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Design of Reinforced Concrete Masonry Lintel

• Span length and bearing length
• Allowable compressive stresses
• Allowable tensile stress
• Allowable Shear stress without shear reinforcement
• Allowable shear stress with reinforcement
• Spacing for shear reinforcement
• Reinforcement parallel to shear reinforcement
• Location of maximum shear
• Theory
• Design Example

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

f_c£   F_c = f_m’ / 3           ACI 530 Sec. 2.3.3.2.2

where f_m’ is compressive strength of masonry.

2. Compressive reinforcement can be used when lateral ties are provided according ACI 530 section, 2.1.4.6. Compressive stress of  in compressive reinforcement shall not exceed 0.4 f_y or 24000 psi, where f_y is yield stress of steel.

Allowable tensile stress

3. The tensile stress, f_s  must be less than the allowable tensile stress, F_s 

f_s£   F_s                     ACI 530 Sec 2.3.2

where F_s = 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, f_v shall be calculated as

            f_v = 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

            F_v = Öf_m     £ 50 psi         ACI 530 Eq. (2-20)

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

            F_v = 3.0 Öf_m    £ 150 psi             ACI 530 Eq. (2-23)

and minimum area of shear reinforcement shall be

            A_v = V s / (F_s 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. 2.3.5.3.1)

Reinforcement parallel to shear reinforcement

8. Reinforcement equal to 1/3 A_v 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.

Theory:

Assumptions:

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 = (E_s/E_m)A_s

where n = E_s/E_m is transformation factor, E_s (= 29,000 ksi) ,is elastic modulus of steel, E_m (= 900f_m'), is elastic modulus of concrete masonry and grout.  f_m' 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 = nA_s.

Take first moment of inertia about neutral axis

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

Rewrite the equation to 

k² + 2k(A_s/bd) - 2 (A_s/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) = A_sf_s jd   or     M = C jd = [(f_b kd b)/2] jd

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

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

f_s = M/(A_s jd)                                    [Eq. 3]

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

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

Design Example