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Masonry Design

Design of Reinforced Concrete Masonry Lintel

Design Code: 

ACI 530 - 05 Building Code Requirements for Masonry Structures

Working stress design

Code requirements:

Span length and bearing length:

1.  The span length for member NOT built integrally with supports shall be taken as clear span plus depth of member but not exceed the distance between center of support.

2. In determine moment of continuous member over supports, the span length shall be taken as the distance between center of supports.

3. Minimum bearing length shall be 4 inches in the direction of member.

Allowable compressive stresses

1. The maximum compression stress in masonry, fc  must be less than allowable stress, Fc

fc£   Fc = fm / 3           ACI 530 Sec.

where fm’ is compressive strength of masonry.

2. Compressive reinforcement can be used when lateral ties are provided according ACI 530 section, Compressive stress of  in compressive reinforcement shall not exceed 0.4 fy or 24000 psi, where fy is yield stress of steel.

Allowable tensile stress

3. The tensile stress, fs  must be less than the allowable tensile stress, Fs 

fs£   Fs                     ACI 530 Sec 2.3.2

where Fs = 20,000 psi for grade 40 and 50 steel, 24000 psi for grade 60 steel, 30,000 psi for wire joint reinforcement.

Allowable Shear stress without shear reinforcement

4. Shear stres, fv shall be calculated as

            fv = V/bd                    ACI 530 Eq. (2-19)

where V is shear force, b is width of masonry wall, d is effective depth.

Allowable shear stress with reinforcement

5. Where shear reinforcement is not provided, allowable shear stress shall be calculated as

            Fv = Öfm     £ 50 psi         ACI 530 Eq. (2-20)

6. Where shear reinforcement is provided, allowable sheare stress shall be calculated as

            Fv = 3.0 Öfm    £ 150 psi             ACI 530 Eq. (2-23)

and minimum area of shear reinforcement shall be

            Av = V s / (Fs d)        ACI 530 Eq. (2-26)

where s is spacing of shear reinforcements.

Spacing for shear reinforcement

7. Spacing for shear reinforcement shall not exceed d/2 or 48 in. (ACI 530 Sec.

Reinforcement parallel to shear reinforcement

8. Reinforcement equal to 1/3 Av shall be provided parallel to the direction of shear reinforcement and shall be uniformly distributed.

Location of maximum shear

The location of maximum shear shall be calculated at d/2 from the support if the following conditions are met:

1. Support reaction introduces compression into the end regions of member and

2. no concentrated load located within d/2 distance from support.



1. Stress distributes linearly across depth of the section.

2. Tensile strength of CMU and grout is neglected.

3. Elastic theory apply.

Transform Section:

For calculation perpurse, tension area of steel is transformed to equivalent concrete area as

nAs = (Es/Em)As

where n = Es/Em is transformation factor, Es (= 29,000 ksi) ,is elastic modulus of steel, Em (= 900fm'), is elastic modulus of concrete masonry and grout.  fm' is strength of concrete masonry.

Location of neutral axis, (N.A.)

The location of neutral axis can be located by taking first moment of inertial about neutral axis.

Compression area = kd b

Tension transform section = nAs.

Take first moment of inertia about neutral axis

kd b (kd/2) - n As (d - kd ) = 0

Rewrite the equation to 

k2 + 2k(As/bd) - 2 (As/bd) = 0

or k2 + 2kr- 2r = 0

Solve the equation, 

k = Ö[2rn + (rn)2 ]- rn                        [Eq. 1]

Calculate tensile and compressive stresses

The applied moment is equal to the tensile force T or compressive force multiplies the moment arm jd,

M = T (jd) = Asfs jd   or     M = C jd = [(fb kd b)/2] jd

where j = 1 -k/3                                [Eq. 2]

The tensile stress in steel, fs,  can be calculate as

fs = M/(As jd)                                    [Eq. 3]

The compressive stress, fc,  in masonry can be calculated as

fc = 2 M / [jkbd2]                                [Eq.4]

Design Example