01 enb276 column_i single

5/25/2018 01 ENB276 Column_I Single

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Queensland University of Technology

CRICOS No. 000213JENB276- Structural Engineering 1ENB276 Structural Engineering 1


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Reinforced Concrete Basics 2E(2010) by

Foster, Kilpatrick & Warner, Pearson Education AS3600:2009 (current BCA still references

AS3600:2001

Others

and Chowdhury, Cambridge University Press

Concrete Structures 1998 b Warner Ran an Hall& Faulkes, Longman

CRICOS No. 000213Ja university for the worldrealR ENB276- Structural Engineering 1


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COLUMNS AND WALLS

floor systems.

, ,

the foundation.

A Wall is a thin planar member with smallwidth compared to the its length.

Both may subjected to pure compression, or

compression with uniaxial orbiaxialbending.


CRICOS No. 000213JENB276- Structural Engineering 1


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-



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max



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Short Column Slender Column



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A column is classified as short if Le/r

for a braced column, less than or equal to the greatero an

*

2

*

1115

'38

M

Mf cC 2001:AS3600for

N

N

M

Mor

u

0

*

*

2

*

1

6.01160

Le = effective length

,

where

r = Ra us o gyrat on o t e co umn's cross sect on

15.06.0/for6.0/5.225.2**

uouoC NNNN

15.06.0/for6.0/5.3/1 ** uouoC NNNN



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A column is considered

braced in a given planee a era s a y o

the structure as a whole

other suitable bracing toresist all lateral forces in

that plane.

in a given plane where lateral

stabilit in that lane is rovidedby the column.



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Find design axial force N* and design maximum moment M*max(factored for load combinations)

design capacities Mu and Nu satisfy

Maxu

*

NNu



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es gn o or o umns

N


CRICOS No. 000213JENB276- Structural Engineering 1


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As : Area of Steel

f : Concrete com ressive stren th

fsy : Yield strength of the steel

Tie Diameter

Spacing and Arrangement



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Behaviour and Load Capacity of Columns

affected by Lateral Reinforcements

Helical

Reinf.

Moderate

Spacing

CloserSpacing

dependent

effects



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u

85.072.0

'

sysgcuo 1

yield stress of steelcross sectional

'1 003.00.1 cf

area of concrete

concrete strength

total area of longitudinal reinforcement



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must be applied to ensure a uniform compressive strain.

dpc= Ag x fc x D/2 + A1 x fsy x d1 +A2 x fsy x d2

Nuo

CRICOS No. 000213Ja university for the worldrealR

ENB276- Structural Engineering 1


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Locate the Plastic Centroid of the following section

Design Parameters

fc = 25 MPa

250100

R10 fitments20 cover

150Plastic

centroid




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-

Calculate 1'

1

003.00.1c

f

925.0

..

85.072.0 1But within the limits Thus 1 = 0.85




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Example 1-(c) d

Nu0Cs2

Cc1

Cc2

Cs3

Cc1 = 0.85x25x600x150 = 1912 kN= = dc1 = 150/2 = 75 mm .

Cs1 = 5x310x500 = 775 kN

Cs2 = 5x310x500 = 775 kN

Cs3 = 4x310x500 = 620 kN

ds1 = 20 + 10 + 20/2 = 40 mm

ds2 = 150 - (20 + 10 + 20/2) = 110 mm

= - =

Total Nu0 = 4720 kN

Taking moments about the top flange and noting that the

internal forces are e uivalent to the external forces ives:

Cc1.dc1 + Cc2.dc2 + Cs1.ds1 + Cs2.ds2 + Cs3.ds3 = Nu0 .dp

1912x75 + 638x200 + 775x40 + 775x110 + 620x210 = 4720 dpHence dp = 109.2 mm

i.e. the plastic centroid is located 109 mm below the top of the Column.

Due to symmetry the plastic centroid is located 300 mm from left edge.




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Ne

N

FailureNu

Loadingline

M = N*e M = NeMu




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Section Capacity Line

(Interaction Diagram)

N

FailureFailure

oes nofail

M = Ne




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0.003 Principal

compression

0.003

yield strain

a ance

failure

M0.003

tensionfailure

> yield strain




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Variation for Columnswhich are under-reinforced in pure bending

N

Principalcom ression

N

failure, uu u0.c

Balancedfailure

uN ,M

u Nu.bal

MPrincipaltension N

u0.t

0.6 0.8

Variation for columns which are under-reinforced in pure bendin




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Exam leExam le 22 aa

Determine the key points A,B,C,D and E on the section capacity line of

the column section shown in figure below for the case where there is

mm o re n orcemen n eac ace.

Bending takes place about the x- axis, fc=40 MPa and fsy=500 MPa




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As cross-section and reinforcement are symmetrical about x axis,

p as c cen ro es on e cen ro a ax s.

d = do = 600-74 mm=526 mm

=As/bd = 2 x 1200 / 400x 600 = 0.01

Calculation of the squash load Nuo (Point A Cl 10.6.2.2)

=1.0-0.003 x 40=0.88

But 0.72


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Calculate the point for ku =1.0 (Point B Decompression Point Cl10.6.2.3)

d = d= 526 mm, =0 and T=0

Calculate the rectangular stress block parameters

= 1.05-0.007 x 40= 0.77 & 2= 0.85

Concrete Compressive region

Cc = 0.85 x 40x 400 x 0.77 x 526 = 5508 x1000 N

sc = 0.003 x (526-74)/526 = 0.00258sc = 200 x1000 x 0.00258 = 515 MPa but >fsy= 500MPa sc = fsy= 500 MPa

en sc= x = x

Nu = 5508 x 103+600 x 103-0 = 6108 kN

Taking moment about plastic centroid,

u - . . -




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Calculating the Balanced Point C (ku=kub)

ub . , u ub, nb .

Cc= 0.85 x 40 x 400 x 0.77 x 287 =3005 x 103kN

sc= 0.003 x (287-74)/287 = 0.00223


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Pure Bending (Point D) , u n

iterating

until Cc+ Csc-T= 0

dn(mm)

ku Cc (kN) sc sc(MPa)

Csc(kN)

T (kN) Cc+Csc-T

(kN)

. .

74 0.140 775 0 0 0 600 +175

dn = 65 mm ku= kuo = 0.124

66 0.125 691 -0.00036 -73 -87 600 -4

Muo = 691 x 103 x (300 0.5 x 0.77 x 66) 87 x 103 x (300-74)

+ 600 x 103 x (526 300) mm = 306 kNm




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a cu a ng pure ens on o n

,

the concrete is cracked and has no tensile capacity:

Nuo.t= (1200 + 1200 ) x 500 N = 1200 kN




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1. Calculate Nu0, Nu0tand locate the plastic centroid

2. Select a range of strain diagrams as defined by theep o e neu ra ax s u 0an en or eac :

3. Calculate the corresponding forces in the concrete (C)

4. Calculate the equivalent axial force and bending

u, u.

5. Calculate and Mu, Nu.

.




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From first principles, derive the Section Capacity Lines

(interaction diagram) for bending about the X-X axis of

600 Design Parameters

300 fc= 32 MPa

cover = 50 mmR10 ties

XX

(fsy= 400 MPa)

'

003.00.1 904.032003.00.11 300

85.072.0 1But within the limits Thus 1 = 0.85




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Exam le 3 b

Pure Axial Compression Load

Nu0 = 0.85fcbD + Asfsy

= 0.85326001000 + 10800400 N= , ,Nu0 = 19,520 kN

= 0.6 (refer AS3600 Table 2.2.2)

Nu0 = 11,712 kN

acts through plastic centroid, dp below top.

The plastic centroid for a symmetrical section corresponds to the

elastic centroid.

Hence d = 500.




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Example 3 (c)

Pure Axial Tension Load

Nu0t = -Asfsy-

= -3,200 kN

0.8= -u ,




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Exam le 3 dCompression Force in Rectangular Stress Block 0.003

Ast4 d4

T

0.85f'c

fs4

Ast3 d3C

T3

fs3d3 Mu

Nu

dc

Ast2 d2 T2

fs1

fs2 d2

Strain

1

Stresses Forces Internal forces

= . 5 c u 1

= 1.05 - 0.007 32 = 0.826

d1 = 1000 - cover - tie dia -/2= 1000 - 50 - 10 - 32/2 = 924mm

C = 0.85326000.826ku924 N

=



, u


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Exam le 3 eTension Force in Typical Layer of Reinforcement - I

Reinforcement may be in tension or compression

It may be either elastic or yielded.

In general the tension force (+ve) in the ith layer Tiis given by

i = st.is.i

= Ast.is.iEs but |Ti| Ast.ifsyApplying similar triangles to strain diag:

s.i = (0.003/kud1)( di-kud1)After noting that Es = 200,000 MPa

i = i- u 1 u 1 st.i u

|Ti| 400Ast.i(the yield force)




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Exam le 3 fTension Force in Typical Layer of Reinforcement - II

Tension Force T

Ast.1 = 3800 = 2,400 mm2

d1

= 924mm

1 - u u ,

T1 = 1,440 (1-ku)/ku kN provided

|T1| 4002400 N = 960 kN

Tension Force T2

Ast.2 = 2800 = 1,600 mm2

d2 = 1000 - 300 = 700

2 - u u ,

T2 = 960 (700-ku924) / (ku924) kN provided

|T2| 4001600 N = 640 kN




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Exam le 3Tension Force in Typical Layer of Reinforcement - III

Tension Force T


|T3

| 640kN (as for T2

above)

Tension Force T4


4




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Exam le 3 h

u v x u

Nu = C- sum of tension forces

= C-(T1 + T2 + T3 + T4)T

4d4 dc

Nu = C-Ti CT

3d3 Mu

Nu Equivalent Bending Moment Mu

T2

d2

Taking moments about the comp. flange:

Mu-Nudp = (T1d1 + T2d2 + T3d3 + T4d4)

-Cdc

T1

d1

Forces Internal forces

Mu = (Tidi) -Ckud1/2 + Nudp




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Exam le 3 i

u = .

C = 12,4561.0

=

T3 = 960(300-1.0924)/(1.0924)

= -,

T1 = 1,440 (1-1.0)/1.0 = 0.0 kN

i.e. zero strain in this layer

|T3| = 648 640 kN

T3 = -640 kN (i.e. yielded in comp.)

T2

= 960(700-1.0924)/(1.0924)

= -233 kN (i.e. comp.) provided

= -233 kN (i.e. comp.) provided

T4 = 1,440(76-1.0924)/(1.0924)

2

T2 = -233 kN

- ,

|T4| = 1,321 960 kN

T4 = -960 kN (i.e. yielded in comp.)




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Exam le 3

N = 12,456 - 0.0 + -233 + -640 + -960

Nu = 14,289 kN

and Mu = (Tidi) -Ckud1/2 + Nudp= (0.00.924 + -2330.700 + -6400.300 + -9600.076)

-12,4560.8261.00.924/2+14,2890.5 kNm

= - - - -. . , . , .

Mu = 1,963 kNm

Hence (Mu,Nu) = (1,963 kNm, 14,289 kN) is one point on the sectioncapacity lines (interaction diagram) for the column.

A similar procedure can be adopted for a range of values of ku

to complete.

This can readily be achieved in a table.

Spreadsheets are useful.




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Section Capacity Lines - I

ku C T1 T2 T3 T4 T1 T2 T3 T4 Nu

assumin steel elastic after considerin ield

Tension 0 960 640 640 960 960 640 640 960 -3200

- - -.

0.2 2491 5760 2676 598 -848 960 640 598 -848 1141

0.3 3737 3360 1464 79 -1045 960 640 79 -960 3018

0.4 4982 2160 858 -181 -1144 960 640 -181 -960 4523

0.5 6228 1440 495 -337 -1203 960 495 -337 -960 6070

Bal 0.6 7473 960 252 -441 -1243 960 252 -441 -960 7662

0.7 8719 617 79 -515 -1271 617 79 -515 -960 9498

0.8 9965 360 -51 -570 -1292 360 -51 -570 -960 11186

0.9 11210 160 -152 -614 -1308 160 -152 -614 -960 12776

1.0 12456 0 -233 -648 -1322 0 -233 -640 -960 14289

Com 16320 -960 -640 -640 -960 -960 -640 -640 -960 19520




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ku dc T1d1 T2d2 T3d3 T4d4 Tid i Cdc Nudp MuTension 0.000 887 448 192 73 1600 0 -1600 0

0.1 0.038 887 448 192 -19 1508 47.5 -369 1091

. . - .

0.3 0.114 887 448 24 -73 1286 427.8 1509 2367

0.4 0.153 887 448 -54 -73 1208 760.5 2262 2709

0.5 0.191 887 346 -101 -73 1059 1188.3 3035 2906

0.6 0.229 887 176 -132 -73 858 1711.2 3831 2978

- -. . .

0.8 0.305 333 -36 -171 -73 53 3042.1 5593 2604

0.9 0.343 148 -106 -184 -73 -216 3850.2 6388 2322

1.0 0.381 0 -163 -192 -73 -428 4753.3 7144 1963

Comp 0.000 -887 -448 -192 -73 -1600 0 1600 0




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-

15000

Mu, Nu

Mu.bal, Nu.bal

0.7

0.8

0.9

.

10000)

0.5

0.4

0.6

5000N(k

.

0.1

0.2

0

-5000

M (kNm) ku as a beam ~ 0.12




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Interaction Dia ram

Mu, NuMub, Nub

hiMu, hiNu

Section Capacity 0.80.9

1.015000

Lines IVApplication of the Reduction Factors 0.6

0.710000

N kN

0.3

.

0.45000

0.1

0.2

0

0 1000 2000 3000

-5000

M kNm ku as a beam ~ 0.1




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Stren th Reduction Factors

(Extracted from AS3600 Table 2.2.2)



Eff ti l thEff ti l th LL


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Effective lengthEffective length LLee

Le=k L

u



01 enb276 column_i single

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