Masonry Magazine June 1973 Page. 32
block wythe. The abrupt break in the fiber stresses at the mid-point of the wall section corresponds to the change from the low modulus block to the higher modulus brick.
Structural design requirements included in "Specification for the Design and Construction of Load-Bearing Concrete Masonry (NCMA 1970)" which are particularly pertinent to composite walls are as follows:
"3.1.4 Combination of Dissimilar Units. In composite walls or other structural members composed of different kinds or grades of units or mortars, the maximum stresses shall not exceed the allowable for the weakest of the combination of units and mortars of which the member is composed except as noted below."
"3.1.4.1 Exception. The maximum compressive stress permitted in composite walls or other structural members may be based upon prism tests."
"3.8.5 Eccentricity Normal to Plane of Member. 3.8.5.1 In walls, columns and cavity walls loaded on both wythes, the eccentricity of the load shall be considered with respect to the center of resistence of the member, taking into account variation in moduli of elasticity of members composed of different materials."
Example Problem. The following example problem illustrates how the transformed area principle can be used in the structural design of a composite concrete masonry wall. Allowable stresses and other design requirements used in the example are based on "Specification for the Design and Construction of Load-Bearing Concrete Masonry."
Given: 10-inch composite wall constructed with 4-inch concrete brick and 6-inch solid concrete block; concrete brick compressive strength 6000 psi; concrete block compressive strength 3000 psi: Type S mortar; wall height 10 ft.
Find: Neglecting weight of wall, compute wall stresses for vertical loads shown in Figure 3 where (a) P₁ 10 kips/ft and P. 2 kips/ft and (b) P₁ = 0 and P₂ = 2 kips/ft.
Solution:
(a) Determine sectional properties of wall.
From Table 3-1 of "Specification for the Design and Construction of Load-Bearing Concrete Masonry."
f'm (concrete brick) = 2400 psi
f'm (solid block) = 1700 psi
E brick
n= E block
=
2400(1000) -1.41
1700(1000)
Moments about A-A:
Block + Joint:
Area Moment Arm
(6)(12)
(3)
Concrete brick:
X
(1.41)(12)(3.62) x (7.81)=
133.5 sq. in.
696
= 5.2 in.
133.5
X
216
e=
Virtual eccentricity (e):
Pie + Pe
M Pie
P
=
P1 + P
5.62
(10)(5.2-4.81)+(2)(5.2-3)
e=0.88 in.
9.62
12
1/6 = = 1.6 in.
6
Therefore, et/6 and entire wall section is subjected to compressive stress.
Determination of computed and allowable stresses:
P 12000
A 133.5
= 89.8 psi
F. (solid block) =
0.2 fm 1-40t
[1]
[1-(409.62)
0.2 (1700) 1-(に
340 (0.97) 330 psi
Pe (12000) (0.88)
480
Therefore, f, F.
696 in.
fm==
= 52.3 psi
202
Fm solid block) = 0.3 f'm =
Moment of inertia about centroidal axis, x-x:
Block + Joint:
12(6)
12
+(72)(5.2-3.0)2 = 564
Concrete Brick:
(1.41)(12)(3.62)+
12
61.5 (4.42-1.81)² = 487
Ix-x564+4871051 in.
Section modulus:
S (block face) =
S (brick face) =
C
I 1051
5.2
I1051
4.42
C
P
P
= 202 in.
= 238 in.
0.3 (1700) = 510 psi
Therefore, fm Check:+
fm
1
F Fm
89.8 52.3
+
330 510
0.272+0.102
0.3741.. OK
Combined Stress (extreme block fiber)
fa+f=89.8+52.3=
142.1 psi (compression)
Combined Stress (extreme brick fiber)
= n (f-f) =
(1.41) 89.8-
(12000) (0.88)
(1.41) (89.8-44.5)
64 psi (compression)
Combined Stress (at block-brick interface):
Combined Stress (block) =
(12000) (0.88)
1051/(6.0-5.2)
82 psi (Comp,; block)
Combined Stress (brick) =
n(82) (1.41)(82) = 115 psi comp.
89.8-
=