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CEE 142L
Reinforced Concrete Structures
Laboratory
Slender Column Experiment
CEE 142L Slender Column Experiment Spring 2002
The column experiment is designed to examine the influence of slenderness on the
axial load capacity of the column. A brief summary of axial load-bending moment relations for
a rectangular column and length effects such as buckling, moment magnification and
slenderness ratio are presented in the following pages.
P-M Interaction Diagram:
The column specimen is subjected to an eccentric axial load which causes combined
bending and axial load. This eccentric axial load can be replaced with a centered axial load and
a bending moment (M=Pe) couple as shown in Figure 1 where e is the eccentricity.
Axial load – bending moment
interaction diagram represents the strength of
analyzed column against axial load – bending
moment configurations. Evaluation of
interaction diagram (Figure 2) is needed to
predict the behavior of the specimen with or
without slenderness effects. Calculations
required to determine the P-M diagram is
presented below.
P e P
Pe
Figure 1
M
P
(0,-T0)
(Mb,Pb)
(M0,0)
(0,P0)
e1
(Mi,Pi)
Figure 2 – P-M Interaction Diagram
CEE 142L Slender Column Experiment Spring 2002
b
d'
dh
d-x
xAs'
As h/2
a
εy
εc = 0.003 0.85fc´
fs′As′ = Cs
fyAs = T
εs′
Pb
Mb
P0 – Pure Axial Compression:
P f A A f Ac g st y st0 085= ⋅ − + ⋅. ' ( )
Balanced Point: Pb-Mb
P C C Tb c s= + −
where
C x f bc c= ⋅ ⋅ ⋅ ⋅β1 085. '
y
xdxε−
=003.0
C A fs s s= ⋅' '
)(003.0 '' dxxs −⋅=ε ⇒ f E fs s s y' '= ⋅ ≤ε
T A fs y= ⋅
2)()()
2(
'' ddPddCadCM bscb
−⋅+−⋅+−⋅=
Pure Flexure - M0 :
The procedure used to calculate M0 is the same as the ultimate moment calculation for
a beam (see beam experiment).
CEE 142L Slender Column Experiment Spring 2002
Pure Tension - T0 :
T A fst y0 = ⋅
Mi,Pi:
In order to get a more precise P-M diagram, number of intermediate point must be
increased. For this purpose Pi, Mi points should be determined for different eccentricities
(e).
Getting Pi,Mi points is similar to the balanced point calculations. However this time an
iterative procedure is required to get x and strain values. The procedure can be summarized
in a few steps:
• Guess an initial x
• Calculate steel strains using similar triangles (εc = 0.003).
• Using the concrete and steel models, determine the stresses corresponding to the
strains.
• In order to satisfy equilibrium following relations should be satisfied:
P C C Ti c s= + −
and
2)()()
2(
'' ddPddCadCM isci
−⋅+−⋅+−⋅=
where
M P ei i i= ⋅
so moment equilibrium equation becomes:
2
)()()2
('
' ddPddCadCeP iscii−
⋅+−⋅+−⋅=⋅
since e is known we can compare Pi’s obtained from two equilibrium equations.
• If equilibrium is satisfied keep x, otherwise change the guessed value.
CEE 142L Slender Column Experiment Spring 2002
Long Column Effects:
Slenderness effect: A slender column is defined as a column that has significant reduction in
its axial-load capacity due to moments resulting from lateral deflections of the column.
Internal bending moment (Mc=P(e+∆)) is equal to the sum of 1st and 2nd order
moments. M0 = Pe is the 1st order moment and 2nd order moment produced by the axial load
acting through the lateral displacement (∆) of the column.
To develop expressions for lateral displacement due to buckling of slender reinforced
concrete columns, it is necessary to review buckling. Euler (1759), developed expressions for
buckling of axially loaded elastic columns.
Assumptions:
1. Pinned - pinned column
2. Pure axial load (no bending)
3. Column is straight
4. Linear elastic material
a. No residual stresses
Figure 3 - Slenderness Effect
CEE 142L Slender Column Experiment Spring 2002
b. σy > σbuckling , no yielding anywhere on cross-section prior to buckling.
5. Column buckles about principal axis (Imin , rmin) and does not twist.
for small strains:
d vdx
MEI
2
2 ≈
since M P v x= − ⋅ ( ) , equation becomes
d vdx
P v xEI
2
2 =− ⋅ ( )
Differential equation:
d vdx
PEI
v2
2 0+ ⋅ = λ2 = P EI
d vdx
v2
22 0+ ⋅ =λ
solution is: v x A x B x( ) sin( ) cos( )= ⋅ + ⋅λ λ
using boundary conditions:
( ) ( ) 010sin)0( =⋅+⋅== BAxv
B = 0
v x L( )= = 0 ⇒ ( ) 0sin)( =⋅== LALxv λ
P P P
M
P
P
y, v
x
y, v
x
Pinned-Pinned Column
Column Bends (Bows) Out
Moment Due to Lateral Deflection
M = -Pv(x)
v(x)
CEE 142L Slender Column Experiment Spring 2002
A = 0 is trivial solution v(x)=0 for all x
( ) 0sin0 =⇒≠ LA λ
therefore λ π π π πL n= , , ,...2 3
λ πL n= n=1 to ∞
but λ2 = P EI and λ π= n L
nL
PEI
2 2
2
π= ⇒ P n
LEI= ⋅
2 2
2
π ⇒ P EILcr =⋅π 2
2 (n=1)
Comments on Buckling:
• Buckling controlled by moment of inertia for weak axis (Imin)
• As L increases, the load carrying capacity (Pcr) is reduced.
• If restrain column at mid height, it can carry 4 times as much load.
Pcr
Pcr
n=14Pcr
4Pcr
n=29Pcr
9Pcr
n=316Pcr
16Pcr
n=4
Figure 4 - Buckling Modes
Pcr
Pinned-Pinnedk=1
Pcr
Fixed-Pink=0.7
Fixed-Fixedk=0.5
Pcr
Fixed-Freek=2.0
L
0.7L
0.5L
Figure 5 - Columns With Different End Restraints
CEE 142L Slender Column Experiment Spring 2002
Moment Magnifier For Symmetric Loaded Columns:
As mentioned in the previous pages, lateral deflection of the slender column creates
internal second order moments besides the first order bending moment which is equal to
ePM ⋅=0 . As a result total internal bending moment becomes:
M=M0+P(∆0+∆a)
It is possible to express internal bending moment, in terms of first order moments. For
this purpose lateral deflections ∆0 and ∆a must be evaluated. For a pin-ended column 1st and 2nd
order deflection can be calculated by using moment-area method. Bending moment diagrams
of the column is given on Figure 6.
EILMLL
EIM
⋅⋅
=⋅
⋅=∆
842
20
20
0 (1st order deflection)
( )aa
a EILPLL
EIP
∆+∆⋅
⋅
⋅=⋅⋅
⋅⋅
∆+∆⋅=∆ 02
20 2
22
2)(
πππ (2nd order deflection)
M
M
P
P
P M
∆0+∆a ∆0 ∆a
M=M0+P(∆0+∆a)
M0
M0
L/4
M0
L/π
P(∆0+∆a)
Figure 6 - Bending Moment Diagrams
b) Total a) First order
CEE 142L Slender Column Experiment Spring 2002
using the equation for Euler Buckling Load:
( )2
2
LkEIPE ⋅
⋅=
π where k=1 for pin-ended column
2nd order deflection can be modified as:
( )α
α−
⋅∆=−
⋅∆=∆+∆⋅=∆11 000
E
Ea
Ea PP
PPPP where α = P PE
so total lateral displacement at midheight can be expressed as:
atotal ∆+∆=∆ 0
−⋅∆=∆
α11
0total
The term in brackets is considered as magnifier of the first order displacement. The
moment is calculated from the displacement as:
( ) totalac PMPMM ∆⋅+=∆+∆⋅+= 000
M M Pc = + ⋅ ⋅−0 01
1∆
α
P PP
P EILE
E= ⋅ = ⋅⋅α π 2
2
απα
−⋅
⋅
⋅⋅
⋅⋅+=
11
8
20
2
2
0 EILM
LEIMM c
−
⋅+⋅=α
απ18
12
0MM c
−
⋅+−⋅=
ααπα
181 2
0MM c
CEE 142L Slender Column Experiment Spring 2002
−⋅+
⋅=α
α1
23.010MM c
Here 0.23 factor, depends on the shape of the moment diagram. Since 0.23α tends to be
very small, ACI 318 omits 1+0.23α term. So the equation becomes:
M Mc = ⋅−01
1 α
where 11−α
is the moment magnifier
ACI Code – Moment Magnifier Approximate Method:
10-12 Magnified Moments – Nonsway Frames
Moment magnifier equations are derived for columns with pinned-pinned ends.
Because of this reason the length of the column should be multiplied with effective length
factor. This factor depends on the stiffness of columns and beams framing into the ends of the
column. Jackson and Moreland alignment charts presented in ACI 318 S10.12 (Figure 7) can
be used to determine k.
10.12.2 Slenderness effects can be neglected (k=1) if
( )211234 MMrlk u ⋅−≤
⋅
k lr
u⋅< 40
M1/M2 is positive for single curvature and negative for double curvature.
M1 and M2 are the end moments of the column where M2 > M1.
lu is the unsupported length (clean distance between floors, i.e. pin to pin center
line).
CEE 142L Slender Column Experiment Spring 2002
r is the radius gyration of cross section of the column and it is given in 10.11.2
as 0.30h for rectangular and 0.25D for circular columns.
10.12.3 Design moment is calculated by magnifying the largest end moment by δns:
M Mc ns= ⋅δ 2 where M2 is the largest end moment
δ nsm
u
c
CP
P
=−
⋅
≥1
0 75
10
.
.
Here C MMm = + ≥0 6 0 4 0 41
2
. . . accounts for variation in moment diagram.
and as derived in preceding section, Pc is the Euler buckling equation which is
Figure 7 - Jackson and Moreland Alignment Chart
CEE 142L Slender Column Experiment Spring 2002
given as:
P EIk lc
u
=⋅⋅
π 2
2( )
The principal difficulty with the magnifier method is that it requires a value for EI. The
code provides an analytical expression when more precise values are not available.
ACI code equations are:
(10-12) EIE I E I
dc g s se=
⋅ ⋅ + ⋅
+ ⋅0 2
1.
β
Ise = moment of inertia of steel about centroidal axis (in4)
I r A dsei
bi i=⋅
+ ⋅∑∑ π 42
4
Ig = gross concrete section
E fc c= 57000 ' psi
Es = 29000 psi
β dsustained
u
PP
=,max
(10-13) EIE Ic g
d
=+
0 41.
β
Equation 10-12 is derived for small eccentricity and high axial load where slenderness
is most important. 10-13 is a simplified version of 10-12, and thus it is less accurate.
Creep is the deformation under stress in time excluding instantaneous deformation and
it is handled with βd coefficient. The effects of creep are difficult to evaluate. Qualitatively it is
known that creep increases magnification factor.
di Abi
CEE 142L Slender Column Experiment Spring 2002
Here, βd is the proportion of the total design load that is constrained so as to contribute
to time dependent deformations – usually the ratio of factored dead load to total factored loads.
(ratio of maximum design dead load moment to maximum design total load moment, 0.0 < βd
< 1.0)
Another way to calculate I C Ise t g= ⋅ ⋅ ⋅ρ γ 2 is presented in MacGregor. Constant C is
different for each column and it is given in Table 12-1 (MacGregor). Here γ is the ratio of
distance between bars to outside dimension of column.
Example calculations for specimen C4 is given in Appendix A.
CEE 142L Slender Column Experiment Spring 2002
Dial Gage
LVDT LVDT
hinge support
Hydraulic Jack
Figure 8 - Test Setup
Specimen Properties:
In choosing the concrete beam-column section, the main motive is to demonstrate the
moment magnifier concept. The starting point in the design was to make the slenderness ratio,
kl/r=70.
The column to be tested is 119 in. long and attached to the clevis such that it will have
an effective length of 119 in. (k=1). The area of the column is 36 in2 which makes the column
have slenderness ratio of 59.4 which is large enough to demonstrate the moment magnifier
concept.
The section was designed with a design compressive strength of 3500 psi and 4-#4
grade 40 reinforcing bars. However for analysis purposes, concrete compressive and steel
tensile strength test results must be used besides design values.
Test Setup and Instrumentation:
Column specimen is subjected to axial load with an eccentricity of 1” with the help of a
hydraulic jack. The jack has a stroke of 6” and controlled by a remote switch. Clockwise
rotation of the switch button extends the jack whereas counterclockwise rotation contracts the
hydraulic jack. Test setup is presented on Figure 8.
CEE 142L Slender Column Experiment Spring 2002
Specimen is instrumented with two LVDT’s, one pressure cell and one dial gage.
Displacement of hydraulic jack and midheight lateral deformation is monitored with LVDT’s.
The dial gage is used to verify the lateral deformation readings obtained from LVDT. Applied
axial load is measured with the pressure cell. Measurements taken during the test are recorded
in applicable scaled units (inches, kips).
It should be noted that the total change in the column length (λ) is due to the bowing of
column (δ) and axial shortening (∆) because of the axial load. So data obtained from the LVDT
which is measuring the total column shortening must be modified in order to get the length
change due to bowing (buckling).
Figure 9 - Axial Shortening of Column
Since axial LVDT reads the total displacement:
δλ +∆=
actual axial deformation due to buckling can be calculated by:
∆−= λδ
EALP
⋅⋅
−= λδ
L
δ
∆
a
CEE 142L Slender Column Experiment Spring 2002
From Timoshenko and Gere, for an assumed deflected shape y a xl
= ⋅sin π , the shortening due
to buckling is:
dxdxdyl
⋅
⋅= ∫
2
021δ
where,
δ = change in length of the beam-column due to bowing action
a = midspan (lateral) deflection
y a xl
= ⋅sin π
⋅
=lx
la
dxdy ππ cos
dxlx
lal
⋅
⋅
⋅= ∫2
0
cos21 ππδ
dxxll
a l
⋅
⋅= ∫
0
22
22
cos21 ππδ
( )l
axlxl
a0
2
22
2sin42
121
+⋅=
ππδ
la 422πδ =
so lateral displacement can be formulated in terms of axial deflection and column length as:
δπ
⋅= la 2
CEE 142L Slender Column Experiment Spring 2002
3.3750"
1.3125"
1.3125"
6.0000"
3.21"
0.85fc’
fyAs εy
εc=0.003
εs’ fs’As’
Pb
Mb
Example Solution:
Specimen C4
Section and Material Properties:
Ag = 36 in2
As = 2x0.2 = 0.4 in2
As’= 0.4 in2
fy = 40 ksi
fc’= 3500 psi
Es = 29 000 ksi
Ec= 57000(fc’)1/2 = 3372 ksi
P-M Diagram:
• [ ] ySSSSgc fAAAAAfP ⋅+++−⋅⋅= )'()'('85.00
ksiinininksiP 408.0)8.036(5.385.0 2220 ⋅+−⋅⋅=
72.1360 =P kips
• Mb-Pb
21.3003.0
=⇒−
= xxdx
yε in
3'' 1077.1)(003.0 −⋅=−⋅= dxxsε > εy compression steel yielded
3.3750"
1.3125"
1.3125"
6.0000"
4#4
CEE 142L Slender Column Experiment Spring 2002
since εs’< 0.01, fs’ = 40 ksi
TCCP scb −+=
yssscb fAfAfbxP ⋅−⋅+⋅⋅⋅⋅= '''85.085.0
ksiinksiinksiPb 402.02402.025.385.0"6"21.385.0 22 ⋅⋅−⋅⋅+⋅⋅⋅⋅=
70.48=bP kips
2)()()
2(
'' ddPddCadCM bscb
−⋅+−⋅+−⋅=
2)"31.1"69.4(70.48)"31.1"69.4(16)
2"21.385.0"69.4(70.48 −
⋅−−⋅+⋅
−⋅= kipskipskipsMb
74.133=bM in.kips
• M0
0=−+ TCC cs
xksixbfxC wcc ⋅=⋅⋅⋅⋅=⋅⋅⋅⋅= 17.15"65.385.085.0'85.085.0
)31.1(003.0290002.02'' 2 inxx
ksiinfAC sss −⋅⋅⋅⋅=⋅=
16=T kips
x (in) εs fs (ksi) T (kips) εs' fs' (ksi) Cs (kips) Cc (kips) Cc+Cs-T M0 (in.kips) 1.22 0.00850 40.00 16 -0.00022 -6.38 -2.55 18.55 0.00 68.71
• 3240)2.04()'( 2 =⋅⋅=⋅+= ksiinfAAT yss kips (pure tension, M=0)
• P-M values are calculated for various eccentricities. M = Pe
Compression steel in tension
CEE 142L Slender Column Experiment Spring 2002
e (in) x (in) εs fs (ksi) T (kips) εs' fs' (ksi) Cs (kips) Cc (kips) P (k) M (in.kips) 0.5 5.96 -0.00064 -18.52 -7.41 0.00234 40.00 16.00 90.35 113.76 56.88 1.0 4.92 -0.00014 -4.04 -1.62 0.00220 40.00 16.00 74.59 92.20 92.20 1.5 4.16 0.00038 10.97 4.39 0.00205 40.00 16.00 63.16 74.77 112.15 2.0 3.66 0.00084 24.35 9.74 0.00192 40.00 16.00 55.57 61.82 123.65 3.0 2.89 0.00186 40 16 0.00163 40.00 16.00 43.91 43.91 131.72 4.0 2.11 0.00367 40 16 0.00113 32.83 13.13 31.98 29.11 116.45 5.0 1.80 0.00480 40 16 0.00082 23.66 9.46 27.35 20.81 104.07 8.0 1.50 0.00640 40 16 0.00037 10.70 4.28 22.71 10.99 87.90
P-M diagram is given on following pages. The diagram is plotted according to the values
given below
Slenderness effect:
M P eM L EI
0
0 02 8
= ⋅
= ⋅ ⋅∆ / ( )
where
119=L in
39.6976911
3.22900010833722.01
2.0=
+⋅+⋅⋅
=+
⋅+⋅⋅=
d
sesgc IEIEEI
β in4
P (kips) M (in.kips) 136.72 0 pure compression 113.76 56.88 e = 0.5 in 92.20 92.20 e = 1.0 in 74.77 112.15 e = 1.5 in 61.82 123.65 e = 2.0 in 48.70 133.74 balanced 43.91 131.72 e = 3.0 in 29.11 116.45 e = 4.0 in 20.81 104.07 e = 5.0 in 10.99 87.90 e = 8.0 in
0 68.71 pure flexure -32 0 pure tension
CEE 142L Slender Column Experiment Spring 2002
I b h in
I r A d in
g
sei
bi ii
= ⋅ ⋅ = ⋅ ⋅ =
=⋅
+ ⋅ = ⋅⋅
+ ⋅ ⋅ =∑∑=
112
112
6 6 108
44 0 25
44 0 2 16875 2 3
3 3 4
42
42 4
1
4 π π . . . .
or
from MacGregor table 12-1 (pg. 486)
422
22stse in3.26
63750.32.0425.0hA25.0I =⋅
⋅⋅⋅=⋅γ⋅⋅=
1=dβ
α =PPE
63.48119
39.697692
2
2
2
=⋅
=⋅
=ππ
LEIPE kips
∆ ∆a = ⋅−0 1α
α
∆ ∆ ∆total a= + 0
M M Pc total= + ⋅0 ∆
P (kips) M0 (in.kips) ∆0 (in) α ∆a (in) ∆total (in) Mc (in.kips) 0 0 0.00 0.00 0.00 0.00 0.00 5 5 0.13 0.10 0.01 0.14 5.71
10 10 0.25 0.21 0.07 0.32 13.19 15 15 0.38 0.31 0.17 0.55 23.25 20 20 0.51 0.41 0.35 0.86 37.24 25 25 0.63 0.51 0.67 1.31 57.64 30 30 0.76 0.62 1.23 1.99 89.61 35 35 0.89 0.72 2.28 3.17 145.91 40 40 1.01 0.82 4.71 5.72 268.83 45 45 1.14 0.93 14.17 15.31 733.93
CEE 142L Slender Column Experiment Spring 2002
P-M Diagram (Specimen C4)
-60
-40
-20
0
20
40
60
80
100
120
140
160
0 20 40 60 80 100 120 140 160
M (in.kips)
P (k
ips)
Slenderness neglectedSlenderness considered