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Reinforced Concrete Design

Design of load bearing RC wall with empirical method

Limitation:   ACI 318-05 Section 14.5

1. solid rectangular cross section.

2. resultant of all factored loads is located with the middle third of the overall thickness of the wall

3. Minimum thickness of wall shall not  be less than 1/25 the support height of length or 4 inches.

4. Minimum thickness of exterior basement walls and foundation walls shall not be less than 7.5 inches.

RC wall strength by empirical design method:

The empirical equation for calculate axial strength of RC wall is

    fPn = 0.55 f fc' Ag [ 1 - ( k lc / 32 h)2]                                     ACI 318 Eq (14.1)

Where f = 0.65, lc is height of wall, h is thickness of wall, Ag is gross section area,

fc' is compression strength of concrete, k is effective length factor.

The effective length factor, k shall be

For wall braced top and bottom against lateral translation, and

1. restrained against rotation at one or both ends, k = 0.8.

2. unrestrained against rotation at both ends, k = 1.0

For walls not braced against lateral translation, k = 2.0

Example 1:

Design code: ACI 318-05 Section 14.5

Situation: A RC exterior wall supporting a two story building.

Design data:

Width of wall: h = 8 in

Height of wall: lc = 12 ft.

Compression strength of concrete: fc' = 3000 psi

Yield strength of concrete: fy = 60000 psi

Tributary width: 15 ft

Roof live load: RL = 20 psf

Roof dead load: RD = 80 psf

Floor live load: FL = 100 psf

Floor dead load: FD = 80 psf

Lateral wind pressure: pw = 15 psf

Lateral seismic force: pE = 10 psf

Assumption: wall is continuous and restrained against rotation at top and bottom.

Requirement: Design wall reinforcement using empirical design method

Solution:

Design for 1 ft width of wall: b = 1 ft

Axial dead load: PD = FD *Trib*2*b + RD * Trib. b = 3.6 kips

Axial roof live load:  PRL = RL * Trib*b = 0.3 kips

Axial floor live load: PFL = FL*Trib*b = 1.5 kip

Axial wind load: PW = 0 

Axial seismic load: PE = 0

Out-of-plan bending dead load moment: MD = 0

Out-of-plan bending Roof ive load moment:: MRL = 0

Out-of-plan bending floor live load moment: MFL = 0

From the continuous beam formula, the maximum negative moment is 0.1time span length lc.

Out-of-plan bending wind load moment: MW = 0.1 * pw*lc2*b = 216 lb-ft

Out-of-plan bending seismic load moment: ME = 0.1 * pE*lc2*b = 144 lb-ft

Load combination:

Case 1: Pu1 = 1.4 PD = 5.04 kip, Mu1  = 0

Eccentricity: eu1 = 0

Case 2:

Pu2 = 1.2 PD + 1.6 PFL + 0.5 PRL = 6.87 kip

Mu2 = 1.2 MD + 1.6 MFL + 0.5 MRL = 0

Eccentricity: eu2 = 0

Case 3:

Pu3 = 1.2 PD + 1.6 PW + 1.0 PFL + 0.5 PRL = 5.97 kip

Mu3 = 1.2 MD + 1.6 MW + + 1.0 MFL + 0.5 MRL = 0.346 ft-kip

Eccentricity: eu3 = Mu3/Pu3 = 0.695 in < 8 in /6 = 1.33 in (O.K.)

Case 4:

Pu4 = 1.2 PD + 1.0 PE + 1.0 PFL = 5.82 kip

Mu4 = 1.2 MD + 1.0 ME + + 1.0 MFL = 0.144 ft-kip

Eccentricity: eu4 = Mu4/Pu4 = 0.297 in < 8 in /6 = 1.33 in (O.K.)

Case 5:

Pu5 = 0.9 PD + 1.6 PW = 3.24 kip

Mu5 = 0.9 MD + 1.6 MW  = 0.346 ft-kip

Eccentricity: eu4 = Mu4/Pu4 = 1.28 in < 8 in /6 = 1.33 in (O.K.)

Strength reduction factor: f = 0.65

Gross section: Ag = b* h = 96 in2

Effective length factor: k = 0.8

Design axial strength of wall:

fPn = 0.55 f fc' Ag [ 1 - ( k lc / 32 h)2] = 82.1 kips

Maximum axial load: Pu3 = 5.97 kips   O.K.

Check vertical reinforcement for bending moment,

Maximum moment per foot width: Mu = 0.346 ft kip

Use one layer of reinforcement, place at middle of the section:

Effective depth: d = 4 in

Factor: Rn = Mu / (0.9*b*d2) = 24 psi, m = fy / (0.85 fc' ) = 23.5

Reinforcement ratio:  r = (1/m)*[1-Ö(1-2 m Rn / fy )] = 0.0004

Less than minimum vertical reinforcement ratio, use r  = 0.0012

Vertical reinforcement: As = 0.0012 Ag = 0.115 in2

Use #4 at 18 in o.c. area of reinforcement, 0.133 in2.

Minimum horizontal reinforcement: rt = 0.002, area of reinforcement, As = 0.192 in2.

Use #4 at 12" o.c. area of reinforcement, 0.2 in2.