perhitungan design balok dengan mathcad

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 4000psiif
0.65 f'c 8000psiif
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 ρminif
"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 1if
"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