RC- Shear Wall Design
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example of shear wall design in reinforced concrete
Transcript of RC- Shear Wall Design
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RC IV
For more info., please contact: Eng. MHD. [email protected]
AIU 2014-2Arab International University
SHEAR WALL DESIGN
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Shear Wall Reinforcement
14
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Failure Modes of RC shear wall
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Layout of Main Reinforcement in Shear Walls
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Layout of Main Reinforcement in Shear Walls
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Layout of Main Reinforcement in Shear Walls
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Shear Wall Design
15
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Shear Wall Design PROCEDURES
Internal ForcesCombinations of LoadsVertical and Horizontal Steel Requirements: Curtains of Reinforcement Minimum Steel Ratio of V & H ReinforcementsMinimum Steel Area of V & H ReinforcementsDesign for Shear - Shear StrengthDesign for Flexure & Axial LoadsBoundary Elements/ NecessityBoundary Elements/ DimensionsBoundary Elements/ Efficiency Boundary Elements/ Additional ReinforcementBoundary Elements/ Confinement Reinforcement
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Design Loads
Shear Force: Maximum at the baseBending Moment: maximum at the baseAxial Load: Including its own weight
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Curtains of Reinforcement
Special reinforced concrete walls are to be provided with reinforcement in two orthogonal directions in the plane of the wall.At least two curtains of reinforcement shall be used in wall if:
Reinforcement spacing each way in structure wall shall not exceed the large of 450mm and 3 wall thickness
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Minimum Steel Area of V & H Reinforcements
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Shear Strength
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Design for Flexure and Axial Loads
To determine the required longitudinal reinforcement, the wall is treated like a column. An interaction diagram can be developed for the selected reinforcing layout, and checked against combinations of axial load and moment as determined from analysis.
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Boundary Elements
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Boundary Elements/ Necessity
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Boundary Elements/ Necessity
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Boundary Elements/ Dimensions
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Boundary Elements/ Efficiency
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Boundary Elements/ Efficiency
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Boundary Elements/ Confinement
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Boundary Elements/ Confinement
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Example 4
E.4
Note:Review example 3 before solving example 4
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Given
Building: Five-story Residential BuildingPlace: SaydnayaLateral Force-Resisting System: Shear WallsBearing Capacity of the Soil: 3.5kg/cm2Live Load: 2kN/m2Total Dead Load Typical Floor: 12kN/m2Total Dead Load Last Floor: 8kN/m2Story Height: 3.25mShear Walls - Thickness: 25cm
Required
Design ALL shear walls for the applied forces
Given
Strengths of Materials: fc=25MPa & fy=420MPa
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Design of the Shear Wall: W1
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Design of the Shear Wall: W1
StoryFyMtFFF
5373.33233.33124.44-5.60118.84
4447.99279.99149.33-6.72142.61
3335.99210.00112.00-5.04106.96
2224.00140.0074.67-3.3671.31
1112.0070.0037.33-1.6835.65
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Design of the Shear Wall: W1
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W1: Combinations of Loads
Comb 3 : 1.43 D + 0.55 L + 1.1 Eh
Comb 4 : 1.1 D + 1.1 Eh
Comb 5 : 0.88 D 1.1 Eh
Comb 2 : 1.4 D + 1.7 L
Comb 1 : 1.4 D
CombNu (kN)Mu (kN.m)Vu (kN)
1156800
2190800
31711.65948.05522.91
412325948.05522.91
5985.6- 5948.05- 522.91
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W1: 3- Vertical & Horizontal Rein.
Curtains
Curtains
Boundary Element
Boundary Element
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W1: 3- Vertical & Horizontal Rein.
Two Curtains
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W1: 3- Vertical & Horizontal Rein.
Acv = 250x4000 = 1000000mm2
f'c = 25MPa
0.083x1000000x(25)0.5 /1000= 415kN
Vu = 522.91kN
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W1: 3- Vertical & Horizontal Rein.
rS = As/Ac
As = Ac x rS
Asv = 4000 x 250 x 0.0025 = 2500mm2 = 25cm2
Asv (m) = As /lw = 25cm2 /4m = 6.25cm2 /m
Ash = 3250 x 250 x 0.0025 = 2031mm2 = 20.31cm2
Ash (m) = As /lw = 20.31cm2 /3.25m = 6.25cm2 /m
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Study of W1 Reinforcement
Two Curtains
Asv = 6.25cm2 /m
Ash = 6.25cm2 /m
fv 10mm
fh 6mm
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Study of W1 Reinforcement
Two Curtains
Asv = 6.25cm2 /m
Ash = 6.25cm2 /m
fv 10mm
fh 6mm
Asv = 2x 5T10/m
Ash = 2x 5T10/m
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W1: 4- Design for Shear
Ach = 250 x 3250 mm2
hw=3.25m
Vn = 2569kN
Lw=4m
3.25/4=0.81
f'c = 25MPa
fy = 420MPa
rh = As/Ac
Vu = 522.91kN
As (m) = 6.25cm2 /m
rh = 2x 5T10 x3.25/Ac
rh = 2551.25/812500
rh = 0.00314
Ach = 812500 mm2
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W1: 5- Design for Flexure&Axial Load
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W1: 5- Design for Flexure&Axial Load
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W1: 5- Design for Flexure&Axial Load
Pu = Nu = 1711.6 kN
Mu = 5948.05 kN.m
Ag = 1000000 mm2
Ig = 250x40003 /12 = 1333.33x109 mm4
Lw = 4000 mm
(1711.6x103/ 1000000) + (5948.05x106/ 1333.33x109)x4000/2 = 10.63
0.2fc = 0.2x25 = 5
Boundary Elements are Needed
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W1: 5- Design for Flexure&Axial Load
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W1: 5- Design for Flexure&Axial Load
Ast = (2x 5T10/m) x4m
w = 0.053
a = 0.068
C = 584mm
t=250mm
f'c = 25MPa
fy = 420MPa
lw=4000mm
Pu = 1711.6 kN
t=250mm
lw=4000mm
f'c = 25MPa
Ast = 3140 mm2
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W1: 5- Design for Flexure&Axial Load
584 0.1x4000 = 184mm
584/2=292mm=29.2cm
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W1: 5- Design for Flexure&Axial Load
(Height of the Wall / 16) = 325/16 = 20.3cm
25cm
29.2cm
Boundary Element = (45 x 25)cm2
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W1: 5- Design for Flexure&Axial Load
Ast = 3140 mm2
Mn = 5175.92 kN.m
Lw= 4000 mm
fy = 420 MPa
Pu= 1711.6 kN
C = 584 mm
Mu = 5948.05
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W1: 5- Design for Flexure&Axial Load
As = 1066 mm2 = 10.66 cm2
Mn = 5175.92 kN.m
Lw= 4000 mm
fy = 420 MPa
d = 0.8x4000 = 3200 mm
f = 0.9
Mu = 5948.05 kN.m
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Study of W1 Reinforcement
Two Curtains
As = 10.66 cm2
Boundary Element = (45 x 25)cm2
Asv = 6.25cm2 /m
Ash = 6.25cm2 /m
fv 10mm
fh 6mm
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Study of W1 Reinforcement
Two Curtains
As = 10.66 cm2
Boundary Element = (45 x 25)cm2
Asv = 6.25cm2 /m
Ash = 6.25cm2 /m
fv 10mm
fh 6mm
Asv = 2x 5T10/m
Ash = 2x 5T10/m
Boundary Element = (50 x 25)cm2
+ 2x 3T10
As = 10T14
= 15.37cm2
rs= 1.23%
10T14
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Study of W1 Reinforcement
Boundary Element = (45 x 25)cm2
fv 10mm
fh 10mm
Asv = 2x 5T10/m
Ash = 2x 5T10/m
Boundary Element = (50 x 25)cm2
Boundary Element = (50 x 25)cm2
As = 10T14
6x14 = 84mm
S = 80mm
Ash > 0.09x8x(50-5)x25/420 = 2.68cm2
Ash > 0.09x8x(25-5)x25/420 = 1.19cm2
Ash = 2.68cm2
Ash = 1.19cm2
3T8/80mm
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Shear wall W1 Detailing
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END OF PR. SESSION 08
TO BE CONTINUED NEXT SESSION
For more info., please contact: Eng. MHD. [email protected]
AIU 2014-2Arab International University
RC IV
SHEAR WALL DESIGN
