This is a structural member of RCC structures to provide lateral stability against horizontal forces caused by Wind pressures and Earthquakes.
Shapes of Shear Wall.
Shear walls are designed in different shapes using Reinforced cement concrete materials which produce resistance against the horizontal forces that act to over-turn the structures.
Placement of Shear Walls.
Shear walls are placed at three
locations
1- On the edges of the building.
2-Inner side of Building on various
locations.
3-On the Center of High rise buildings.
The function of Shear Wall.
There are
various useful functions of shear walls.
1-Shear Wall
provides resistance against self-load of structure, movement loading, and
resultant loading such as uplift loads acting on shear walls.
2-Shear Wall
holds gravity lines of all areas of the building that can create extra stresses during movements and may damage nonstructural elements in the building.
3-Shear wall
keeps stiffness of junctions (on column and beams) within the design limit.
4-Shear wall
provides protection in lateral displacement and sway of the building.
5-Shear wall
provides safe structural movement under earthquake and wind pressures.
The Calculation for Loading Function.
Let's have a
calculation session on design of shear walls as shown in this building plan.
Suppose the
each span is 20 ft and slab design load is 300 Pound/Sq.Ft.(Including weight of
walls ).
Total Load
of one floor = (80'x40'x300)= 960000
Pounds
Factored load =(960000x1.5) =1440000 Pounds
Now whole
Plan will be divided 5 Parts as showing (Area
1=800 Sft), (Area 2, =600 Sft + 600 Sft ),(Area 3,= 300 Sft+300 Sft).
Now load can
be calculated for each shear force
Load on Area
1 Shear Wall =(800x1440000)/(40x80) =360000 Pounds
Load on Area
2 Shear wall = (600x1440000)/(40x80)=270000 Pounds
Load on Area
3 Shear Wall =(300x1440000)/(40x80) =135000 Pounds
Uniformly Distributed Load
Calculation
Considering
shear wall as a beam between two junctions of the slab and beams vertical height is
12 ft.
UDL For Area 1, Shear Wall = 360000/12= 30,000
Pound/Ft
UDL For Area 2, Shear Wall = 270000/12= 22,500 Pound/Ft
UDL For Area 1, Shear Wall = 135000/12= 11,250 Pound/Ft
Moment of
Area 1, Shear Wall=( w.L.L.x12) /8=
= 30000x12x12x12/8=6,480,000 (Lbs-Inch)
Moment of
Area 2, Shear Wall=( w.L.L.x12) /8=
= 22500x12x12x12/8=4,860,000 (Lbs-Inch)
Moment of
Area 3, Shear Wall=( w.L.L.x12) /8=
= 11250x12x12x12/8=2,430,000 (Lbs-Inch)
Moment of Resistance = 250xbxd1^2 ................ EQ.1
we knows
that M.r = Moment
Putting values into Eq.1 for Area 1 Shear Wall
Longer side in Inches =10x12= 120 Inch
Shorter side
in Inches =8x12= 96 Inch.
(6,480,000/2)
(Lbs-Inch) = 250x 120xd1^2
d1 = 7.34
say 7.5" Adding cover 1" on both faces = 9.5"
Repeating
for Shorter Side
(6,480,000/2)
(Lbs-Inch) = 250x 96 xd1^2
d1= 11.61
Say 12" Adding cover of 1" at
both faces = 14"
Maximum size of wall thickness is
14" so adopting this
Area of
Steel = M/Fst.Lever Arm
6,480,000/2x20000x.75x12
=18 Sq.Inch
For Spacing Using
1 " Dia Bars = Area of Bar x width /Area of Steel
Spacing =(0.785x96)/18 =4.18 say 4" C/C (Vertical Bars)
Now
Calculating the stirrups spacing
Spacing of
1/2" Dia bar =Area of 6-Leg Stirrups x Fst x Lever Arm/Shear Force
6x0.302x20000x0.75x12/(30000x12x.5x.5)
=3.5" C/c
Shear Wall Area 2
Longer side in Inches =15x12= 180 Inch
(4,860,000 (Lbs-Inch)= 250x 180xd1^2
d1 = 10.39 say
10.5" Adding cover 1" on both faces = 12.5" (Shear wall Thickness)
Area of
Steel = M/Fst.Lever Arm
4,860,000 /20000x.75x10.5=
30.85 Sq.Inch
For Spacing Using
1 " Dia Bars = Area of Bar x width /Area of Steel
Spacing =(.785x180)/30.85 =4.5 "c/c (Vertical Bars)
Now
Calculating the stirrups spacing
Spacing of 1/2
" Dia bar =Area of 12-Leg Stirrups x Fst x Lever Arm/Shear Force
=
12x.302 x20000x0.75x10.5/(22500x12x.5) =4.25" C/c
Shear Wall Area 3
Longer side in Inches =8x12"= 96 Inch
2,430,000 (Lbs-Inch)= 250x 96xd1^2
d1 = 10 Adding cover 1" on both faces = 12" (Shear Wall Thickness)
Area of
Steel = M/Fst.Lever Arm
2,430,000 /20000x.75x10=
16.20 Sq.Inch
For Spacing Using
1 " Dia Bars = Area of Bar x width /Area of Steel
Spacing =(.785x96)/16.2=4.5 "c/c (Vertical
Bars)
Now
Calculating the stirrups spacing
Spacing of 1/2
" Dia bar =Area of 6-Leg Stirrups x Fst x Lever Arm/Shear Force
=
6x.302 x20000x0.75x10/(11,250 x12x.5) =4" C/c
