Page 469 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.31

= 1 [( . 3 + 3 0 2778)]
0 2778)
( .
( 1800 7 625)
)( .
5 −
=6 2883((10 )in.
.

Calculate the deflection of Pier 2 assuming it as fixed-ended.
−4
∆ = 4.1553 (10 ) in. (calculated earlier)
2
Calculate the deflection of the solid strip containing piers 3, 4, and 5 (ignore
openings), assuming it as fixed-ended. This strip contains piers 3, 4, and 5.
h 10
= = 05 .
d 20
h h ⎞ ⎤
3
⎛
h ⎞
∆ = 1 ⎢ ⎛ ⎡ ⎜ ⎟ + 3 ⎜ ⎟ ⎥
++5
solidstrip 34 Et ⎝ ⎠ ⎝ d ⎠ ⎦
m ⎣
d
(
= 1 [( 05 . ) 3 + 3 05 . ))]
)( .
( 1800 7 625)
=1 .184 (10 −4 )in.
Calculate the deflection of the solid strip containing piers 3 and 4, assuming it as
fixed-ended.
h 6
= = 03 .
d 20
h ⎞ ⎤
3
⎛ ⎛
h ⎞
∆ = 1 ⎢ ⎛ ⎡ ⎜ ⎟ + 3 ⎜ ⎟ ⎥
+
solidstriop 34 Et ⎝ ⎠ ⎝ d ⎠ ⎦
m ⎣
d
= 1 [( 03 . ) 3 + 3 03 . )]
(
)( .
( 1800 7 625)
= = 6 7541 10. ( −5 )in.
Calculate deflections of individual piers 3 and 4, assuming each as fixed-ended.
−4
∆ = 2.4984 (10 ) in. (calculated earlier)
3
−4
∆ = 2.4551 (10 ) in. (calculated earlier)
4

Calculate the net deflection of the Pier 3-4-5.
−
∆ = 1 = 1 = 1 2381 10 )in.
4
.
(
1
+
34 1 + 1 1 + 1
−
−
∆ ∆ 2 4984 10 ) 2 4551 10 )
4
4
(
.
.
(
3 4
∆ 3 + 4 + 5 = ∆ solid − ∆ solid 3 + 4 + ∆ 3 + 4 = (1.184 – 0.6754 + 1.2381)(10 ) = 1.7647(10 ) in.
−4
−4
∆ = 1 = 1 = 1 2387 10 )in.
−
4
(
.
+
++
234 5 1 + 1 1 + 1
∆ ∆ 4 1553 10 ) 1 7647710( − 4 )
−
4
(
.
.
++
2 3 4 5