Doubly Reinforced Beams

cypriotcamelUrban and Civil

Nov 29, 2013 (3 years and 6 months ago)

59 views

雙筋樑

Doubly Reinforced Beams

-

Strength of Doubly Reinforced Beam

-

Compression Steel Yield Condition

-

Design of Double RC Beams

-

Investigation of Double RC Beams

3
-
3

2
RC
-
01

雙筋梁之強度
Strength of Doubly Reinforced Beam

b

d

h

d’

A’
s

A
s

a

e
cu

=
0.003

e
s

T=A
s

f
y

x

s
e

c
f

85
.
0
0.85
c c
C f ba

=
s s s
C A f
  
=
1 2
( )
2
0.85 ( )
2
n
c s
c s s
M M M
a
C d C d d
a
f ba d A f d d
= 
 
 
=   
 
 
 
   
=   
 
 
Moment
:

Force:

1 2
1 2
1 1
2 2
0.85
0.85
c s
s y c s s
s s s
s y c c
s y s s s
T T T C C
A f f ba A f
A A A
T A f C f ba
T A f C A f

=  = 
  
= 
= 

= = =
  
= = =
2
RC
-
02

壓力鋼筋降伏之條件
Compression Steel Yield Condition

b

d

h

d’

A’
s

A
s

e
cu

=
0.003

e
s

x

s
e

0.003
6,120
s
s
s
s y
f
x d
x E
x d
f f
x
e



 

= =
 
 


 

= 
 
 
Compression steel yield condition:

or
s y s y
f f
e e
 
= =
1
6,120
6,120
6,120
y
s y
y
y y
y
x d
f f
x
d
x a
f

 



= =
 
 
 

= =

0.85
y sy y cy sy c y s y
T A f C C f ba A f
  
= =  = 
1
0.85
6120
1 0.85
6120
/ = reinforcement ratio of compression s
teel
c c
y y y
s
f f d
f f d f
A bd
  

     
  

  
     
     

     
 
=
2
RC
-
03

雙筋樑之設計步驟
Design Procedure of Double RC Beam

STEP 1:
Moment strength from single RC beam

1 1 1
1
2
1 1
Choose 0.75/
1
1.7
s sb s
y
n y
c
A A A bd
f
M f bd
f



  =
 
= 
 

 
STEP 2:
Addition moment strength required

2 1
/
n u n
M M M

= 
STEP 3:
Addition tension steel
A
s
2

2 2 2
( ) ( )
n s y
M T d d A f d d
 
=  = 
2
RC
-
04

STEP 4:
Total tension steel
A
s

=
A
s
1

+
A
s
2

STEP 5:
Stress in compression steel

1
1
,/
0.85
s y
c
A f
a x a
f b

= =

0.003 6,120
s s y
x d x d
f E f
x x
 
 
   

= = 
   
   
STEP 6:
Compression steel

2
s y s s
A f A f
 
=
2
RC
-
05

Example
1
:
Determine
A
s

and
A’
s

required.
M
LL

= 32 t
-
m,
M
DL

= 18 t
-
m


f’
c

= 240 ksc,
f
y

= 4,000 ksc

40 cm

50 cm

60 cm

d’
= 6 cm

A’
s

A
s

a

e
cu

=
0.003

e
s

C
c

T

x

C’
s

s
e

c
f

85
.
0
M
u

= 1.4 (18) + 1.7 (32) = 80 t
-
m

M
n

= M
u
/

= 80/0.9 = 89 t
-
m

2
RC
-
06

Moment strength from single RC beam

A
s1

= 0.75

b
bd
= 0.75(0.0262)(40)(50) = 39.3 cm
2
( )

1
2
1 1
2
1
1.7
0.0197 4.0
0.0197 4.0 40 50 1/100 63.6 t-m
1.7 0.24
y
n y
c
f
M f bd
f


 
= 
 

 

 
=     =
 

 
Addition moment strength required

M
n2

=
M
n

-

M
n1

= 89
-

63.9 = 25.4 t
-
m

Addition tension steel
A
s
2

2
2
2
25.4 100
14.4 cm
( ) 4.0(50 6)
n
s
y
M
A
f d d

= = =

 
2
RC
-
07

Total tension steel
:
A
s

=
A
s1

+
A
s2

= 39.3 + 14.4 = 53.7 cm
2

7
DB32(
A
s

= 56.3 cm
2
)

Stress in compression steel

1
1
39.3 4.0
19.3 cm,
0.85 0.85 0.24 40
/19.3/0.85 22.7 cm
s y
c
A f
a
f b
x a


= = =

 
= = =
2
22.7 6
6,120 6,120 4,500 kg/cm
22.7
s y
x d
f f
x

 
   

= = = 
   
   
Compression steel

2 2
2
14.4 cm USE 3DB25 ( 14.73 cm )
s y s s s
f f A A A
  
=  = = =
2
RC
-
08

Investigation of Double RC Beam

for under RC ( f
s

= f
y

before
e
c

=
0
.
003
)


Check compression steel yield condition:

Tension:
T
y
=
C
cy
+
C’
sy
,
A
sy
=
T
y
/
f
y

If Comp. steel yields
If Comp. steel not yields 6,120
s sy s y
s sy s y
A A f f
x d
A A f f
x

 =


 

 = 
 
 
1
0.85 6,120
s y c s
x d
A f f b x A
x



 
 
= 
 
 
0.85,
cy c sy s y
C f ba C A f
  
= =
Compression:

1
6,120
,
6,120
y y y
y
d
x a x
f


= =

Neutral axis:

Real neutral axis can be obtained from
T

=
C
c

+ C’
s

2
RC
-
09

拉力鋼筋降伏之條件
Tension Steel Yield Condition

1
6,120
,
(6,120 )
b b b
y
d
x a x
f

= =

Balance strain condition:

0.85,
cb c b sb s s
C f ba C A f
   
= =
Compression:

Stress in comp. steel at balance condition:

6,120
b
s y
b
x d
f f
x
 



= 
 
 
Tension:
T
b
=
C
cb
+
C’
sb

,
A
sb
=
T
b
/
f
y

2
RC
-
10

If Tension steel yields
If Tension steel not yields 6,120
s sb s y
s sb s y
A A f f
d x
A A f f
x
 =

 
 = 
 
 
1
1
0.85 6,120
6,120 0.85 6,120
s s c s
s c s
x d
A f f b x A
x
d x x d
A f b x A
x x




 
 
= 
 
 

 
   
 
= 
   
   
Real neutral axis can be obtained from
T

=
C
c

+ C’
s

2
RC
-
11

Investigation Procedure


STEP 2:
Determine
A
sb

in balance strain condition



Compare
A
s

with
A
sb

and compute
f
s

s
f

f
y

A
sy

A
s

STEP 1:
Determine
A
sy

in comp. steel yield condition



Compare
A
s

with
A
sy

and compute

s
f

f
y

A
sb

A
s

f
s

over RC

under RC

2
RC
-
12

STEP 3:
Determine
M
n

in actual condition



0.85
0.85
2
s s
s s s
c c s
s
c
n c s
T A f
C A f
C f ba T C
T C
a
f b
a
M C d C d d
=
 
=

= = 

=

 

=   
 
 
2
RC
-
13