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: f_y = 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: 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 / (bd²jk) = 269 psi

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

Check shear stress:

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.