column design to bs8110 template

CLIENT UNITED ENGINEERING SERVICES LTD. Date 9-Jun-15PROJECT PIARCO GENERATOR BUILDING By M RampersadLocation Structural FrameSub-Location RC Column Design

REF OUTPUTDESCRIPTIONRC COLUMN DESIGN


REF OUTPUTDESCRIPTION

References Used

1 Reinforced Concrete Design, 4th ed, Moseley & Bungey

2 BS8110: 1997, Structural Use of Concrete

3 STAAD Model analysis

4 ASTM A615: 04, Std. Spec for Deformed & Plain Carbon Steel Bars etc.

MEMBER GEOMETRYGeometric data

Member width, b = 450.0 mm 450 Member depth, h = 450.0 mm 450

Cover to reinforcement = 40.0 mmConc. compressive strength, fcu = 30.0 N/mm2

4 Tension rfct strength, fy = 420.0 N/mm24 Shear rfct strength, fyv = 280.0 N/mm2

Length between restraints, LO = 3500.0 mm

450

450

00

Nominal aggregate size 20.0 mm OK against coverMin. allowable bar spacing = 25.0 mm

Effective depth, h' = 400.0 mm 400 Effective width, b' = 410.0 mm 410

bd2fcu term = 2,160,000,000 NmGross cross-sectional area, Ag = 202,500 mm2

40

410

Check effective height & slendernessX-X axis

Support condition at TOP 1.0Support condition at BOTTOM 2.0

2 - Tbl 3.19 end restraint beta, 0.802 - Cl. 3.8.1.6.1 Effective height (major axis), lex = LO = 2,800 mm2 - Cl. 3.8.1.3. Classification (short or slender) = 6.2 Short

Y-Y axisSupport condition at TOP 3.0

Support condition at BOTTOM 2.02 - Tbl 3.19 end restraint beta, 0.952 - Cl. 3.8.1.6.1 Effective height (minor axis), ley = LO = 3,325 mm2 -Cl. 3.8.1.3. Classification (short or slender) = 7.4 Short

Slenderness limit2 - Cl. 3.8.1.7 Slenderness limit (LO should be



ULS LOAD DATAAxial load, N = 150.0 kN 150kN

Moment about major axis, MX = 50.0 kNm 50kNmMoment about minor axis, MY = 20.0 kNm 20kNm

Max shear force, V = 75.0 kNSTAAD Member Ref = 115

Load Case = 114

1 - Sxn 9.4.4 Check MX/h' = 125.0 kNkN

3

150kN

50kNmCheck MY/b' = 48.8 kN

N/bdfcu = 0.02 [unitless]1 - Tbl 9.4 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

20kNm

REINFORCEMENT CHECKS2 - Cl. 3.4.4.4 Calculation for reinforcement

Total steel required, As req = (see supplemental calcs) 1,040 mm2

Select bar diameter = 20 mmSpecify number of bars = 4 Nr

C i t l id d A 2Compression steel provided, As prov = 1,260 mm2

Number of layers = 2 Min. spacing between bars = 330.0 mm Spacing OK

Provide 4Nr T20 bars in 2 layer(s)Check for minimum steel

2 - Tbl. 3.25 100 AS/AC = 0.62 %2 - Cl. 3.12.5.3 Recommended value (Minimum) = 0.40 % Min. steel OK2 - Cl. 3.12.6.2 Recommended value (Maximum) = 4.00 % Max. steel OK

SHEAR REINFORCEMENT CHECKS2 - Cl. 3.4.5.2 Design (Actual) shear stress, = V/bvd

V = 75.0 kNb = b = 450 0 mmbv = b = 450.0 mm

d = 400.0 mm Design (actual) shear stress, = 0.42 N/mm2


REF OUTPUTDESCRIPTION2 - Cl. 3.4.5.2 Check limits, < lesser of 0.8fcu or 5N/mm2

0.8fcu = 4.38 N/mm2 Shear limit OKDesign concrete shear stress, c

=0.79*(100AS/bvd)1/3*(400/d)1/4*fcu multiplier /m 1/32-Tbl 3.8 Note 2 (100AS prov/bvd)1/3 = 0.89 [unitless]

2-Tbl 3.8 Note 2 (400/d)1/4 = 1.00 [unitless]2-Tbl 3.8 Note 2 fcu multiplier = 1.06 N/mm2

c = 0.60 N/mm22 - Tbl 3.16 Check range of to c

< c Yes 0.74 Use nominal links0 4 c < < (c+0.4): Use Asv/sv = 0.4b/0.87fyv No 0.74

(c+0.4) < < 0.8fcu (or 5.0): Use Asv/sv = b(-c)/0.87fyv No -0.33 2 - Tbl 3.7 Shear reinforcement requirement

Shear reinforcement ratio required, Asv/sv = 0.74Select bar diameter = 10Specify bar spacing = 150

Shear reinforcement ratio provided, Asv/sv = 1.05Provide R10 at 150mm centres


REF OUTPUTDESCRIPTIONSupplemental Calculation for A S REQNote: 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.1. Guess an initial value of A' S2. See if this results in a point exactly on the curve of the M-N p yinteraction diagram3. If this does, then use Excel Goal-Seek to derive the depth of the neutral axis (x)

Try A S = 520 mm 2

1-Eg 9.3 Depth of neutral axis, x = 44.0 mmcompressive steel strain, sc = 0.0003

tensile steel strain, s = 0.0283Design yield strain, y = f Y / M /E S = 0.0018 strains OK

balanced neutral axis, x bal = 262.9 mm neutral axis depth OK

Depth of stress block, s = 0.9x = 39.6 mm

From yield strain curvef SC (compression) = E S SC 63.6 N/mm 2

f SC (tension) = E S S 5663.7 N/mm 2For 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) = 91463239 Nmm

e = M/N = 463.4 mmN(e+h/2-d 2 ) = 97262195 Nmm

To allow for the area of concrete displaced, f SC = 351.7 N/mm2

f SC (d-d') = 126618 Nmm

IF((C117-C115)/C119



N/bh = 0.74M/bh 2 = 0.76

18.0MNinteractiondiagram

8 0

10.0

12.0

14.0

16.0

N/bh

g

0.0

2.0

4.0

6.0

8.0N

M/bh2

MNInteractionDiagramLoadpoint

0.00 0.50 1.00 1.50 2.00 2.50 3.00

column design to bs8110 template

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