# Strength of Rectangular Section in Bending

Urban and Civil

Nov 26, 2013 (5 years and 2 months ago)

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RC03
-
24

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

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

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

Strength Design Method (SDM)

Ultimate Stress Design (USD)

1
) Consider mode of failure

2
) Nonlinear behavior of concrete

3
) More realistic F.S.

4

5
%

5
) Saving (lower F.S.)

RC03
-
34

Uncertainties in analysis, design and construction of

RC Structures

1

2

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

Safety Margin

M

=
S

-

Q

>
0

strength

n d
S Q
 

nominal strength

strength reduction

factor < 1.0

n D L
S DL LL
  
 

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

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

1
) Internal forces

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

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

Before crack

RC03
-
38

A
s

After crack

c
e
s
e
f
s

f
c

cu
e
s
e
f
s

f
c

crushing strain
cu
e

RC03
-
39

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

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

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

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

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

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
 

 
 
 

B
2
weight (assume section
30

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

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