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ACI 530 - 05 Building Code Requirements for Masonry Structures

Design load: W = 1000 lb/ft

Length of lintel opening: L = 8 ft

Depth of lintel: D = 24 in

Thickness of wall: 8 inch nominal

Compressive strength of masonry: f'_{m} =
1500 psi

Yield strength of steel reinforcement: f_{y} =
60000 psi

Design reinforcement for masonry lintel

1. Design flexural reinforcement

Calculate design moment: M = WL^{2}/8 = 8000 lb-ft

Actual Thickness of wall: b = 7.625 in.

Effective depth: d = 24 in -7.625 in /2 = 20.187 in

Elastic modulus of concrete masonry: E_{m} =
900 f'_{m} =
(900)(1500 psi)/1000 = 1350 ksi

Elastic modulus of steel: E_{s} =
29000 ksi

Transformation factor: n = E_{s} /
E_{m }= 21.5

Try one #5 bar, reinforcement area, As = 0.3 in2.

Reinforcement ratio r =
A_{s}/bd = (0.3)/(7.625)(20.187) = 0.002

Calculate Factor, k = Ö[2 r n
+ (r n)^{2} ]
- r n
= 0.251

Calculate Factor j = 1 - k/3 = 0.916

Check Tensile stress in steel, f_{s} =
M / (A_{s }j d )=
17.3 ksi

Allowable tensile stress: F_{s} =
24000 psi O.K.

Check Stress in masonry: f_{c} =
2 M / (bd^{2}jk) = 269 psi

Allowable stress in masonry: F_{c} =
f'm /3 = 500 psi O.K.

Calculate shear force: V = W L /2 = 4100 lbs

Shear stress: f_{v} =
V/bd = 26 psi

Allowable shear stress, Fv = Öf'm = 38.7 psi O.K.

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