Function of Shear Wall

 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