E3 Column Buckling

CA 3632 Mechanics of Structures and MaterialsExperiment 3 : Column Buckling

Department of Civil and ArchitecturalEngineering (CA)

CA3632 Mechanics of Structures andMaterials

Title: Column Buckling

No.: E3

Name: Au Yuk Kit (53307091)Chan Chun Tong Tony (53238775)Chan Man Ching (53038243)

Group No.: G1

Date of Experiment: 07 OCT 2013

Date Submitted: 28 OCT 2013

Page 1


Laboratory Test: Column buckling A: Determination of critical load of Rods

GeneralA fundamental condition in all problems isthe equilibrium of internal and externalforces. If the system of forces isdisturbed owing to a small displacement ofa body, two principal situations arepossible: either the body will return toits original configuration owing torestoring forces during displacement, orthe body will accelerate farther away from its original state owing to displacing forces. The latter situation istermed instable equilibrium.

The instability of structural members subjected to compressive loading (see Fig. 1(b)) may be regarded as a mode of failure, even though stress may remain elastic, owing to excessive deformation and distortion of the structure. This mode of failure is termed buckling and isprevalent in members for which the transverse dimension is small compared with the overall length.

ObjectiveTo study the buckling of columns under axial load

Page 2


Theory

The critical buckling load c P of a simply supported column is given by where E is the Young’s modulus; L is the length of the column and I is the second moment of area with b and t are the width and height of the column,respectively.

Apparatus1. Vernier calipers and ruler.2.

3. Several steel and aluminum specimens with different slenderness ratios.

4. Loading frame with a set of standard weights.

Page 3


Procedure1. Measure the dimensions of a column, and calculate the

second moment of area.2. Place the column in the position of the loading frame.3. Load the weight hanger with the necessary weights until

the column is buckled (Fig.2).4. The data was recorded.5. Repeat the above process for the other specimens.

Results and calculations:

Experimental results:

The weight of the frame adding to the column is 4110g.

Page 4


Specimen Width/Diameterb (mm)

Heightt (mm)

LengthL (mm)

E (GPa)

Weight added(kg)

Middleconstrain(Yes or No)

Experi-mental Critical loadPc (N)

1. Short circular aluminum column

4.98 570 70 0.9 No 63.3

2. Long circular aluminum column

4.98 967 70 2.5 Yes 88

3. Long rectangular aluminum column

4.62 6.73 970 70 0.47 No 84.9

4. Short rectangular steel column

5.01 6.96 570 210 8.5 No 89.7

5. Long rectangular steel column

4.68 6.69 970 210 2.5 No 25.7

Page 5


Theoretical results:The Dimension and Second Moment of Area of the Specimens

For rectangular section, Ixx = 123bh , Iyy = 12

3hb .For

circular section, Ixx = Iyy = 44r

Specimen Width, b(mm)

Height,h (mm)

Diameter,2r (mm)

Ixx(mm4)

Iyy(mm4)

1 4.98 30.19 30.192 4.98 30.19 30.193 4.62 6.73 117.36 55.304 5.01 6.96 140.76 72.935 4.68 6.69 116.77 57.15

The Theoretical Critical Buckling Load, PcBy using the Euler formula

2

2

LEIPC

For Pc of the specimen 1 (Aluminum short circular), thebuckling occur in the minor axis, Iyy

2

2

LEIPC

= π2 x 70 x 103 x 30.19/ 5702

= 64.20N

For Pc of the specimen 2 (Aluminum long circular), the buckling occur in the minor axis, Iyy

2

2

LEIPC

= π2 x 70 x 103 x 30.19 /483.5 2

Page 6


= 89.22N

For Pc of the specimen 3 (Aluminum long rectangular), thebuckling occur in the minor axis, Iyy

2

2

LEIPC

= π2 x 70 x 103 x 55.3/ 9702

= 40.6N

For Pc of the specimen 4 (Steel short circular)), the buckling occur in the minor axis, Iyy

2

2

LEIPC

= π2 x 210 x 103 x 72.93 / 5702

= 465.23NFor Pc of the specimen 5 (Steel long circular), the buckling occur in the minor axis, Iyy

2

2

LEIPC

= π2 x 210 x 103 x 57.15/ 9702

= 125.89N

Summary: Specimen 1.

Short circular aluminum column

2. Longcircular aluminum column

3. Long rectangular aluminumcolumn

4. Short rectangular steel column

5. Long rectangular steel column

Experimental Critical load

63.3 88 84.9 87.9 25.7

Page 7


Pc (N)Theoretical Critical loadPc (N)

64.20 89.22 40.6 465.23 125.89

Laboratory Test: Column buckling B: Determination of critical load of Long Column and Short Column

GeneralThe load-deflection behavior for an ideal Euler column isillustrated in Fig 1(a). For applied load up to the critical value c P small transverse displacement u can bemaintained under load in a stable-equilibrium state. In

Page 8


the case of a real column, which incorporates some deficiency such as eccentricity of loading, deflection will occur from the moment when load is applied as shown in Fig 1(b). The curve becomes asymptotic to the criticalload at large deflection. The response of the column is almost linear when the ratio P/ Pc is small. According tothe linear relationship between the axial load and the deflection of the column, the critical load Pc can also be obtained.

ObjectiveTo determine the critical load and the relationship between the axial load and the deflection of the column

TheoryThe column is illustrated in Fig. 2, in which the initialmaximum deflection is ao, the value of the deflection at Ddistance y from P is uo, and

Page 9


Apparatus1. Vernier calipers.2. Buckling apparatus with a hydraulic jack.

Page 10


5. Two aluminum box section columns with different lengths.

6. Displacement transducer.

7. Force transducer.

Procedure1. Measure the dimensions of the column.2. The test setup of the column is shown in Fig. 4. The

force and displacement transducers are installed with the hydraulic jack and in the middle of the column, respectively.

Page 11


3. Apply axial load at one end of the column. For each load increment (0.2kN), record the load-displacement data in Form. It should avoid the occurrence of plasticdeformation in the long column.

4. Repeat the above steps for the short column, and the

column is loaded to failure condition.

Results and calculations:

Dimensions of the specimen:Short Column (mm) Long Column (mm)

Width b1 (mm) 25.2 25.23

Page 12


b2(mm) 19.46 20.03Length L (mm) 1800 2664Height h1 (mm) 50 50.65h2 (mm) 44.42 45.09tb (mm) 2.79 2.78th (mm) 2.87 2.6

Values of I and E of the specimenBecause the buckling is along the yy-axis, so, the valueof I is about the yy-axis. We use Iyy.

For the short column:Iyy = b1

3h1/12 – b23h2/12

= 25.23(50)/12 – 19.463(44.42)/12 = 39400mm4

For the long column:Iyy = b1

3 (h1) / 12 – b23 (h2) / 12

= 25.233(50.65)/12 – 20.033(45.09)/12 = 37592.08mm4

Short column Long columnIyy (mm4) 39400 37592.08E (GPa) 70 70Theoretical critical load of the specimenTheoretical Critical Buckling Load Pc =π2 EI / L2

For the short column:Pc = π2 (70) (39400)/18002

=8.4kN

For the long column:Pc = π2 (70) (37592.08)/26642) = 3.66kN

Short Column Long ColumnTheoreticalCritical

8.4kN 3.66N

Page 13


Buckling LoadsPc

P value and u’ value of the specimenLong column

Load P(kN)

Deflection u'(mm)

u' / P(mm/kN)

1 0.2 0.090 0.52 0.4 0.218 0.53 0.6 0.342 0.64 0.8 0.478 0.65 1.0 0.656 0.76 1.2 0.875 0.77 1.4 1.111 0.88 1.6 1.335 0.89 1.8 1.607 0.910 2.0 1.993 1.011 2.2 2.480 1.112 2.4 3.196 1.313 2.6 4.140 1.614 2.8 5.258 1.915 3.0 7.485 2.516 3.2 11.646 3.6

Short column

LoadP

(kN)

Deflection u'(mm)

u' / P(mm/kN)

Load P(kN)

Deflection u'(mm)

u' / P(mm/kN)

1 0.6 0.124 0.2 21 4.6 1.92 0.42 0.8 0.152 0.2 22 5.0 2.38 0.53 1.0 0.191 0.2 23 5.4 2.914 0.54 1.2 0.238 0.2 24 5.6 3.494 0.65 1.4 0.289 0.2 25 5.8 3.824 0.7

Page 14


6 1.6 0.328 0.2 26 6.2 5.234 0.87 1.8 0.391 0.2 27 6.6 7.023 1.18 2.0 0.441 0.2 28 7.0 9.32 1.39 2.2 0.504 0.2 29 7.4 15.825 2.110 2.4 0.557 0.211 2.6 0.647 0.212 2.8 0.737 0.313 3.0 0.78 0.314 3.2 0.881 0.315 3.4 0.961 0.316 3.6 1.137 0.317 3.8 1.335 0.418 4.0 1.459 0.419 4.2 1.568 0.420 4.4 1.799 0.4

0.0 0.5 1.0 1.5 2.0 2.5 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 16.000 18.000

f(x) = 7.99373755437816 x − 1.38423830684382

f(x) = NaN x + NaNu' against u'/P

Long ColumnLinear (Long Column)Short ColumnLinear (Short Column)

u'/P

u'

For short column:From the equation, (Pc/P)u’-u’=ao => u’=Pc(u’/P)- ao

From the graph, y-axis is u’ and x-axis is u’/P.

Page 15


Therefore, the slope is Pc.The slope of the graph is 7.99 ∴The Pc of the short column is 8.4 kN

For long column:From the equation, (Pc/P)u’-u’=ao => u’=Pc(u’/P)- ao

From the graph, y-axis is u’ and x-axis is u’/P.Therefore, the slope is Pc.The slope of the graph is 3.68∴The Pc of the long column is 3.66 kN

Summary:Long Column Short Column

TheoreticalCritical Buckling

Loads Pc,

3.66 kN 8.4 kN

ExperimentalCritical Buckling

Loads Pc

3.68 kN 7.99 kN

Discussion

Laboratory Test: Column buckling A: Determination of critical load of Rods The test is going to determine the critical buckling loadof rods come with different shape and material, aluminum and steel. Each shape has different value of critical buckling load.

Page 16


Summary: Specimen 1.

Shortcircula

raluminu

mcolumn

2. Longcircula

raluminu

mcolumn

3. Longrectangu

laraluminumcolumn

4.Shortrectangularsteelcolumn

5.Longrectangularsteelcolumn

Experimental

CriticalloadPc (N)

63.3 88 84.9 87.9 25.7

Theoretical

CriticalloadPc (N)

64.20 89.22 40.6 465.23 125.89

Above is a table summarizing the result of the experiment. It indicates that the difference between experimental and theoretical result of first two specimens are minor but big difference of last three specimens remain high. This is because errors have been happened in the experiment.

There are some errors may encounter during the experiment:

1. The materials have been deterioratedThe materials have not been replaced and use for more time so that it deteriorates so that the materials properties of them vary.

2. The columns have been permanently deformed

Page 17


Due to repeatly use, the columns have been deform in some lateral displacement because the columns have beenexceeded plastic limit and the column is not going to restore to its original position.

3. The loadings are not acting on top of the columnThe loadings are put on the crane of the device so thatthere is some distance between column and the crane, moment induced due to eccentricity. As the result, thefull magnitude of loadings are not completely transfer to the rods and the result may not exactly 100% the actual values.

4. The centroid of the horizontal transferring rod is not the mid- span of itThe centroid of the rod is taken as the mid-span of it,but, actually, it is not, because there are lots of hollow circle exist in it so that the location of centroid must not be min-span. Therefore, calculation and result may not accurate.

Recommendations

1. Use new rods instead of old batch.2. Replace the hollow circle included transferring rod to

solid rod.3. Modify the device, amend to position of putting

loadings from the crane to the top of the column.4. Use more different types of materials to obtain results

of diversification (e.g.: Brass)

Laboratory Test: Column buckling B: Determination of critical load of Long Column and Short Column

This experiment is to find the critical buckling load of

Page 18


short column and long column by using dial gauge reading and exert pressure to hydraulic jack to control the force.

Summary:Long Column Short Column

TheoreticalCritical Buckling

Loads Pc,

3.66 kN 8.4 kN

ExperimentalCritical Buckling

Loads Pc

3.68 kN 7.99 kN

The summary shows that the experimental result is quite near to theoretical result, which is a successful result.And the relationship between length of the column and thedisplacement. It shows that the length of the column is inversely proportional to the exerted load and less lateral displacement.

0.0 0.5 1.0 1.5 2.0 2.5 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 16.000 18.000

f(x) = 7.99373755437816 x − 1.38423830684382

f(x) = NaN x + NaNu' against u'/P

Long ColumnLinear (Long Column)Short ColumnLinear (Short Column)

u'/P

u'

According to the graph, the slope if positive and

Page 19


illustrate the relationship between displacement and acting loads. And the theory is proved.

By the way, errors are occurred during the test because no perfect performance happens. They are in below.

1. The materials may use for a long time, so it yieldsThe column is made by aluminum so that the shear modulus, such properties are less than steel. So that it is likely to yield if aluminum column is used repeatly. When it yields permanently, initial lateral displacement is happened because of exceeded plastic limit.

2. The reading of dial gauge is not accurateThe dial gauge could read the displacement of the column because the contact point may not exactly touch the column so that the result may also not read accurately. As the result, the difference between experiment and theoretical becomes major and the resultcould not obtained accordingly. The column moves when several loadings are added, so that dial gauge may be reading the lateral displacement obviously. As the result, the result may not as accurate as the experiment.

3. The exerted force form the hydraulic jack may not actually the required values

Force is exerted by means of hydraulic jack, and it is formed by exerting pressure to the jack by manual. So that, the reading of dial gauge always changes because of its sensitivity. Therefore, the result is just guessafter the exerted force has been contact to the required load.

Page 20


4. The dial gauge may not exactly contact the column

Recommendations

1. Use new columns instead of the old column.2. Use new dial gauges with sucker contact so that the

contact surface area is greater than the traditional contact point base

3. Use other types of fore exerting system to stabilize the loading required.

Page 21

E3 Column Buckling

Documents