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DESIGN OF SHEAR-KEY BELOW COLUMN BASE-PLATE
Actual longitudinal shear force = Actual transverse shear force =
300 93
kN kN
Grade of concrete =
30
MPa
L/C : node :
Crushing strength (per B !"!#$ Cl% &%'%)%'* = #%+ , # MPa 18 N622+
= hear strength (per AC-.'* = =
& , #%! , ( # , '&!%# , #% * 0 #%! psi )##%+ psi 1.38 N622+
= Considering φshear =
0.85
(Concrete characteristic cu1e strength converted to c2linder strength 12 3ultipl2ing 3ultipl2ing 4ith a factor of #%*
;trans
7
BP
;long
LP
S!"# K!$ S!%&'() * Me31er = UC+03x+03x, 5= )#%) 33 B= )#%+ 33 tf = '' 33 t4 = 6%) 33
7, = 78 = , = 8 = p2 =
5epth of shear.ke2 considered = Grout thickness = 9ffective e31ed3ent of shear.ke2 = i8e of 4eld 1et4een shear.ke2 1ase.plate =
Pedestal si8e :
LP = BP =
&!# '!) &"6 )' )!#
c3 c3 c3 c3 Mpa
+00 33 +5 33 '6! 33 33 8 33 33 50 33 50 33
C""%'&$ (/ &! !"# !$ /#(2 %()%#!&! %#')4 %()'!#"&'() * Along longitudin longitudinal al direc direction tion = )#%)# )#%)# , '6! , '%## '%## = +&### N = ,0. ,0.08 08 N
Alon Along g tra trans nsve vers rsee dir direc ecti tion on = = =
)#% )#%+# +# , '6! '6! , '% '%## ## +&'&# N ,1. ,1.3, 3, N
(1et (1et4e 4eeen fla flang ngee and and 1ase ase pla plate* te* (1e (1et4een 4e 4e1 an and 1a 1ase pl plate*
1
C""%'&$ (/ &! !"# !$ /#(2 !: &#!)4& %()'!#"&'() *
) , () , ''.6%)# . ) , +* , * , #%6 = (effective in resisting direct shear* 11,.08 22+
M%- of 4eld a1out 7 = ( ('/')* , )#%+# 0 , & . ('/')* , ( 6%)# ) , + * 0 , ) & , ''%## , ( )#%+# /) * 0) * , #%6 , ( ) , ( )#%)# . ) , ''%## * , ( 6%)# / ) * 0 ) * * , ( #%6 , + * = 183+0958.1 22, M%- of 4eld a1out ? = ( )#%+# , (( )#%)# / ) * 0) * , ) ( )#%+# . 6%)# . ) , + * , ( ( )#%)# / ) . ''%## * 0 ) * , )* , #%6 , ( ' / ') * , ( ( )#%)#.) , ''* 0 * , ) , ( #%6 , + * ( & , '/') , '' 0 & , '' , ( ()#%).'' * / )* 0) * , ( #%6 , * = ,933935. 22, Let the 3a,i3u3 force along longitudinal direction = Let the 3a,i3u3 force along transverse direction =
3,0.,8 kN ;,9.93 kN
Ma,i3u3 direct longitudinal shear stress in 4eld = &#%& , '### / ( +''&%' * = !!%+6! N/33) Ma,i3u3 direct transverse shear stress in 4eld = 6&"%" , '### / ( +''&%' * = '))%+!+" N/33) Ma,i3u3 tension in 4eld due to 3o3ent created 12 long% shear = @ &#%! , '### , ( '6! / ) )! * , ( )#%+ / ) * / ')#"!%' = )')%! N/33) Ma,i3u3 tension in 4eld due to 3o3ent created 12 transv% shear = @ 6&"%" , '### , ( '6! / ) )! * , ( )#%) / ) * / &+""!%+ = ')%+& N/33) esultant longitudinal shear in 4eld = =
@ ( !!%6 * 0 ) * ( )')% * 0 ) * 0 #%! ))# N/33) = ))# N/33)
esultant transverse shear in 4eld = =
@ ( '))%6 * 0 ) * ( ')%+ * 0 ) * 0 #%! ))# N/33) = ))# N/33)
C!%')4 /(# "%&": #!:&")& !"# ") 2(2!)& *
Ma,i3u3 direct longitudinal shear stress in 4eld = ##%## , '### / ( +''&%' * = &"%#6 N/33)
1
Ma,i3u3 direct transverse shear stress in 4eld = "%## , '### / ( +''&%' * = '!%)' N/33) Ma,i3u3 tension in 4eld due to 3o3ent created 12 long% shear = @ ##%# , '### , ( '6! / ) )! * , ( )#%+ / ) * / ')#"!%' = '6%! N/33) Ma,i3u3 tension in 4eld due to 3o3ent created 12 transv% shear = @ "%# , '### , ( '6! / ) )! * , ( )#%) / ) * / &+""!%+ = ))%+!# N/33) Ma,i3u3 resultant shear in 4eld = (&"%#60)'!%)'0)('6%!))%+!*0)*0#%! = )'+%+6& O.K.
C""%'&$ (/ &! !"# !$ /#(2 %()%#!&! &!)'() %()'!#"&'() *
@ As per AC-.'$ considered that the stress distri1ution in concrete is in &! deg% A:()4 :()4'&')": '#!%&'() *
At side surface of the concrete pedestal$ height of the proDected area = = 4idth of the proDected area = = =
'6! ( +!# . )#%+ * / ) "%) 33 s3aller of )#%) ( +!# . )#%+ * and +!# +&"%+ 33
Area of concrete under tension =
=
"%)# , +&"%+# . )#%) , '6! ))''#%6) E%33
= =
))''#%6) , '%# #6")% 30;.893
Corresponding tensile capacit2 =
1
N N
A:()4 &#")
At side surface of the concrete pedestal$ height of the proDected area = = 4idth of the proDected area = = = Area of concrete under tension = = Corresponding tensile capacit2 =
= =
'6! ( +!# . )#%) * / ) "%& 33 s3aller of )#%+ ( +!# . )#%) * and +!# +!# 33 "%&# , +!#%## . )#%+ , '6! ))# E%33 ))#%## , '%# #'"!%& 308.195
1
N N
C""%'&$ (/ &! !"# !$ !%&'() /(# !"# ") =!)')4 * S!%&'() %:"'/'%"&'()*
1 = <=
'#'% '' "%)! '%#!
33 33 Class 1 Plastic (As per
F(# !"# ":()4 :()4'&')": '#!%&'() *
Ma, shear along this direction = ## kN Bending 3o3ent (M88* due to this shear = ## , ('6! / ) )! * > %6! kN.M Ma, shear capacit2 along long direction = '!# , Av Av = #%" , )#%+ , '' , ) > &#'%) 33) Fence long shear capcit2 = +#&%+" kN low shear Mo3ent capacit2 a1out 7.7 = )!# , )' / '### > !6%6! kN.M
ρ = =
o.k.
d v = =
F(# !"# ":()4 &#")
Ma, shear along this direction = " kN Bending 3o3ent (M,,* due to this shear = " , ('6! / ) )! * > '#%&+ kN.M Ma, shear capacit2 along transverse direction = '!# , Av Av = )#%) , 6%) > '&+%#& 33) Fence transverse shear capcit2 = )'"%&+ kN low shear Mo3ent capacit2 a1out ?.? = )!# , &"6 / '### > ')&%)! kN.M
ρ = =
o.k.
d v = =
I)&!#"%&'() #"&'( /(# 2(2!)&*
Mu8/Mall Mu,/Mall #%! #%# > H!)%! !"# !$ %""%'&$ * A:()4 :()4'&')": '#!%&'() > 3aller of +&#%# $ &#%&6"" $ #6%" = 30;.89 N A:()4 &#") 3aller of +&'%& $ 6&"%"&' = +19., N
#%'"!
I)&!#"%&'() C!%
Actual Longitudinal hear / Longitudinal hear Capacit2 Actual
## / # " / )'" '%&#
'%&
=
#%++
$
+#&%+"
kN
$
)'"%&+
kN
='$o%k%
hence$ o%k%
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10.0 DESIGN OF BASE-PLATE U)=#"%! ="$ S!42!)& 1 + Sp
Me31er = UC+03x+03x, LP5 = ,50 33 BP5 = ,50 33
tw
t f D BPL BPD
LPL =
300 33
BPL =
300 33 1 33
dia =
dia
B LPL
nos = 5= B= tf =
, 9;H 33 9;H 33
t4 =
9;H 33 ')# 33
9;H 33
p =
LPD
5esign against crushing of concrete Ma,i3u3 vertical co3pressive load on 1ase = Ma,i3u3 vertical tensile load on 1ase =
+38 kN 9;H kN
Assu3e thickness of 1ase.plate =
+0
33
Considering a &!o dispersion of load on pedestal$ the total 1earing area = 9;H =
9;H 33)
(efer Cl% &%'%%) of Bs!"!#:)###*
Bearing Pressure (s* = 9;H N/33) ' N/33)
( ;cu =
30
N/33) *
#%+ , Grade of Concrete @#Mpa @efer Cl%&%'%)%' of B !"!#.':)### 9;H Iield tress of tructural teel =
+50 Mpa
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Calculation of 1aseplate thickness fro3 anchor.1olt tension criteria
9;H
= Load tranferred to each flange =
9;H kN 9;H
= Bending Mo3ent at 4e1 edge = Bending Mo3ent at flange edge
9;H kN 9;H 9;H
9;H kN.3 9;H kN.3
Considering a &!o dispersion$ 9ffective 4idth of 1ase.plate section resisting 1ending 3o3ent at 4e1.edge = 9;H = 9;H 33 9ffective 4idth of 1ase.plate section resisting 1ending 3o3ent at flange.edge = 9;H = 9;H 33 Baseplate thickness fro3 anchor.1olt tension criteria due to 1ending 3o3ent at 4e1 edge =
9;H
=
9;H
33
Baseplate thickness fro3 anchor.1olt tension criteria due to 1ending 3o3ent at flange edge =
9;H
=
Fence$ reEuired thickness of Base.Plate = 9;H
9;H
33
9;H 33 )# 33 9;H
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DESIGN OF BASE-PLATE BRACED BAY SEGMENT -1 Sp
Me31er = UC+5,x+5,x;3 LP5 = 500 33 BP5 = 500 33
tw
t f
dia
D BPL BPD
LPL =
350 33
BPL =
350 33 +0 33
dia =
B LPL
nos = 5= B= tf =
, 9;H 33 9;H 33
t4 =
9;H 33 '&# 33
9;H 33
p =
LPD
5esign against crushing of concrete Ma,i3u3 vertical co3pressive load on 1ase = Ma,i3u3 vertical tensile load on 1ase =
335., kN 9;H kN
Assu3e thickness of 1ase.plate =
+0
33
Considering a &!o dispersion of load on pedestal$ the total 1earing area = 9;H =
9;H 33)
Bearing Pressure (s* = 9;H N/33)
( ;cu =
# N/33) *
' N/33) #%+ , Grade of Concrete @#Mpa @efer Cl%&%'%)%' of B !"!#.':)### 9;H Iield tress of tructural teel =
+50 Mpa
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Calculation of 1aseplate thickness fro3 anchor.1olt tension criteria
9;H kN 9;H 33
5istance of anchor 1olt centre.line fro3 the flange edge = 9;H 33 The loads transferred to web and flanges are considered to be inversely proportional to the perpendicular distance from centre of anchor bolt. Fence$ Load tranferred to 4e1 =
9;H
= Load tranferred to each flange =
9;H kN 9;H
= Bending Mo3ent at 4e1 edge = Bending Mo3ent at flange edge
9;H kN 9;H 9;H
9;H kN.3 9;H kN.3
Considering a &!o dispersion$ 9ffective 4idth of 1ase.plate section resisting 1ending 3o3ent at 4e1.edge = 9;H = 9;H 33 9ffective 4idth of 1ase.plate section resisting 1ending 3o3ent at flange.edge = 9;H = 9;H 33 Baseplate thickness fro3 anchor.1olt tension criteria due to 1ending 3o3ent at 4e1 edge =
9;H
=
9;H
33
Baseplate thickness fro3 anchor.1olt tension criteria due to 1ending 3o3ent at flange edge =
9;H
=
Fence$ reEuired thickness of Base.Plate = 9;H
9;H
33
9;H 33 )# 33 9;H
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DESIGN OF BASE-PLATE BRACED BAY SEGMENT -+ Sp
Me31er = UC+03x+03x, LP5 = ,50 33 BP5 = ,50 33
tw
t f
dia
D BPL BPD
LPL =
300 33
BPL =
300 33 1 33
dia =
B LPL
nos = 5= B= tf =
, 9;H 33 9;H 33
t4 =
9;H 33 ')# 33
9;H 33
p =
LPD
5esign against crushing of concrete Ma,i3u3 vertical co3pressive load on 1ase = Ma,i3u3 vertical tensile load on 1ase =
1;.; kN 9;H kN
Assu3e thickness of 1ase.plate =
+0
33
Considering a &!o dispersion of load on pedestal$ the total 1earing area = 9;H =
9;H 33)
Bearing Pressure (s* = 9;H N/33)
( ;cu =
# N/33) *
' N/33) #%+ , Grade of Concrete @#Mpa @efer Cl%&%'%)%' of B !"!#.':)### 9;H Iield tress of tructural teel =
+50 Mpa
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Calculation of 1aseplate thickness fro3 anchor.1olt tension criteria
9;H kN 9;H 33
5istance of anchor 1olt centre.line fro3 the flange edge = 9;H 33 The loads transferred to web and flanges are considered to be inversely proportional to the perpendicular distance from centre of anchor bolt. Fence$ Load tranferred to 4e1 =
9;H
= Load tranferred to each flange =
9;H kN 9;H
= Bending Mo3ent at 4e1 edge = Bending Mo3ent at flange edge
9;H kN 9;H 9;H
9;H kN.3 9;H kN.3
Considering a &!o dispersion$ 9ffective 4idth of 1ase.plate section resisting 1ending 3o3ent at 4e1.edge = 9;H = 9;H 33 9ffective 4idth of 1ase.plate section resisting 1ending 3o3ent at flange.edge = 9;H = 9;H 33 Baseplate thickness fro3 anchor.1olt tension criteria due to 1ending 3o3ent at 4e1 edge =
9;H
=
9;H
33
Baseplate thickness fro3 anchor.1olt tension criteria due to 1ending 3o3ent at flange edge =
9;H
=
Fence$ reEuired thickness of Base.Plate = 9;H
9;H
33
9;H 33 )# 33 9;H
