Engr 244 lab 6 - engr 244 lab 6 academic year 2016/2017 PDF

Title Engr 244 lab 6 - engr 244 lab 6 academic year 2016/2017
Course Mechanics of Materials
Institution Concordia University
Pages 5
File Size 169.1 KB
File Type PDF
Total Downloads 76
Total Views 132

Summary

engr 244 lab 6 academic year 2016/2017...


Description

Objective:

The purpose of this experiment is to determine the load at which at which different column buckles. Introduction: Columns are an imminent component of all structures, for just about every branch of civil engineering, mechanical engineering, but especially for structural purposes. The application of load that increases on a support column causes the material bending at a point called the critical load. The load is called critical buckling load. The magnitude of the critical buckling load depends on how the column is supported, the shape of it cross-section, the column length and the applied force location relative to the cross-section’s centroid. M ⅆ2 y = . Say the load is centric, 2 EI ⅆx and is acting along the cross section’s centroidal axis, the ends of the column are pinned and the column simply experiences deformation that is elastic. From this, we obtain the equation of the critical load also known as Euler equation

The equation for deformation in an elastic member is

Pcr =

π 2 EI L2

and the critical stress is σ

cr

= Pcr / A.

Procedure: 1- Place the loading device into compression mode. 2- Using the Vernier caliper, record the diameter (outer and inner if hollow), the length making sure to include the end fittings 3- With each specimen, we perform the following supports: a- Both ends fixed b- One end fixed one end pinned c- Both ends pinned 4- Slowly and progressively increase the compressive load for every specimen and observe the bending to occur. Record the load at that point which the maximum load. 5- For the 125mm and 75mm, we perform the test for the case of both ends being pinned. 6- Sketch the mode of failure of the specimen tested. Analysis of results: 1. - Cross sectional area: a- Hollow

A=

π (douter2-dinner2) = 4

π ((6.36)2-(4.63)2) = 14.93 mm2 4

b- Solid π 2 π d = (6.45)2 = 32.7 mm2 4 4 Moment of inertia A=

-

π 4 r 4 a- Hollow

I=

π 4

Ihollow = Iouter – Iinner =

(router4-rinner4) =

π ((3.18)4-(2.32)4) = 57.56 mm4 4

b- Solid π 4 4 r = Radius of gyration a- Hollow

π 4 4 4 (3.23) = 85.5 mm

Isolid =

-



rhollow =

I A



=

57.6 14.93

= 1.96 mm

b- Solid rsolid=



I A

=



85.5 32.7

= 1.62 mm

2- critical load hollow 75mm pin-pin column Pcr =

π 2 EI L2

Pcr = π2 (70 x 109) x 5.775766 x 10-11 / (0.075)2 Pcr = 7093.89 N 3- critical stress (experimental value) Hollow 75mm pin-pin column σ=P/A σ = 7093.89 / 1.493 x 10-5 = 475.1 MPa slenderness ratio, λ= KL/r hollow 75 mm pin-pin column k = 1, L = 0.075m and r = 0.00196 m

λ = 38.26 4- theoretical stress Hollow 75 mm pin-pin column σ = π2E/λ2 cr σ=

π

2

(70x109) / (38.26)2

σ = 471.96 MPa Report Type of colum n

Hollow

solid

Type of column

L End (mm connectio ) n 75 PinPin 125 PinPin 225 PinPin 225 PinFix

K

Lef mm

75 4.63 125 4.25 225 4.46 159.075 4.58

225 225 225

FixFix PinPin PinFix

225

FixFix

1 1 1 0.70 7 0.5 1 0.70 7 0.5

λ

σ

all

din(mm) dout(mm)

112.5 225 159.075

4.37

112.5

MPa

Theo Pcr N

Hollow

38.3 65.4 115.9 81.2 58.6

105.8 82.3 26.1 53.3 88.1

7069.88 2865.48 824.116 808.204 844.871

Solid

139.7 98.2 69.8

18.0 36.4 72.04

1159.4 1195.8 1159.4

A(mm2) I(mm4 )

r(mm)

6.36 6.37 6.35 6.37

14.93 17.68 16.05 15.40

57.56 1.96 64.81 1.91 60.39 1.94 59.22 1.96

6.35 6.45 6.50

16.67 32.67 33.18

61.91 84.96 87.62

1.92 1.61 1.62

6.45

32.67

84.96

1.61

σ cr MPa theo 470.98 161.53 51.431 104.78 201.19 35.400 71.643 141.80

Experi Pcr N 3903 3629 1774 2036 1876

σ cr MPa experi 261.38 205.23 110.55 132.26 112.53

2362 2094 3198

72.289 63.104 97.8774

stress vs slederness graph 160 140 stress (mpa)

120 100 80 60 40 20 0 60

70

80

90

100

110

120

130

140

150

slenderness ratio

Series 1: allowable stress Series 2: experimental stress Series 3: theoretical stress

Discussion 1. From the value of the different stress calculated it can be seen that the error between the theoretical value and the experimental value of the 225 mm pin-pin is greater than 100% which shows that there is a huge mistake. The 225mm fix-fix and the 225mm pin-fix has an error 31.15% and 12.06% respectively. The main explanation to these errors are because of the the method of viewing the bending onset. Another reason for these errors maybe the displacement of the specimen of the loading machine was not properly calibrated. These errors can lead to inaccurate measurement of the maximum force 2. Both theoretical and experimental stresses are greater than the allowable stress of the aluminium obtained from the aluminium association. Moreover, the allowable stress take into consideration small discrepancies that Euler’s equation does not consider in real column and additionally it contains a safety factor. 3. The following are factors that may affect the buckling strength of real columns. a. The type of material used, b. The shape of the cross-section. c. The tube length. d. The support type. e. The location of the load relative to the centroid axis of the column. In the real world, we deal with forces in three dimensional space not two dimensional space. Also, temperature would be a factor and the melting point of the material in case an explosion where to occur like the world trade center in 2001. The ground in which the columns are supported or the other components of the structure will also play a role in the buckling strength. The age of material and

conditions that the columns experience will play a role, such as vibrations along with loading and unloading.

Pcr 

 2 EI L2e and we are dealing with only aluminum

4. Since buckling load is calculated by: samples in this laboratory experiment, the two factors that affect the buckling load are the second moment of inertia (I) and the effective length. Assuming that they are both the same length, then the second moment of inertia would be the determining factor in the buckling load.

1  (c )4 I Ibig  Ismall 4 I  If we have a solid circular column then , but if it is hollow . The larger I would generate a larger bucking load. Therefore a solid circular column results in a larger buckling strength....


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