The document reports on an experiment to determine the critical compression load of an aluminum alloy bar. The bar was 300mm long with a 25mm diameter and 1mm thickness. It was tested on a machine that applied increasing loads until failure. The bar failed at a maximum load of 2404N. Calculations determined the theoretical critical load to be 42936N and theoretical critical stress to be 327MPa, which is close to the experimental value of 318MPa with a 2.83% error.

1. Islam Mansour | OPK-A Aircraft Structure I | November 12, 2015
Report I
CRTICAL COMPERSSION LOAD FOR ALUIMUALN BAR
SUBMITTED TO: PROF. ING. ANTONÍN PÍŠTĚK.

2. PAGE 1
Contents
Defenation of the sample .............................................................................................................................2
Experminent ..................................................................................................................................................3
Resutls from test............................................................................................................................................5
Thertoical calcuatlion .................................................................................................................................. 6
Figure 1: The length of specimen .................................................................................................................2
Figure 2: The diameter of specimen ............................................................................................................2
Figure 3: The specimen before the test .......................................................................................................3
Figure 4: The testing machine......................................................................................................................3
Figure 5: The testing machine..................................................................................................................... 4
Figure 6: The Load vs. Displacement ..........................................................................................................5
Figure 7: The specimen failure.....................................................................................................................5

3. PAGE 2
Defenation of the sample
The sample is a hollow cylinder tube of Duralumin alloy which has charities as followings:
L : Length = 300 mm
t : Thickness = 1 mm
D : Diameter = 25 mm
Figure 1: The length of specimen
Figure 2: The diameter of specimen
From the table 1.5 for this section:
A: Area = 75.4 mm2
J: Moment of inertia = 5438 mm4
i: Radius of gyration = 8.49 mm
Rm: = 440 MPa
RP02: = 280 MPa
E: = 72000 MPa

4. PAGE 3
Experminent
The test machine had been used to load the specimen and calculate the critical force. It is
shown that the specimen has free end.
Figure 3: The specimen before the test
Figure 4: The testing machine

5. PAGE 4
Figure 5: The testing machine

6. PAGE 5
Resutls from test
It is concluded from the test that the maximum critical force was FCR = 2404 N at
Displacement D = 2.8168 mm, where the critical force has been taken as the maximum applied
force to the specimen.
𝐹𝐶𝑅 = 2404 𝑁, 𝐴 = 75.4 𝑚𝑚2
𝜎𝑐𝑟(𝑒𝑥𝑝) =
𝐹𝑐𝑟(𝑒𝑥𝑝)
𝐴
=
24040
75.4
= 318 𝑀𝑃𝑎
Figure 6: The Load vs. Displacement
Figure 7: The specimen failure

7. PAGE 6
Thertoical calcuatlion
It is shown that specimen has a free end which lead to C = 1
𝐹𝐶𝑅(𝑡ℎ𝑒𝑟𝑜) = 𝐶 𝜋2
𝐸 𝐽
𝑙2
= 1 ∗ 𝜋2
∗
72000 ∗ 5438
3002
= 42936 𝑁 > 𝐹𝐶𝑅(exp)
𝜆 𝑚 = √𝑐
𝜋2 𝐸
𝑅 𝑝𝑜2
= √1
𝜋2 ∗ 72000
280
= 50.3
𝜆 =
𝑙
𝑖
=
300
8.49
= 35.3 < 𝜆 𝑚
𝜎𝑘𝑟(𝑡ℎ𝑒𝑟𝑜) = 𝐶 𝜋2
𝐸
𝜆2
= 1 ∗ 𝜋2
∗
72000
35.52
= 570 𝑀𝑃𝑎
𝜎𝑐𝑟(𝑡ℎ𝑒𝑟𝑜) = 𝑅 𝑒 +
𝑅 𝑚 − 𝑅 𝑒
𝜆 𝑚
(𝜆 𝑚 − 𝜆)
𝜎𝑐𝑟(𝑡ℎ𝑒𝑟𝑜) = 280 +
440 − 280
50.3
(50.3 − 35.3) = 327 𝑀𝑃𝑎
It is conclude that the theoretical value is near to the experimental one with 𝑒𝑟𝑟𝑜𝑟 = 2.83 %
𝜎𝑐𝑟(𝑡ℎ𝑒𝑟𝑜) = 327 𝑀𝑃𝑎 ≅ 𝜎𝑐𝑟(𝑒𝑥𝑝) = 318 𝑀𝑃𝑎