Page 162 - Design of Reinforced Masonry Structures

4.26 CHAPTER FOUR

Solution
a. Span, L = 10 ft 8 in. = 10.67 ft From Example 4.3
φM = M = 50 78 k-ft
.
n
u
2
For a uniformly loaded beam, M = wL 8/, from which we obtain
u
w = 8 M u
u 2
L
= ( 850 8 . ) = 357 k/ft
.
(10 67 ) 2
.
b. Uniform live load on the beam
D/L = 0.7, therefore, D = 0.7L
Because D/L < 8, U = 1.2 D + 1.6L load combination governs.
w = 1.2D + 1.6L = 3.57 kips/ft
u
1.2(0.7L) + 1.6(L) = 3.57 kips
L = 1.463 kips/ft
D = 0.7L = 0.7(1.463) = 1.024 kip/ft
(Check: 1.2D + 1.6L = 1.2(1.024) + 1.6(1.463) ≈ 3.57 kips/ft)

3
For an 8-in. nominal normal weight concrete masonry (grout weight 140 lb/ft ),
average weight = 84 lb/ft depth (h) of beam (Table A.19).
⎛ 24 ⎞
Self-weight of beam = 84 ⎜ ⎟ = 168 lb/ft
⎝ 12 ⎠

Superimposed uniform load = 1024 – 168 = 856 lb/ft

4.6 MODULUS OF RUPTURE AND NOMINAL
CRACKING MOMENT OF A MASONRY BEAM

4.6.1 Modulus of Rupture

Modulus of rupture refers to the tensile strength of an unreinforced beam, that is, maximum
stress in an unreinforced beam when it cracks under bending. Beams are required to be fully
grouted (code requirement). The modulus of rupture for masonry elements subjected to in-
plane or out-of-plane bending, f , is specified in MSJC-08 Code Section 3.1.8.2. Modulus
r
of rupture for masonry depends on several variables:
a. Layout of masonry units: stack or running bond.
b. Types of masonry units: solid or hollow.
c. Units are ungrouted, partially grouted, or fully grouted.
d. Type of mortar used in construction.