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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
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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.
Queensland University of Technology
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CRICOS No. 000213Ja university for the worldrealR ENB276- Structural Engineering 1
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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
Queensland University of Technology
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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
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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
=
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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 = 960 (300-ku924) / (ku924) kN provided
|T3
| 640kN (as for T2
above)
Tension Force T4
T4 = 1440 (76-ku924) / (ku924) kN provided
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)
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Eff ti l thEff ti l th LL
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Effective lengthEffective length LLee
Le=k L
u
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