1. 10. Design of Singly Reinforced Beams
A. Concrete Stress Distribution
In actual distribution
Resultant C α f'c c b=
Location β c
In equivalent distribution
Location β c
a
2
=
Resultant C α f'c c b= γ f'c a b=
Thus, a 2 β c= β1 c= where β1 2 β=
γ α
c
a
=
α
β1
=
f'c 4000psi 5000psi 6000psi 7000psi 8000psi
α 0.72 0.68 0.64 0.60 0.56
β 0.425 0.400 0.375 0.350 0.325
β1 2 β= 0.85 0.80 0.75 0.70 0.65
γ
α
β1
=
0.72
0.85
0.847
0.68
0.80
0.85
0.64
0.75
0.853
0.60
0.70
0.857
0.56
0.65
0.862
Page 28
2. Conclusion: γ 0.85=
β1 0.85 f'c 4000psiif
0.65 f'c 8000psiif
0.85 0.05
f'c 4000psi
1000psi
otherwise
= 4000psi 27.6 MPa
8000psi 55.2 MPa
1000psi 6.9 MPa
B. Strength Analysis
Equilibrium in forces
X
0=
C T=
0.85 f'c a b As fs= (1)
Equilibrium in moments
M
0=
Mn C d
a
2
= T d
a
2
=
Mn 0.85 f'c a b d
a
2
= (2.1)
Mn As fs d
a
2
= (2.2)
Conditions of strain compatibility
εs
εu
d c
c
=
εs εu
d c
c
= or εt εu
dt c
c
= (3.1)
c d
εu
εu εs
= or c dt
εu
εu εt
= (3.2)
Unknowns = 3 a As fs
Equations = 2 X
0= M
0=
Additional condition fs fy= (From economic criteria)
Page 29
3. C. Steel Ratios
ρ
As
b d
=
As fy
b d fy
=
0.85 f'c a b
b d fy
= 0.85 β1
f'c
fy
c
d
= 0.85 β1
f'c
fy
c
dt
dt
d
=
ρ 0.85 β1
f'c
fy
εu
εu εs
= 0.85 β1
f'c
fy
εu
εu εt
dt
d
=
Balanced steel ratio
fc f'c= fs fy= εs εy=
fy
Es
=
ρb 0.85 β1
f'c
fy
εu
εu εy
= 0.85 β1
f'c
fy
600MPa
600MPa fy
=
εu 0.003 Es 2 10
5
MPa εu Es 600 MPa
Maximum steel ratio
ACI 318-99 ρmax 0.75 ρb=
ACI 318-02 and later ρmax 0.85 β1
f'c
fy
εu
εu εt
= with εt 0.004
For fy 390MPa εs
fy
Es
0.002
For εt 0.004
ρmax
ρb
εu εy
εu 0.004
=
5
7
= 0.714=
For εt 0.005
ρmax
ρb
εu εy
εu 0.005
=
5
8
= 0.625=
Minimum steel ratio
ρmin
3 f'c
fy
200
fy
= (in psi)
ρmin
0.249 f'c
fy
1.379
fy
= (in MPa)
Page 30
4. D. Determination of Flexural Strength
Given: b d As f'c fy
Find: ϕMn
Step 1. Checking for steel ratio
ρ
As
b d
=
ρ ρmin : Steel reinforcement is not enough
ρmin ρ ρmax : the beam is singly reinforced
ρ ρmax : the beam is doubly reinforced
ρ ρmax= As ρ b d=
Step 2. Calculation of flexural strength
a
As fy
0.85 f'c b
= c
a
β1
=
Mn As fy d
a
2
=
εt εu
dt c
c
= ϕ ϕ εt =
The design flexural strength is ϕ Mn
Example 10.1
Page 31
5. Concrete dimension b 200mm h 350mm
Steel reinforcements As 5
π 16mm( )
2
4
10.053 cm
2
d h 30mm 6mm 16mm
40mm
2
278 mm
dt h 30mm 6mm
16mm
2
306 mm
Materials f'c 25MPa fy 390MPa
Solution
Checking for steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1
f'c
fy
εu
εu 0.004
0.02
ρmin max
0.249MPa
f'c
MPa
fy
1.379MPa
fy
0.00354
ρ
As
b d
0.018
Steel_Reinforcement "is Enough" ρ ρminif
"is not Enough" otherwise
Steel_Reinforcement "is Enough"
As min ρ ρmax b d 10.053 cm
2
Calculation of flexural strength
a
As fy
0.85 f'c b
92.252 mm c
a
β1
108.532 mm
Mn As fy d
a
2
90.911 kN m
Page 32
6. εt εu
dt c
c
0.00546
ϕ 0.65 max
1.45 250 εt
3
min 0.9
0.9
The design flexural strength is ϕ Mn 81.82 kN m
E. Determination of Steel Area
Given: Mu b d f'c fy
Find: As
Relative depth of compression concrete
w
a
d
=
0.85 f'c a b
0.85 f'c b d
=
As fy
0.85 f'c b d
=
ρ fy
0.85 f'c
1=
Flexural resistance factor
R
Mn
b d
2
=
As fy d
a
2
b d
2
=
As
b d
fy
d
a
2
d
= ρ fy 1
1
2
w
=
R ρ fy 1
ρ fy
1.7 f'c
= 0.85 f'c w 1
1
2
w
=
Quadratic equation relative w
R
0.85 f'c
w 1
1
2
w
=
w
2
2 w 2
R
0.85 f'c
0=
w1 1 1 2
R
0.85 f'c
1= w2 1 1 2
R
0.85 f'c
1=
w 1 1 2
R
0.85 f'c
=
ρ 0.85
f'c
fy
w= 0.85
f'c
fy
1 1 2
R
0.85 f'c
=
Page 33
7. Step 1. Assume ϕ 0.9=
Mn
Mu
ϕ
=
Step 2. Calculation of steel area
R
Mn
b d
2
=
ρ 0.85
f'c
fy
1 1 2
R
0.85 f'c
=
ρ ρmax : the beam is doubly reinforced
(concrete is not enough)
ρ ρmax : the beam is singly reinforced
As max ρ ρmin b d= (this is a required steel area)
Step 3. Checking for flexural strength
a
As fy
0.85 f'c b
= (As is a provided steel area)
Mn As fy d
a
2
=
c
a
β1
= εt εu
dt c
c
= ϕ ϕ εt =
FS
Mu
ϕ Mn
= (usage percentage)
FS 1 : the beam is safe
FS 1 : the beam is not safe
Example 10.2
Required strength Mu 153kN m
Concrete section b 200mm h 500mm
d h 30mm 8mm 18mm
40mm
2
424 mm
Page 34
8. dt h 30mm 8mm
18mm
2
453 mm
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1
f'c
fy
εu
εu 0.004
0.02
ρmin max
0.249MPa
f'c
MPa
fy
1.379MPa
fy
0.00354
Assume ϕ 0.9
Mn
Mu
ϕ
170 kN m
Steel area
R
Mn
b d
2
4.728 MPa
ρ 0.85
f'c
fy
1 1 2
R
0.85 f'c
0.014
ρmin ρ ρmax 1
As ρ b d 11.783 cm
2
As 6
π 16mm( )
2
4
12.064 cm
2
Checking for flexural strength
a
As fy
0.85 f'c b
110.702 mm c
a
β1
130.238 mm
Mn As fy d
a
2
173.444 kN m
Page 35
9. εt εu
dt c
c
0.00743
ϕ 0.65 max
1.45 250 εt
3
min 0.9
0.9
FS
Mu
ϕ Mn
0.98
The_beam "is safe" FS 1if
"is not safe" otherwise
The_beam "is safe"
F. Determination of Concrete Dimension and Steel Area
Given: Mu f'c fy
Find: b d As
Step 1. Determination of concrete dimension
Assume εt 0.004 (Usually εt 0.005 )
ρ 0.85 β1
f'c
fy
εu
εu εt
= R ρ fy 1
ρ fy
1.7 f'c
=
ϕ ϕ εt = Mn
Mu
ϕ
=
bd
2
Mn
R
=
Option 1: b
Mn
R
d
2
=
Option 2: d
Mn
R
b
=
Option 3: k
b
d
= d
3
Mn
R
k
= b k d=
Step 2. Calculation of steel area
R
Mn
b d
2
=
Page 36
10. ρ 0.85
f'c
fy
1 1 2
R
0.85 f'c
=
As max ρ ρmin b d=
Step 3. Checking for flexural strength
a
As fy
0.85 f'c b
= c
a
β1
=
Mn As fy d
a
2
=
εt εu
dt c
c
= ϕ ϕ εt =
FS
Mu
ϕ Mn
=
Example 10.3
Required strength Mu 700kN m
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1
f'c
fy
εu
εu 0.004
0.02
ρmin max
0.249MPa
f'c
MPa
fy
1.379MPa
fy
0.00354
Assume εt 0.007
ρ 0.85 β1
f'c
fy
εu
εu εt
0.014 R ρ fy 1
ρ fy
1.7 f'c
4.728 MPa
Page 37
11. ϕ 0.65 max
1.45 250 εt
3
min 0.9
0.9 Mn
Mu
ϕ
777.778 kN m
Concrete dimension
k
b
d
= k
400
600
Cover 30mm 10mm 25mm
40mm
2
Cover 85 mm
d
3
Mn
R
k
627.231 mm b k d 418.154 mm
h Round d Cover 50mm( ) 700 mm b Round b 50mm( ) 400 mm
d h Cover 615 mm
b
h
400
700
mm
Steel area
R
Mn
b d
2
5.141 MPa
ρ 0.85
f'c
fy
1 1 2
R
0.85 f'c
0.015
As max ρ ρmin b d 37.741 cm
2
As 8
π 25mm( )
2
4
39.27 cm
2
dt h 30mm 10mm
25mm
2
dt 647.5 mm
Checking for flexural strength
a
As fy
0.85 f'c b
180.18 mm c
a
β1
211.976 mm
Mn As fy d
a
2
803.914 kN m
εt εu
dt c
c
0.00616
ϕ 0.65 max
1.45 250 εt
3
min 0.9
0.9
FS
Mu
ϕ Mn
96.749 %
Page 38