Page 470 - Design of Reinforced Masonry Structures

7.32 CHAPTER SEVEN

∆ = ∆ gross − ∆ solid + ∆ 2 + 3 + 4 + 5 = (1.2271 − 0.6288 + 1.2387)(10 ) = 1.837(10 ) in.
−4
−4
net
R = 1 = 1 = 5444kips/in.
∆ (. )( −4
1 837 10 )
net
Method C: In Fig. E7.6, all piers except Pier 1 can be considered as fixed against
rotation at top and bottom. Pier 1 would be considered as a cantilever, fixed at
the base and free to rotate at the top. With these assumptions, rigidities of various
piers can be calculated as follows:
Pier 1:
h 6
= = 0 167
.
d 36
∆ = 1
1 ⎡ ⎛ h ⎞ 3 ⎛ h ⎞ ⎤
Et 4 ⎝ + 3 ⎝ ⎥
m ⎢
⎣ d ⎠ d ⎠ ⎦
= 1 [( .167 + 3 ( .167 )]
3
0
0
)
4
( (1800 )( .625 )
7
= . 3 786 (10 )in.
− −5

,
R = 1 = 1 = 26 413kips/in .
∆ 1 (. )( −5
1
3 786 10 )
Pier 2:
h 10
= = 125
.
d 8
⎛ ⎡
3
h ⎞
⎛
h ⎞
∆ = 1 = ⎜ ⎟ + ⎜ ⎟ ⎥ ⎤ ⎥
⎢
3
2 ⎝ d ⎠ ⎝ d ⎠
Et ⎣ ⎦
m
= 1 [( . 3 +3 125)]
( .
125)
( 1800 7 625)
)( .
4 −
0
.
= 4 1553 1 ( 0 )in.

1 1
R 2 = ∆ = 4 − = 2407kips/in.
4 1553 10 )
2 (. )(
Pier 3:
h = 6 = 09 .
d 667
.
3
⎛
h ⎞
h ⎞
∆ = 1 ⎢ ⎛ ⎡ ⎜ ⎟ + ⎜ ⎟ ⎥ ⎤ ⎥
3
3 Et ⎝ ⎠ ⎝ d ⎠ ⎦
m ⎣
d
1
= [( 09 . ) 3 + 3 09 . )]
(
( 1800 7 625)
)( .
−4
.
)
= 2 498 10 )in.
(

R = 1 = 1 = 4003 kips/in .
3 ∆ (.498 )(10 4 − )
2
3