Strain Measurement Of Cantilever Beam 3s3vy

APPLIED MECHANICS

Strain Measurement of Cantilever Beam FACULTY OF ENGINEERING & INFORMATION TECHNOLOGY (Sep-Dec 2015)

Introduction: Strain gauges are used in many systems as sensors to measure forces, moments, and the deformations of structures and materials. This experiment deals with measuring the strain in a cantilever beam through the use of four resistance strain gages; two mounted on top of the beam and two mounted below. A static load will be incremented at different locations along the beam to produce measurable strains.

Beam Data: L= 21 cm l = 23 cm b = 2.484 cm t = 0.312 cm

We are going to use the following equations in order to find: Bending stress,

σ , at distance y from the neutral axis of the beam is

given by σ=

My I

(Equation 1)

Where;

M = bending moment, N.mm

Y = distance from the neutral axis, mm I

= second moment of area m4

Since the Elastic modulus, E =

σ ε

, so

σ =E x ε

(Equation 2) ε

Where,

= strain

Substitute Equation (2) into equation (1), therefore  M =ε x

EI y

Results and Discussion: M=Fd M-Fd=0 When force is 5N, the formula M + 1150−5 x=0

M =5 x−1150 When a force of 10N is used the formula is M =10 x−2300

When a force of 15N is used the formula is: M =15−3450

(Equation 3)

Apply the same concept to the other forces. M =(5 x 22)−1150=−1040

M =(10 x 22)−2300=−2080 M =(15 x 22)−3450=−3120

M =(20 x 22)−4600=−4160 M =(25 x 22)−5750=−5200

When a force of 5N is used the strain is found using the method below: I= second moment of inertia of area

I=

bh 3 12

σ =

ε=

=

1 x 24.84 x 3.123 =62.87 12

My 1040 x 1.56 = I 62.87

mm4

= 25.81

stress σ 25.81 = = =1.291 E−10 elastic modulus E 200 x 109

When a force 10N is used the strain is found using the method below:

I=

bh 3 12

σ =

ε=

=

1 3 x 24.84 x 3.12 =62.87 12

My 2080 x 1.56 = I 62.87

mm4

= 51.62

stress σ 51.62 = = =2.581 E−10 elastic modulus E 200 x 109

Repeat the same steps for the remaining results. For a force of 15N = 3.87x10-10 For a force of 20N = 5.162 x10-10

For a force of 25N= 6.453 x10-10 Data for tension and compression Note: AVTI = average tension for initial value (load 0N) AVT = average tension value during loading

Initial (0N) 0:16:3 3 0:16:3 4 0:16:3 6 0:16:3 8 0:16:4 0 0:16:4 2 0:16:4 4 0:16:4 6 0:16:4 8 0:16:5 0 0:16:5 2 0:16:5 4 0:16:5 6 0:16:5 8 0:17:0 0 0:17:0 2 Avg :

-5.44 -5.44 -5.44 -5.44 -5.45 -5.44 -5.44 -5.44 -5.44 -5.44 -5.44 -5.44 -5.44 -5.44 -5.45 -5.44 5.4412 5

14.4 4 14.4 4 14.4 4 14.4 4 14.4 4 14.4 4 14.4 4 14.4 4 14.4 4 14.4 5 14.4 4 14.4 4 14.4 5 14.4 4 14.4 4 14.4 4

Force 5N 0:18:0 6 0:18:0 8 0:18:1 0 0:18:1 2 0:18:1 4 0:18:1 6 0:18:1 8 0:18:2 0 0:18:2 2 0:18:2 4 0:18:2 6 0:18:2 8 0:18:3 0 0:18:3 2 0:18:3 4 Avg :

Force 10N

-5.22 -5.22 -5.22 -5.22 -5.22 -5.21 -5.22 -5.22 -5.22 -5.22 -5.22 -5.22 -5.22 -5.22 -5.22 5.2193 3

14.6 5 14.6 6 14.6 6 14.6 6 14.6 6 14.6 6 14.6 6 14.6 6 14.6 5 14.6 6 14.6 6 14.6 6 14.6 6 14.6 6 14.6 6

0:19:2 1 0:19:2 3 0:19:2 5 0:19:2 7 0:19:2 9 0:19:3 1 0:19:3 3 0:19:3 5 0:19:3 7 0:19:3 9 0:19:4 1 0:19:4 3 0:19:4 5 0:19:4 7 Avg :

-5 -5 -5 -5 -4.99 -4.99 -4.99 -5 -5 -5 -5 -5 -5 -4.99

14.8 7 14.8 7 14.8 7 14.8 7 14.8 7 14.8 7 14.8 7 14.8 7 14.8 7 14.8 8 14.8 7 14.8 7 14.8 7 14.8 7

4.997 14

Force 15N 0:20:3 7 0:20:3 9 0:20:4 1 0:20:4

-4.77 -4.77 -4.77 -4.77

15.0 8 15.0 8 15.0 8 15.0

3 0:20:4 5 0:20:4 7 0:20:4 9 0:20:5 1 0:20:5 3 0:20:5 5 0:20:5 7 0:20:5 9 Avg:

-4.77 -4.77 -4.77 -4.77 -4.78 -4.77 -4.77 -4.77

8 15.0 8 15.0 8 15.0 8 15.0 8 15.0 8 15.0 8 15.0 8 15.0 8

4.770 83

Force 20N 0:21:3 2 0:21:3 3 0:21:3 5 0:21:3 7 0:21:3 9 0:21:4 1 0:21:4 3 0:21:4 5 0:21:4 7 0:21:4 9 0:21:5 1 0:21:5

-4.54 15.3 -4.54 15.3 -4.54 15.3 -4.55 15.3 -4.54

15.2 9

-4.54 15.3 -4.54 15.3 -4.54 15.3 -4.54 15.3 -4.54 15.3 -4.55 15.3 -4.54 15.2

3 0:21:5 5 Avg :

9 15.2 -4.54 9 4.5415 4

Force 25N 0:22:1 9 0:22:2 1 0:22:2 3 0:22:2 5 0:22:2 7 0:22:2 9 0:22:3 1 0:22:3 3 0:22:3 5 0:22:3 7 0:22:3 9 0:22:4 1

-4.31 -4.31 -4.31 -4.31 -4.32 -4.31 -4.32 -4.31 -4.32 -4.31 -4.32 -4.32

4.3141 Avg : 7

15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 1 15.5 2

Loading Number 0 1 2 3 4 5

Load(N)

Distance from L(mm)

Bending moment at XX(N-mm)

0 5 10 15 20 25

230 230 230 230 230 230

0 1150 2300 3450 4600 5750

Bending Moment(Nmm)

Theoretical Strain(micro strain)

Experimental Strain (micro strain)

1150

129

76

Percent age Error (%) 41.1

2300

258

156

39.5

3450

387

234

39.5

4600

516

309

40.1

5750

645

386

40.1

7000 6000 5000 4000 Bending moment (N-mm)

3000

Y-Values Linear (Y-Values)

2000 1000 0 0

200 400

600

experimental strain (microstrain)

7000 6000 5000 4000 Bending moment (N-mm)

Y-Values

3000

Linear (Y-Values) 2000 1000 0 0

200 400 600 800

Theoretical strain (microstrain)

Calculations: Second moment of area of the steel beam I is;

I=

bh 3 12

=

1 x 24.84 x 3.123 =62.87 12

mm4

Then from the experimental graph, the young modulus of the steel is determined by; m=

So,

y 2− y 1 5750−1150 = =14.84 x 2−x 1 386−76 M EI = ε y

14.84=

E x 62.87 mm4 1.56

E=





23.15 =0.3682 62.87 The relation between stress and the bending moment is directly proportional because we had a linear graph. It is shown that the position of the load affects the magnitude of the strain. This is because difference in distance between the load and the strain gauge, the longer the distance, the bigger the strain. The increase in distance leads to an increase in the bending moment. Based on the calculated result of experimental and theoretical value, there is a 39-41% percentage of error, this quite a huge mistake. This might be as a result of the change in the temperature of the wire. Changes in temperature lead to the thermal expansion of the wire as the strain gauge is left on for some time. It will make the thermal sensitivity error. The errors could be minimized using a halfbridge or a full bridge configuration. With all stain gauges in the bridge at the same temperature, any change in temperature affects the gauges at the same ratio. Because of the temperature changes is similar for all of the gauges, both the ratio of their change in resistance and their output voltage didn’t change.

 

In conclusion:



We get greater strain when we apply greater force. But there was an error due to the temperature difference. It tells us that temperature plays an important part in affecting the result which can be avoided next time.