7. 2. Objective
• To know the process of column design procedure
• To understand the mechanical properties of reinforcing concrete
• To learn the behavior of structural members; beam,column,slab
• To acknowledge how loadings (live load,dead load,wind load,super
dead load,earthquake load etc) transmissions work
• To be familiar with building design codes
• To cultivate the teamwork spirit
2
8. 3. Scope of The Study
Collecting Required Information
- Site Location TU(YAMETHIN)
- Building type Residential Building ( Two Storeied Apartment)
- Material Strength Fy = 40 ksi
fc’ = 3 ksi
- Design Code MNBC -2020
ACI 318 – 08
- Design Method LRFD
3
9. Design Procedure
Basic steps;
• Determination of fc’ and fy
• Structural Analysis
• Member size
• Selection area of Reinforcing Steel
6
10. Advantages of Reinforced Concrete
Combine performance for ( compression and tension)
Flexible section
Fire-resistant properties
Resistance to weathering, providing a longer lifespan
Structural Stability
4
11. Disadvantages of Reinforced concrete
The tensile strength of reinforced concrete ≅
1
10
compressive
strength.
mixing, casting, and curing.
The cost of the forms.
Shrinkage causes crack
5
12. Design Method
Load And Resistance Factored Design Method (LRFD)
o Reducing Strength
o Increasing Load by factor
8
13. Values of the strength reduction factor Φ (Phi) (MNBC
2020)
For flexure of tension controlled sections Phi = 0.9
For shear and torsion Phi = 0.75
For compression members with spiral reinforcement Phi = 0.70
For compression members with lateral ties Phi = 0.65
14. Design Load
Load combination for Load and resistance factor design
1.4D
1.2D+1.6L+0.5(Lr or S or R)
1.2D+1.6(Lr or S or R)+(L or 0.5W)
1.2D+1.0W+L+0.5(Lr or S or R)
1.2D+1.0E+L+0.2S
0.9D+1.0W
0.9D+1.0E 7
17. Specificied Compressive strength of concrete
fcu’= 28 days compressive strength of concrete (standard cube strength)
fc’= 28 days compressive strength of concrete (standard cylinder strength)
fc’=fcu’x0.78 (fc’ ≤ 3500psi)
fc’=fcu’x0.8 (3500 < fc’ ≤ 5000psi)
fc’=fcu’x0.81 (5000 < fc’ ≤ 6000psi)
fc’=fcu’x0.83 (fc’ > 6000psi)
Unit Weight of Concrete → 150 lb/ft3
Water Cement Ratio →Minimum water cement ratio = 0.25 (for hydration)
18. Table -1. Minimum Concrete Cover
Particulars Not exposed directly
to weather or not in
contact with ground
Exposed directly to
weather or in contact
with ground
Slab and Walls 0.75 in 2 in but 1.5 in for
16mm and smaller bars
Beams and Columns 1.5 in 2 in
19. Table - 2. Bar Spacing (For Maximum Size of Aggregate 1”)
Type of member Minimum Clear
Spacing
Maximum c/c Spacing
Beam 1.33 in
Column 1.5 d or 1.5 in
Slab or Stair 3 in ( to save labor) 3h or 18” (one way)
2h or 18” (two way)
Temp & Shrinkage Steel
5h or 18”
For practical work
Slab or Stair 4 in 2h ( one way and two way)
22. Table – 3. Minimum Uniformly Distributed Live Load
23. Types of Column
1. Base on Loading
• Axial Load
• Uniaxial Load
• Biaxial Load
2. Base on Length
• Short Column => l/b < 12
• Slender Column => l/b ≥ 12
3. Base on Shape of section
Square column Rectangular column Circular column
24. 4. Base on Column Ties
5. Base on Material
• RC column
• Prestress column
• Composite column
Tie column Spiral column
25. Axial Load on Column
Pn = ∅ K(0.85fc’(Ag – As) + (As.Fy) Eq. 1
ACI Code Limitation
1. For tie column; K = 0.8 , ∅ = 0.65
For spiral column; K = 0.85 , ∅ = 0.75
ACI 318 - 08
Behavior of tied and spiral columns
26. 2. Longitudinal Steel
• Minimum steel = 1%
• Maximum steel = 8%
• Working maximum steel = 3%
• At least 4 bars in rectangular column
• At least 6 bars in circular column
3. Tie of column
• Minimum ∅ = 3/8 (for #10 and smaller bars)
• Minimum ∅ = 1/2 (for bars larger than #10)
27. 4. Spacing of ties
Smax = 16 db
= 48 dt
= b or d
Smaller control
Table - 4. Maximum Spacing of Ties
29. Approximate Solution
Whitney’s formula
Pn =
𝐴𝑔 𝐹𝑐′
3ℎ𝑒
𝑑2
+1.18
+
𝐴𝑠′𝐹𝑦
𝑒
𝑑−𝑑′+0.5
Ag = b.h = gross area of the section
Pn = nominal axial strength of the column section
As = total area of nonprestressed longitudinal reinforcement
d = effective depth = h - d’
30. Column A’-2’ ( From 1st Floor to Roof Floor )
Material Properties
Concrete Strength , fc’ = 3 ksi
Rebar Strength , fy = 40 ksi
Section Properties
Column Size = 9” x 9”
Gross Area , Ag = 81 in2
d’ = 2.5 in
d = h – d’ = 9 – 2.5 = 6.5 in
32. Steel Area
As(min) = 0.01 Ag = 0.01 x 81 = 0.81 in2
As(prov) = 1.24 in2 ( 4 #5 bars )
As’ =
𝐴s
(
prov
)
2
=
1.24
2
= 0.62 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
=
4.85 × 12
7.16
= 8.13 in2 > 0.1 h = 0.9 in
Design Compressive Strength
Ø = 0.65 , K = 0.8
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 0.65 x 0.8 ( 0.85 x 3 x ( 81 – 1.24 ) + 1.24 x 40)
= 131.55 kips > Pu = 7.16 kips
∴ 𝐎𝐊
33. Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 9 in (or) 9 in = 9 in
∴ Smax = 9 in 𝑐
𝑐 spacing.
35. Column B-1 ( From 1st Floor to Roof Floor )
Section Properties
Column Size = 10” x 10”
Gross Area , Ag = 100 in2
d’ = 2.5 in
Axial Load
P(total) = 16.45 kips
Moments
Mux = 11.7 – 0 = 11.7 kip.ft
Muy = 27.14 – 0.6 = 26.54 kip.ft
Mu(max) = Muy = 26.54 kip.ft
36. Steel Area
As(min) = 0.01 Ag = 0.01 x 100 = 1 in2
As(prov) = 1.24 in2 ( 4 #5 bars )
As’ =
𝐴s
(
prov
)
2
=
1.24
2
= 0.62 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
=
26.54 × 12
16.45
= 19.36 in2 > 0.1 h = 1 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 156.75 kips > Pu = 16.45 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, Pu =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
37. Pn = 31.74 kips
Ø Pn = 20.63 kips > Pu = 16.45 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 10 in (or) 10 in = 10 in
∴ Smax = 10 in 𝑐
𝑐 spacing.
38. Column C-2 ( From 1st Floor to Roof Floor )
Section Properties
Column Size = 14” x 14”
Gross Area , Ag = 196 in2
d’ = 2.5 in
Axial Load
P(total) = 48.24 kips
Moments
Mux = 16 -13 = 3 kip.ft
Muy = 72.1 - 4.2 = 67.9 kip.ft
Mu(max) = Muy = 67.9 kip.ft
39. Steel Area
As(min) = 0.01 Ag = 0.01 x 196 = 1.96 in2
As(prov) = 2.48 in2 ( 8 #5 bars )
As’ =
𝐴s
(
prov
)
2
= 1.24 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
=
67.9 × 12
48.24
= 16.89 in2 > 0.1 h = 1.4 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 308.19 kips > Pu = 48.24 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, Pu =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
40. Pn = 110.72 kips
Ø Pn = 71.97 kips > Pu = 48.24 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 14 in (or) 14 in = 14 in
∴ Smax = 10 in 𝑐
𝑐 spacing.
41. Column C-3 ( From 1st Floor to Roof Floor )
Section Properties
Column Size = 12” x 12”
Gross Area , Ag = 144 in2
d’ = 2.5 in
Axial Load
P(total) = 35.02 kips
Moments
Mux = 16.48 -16 = 0.48 kip.ft
Muy = 39.5 – 0 = 39.5 kip.ft
Mu(max) = Muy = 39.5 kip.ft
42. Steel Area
As(min) = 0.01 Ag = 0.01 x 144 = 1.44 in2
As(prov) = 1.86 in2 ( 6 #5 bars )
As’ =
𝐴s
(
prov
)
2
= 0.93 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
= 13.54 in2 > 0.1 h = 1.2 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 227.17 kips > Pu = 35.02 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, PU =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
43. Pn = 80.94 kips
Ø Pn = 52.61 kips > Pu = 35.05 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 12 in (or) 12 in = 12 in
∴ Smax = 10 in 𝑐
𝑐 spacing.
44. Column A’-2’ ( From Ground Floor to 1st Floor )
Section Properties
Column Size = 9” x 9”
Gross Area , Ag = 81 in2
d’ = 2.5 in
Axial Load
P(total) = 21.14 kips
Moments
Mux = 11.21 - 0 = 11.21 kip.ft
Muy = 9.47 – 0 = 9.47 kip.ft
Mu(max) = Mux = 11.21 kip.ft
45. Steel Area
As(min) = 0.01 Ag = 0.01 x 81 = 0.81 in2
As(prov) = 1.24 in2 ( 4 # 5 bars )
As’ =
𝐴s
(
prov
)
2
= 0.62 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
= 6.36 in2 > 0.1 h = 0.9 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 131.55 kips > Pu = 21.14 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, PU =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
46. Pn = 58.24 kips
Ø Pn = 37.86 kips > Pu = 7.16 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 9 in (or) 9 in = 9 in
∴ Smax = 9 in 𝑐
𝑐 spacing.
47. Column B - 1 ( From Ground Floor to 1st Floor )
Section Properties
Column Size = 12” x 12”
Gross Area , Ag = 144 in2
d’ = 2.5 in
Axial Load
P(total) = 43.29 kips
Moments
Mux = 21.3- 0 = 21.3 kip.ft
Muy = 43.7 – 0.81 = 42.89 kip.ft
Mu(max) = Muy = 42.89 kip.ft
48. Steel Area
As(min) = 0.01 Ag = 0.01 x 144 = 1.44 in2
As(prov) = 1.86 in2 ( 6 # 5 bars )
As’ =
𝐴s
(
prov
)
2
= 0.93 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
= 11.89 in2 > 0.1 h = 1.2 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 227.17 kips > Pu = 43.29 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, PU =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
49. Pn = 89.88 kips
Ø Pn = 58.42 kips > Pu = 43.29 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 12 in (or) 12 in = 12 in
∴ Smax = 10 in 𝑐
𝑐 spacing.
50. Column C - 2 ( From Ground Floor to 1st Floor )
Section Properties
Column Size = 15” x 15”
Gross Area , Ag = 225 in2
d’ = 2.5 in
Axial Load
P(total) = 110.37 kips
Moments
Mux = 24.5- 18.6 = 5.9 kip.ft
Muy = 101.2 – 67.2 = 34 kip.ft
Mu(max) = Muy = 34 kip.ft
51. Steel Area
As(min) = 0.01 Ag = 0.01 x 225 = 2.25 in2
As(prov) = 1.24 in2 ( 8 # 5 bars )
As’ =
𝐴s
(
prov
)
2
= 1.24 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
= 3.7 in2 > 0.1 h = 1.5 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 346.65 kips > Pu = 110.37 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, PU =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
52. Pn = 357.76 kips
Ø Pn = 232.54 kips > Pu = 110.37 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 15 in (or) 15 in = 15 in
∴ Smax = 10 in 𝑐
𝑐 spacing.
53. Column C - 3 ( From Ground Floor to 1st Floor )
Section Properties
Column Size = 15” x 15”
Gross Area , Ag = 225 in2
d’ = 2.5 in
Axial Load
P(total) = 81.19 kips
Moments
Mux = 19.7 - 18.6 = 1.1 kip.ft
Muy = 70.1 – 0 = 70.1 kip.ft
Mu(max) = Muy = 70.1 kip.ft
54. Steel Area
As(min) = 0.01 Ag = 0.01 x 225 = 2.25 in2
As(prov) = 2.48 in2 ( 8 # 5 bars )
As’ =
𝐴s
(
prov
)
2
= 1.24 in2
Eccentricity
e =
𝑀𝑢
𝑃𝑢
= 10.36 in2 > 0.1 h = 1.5 in
Design Compressive Strength
Ø Pn = Ø K ( 0.85 fc’ ( Ag - As ) + As Fy )
= 346.65 kips > Pu = 81.19 kips ∴ 𝐎𝐊
By Whitney’s Approximate Formula, PU =
𝐴𝑔 𝑓𝑐′
3 ℎ𝑒
𝑑2 + 1.18
+
𝐴𝑠′ 𝐹𝑦
𝑒
𝑑−𝑑′ +0.5
55. Pn = 194.4 kips
Ø Pn = 126.36 kips > Pu = 81.19 kips ∴ 𝐎𝐊
Section of ties
Reinforcement for tie should be selected.
Use #3 bar for tie.
Determination of vertical spacing.
Smax = 16 db = 16 x
5
8
= 10 in
Smax = 48 dt = 48 x
3
8
= 18 in
Smax = b (or) h = 15 in (or) 15 in = 15 in
∴ Smax = 10 in 𝑐
𝑐 spacing.
![( INTEGRADTED DESIGN PROJECT )
TECHNOLOGICAL UNIVERSITY
[YAMETHIN]
DEPARTMENT OF CIVIL
ENGINEERING
Group - 1](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)