l = 3.51 from 20.86 to 24.37
Transversal reinforcement: main
stirrups 11 R8 l = 0.81 e = 1*0.05 + 10*0.18 (m)
6.1 Beam: Beam220
6.1.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
26
Transversal reinforcement : R fy = 250.00 (MPa)
Additional reinforcement: : T fy = 460.00 (MPa)
6.1.2 Calculation results:
6.1.2.1 Internal forces in ULS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 16.36 -0.00 -16.77 -25.20 35.58 -37.68 P2 8.18 -0.00 -18.24 -17.45 28.43 -29.29 P3 9.03 -0.00 -16.35 -18.49 27.71 -29.48 P4 8.67 -0.00 -16.21 -19.69 28.07 -31.52 P5 12.24 -0.00 -20.66 -18.72 32.74 -32.29 P6 5.26 -0.87 -7.21 -5.34 13.28 -12.76
0 5 10 15 2040
30
20
10
0
-10
-20
-30
-40
-50
-60
[m]
[kN*m]
Bending Moment ULS: M Mr Mc
0 5 10 15 20-80
-60
-40
-20
0
20
40
60
80
[m]
[kN]
Shear Force ULS: V Vr Vc(stirrups) Vc(total)
27
6.1.2.2 Internal forces in SLS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 16.36 0.00 -16.77 -25.20 35.58 -37.68
P2 8.18 0.00 -18.24 -17.45 28.43 -29.29
P3 9.03 0.00 -16.35 -18.49 27.71 -29.48
P4 8.67 0.00 -16.21 -19.69 28.07 -31.52
P5 12.24 0.00 -20.66 -18.72 32.74 -32.29
P6 5.26 -0.87 -7.21 -5.34 13.28 -12.76
6.1. Required reinforcement area
0 5 10 15 2020
15
10
5
0
-5
-10
-15
-20
-25
-30
[m]
[kN*m]
Bending Moment SLS: M Mr Md
0 5 10 15 20-40
-30
-20
-10
0
10
20
30
40
[m]
[kN]
Shear Force SLS: V Vr Vd
0 5 10 15 2010
8
6
4
2
0
2
4
6
8
10
[m]
[cm2]
Reinforcement Area for Bending: Abt Abr Abmin Ades Aver_gross
0 5 10 15 206
4
2
0
2
4
6
[m]
[cm2/m]
Reinforcement Area for Shear: Ast Asr AsHang
28
6.1.2. Deflection and cracking
6.1.2. Reinforcement:
2..1 P1 : Span from 0.30 to 4.70 (m)
Longitudinal reinforcement: bottom
2 T16 l = 4.13 from 0.03 to 4.16 2 T16 l = 2.63 from 3.48 to 6.12
assembling (top)
2 T16 l = 4.16 from 0.42 to 4.58
support (T)
2 T16 l = 1.42 from 0.03 to 1.36
Transversal reinforcement: main
stirrups 25 R8 l = 0.81 e = 1*0.04 + 24*0.18 (m)
.2 P2 : Span from 5.00 to 8.90 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.87 from 5.44 to 10.32
assembling (top)
2 T16 l = 3.44 from 5.44 to 8.88
support (T)
2 T16 l = 2.74 from 3.64 to 6.38 2 T16 l = 2.22 from 7.94 to 10.16
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.3 P3 : Span from 9.20 to 13.10 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.01 from 9.64 to 12.66
assembling (top)
2 T16 l = 3.86 from 9.22 to 13.08
support (T)
2 T16 l = 2.22 from 12.14 to 14.36
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
0 5 10 15 202
1.5
1
0.5
0
-0.5
-1
-1.5
-2
[m]
[cm]
Deflections: ap(s-t) ap(l-t) at(s-t) a a adm
0 5 10 15 200.3
0.2
0.1
0
0.1
0.2
0.3
[m]
[mm]
Cracking: Cw Cw adm
29
.4 P4 : Span from 13.40 to 17.30 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.87 from 11.98 to 16.86
assembling (top)
2 T16 l = 3.86 from 13.42 to 17.28
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.5 P5 : Span from 17.60 to 21.90 (m) Longitudinal reinforcement: bottom
2 T16 l = 5.19 from 16.18 to 21.38
assembling (top)
2 T16 l = 4.10 from 17.70 to 21.80
support (T)
2 T16 l = 2.30 from 16.34 to 18.64
Transversal reinforcement: main
stirrups 24 R8 l = 0.81 e = 1*0.08 + 23*0.18 (m)
.6 P6 : Span from 22.20 to 24.10 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.67 from 20.70 to 24.37
support (T)
2 T16 l = 3.51 from 20.86 to 24.37
Transversal reinforcement: main
stirrups 11 R8 l = 0.81 e = 1*0.05 + 10*0.18 (m)
6.2 Beam: Beam234 Number: 1
6.2.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
Additional reinforcement: : T fy = 460.00 (MPa)
30
6.2.2 Calculation results:
6.2.1 Internal forces in ULS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 3.28 -0.00 -6.84 -4.61 8.61 -7.60 P2 3.02 -0.00 -2.59 -5.52 6.43 -7.93 P3 2.57 -0.00 -4.25 -4.77 7.05 -7.32 P4 2.73 -0.00 -4.21 -4.50 7.11 -7.26 P5 3.78 -0.00 -4.79 -4.85 7.90 -7.93 P6 2.56 -1.64 2.56 -4.99 -0.50 -7.47
6.2.2 Internal forces in SLS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 3.28 0.00 -6.84 -4.61 8.61 -7.60
P2 3.02 0.00 -2.59 -5.52 6.43 -7.93
P3 2.57 0.00 -4.25 -4.77 7.05 -7.32
P4 2.73 0.00 -4.21 -4.50 7.11 -7.26
P5 3.78 0.00 -4.79 -4.85 7.90 -7.93
P6 2.56 -1.64 2.56 -4.99 -0.50 -7.47
0 5 10 15 2040
30
20
10
0
-10
-20
-30
-40
-50
-60
[m]
[kN*m]
Bending Moment ULS: M Mr Mc
0 5 10 15 20-80
-60
-40
-20
0
20
40
60
80
[m]
[kN]
Shear Force ULS: V Vr Vc(stirrups) Vc(total)
0 5 10 15 204
2
0
-2
-4
-6
-8
[m]
[kN*m]
Bending Moment SLS: M Mr Md
0 5 10 15 20-8
-6
-4
-2
0
2
4
6
8
10
[m]
[kN]
Shear Force SLS: V Vr Vd
31
6.2. Required reinforcement area
6.2.5 Deflection and cracking
2. Reinforcement:
2..1 P1 : Span from 0.30 to 4.70 (m)
Longitudinal reinforcement: bottom
2 T16 l = 4.13 from 0.03 to 4.16 2 T16 l = 2.63 from 3.48 to 6.12
assembling (top)
2 T16 l = 3.69 from 0.89 to 4.58
support (T)
2 T16 l = 1.80 from 0.03 to 1.83 2 T16 l = 2.32 from 3.64 to 5.96
Transversal reinforcement: main
stirrups 25 R8 l = 0.81 e = 1*0.04 + 24*0.18 (m)
.2 P2 : Span from 5.00 to 8.90 (m)
0 5 10 15 2010
8
6
4
2
0
2
4
6
8
10
[m]
[cm2]
Reinforcement Area for Bending: Abt Abr Abmin Ades Aver_gross
0 5 10 15 206
4
2
0
2
4
6
[m]
[cm2/m]
Reinforcement Area for Shear: Ast Asr AsHang
0 5 10 15 202
1.5
1
0.5
0
-0.5
-1
-1.5
-2
[m]
[cm]
Deflections: ap(s-t) ap(l-t) at(s-t) a a adm
0 5 10 15 200.3
0.2
0.1
0
0.1
0.2
0.3
[m]
[mm]
Cracking: Cw Cw adm
32
Longitudinal reinforcement: bottom
2 T16 l = 4.87 from 5.44 to 10.32
assembling (top)
2 T16 l = 3.44 from 5.02 to 8.46
support (T)
2 T16 l = 3.06 from 7.52 to 10.58
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.3 P3 : Span from 9.20 to 13.10 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.01 from 9.64 to 12.66
assembling (top)
2 T16 l = 3.02 from 9.64 to 12.66
support (T)
2 T16 l = 3.06 from 11.72 to 14.78
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.4 P4 : Span from 13.40 to 17.30 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.87 from 11.98 to 16.86
assembling (top)
2 T16 l = 3.02 from 13.84 to 16.86
support (T)
2 T16 l = 2.72 from 15.92 to 18.64
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.06 + 21*0.18 (m)
.5 P5 : Span from 17.60 to 21.90 (m) Longitudinal reinforcement: bottom
2 T16 l = 5.19 from 16.18 to 21.38
assembling (top)
2 T16 l = 4.10 from 17.70 to 21.80
Transversal reinforcement: main
stirrups 24 R8 l = 0.81 e = 1*0.08 + 23*0.18 (m)
.6 P6 : Span from 22.20 to 24.10 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.67 from 20.70 to 24.37
support (T)
2 T16 l = 3.51 from 20.86 to 24.37
Transversal reinforcement: main
stirrups 11 R8 l = 0.81 e = 1*0.05 + 10*0.18 (m)
33
VII. TRANSVERSAL BEAM (BEAM 133, 212 and 226
2 Beam: Beam133 Number: 1
2.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
Additional reinforcement: : T fy = 460.00 (MPa)
2.2 Calculation results:
2.2.1 Internal forces in ULS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 14.90 -0.00 -5.73 -25.93 24.16 -40.91 P2 12.89 -0.00 -22.71 -23.48 35.50 -34.89 P3 10.18 -0.00 -18.55 -18.30 32.22 -31.99 P4 6.69 -0.73 -13.95 -33.37 25.67 -38.06 P5 0.00 -19.80 -39.32 -0.24 43.63 4.56
34
2.2.2 Internal forces in SLS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 14.90 0.00 -5.73 -25.93 24.16 -40.91
P2 12.89 0.00 -22.71 -23.48 35.50 -34.89
P3 10.18 0.00 -18.55 -18.30 32.22 -31.99
P4 6.69 -0.73 -13.95 -33.37 25.67 -38.06
P5 0.00 -19.80 -39.32 -0.24 43.63 4.56
2.. Required reinforcement area
0 2 4 6 8 10 12 14 1660
40
20
0
-20
-40
-60
-80
[m]
[kN*m]
Bending Moment ULS: M Mr Mc
0 2 4 6 8 10 12 14 16-80
-60
-40
-20
0
20
40
60
80
[m]
[kN]
Shear Force ULS: V Vr Vc(stirrups) Vc(total)
0 2 4 6 8 10 12 14 1620
10
0
-10
-20
-30
-40
[m]
[kN*m]
Bending Moment SLS: M Mr Md
0 2 4 6 8 10 12 14 16-50
-40
-30
-20
-10
0
10
20
30
40
50
[m]
[kN]
Shear Force SLS: V Vr Vd
35
2.. Deflection and cracking
2. Reinforcement:
2..1 P1 : Span from 0.20 to 3.70 (m)
Longitudinal reinforcement: bottom
2 T16 l = 3.59 from 0.03 to 3.34
assembling (top)
2 T16 l = 3.33 from 0.03 to 3.36
support (T)
2 T16 l = 1.13 from 0.03 to 0.87 2 T16 l = 2.45 from 2.42 to 4.87
Transversal reinforcement: main
stirrups 20 R8 l = 0.81 e = 1*0.04 + 19*0.18 (m)
.2 P2 : Span from 3.90 to 7.69 (m) Longitudinal reinforcement: bottom
2 T16 l = 2.33 from 2.66 to 4.99 2 T16 l = 4.56 from 4.32 to 8.88
0 2 4 6 8 10 12 14 1610
8
6
4
2
0
2
4
6
8
10
[m]
[cm2]
Reinforcement Area for Bending: Abt Abr Abmin Ades Aver_gross
0 2 4 6 8 10 12 14 166
4
2
0
2
4
6
[m]
[cm2/m]
Reinforcement Area for Shear: Ast Asr AsHang
0 2 4 6 8 10 12 14 162
1.5
1
0.5
0
-0.5
-1
-1.5
-2
[m]
[cm]
Deflections: ap(s-t) ap(l-t) at(s-t) a a adm
0 2 4 6 8 10 12 14 160.3
0.2
0.1
0
0.1
0.2
0.3
[m]
[mm]
Cracking: Cw Cw adm
36
assembling (top)
2 T16 l = 3.74 from 3.93 to 7.66
support (T)
2 T16 l = 2.04 from 6.72 to 8.76
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.01 + 21*0.18 (m)
.3 P3 : Span from 7.89 to 11.18 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.20 from 8.21 to 12.42
assembling (top)
2 T16 l = 3.44 from 7.82 to 11.25
Transversal reinforcement: main
stirrups 19 R8 l = 0.81 e = 1*0.03 + 18*0.18 (m)
.4 P4 : Span from 11.38 to 14.88 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.61 from 11.74 to 15.35
assembling (top)
2 T16 l = 2.82 from 11.35 to 14.17
support (T)
2 T16 l = 1.98 from 10.31 to 12.29
Transversal reinforcement: main
stirrups 20 R8 l = 0.81 e = 1*0.04 + 19*0.18 (m)
.5 P5 : Right cantilever from 15.08 to 16.83 (m) Longitudinal reinforcement: assembling (bottom)
2 T16 l = 2.39 from 14.41 to 16.80
support (T)
2 T16 l = 3.85 from 13.23 to 16.80 2 T16 l = 1.35 from 14.40 to 15.74
Transversal reinforcement: main
stirrups 10 R8 l = 0.81 e = 1*0.07 + 9*0.18 (m)
2 Beam: Beam212 Number: 1
2.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
Additional reinforcement: : T fy = 460.00 (MPa)
2.2 Calculation results:
2.2.1 Internal forces in ULS
37
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 15.08 -0.00 -6.93 -25.41 25.78 -40.80 P2 13.28 -0.00 -22.38 -23.84 35.78 -35.39 P3 9.90 -0.00 -18.20 -18.69 31.56 -32.05 P4 6.61 -0.62 -14.18 -32.48 25.69 -37.17 P5 0.00 -19.84 -39.34 -0.27 43.61 4.60
2.2.2 Internal forces in SLS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 15.08 0.00 -6.93 -25.41 25.78 -40.80
P2 13.28 0.00 -22.38 -23.84 35.78 -35.39
P3 9.90 0.00 -18.20 -18.69 31.56 -32.05
P4 6.61 -0.62 -14.18 -32.48 25.69 -37.17
P5 0.00 -19.84 -39.34 -0.27 43.61 4.60
2.. Required reinforcement area
0 2 4 6 8 10 12 14 1660
40
20
0
-20
-40
-60
-80
[m]
[kN*m]
Bending Moment ULS: M Mr Mc
0 2 4 6 8 10 12 14 16-80
-60
-40
-20
0
20
40
60
80
[m]
[kN]
Shear Force ULS: V Vr Vc(stirrups) Vc(total)
0 2 4 6 8 10 12 14 1620
10
0
-10
-20
-30
-40
[m]
[kN*m]
Bending Moment SLS: M Mr Md
0 2 4 6 8 10 12 14 16-50
-40
-30
-20
-10
0
10
20
30
40
50
[m]
[kN]
Shear Force SLS: V Vr Vd
38
2.. Deflection and cracking
2. Reinforcement:
2..1 P1 : Span from 0.20 to 3.70 (m)
Longitudinal reinforcement: bottom
2 T16 l = 3.31 from 0.03 to 3.34
assembling (top)
2 T16 l = 3.56 from 0.17 to 3.73
support (T)
2 T16 l = 1.37 from 0.03 to 1.11
Transversal reinforcement: main
stirrups 20 R8 l = 0.81 e = 1*0.04 + 19*0.18 (m)
.2 P2 : Span from 3.90 to 7.69 (m) Longitudinal reinforcement: bottom
2 T16 l = 2.33 from 2.66 to 4.99 2 T16 l = 4.56 from 4.32 to 8.88
assembling (top)
0 2 4 6 8 10 12 14 1610
8
6
4
2
0
2
4
6
8
10
[m]
[cm2]
Reinforcement Area for Bending: Abt Abr Abmin Ades Aver_gross
0 2 4 6 8 10 12 14 166
4
2
0
2
4
6
[m]
[cm2/m]
Reinforcement Area for Shear: Ast Asr AsHang
0 2 4 6 8 10 12 14 162
1.5
1
0.5
0
-0.5
-1
-1.5
-2
[m]
[cm]
Deflections: ap(s-t) ap(l-t) at(s-t) a a adm
0 2 4 6 8 10 12 14 160.3
0.2
0.1
0
0.1
0.2
0.3
[m]
[mm]
Cracking: Cw Cw adm
39
2 T16 l = 3.74 from 3.93 to 7.66
support (T)
2 T16 l = 2.08 from 2.79 to 4.87 2 T16 l = 2.04 from 6.72 to 8.76
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.01 + 21*0.18 (m)
.3 P3 : Span from 7.89 to 11.18 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.20 from 8.21 to 12.42
assembling (top)
2 T16 l = 3.09 from 7.82 to 10.91
support (T)
2 T16 l = 2.33 from 9.96 to 12.29
Transversal reinforcement: main
stirrups 19 R8 l = 0.81 e = 1*0.03 + 18*0.18 (m)
.4 P4 : Span from 11.38 to 14.88 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.61 from 11.74 to 15.35
assembling (top)
2 T16 l = 2.82 from 11.35 to 14.17
Transversal reinforcement: main
stirrups 20 R8 l = 0.81 e = 1*0.04 + 19*0.18 (m)
.5 P5 : Right cantilever from 15.08 to 16.83 (m) Longitudinal reinforcement: assembling (bottom)
2 T16 l = 2.39 from 14.41 to 16.80
support (T)
2 T16 l = 3.85 from 13.23 to 16.80 2 T16 l = 1.34 from 14.41 to 15.74
Transversal reinforcement: main
stirrups 10 R8 l = 0.81 e = 1*0.07 + 9*0.18 (m)
2 Beam: Beam226
2.1 Material properties:
Concrete : C25 fcu = 25.00 (MPa) Unit weight : 2501.36 (kG/m3)
Longitudinal reinforcement : T fy = 460.00 (MPa)
Transversal reinforcement : R fy = 250.00 (MPa)
Additional reinforcement: : T fy = 460.00 (MPa)
2.2 Calculation results:
2.2.1 Internal forces in ULS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
40
P1 2.62 -0.00 -3.20 -2.94 6.52 -6.37 P2 3.28 -0.00 -1.60 -5.43 5.96 -8.00 P3 1.47 -0.48 -1.74 -5.78 4.83 -7.29 P4 2.09 -0.00 -3.46 -3.75 6.36 -6.53 P5 0.00 -3.09 -5.66 -0.00 6.45 0.00
2.2.2 Internal forces in SLS
Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 2.62 0.00 -3.20 -2.94 6.52 -6.37
P2 3.28 0.00 -1.60 -5.43 5.96 -8.00
P3 1.47 -0.48 -1.74 -5.78 4.83 -7.29
P4 2.09 0.00 -3.46 -3.75 6.36 -6.53
P5 0.00 -3.09 -5.66 -0.00 6.45 0.00
2.. Required reinforcement area
0 2 4 6 8 10 12 14 1660
40
20
0
-20
-40
-60
[m]
[kN*m]
Bending Moment ULS: M Mr Mc
0 2 4 6 8 10 12 14 16-80
-60
-40
-20
0
20
40
60
80
[m]
[kN]
Shear Force ULS: V Vr Vc(stirrups) Vc(total)
0 2 4 6 8 10 12 14 164
3
2
1
0
-1
-2
-3
-4
-5
-6
[m]
[kN*m]
Bending Moment SLS: M Mr Md
0 2 4 6 8 10 12 14 16-8
-6
-4
-2
0
2
4
6
8
[m]
[kN]
Shear Force SLS: V Vr Vd
41
2.. Deflection and cracking
2. Reinforcement: 2..1 P1 : Span from 0.20 to 3.70 (m)
Longitudinal reinforcement: bottom
2 T16 l = 3.31 from 0.03 to 3.34
assembling (top)
2 T16 l = 3.56 from 0.17 to 3.73
support (T)
2 T16 l = 1.37 from 0.03 to 1.11 2 T16 l = 1.75 from 2.79 to 4.54
Transversal reinforcement: main
stirrups 20 R8 l = 0.81 e = 1*0.04 + 19*0.18 (m)
.2 P2 : Span from 3.90 to 7.69 (m) Longitudinal reinforcement:
0 2 4 6 8 10 12 14 1610
5
0
5
10
15
[m]
[cm2]
Reinforcement Area for Bending: Abt Abr Abmin Ades Aver_gross
0 2 4 6 8 10 12 14 166
4
2
0
2
4
6
[m]
[cm2/m]
Reinforcement Area for Shear: Ast Asr AsHang
0 2 4 6 8 10 12 14 162
1.5
1
0.5
0
-0.5
-1
-1.5
-2
[m]
[cm]
Deflections: ap(s-t) ap(l-t) at(s-t) a a adm
0 2 4 6 8 10 12 14 160.3
0.2
0.1
0
0.1
0.2
0.3
[m]
[mm]
Cracking: Cw Cw adm
42
bottom
2 T16 l = 2.33 from 2.66 to 4.99 2 T16 l = 4.56 from 4.32 to 8.88
assembling (top)
2 T16 l = 3.74 from 3.53 to 7.27
support (T)
2 T16 l = 2.48 from 6.32 to 8.80
Transversal reinforcement: main
stirrups 22 R8 l = 0.81 e = 1*0.01 + 21*0.18 (m)
.3 P3 : Span from 7.89 to 11.18 (m) Longitudinal reinforcement: bottom
2 T16 l = 4.20 from 8.21 to 12.42
assembling (top)
2 T16 l = 2.74 from 7.82 to 10.56
support (T)
2 T16 l = 3.09 from 9.57 to 12.66
Transversal reinforcement: main
stirrups 19 R8 l = 0.81 e = 1*0.03 + 18*0.18 (m)
.4 P4 : Span from 11.38 to 14.88 (m) Longitudinal reinforcement: bottom
2 T16 l = 3.61 from 11.74 to 15.35
assembling (top)
2 T16 l = 2.82 from 11.72 to 14.54
Transversal reinforcement: main
stirrups 20 R8 l = 0.81 e = 1*0.04 + 19*0.18 (m)
.5 P5 : Right cantilever from 15.08 to 16.83 (m) Longitudinal reinforcement: assembling (bottom)
2 T16 l = 2.39 from 14.41 to 16.80
support (T)
2 T16 l = 3.49 from 13.60 to 16.80
Transversal reinforcement: main
stirrups 10 R8 l = 0.81 e = 1*0.07 + 9*0.18 (m)
44
1. Slab: Plate147 and 241
1.1. Reinforcement:
Type : RC floor
Main reinforcement direction : 0°
Main reinforcement grade : T; Characteristic strength = 460.00 MPa
Bar diameters bottom d1 = 1.2 (cm) d2 = 1.2 (cm) top d1 = 1.2 (cm) d2 = 1.2 (cm)
Cover bottom c1 = 3.0 (cm) top c2 = 3.0 (cm)
1.2. Concrete
Class : CONCR; Characteristic strength = 25.00 MPa
Density : 2501.36 (kG/m3)
Concrete creep coefficient : 1.87
1.3. Hypothesis
Calculations according to :
Method of reinforcement area calculations : N/A
Allowable cracking width - upper layer : 0.40 (mm) - lower layer : 0.40 (mm)
Allowable deflection : 3.0 (cm)
Verification of punching : yes
Fire rating : 0 h
Exposure - upper layer : X0 - lower layer : X0
Calculation type : simple bending
1.4. Slab geometry Thickness 0.17 (m) Contour: edge beginning end length x1 y1 x2 y2 (m) 1 0.00 -8.90 16.73 -8.90 16.73 2 16.73 -8.90 16.91 -8.45 0.48 3 16.91 -8.45 17.05 -7.99 0.48 4 17.05 -7.99 17.15 -7.51 0.48 5 17.15 -7.51 17.21 -7.03 0.48 6 17.21 -7.03 17.23 -6.55 0.48 7 17.23 -6.55 17.21 -6.07 0.48 8 17.21 -6.07 17.15 -5.59 0.48 9 17.15 -5.59 17.05 -5.11 0.48 10 17.05 -5.11 16.91 -4.65 0.48 11 16.91 -4.65 16.73 -4.20 0.48 12 16.73 -4.20 16.89 -3.84 0.40 13 16.89 -3.84 17.03 -3.47 0.40 14 17.03 -3.47 17.13 -3.08 0.40
45
Support: n° Name dimensions coordinates (m) x y 71 point 0.30 / 0.20 0.00 -8.90 71 point 0.30 / 0.20 0.00 -8.90 71 linear 0.20 / 16.73 8.37 -8.90 71 linear 8.90 / 0.20 0.00 -4.45 72 point 0.30 / 0.20 0.00 -4.20 72 point 0.30 / 0.20 0.00 -4.20 72 linear 0.20 / 16.73 8.37 -4.20 73 point 0.30 / 0.20 0.00 0.00 73 point 0.30 / 0.20 0.00 0.00 73 linear 0.20 / 16.73 8.37 0.00 75 point 0.30 / 0.20 3.70 -8.90 75 point 0.30 / 0.20 3.70 -8.90 75 linear 8.90 / 0.20 3.70 -4.45 76 point 0.30 / 0.20 3.70 -4.20 76 point 0.30 / 0.20 3.70 -4.20 77 point 0.30 / 0.20 3.70 0.00 77 point 0.30 / 0.20 3.70 0.00 79 point 0.30 / 0.20 7.69 -8.90 79 point 0.30 / 0.20 7.69 -8.90
1.5. Calculation results:
1.5.1. Maximum moments + reinforcement for bending Ax(+) Ax(-) Ay(+) Ay(-) Provided reinforcement (cm2/m): 5.24 5.24 5.24 5.24 Modified required reinforcement (cm2/m): 0.00 4.52 0.00 4.52 Original required reinforcement (cm2/m): 0.00 0.00 0.00 0.00 Coordinates (m): 16.73;-8.90 16.73;-8.90 16.73;-8.90
16.73;-8.90 1.5.2. Maximum moments + reinforcement for bending Ax(+) Ax(-) Ay(+) Ay(-) Symbol: required area/provided area Ax(+) (cm2/m) 0.00/5.24 0.00/5.24 0.00/5.24 0.00/5.24 Ax(-) (cm2/m) 4.52/5.24 4.52/5.24 4.52/5.24 4.52/5.24 Ay(+) (cm2/m) 0.00/5.24 0.00/5.24 0.00/5.24 0.00/5.24 Ay(-) (cm2/m) 4.52/5.24 4.52/5.24 4.52/5.24 4.52/5.24 Coordinates (m) 16.73;-8.90 16.73;-8.90 16.73;-8.90
16.73;-8.90 Coordinates* (m) 0.00;0.00;0.00 0.00;0.00;0.00 0.00;0.00;0.00
0.00;0.00;0.00 * - Coordinates in the structure global coordinate system
1.5.4. Deflection
46
|f(+)| = 0.0 (cm) <= fdop(+) = 3.0 (cm) |f(-)| = 3.3 (cm) > fdop(-) = 3.0 (cm) 1.5.5. Cracking upper layer ax = 0.40 (mm) <= adop = 0.40 (mm) ay = 0.40 (mm) <= adop = 0.40 (mm) lower layer ax = 0.26 (mm) <= adop = 0.40 (mm) ay = 0.40 (mm) <= adop = 0.40 (mm)
2. Loads: Case Type List Value 1 self-weight 147 241 PZ Negative 2 uniform load PZ=-0.48(kN/m) 3 uniform load PZ=-0.26(kN/m) 4 (FE) uniform 147 241 PZ=-0.24(kN/m2) 5 self-weight 147 241 PZ Negative 5 (FE) uniform 147 241 PZ=-4.79(kN/m2) Combination/Component Definition ULS/6 (1+2+3+4+5)*1.00 SLS/7 (1+2+3+4+5)*1.00
3. Results - detailing List of solutions: Reinforcement: bars Solution no. Reinforcement range Total weight Diameter / Weight (kG) 1 - 5157.53 Results for the solution no. 1 Reinforcement zones Bottom reinforcement Name coordinates Provided reinforcement At Ar
x1 y1 x2 y2 (mm) / (cm) (cm2/m) (cm2/m) 1/1- Ax Main 0.00 -8.90 17.24 15.20 10.0 / 15.0 20.45 > 5.24 1/2- Ay Perpendicular 0.00 -8.90 17.24 15.20 10.0 / 15.0 27.13 > 5.24 Top reinforcement Name coordinates Provided reinforcement At Ar
x1 y1 x2 y2 (mm) / (cm) (cm2/m) (cm2/m) 1/1+(1/3+) Ax Main 0.00 -8.90 17.24 0.00 10.0 / 15.0 38.50 > 5.24 1/2+(1/3+) Ax Main 4.29 13.00 17.24 15.20 10.0 / 15.0 28.96 > 5.24 1/3+ Ax Main 6.09 0.00 17.24 13.00 10.0 / 15.0 36.22 > 5.24 1/4+(1/6+) Ay 0.00 -8.90 17.24 0.00 10.0 / 15.0 60.03 > 5.24 1/5+(1/6+) Ay 4.29 13.00 17.24 15.20 10.0 / 15.0 47.48 > 5.24 1/6+ Ay Perpendicular 6.09 0.00 17.24 13.00 10.0 / 15.0 52.97 > 5.24
4. Material survey
47
Concrete volume = 56.02 (m3)
Formwork = 329.52 (m2)
Slab circumference = 85.23 (m)
Area of openings = 15.64 (m2)
Steel T
Total weight = 5270.52 (kG)
Density = 94.09 (kG/m3)
Average diameter = 10.0 (mm)
Survey according to diameters:
Done at KIGALI, 6th December 2018
Eng. Dieudonne NTEZIYAREMYE
No A 298/EC/IER/2015