Strength of Rectangular Section in Bending

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RC03
-
24

矩形斷面之撓曲強度

Strength of Rectangular Section in Bending

3
-
2

-

Analysis of RC Beam (WSD)

-

Double Reinforcement Beam (WSD)

-

Strength Design Method (SDM)

-

Nominal Moment Strength (M
n
)

-

Balance Steel Ratio (
r
b
)

-

Coefficient of Resistance (R
n
)

RC
樑之分析

Analysis

of RC Beam

RC03
-
25

Given:

Section
A
s
,
b
,
d


Materials
f
c

,
f
s


Find:

M
allow
= Moment capacity of section

STEP 1

: Locate Neutral Axis (
kd
)



n
n
n
k
r
r
r



2
2
3
/
1
k
j


where

ratio
ent
Reinforcem


bd
A
s
r
6
2.04 10 134
15,100
s
c
c c
E
n
E
f f

  
 
STEP 2

: Resisting Moment

Concrete:

2
2
1
d
b
j
k
f
M
c
c

Steel:

d
j
f
A
M
s
s
s

If
M
c

>
M
s

, Over reinforcement


M
allow

=
M
s

If
M
c

<
M
s

, Under reinforcement



M
allow

=
M
c

Under reinforcement is preferable because steel is weaker

than concrete. The RC beam would fail in ductile mode.

RC03
-
26

Example 3.3

Determine the moment strength of beam

4
DB 20

A
s

= 12.57 cm
2

80
cm

40
cm

f
c

= 65 ksc, f
s

= 1,700 ksc,

n = 10, d = 75 cm

0419
.
0
,
00419
.
0
75
40
57
.
12





n
bd
A
s
r
r
916
.
0
3
/
251
.
0
1
251
.
0
0419
.
0
)
0419
.
0
(
0419
.
0
2
2









j
k
M
c

=
0
.
5
(
65
)(
0
.
251
)(
0
.
916
)(
40
)(
75
)
2
/
10
5

= 16.81 t
-
m

M
s

= (
12
.
57
)(
1
,
700
)(
0
.
916
)(
75
)/
10
5

= 14.68 t
-
m (control)

RC03
-
27

雙筋樑
Double Reinforcement

When M
req’d

> M
allow

-

Increase steel area

-

Enlarge section

-

Double RC


only when no choice

e
s

e
c

M

T = A
s
f
s

C =

f
c
k b d

1

2

A
s

A’
s

e

s

d’

T’ = A’
s

f’
s

A
s1

f
s

A
s2

f
s

RC03
-
28

T = A
s
f
s

C =

f
c
kbd

1

2

T’ = A’
s

f’
s

T
1

= A
s1
f
s

C =

f
c
kbd

1

2

T
2

= A
s2
f
s

T’ = A’
s

f’
s

jd

d
-
d’

2
1
1
1
2
c c
s s
M M f kjbd
A f jd
 

2
2
( )
( )
c
s s
s s
M M M
A f d d
A f d d
 

 
  
 
Steel area A
s

=

1
c
s
s
M
A
f jd

+

2
( )
c
s
s
M M
A
f d d




M = M
1

+
M
2

Moment strength

RC03
-
29

Compatibility Condition

e
c

d

d’

kd

e

s

e
s

s
s
d kd
kd d
e
e


 

From Hook’s law:
e
s
=
E
s
f
s
,
e

s
=
E
s
f’
s

s s s
s s s
E f f
d kd
E f f kd d

 
  

1
s s
k d d
f f
k





2
1
s s
k d d
f f
k





RC03
-
30

(
A’
s

)

T
2

= A
s2
f
s

T’ = A’
s

f’
s

d
-
d’

Force equilibrium [
S
F
x
=
0

]

T’ = T
2

A’
s
f’
s

= A
s2
f
s

Substitute

2
1
s s
k d d
f f
k





2
1 1
2
s s
k
A A
k d d





RC03
-
31

Example 3.4

Design 40x80 cm beam using double RC

f
c

= 65 ksc, f
s

= 1,700 ksc,

n = 10, d = 75 cm

k = 0.277, j = 0.908, R = 8.161 ksc

w = 6 t/m

5
.0 m

Required M = (6.768) (5)
2
/ 8 = 21.15 t
-
m

Beam weight
w
bm

=
0
.
4

0
.
8


2
.4
(t/m
3
) =
0.768
t/m

M
c

= Rbd
2

=
8
.
161
(
40
)(
75
)
2
/
10
5

= 18.36 t
-
m < req’d
M

Double RC

5
2
1
18.36 10
15.86 cm
1,700 0.908 75
c
s
s
M
A
f jd

  
 
5
2
2
(21.15 18.36) 10
2.34 cm
( ) 1,700 (75 5)
c
s
s
M M
A
f d d

 
  

  
RC03
-
32

Tension steel
A
s

=
A
s1

+
A
s2

= 15.86 + 2.34 = 18.20 cm
2

USE 6DB20 (
A
s

=

18
.85 cm
2
)

Compression steel

2
2
1 1 1 1 0.277
2.34 4.02 cm
2 2 0.277 5/75
s s
k
A A
k d d
 

    

 
USE
2
DB
20
(
A
s

=

6
.28
cm
2
)

6
DB20

2
DB20

0
.80 m

0
.40 m

RC03
-
33

強度設計法
Strength Design Method (SDM)

Ultimate Stress Design (USD)

Advantage of SDM over WSD:

1
) Consider mode of failure

2
) Nonlinear behavior of concrete

3
) More realistic F.S.

4
) Ultimate load prediction


5
%

5
) Saving (lower F.S.)

RC03
-
34

Uncertainties in analysis, design and construction of

RC Structures

1
) Actual load


Assumed load

2
) Load distribution may differ

3
) Simplification of analysis

4
) Actual structure behavior not known perfectly

5
) Actual dimension may differ

6
) Steel position may not proper

7
) Actual material strength may differ

RC03
-
35

安全考量
Safety Margin

M

=
S

-

Q

>
0

load

strength

n d
S Q
 

design load

load factor > 1.0

nominal strength

strength reduction

factor < 1.0

n D L
S DL LL
  
 


n D L W
S DL LL W
   
   
General:

Load combination:

RC03
-
36

ACI Provision

Design strength



Required strength

Assumptions:

Moment:

n u
M M


Shear:

n u
V V


Trust:

n u
P P


n = Nominal strength

u = Ultimate load

1
) Internal forces


External loads

2
) No slip between concrete and steel

3
) Cross section remain plain

4
) Concrete can resist no tensile stress

1.4 1.7
n u D L
S S

 

RC03
-
37

荷重作用下之混凝土粱行為

Behavior of Concrete Beam under increasing load

A
s

h

b

A
s

d

c
e
s
e
f
s

f
c

ct
e
f
ct

開裂前
Before crack

RC03
-
38

A
s

開裂後
After crack

c
e
s
e
f
s

f
c

cu
e
s
e
f
s

f
c

crushing strain
cu
e

RC03
-
39

標稱彎矩強度
Nominal Moment Strength (M
n
)

c

b

c

c
C f cb



c
f

f
y

T=A
s

f
y

e
s

e
cu



0
x
F C T
S  
c s y
s y y
c
f cb A f
A f f d
c
f b f

r
 


 
 
2
2
( ) 1
1
1.7
y y
n s y y
c c
y
n y
c
f d f
M T d c A f d f bd
f f
f
M f bd
f
br br
b r
 
r
r
   
     
   
 
   
 
 
 

 
RC03
-
40

等值矩形應力塊
Equivalent Stress Distribution

(Whitney stress block)

c

b

c

c
C f cb



c
f

f
y

T=A
s

f
y

a=
b
1
c

a/2

c
C f ab



0.85
c c
f f

 

f
y

T=A
s

f
y

1
280
0.85 0.05
70
c
f
b


 
1
0.65 0.85
b
 


0
x
F C T
S  
0.85
0.85 0.85
c s y
s y y
c c
f ab A f
A f f d
a
f b f
r


 
 
2
(/2)
2(0.85)
1
1.7
y
n s y
c
y
n y
c
f d
M T d a A f d
f
f
M f bd
f
r
r
r
 
   
 

 
 
 
 

 
RC
03
-
41

平衡鋼筋比
Balance Steel Ratio
(
r
b
)

cu
cu y
c
d
e
e e


cu
cu y
c d
e
e e
 

 
 

 
0.85
c
C f ab


T=A
s

f
y

0.003
cu
e

c

d

/
s y y s
f E
e e
 
Concrete crushing

Steel yielding



0;
x
F C T
S  
1
0.85
c s y b y
f cb A f f bd
b r

 
1
0.85
c cu
b
y cu y
f
f
e
r b
e e
 


 
 

 
RC
03
-
42

r

=
r
b

: balance,
r

>
r
b

: over RC,
r

<
r
b

: under RC

Design:
r

<
[
r
max

=
0
.
75

r
b
]


Conservative design:
r

=
0
.
5
r
max

=
0
.
375

r
b


1
0.85
6,120
6,120
c
b
y y
f
f f
r b
 


 
 

 
6
0.003,/2.04 10
cu y y
f
e e
  
Substitute

RC03
-
43

Minimum steel ratio:
r
min

= 14 /
f
y

(Concrete first crack)


Coefficient of Resistance (
R
n
)

M
u
/


=

M
n

= R
n

b
d
2

/0.85
y c
m f f


1
1.7
y
n y
c
f
R f
f
r
r
 
 
 

 
1
1
2
n y
R f m
r r
 
 
 
 
2
2 2 0
y y n
m f f R
r r
  
2
2 4 8
2
1
1 1
2
y y n y
n
y y
f f mR f
mR
m f m f
r
 
 
   
 
 
 
2
1
1 1
n
y
mR
m f
r
 
  
 
 
 
RC03
-
44

f’
c

(ksc)

r
min

r
b

r
max

m

R
n

(ksc)

180

210

240

280

320

350

0.0035

0.0035

0.0035

0.0035

0.0035

0.0035

0.0197

0.0229

0.0262

0.0306

0.0338

0.0360

0.0147

0.0172

0.0197

0.0229

0.0253

0.0270

26.1

22.4

19.6

16.8

14.7

13.4

47.62

55.55

63.49

74.07

82.46

88.36

0
0.005
0.01
0.015
0.02
0.025
0
10
20
30
40
50
60
70
80
Strength Curve (
R
n

vs.
r
⤠景f卄㐰4剥楮景R捥敮e

Coefficient of resistance
R
n

(kg/cm
2
)

Reinforcement ratio
r

=
A
s
/
bd

f’
c

=
180
ksc

f’
c

= 210 ksc

f’
c

=
240
ksc

f’
c

= 280 ksc

Upper limit at 0.75
r
b

單筋樑
Design Procedure for Section with Tension Reinforcement only

STEP 1

Select approximate tension reinforcement ratio

min max
r r r
 
min max
1
1
14/0.75
0.85
6,120
6,120
0.85;280 ksc
280
0.85 0.05;280 560 ksc
70
0.65;560 ksc
y b
c
b
y y
c
c
c
c
f
f
f f
f
f
f
f
r r r
r b
b
 
 


 
 

 







 

   

 
 





Conservative design select
r

=
0
.
5
r
max

=
0
.
375

r
b


STEP 2

With
r
灲敳e琠捯浰畴攠
bd
2

required:

2
Required
n u
n n
M M
bd
R R

 
1
1 1
1.7 2
y
n y y
c
f
R f f m
f
r
r r r
 
 
   
 
 

 
 
where



=
0
.90
for flexure

STEP 3

Select
d

from recommended
h

oneway

slab

L/24

L/28

L/
10

L/20

BEAM

L/18.5

L/
21

L/8

L/16

STEP 4

Revise steel ratio
r

based on
bd
2

2
1
Compute 1 1
n
y
mR
m f
r
 
  
 
 
 
min max
Check
r r r
 
2 2
n u
n
M M
R
bd bd

 
4
.1) Exact method:

4
.2
) Approximate proportion:

(revised )
(original )
(original )
n
n
R
R
r r
 
*
Relationship between
R
n

and
r

is approximately linear.

STEP
5

Select steel reinforcement

s
A bd
r

STEP 6

Check strength of section

2
1
1.7
n u
y
n y
c
M M
f
M f bd
f

r
r

 
 
 

 
Example
Design B1 in the floor plan shown below.

Slab thickness = 12 cm

LL = 300 kg/m
2


= 280 kg/cm
2

Steel: SD40

c
f

B1

B
2

8.00

4.00

2.00

5.00

3.00

Reaction at B2’s ends =
wL
/2 = (2,331+504)(4)/2 = 5,670 kg

Slab DL = 0.12(2,400) = 288 kg/m
2

Ultimate load =
1
.4
(
288
) +
1
.7
(
300
) =
913
.2
kg/m
2

2
913.2(4) 913.2(3) 3 0.75
2,331 kg/m
3 3 2
 

 
 
 
Load on B2 =

B
2
weight (assume section
30


50
cm) =
1
.4
(
0
.3
)(
0
.5
)(
2
,
400
) =
504
kg/m

Load on B
1
:

B1

B
2
:
5
,
670
kg

2
,350 kg/m

1
,826 kg/m

5
,670 kg

5
.00
m

3
.00 m

5
,
670
kg

913
.2 kg/m

913.2

1
,437 kg/m

B
1
weight: simply support min. depth =
800
/
16
=
50
cm

Try section
30

60
cm,
w
u
=
1
.4
(
0
.3
)(
0
.6
)(
2400
) =
605
kg/m

M
max

=
2
,
431
(
8
.
0
)
2
/
8


=
19
,
448
kg
-
m

1
max
524(5)(5/2)
819 kg
8
819(3) 2,456 kg-m
R
M
 
 
1
,826 + 605 = 2,431 kg/m

8.00

5.00

2
,
350
-
1
,
826
=
524
kg/m

3.00

R
1

5.00

5
,
670
kg

3.00

max
5,670(5.0)(3.0)
8
10,631 kg-m
M


M
u

=
19
,
448
+
2
,
456
+
10
,
631
=
32
,
535
kg
-
m

Max. moment on B1:

USE DB
20
:
d

=
60
-

4
-

2
.0
/
2
-

0
.9
=
54
cm

0.85(280) 6120
(0.85) 0.0306
4,000 6120 4000
b
r
 
 
 

 
r
max
=
0
.
75
r
b

=
0
.
75
(
0
.
0306
) =
0
.
0230

2 2
4,000
16.81
0.85 0.85(280)
32,535(100)
41.32
0.9(30)(54)
y
c
u
n
f
m
f
M
R
bd

  

  
r
min

=
14
/
f
y

=
14
/
4
,
000
=
0.0035

2
1
Required 1 1
1 2(16.81) (41.32)
1 1
16.81 (4,000)
0.0114
n
y
mR
m f
r
 
  
 
 
 
 
  
 
 
 

r
min

=
0
.
0035
<
r






r
max

=
0.0230

OK

0.60

0.30

6
DB
20

BUT
6
DB20 need

b
min

= 35.7 cm

NG

Home work: redesign section

A
s

=
r
bd

=
0
.0114
(
30
)(
54
) =
18
.51
cm
2

USE 6DB20 (
A
s

=
18
.85 cm
2
)

Tension Steel Position in Beam

w

L

+ M
max

= wL
2
/
8

Bending Moment Diagram

d

Effective depth

Compression face

Centroid of

steel area

Elastic curve

Need reinforcement

L

L

wL
2
/
8

wL
2
/
14

wL
2
/
14

d

d

d

L

L

<< L

d

d

d

Small
-
M

Critical section

at face of supports

3
m

2
m

6
m

w

w

w =
1
t/m

1
.04 t
-
m

-
3
.48
t
-
m

2
.93 t
-
m

2
DB
20

7
DB20

6
DB
20

2
DB
20

2
DB20

A

A

7
DB
20

2
DB20

B

B

2
DB
20

6
DB20

C

C

A

A

B

B

C

C

3
m

2
m

6
m

ACI Moment Coefficients

(
a
)

1
/
24

1/14

0

1
/
11

1
/
16

1/14

1
/
10

1/11

1
/
16

1/11

1
/
11

(
Spandrel)

(
b
)

1
/
24

1/14

0

1
/
11

1
/
16

1/14

1
/
9

1/9

1
/
14

1/16

(
c
)

1/12

1
/
14

1/12

1
/
12

1/16

1
/
12

1/12

(
d
)

1
/
12

1/14

1
/
12

1/12

1
/
16

1/12

1
/
12