Page 471 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.33

Pier 4:
h 6
= = 1 125
.
.
d 5 333
1 ⎛ ⎡ h ⎞ 3 ⎛ h ⎞ ⎤
∆ = = ⎜ ⎟ + ⎜ ⎟ ⎥
3
⎢
4
Et ⎣ ⎣ ⎝ d ⎠ ⎝ d ⎠ ⎦
m
= 1 [( . 3 +3 1 125)]
1 125)
( .
)( .
(1800 7 625 )
4 −
= 3.4496 10( )in.

R = 1 = 1 = 2860 kips//in.
4 ∆ (. ) ( 4 −
3 496 10 )
4
Pier 5:
h 4
= = 02 .
d 20
⎛ ⎡
3
h ⎞
⎛
h ⎞
∆ = 1 = ⎜ ⎟ + ⎜ ⎟ ⎥ ⎤
⎢
3
5 ⎝ d ⎠ ⎝ d ⎠
Et ⎣ ⎦
m
= = 1 [ (02 . ) + ( 3 02 . )]
3
)( .
( 1800 7 625)
−5
= 443 10 )inn.
(
.
R = 1 = 1 = 22 573 kips/in.
,
∆ (. )(
5 5 −
443 10 )
5
Piers 3 and 4 are in parallel. Therefore, the combined rigidity of these two piers
is the sum of their rigidities. Thus,
R 3+4 = R + R = 4003 + 2860 = 6863 kips/in.
4
3
Piers 3-4 and 5 are in series. Therefore, the reciprocal of the rigidity of Pier 3-4-5
is given by the sum of the reciprocals of the rigidities of piers 3-4 and 5. Thus,
R 3+4+5 = 5263 kips/in.
1 1 1 1 1 4 −
= + = + = . (
1 900 10 )
R 34 5 R 34 R 5 6863 22 573
,
++
+
Piers 2 and 3-4-5 are in parallel. Therefore, their combined rigidity is given by
the sum of their rigidities. Thus,

R 2+3+4+5 = R + R 3+4+5 = 2407 + 5263 = 7670 kips/in.
2
Pier 1 and Pier 2-3-4-5 are in series. Thus, the combined rigidity of these piers is
given by
1 1 1
= +
R 1 234 5 R 1 R 234 5
+
+
+
+
++
+

4 −
= 1 + 1 = 16 .8842 10 )
(
,
26 413 7670
1
R 1 234 5 = 4 − = 5944 kips/in .
++
+
+
(
.
1 6842 10 )