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1. Welcome To CE 413
J. M. Raisul Islam Shohag
B.Sc. in Civil Engineering (RUET)
Lecturer
Department of Civil Engineering
Daffodil International University.
Lecture-10

2. Submitted By
Mehedi Hasan
ID-183-47-817
Evening Shift, E1 Section
Department of Civil Engineering
Daffodil International University
Aynul Fahad Peal
ID-183-47-828
Evening Shift, E1 Section
Department of Civil Engineering
Daffodil International University
Emtiaz Ahmmed
ID-183-47-824
Evening Shift, E1 Section
Department of Civil Engineering
Daffodil International University

3. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
Design of (CFT) Column
Problem-1 :
Figure given below is Concrete Filled Tube Column (CFT).
Given data : Span = 14’
x
y
B=6’’
H=10’’
L=14’
PD = 32.0 kips
PL = 84.0 kips
HSS
HSS=10*6*
3
8

4. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
Given data :
Normal wt of concrete =145 lb/𝑓𝑡3
Concrete Compressive strength , f’c = 5 ksi
ASTM GA 500 Grade B , Fy = 46 ksi , Fy = 58 ksi
Geometric Properties of HSS 10*6*3/8 are follows
As = 10.4 ⅈ𝑛2 , H = 10 in , B = 6.00 in
l min = 0.349 in ( Design Wall Thickness )
l/t = 25.7 , b/t = 14.2
I sx = 137 ⅈ𝑛4
I sy = 661.8 ⅈ𝑛4
Design of (CFT) Column

5. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
Solution:
Limitation of AISC specification:
1) Concrete strength: 3 ksi ≤ f’c ≤ 10 ksi
Here , f’c = 5 ksi → ok
2) Specified minimum yield stress of structural steel :
Fy ≤ 75 Ksi
Here, Fy = 46 ksi → ok
3) Cross Sectional area of steel section : As ≥ 0.01 Ag
Here, As = 10.4 ⅈ𝑛2
Design of (CFT) Column

6. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
Internal Clear distance :
h i = h - 2t =10-2×0.349 = 9.3 in
b i = B - 2t = 6 – 2 × 0.349 = 5.3 in
Area of concrete, Ac = 9.3 ×5.3 = 49.29 ⅈ𝑛2
Ag = Ac + As = 49.29 + 10.4 = 59.69 ⅈ𝑛2
0.01Ag = 0.5969 ⅈ𝑛2 < As ∴ As > 0.01 Ag → ok
Check For local buckling:
𝜆b = 2.26
𝐸
𝐹𝑦
[h = 10 – 3 × 0.349]
Design of (CFT) Column
Solution:

7. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
= 2.26
29000
46
=56.7
ℎ
𝑡
=
10−3× 0.349
9.349
= 25.7
𝜆 Controlling = Max
ℎ
𝑡
= 25.7
𝑏
𝑡
= 14.2
= 25.7
𝜆 Controlling < 𝜆p → ∴ Section is compact
Available compressive Strength:
Pno = Fy As + C2 f’c ( Ac + Asr
𝐸𝑠
𝐸𝑐
) Here,C2 = 0.85 for rectangular section
Pno = 46 × 10.4 + 0.85 × 5 × (49.29+0) = 687.88 kip ≈ 688 kip
Design of (CFT) Column
Solution:

8. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
Now, EI eff = Es Isy + Es Isr + C3 Ec Icy
C3 = 0.6 + 2 (
𝐴𝑠
𝐴𝑐+𝐴𝑠
) ≤ 0.9
= 0.6 + 2 ×
10.4
49.29+10.4
= 0.947 ≤ 0.9 = 0.9
Ec = 𝑤𝑐1.5
f’c = (145)1.5
5 = 3900 ksi
Icy =
𝐻−4𝑡 𝑏𝑖3
12
+
𝑡(𝐵−4𝑡)3
6
+
9𝜋2−64 𝑡4
36𝜋
+ π𝑡2(
𝐵−4𝑡
2
+
4𝑡
3𝜋
)2
=
10−4∗0.349 ∗5.33
12
+
0.349∗ (6−4∗0.349)3
6
+
9𝜋2−64 ∗0.3494
36𝜋
+
𝜋 ∗ 0.3492
(
6−4∗0.349
2
+
4∗0.349
3𝜋
)
2
= 116.7𝑖𝑛4
≈ 115𝑖𝑛4
Design of (CFT) Column
Solution:

9. Design of (CFT) Column
Alternative:
Icy = Ig – Isy = 10∗63
12
− 61.8 = 118.2 ⅈ𝑛4
Alternative
Icy =
H−2t ∗(B−2t)3
12
= 115 .5 ⅈ𝑛4
Solution:

10. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
∴ EI eff = 29000 * 61.8 + 0 + 0.9 * 3900 * 115
= 2195850 ≈ 2200000 kip-𝑖𝑛2
Pe =
𝜋2(EI cft)
(𝐾𝐿)2 =
𝜋2∗2200000
(1∗14∗12)2 = 769 kips
Pn𝑜
Pe
=
688
769
= 0.895 < 2.25
∴ Pn = Pno [ 0.658
Pn𝑜
Pe ] = 688*0.6580.895
= 473 kips
LRFD
Pu = 1.2 PDL + 1.6 PLL
= (1.2*32) + (1.6*84) = 173 kips
Design of (CFT) Column
Solution:

11. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU
∅𝑃 𝑛 = 0.75*473 = 355 kips (∅𝑃𝑛 > 𝑃𝑢)→ 𝐨𝐤
ASD
Pu = P DL + P LL
= 32 + 84
= 116 kips
𝑃𝑛
Ω
=
473
2.00
= 236.5 kips
∴
𝑃𝑛
Ω
> Pu → ok
Design of (CFT) Column
Solution:

12. J. M. Raisul Islam Shohag, Lecturer, Civil Engineering Department, DIU