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Pengiraan Tiang

1) The column is 350mm x 450mm with an effective length of 3800mm carrying an axial load of 2200kN. 2) The required reinforcement is calculated to be 2 H20 bars and 2 H12 bars, with H8 links at 200mm spacing. 3) The total quantity of materials estimated are 0.65 bars of H20, 0.65 bars of H12, 1.28 bars of H8 links, and 0.63m3 of concrete.

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0% found this document useful (0 votes)
1K views7 pages

Pengiraan Tiang

1) The column is 350mm x 450mm with an effective length of 3800mm carrying an axial load of 2200kN. 2) The required reinforcement is calculated to be 2 H20 bars and 2 H12 bars, with H8 links at 200mm spacing. 3) The total quantity of materials estimated are 0.65 bars of H20, 0.65 bars of H12, 1.28 bars of H8 links, and 0.63m3 of concrete.

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Ref

Calculations
Classification : Brace non-slender column
Design the longitudal and transverse reinforcement for the column shown
below. The column are objected to 50 years working life, 1 hour fire
resistance and build inside building. Use grade C25 concrete and grade
500 steel reinforcement.
Specification
Concrete, fck = 25 N/mm2
Reinforcement, fyk = 500 N/mm2
Exposure class = XC1
Fire Resistance = 1.0 hours
Design life = 50 years
Size, b X h = 350mm x 450mm
Effective length, l0 = 3800mm
Assumed : link = 8 mm
bar = 20 mm
Axial force, Ned = 2200 kN

Durability, Bond & Fire Resistance


Table 4.2

Min.cover with regard to bond, Cmin,b = 20mm

Table 4.4N

Min.cover with regard to durability, Cmin,dur = 15mm


Min.required axis distance for R60 fire resistance

Table 5.2A

Resistance sd = 40 mm
Min.concrete cover with regard to fire,
Cmin = sd - link - bar/2
= 40 - 8 - (20/2)
= 22 mm

4.4.13

Allowance in design for deviation, C dev = 10 mm

Output

4.4.1.4 (2)

Nominal Cover
Cnom = Cmin + C dev
= 22 + 10
= 32 mm

5.8.8.2

Design Moment
For non-slender column the design moment,
Med = Max ( Mo2 , M min ) (choose highest )
Where,
Mo2 = M = Ned, ei
M = max ( M bot , M top ) = 55 kNm

5.2 (7)

ei = ( lo/ 400) = 3800/400 = 9.5mm


Mo2 = 55 + ( 2200 x 0.0095 ) = 75.90 kNm

6.1 (4)

M min = Ned, eo
eo = h/30 20 mm
= 450/30 = 15 mm
= 15 < 20 mm ( use 20mm) @ 0.02

M min = 2200 x 0.02 = 44 kNm


Mo2 > Min
75.90 > 44
Med = 75.90 kNm

Use :
Cnom = 35mm

Reinforcement
d2 = Cnom + link + bar/2
= 35 + 8 + 10
= 53 mm
d2 /h = 53/450
= 0.12
Refer
Design
Chart
9.5.2 (2)

NEd/bh x fck = 2200 x 103 / 350 x 450 x 25


= 0.56
MEd/bh2 x fck = 75.90 x 106 / 350 x 4502 x 25
= 0.04

As x fyk / bh x fck = 0.09

As = 0.09 bh x fck / fyk


= 0.09 (350 x 450) (25) / 500
= 708.75 mm2

Use 2H20 (628 mm)


(+)
2H12 (226 mm)
= (854 mm)

9.5.2 (2)

As,min = 0.1 Ned /fyk(0.87)


= 0.1 (2200 x 103) /(500x0.87)
= 506 mm2 or 0.002 Ac = 315 mm2

9.5.2 9 (3)

As,max = 0.04 Ac
= 0.04 x 350 x 450
= 6300 mm2

Use : 506

Proven : As min < As prov < As max

OK!

500 mm2 < 854 mm2 < 6000 mm2

9.5.3

Links, min = 0.25 x the larger of bar

Use: link
H8

= 0.25 x 20
= 5 mm or 6 mm

Sv max = 20 x the lesser of bar


= 20 (12)
= 240 or 350 or 400
use H8 - 200

At section 300 mm below & above beam and at lapped joints ,


Sv max = 0.6 x 200
= 120

Use H8-100

Bengkok Link @ Link Tambahan


35

280

35

35

380

35

1) 350 2(35) - bar 2(link bar)


= 350 70 20 2 (8)
= 244 mm
2) 450 2(35) - bar 2(link bar)
= 450 70 10 - 6 2(8)
= 348mm
3) 350- 2(35) - bar 2(link bar)
=350 70 12 16
= 252mm
Tambatan Tetulang
= (244 + 348 + 348 + 252) + 2(2 r bar) + 2(2 r bar) + 2(4 link)
= 1192 + 2(2 (10))+ 2(2 (6)) + 64
= 1192 + 62.83 + 37.7 + 64
= 1.36m

Bilangan Tetulang
Lapping = (h / b) + 1
(450/350) + 1 = 2.29 ~ 3 batang

Panjang tiang - overlapping


3800 600 = 3200
3200 / h = 3200/450 =7.1 ~ 8 batang

3 + 8 = 11 nos

Anggaran Bahan
Jumlah link
= 1.36m 11nos
= 14.96+ 3% pembaziran =15.41
=15.41/12 = 1.28 ~ 2 batang (H8)
Besi tetulang
(H20)
= 3.8m 2btg
= 7.6 / 12 = 0.63 batang + 3% pembaziran
= 0.65
(H12)
= 7.6 / 12 = 0.63 batang + 3% pembaziran
= 0.65
Konkrit
Lhb
= 3.8 0.45 0.35
= 0.6m3 + 5% pembaziran
= 0.63m3

Drawing

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