Strain Measurement of Cantilever Beam

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 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 dierent locations along the beam to produce measurable measurable strains.

Beam ata! "# $% cm l # $& cm b # $.'(' cm t # ).&%$ cm

*e are going to use the following e+uations in order to nd! Bending stress,

σ 

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

given by  M y σ =  I 

*here;

-+uation %/ 0 # bending moment, 1.mm

 2 # distance from the neutral axis, mm I

# second moment of area m'

Since the lastic modulus,  #

σ  ε  , so

σ 

= E x ε

-+uation $/ ε

*here,

 # strain

Substitute +uation -$/ into e+uation -%/, therefore 3  M =ε x

 EI   y

Results and Discussion: 0#4d 054d#) *hen force is 61, the formula

+ 1150−5  x = 0

 M 

 M 

=

5 x −1150

*hen a force of %)1 is used the formula is

=10 x −2300

 M 

*hen a force of %61 is used the formula is!  M 

=

15 −3450

-+uation &/

Apply the same concept to the other forces.

=(5 x 22)− 1150=−1040

 M 

 M =( 10 x 22)− 2300 =−2080

=(15 x 22)− 3450=−3120

 M 

 M =( 20 x 22)− 4600 =−4160

=(25 x 22)− 5750=−5200

 M 

*hen a force of 61 is used the strain is found using the method below! 7# second moment of inertia of area bh

3

1 12

12

I #

 x 24.84  x 3.12

3

=62.87 mm'

 #  My  I 

σ  #

ε=

=

1040  x 1.56

 # $6.(%

62.87

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

*hen a force %)1 is used the strain is found using the method below! bh

3

1 12

12

I # σ 

ε

=

 x 24.84  x 3.12

3

=62.87 mm'

 # #

 My 2080  x 1.56 =  # 6%.8$ 62.87  I 

stress σ  elastic modulus  E =

51.62 =

200 x 10

9

=

2.581 E − 10

9epeat the same steps for the remaining results. 4or a force of %61 # &.(:x%)5%)  4or a force of $)1 # 6.%8$ x%)5%)

 4or a force of $61# 8.'6& x%)5%) ata for tension and compression 1ote! AT7 # average tension for initial value -load )1/ AT # average tension value during loading

7nitial -)1/ )!%8!& & )!%8!& ' )!%8!& 8 )!%8!& ( )!%8!' ) )!%8!' $ )!%8!' ' )!%8!' 8 )!%8!' ( )!%8!6 ) )!%8!6 $ )!%8!6 ' )!%8!6 8 )!%8!6 ( )!%:!) ) )!%:!) $

Avg !

56.'' 56.'' 56.'' 56.'' 56.'6 56.'' 56.'' 56.'' 56.'' 56.'' 56.'' 56.'' 56.'' 56.'' 56.'6 56.'' 5 6.''%$ 6

%'.' ' %'.' ' %'.' ' %'.' ' %'.' ' %'.' ' %'.' ' %'.' ' %'.' ' %'.' 6 %'.' ' %'.' ' %'.' 6 %'.' ' %'.' ' %'.' '

4orce 61 )!%(!) 8 )!%(!) ( )!%(!% ) )!%(!% $ )!%(!% ' )!%(!% 8 )!%(!% ( )!%(!$ ) )!%(!$ $ )!%(!$ ' )!%(!$ 8 )!%(!$ ( )!%(!& ) )!%(!& $ )!%(!& '

Avg !

4orce %)1

56.$$ 56.$$ 56.$$ 56.$$ 56.$$ 56.$% 56.$$ 56.$$ 56.$$ 56.$$ 56.$$ 56.$$ 56.$$ 56.$$ 56.$$ 5 6.$%<& &

%'.8 6 %'.8 8 %'.8 8 %'.8 8 %'.8 8 %'.8 8 %'.8 8 %'.8 8 %'.8 6 %'.8 8 %'.8 8 %'.8 8 %'.8 8 %'.8 8 %'.8 8

)!%
Avg !

56 56 56 56 5'.<< 5'.<< 5'.<< 56 56 56 56 56 56 5'.<<

%'.( : %'.( : %'.( : %'.( : %'.( : %'.( : %'.( : %'.( : %'.( : %'.( ( %'.( : %'.( : %'.( : %'.( :

5 '.<<: %'

4orce %61 )!$)!& : )!$)!& < )!$)!' % )!$)!'

5'.:: 5'.:: 5'.:: 5'.::

%6.) ( %6.) ( %6.) ( %6.)

& )!$)!' 6 )!$)!' : )!$)!' < )!$)!6 % )!$)!6 & )!$)!6 6 )!$)!6 : )!$)!6 <

Avg!

5'.:: 5'.:: 5'.:: 5'.:: 5'.:( 5'.:: 5'.:: 5'.::

( %6.) ( %6.) ( %6.) ( %6.) ( %6.) ( %6.) ( %6.) ( %6.) (

5 '.::) (&

4orce $)1 )!$%!& $ )!$%!& & )!$%!& 6 )!$%!& : )!$%!& < )!$%!' % )!$%!' & )!$%!' 6 )!$%!' : )!$%!' < )!$%!6 % )!$%!6

5'.6' %6.& 5'.6' %6.& 5'.6' %6.& 5'.66 %6.& 5'.6'

%6.$ <

5'.6' %6.& 5'.6' %6.& 5'.6' %6.& 5'.6' %6.& 5'.6' %6.& 5'.66 %6.& 5'.6' %6.$

& )!$%!6 6

Avg !

< %6.$ 5'.6' < 5 '.6'%6 '

4orce $61 )!$$!% < )!$$!$ % )!$$!$ & )!$$!$ 6 )!$$!$ : )!$$!$ < )!$$!& % )!$$!& & )!$$!& 6 )!$$!& : )!$$!& < )!$$!' %

5'.&% 5'.&% 5'.&% 5'.&% 5'.&$ 5'.&% 5'.&$ 5'.&% 5'.&$ 5'.&% 5'.&$ 5'.&$

5 '.&%'% Avg ! :

%6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 % %6.6 $

Loading Number ) % $ & ' 6

Load(N)

Distance from support L(mm)

Bending moment at ! (N!mm)

) 6 %) %6 $) $6

$&) $&) $&) $&) $&) $&)

) %%6) $&)) &'6) '8)) 6:6)

Bending Moment(N! mm)

"#eoretical Strain(micro strain)

$%perimental Strain (micro strain)

%%6)

%$<

:8

&ercent age $rror (') '%.%

$&))

$6(

%68

&<.6

&'6)

&(:

$&'

&<.6

'8))

6%8

&)<

').%

6:6)

8'6

&(8

').%

:))) 8))) 6))) '))) Bending moment (N!mm)

&)))

 25alues "inear -25alues/

$))) %))) ) )

$))

'))

8))

e%perimental strain (microstrain)

:))) 8))) 6))) '))) Bending moment (N!mm)

 25alues

&)))

"inear -25alues/ $))) %))) ) )

$)) ')) 8)) ())

"#eoretical strain (microstrain)

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

3

1 12

12

I #

 x 24.84  x 3.12

3

=62.87 mm'

 #

 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 =

4

 E x 62.87 mm

14.84 =

1.56

 E=

•

•

23.15 62.87

= 0.3682

 The relation between stress and the bending moment is directly proportional because we had a linear graph. 7t is shown that the position of the load aects the magnitude of the strain. This is because dierence 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 &<5'%= percentage of error, this +uite a huge mista>e.  This might be as a result of the change in the temperature of the wire. ?hanges in temperature lead to the thermal expansion of the wire as the strain gauge is left on for some time. 7t will ma>e the thermal sensitivity error. The errors could be minimi@ed using a half5 bridge or a full bridge conguration. *ith all stain gauges in the bridge at the same temperature, any change in temperature aects 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! *e get greater strain when we apply greater force. But there was an error due to the temperature dierence. 7t tells us that temperature plays an important part in aecting the result which can be avoided next time.