Page 237 - Design of Reinforced Masonry Structures

4.100 CHAPTER FOUR

Moment due to distributed load on the end segments of the lintel is,
⎛ 131 ⎞
.
M = 1431 1 3(. ) −1092 1 31 3(. ) ⎜ − ⎟ = 938 lb-fft
u ⎝ 2 ⎠
Total factored moment due to Load Combination 1 is, M = 983 + 938 = 1921 lb-ft
u1
Load Combination 2: U = 1.2D + 1.6L r
The triangular load is all dead load due to the wall.

U = 1.2 (702) = 842 lb = W u
V = ½W = 842/2 = 421 lb
u
u
)
(
M = WL e = ( 842 6) = 842 lb-ft
u
u
6 6
Due to the distributed load on the end segments of the beam,
U = 1.2D + 1.6L = 1.2 (780) + 1.6 (1820) = 3848 lb/ft
Support reaction,
V = 3848 (1.31) = 5041 lb
u
⎛ 131 ⎞
.
M = 5041 3() − 3848 1 31 3( . ) ⎜ − ⎟ = 3302 lb-ftt
u ⎝ 2 ⎠
Total factored moment due to Load Combination 2 is
M = 842 + 3302 = 4144 lb-ft > M = 1921 lb-ft
u2 u1
Total factored shear is

V = 421 + 5041 = 5462 lb
u2
Thus, the design forces for the lintel are:
V = 5462 lb
u
M = 4144 lb-ft
u
b. Strength design of lintel:
Calculate the depth of lintel to resist the entire shear without any transverse rein-

forcement. From Eq. (4.127), the depth of lintel required is
d = V u = 5462 = 712. in..
reqd bf ′
.(
.
.
225 2 25 7 625) 2000
m
The overall depth of lintel required is, h = 7.12 + 4 = 11.12 in., say 12 in.
Therefore, of the total height of wall above the opening, only 12 in. depth of masonry
is required to resist the entire shear. This is much smaller than 7.25 ft height of masonry
available above the opening. Therefore, no transverse reinforcement is required in the
lintel to resist shear.
With h = 12 in., it is observed that the apex of the 45° traingle is at the same level
as the superimposed loads; arching action is justified. Calculate the amount of tensile