Design of Reinforced Concrete Structures

Lecture 10: Design of Reinforced Concrete Columns

Dr.-Ing. Mahmoud Elshahawi

1 Dr.-Ing. Mahmoud Elshahawi

 ‫»‪«Columns‬‬
‫‪ -‬األعمدة‬

‫• عنصر إنشائي رأسي يستخدم لنقل ردود أفعال الكمرات إلى القواعد‪.‬‬

‫• في حالة وجود كمرات ترتكز على العمود فأن حمل العمود عند ظهر االساسات ‪ = Pu‬ردود أفعال‬

‫الكمرات‪.‬‬

‫‪Pu = 1.10 ∗ n ∗ ∑ R , where n is the number of stories‬‬

‫• في حالة البالطات الالكمرية فأن حمل العمود عند ظهر االساسات ‪ = Pu‬يحسب بقانون المساحات‪.‬‬

‫‪Pu = 1.10 ∗ n ∗ A ∗ Wsu‬‬

‫‪Dimensions:‬‬
‫‪• b ≮ 200 mm‬‬
‫‪• t ≯ 5b‬‬
‫‪• t > 5b → Shear Walls‬‬
‫‪ -‬تستخدم في المباني العالية لمقاومة االحمال الجانبية (أحمال الزالزل والرياح)‪.‬‬
‫‪❖ Classifications of Columns:‬‬
‫‪1 . Stirrups‬‬
‫‪2 . Slenderness Ratio‬‬
‫)‪3 . Lateral Load (Braced or Unbraced‬‬
‫)‪4 . Loads (Stresses‬‬

‫‪2‬‬ ‫‪Dr.-Ing. Mahmoud Elshahawi‬‬

1. Stirrups

A. Tied columns B. Spiral columns

‫عمود بكانات منفصلة‬
‫ عمود بكانات حلزونية‬-
‫ الحديد‬+ ‫• الذي ينقل الحمل هو (الخرسانة‬
+ ‫• الذي ينقل الحمل الرأسي (الخرسانة‬
.)‫الرأسي‬
)‫ الكانات‬+ ‫الحديد‬
.‫• الكانات للتربيط فقط‬

❖ For tied columns:

Pu = 0.35 ∗ fcu ∗ (Ac − As ) + 0.67 ∗ fy ∗ As
Pu = Ac [0.35 ∗ fcu ∗ (1 − μ) + 0.67 ∗ fy ∗ μ]
Pu = ‫حمل العمود‬
Ac = ‫مساحة الخرسانة‬
As = ‫مساحة الحديد‬
As
μ= → As = μ ∗ Ac
Ac
0.008 Ac,required
As,min = the largest from {
0.006 Ac, chosen
0.04 for interior columns
μmax = { 0.05 for edge columns
0.06 for corner columns

3 Dr.-Ing. Mahmoud Elshahawi

 ❖ For spiral columns:

Required: As , D , øspiral , Pitch (P)

Pitch (P) = (30 – 80) mm

Dk = D − 50 mm

Ak = πD2k /4

Ac = πD2 /4

Pu = the least from:

▪ Pu1 = 1.14(0.35 ∗ fcu ∗ (Ac − As ) + 0.67 ∗ fy ∗ As )

▪ Pu1 = 0.4 ∗ fcu ∗ (Ac − As ) + 0.76 ∗ fy ∗ As

▪ Pu2 = 0.35 ∗ fcu ∗ (Ak − As ) + 0.67 ∗ fy ∗ As + 1.38 ∗ Vsp ∗ fysp

0.01 Ac
As = the largest from {
0.012 Ak

Asp ∗ π ∗ Dk Vsp fcu Ac

Vsp = , μsp = ≥ μsp,min = 0.36 ∗ ( − 1)
P Ak fys Ak

4 Dr.-Ing. Mahmoud Elshahawi

 2. Slenderness Raito “𝛌” ‫نسبة النحافة‬
Short column Slender column

Madd = 0.0

Madd = Pu ∗ δ
---------------------------------------------------------------------------

3. Lateral Loads ‫األحمال الجانبية‬

---------------------------------------------------------------------------

4. Loads (stresses)

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 ❖ Example:
Pu = 3300 kN
fcu = 25 N⁄mm2
fy = 350 N⁄mm2
fy,st = 240 N⁄mm2
➢ Using the ultimate limit state design method, design and draw cross section
details for the following “Short Braced”.
a. Square tied column.
b. Rectangular tied column of width equal to 300 mm.
c. Circular tied column.
d. Circular spiral column.
------------------------------------------------------------------------------------------------------
➢ For Tied Columns
Pu = 0.35 ∗ fcu ∗ (Ac − As ) + 0.67 ∗ fy ∗ As
3300 ∗ 103 = 0.35 ∗ 25 ∗ (0.992Ac ) + 0.67 ∗ 350 ∗ 0.008Ac
As = 0.008 ∗ Ac
Acreq = 312618.42 mm2
-------------------------------------------------------------------------------------------------------
a. For Square Tied Column
A c = b2
b = √Ac = 559 mm
use b ∗ b = (600 ∗ 600)mm
As = ‫األكبر‬
o 0.008 ∗ Ac Req = 0.008 ∗ 312618.42 = 2500.95 mm2
o 0.006 ∗ Ac actual = 0.006 ∗ 6002 = 2160 mm2
As = 2500.95 mm2
Use 16D16

6 Dr.-Ing. Mahmoud Elshahawi

 ‫❖ شروط رص األسياخ والكانات‪:‬‬
‫‪ . 1‬المسافة بين األسياخ الرأسية ال تزيد عن ‪ 250‬مم‪.‬‬
‫‪ . 2‬المسافة بين األسياخ المربوطة والغير مربوطة ال تزيد عن ‪ 150‬مم‪.‬‬
‫حجم الكانات‬ ‫‪Vstip‬‬ ‫‪0.25‬‬
‫=‬ ‫≮‬ ‫‪. 3‬‬
‫حجم الخرسانة‬ ‫‪Vconc‬‬ ‫‪100‬‬
‫‪Vst‬‬ ‫‪0.25‬‬
‫=‬
‫‪Vconc‬‬ ‫‪100‬‬
‫‪n∗Ast ∗∑ L‬‬ ‫‪0.25‬‬
‫=‬
‫‪b∗b∗1000‬‬ ‫‪100‬‬
‫‪550‬‬
‫∗‪n∗50.3∗(4∗550+4‬‬ ‫)‪√2‬‬ ‫‪0.25‬‬
‫‪2‬‬
‫=‬
‫‪600∗600∗1000‬‬ ‫‪100‬‬
‫‪′‬‬
‫‪n = 4.76 → Use 5∅8/m‬‬ ‫عدد األسياخ ال يقل عن ‪ 5‬في المتر‬

‫‪-------------------------------------------------------------------------------------------------------‬‬
‫‪b. Rectangular column‬‬
‫‪Ac = b ∗ t‬‬
‫‪312618.42 = 300 ∗ t‬‬
‫‪t = 1042.1 mm‬‬
‫‪Use b ∗ t = (300 ∗ 1100)mm‬‬
‫األكبر = ‪As‬‬
‫‪o 0.008 ∗ Ac Req = 0.008 ∗ 312618.42 = 2500.95mm2‬‬
‫‪o 0.006 ∗ Ac actual = 0.006 ∗ 300 ∗ 1100 = 1980 mm2‬‬
‫‪As = 2500.95 mm2 → Use 16D16‬‬

‫‪7‬‬ ‫‪Dr.-Ing. Mahmoud Elshahawi‬‬

 Vst 0.25
=
Vconc 100
n∗50.3∗∑ L 0.25
=
b∗t∗1000 100
∑L = 6 ∗ 250 + 2 ∗ 1050 + 4 ∗ 150 = 4200 mm
n∗50.3∗4200 0.25
=
300∗1100∗1000 100
n = 3.9 → Use 5∅8/m′
------------------------------------------------------------------------------------------------
c. For Circular Tied Column
D2
Ac = π ∗
4
D2
312618.42 = π ∗
4
D = 630 mm
Use D = 650 mm
As = ‫األكبر‬
o 0.008 ∗ Ac Req = 0.008 ∗ 312618.42 = 2500.95 mm2
6502
o 0.006 ∗ Ac actual = 0.006 ∗ π ∗ = 1990 mm2
4
As = 2500.95 mm2 → Use 16D16
▪ Check Volume of Stirrups:
n∗50.3∗π∗600 0.25
6502
=
π∗ ∗1000 100
4

n = 8.74 → Use 9∅8/m′

8 Dr.-Ing. Mahmoud Elshahawi

 ❖ Spiral Columns

D2c⁄
Ac = π ∗ 4
2
D
Ak = π ∗ k⁄4
π∗Asp ∗Dk
Vsp =
P
As = the largest from
▪ 0.01 ∗ Ac
▪ 0.012 ∗ Ak
❖ Solution:
3300 ∗ 103 = 0.4 ∗ 25 ∗ (Ac ∗ 0.99) + 0.76 ∗ 350 ∗ 0.01Ac
D2
▪ Ac = 262738.85 mm2 = π ∗ c⁄4
Dc = 578 mm
take Dc = 600 mm
Dk = Dc − 50 = 550 mm
▪ As = ‫األكبر‬
2
o 0.01* π ∗ 600 ⁄4 = 2827 mm2
2
o 0.012* π ∗ 550 ⁄4 = 𝟐𝟖𝟓𝟏 𝐦𝐦𝟐
Use 12D18 with Asact = 3048 mm2
--------------------------------------------------------------------------------------------

9 Dr.-Ing. Mahmoud Elshahawi

 Pu = 0.35 ∗ fcu ∗ (Ak − As ) + 0.67 ∗ fy ∗ As + 1.38 ∗ Vsp ∗ fysp
5502
3300 ∗ 103 = 0.35 ∗ 25 ∗ (π ∗ − 3048) + 0.67 ∗ 350 ∗ 3048 + 1.38 ∗
4
Vsp ∗ 240
Vsp = 1609.5 mm3
➢ Check μsp :
min
Vsp 1609.5
μsp = = 2 = 6.77 ∗ 10−3
Ak π∗550 ⁄4
fcu A
μsp = 0.36 ∗ (
fysp
) ∗ (Ac − 1)
min k
2
25 π∗600 ⁄4
μsp = 0.35 ∗ ( )∗( 2 − 1) = 7.128 ∗ 10−3
min 240 π∗550 ⁄4

Take μsp = 7.128 ∗ 10−3

Vsp
7.128 ∗ 10−3 =
Ak
Vsp = 1693.5 mm3
Vsp = π ∗ Dk ∗ Asp ⁄P
Asp → ∅8 → 50.3
1693.5 = π ∗ 550 ∗ 50.3⁄P
P = 51 mm → take P = 50 mm

10 Dr.-Ing. Mahmoud Elshahawi

 «Pu, Mu» “Uniaxial”
❖ Design
• Given Mu, Pu, b, t
• Required As
--------------------------------------------------------
.‫ ال بد من عمل فحص هل يمكن اهمال أي من العزم او القوة‬:‫أوال‬
Pu
if < 0.04 → neglect Pu
fcu ∗b∗t

Mu 0.05 t
if e = < emin = the largest from { → neglect Mu
Pu 20 mm

In case of both Pu and Mu are considered, there are two cases:

Mu t
a. e = > → big eccentricity → design using Mus approach
Pu 2
Mu t
b. e = < → small eccentricity → design using the interaction diagrams
Pu 2

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Interaction diagrams:
1- Symmetric Reinforcement
A′s = α ∗ As
α = 1.0
eccentricity
M
▪ e = u⁄P
u
▪ e⁄t > 0.2

2- Un symmetric Reinforcement
A′s = α ∗ As
α = 0.8

3- Uniform Distribution ‫حديد موزع بانتظام‬

M
▪ e = u⁄P
u
▪ e⁄t < 0.2

Note:
‫ القطاع المستطيل يتحمل عزم أكبر من القطاع‬-
‫ والقطاع المربع يتحمل عزم أكبر من‬،‫المربع‬
.‫ي‬
ّ ‫القطاع الدائر‬

12 Dr.-Ing. Mahmoud Elshahawi

 ‫❖ خطوات دخول الـ ‪:I.D‬‬
‫‪ )1‬يتم تحديد الصفحة عن طريق ‪fy , α, ξ‬‬
‫‪(d − d′ )⁄‬‬
‫=‪ξ‬‬ ‫‪t‬‬ ‫‪d = t − cover, d′ = 50 mm‬‬
‫‪ξ = 0.8 or 0.9‬‬
‫‪ )2‬يتم حساب كل من دليل الدخول األفقي والرأسي‬
‫‪Pu‬‬ ‫‪Mu‬‬
‫& ‪⁄f ∗ b ∗ t‬‬ ‫‪⁄f ∗ b ∗ t 2‬‬
‫‪cu‬‬ ‫‪cu‬‬

‫‪ )3‬يتم دخول الـ ‪ Curves‬وحساب قيمة ‪ρ‬‬

‫‪μ = ρ ∗ fcu ∗ 10−4‬‬
‫‪As = μ ∗ b ∗ t‬‬
‫‪--------------------------------------------------------‬‬
‫”‪➢ Point “1‬‬
‫نزود األبعاد ‪Unsafe‬‬

‫”‪➢ Point “2‬‬

‫⋯=‪ρ‬‬
‫⋯=‪μ‬‬
‫⋯ = ‪As‬‬

‫”‪➢ Point “3‬‬

‫نأخذ قيمة ‪ ρ‬األقل‬

‫‪13‬‬ ‫‪Dr.-Ing. Mahmoud Elshahawi‬‬

Example:
• Pu = 2700 kN
• Mux = 675 kN.m
• b = 300 mm, t = 600 mm
• fcu = 25 MPa, fy = 350 MPa
• ξ = 0.9
❖ Solution
Check:
Pu
> 0.04 consider
fcu ∗ b ∗ t
Mu 675
e= = = 0.25 m → consider
Pu 2700
e⁄ = 0.25
t 0.6 = 0.42 > 0.2
Symmetric
✓ fy = 360 MPa
✓ α=1
✓ ξ = 0.9
Pu 3
⁄f ∗ b ∗ t = 2700 ∗ 10 ⁄25 ∗ 300 ∗ 600 = 0.6
cu
Mu 6
⁄f ∗ b ∗ t 2 = 675 ∗ 10 ⁄25 ∗ 300 ∗ 6002 = 0.25
cu
unsafe
use b * t = 300 * 800
e⁄ > 0.2 → Symmetric
t
Pu 3
⁄f ∗ b ∗ t = 2700 ∗ 10 ⁄25 ∗ 300 ∗ 800 = 0.45
cu
Mu 6
⁄f ∗ b ∗ t 2 = 675 ∗ 10 ⁄25 ∗ 300 ∗ 8002 = 0.14
cu

ρ = 6.4
μ = ρ ∗ fcu ∗ 10−4 = 6.4 ∗ 25 ∗ 10−4 = 0.016
As = μ ∗ b ∗ t = 0.016 ∗ 300 ∗ 800 = 3840 mm2
Use 8D25
A′s = α ∗ As = 3840 mm2
Use 8D25

14 Dr.-Ing. Mahmoud Elshahawi

 ❖ Example:
For the shown-braced rectangular column section, it is required to determine the
required reinforcement if Pu = 2000kN and Muy = 200kN.m.

• Pu = 2000 kN
• Mu = 200 kN.m
• b = 600 mm , t = 300 mm
• fcu = 25 MPa , fy = 350 MPa
• ξ = 0.9
Solution
Pu
check: > 0.04 consider
fcu ∗ b ∗ t
Mu 200
e= = = 0.1 m
Pu 2000
e⁄ = 0.1
t 0.3 = 0.33 > 0.2
Pu 3
⁄f ∗ b ∗ t = 2200 ∗ 10 ⁄25 ∗ 600 ∗ 300 = 0.44
cu
Mu 6
⁄f ∗ b ∗ t 2 = 200 ∗ 10 ⁄25 ∗ 600 ∗ 3002 = 0.148
cu
Symmetric
ρ = 6.4
ρ = 6.4
μ = ρ ∗ fcu ∗ 10−4 = 6.4 ∗ 25 ∗ 10−4 = 0.016
As = μ ∗ b ∗ t = 0.016 ∗ 600 ∗ 300 = 2880 mm2
Use 8D22
A′s = α ∗ As = 2880 mm2
Use 8D22

15 Dr.-Ing. Mahmoud Elshahawi

 «Pu, Mux, Muy» “Biaxial”
Using interaction diagrams:

(R b , fy ) ‫ يتم تحديد رقم الصفحة عن طريق‬-

Pu
Rb = ⁄f ∗ b ∗ t = 0.3, 𝑜𝑟 0.40, 𝑜𝑟 0.50, 𝑜𝑟 0.60
cu

get ρ = ⋯
μ = ρ ∗ fcu ∗ 10−4
As,total = μ ∗ b ∗ t

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 Interaction diagrams for uniaxially loaded columns

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 Interaction diagrams for biaxially loaded columns

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