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    MECHANICS AND MATERIALS LABORATORY (MEMB221)

    SEMESTER 2, 2015/2016

    EXPERIMENT 2 : TORSION TEST

    DATE PERFORMED : 3rdDECEMBER 2015

    DUE DATE : 11thDECEMBER 2015

    SECTION : 05

    GROUP : 04

    GROUP MEMBERS I / D NUMBER

    Mithradassa Nair A! G Dha"#dhara$ ME0%5512

    S#r$a &ai'ash A! &a$$a$ &(r())a' ME0%55*%

    Ra+,,$ A! Thi-a.(.#/a' ME0%552

    Pra.at,sh &("ar A! Ash#) &("ar ME0%55

    Sar+aisa$ A! M($ia$d- ME0%55*0

    LAB INSTRUCTOR: Ci) N(ras'i$da Bi$ti A$(ar

    TABLE OF CONTENT

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    SUMMAR ABSTRACT 1

    OBECTIE 2

    TEOR 263

    E7PERIMENT E8UIPMENT 36

    PROCEDURE 6*

    DATA AND OBSERATIONS 106%

    ANA!SIS AND RESU!TS 11612

    DISCUSSIONS 13614

    CONC!USIONS 14615

    REFERENCES 15

    ABSTRACT

    This experiment is performed to study the principle of torsion test and

    also to determine the modulus of shear, G through measurement of the

    applied torque and angle of twist. The variation of pure shear when a

    structural member is twisted is called torsion. The torsional forces producea rotating motion about one end to another end of the member.

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    Two dierent specimens has been used in this experiment, specimen A

    and specimen B. pecimen A is bright gold in colour whereas pecimen B

    is silver and much more lighter than specimen A. The dimension for the

    both specimen is measured and recorded before the experiment begin.The torque measuring unit is calibrated !rst by inspecting the read out

    torque from ampli!er to be similar with the applied torque. The test is

    then performed. To avoid some measurement errors several measure were

    ta"en which can aect the results.

    The test specimen is place between the loading device and the torque

    measuring unit. The reading from the ampli!er is ta"en out each time

    when the load is applied. The results were ta"en and some calculations is

    performed using the formula given in the lab manual which is the applied

    torque, angle of twist, number of revolutions and the percentage error and

    from the results obtain a graph is plotted. The modulus can be determined

    when the specimen is still wor"ing under the elastic limit.

    Based on the results obtained, it is concluded that specimen B is more

    ductile than specimen A. The G value for specimen B theoretically is larger

    than specimen A, hence it is harder to twist than specimen B. The torque

    needed to twist specimen A to the same amount degree of rotation as

    specimen B is greater.

    #

    O!"#$%&"'

    To understand the principle of torsion test.

    To determine the modulus of shear, G through measurement of the

    applied torque and angle of twist.

    T"*+

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    Torsion is a force produced when a structural member is twisted , torsional forces

    produces a rotating motion around the ob$ect. %n each test, the torque and

    twisting angle are measured to determine the shear modulus, G.

    The shear modulus G is calculated based on this formula&'

    L

    G

    J

    T =

    where322

    42dr

    J

    ==

    Th, a$.', #9 tist; i$ radia$s; 9#r a s#'id r#($d

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    Figure 1: layout of the torsion apparatus

    T"#%# "'#*%3$% 4 $" 33*$'

    Th, a//arat(s ?#$sists "ai$'- #9:

    1@ !#adi$. d,+i?, ith s?a', a$d r,+#'(ti#$ ?#($t,r 9#r tisti$. a$.', ",as(r,",$t

    2@ T#r=(, ",as(r,",$t ($it

    3@ Ca'i

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    !#adi$. d,+i?,:6

    #r" .,ar r,d(?ti#$ rati#: 2

    R,+#'(ti#$ ?#($t,r: 5 di.it; ith r,s,t

    O(t/(t s?a',: 30

    I$/(t s?a',: 30

    I$di?at#r: Ad(sta

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    T*7" M"'*"8"$ U%$

    Th, s/,?i",$ is "#($t,d at #$, sid, t# th, '#adi$. d,+i?, a$d th, #th,r sid, t# th, t#r=(,

    ",as(r,",$t d,+i?,@ Th, t#r=(, a//'i,d t# th, s/,?i",$ i'' /r#d(?, sh,ar str,ss,s hi?h

    ar, d,t,?t,d

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    S/,?i",$ A H

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    *@ R,+#'(ti#$ ?#($t,r as r,s,t@

    S3"#%8" % $$%

    1@ a$d h,,' at th, i$/(t #9 th, .,ar as t(r$,d ?'#?)is, t# '#ad th,

    ,/,ri",$ta' "at,ria'@ It sh#s #$'- t(r$,d 9#r a d,9i$,d a$.', i$?r,",$t@

    2@ F#r th, 9irst r#tati#$; a$ i$?r,",$t #9 =(art,r r#tati#$ H%0K as ?h#s,$@ F#r

    th, s,?#$d a$d third r#tati#$ #9 a ha'9 =(art,r H1*0K as ?h#s,$ a$d 9#r th,

    9#(rth t# t,$th r#tati#$ H30K as ?h#s,$@

    3@ Th, tist a$.', as ?a'?('at,d at th, s/,?i",$

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    N9 4 R$$%A" $ "* %3$

    ("*"")

    R" $ $*7"

    (N8)

    A" 4 $.%'$,

    ("*"")

    1st %0 1@40 1@45

    1*0 2@45 2@%0

    20 3@%5 4@35

    30 4@5 5@*1

    2$d 540 @25 *@1

    20 *@20 11@1

    3rd %00 *@0 14@52

    10*0 %@00 1@42

    4th 1440 %@40 23@23

    5th 1*00 %@0 2%@03

    th 210 %@*0 34@*4

    th 2520 %@%0 40@5

    *th 2**0 10@15 4@45

    %th 3240 10@35 52@2

    10th 300 10@50 5*@0

    Ta

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    S3"#%8" B (S%&"* C* M$"*%)

    O(t,r !,$.th; !# L 115"" L 0@15"

    I$$,r !,$.th; !i L@2"" L 0@02"

    Dia",t,r; L @1"" L 0@001"

    N9 4 R$$%A" $ "* %3$

    ("*"")

    R" $ $*7"

    (N8)

    A" 4 $.%'$,

    ("*"")

    1st %0 0 1@45

    1*0 0@15 2@%0

    20 0@25 4@35

    30 0@5 5@*1

    2$d 540 1@%5 *@1

    20 3@15 11@1

    3rd %00 5@*5 14@52

    10*0 *@35 1@42

    4th 1440 11@%0 23@23

    5th 1*00 12@25 2%@03

    th 210 12@40 34@*4

    th 2520 12@55 40@5

    *th 2**0 12@0 4@45

    %th 3240 12@5 52@2

    10th 300 12@*0 5*@0

    Ta

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    I@ FROM CA!IBRATION CURE

    Th, .radi,$t +a'(, #

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    P,r?,$ta., ,rr#r; L

    M42@%%10000@2

    15%@000@2=

    11

    * #* (* +* ,* -* .* /**

    (

    ,

    .

    )

    #*

    #(

    Graph of Read Out Torque ! A"#$e of T%&!t

    A,61" 4 T.%'$ (2"6)

    R"02 O5$ T*75" (N8)

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    19 F* S3"#%8" A

    * #* (* +* ,* -* .* /**

    (

    ,

    .

    )

    #*

    #(

    #,

    G*03 4 R"02 O5$ T*75" &' A,61" 4 T.%'$

    A,61" 4 T.%'$ (2"6)

    R"02 O5$ T*75" (N8)

    29 F* S3"#%8" B

    12

    D&!'u!!&o"

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    #. 0rom the graph of read out ampli!er vs applied load torque it can beclearly seen that the graph is linearly proportional, which means as theapplied load increase there will be an increase in read out ampli!er. Theequation of the graph is 12 *.3-#4 5#.#*(.

    (. 0rom the graph of read out torque vs angle of twist it can be seen thatat a angle of * specimen A has a higher torque because it is a morebrittle material while specimen B has a lower torque value because it is amore ductile material.

    +.0rom the results obtained at table #, the shear modulus, G for specimen A is

    MPa%@4

    .6hereas the shear modulus for material B isMPa*5@15

    . The

    theoretical value of shear modulus of specimen A is (/78a and for specimen B is

    +378a. 0rom this we can see that the experimented value of both specimen is

    higher than its theoretical value. This is due to random errors.

    . Based on the results of this experiment, material A and B has an increase of

    torque when the number of rotation of the hand gear increases. The percentage

    error obtained for specimen A isM*3@%%

    whereas for specimen B isM42@%%

    . ,$?,

    th, ,/,ri",$ts ,r, ',ss a??(rat, 9#r specimenA@ Th, +a'(, #9 th, /,r?,$ta., ,rr#r t(r$,d

    #(t t#

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    E***' P*"#$%'

    Gra/hs #9 s/,?i",$ A a$d B ar, ,/,?t,d t#

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    /r#d(?, a$ a$.', #9 tist hi?h is th,$ ?a'?('at,d t# )$# hi?h s/,?i",$ is "#r,

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