Strain Gauge

ENGD2007- Strength of Materials

Strain gauge Rosette Objective (1) To become familiar with the use of a strain gauge rosette and hence determine the magnitude and direction of the principal strains. (2) To become familiar with the construction of Mohr’s strain circle, from which to determine the magnitude and direction of the principal strains. (3) To determine the elastic constants, E, v, for a given steel specimen. Apparatus   

Oslen testing machine. Tensile specimen with 45o rosette oriented at an angle (15o) to the bar axis. Tinsley 10 channel strain gauge conditioning equipment.

Figure 2: Diagram of tensile specimen with 45 o rosette

Figure 1a: Oslen testing machine Tensile specimen with 45o rosette oriented at an angle (15o) to the bar axis)

Data: Cross sectional dimensions of specimen: 50.7 mm x 6 mm Conversion factor: 1 kN = 224.81 lbf

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ENGD2007- Strength of Materials Procedure: (1) Ensure that the Tinsley 10 channel unit is set to to 2.1.

1

4

bridge volts, and the gauge factor set

(2) Zero each gauge with no load using slotted potentiometers. (3) Monitor each gauge in turn, and measure strain gauge reading for increments of load from zero to 4000 lbf in increment of 400 lbf. (4) Plot strain ( 1,

2,

3)

for each gauge against load of the same set of axis’s.

(5) Select a load and pick off three strains from your graph. (6) Using these strains and the gauge arrangement, construct Mohr’s strain circle. Note:

(Eq.(2.21, page 101 of the lecture notes)

Please refer to page 99-100 of the lecture notes for the construction of the Mohr’s strain circle.

(7) From your strain circle, determine (a) The principal strains (b) The direction of the maximum principal strain relative to gauge 1, and check with the gauge alignment on the specimen. (c) E: using the load selected and the bar cross-section, combined with the maximum principal strain. (d) Determine the Poisson’s ratio from the ratio of the principal strains.

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ENGD2007- Strength of Materials Results Load

Strain (x10-6)

Stress (N/m2) On

Gauge B Off Mean

On

Gauge C Off Mean

lbf

N

0

0

0

0

8

4

0

4

2

0

-14

-7

400

1779.29

5849080

36

49

42.5

25

32

28.5

-7

-22

-14.5

800

3558.58

11698159

69

81

75

48

52

50

-14

-30

-22

1200

5337.866

17547226

101

113

107

70

75

72.5

-22

-33

-27.5

1600

7117.155

23396302

133

142

137.5

91

93

92

-28

-36

-32

2000

8896.443

29245375

162

170

166

112

113

112.5

-35

-39

-37

2400

10675.73

35094444

193

199

196

132

132

132

-41

-42

-41.5

2800

12455.02

40943524

222

225

223.5

152

151

151.5

-48

-44

-46

3200

14234.31

46792604

248

250

249

171

168

169.5

-44

-49

-46.5

3600

16013.6

52641683

275

277

276

187

186

186.5

-51

-53

-52

4000

17792.89

58490763

300

300

300

204

204

204

-58

-58

-58

Nizamuddin Patel

On

Gauge A Off Mean

Mohr Circle Results (only for the selected load)

P15219444

Emax (x10-6)

Emin (x10-6)

E (GPa)

v

310

-50

169.81

0.19

ENGD2007- Strength of Materials

Gauge A 350 y = 0.0177x

Strain (x10-6)

300 250 200 150 100 50 0 0

5000

10000

15000

20000

Load (N) Figure 3: The relaitionship between strain and load using mean figures for gauge a

Gauge B 250 y = 0.012x

Strain (x10-6)

200 150 100 50 0 0

5000

10000

15000

20000

Load (N) Figure 4 The relationship between strain and load using mean figures for gauge b

Gauge C 0 0

5000

10000

15000

20000

-10

Strain (x10-6)

-20

-30 -40 -50 -60 -70

y = -0.0036x

Load (N)

Figure 5 The relationship between strain and load using mean figures for gauge c

Nizamuddin Patel

P15219444

ENGD2007- Strength of Materials For 3600 lbf load, A= εx = 276 × 10-6 B = εb = 186.5 × 10-6 C = εy = -52 × 10-6 𝜸𝒙𝒚 = 𝟐𝒆𝒙 − 𝒆𝒙 − 𝒆𝒚 𝛾𝑥𝑦 = (2 × 186.5) − 276 − (−52) 𝛾𝑥𝑦 = (2 × 186.5) − 276 + 54 𝜸𝒙𝒚 = 𝟏𝟒𝟗 × 𝟏𝟎−𝟔 Maximum Shear Stress 𝜺𝒙 − 𝜺𝒚 𝟐 𝜸𝒙𝒚 𝟐 𝜸𝒎𝒂𝒙 = √( ) +( ) 𝟐 𝟐 𝟐 2

𝛾𝑚𝑎𝑥 276 × 10−6 + 52 × 10−6 ) 149 × 10−6 = √( ) +( ) 2 2 2

2

𝛾𝑚𝑎𝑥 = 180.13 × 10−6 2 𝛾𝑚𝑎𝑥 = (180.13 × 10−6 )2 𝜸𝒎𝒂𝒙 = 𝟑𝟔𝟎. 𝟐𝟔 × 𝟏𝟎−𝟔 Principle Strain 𝜺𝟏,𝟐 =

𝜺𝒙 + 𝜺𝒚 𝜺𝒙 − 𝜺𝒚 𝟐 𝜸𝒙𝒚 𝟐 ± √( ) +( ) 𝟐 𝟐 𝟐 2

𝜀1,2

2

276 × 10−6 + (−52 × 10−6 ) 276 × 10−6 − (−52 × 10−6 ) 149 × 10−6 = ± √( ) +( ) 2 2 2 2

𝜀1,2

276 × 10−6 + 52 × 10−6 ) 276 × 10−6 + 52 × 10−6 ) 149 × 10−6 = ± √( ) +( ) 2 2 2

2

𝜺𝟏 = 𝟑𝟒𝟒. 𝟏𝟑 × 𝟏𝟎−𝟔 𝜺𝟐 = −𝟏𝟔. 𝟏𝟑 × 𝟏𝟎−𝟔

Nizamuddin Patel

P15219444

ENGD2007- Strength of Materials Mohr’s Strain Circle (for 3600 lbf load)

Point A: (276, 74.5) Point B: (-52, -74.5)

Figure 6: Mohr’s Circle

𝑷𝒐𝒊𝒏𝒕 𝑨 = (𝜺𝒙 ,

𝜸𝒙𝒚 ) 𝟐

𝑷𝒐𝒊𝒏𝒕 𝑨 = (𝟐𝟕𝟔, 𝟕𝟒. 𝟓)

Then you need to work out point B: 𝑷𝒐𝒊𝒏𝒕 𝑩 = (𝜺𝒚 , −

𝜸𝒙𝒚 ) 𝟐

𝑷𝒐𝒊𝒏𝒕 𝑩 = (−𝟓𝟐, −𝟕𝟒. 𝟓) 𝑪𝒆𝒏𝒕𝒓𝒆 𝒐𝒇 𝑪𝒊𝒓𝒍𝒄𝒆, 𝑪 = 𝐶=

𝜺𝒙 + 𝜺𝒚 𝟐

276 + (−52) 2

𝑪 = 𝟏𝟏𝟐, (𝒉𝒆𝒏𝒄𝒆 𝟎, 𝟏𝟏𝟐)

  

From the diagram, εmax and εmin are estimated at 310 and -50 respectively 𝛾𝑚𝑎𝑥 is 320 x 10-6 The angle I measured from the Mohr’s Circle was 𝟐𝟖𝒐 . Hence 𝟐𝜽 = 𝟐𝟖𝒐, 𝜽 = 𝟏𝟒°

Nizamuddin Patel

P15219444

ENGD2007- Strength of Materials

Calculating the Stress (𝝈): 𝑭 𝑺𝒕𝒓𝒆𝒔𝒔 = 𝑨 16013.6𝑁 = 304.2 × 10−6 𝑺𝒕𝒓𝒆𝒔𝒔 = 𝟓𝟐. 𝟔𝟒 × 𝟏𝟎𝟔 𝑵/𝒎𝟐

Calculating the Young’s Modulus:

𝒔𝒕𝒓𝒆𝒔𝒔 𝒀𝒐𝒖𝒏𝒈′ 𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 (𝑬) = 𝒔𝒕𝒓𝒂𝒊𝒏 𝜎 = 𝜀𝑚𝑎𝑥 52.64 × 106 = 310 × 10−6 = 𝟏𝟔𝟗. 𝟖𝟏 𝑮𝒑𝒂

Poisson’s Ratio: 𝝐𝒚 = −𝒗𝝐𝒙 𝜖𝑦 𝑣=− 𝜖𝑥 −52 × 10−6 𝑣=− 276 × 10−6 52 × 10−6 𝑣= 276 × 10−6 𝒗 = 𝟎. 𝟏𝟗

Determining the Poisson Ratio using the Principle Strains: 𝜺𝟏 = 𝟑𝟒𝟒. 𝟏𝟑 × 𝟏𝟎−𝟔 𝜺𝟐 = −𝟏𝟔. 𝟏𝟑 × 𝟏𝟎−𝟔 −(−𝟏𝟔. 𝟏𝟑 × 𝟏𝟎−𝟔 ) 𝒗 = 𝟑𝟒𝟒. 𝟏𝟑 × 𝟏𝟎−𝟔 𝒗 = 𝟎. 𝟎𝟓

Nizamuddin Patel

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ENGD2007- Strength of Materials Conclusion The experimental values for 𝛾𝑚𝑎𝑥 is calculated as 𝟑𝟔𝟎. 𝟐𝟔 × 𝟏𝟎−𝟔 . The 𝛾𝑚𝑎𝑥 value obtained from Mohr’s circle is 320 x 10-6. This has resulted in a difference of 40.26, with a percentage error of 11.18%. The Poisson’s ratio value obtained using the plane stress results is 0.19. Using principal strain the Poisson’s ratio was calculated as 0.05. These results are different as one occurs under perpendicular conditions and the other at an angle. Another value to consider are the angles obtained from Mohr’s circle. The value of the angle is 28 o, resulting to 𝜽 = 14o. This should be 45o, thus resulting in an error of 68.89%. The errors above could be the cause of many results not matching the theoretical values. One of the main errors could be the creation of the Mohr’s circle diagram. As this was drawn by hand, this method could result in many errors to the end result if not drawn carefully. The problem could have developed from many parts of the diagram like from the circle itself to the spacing of the axis and its straightness. The circle could have been created using a computerised tool to model the circle with more precision and reduce the errors caused by drawing it free hand. The results could have been very different to the ones obtained by the diagram. Apparatus error could have been caused by the manufacturer of the mechanisms used whilst carrying out the experiment, as well as the person/people looking after the equipment. The tools used could have been damaged or not to the specification required for the experiment to be carried out successfully. The manufacturer could have certain tolerances to the parts of the Olsen testing machine and therefore the weights within the mechanism could have been different to what the machine displays to the user. Random error could be caused by the user making a mistake whilst carrying out the experiment and therefore could result in very large anomalies. A mistake could cause a missed result or not balancing the device properly and therefore leading to an error in the results. To avoid this type of error, the experiment could be repeated at least 3 times for each gauge, however this would lead to a very long experiment which could increase the error occurrence. Hysteresis error would occur if the apparatus was not maintained and was damaged due to this. The Olsen testing machine could have internal damage to which could not be seen and accessed. Another part of the machine could be the belt used to turn the device; the belt could have been loosened/stretched by overall use in the past and therefore could be unsynchronised with the weights to the display, and therefore the user would be unable to carry out an experiment with correct values. Parallax error is a big possibility to the large error occurrence as the experiment heavily depended on the user reading the figures from the machine, and the figures were displayed to the person’s eyes in different angles to where he/she was standing; reading values for the force applied could have been an issue as the marks and text were small, yet was read at an angle causing the parallax. Another huge parallax error could have been caused whilst levelling the device as per the instructions. This was tricky and resulted in many twists of the lever. The person making sure the device was level could have been obstructed from a clear view etc. All the errors mentioned above could have had an adverse effect towards the results, causing the significant percentage error.

Nizamuddin Patel

P15219444