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- 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

ACI 530 - 05 Building Code Requirements for Masonry Structures

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.

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.

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.

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.

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.

7. Spacing for shear reinforcement shall not exceed d/2 or 48 in. (ACI 530 Sec. 2.3.5.3.1)

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

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.

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.

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

k^{2 }+ 2k(A_{s}/bd)
- 2 (A_{s}/bd) = 0

or k^{2 }+ 2kr-
2r =
0

Solve the equation,

k = Ö[2rn +
(rn)_{2} ]-
rn
[Eq. 1]

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

M = T (jd) = A_{s}f_{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 /
[jkbd^{2}]
[Eq.4]