Page 265 - Design of Reinforced Masonry Structures

4.128 CHAPTER FOUR

calculate the deflection in concrete beams [4.2]. In using either Eq. (4.148) or Eq. (4.149),
M needs to be calculated at the stage of loading of the beam. Consider, for example, a
a
reinforced masonry beam. When the beam is supporting its own weight, the moment would
likely be smaller than its cracking moment (M ) so the beam would remain uncracked. The
cr
deflection under this loading condition (call it stage 1) can be calculated using the gross
moment of inertia, I . When superimposed dead load is applied, the moment increases. If the
g
total moment at this stage (call it stage 2) is greater than M , the beam would have cracked;
cr
therefore, the deflection due to the superimposed load should be calculated using the effec-
tive moment of inertia of the beam (I ). The applied moment in stage 2 loading, M , to be
a
eff
used in Eq. (6.148) or Eq. (4.149), equals the sum of moments due to self-weight of the
beam and the superimposed dead load. Next, when the live load is applied, the moment
increases again (call it stage 3), which causes increased cracking of the beam. The applied
moment, M , now equals the sum of moments due to (1) self-weight of the beam, (2) super-
a
imposed dead load, and (3) service load. The deflection due to service live load should be
calculated using the effective moment of inertia, I , calculated based on the value of M at
a
eff
stage 3 (which would be different than M at stage 2). The total deflection in the beam would
a
be the sum of deflections at the three loading stages just described. See Example 4.34.
Equation (4.148) or Eq. (4.149) is applicable to simply supported beams. For continu-
ous beams, the value of I would be different. Reference 4.26 suggests the following aver-
e
age values of I :
eff
Beams with two ends continuous:
I eff, avg = 0.70I + 0.15(I + I ) (4.150)
e1
em
e2
With one end continuous:
I eff,avg = 0.85I + 0.15(I e, continuous end ) (4.151)
em
where I = value of I at the midspan
em eff
I , I = values of I at the two ends
e1 e2 eff
For deflection of masonry members, MSJC-08 Section 1.13.3.2 requires deflection of rein-
forced masonry beams to be computed from Eq. (4.152) [a modified version of Eq. (4.148)]
unless effective stiffness properties are determined from a more comprehensive analysis:
⎛ M ⎞ 3 ⎡ ⎛ M ⎞ 3 ⎤
eff ⎜
.
I = cr ⎟ I + ⎢1 − ⎜ cr ⎟ I ⎥ ≤ I ≤ 05 I (4.152)
⎝ M ⎠ n ⎢ ⎝ M ⎠ ⎥ c cr n g
a ⎣ a ⎦
where I = moment of inertia of the net masonry section.
n
The upper limit (0.5I ) in Eq. (4.152) applies to reinforced masonry designed under the
g
strength design provisions of MSJC-08 code (Section 3.5.1.2).

For continuous members, I may be computed as the average of values obtained from
eff
Eq. (4.152) for the critical positive and negative moment regions.
More accurate deflections can be calculated by performing moment-curvature analysis.
This is particularly true for lightly reinforced sections with service load moments close to
the cracking moment. If the curvatures f are known at the ends and midspan of a concrete
beam, and if they are assumed to be parabolically distributed along the span, the midspan
deflection, ∆, of a simple or a continuous beam can be calculated from Eq. (4.153) [4.25]:
)
∆= L 2 (φ 1 +10φ 2 +φ (4.153)
3
96
where L = length of span, and f , f , and f are curvatures at the one end, midspan, and
1
2
3
at the other end, respectively. A discussion on the use of Eq. (4.153) can be found in
Refs. 4.27 and 4.28.