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## Design of Unreinforced Masonry Wall

### General:

Unreinforced masonry walls are often used as load bearing or non-loading interior wall in one story building. Although it is called “unreinforced”, the masonry wall still needs to be reinforced with joint reinforcements.  In addition, ordinary and detailed plan masonry walls are allowed as shear walls in seismic design category A & B.  But it still needs to meet code required minimum reinforcement requirements.  By definition of ACI 530 Section 1.6, the tensile strength of masonry is considered, but the strength of reinforcing steel is neglected.

2. Lateral load from wind, seismic, earth pressure etc. ### Stresses in concrete masonry wall:

1. Compressive stress from vertical load
2. Compressive stress from flexural moment due to lateral load.
3. Tensile stress from flexural moment due to lateral load, eccentric moment, etc. ### Design requirements:

When the wall, pilaster, and column is subjected to axial compression and flexure.

1. The maximum compression stress shall satisfy the following equation

fa/Fa + fb/Fb £ 1                                               (ACI 530 Eq. 2-10)

Where, fa is compressive stress from axial load, fb is compressive stress from flexure; Fa and Fb, Fv  are allowable compressive stress and tensile stress calculation from equation below:

For member with h/r £ 99:

Fa = (1/4) fm’ {1-[h/(140r)]2}                         (ACI 530 2-12)

For member with h/r > 99:

Fa = (1/4) fm’ (70r/h)2                                               (ACI 530 2-13)

Where h is effective height of wall, column or pilaster, r is radius of gyration, fm’ is compressive strength of masonry

Allowable compressive stress from flexure:

Fb =  (1/3) fm                                                   (ACI 530 2-14)

1. The maximum axial force P shall satisfy the following equation

P£ (1/4) Pe                                                       (ACI 530 Eq. 2-11)

Where Pe is calculated as

Pe = [p2EmI/h2][1-0.577*e/r)2                          (ACI 530 2-15)

Where Em is elastic modulus of masonry, I is moment of inertia, h is the height of wall, column or plaster, e is eccentricity of axial load, r is radius of gyration.

1. The tensile stress due to flexure shall not exceed the value listed in ACI 530 Table 2.2.3.2. as shown below:

Allowable flexural tension stress for hollow core concrete masonry unit, psi.

 Masonry Mortar Portland cement/lime or mortar cement Masonry cement or air entrained Portland cement/lime Normal to bed joints M or S N M or S N Ungrouted hollow units 25 19 15 9 Grout hollow units 68 58 41 29 Parallel to bed joints in running bond M or S N M or S N Ungrouted or partially grout hollow units 50 38 30 19 Fully grouted hollow units 80 60 48 30

Example 1: Design of an interior unreinforced load bearing masonry wall Design data:

Tributary width: 30 ft

Height of wall: 12 ft

Normal width of wall: 8 in

Assume minimum eccentricity: 0.8 in

Seismic load from ASCE 7: 4 psf

Requirement: Check if an 8 in unreinforced masonry wall is adequate

Solution:

Axial load per foot width of wall from roof, P = (20 psf+20 psf)*30 psf = 1200 lb/ft

Eccentricity: e = 0.8 in

Eccentric moment: Mc = 1200*0.8/12 = 80 lb/ft

Seismic moment: Ms = 5 psf * (12ft)2/8 = 90 lb-ft/ft

Moment per foot width of wall, M = (80 + 90) lb-ft/ft = 170 lb-ft/ft

Width of wall: 7.625 in

Cross section area: A = 42.8 in2.

Moment of inertia: I = 330.9 in4.

Section modulus: S = 86.8 in3.

Radius of gyration: r = 2.78 in

Check flexural tensile stress: fb = M/S = 170*12/86.8 = 23.5 psf

Less than allowable tensile stress 25 psi for ungrouted hollow unit O.K.

Check compressive stress:

Weight of wall at mid-height: W = 55 psf * 6 ft = 330 lb/ft

Axial compressive stress: fa = (P+W)/A = 35.8 psi/ft

Slenderness ratio: h/r = 12*12/2.78 = 51.7   < 99

Use 1900 psi concrete masonry units with type S mortar,

Compressive strength of concrete masonry, fm’ = 1500 psi

Fa = (1/4) fm’ {1-[h/(140r)]2} = 323.7 psi

Allowable flexural strength:

Fb =  (1/3) fm = 500 psi

Combined stress equation:

fa/Fa + fb/Fb = 0.158 < 1    O.K.

Check axial force:

Elastic modulus, Em = 900 fm’ = 1.35x106 psi

Pe = [p2EmI/h2][1-0.577*e/r)2= 1.23x105 lb

(1/4) Pe = 3.09 x 104 lb > 1200+330 = 1530 lb     O.K.