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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


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).


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.


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


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)


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


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


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


−⋅+

⋅=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).


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


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


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.


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.


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


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


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


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


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


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


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

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