Advance Design of RC Structure

Lecture 4

University of Palestine

Shear Wall Design

Dr. Ali Tayeh

Design LoadsShear force – maximum at the baseBending moment – maximum at the baseAxial load – including its own weight

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:

Vu exceeds

The wall thickness is more than 250 mm

Reinforcement spacing each way in structure wall shall not exceed the large of 450 mm & 3 wall thickness

0.17 'cv cA f

The minimum reinforcement ratio for both the vertical & horizontal reinforcement is: 0.0025v h

If

The minimum vertical reinforcement ratioV = 0.0012 for No. 16 bars or smaller, fy 420 MPa

= 0.0015 for other bars, fy 420 MPa

The minimum horizontal reinforcement ratioh = 0.002 for No. 16 bars or smaller, fy 420 MPa

= 0.0025 for other bars, fy 420 MPa

0.083 'u cv cV A f

Vu is obtained from the lateral load analysis in accordance with the factored load combinations.

Acv is the gross area of concrete section bounded by the web thickness & length of section in the direction of the shear force considered

Shear Strength

The nominal shear strength Vn

Where c is 0.25 for hw/Lw 1.5

0.17 for hw/Lw 2

►varies linearly between 0.25 & 0.17 for hw/Lw between 1.5 & 2

►hw = height of entire wall of wall considered

►Lw = length of entire wall in the direction of shear force

For all wall piers sharing a common lateral force Vn shall not be taken larger than

'n cv c c v yV A f f

0.66 'cv cA f

For individual wall piers Vn shall not be taken larger than

►Acw is the area of concrete section of the individual pier considered

0.83 'cw cA f

Design for Flexural & Axial LoadsThe compression zones must include special boundary elements when

600 /w

u w

LC

h / 0.007u wh

Where

C = distance from the extreme compression fiber to the neutral axis calculated from the factored axial force & nominal moment strength, consistent with the design displacement u, resulting in the largest neutral axis

Lw = length of the entire wall considered in the direction of the shear force

δu = design displacement & it is determined by multiplying the deflections from an elastic analysis under the prescribed seismic forces by a deflection amplification

hw = height of entire wall or of the segment of wall considered

When special boundary elements are required, they must extend horizontally from the extreme compression fiber a distance not less than the large of C-0.1Lw & C/2

In the vertical direction the special boundary elements must extend from the critical section a distance greater than or equal to the large of Lw or Mu/4Vu.

►This distance is based on upper bound estimates of plastic hinge lengths.

The compression zone shall include special boundary elements where the maximum extreme fiber stress corresponding to the factored forces, including earthquake effects, exceeds 0.2fc’

Special boundary elements can be discontinued where the compressive stress is less than 0.15fc’.

0.2 '2

u u wc

g g

P M Lf

A I

When special boundary elements are not required & if the longitudinal reinforcement ratio at the wall boundary is greater than 2.8/fy, the maximum longitudinal spacing of transverse reinforcement in the boundary shall not exceed 200mm. Except when 0.083 'u cv cV A f

If the wall subjected to axial load smaller than the producing balanced failure, the following approximate equation can be used to determine the design moment capacity of the wall:

0.5 1 1un st y w

st y w

P cM A f

A f

Ast = total area of vertical reinforcement

lw = horizontal length of wall

Pu = factored axial compressive load

fy = yield strength of reinforcement

0.5 1 1un st y w

st y w

P cM A f

A f

12 0.85w

C

β1 = factor relating depth of equivalent rectangular compressive stress block to the neutral axis depth= 0.85 for fc’= 17 to 28 MPa & = 0.8 for fc’ > 28 MPa

'yst

w c

fA

h f

fc ' = compressive strength of concrete

'u

w c

P

hf

h = thickness of wall

φ = 0.90 (strength primarily controlled by flexure with low axial load)

Detailing reinforcement for boundary ZoneDimension

The minimum section dimension of the boundary zone shall be lw/16.Boundary zones shall have a minimum length of 450mm (measured along the length) at each end of the wall or portion of wall.In I-, L-, C- or T-section walls, the boundary zone at each end shall include the effective flange width and shall extend at least 250mm. into the web.

Confinement ReinforcementAll vertical reinforcement within the boundary zone shall be confined by hoops or cross-ties having a steel cross sectional area Ash> 0.09S.bc. fc′ / fy

All Hoops and cross-ties shall have a vertical spacing,

max

150

6

mmS

)diameter of largest vertical bar within boundary zone)

The length-to-width ratio of the hoops shall not exceed 3 and all adjacent hoops shall be overlapping.Cross-ties or legs of overlapping hoops shall not be spaced farther apart than 250mm along the wall.Alternate vertical bars shall be confined by the corner of a hoop or cross-tie.

Minimum for vertical reinforcement

Av > 0.005 area of boundary zone or two no. 16 bars at each edge of the boundary zone

Reinforcement details for special boundary elements

Load CombinationACI 318-05 required that structure, their components, & their foundation be designed to have strength not less than the most severe of the following combination of loads:

U = 1.4(D + F )U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or

R)U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W) U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)U = 1.2D + 1.0E + 1.0L + 0.2S U = 0.9D + 1.6W + 1.6H U = 0.9D + 1.0E + 1.6H

Where:U = required strength to resist the factored loads

D = dead load L = live load

W = wind loadE = earthquake load

F = load due to fluids with and maximum heights well-defined pressuresH = load due to soil pressureT = load due to effects of temperature, shrinkage,

expansion of shrinkage compensating concrete, creep, differential settlement, or combinations thereof. Lr = roof live load

S = snow loadR = rain load

The effect of seismic load E is defined as follows:

0.2E dsE Q S D

E = the effect of horizontal and vertical earthquake-induced forces,

SDS = the design spectral response acceleration at short periodsD = the effect of dead load

ρ = the reliability factorQE = the effect of horizontal seismic forces

If the there is a vertical discontinuities of the structure system that resisting the seismic load the effect of seismic load E is defined as follows:

0.2o E dsE Q S D

where Ωo is the system over strength factor. The value of Ωo varies between 2 to 3 as give in table depending on the type of

lateral force resisting system

ExampleDesign the shear wall section at the basement

100t

784t.m

5m0.25m

D.L = 120tL.L = 50t

980kN

7840kN.m

1176kN490kN

Determine minimum vertical & horizontal reinforcement requirements in the wall

Check if two curtains of reinforcement are required.

0.17 'u cv cV A f

25 0.25 1.25cvA m

0.98 0.17 1.25 28 1.124uV MN MN

According to the above equation, no need for two curtains but as the wall thickness is 250mm the code recommend to use two curtains of reinforcement.

The minimum reinforcement ratio

0.083 'u cv cV A f

0.083 1.25 28 0.549 0.98MN Vu MN

Therefore the minimum reinforcements ration are

0.0025v 0.0025h

Minimum steel area

225 100 0.0025 6.25 /sA cm m

Use 612 @ 100cm in both sides of the wall & in the vertical & horizontal directions

21.2 2

4 0.0025)25) 36or S cmS

Smax = the smaller of 325 = 75cm & 45cm

Smax = 45cm > S = 36 cm O.K. 35cm

35cm

12@35cm

12@35cm

Use 112 @ 35cm in each side of the wall in the vertical & horizontal direction

Determine reinforcement requirements for shear

Shear strength of wall

'n cv c c v yV A f f

c = 0.17 for hw/Lw = 18/5 = 3.6 2

1.13 20.0026

35 25v

1.25 0.17 28 0.0026 420 2.489 0.98nV MN Vu MN

Therefore, the 112 @ 35cm in the vertical & horizontal direction would be enough for the shear

Determine reinforcement requirements for combined flexural & axial loads

Determine if special boundary elements are required

0.2 '2

u u wc

g g

P M Lf

A I

Pu = 1.2D + 1.0E + 1.0L orPu = 1.2D + 1.6L

There is a vertical load comes from the earthquake which can be computed by Ev = 0.2Sds*D )D is the total dead load)

Pu = 1.2D + 1.0)0.2Sds*D) +1.0 L

= 1.21176 + 0.20.3331176 + 490 = 1980 kN

Pu = 1.21176 + 1.6490 = 2195 kN

Therefore Pu = 2195 kN

340.25 5

2.612gI m

25 0.25 1.25gA m

22195 7840 59295 0.2 ' 5600 /

1.25 2.6 2 ckN f kN m

Therefore, special boundary elements are required

Special boundary length

Lager of C/2, C – 0.1Lw 45cm or Lw/16

12 0.85w

C

1.13 2 50042035 0.039

' 500 25 28yst

w c

fA

h f

2195 /10000.063

' 5 0.25 28u

w c

P

hf

0.039 0.0630.13

2 0.039 0.85 0.85w

C

500 0.13 65C cm

6532.5

2 2

Ccm

0.1 65 0.1 500 15wC L cm

Lw/16 = 500/16 = 31cm

Take the boundary length as 45cm as it’s the minimum

Check moment capacity Mn

0.5 1 1un st y w

st y w

P cM A f

A f

32.29 2195 /10000.9 0.5 420 5 1 1 0.13 6.95

32.2910000 42010000

nM MPa

Mn = 6.95MPa < Mu = 7.84MPa Therefore additional reinforcements are required for flexural

0.9u n

sy

M MA

f d

0.8 0.8 5 4wd L m

27.84 6.956

0.9 420 4sA cm

=316

216 at each edge of the boundary zone is the minimum vertical reinforcement in the boundary zone

Confinement Reinforcement

Confinement of the boundary elements

Maximum Hoops and cross-ties vertical spacing

Smax = smaller of 150mm or 6largest bar diameter)16mm) = 96mm

Smax = 90mm

Reinforcement cross-section area long direction

20.09 ' 0.09 9 )25 5 1.6) 281

420c c

shy

Sh fA cm

f

Use 28 @ 9cm parallel to the wall

Reinforcement cross-section area short direction

20.09 ' 0.09 9 )45 5 1.6) 282

420c c

shy

Sh fA cm

f

Use 48 @ 9cm perpendicular to the wall

45cm

416

212212

45cm

212@35cm

112@35cm

112@35cm

35cm

18@9cm

18@9cm