1-Reinforced Concrete Equations1
1-Reinforced Concrete Equations1
1-Reinforced Concrete Equations1
a
mρ
2
,
a
' '
0.85 f c b 0.85 f c bd ( )
M n= A s f s d− , M n=ρbd f s d−
2 2 ( )
2
'
a 0.85 f c β1∗0.003
'
M n=0.85 f ab d−c
2 ( ) ρ0.005=
f y 0.003+0.005
+ ρ'
ϵ t =ϵ y → ∅=0.65, ϵ t >0.005 → ∅=0.9, 0.85 f ' c β 1∗0.003 '
ϵ y <ϵ t < 0.005 → ∅=0.483+83.3 ϵ t ρb= +ρ ,
f y 0.003+ ϵ y
' '
Strength Method: T-beams 0.85 f c β 1∗d 0.003
ρ ' cy = +ρ '
fy d 0.003−ϵ y
Interior Exterior Isolated ( A s− A s 2 ) fy
b=bw +16 h f b=bw +6 h f b ≤ 4 bw a= '
, A s 2= A's
0.85 f c b
L L 1
b= b=bw + hf ≥ b w a
4
b=c /c
12
L
2
L
( )
M n 1=( A s−A s 2 ) fy d −
2
b=bw + n b=bw + n a
c / c spacing of beams
2 2 M n 1=0.85 f ' c ab d− ( ) 2
Ln clear distance ¿ next beam M n 2=A s 2 fy ( d−d ' )= A s ' fy ( d −d ' )
(ACI8.12.2),(ACI8.12.3)
M n=M n 1+ M n 2
0.85 f ' c h f (b−b w ) A Case(2) :Compression Steel not yielded
A sf = , ρ f = sf A
fy bw d
ρ= s < ρ' cy
h h bd h
( 2 ) ( 2 )
M nf = A sf f y d− f , M nf = ρf b w d f y d− f , M nf =0.85 f ' c hf ( b−b w0.85
ρ ' max =
) ( c2)
d−f ' f ,βM∗0.003
1
n=M nf + Mρnw∗f
+
' '
s
A sw = A s− A sf f y 0.003+ 0.004 f y
A sw fy ρw dfy A sw 0.85 f ' c β 1∗0.003 '∗f 's
a= = , ρ w= ρ ' 0.005= +ρ
0.85 f ' c b w 0.85 f 'c bw d f y 0.003+0.005 fy
' ' '
0.85 f c β 1∗0.003 ρ ∗f s
a a ρ b= + ,
f ya 0.003+ ϵ y
( ) ( )
M nw =A sw f y d− , M nw =ρw bw d f y d− , M nw=0.85 f ' c a bw d−
2 2 ( )
A s 2 ≠ A 's , 2A s 2 fy=A 's f 's
fy
ρf
M nw =f ' c b w d 2 ω ( 1−.59 ω ) , ω= ' y '
fc 2 600 A s− A s f y 600 A 's d '
c+ c− =0
1 ρ w fy 0.85 f ' c b β 1 0.85 f 'c b β 1
M nw =ρw f y b w d 1− 2
( )
2 0.85 f 'c f 's=600
c−d
c
'
, a=β 1 c
M nw M uw
Rnw = 2
= 2 '
bw d ∅ b w d fs
A s 2= A 's
fy
ρw =
0.85 f 'c
fy ( √
1− 1−
)
2 Rnw
0.85 f 'c
M n 1=( A s−A s 2 ) fy d − ( ) a
As 2
ρT =
a
bw d
'
0.85 f c β 1∗0.003
M n 1=0.85 f ' c ab d− ( ) 2
ρT . ϵ = + ρf M n 2=A s 2 fy ( d−d ' )= A s ' fs ' ( d−d ' )
¿0.005
f y 0.003+0.005
'
0.85 f c β 1∗0.003 M n=M n 1+ M n 2
ρT . max = +ρ
f y 0.003+0.004 f
Desing of shear force stirups:
0.85 f ' c β 1∗0.003 f fy Vu
ρ T . b= + ρf , ϵ y = y =
f y 0.003+ϵ y E s 200000 V n= , ∅=0.75
∅
Strength Method:
Rectangular Section/Doubly Reinforced: 1
V c= f' b d
Case(1) :Compression Steel yielded 6√ c w
A A '
ρ= s ≥ ρ' cy , ρ'= s 3 Av f y
bd bd V n @ x <3 V c smax ≤ d /2,600,
'
0.85 f c β1∗0.003 bw
ρmax = +ρ '
f y 0.003+ 0.004
3
3 Av f y Negative moment at interior face of
V n @ x >3 V c smax ≤ d / 4,300,
bw exterior support for members built
V n=V s+ V c , V s =V n−V c integrally with supports
For vertical stirrups Where support is spandrel beam
2
Av f y d Av f y d W u l n /24
s= =
Vs V n−V c Where support is a column
For equally spaced inclined bar or W u l 2n /16
Shear
stirrups
Shear in end members at face of
A v f y (cos α +sin α )d
s= first interior
Vs
support 1.15W u l n /2
For single bar or a group of Shear at face of all other supports
parallel bars at same place W u l n /2
V s= A v f y sin α < 1.5V c
ρf y
M n=f ' c bd 2 ω ( 1−.59 ω ) , ω= '
f c
Mn Mu
Slabs Rn= =
b d2 ∅ b d2
{
1.5 d b .
S max for Tepm .∧Shrinkage Rein . S ver .min = 1 1 max . Agg .
3
S max <5 h<460 mm 40
Reinforcement Limits
Two Way Slabs 0.01 A g < A st <0.08 A g ,
peremeter ofpanel A
hmin = ≥ 90 mm 0.01<ρt < 0.08, ρt = st
180 Ag
S max <2 h<460 mm e min =15+0.03 h∈mm
e=M u / Pu
short span l a Concentric Columns:
m= =
long span l b
Tied Columns:
Bending Moments Pn=( 0.85 f ' c ( A g −A st ) + A st f y ) ,
For short Span Pn max =0.8 P n ,
a , ¬¿ w l2a Pu=0.65 Pn max =0.52 P n
Continuous Edge a , ¬¿=C¿
M¿ Pu=0.52( 0.85 f ' c (A g −A st )+ A st f y )
{
a . pos ,≪¿ 16 d b .
Mid span M S max = 48 dtie
a , pos =M a , pos ,dl +M ¿
2
M a , pos ,dl =C a , dl w dl l a Least dimesion
2 1.92 Pu
a ,≪¿ w¿ l a A g=
0.85 f c +(f y −0.85 f ' c ) ρt
'
a , pos ,≪¿=C ¿
'
M¿ 1.92 Pu−0.85 f c A g
M a , pos A st = '
Discontinuous Edge −M a , dis = f y −0.85 f c
3
Spiral Columns:
For Long Span
Pn=( 0.85 f ' c ( A g −A st ) + A st f y ) ,
b , ¬¿ w l2b
Continuous Edge b , ¬¿=C¿ Pn max =0.85 P n ,
M¿ Pu=0.7 Pn max =0.6 Pn
b . pos ,≪¿ Pu=0 .6 ( 0.85 f ' c ( A g− A st ) + A st f y )
Mid span
M b , pos =M b , pos ,dl +M ¿
1.66 Pu
M b , pos ,dl =C b , dl w dl l 2b A g= ' '
0.85 f c +(f y −0.85 f c ) ρt
b ,≪¿ w¿ l 2b '
1.66 Pu −0.85 f c A g
b , pos ,≪¿=C ¿ A st = '
M¿ f y −0.85 f c
ρmin = → other f y fy
fy
25 mm ≤ S sp ≤ 80 mm
ρmin =0.002 → f y =280∨350 Mpa
5
π ( D−2 cover−2tie−d b ) −db∗No .bars 12V u
S ver = d=
No . of bars αs d
∅
( b0
+2
)√ f 'c b0
Eccentric Columns 6Vu
600 d= , for β c > 2
8
ab =β 1
600+f y
d=0.85 h
d ∅
βc( )√ f
+1 '
c b0
, b0 =4 ( a+d )
For symmetric columns
d−d́
é=e +
2
Pn=0.85 f́ c ab + Á s f́ s− A s f y
a−β1 d́
f́ s= ∗600 ≤ f y
a
β d−a
f s= 1 ∗600 ≤ f y
a
Á s f́ s ( d−d́ )
a
( )
Pn é=( 0.85 f´c a b ) d− +¿
2
Á s =
( a2 )
Pn é−( 0.85 f́ c ab ) d−
f́ s (d− d́)
Pn − Á s f´ s + A s f s P n− Ás ( f́ s−f s )
a= =
0.85 f́ c b 0.85 f́ c b
Footing
Wide Beam Shear
Vu
d=
1 '
∅
6
√ f c bw
Punching shear
Vu
d=
1 '
∅
6
√ f c b0