13. 13. Col Colum um n Design
Columns: - Axially loaded columns:
M=0
- Eccentric columns:
M
≠
0
Short columns:
Pu , M u
Long (slender) columns:
Pu , M u ⋅ δns
13.1 13.1.. Ax ially Lo aded Colu mns Safety provision provision Pu
≤ ϕPn.max
where Pu
= required strength
ϕPn.max
= design strength
ϕ
= strength reduction factor
Pn
= nominal strength
Pn.max
= maximum (allowable) strength
For tied columns:
ϕPn.max = 0.80⋅ ϕ ⋅ 0.85⋅ f'c⋅ Ag
(
with
−
Ast
) + Ast⋅ f y
−
Ast
) + Ast⋅ f y
ϕ = 0.65
For spirally reinforced reinforced c olumns:
ϕPn.max = 0.85⋅ ϕ ⋅ 0.85⋅ f'c⋅ Ag
(
with
ϕ = 0.70
where Ast
= area of s teel reinforcement
Page 76
Ag = b c⋅ h c Ac = Ag
−
= area of gross section of column Ast
= area of concrete section
Column steel ratio: Ast ρg = = 1% .. 8% Ag
Lateral Lateral ties Acc ording ording ACI ACI 7.10.5 t he diameters diameters of lateral tie are For longitudinal bar D ≤ 32mm:
Dv ≥ 10mm
For longitudinal bar D > 32mm:
Dv ≥ 12mm
The spacing of tie s
≤ 16D
s
≤ 48Dv
s
≤ bc
In practice Dv =
⎛ 1 .. 1 ⎞ D ⎝ 3 4 ⎠
s = 100 mm m m .. 150mm .. 300mm
Page 77
Spirals Dv ≥ 10mm
Acc ording ording ACI ACI 7.10.4 t he diameter of of s piral: Clear spacing of spiral:
25mm
≤ s ≤ 75mm
Determination of concrete section Pu 0.80⋅ ϕ
Ag = 0.85⋅ f'c⋅ 1
(
− ρg ) +
f y ⋅ ρg
−
0.85⋅ f'c⋅ Ag
Determi Determi natio n of steel steel area Pu Ast =
0.80ϕ
−0.85⋅ f'c +
f y
Example Examp le 13.1 13.1 Tributary area
B := 4m
L := 8m
Materials
f'c := 25MPa
f y := 390MPa
Loads on slab DL := 50mm⋅ 22
kN 3
+
100mm⋅ 25
m LL := 2.00
kN 3
m
kN
kN
+ 0.40
2
m
+ 1.00
kN 2
= 5⋅
m
(for classroom)
2
Reduction of live load
=
2
(tributary area)
32 m
K LL := 4
(for interior column)
AI := K LL⋅ AT
αLL := 0.25
=
2
128⋅ m
4.572 AI
+
(influence area)
= 0.654
αLL⋅ LL = 1.308 ⋅
kN 2
m
2
m
Loads of wall Void oid := 30m 30mm⋅ 30mm⋅ 190 mm⋅4 Page 78
2
m
m
AT := B⋅L
kN
⎛ 120mm − Void ⋅ 55 ⎞ ⋅ 20 kN = 1.648 ⋅ kN 2 3 2 1m ⎠ m m ⎝
Brick hollow.10 :=
Loads on column PD.slab := DL⋅ B ⋅ L
=
160⋅ kN
PL.slab := LL⋅ αLL⋅ B ⋅ L
=
41.863⋅ kN
PB1 := 25cm⋅ ( 65cm − 100mm) ⋅ 25
kN 3
⋅L =
27.5⋅ kN
m PB2 := 20cm⋅ ( 35cm − 100mm) ⋅ 25
kN 3
⋅ B = 5 ⋅ kN
m Pwall.1 := Brick hollow.10⋅ ( 3.5m
−
65cm) ⋅ ( L
Pwall.2 := Brick hollow.10⋅ ( 3.5m
−
35cm) ⋅ B
Pcolumn := 35cm⋅ 45cm⋅ 25
kN 3
⋅ ( 3.5m −
−
=
65cm)
2m )
=
28.174⋅ kN
20.76⋅ kN
=
11.222⋅ kN
m PD := PD.slab
+
PB1
PL := PL.slab⋅ 5
=
209.316 ⋅ kN
(
Pu := 1.2⋅ PD + 1.6⋅ PL
+
=
PB2⋅ 2.5
+
Pwall.1
+
Pwall.2⋅ 2
+
Pcolumn ⋅ 5
)
=
1404.577⋅ kN
2020.397⋅ kN
Verification PL
=
PD + PL
PD + PL kN = 10.087⋅ B⋅L ⋅ 5 2 m
12.97⋅ %
Pcolumn⋅ 5
=
PD + PL
3.477⋅ %
Column section 0.02 ρg :=
ϕ := 0.65 Pu 0.80⋅ ϕ
Ag := 0.85⋅ f'c⋅ 1
(
k :=
35cm 45cm
− ρg ) + h c :=
f y⋅ ρ g
Ag k
=
=
417.75⋅ mm
h c := Ceil h c , 50mm
(
2
1357.338⋅ cm
) = 450⋅ mm
Page 79
b c := k ⋅ h c
=
324.916⋅ mm
⎛ bc
⎛ 350 ⎞ mm ⋅ hc 450 ⎠ ⎝ ⎝ ⎠
b c := Ceil b c , 50mm
) = 350⋅ mm
(
Ag := b c⋅ h c
=
=
2
1575⋅ cm
Steel area Pu Ast :=
0.80ϕ
−
0.85⋅ f'c⋅ Ag
−0.85⋅ f'c +
450mm − 50mm⋅ 2
f y
=
− 5 ⋅ 16mm
4
2
14.604⋅ cm
=
67.5⋅ mm
Lateral t ies D := 16mm
Dv := 10mm
s := min 16⋅ D , 48⋅Dv , b c
(
s := Floor( s , 50mm)
=
) = 256⋅ mm
250 ⋅ mm
13.2. Design o f Short Co lumn s Safety provision Pu
≤ ϕPn
Mu
≤ ϕMn
Equilibrium in forces
∑
X=0
Page 80
12⋅
π⋅ ( 16mm)
2
=
4
350mm − 50mm⋅ 2 2
2
24.127⋅ cm
− 3 ⋅ 16mm
=
101⋅ m
Pn = C
+
Cs
−
T
where C = 0.85⋅ f'c⋅ a⋅ b Cs = A' s⋅ f's T = As⋅ f s
∑
Equilibrium in moments h Mn = Pn e ⋅ = C⋅ ⎛ ⎝ 2
−
h Mn = 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ 2
a ⎞ 2 ⎠
−
⎝
+
a ⎞ 2 ⎠
M=0
h C s⋅ ⎛ ⎝ 2
+
−
d' ⎞
+ T⋅ ⎛ d −
⎠
h A's⋅ f's⋅ ⎛ ⎝ 2
⎝
−
d' ⎞
⎠
+
h ⎞ 2 ⎠
h As⋅ f s⋅ ⎛ d − ⎞ ⎝ 2 ⎠
Conditions of strain compatibility
εs εu
=
d
−
c
c d
ε s = ε u⋅ ε's εu
=
c
−
−
c
f s = Es⋅ ε s = Es⋅ ε u ⋅
c
d
−
c
c
≤ f y
d'
c
ε's = ε u ⋅
c
−
d'
f's = Es⋅ ε 's = Es⋅ ε u ⋅
c
Unknowns = 5:
a , As , A's , f s , f's
Equations = 4:
∑
∑
X=0
Ast
Case of symmetrical column:
As = A's = 2
Case of unsymmetrical column:
f s = f y
Given:
Pu , Mu , b × h , f'c , f y
Find:
As = A's =
Ast 2
Page 81
− c
d'
≤ f y
2 conditions of strain compatibility
M=0
A. Determination o f Steel Area
c
Pu
Functions
AsN( a ) =
−
ϕ
0.85⋅ f'c⋅ a⋅ b
f's( a ) Mu
−
ϕ
AsM( a) =
−
h 0.85⋅ f'c⋅ a ⋅ b ⋅ ⎛ ⎝ 2
h f's⋅ ⎛ 2
⎝
Answer:
f s( a )
−
d' ⎞
⎠
−
+ f s⋅ ⎛ d −
⎝
a ⎞ 2 ⎠
h ⎞ 2 ⎠
As = AsN( a ) = AsM( a)
Example 13.2 Pu := 1200kN
Required strength
Mu := 30kN⋅ m b := 300mm
Concrete dimension
h := 300mm f'c := 25MPa
Materials
f y := 390MPa
Solution d := h
−
⎛ 30mm + 10mm + ⎝
d' := 30mm
+
20mm ⎞
⎠
2
20mm 10mm + 2
=
=
250⋅ mm
50⋅ mm
f'c − 27.6MPa ⎞ ⎡⎛ ⎤ min 0.85 = 0.85 β1 := 0.65 max 0.85 − 0.05⋅ 6.9MPa ⎠ ⎣⎝ ⎦ 5
Es := 2 ⋅ 10 MPa
c ( a)
:=
f s( a )
:=
0.003 ε u :=
a
β1 min⎛ Es⋅ ε u ⋅
d
−
c ( a)
, f y ⎞
c ( a) ⎝ ⎠ c( a) − d' ⎞ f's( a ) := min⎛ Es⋅ ε u ⋅ , f y c ( a) ⎝ ⎠ d t := d
Page 82
f s( 150mm)
=
250 ⋅ MPa
f's( 100mm)
=
345 ⋅ MPa
ϕ( a ) :=
← ε u⋅
εt
0.65 max Pu
:=
AsN( a )
−
ϕ( a )
AsM( a )
:=
ϕ( a )
⎡⎛ 1.45 + 250⋅ εt ⎞ ⎤ min 0.9 3 ⎣⎝ ⎠ ⎦
−
−
h f's( a ) ⋅ ⎛ 2
AsM( a )
f s( a )
h 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ ⎝ 2
−
d' ⎞
+
⎠
−
a ⎞ 2 ⎠
h f s( a ) ⋅ ⎛ d − ⎞ ⎝ 2 ⎠
a2 := 258 mm
4.09×10
−4
4.08×10
−4
4.07×10
−4
4.06×10
−4
4.05×10
−4 0.2579
AsN( a )
ϕ( 90mm)
c ( a)
⎝
a1 := 257.9mm
c ( a)
0.85⋅ f'c⋅ a⋅ b
f's( a ) Mu
−
d t
0.25795 a
a
a := 257.94mm AsN( a )
=
4.071 ⋅ cm
Ast := AsN( a ) 6⋅
Ag := b ⋅ h
π⋅ ( 14mm) 4
h 2
AsM( a )
+
AsM( a )
=
9.236⋅ cm
2
=
=
=
0.86
4.071 ⋅ cm
2
8.141⋅ cm
2
Ag
=
2
900⋅ cm
Page 83
2
=
0.824
Asteel( N)
:=
←
∆a
h N
← ORIGIN for a ∈ ∆a , ∆a + ∆a .. h f ← f's( a ) − f s( a )
i
( continue ) if f = 0 Pu AsN ←
ϕ( a )
−
0.85⋅ f'c⋅ a⋅ b f
( continue ) if AsN ≤ 0 fd
← f's( a ) ⋅ ⎛ −
d' ⎞
h
⎝ 2
⎠
+
h f s( a ) ⋅ ⎛ d − ⎞ ⎝ 2 ⎠
( continue ) if fd = 0 Mu AsM
←
ϕ( a )
h 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ ⎝ 2
−
fd
( continue ) if AsM
∆A ←
AsM
−
≤0
AsN
AsN
⎛ a ⎞ ⎜ h ⎟ ⎜ AsN ⎟ 〈 i〉 ⎜ A ⎟ Z ←⎜ g ⎟ ⎜ AsM ⎟ ⎜ ⎟ A ⎜ g ⎟ ⎝ ∆A ⎠ i
←i+
(
T
csort Z
1
, ORIGIN +
3
)
Z := Asteel( 10000) a := Z
⋅h =
0, 0
257.94⋅ mm 2
AsN := Z ⋅A 0, 1 g
=
4.071⋅ cm
AsM := Z ⋅A 0, 2 g
=
4.071⋅ cm
Ast := AsN + AsM
2
=
2
8.141⋅ cm
Page 84
−
a ⎞ 2 ⎠
0
Z
=
1
2
3
0
0.86
4.523·10-3
4.523·10-3
3.169·10-5
1
0.86
4.518·10-3
4.532·10-3
3.022·10-3
2
0.86
4.528·10-3
4.514·10-3
3.078·10-3
3
0.86
4.513·10-3
4.541·10-3
6.083·10-3
4
0.86
4.533·10-3
4.505·10-3
6.116·10-3
5
0.859
4.537·10-3
4.496·10-3
9.147·10-3
6
0.86
4.509·10-3
4.55·10-3
9.151·10-3
7
0.859
4.542·10-3
4.487·10-3
0.012
8
0.86
4.504·10-3
4.559·10-3
0.012
9
0.859
4.547·10-3
4.478·10-3
0.015
10
0.86
4.499·10-3
4.568·10-3
0.015
11
0.859
4.552·10-3
4.469·10-3
0.018
12
0.86
4.494·10-3
4.577·10-3
0.018
13
0.859
4.557·10-3
4.46·10-3
0.021
14
0.86
4.489·10-3
4.586·10-3
0.022
15
0.859
4.562·10-3
4.451·10-3
...
B. Interaction Diagram for Colu mn Strength
Interaction diagram is a graph of parametric funct ion Mn , Pn
(
ϕPn ( a) = ϕ( a) ⋅ 0.85⋅ f'c⋅ a⋅ b
(
+
A's⋅ f's
h ϕMn( a) = ϕ( a) ⋅ ⎡0.85⋅ f'c⋅ a⋅ b⋅ ⎛ ⎣ ⎝ 2
where
f s( a ) = Es⋅ ε u⋅
d
f's( a ) = Es⋅ ε u⋅
c
−
c
c
− c
d'
−
−
a ⎞ 2 ⎠
)
As⋅ f s
) ≤ ϕPn.max
+
h A's⋅ f's⋅ ⎛ ⎝ 2
−
d' ⎞
⎠
+
As⋅ f s⋅ ⎛ d −
⎝
h ⎞⎤ 2 ⎠⎦
≤ f y ≤ f y
Example 13.3 Concrete dimension
b := 300mm
h := 300mm
16mm d' := 30mm + 8mm + 2 d := h
−
⎛ 30mm + 8mm + ⎝
Page 85
=
46⋅ mm
16mm ⎞ 2
⎠
=
254⋅ mm
As := 3 ⋅
Steel reinforcements
π⋅ ( 16mm)
A's := 3 ⋅
2
4
π⋅ (16mm)
2
=
6.032⋅ cm
=
6.032⋅ cm
2
4
2
f'c := 20MPa
Materials
f y := 390MPa
Solution 2
Ag := b ⋅ h
=
900⋅ cm
Ast := As
+
A's
=
2
12.064⋅ cm
Case of axially loaded column
ϕ := 0.65 ϕPn.max := 0.80⋅ ϕ ⋅ 0.85⋅ f'c⋅ Ag
(
−
Ast
) + f y⋅ Ast
ϕPn.max = 1029.588⋅ kN Case of eccentric column
β1 := 0.65 max c ( a)
:=
f'c − 27.6MPa ⎞ ⎡⎛ ⎤ 0.85 − 0.05⋅ min 0.85 = 0.85 6.9MPa ⎠ ⎣⎝ ⎦
a
β1 5
0.003 ε u :=
f s( a )
:=
Es := 2 ⋅ 10 MPa
min⎛ Es⋅ ε u ⋅
d
−
c ( a)
, f y ⎞
c ( a) ⎝ ⎠ c( a) − d' ⎞ f's( a ) := min⎛ Es⋅ ε u ⋅ , f y c ( a) ⎝ ⎠
f s( 200mm)
=
47.7 ⋅ MPa
f's( 100mm)
=
365.4 ⋅ MPa
d t := d
ϕ( a ) :=
−
c ( a)
← ε u⋅
ϕ
← 0.65 max
ϕ( 100mm)
c ( a)
=
0.773
⎡⎛ 1.45 + 250⋅ εt ⎞ ⎤ min 0.9 3 ⎣⎝ ⎠ ⎦
min ϕ( a) ⋅ 0.85⋅ f'c⋅ a⋅ b
+
h ϕMn( a) := ϕ( a) ⋅ ⎡0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ ⎣ ⎝ 2
−
ϕPn ( a)
:=
d t
εt
(
A's⋅ f's( a ) a ⎞ 2 ⎠
+
Page 86
−
As⋅ f s( a ) , ϕPn.max
h A's⋅ f's( a ) ⋅ ⎛ ⎝ 2
)
−
d' ⎞
⎠
+
As⋅ f s( a ) ⋅ ⎛ d −
⎝
h ⎞⎤ 2 ⎠⎦
a := 0 ,
h 100
.. h
Interaction diagram for column strength 1200 1100 1000 900 800 700
ϕPn( a )
600
kN 500 400 300 200 100 0
0
10
20
30
40
ϕMn( a) kN⋅m
Page 87
50
60
70
80
C. Case of Distribu ted Reinforc ements
Equilibrium in forces n
Pn = C
−
∑
n
T = 0.85⋅ f'c⋅ a⋅ b i
∑
−
i= 1
i= 1
(As , i⋅f s , i)
Equilibrium in moments n
h a Mn = Pn e ⋅ = C⋅ ⎛ − ⎞ + ⎝ 2 2 ⎠
∑
h Mn = 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ ⎝ 2
+
−
a ⎞ 2 ⎠
i= 1 n
⎡T ⋅ ⎛ d − h ⎞⎤ ⎣ i ⎝ i 2 ⎠⎦
∑
i= 1
⎡A ⋅ f ⋅ ⎛ d − h ⎞⎤ ⎣ s , i s , i ⎝ i 2 ⎠⎦
Condition of strain compatibility
ε
s, i
εu
−
d
i
=
c
c
d
ε = εu ⋅ s, i
−
c
c d
= Es⋅ ε = Es⋅ ε u ⋅ s, i
f
s, i
Unknows = 2⋅ n Equations = n
i
+
+
1
2
a,A
s, i
∑
i
− c
c
with
, f s , i
X=0
∑
M=0
n Conditions of strain compatibility Page 88
f
s, i
≤ f y
Additional conditions A
s, i
= n ⋅ A i
s0
Interaction Diagram for Colu mn Strength
⎡
n h a ⎞ h ⎞⎤⎤ ⎛ ⎡ ⎛ − + A ⋅ f ⋅ d − ϕMn ( a) = ϕ⋅ 0.85 ⋅ f'c⋅ a⋅ b ⋅ ⎝ 2 2 ⎠ ⎣ s , i s , i ⎝ i 2 ⎠⎦ i= 1 ⎣ ⎦ n ⎡ ⎤
Abscis sa
∑
ϕPn ( a) = ϕ⋅ 0.85 ⋅f'c⋅ a⋅ b
Ordinate
⎣
−
∑
i= 1
(As , i⋅ f s , i)
≤ ϕPn.max
⎦
Example 13.4 Concrete dimension
b := 700mm
h := 1000mm
Steel reinforcements
⎛ 7 ⎞ ⎜7⎟ ⎜4⎟ ⎜4⎟ ⎜ ⎟ (32mm) 2 As := ⎜ 4 ⎟ ⋅ π⋅ 4 ⎜4⎟ ⎜ ⎟ ⎜4⎟ ⎜7⎟ ⎝ 7 ⎠
⎛ 70 ⎞ ⎜ 177.5 ⎟ ⎜ 285 ⎟ ⎜ 392.5 ⎟ ⎜ ⎟ d := ⎜ 500 ⎟ mm ⎜ 607.5 ⎟ ⎜ ⎟ ⎜ 715 ⎟ ⎜ 822.5 ⎟ ⎝ 930 ⎠
f'c := 45MPa
Materials
f y := 390MPa
Solution ORIGIN
:=
1
Case of axially load column Ag := b ⋅ h Ast :=
∑
n s := rows As
( )
As
Ast
=
ns
=
Ast ρg := Ag
2
386.039⋅ cm
ϕ := 0.65 ϕPn.max := 0.80⋅ ϕ ⋅ 0.85⋅ f'c⋅ Ag
(
−
Ast
9
) + f y⋅ Ast
ϕPn.max = 20984.038⋅ kN Page 89
=
0.055
Case of eccentric column f'c − 27.6MPa ⎞ ⎡⎛ ⎤ min 0.85 = 0.724 β1 := 0.65 max 0.85 − 0.05⋅ 6.9MPa ⎠ ⎣⎝ ⎦ c ( a)
a
:=
β1 5
0.003 ε u :=
Es := 2 ⋅ 10 MPa d
f s( i , a)
:=
εs
← εu⋅
i
−
c ( a) f s( 2 , 200mm)
c( a )
sign ε s min Es⋅ ε s , f y ⋅
( )
(
d t := max( d )
ϕ( a ) :=
d t
:=
−
← ε u⋅
ϕ
← 0.65 max
930⋅ mm
ϕ( 300mm)
c ( a)
⎡⎛ 1.45 + 250⋅ εt ⎞ ⎤ min 0.9 3 ⎣⎝ ⎠ ⎦
⎡
min ϕ( a) ⋅ 0.85⋅ f'c⋅ a⋅ b ⎢ ⎢
⎣
)
c ( a)
εt
⎡ ϕPn ( a)
d t
=
⎣
= −214.516 ⋅ MPa
ns
−
∑
i= 1
⎡
As ⋅ f s( i , a) i
⎤
⎤
⎥ ⎦
, ϕPn.max
⎥ ⎦
⎤ h a ⎞ h ⎞⎤ ⎛ ⎡ ⎛ A ⋅ f ( i , a) ⋅ d − ϕMn( a) := ϕ( a) ⋅ 0.85⋅ f'c⋅ a⋅ b ⋅ − + ⎢ ⎝ 2 2 ⎠ ⎣ si s ⎝ i 2 ⎠⎦⎥ i= 1 ⎣ ⎦ a := 0 ,
h 100
ns
∑
.. h
Page 90
=
0.794
Interaction diagram for column strength 25000
20000
15000
ϕPn( a ) kN 10000
5000
0
0
1000
2000
3000
4000
5000
6000
ϕMn( a) kN⋅ m
Determination of Steel Area
∑
n
X=0
Pn = 0.85⋅ f'c⋅ a⋅ b
−
∑
i= 1
0.85⋅ f'c⋅ a⋅ b AsN( a ) =
i= 1
∑
(As , i⋅ f s , i) = 0.85⋅f'c⋅a⋅ b − As0⋅ ∑ (ns , i⋅ f s , i) i= 1
Pu
ϕ
n
∑ M=0
−
n
(ns , i⋅ f s , i)
h a Mn = 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ − ⎞ + ⎝ 2 2 ⎠
n
∑
i= 1
h a Mn = 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ − ⎞ + As0⋅ ⎝ 2 2 ⎠
Page 91
⎡A ⋅ f ⋅ ⎛ d − h ⎞⎤ ⎣ s , i s , i ⎝ i 2 ⎠⎦ n
∑
i= 1
⎡n ⋅ f ⋅ ⎛ d − h ⎞⎤ ⎣ s , i s , i ⎝ i 2 ⎠⎦
Mu AsM( a ) =
ϕ
−
n
∑
i= 1
h 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ ⎝ 2
−
a ⎞ 2 ⎠
⎡n ⋅ f ⋅ ⎛ d − h ⎞⎤ ⎣ s , i s , i ⎝ i 2 ⎠⎦
Example 13.5 Pu := 19000kN
Required strength
Mu := 300kN⋅ m
Concrete dimension
b := 700mm
Concrete cover
Cover := 40mm
Diameter of stirrup
Dv := 10mm
Diameter of coner bar
D0 := 32mm
h := 1000mm
Distribution of reinforcements
⎛ 7 ⎞ ⎜7 ⎟ ⎜4 ⎟ ⎜4 ⎟ ⎜ ⎟ n := ⎜ 4 ⎟ ⎜4 ⎟ ⎜ ⎟ ⎜4 ⎟ ⎜7 ⎟ ⎝ 7 ⎠ f'c := 45MPa
Materials
f y := 390MPa
Solution Case of axially loaded column 2
Ag := b ⋅ h = 0.7m
ϕ := 0.65 Pu Ast :=
0.80⋅ ϕ
−
0.85⋅ f'c⋅ Ag
−0.85⋅ f'c +
f y
=
2
277.568 ⋅ cm
Case of eccentric column
Page 92
48⋅
π⋅ ( 32mm) 4
2
=
2
386.039 ⋅ cm
d := C over
+
1
Dv
+
D0 2
=
66⋅ mm h
n s := rows( n)
=
∆ :=
9
ns
i := 2 .. n s
d := d
d t := max( d )
d t
c ( a)
i−1
i
β1 := 0.65 max
−
=
d ⋅ 2 1
−
1
=
108.5 ⋅ mm
+∆
934⋅ mm
f'c − 27.6MPa ⎞ ⎡⎛ ⎤ 0.85 − 0.05⋅ min 0.9 = 0.724 6.9MPa ⎠ ⎣⎝ ⎦
a
:=
β1
ϕ( a ) :=
εt
← ε u⋅
d t
−
c ( a)
ϕ( 300mm)
c ( a)
=
0.797
⎡⎛ 1.45 + 250⋅ εt ⎞ ⎤ min 0.9 ϕ ← 0.65 max 3 ⎣⎝ ⎠ ⎦ d f s( i , a)
:=
εs
← εu⋅
i
−
c ( a) f s( 2 , 300mm)
c( a )
sign ε s ⋅ min Es⋅ ε s , f y
( )
(
)
Page 93
= −347.354 ⋅ MPa
−
h Asteel(No)
:=
∆
←
d
1
No
←1 for a ∈ d , d + ∆ .. h 1 1
k
ns
∑
nf ←
(ni⋅ f s( i , a) )
i= 1
( continue ) if nf = 0 0.85⋅ f'c⋅ a⋅ b
AsN ←
−
Pu
ϕ( a )
nf
( continue ) if AsN ≤ 0 ns
nfd ←
∑
i= 1
⎡n f ( i , a) ⋅ ⎛ d − h ⎞⎤ ⎣is ⎝ i 2 ⎠⎦
( continue ) if nfd = 0 Mu AsM
←
−
ϕ( a )
h 0.85⋅ f'c⋅ a⋅ b ⋅ ⎛ ⎝ 2
−
a ⎞ 2 ⎠
nfd
( continue ) if AsM
∆A ← AsM
⎛
−
a
≤0
AsN
⎞
h
⎜ ⎟ A ⎜ sN ⎟ ⎜ A ⎟ 〈 k 〉 ⎜ g ⎟ Z ← ⎜ AsM ⎟ ⎜ A ⎟ ⎜ g ⎟ ⎜ ∆A ⎟ ⎜ A ⎟ ⎝ g ⎠ k
(
←k+ T
csort Z
1
, 4)
Z := Asteel( 20000) a := Z
⋅h =
1, 1
rows( Z)
980.806 ⋅ mm
AsN := Z ⋅A 1, 2 g
=
2
1.917⋅ cm
Page 94
=
766
AsM := Z ⋅A 1, 3 g
As :=
2
=
AsN + AsM 2
1.918⋅ cm
∑
⋅
n
Ast As := max As , Ast , 0.01⋅ Ag
(
)
As
2
=
As
92.044⋅ cm
2
= =
277.568 ⋅ cm
2
277.568⋅ cm
D. Design of Circu lar Column
Area an d centro id of segment
⎛ r c − a ⎞
α = acos
⎝
Asector =
r c
Dc
r c = 2
⎠
Radius × Arch 2
=
r c⋅ r c⋅ 2 ⋅ α 2
= α⋅ r c
Page 95
2
Atriangle =
Base × Height 2
r c⋅ sin( α) ⋅ 2 ⋅ r c⋅ cos( α)
=
2
2
= r c ⋅ sin( α) ⋅ cos( α)
2
Asegment = Asector − Atriangle = r c ⋅ ( α − sin( α) ⋅ cos( α) )
2⋅ r c 3
⋅ cos( θ)
⌠ α ⎮ x c d A = ⎮ ⎮ ⌡− α
⌠ ⎮ ⎮ ⌡
⌠ ⎮ ⎮ ⌡ x sector =
x segment =
3
3
3
⋅ cos( θ) d θ =
x c d A
2 ⋅ r c ⋅ sin( α) 3
3
=
2 ⋅ r c ⋅ sin( α) 3
⋅
1
α⋅ r c
2
=
2
⋅ r ⋅ cos( α) 3 c
x sector ⋅ Asector − x triangle⋅ Atriangle Asegment 2 ⋅ r c sin( α)
x segment =
r c
A
x triangle =
x segment =
2
d θ ⋅ r ⋅ r ⋅ d θ = 2 c c 2
dA =
xc =
r c
1
3
⋅
α
2
⋅ α⋅ r c −
2 ⋅ r c 3
2
⋅ cos( α) ⋅ r c ⋅ sin( α) ⋅ cos( α)
2
r c ⋅ ( α − sin( α) ⋅ cos( α) ) 2 ⋅ r c 3
⋅
sin( α)
3
α − sin( α) ⋅ cos( α)
Location of steel re-bars d = r c i
− r s⋅ cos( αi)
r s = r c − Cover
α =
2 ⋅ π
i
ns
Equilibrium in forces
Page 96
ns
⋅ (i −
1)
= number of steel re-bars
2 ⋅ r c sin( α) 3
⋅
α
ns
Pn = C
−
∑
ns
T = 0.85⋅ f'c⋅ Ac i
∑
−
i= 1
i= 1
(As , i⋅ f s , i)
Equilibrium in moments ns
Mn = Pn e ⋅ = C⋅x c
+
∑
T ⋅ d i
i= 1
(
i
−
r c
)
ns
Mn = 0.85⋅ f'c⋅ Ac⋅ x c
+
∑
A
⋅ f ⋅ (d i −
s, i s, i
i= 1
r c
)
2
Ac = r c ⋅ ( α − sin( α) ⋅ cos( α) )
where
2 ⋅ r c
xc =
3
⋅
sin( α)
3
α − sin( α) ⋅ cos( α)
Condition of strain compatibility
ε
s, i
εu
d
i
=
−
c
d
ε
c
s, i
= εu ⋅
i
−
c
c d
f
s, i
= Es⋅ ε = Es⋅ ε u ⋅ s, i
i
− c
c
≤ f y
Example 13.6 Pu := 3437.31kN
Required strength
Mu := 42.53kN⋅ m f'c := 25MPa
Materials
f y := 390MPa
Solution Case of axially l oaded co lum n
ϕ := 0.70 ρg := 0.025
Ass ume
Pu
Ag :=
0.85⋅ ϕ
0.85⋅ f'c⋅ 1
(
− ρg ) +
3
Ag = 1.896 × 10 ⋅ cm
f y⋅ ρ g
Page 97
2
Dc := Ceil
Diameter of column
π⋅ Dc
Ag :=
2
⎛
Ag
⎜ ⎝
π
, 50mm
4
⎟ ⎠
4 Pu 0.85⋅ ϕ
Ast :=
−
3
=
Ast
f y
Ast ρg := Ag
500⋅ mm
Ag = 1.963 × 10 ⋅ cm
0.85⋅ f'c⋅ Ag
−0.85⋅ f'c +
Ds := Dc
=
Dc
2
43.514⋅ cm
ρg = 0.022
− 2 ⋅ ⎛ 30mm +
⎝
n s := 14
20mm ⎞ 8mm + 2 ⎠
π⋅ ( 20mm)
As0 :=
Ds
=
Ast := ns⋅ As0
4
ϕPn.max := 0.85⋅ ϕ ⋅ 0.85⋅ f'c⋅ Ag
(
−
404⋅ mm
2
s :=
Ast
) + f y⋅ Ast
π⋅ Ds ns
=
=
Dc r c := 2
Ds r s := 2
i := 1 .. n s
αs := i
360deg ns
d := r c i
β1 := 0.65 max
:=
90.657⋅ mm
ϕPn.max = 3447.594⋅ kN
:=
i
f'c − 27.6MPa ⎞ ⎡⎛ ⎤ 0.85 − 0.05⋅ min 0.85 = 0.85 6.9MPa ⎠ ⎣⎝ ⎦ d t := max( d )
β1
:=
Ac( a )
:=
d t
−
← ε u⋅
ϕ
← 0.70 max
2 ⋅ r c 3 2
⋅
r c
⎡⎛ 1.7 + 200⋅ εt ⎞ ⎤ min 0.9 3 ⎣⎝ ⎠ ⎦
⎠ sin( α( a ) )
α( a )
r c ⋅ ( α( a )
− −
=
452⋅ mm
ϕ( 150mm)
c ( a)
⎛ r c − a ⎞ ⎝
d t
c ( a)
εt
acos
x c( a )
1)
− r s⋅ cos αs
a
ϕ( a ) :=
α( a )
⋅(i −
3
sin( α( a ) ) ⋅ cos( α( a ) ) sin( α( a ) ) ⋅ cos( α( a ) ) ) Page 98
=
2
43.982⋅ cm
Case of Eccentric Column
c ( a)
2
0.879
d f s( i , a)
:=
εs
← εu⋅
i
−
c ( a)
c( a )
sign ε s min Es⋅ ε s , f y ⋅
( )
⎡ ϕPn ( a)
:=
(
)
⎡
min ϕ( a) ⋅ 0.85⋅ f'c⋅ Ac( a ) ⎢ ⎢
⎣
−
⎣
∑
i= 1
⎡ +
⎣
Dc 100
⎤
( As0⋅ f s( i , a) )⎥ , ϕPn.max⎥ ⎦
⎦ ⎤
ns
ϕMn( a) := ϕ( a) ⋅ 0.85⋅ f'c⋅ Ac( a ) ⋅ xc( a ) ⎢
a := 0 ,
⎤
ns
∑
As0⋅ f s( i , a) ⋅ d i
(
i= 1
.. Dc
Interaction diagram for column strength
3000
ϕPn( a) kN Pu
2000
kN
1000
0
0
100
200
ϕMn( a) kN⋅ m
,
300 Mu kN⋅ m
Page 99
−
r c
)⎥ ⎦
13.3. Lon g (Slender) Column s Stability index Q=
where
ΣPu ⋅ ∆0 Vu ⋅ L c
ΣPu , Vu
= total vertical load and story shear
∆0
= relative deflection between the top and bottom of story
Lc
= center-to-center length of column
Q ≤ 0.05
: the column is nonsway (braced)
Q > 0.05
: the column is sway (unbraced)
Unbraced Frame
Braced Frame
l l a W r a e h S
Braced Frame Brick Wall
Ties
Slenderness o f Column Column is short, if k ⋅ Lu
For nonsway columns:
r k ⋅ Lu
For sway columns:
r
≤ 34 −
12⋅
≤ 22
where Lu
= unsupported length of column
r
= radius of gyration of column section
Page 100
r =
I A
M1 M2
≤ 40
For rectangular column
r =
h 12
I, A
= moment of inertia and area of column section
M1 = min M a , M b
(
Ma , M b
)
M2 = max Ma , M b
(
= moments at the ends of column
k = k ψa , ψ b
(
ψ =
)
is an effective length factor
ΣEIc ΣEI b
is a degree of end restraint (degree of end release)
Page 101
)
= 0.289⋅ h
For nonsway column
ψa⋅ ψ b
⋅ ⎛
π ⎞
2
+
⎝ k ⎠
4
⎛
+ ψ b
ψa
π k
⋅⎜1 −
2
⎞ ⎟+
π tan⎛ ⎞ ⎟ ⎝ k ⎠ ⎠
⎜ ⎝
π 2 ⋅tan⎛ ⎞ ⎝ 2⋅ k ⎠ π k
For sway column 2
π ψa⋅ ψ b ⋅ ⎛ ⎞ − 36 ⎝ k ⎠ 6 ⋅ ψa
(
π
=
+ ψ b )
k
π tan⎛ ⎞ ⎝ k ⎠
Appr ox imate val ues of k In nonsway frames: k = 0.7
+
k = 0.85
0.05⋅ ψA + ψB
(
+
) ≤ 1.0
0.05⋅ ψ min ≤ 1.0
ψmin = min ψA , ψB
(
)
In sway frames: Case ψm < 2 20 k =
− ψm 20
⋅
1
+ ψm
Case ψm ≥ 2 k = 0.9⋅ 1
ψm =
+ ψm
ψA + ψB 2
Case of column is hinged at one end k = 2.0
+
0.3⋅ ψ
ψ
is the value in the restrained end.
Page 102
=1
Case of Slender Colum n Moment on column Mc = M2 ⋅ δns
≥ M 2.min⋅ δ ns
where M2.min = Pu⋅ ( 0.6in + 0.03h ) = Pu⋅ ( 15mm + 0.03h )
δns
= moment magnification factor Cm
δns = 1
−
Cm = 0.6
Pc
≥1
Pu 0.75Pc
+
0.4⋅
M1 M2
≥ 0.4
= Euler's critical load 2
Pc =
EI =
π ⋅ EI
( k ⋅ Lu )
2
0.4⋅ Ec⋅ Ig 1
+ βd
1.2PD βd = 1.2PD + 1.6PL
Ec
= concrete modulus of elasticity
Ig
= moment of inertia of gross section of column
Page 103
Example 13.7
Design column b c := 18in = 45.72 ⋅ cm h c := 18in = 45.72 ⋅ cm Lc := 14ft = 4.267 ⋅ m
Lu := 13ft = 3.962 m
Loads on column PD := 230kip
=
M2.D := 2ft⋅ kip M1.D :=
1023.091⋅ kN
PL := 173kip
=
M2.L := 108ft⋅ kip
=
146.428⋅ kN⋅ m
M1.L := 100ft⋅ kip
=
135.582⋅ kN⋅ m
2.712⋅ kN⋅ m
−2 ft ⋅kip = −2.712 ⋅ kN⋅m
=
769.542⋅ kN
Materials f'c := 4000psi = 27.579⋅ MPa
f y := 60ksi
=
413.685 ⋅ MPa
Upper column and beams b a := 18in
h a := 18in
La := 14ft
b a1 := 48in = 1219.2⋅ mm
h a1 := 12in = 304.8 ⋅ mm
La1 := 24ft = 7.315 ⋅ m
Page 104
b a2 := 48in = 1219.2⋅ mm
h a2 := 12in = 304.8 ⋅ mm
La2 := 24ft = 7.315 ⋅ m
b b := 18in
h b := 18in
L b := 14ft
b b1 := 48in = 1219.2⋅ mm
h b1 := 12in = 304.8 ⋅ mm
L b1 := 24ft
=
7.315 ⋅
b b2 := 48in = 1219.2⋅ mm
h b2 := 12in = 304.8 ⋅ mm
L b2 := 24ft
=
7.315 ⋅
Lower column and beams
Solution Determination o f colum n section (Case of axially l oaded co lum n) Pu := 1.2⋅ PD + 1.6⋅ PL
=
2458.977⋅ kN
0.02 ρg :=
ϕ := 0.65 Pu 0.80⋅ ϕ
Ag := 0.85⋅ f'c⋅ 1
(
b :=
=
Ag
=
=
16⋅ in 3
0.80⋅ ϕ
−
f y
=
2
389.019 ⋅ mm
h := Ceil ( h , 2 in)
=
16⋅ in
2
0.85⋅ f'c⋅ Ag
−0.85⋅ f'c +
3
1.513 × 10 ⋅ cm
h := b
1.652 × 10 ⋅ cm
Pu Ast :=
f y⋅ ρ g
389.019⋅ mm
b := Ceil(b , 2in )
Ag := b ⋅ h
− ρg ) +
=
=
2
21.962⋅ cm
Ast := 8 ⋅
π⋅ ( 20mm)
2
4
=
Slenderness of colum n
r :=
Radius of gyration
h
=
0.117 m
12
Concrete modulus of elasticity 1.5
wc := 24
kN 3
m
⎛ wc ⎞ Ec := 44MPa ⋅ ⋅ ⎜ kN ⎟ ⎝ m3 ⎠ Page 105
f'c MPa
=
4
2
25.133⋅ cm
2.717 × 10 ⋅ MPa
Relative flexural st iffness of column and beam b⋅ h
b a⋅ h a
3
EIc := Ec⋅ 0.70⋅ 12
EIa := Ec⋅ 0.70⋅
3
b b ⋅ h b
EI b := Ec⋅ 0.70⋅
12
3
3
b a1⋅ h a1
EIa1 := Ec⋅ 0.35⋅
b a2⋅ h a2
EIa2 := Ec⋅ 0.35⋅
12 3
3
b b1⋅ h b1
EI b1 := Ec⋅ 0.35⋅
Σi ca :=
EIc Lc
EIa1 i Σ ba := La1
Σi ca
ψa := Σi ba
EI b2 := Ec⋅ 0.35⋅
12
Σi cb :=
La EIa2
+
=
b b2⋅ h b2
EIa
+
12
EIc Lc
La2
Σi cb
ψ b := Σi bb
3.524
EI b
+
EI b1 i Σ bb := L b1
12
L b
+
=
EI b2 L b2
3.524
Effective length factor for nonsway column
F( k )
:=
ψa⋅ ψ b 4
⋅ ⎛
π ⎞
⎝ k ⎠
Bisection( a, b , ε )
:=
2
+
c
ψa
←
+ ψ b 2
a
+
⎛
k
⋅⎜1 −
return
−
k
≤ε Bisection( a, c , ε ) Bisection( c, b , ε ) a
k
Ma := 1.2⋅ M1.D
+
1.6⋅ M1.L
=
213.677 ⋅ kN⋅ m
M b := 1.2⋅ M2.D
+
1.6⋅ M2.L
=
237.539⋅ kN⋅ m
if F( a) ⋅ F( c ) otherwise
=
0.906
) = 213.677⋅ kN⋅ m
M2 := max Ma , M b Column :=
π
b
k := Bisection( 0.5 , 1 , 0.000001)
(
⎟+
2
return
(
π 2 ⋅tan⎛ ⎞ ⎝ 2⋅ k ⎠
π tan⎛ ⎞ ⎟ ⎝ k ⎠ ⎠
⎜ ⎝
return c if b
M1 := min M a , M b
⎞
π
) = 237.539⋅ kN⋅ m
"is Short" if
k ⋅ Lu r
⎛
≤ min
34
⎝
"is Long" otherwise Column = "is Long" Page 106
−
12⋅
M1 M2
⎞
, 40
⎠
≤0
−
1
12
3
Required moment on column 1.2⋅ PD
βd := 1.2⋅ PD + 1.6⋅ PL
=
0.499 1.5
wc := 24
⎛ wc ⎞ Ec := 44MPa ⋅ ⋅ ⎜ kN ⎟ ⎝ m3 ⎠
kN 3
m
Ig :=
b c⋅ h c
3
EI :=
12
⎛
Cm := max 0.6
⎝
+
0.4⋅
M1 M2
0.4⋅ Ec⋅ Ig
+ βd
1
⎞
, 0.4 =
⎠
=
f'c MPa
4
=
4
2.717 × 10 ⋅ MPa
2.639 × 10 ⋅ kN⋅m
2
0.96
2
Pc :=
π ⋅ EI
( k ⋅ Lu )
2
⎛ δns := max ⎜
1
⎝
=
20195.95⋅ kN
⎞ ⎟
Cm
,1 =
Pu
−
0.75⋅ Pc
M 2.min := Pu ⋅ 15mm
(
M c := max M 2
(
+
1.146
⎠ 0.03⋅ h c
) = 70.612⋅ kN⋅ m
, M2.min) ⋅ δns =
272.18⋅ kN⋅ m
Interaction di agram for col umn strength Distribution of reinforcements
⎛ 1 ⎜1 Bars := ⎜ 1 ⎜1
1 1 1 1 ⎞
⎝ 1
1 1 1 1 ⎠
0 0 0 1 0 0 0 0 0 0
n s := cols( Bars) i := 1 .. n s
∑
:= i
n1
:= 40mm + 1
d1
=
⎟ 1⎟ 1⎟
As0 :=
5
〈 i〉
n := n1
Bars
20mm 10mm + 2
=
60⋅ mm
Page 107
π⋅ ( 20mm)
2
4
⎛ 5 ⎞ ⎜2⎟ n = ⎜2⎟ ⎜2⎟ ⎝ 5 ⎠
∆ :=
−
hc
d1 ⋅ 2 1
−
ns
i := 2 .. n s
d1
i
=
1
:=
d t := max( d )
d1
i−1
=
+∆
⎝ 397.2 ⎠
f'c − 27.6MPa ⎞ ⎡⎛ ⎤ 0.85 − 0.05⋅ min 0.85 = 0.85 6.9MPa ⎠ ⎣⎝ ⎦
a
:=
β1
−
d f s( i , a)
d := d1
397.2 ⋅ mm
β1 := 0.65 max c ( a)
⎛ 60 ⎞ ⎜ 144.3 ⎟ d = ⎜ 228.6 ⎟ ⋅ mm ⎜ 312.9 ⎟
84.3⋅ mm
:=
← εu⋅
εs
i
c ( a)
c( a )
sign ε s ⋅ min Es⋅ ε s , f y
( )
∑
Ast := As0 ⋅
n
=
(
)
2
50.265⋅ cm
ϕ := 0.65 ϕPn.max := 0.80⋅ ϕ ⋅ 0.85⋅ f'c⋅ Ag
(
ϕ( a ) :=
−
Ast
) + f y⋅ Ast
=
3033.321⋅ kN
c ( a)
εt
← ε u⋅
ϕ
← 0.65 max
b := b c
c ( a)
⎡⎛ 1.45 + 250⋅ εt ⎞ ⎤ min 0.9 3 ⎣⎝ ⎠ ⎦
h := h c
⎡ ϕPn ( a)
d t
−
:=
⎡
min ϕ( a) ⋅ 0.85⋅ f'c⋅ a⋅ b ⎢ ⎢
⎣
⎣
⎡
⎤
ns
−
∑
i= 1
⎤
(As0⋅ n i⋅ f s( i , a) )⎥ , ϕPn.max⎥ ⎦
⎦
⎤ h ⎞⎤ ⎡ ⎛ − + A ⋅ n ⋅ f ( i , a) ⋅ d − ϕMn( a) := ϕ( a) ⋅ 0.85⋅ f'c⋅ a⋅ b ⋅ ⎢ ⎝ 2 2 ⎠ ⎣ s0 i s ⎝ i 2 ⎠⎦⎥ i= 1 ⎣ ⎦ e :=
a := 0 ,
h 100
Mc Pu
=
110.688⋅ mm
⎛ h
ns
a ⎞
e h
=
∑
0.242
.. h
Page 108
