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## Reinforced Concrete Masonry Lintel Design example ### Design code:

ACI 530 - 05 Building Code Requirements for Masonry Structures

### Design data:

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: fy = 60000 psi

### Requirement:

Design reinforcement for masonry lintel

### Solution:

1. Design flexural reinforcement

Calculate design moment: M = WL2/8 = 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: Em = 900 f'm = (900)(1500 psi)/1000 = 1350 ksi

Elastic modulus of steel: Es = 29000 ksi

Transformation factor: n = Es / Em = 21.5

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

Reinforcement ratio r = As/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, fs = M / (As j d )= 17.3 ksi

Allowable tensile stress: Fs = 24000 psi        O.K.

Check Stress in masonry: fc = 2 M / (bd2jk) = 269 psi

Allowable stress in masonry: Fc = f'm /3 = 500 psi    O.K.

### Check shear stress:

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

Shear stress: fv = V/bd = 26 psi

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