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PPI2PASS SE Exam Review Course Fall 2016 Lecture 23 Structural Engineering Course
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Datasheet for Detonation Flame- Arrestor
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Janin Letak Lintang
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NOTES ON SHIP HANDLING AND EFFECT OF TRANSVERSE THRUST ON MOVEMENT OF SHIP
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seismic inversionDescription complète
OMS INFRA CONSULTANTS PVT. LTD.
5 DESIGN OF SEISMIC ARRESTORS The Seismic arrestor is designed for the seismic force transmitted to it from the superstructure. Total dead load and sdl from the Super structure
=
2
Total live load on the pier cap only only
20 %
x
=
8140.0 KN
=
1000 KN
4070.00
live live load load is to be cons consid ider ered ed in the the tran transv sver erse se seis seismi mic c calc calcul ulat atio ion n
Design of Transverse seismic arrestor Lo Longitudinal Seismic coefficient
=
0.54
Total seismic force in the longitudinal direction
=
0.54
=
4503.60 KN KN
Tota Totall numbe mber of arre rresto stors in the the lon longitud itudin inal al direc irecti tio on Load on each arrestor in the longitudinal direction
=
x
8340.0
=
2
4503.60
x
1
x
1200
2 =
2251.8
Assuming Size of longitudinal longitudinal seismic arrestor
=
950
Allowable stress for
=
11.67
M 35
Concrete
Force on on ea each se seismi c arrestor in in th the lo longitudinal di di re rection =
KN
Mpa
2251.8 KN
Height of seismic arrestor
=
1200 mm
Height of pedastal and bearing
=
800 mm
Z X
Transverse direction
950 Check B1 B1 for for CORBEL :1200 L1 =
1000 mm, ;
D =
400 1200
L1/D= 800
= 950
1000
/
950
1.05
>
1.00
1000
Hence He nce this is Not a Corbel
950
mm
OMS INFRA CONSULTANTS PVT. LTD.
Moment at the base of the seismic arrestor
=
2251.8
x
1000 1000
= d
=
950
=
894 mm
-
2251.8 KNm
40
-
16
Permissibl e stress es vid e IRC:21 - 2000. M 45 Fe 500
=
15.00
N/mm
st
=
240
N/mm
Modular Ratio ,
m
=
2
cbc
2
10
Design Constants n
=
0.385
;
j =
Effective Depth Required, "deff"
0.872
=
;
Q =
M Q x beff
2.515
;
=
864 mm
<
894 mm O.K
Reinforcement Required, Ast reqd.
= 240
Provide
=
8026
12
nos of tor
x mm
=
M st x j x d
2251.80 0.872
x x
1000000 894.0
x
1.5
2
32 mm
=
9651 mm >
2
8026
mm
2
Check for shear Reinforcement provided
Shear stress
=
=
9651
x
100
950
x
1200
2251.8
x
1000
950
x
894
=
0.85
=
2.65
Mpa
=
0.395
Mpa
As per Table 12B of IRC :21-2000, Permissible shear stress for
0.85
% of steel for M45
As per Clause 304.7.1.3.3 of IRC :21-2000, the permissible stress in concrete can be increased by a factor of, d
=
1
+
5P
but not exceeding
1.5
Ag. fck d
=
1
+
5 950
=
1.22
<
1.5
Maximum allowable shear stress on pedestals
=
x x
2251.8 1200
x x
1.22
x
0.395
1000 45
%
OMS INFRA CONSULTANTS PVT. LTD.
=
0.482
<
2.65
Mpa
Hence Shear reinforcement is required. Asw
=
T - Tc
=
= st
=
Assuming
[ { ( T - Tc ) x S } / st. d ] (
2251.80
x
1000
)
-
(
0.482
x
950
x
894.0
1842231 KN 200.0
2
N/mm
8 L - Tor 12
( As per Table 10 of IRC :21-2000 for Fe 500 ) stirrups
