Page 536 - Design of Reinforced Masonry Structures

7.98 CHAPTER SEVEN

wall, have yielded (in tension) so that stress in each of those bars equals f . Accordingly, the
y
maximum tensile force can be expressed as given by Eq. (7.92):
T max = A f (7.92)
s y
Ignoring the concrete displaced by the reinforcing bars in the compression zone as well as
the compression force in the bar in the compression zone of the wall, the maximum com-
pression force in masonry can be expressed as given by Eq. (7.93):

C max = 0.80 ′ f at (7.93)
m
where a = depth of compression zone in masonry. Equating C max to T max , one obtains the
first approximate value of a from Eq. (7.94):

At
a = sy (7.94)
.
080 ft ′
m
Note that the value of a given by Eq. (7.94) is approximate because the compression force
in the reinforcing bar located in the compression zone of the wall has not been accounted
for in Eq. (7.93). As a result, the actual value of a would be smaller than that given by
Eq. (7.94). Therefore, for the second trial, a smaller value of a than the one given by
Eq. (7.94) should be assumed in order to determine the compression force in masonry,
C, compression force in the reinforcing bar, C , and the tensile force resultant in tension
s
bars, T. Note that the forces in compression and tension bars would depend on stresses in
those bars which can be determined from strain distribution across the wall cross section,
which is defined by ultimate strain in masonry (0.0025 and 0.0035 for concrete and clay
masonry, respectively) and the position of neutral axis given by the relationship: a = 0.80c
[Chap. 4, Eq. (4-5a)]. Stresses in various bars can be determined from strains in those bars
(stresses in compression and tension bars nearest the neutral axis, in all likelihood, would
be less than yield stress, f ). The correct value of a must satisfy the force equilibrium condi-
y
tion expressed by Eq. (7.95):
∑ F = 0
y
C + C – T = 0 (7.95)
s
Eq. (7.95) cannot be solved directly because
1. The total compression force (C + C ) depends on the value of a which, in turn, depends
s

on the value of c (because a = 0.80c). Any variation in the value of a affects both
C and C . s
2. The total tensile force T also depends on the value of a. Note that the strain in the tension
bar closest to neutral axis might be smaller than the yield strain.
Accordingly, a few iterations might be required before force equilibrium condition is
satisfied. See Example 7.17. Readers should review Examples 5.9 through 5.11 (Chap. 5)
for an overview of this procedure.
Walls under axial loads can be treated using provisions discussed in Chap. 5. It is reit-
erated that shear walls are invariably subjected to out-of-plane loads. Such walls are sub-
ject to the following restrictions where the factored axial stress exceeds 0.05 ′ f (MSJC-08
m
Section 3.3.5.3).