Page 267 - Design of Reinforced Masonry Structures

4.130 CHAPTER FOUR

Because, the applied moment is less than the cracking moment for the beam, the
beam is not cracked. Use the moment of inertia of gross section for calculating deflec-
tion at this stage of loading.
4
I = 8790 in. (calculated in Example 4.32)
g
6
(
.
E = 700 f ′ = 700 1500( )( ) = 1 05 10 ) psi = 1050 ksi
m
m
2
)
(.
(
∆ = 5 ML 2 = 5 3 02 12) ( 12) 3 = 0008 in.
.
SW
(
(
)
48 EI 48 1050 87990)
me
Loading Stage 2: Moment due to superimposed dead load on the beam (M ).
SD
10 12)
M = wL 2 = (. ) ( 2 = 18 k-ft > M = 12 21. k-ft
SD cr
8 8
Because the moment due to superimposed dead load is greater than the cracking
moment, the beam is cracked, so use the effective moment of inertia to calculate
deflection.
M = M + M = 3.02 + 18.0 = 21.02 k-ft
a
SW
SD
Calculate the effective moment of inertia from Eq. (4.148):
4
4
I = 8790 in , I = 6773 in (calculated in Example 4.32)
g cr
3
⎛ M ⎞ 3 ⎡ ⎛ M ⎞ ⎤
I = cr I + ⎢1 − cr I ⎥
⎟
eff ⎜ ⎝ M ⎠ ⎟ g ⎢ ⎣ ⎜ ⎝ M ⎠ ⎥ ⎦ c cr
a
a
3
⎞
.
.
⎛ 12 21 ⎞ 3 ⎡ ⎛ 12 21 ⎞ ⎤
= ⎜ ⎟ 8790) + 1 − ⎜ ⎟ ⎥ ( 6773)
⎢
(
.
.
⎝ 21 02 ⎠ ⎣ ⎝ 21 02 ⎠ ⎦
= 1722 8 5445 5 7168. + . = in 4
Deflection due to superimposed dead load is Eq. (4.156)
2
(. )(
∆ = 5 ML 2 = 5 18 0 12) ( 12) 3 = 0 062 in.
.
SD
(
)(
48 EI 48 1050 71668)
me
Loading Stage 3: Moment due to service live load on the beam (M ).
L
175 12)
M = wL 2 = (. )( 2 = 31.5 k-ft
L 8
8
M = M + M + M = 3.02 + 18.0 + 31.5 = 52.52 k-ft
SW
SD
L
a
Calculate the effective moment of inertia using M = 52.52 k-ft.
a
3
⎛ M ⎞ 3 ⎡ ⎛ M ⎞ ⎤
I = cr I + ⎢1 − cr I ⎥
⎟
eff ⎜ ⎝ M ⎠ ⎟ g ⎢ ⎣ ⎜ ⎝ M ⎠ ⎥ ⎦ c cr
a
a
3
⎞
.
.
⎛ 12 21 ⎞ 3 ⎡ ⎛ 12 21 ⎞ ⎤
= ⎜ ⎟ 8790) + 1 − ⎜ ⎟ ⎥ ( 6773)
(
⎢
⎝ 52 52 ⎠ ⎣ ⎝ 52 52 ⎠ ⎦
.
.
=
+
= 110 6688 6798 in 4