Stability of Column

7/28/2019 Stability of Column

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HASHEMITE UNIVERSITY

Faculty of EngineeringMechanical Engineering Department

Student Name : T{xw [t| ftx T@bt|

Student Reg. No : 431900

Section No. : 4

Lab. Day : Wednesday

Lab. Date : 4 / 10 / 2006

Experiment # : 2

Experiment Title : Stability of Columns

Submitted to

Instructor: Dr. Ahmed Al-Shyyab

Engineer: Yousef Zakariya

Due Date: 11 / 10 / 2006

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T{xw [t| ftx T@bt| \W 6 GFDLCCCivil Engineering just4just.com/ahmed

:Objective:We want from this experiment to:

1) See & study the behavior of axially loaded columns.2) Determine the critical buckling load.3) To compare the results with Euler's formula.

:Theory:

Figure (1)

Eulers formula:

As we see in Figure (1) the column which has a length L supporting and axial load P,

increased until it reaches a critical value Pcrand the column dont yield, then the column

will fail due to buckling. The critical load is given by Euler formula:

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Where:

E: The modulus of elasticity of the column.

I: The least econ moment of inertia.

L: The length of the column.

K: Factor depends on the end condition of the supporting the

column, where:

k=1 . . . . For( Pinned Pinned ).

k=2. . . . For( Fixed Free ).

k=0.5. . . For( Fixed Fixed ).

k=0.7. . . For( Fixed Pinned ).

- We have in this experiment four cases to study and analysis:

1) Pinned Pinned.2)

Fixed Pinned.

3)Fixed Fixed.4)Fixed Free.

- And we did in the lab two of them ( No. 1 & 2 ).

Figure (2) 2


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:Equipments:o The test device mainly consists of a basic frame, the guide columns and the load cross

bar.

o The basic frame contains the bottom mounting for the rod specimen, consisting of aforce-measuring device for measuring the testing force and an attachment socket,

which can hold different pressure pieces for realizing various storage conditions.

o The height of the load cross bar can be adjusted along the guide columns and it can beclamped in position. This allows rod specimens with different buckling lengths to be

examined.

o The loads cross bar features a load spindle for generating the test force. Using the loadnut, the test force is applied to the rod specimen via guided thrust pieces. An axial

mounting between the load nut and the thrust pieces prevents torsional stresses frombeing applied to the rod specimen.

o Two different thrust pieces are available for the different storage conditions.o The device can be used both vertically as well as horizontally. The device is equipped

with a base foot on one of the guide columns for horizontal set-up. The display

instrument of the force measuring device can be turned 90 for easy readability.

3Figure (3)


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:Data Results and Analysis:Table 1:

Rod material : Steel.

Cross Section dimensions : 4*20 mm.

Length : 650 mm.

Modulus of elasticity E : 210 Gpa.

Cross Section Type : Rectangle.

Ends Condition : fixed-Pinned.

Deflection y (mm) 0.5 1.0 1.5 2.0 2.5Load P (N) 450 550 625 675 725P/y (N/mm) 900 550 416.67 337.5 290

Sample of Calculation :y=0.5 mm , P=450 N P/y = 450/0.5 = 900 N/mm

Fixed - Pinned

y = -0.4292x + 819.12

0

100

200

300

400

500

600

700

800

0 200 400 600 800 1000

( P / L ) N/mm

(P

)N

Figure (4)

I = b h3 / 12 = 20 * 43 / 12 = 106.67 mm4.

From the figure Pcr (@ x = 0 ) = -0.4292 * 0 + 819.12 = 819.12 N

Theoretically: Pcr = 2 * 210 * 103 * 106.67 / (0.7 * 650 )2 = 1067.92 N

4

% Error: % error = ( 1067.92 819.12 ) / 1067.92 * 100 % = 23.298 %


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T{xw [t| ftx T@bt| \W 6 GFDLCCCivil Engineering just4just.com/ahmedTable 2:

Rod material : Steel.

Cross Section dimensions : 4*20mm.Length : 700 mm.

Modulus of elasticity E : 210Gpa.

CrossSection Type : Rectangle.

Ends Condition : Pinned-Pinned.

Deflection y (mm) 0.5 1.0 1.5 2.0 2.5Load P (N) 175 240 260 300 325

P/y (N/mm) 350 240 173.33 150 130

Sample of Calculation :y=0.5 mm , P=175 N P/y = 175/0.5 = 350 N/mm

Pinned - Pinnedy = -0.6314x + 391.74

0

50

100

150

200

250

300

350

0 100 200 300 400

( P / L ) N/mm

(P

)N

Figure (5)

I = b h3 / 12 = 20 * 43 / 12 = 106.67 mm4.

From the figure Pcr (@ x = 0 ) = -0.6314 * 0 + 391.74 = 391.74 N

Theoretically: Pcr = 2 * 210 * 103 * 106.67 / (700 )2 = 451.196 N

% Error: % error = ( 451.196 391.74 ) / 451.196 * 100 % = 13.177 %

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Error sources:There were some sources of errors in the experiment:

1- Reading error.2- Machine error.3- There was a hardness in applying 100% of an axially load on both sides for the beam.4- The vibrations of the bench.5- We put a load a the right side of the beam to have a control on the direction of the

deflection which will make an error in the reading

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Stability of Column