Column Design to Bs8110 Template

CLIENT PROJECT Location Sub-Location REF

UNITED ENGINEERING SERVICES LTD. PIARCO GENERATOR BUILDING

Date By

9-Jun-15 M Rampersad

Structural Frame RC Column Design

DESCRIPTION

RC COLUMN DESIGN

OUTPUT

Trusted by over 1 million members

Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.



CLIENT PROJECT Location Sub-Location

Date By




REF

DESCRIPTION

OUTPUT

References Used 1

Rein Reinfo forc rced ed Con Concr cret ete e Desi Design gn,, 4th 4th ed, ed, Mose Mosele ley y & Bung Bungey ey

2

BS81 BS8110 10:: 199 1997, 7, Str Struc uctu tura rall Use Use of Conc Concre rete te

3

STAAD Model analysis

4

ASTM ASTM A615: A615: 04, 04, Std. Std. Spec Spec for Deform Deformed ed & Plain Plain Carbo Carbon n Steel Steel Bars Bars etc. etc.

MEMBER GEOMETRY

Geometric data Member width, b =

450.0

mm

450

Member depth, h =

450.0

mm

450

Cover to reinforcement = Conc. compressive strength, f cu cu =

40.0

mm N/mm2

450

4 4

Tension rfct strength, fy = Shear rfct strength, f yv yv =

280.0

N/mm2 N/mm2

Length between restraints, LO =

3500.0

mm

20.0

mm

0 5 4

0 0 4

30.0

Nominal ag aggregate si size

410

420.0

Min. allowable bar spacing =

25.0 mm

Effective depth, h' =

400.0 mm

400

Effective width, b' = bd2f cu cu term =

410.0 mm 2,160,000,000 Nm

410

Gross cross-sectional area, Ag =

202,500 mm 2

Check effective height & slenderness X-X axis Support condition at TOP

1.0

Support condition at BOTTOM

2.0 0.80 2,80 2,8000 mm

end restraint beta,  

2 - Tbl 3.19

Effective height (major axis), l ex = LO =

2 - Cl. 3.8.1.6.1

Classification (short or slender) =

2 - Cl. 3.8.1.3.

6.2 Short

Y-Y axis

2 - Tbl 3.19

OK against cover

Support condition at TOP

3.0

Support condition at BOTTOM

2.0

end restraint beta,  

0.95




Date By




REF

DESCRIPTION

OUTPUT

ULS LOAD DATA

Axial load, N = Moment about major axis, M X =

150.0

150kN

50.0

kN kNm

Moment about minor axis, M Y =

20.0

kNm

20kNm

Max shear force, V =

75.0

kN

STAAD Member Ref =

115

Load Case =

114

3

150kN

1 - Sxn 9.4.4 50kNm

Check MX/h' = Check MY/b' =

1 - Tbl 9.4

20kNm

50kNm

125.0 kN 48.8

N/bdf cu cu =

0.02 [unitless]

coefficient beta,  =

1.00 [unitless]

beta*(h'/b') =

0.98 [unitless]

Since Mx/h'>My/b', Use single axis design moment about X-axis Use M'x as the design moment, M =

69.5 kNm

Calculation for reinforcement Total steel required, A s req = (see supplemental calcs)

2 1,040 mm

REINFORCEMENT CHECKS 2 - Cl. 3.4.4.4

Select bar diameter = Specify number of bars = ompress on s ee pro prov e , s prov = Number of layers = Min. spacing between bars =

20 mm 4 Nr 2 1,260 mm 2 330.0 mm

Spacing OK

Provide 4Nr T20 bars in 2 layer(s)

Check for minimum steel 100 AS/AC =

2 - Tbl. 3.25

0.62 %

2 - Cl. 3.12.5.3

Recommended value (Minimum) =

0.40 %

Min. steel OK

2 - Cl. 3.12.6.2

Recommended value (Maximum) =

4.00 %

Max. steel OK

SHEAR REINFORCEMENT CHECKS 2 - Cl. 3.4.5.2

Design (Actual) shear stress,  = V/bvd V=

75.0 kN






REF 2 - Cl. 3.4.5.2

Date By


DESCRIPTION

Check limits,  < lesser of 0.8√f cu cu or 5N/mm

OUTPUT

2

0.8√f cu cu = Design concrete shear stress,

2 4.38 N/mm

Shear limit OK

c

1/3

=0.79*(100AS/bvd) *(400/d)1/4*fcu multiplier / m (100A S prov/bvd)

2-Tbl 3.8 Note 2 2-Tbl 3.8 Note 2 2-Tbl 3.8 Note 2

=

0.89 [unitless]

(400/d)1/4 = f cu cu multiplier =

1.00 [unitless] 2 1.06 N/mm

∴

2 - Tbl 3.16

c =

Check range of  to c

c <  < (c+0.4): Use Asv/sv = 0.4b/0.87f yv yv

Yes 0.74 No .7

(c+0.4) <  < 0.8√f cu cu (or 5.0): Use A sv/sv = b(-c)/0.87f yv yv

No -0.33

 < c

2 - Tbl 3.7

2 0.60 N/mm

Shear reinforcement requirement requirement Shear reinforcement ratio required, A sv/sv =

0.74

Select bar diameter =

10

Specify bar spacing = Shear reinforcement ratio provided, A sv/sv =

150 1.05

Provide R10 at 150mm centres

Use nominal links




Date By




REF

DESCRIPTION

OUTPUT

Supplemental Calculation for A S REQ Note: This is an iterative process where an assumption is made for the steel provided (A' S ) and tested to see if this provides an intersection point on the M-N interaction diagram. diagram. 1. Guess an initial value of A' S 2. See if this results in a point exactly on the curve of the M-N interaction diagram 3. If this does, then use Excel Goal-Seek to derive the depth of the neutral axis (x)

1-Eg 9.3

Try A S =

520 mm

Depth Depth of neutra neutrall axis, axis, x =

44.0 44.0 mm

compressive steel strain,  sc =

0.0003

tensile steel strain,  s =

0.0283

Design yield strain,  y = f Y /  M /E S =

0.0018

balanced neutral axis, x bal =

262.9 mm

Depth Depth of stress stress block, block, s = 0.9x 0.9x =

39.6 39.6 mm

2

strains OK neutral axis depth OK

From yield strain curve f SC (compression) = E S  SC

63.6 N/mm

2

f SC (tension) = E S  S

5663.7 N/mm

2

For compression N(e+h/2-d 2 ) = 0.45f CU bs(d-s/2) + f SC A' S (d-d') 0.45f CU bs(d-s/2) = e = M/N = N(e+h/2-d 2 ) = To allow for the area of concrete displaced, f SC = f SC (d-d') =

914 914632 63239 Nmm Nmm 463.4 mm 972 972621 62195 Nmm Nmm 351.7 N/mm

2

126618 Nmm IF((C117-C115)/C119<0,C73*C28/100,(C117-C115)/C

For tension N = 0.45f CU bs + f SC A' S +f S A S ∴

A S = (0.45f CU bs - f SC A' S -N)/f S 0.45f CU bs =

240566 N

f SC A' S =

182893 N

A' S =

520 mm

Total steel required, A S REQ =

1,040 mm

∴

2




Date By




REF

DESCRIPTION

OUTPUT

N/bh = 2

M/bh = 18.0

0.74 0.76

M‐N interaction dia ram

16.0

14.0

12.0

10.0 h b /

.

6.0

M‐N Interaction Diagram Load point

4.0

2.0

M/bh2

0.0 0.00

0.50

1.00

1.50

2.00

2.50

3.00

Column Design to Bs8110 Template

Recommend Documents