MIDDLE EAST TECHNICAL UNIVERSITY DEPARTMENT OF CIVIL ENGINEERING CE 241: MATERIALS SCIENCE
Determination of Modulus of Elasticity and LABORATORY REPORT # 1: Determination Poisson’s Ratio of Concrete
LAB GROUP: 6
ABDULLAH ATHAR 2125854 SECTION # 6
MAHSUN KAŞCI 20206 SECTION # 6
SUBMISSION DATE: 05!12!2016
Table of Contents
1. Objectives and Scope…………………………………………………………………………. 3 2. Preliminary 3. 4. 5. ". $. %.
Remarks………………………………………………………………………….
3 Test Specimen………………………………………………………………………………... 5 pparat!s…………………………………………………………………………………….. 5 Test Proced!res……………………………………………………………………………..... " #alc!lations…………………………………………………………………………………... " Res!lts………………………………………………………………………………………. 13 &isc!ssion o' Res!lts………………………………………………………………………...
13 (. #oncl!sion…………………………………………………………………………………... 14 Re'erences……………………………………………………………………………………..... 15
1- Objects and Scoe!
The object of this experiment is to observe the behaviour of concrete under the application of stress produced by uniaxial compressive forces. The scope of this experiment is to calculate the Modulus of Elasticity and Poisson’s Ratio of concrete. 2" P$%&'&()* R$')+,: S-$,,:
)t is t*e 'orce applied per !nit area o' t*e material. For this experiment we will consider
compressive stress i.e. the stress resultin from the application of compressive forces. Stress is calc!lated by !sin+ t*e +iven 'orm!la, ! " F#$% where ! " stress F " applied force $ " $rea upon which the force is bein applied Strain!
&t is the deformation per unit lenth of a material and is also called ' a"ial strain ’. For this experiment we will consider compressive strain i.e. the
strain resultin from the application of compressive forces. (train is dimensionless and is calculated by usin the followin formula) ε = ΔL/Lo% where% ε = strain ΔL = change in length Lo " oriinal lenth
*
•
+ateral (train is the ratio of deformation of the dimension of a body perpendicular to the direction of the applied force ,lateral dimensionto the oriinal lateral dimension. / " Δd/d where, / " lateral strain 0d " deformation d " oriinal lateral dimension
Deformation!
&t is the alteration in the dimension or shape of a material as a result of application of stress or the variation in temperature. ,For this experiment we will observe deformation resultin from the application of stressElastic Materials!
The materials which return to their oriinal dimensions and shapes after bein subjected to deformations resultin from the application of external forces. #oo$e’s %a&!
This law states that in an elastic body% the stress and strain are proportional to each other% independent of time.
Modulus of Elasticity '(oun)’s Modulus*!
&t is the ratio between stress and stress and measures the sti1ness of a material. E " !#ε, where E " youn’s modulus ! " stress 2
ε " strain
Poisson’s Ratio!
&t is the ratio between lateral strain and axial strain. + , -/# a% where% + , Poisson’s ratio
/ " lateral strain a " axial strain sotroic Materials!
Materials which exhibit the same properties in all directions. The specimen under observation i.e. concrete is also an isotropic material. ." T$,- S/$&'$(
T*e specimen bein+ !sed is a solid cylinder o' concrete. T*e cylinder *as a diameter o' 1- cm it* a *ei+*t o' 2- cm. 4" A//))-,
/T0 !niversal testin+ mac*ine is !sed 'or t*is eperiment. )t is an apparat!s !sed to meas!re t*e tensile stress and compressive stress o' materials.
3
i+!re 1, /niversal Testin+ 0ac*ine /T0 5" ). )). ))).
T$,- P$3$ Place t*e cylinder vertically beteen to level plates in t*e /T0. oad t*e specimen and t*en !nload it6 repeatin+ t*e process t*rice. Record t*e applied load and displacement data 'rom t*e /T0 contin!o!sly.
6" C)%%)-&(,
4
Time 's*
F&$ 2, aial 'orce 7 time.
1-
2
nd
%oadin) Rate
5
Time 's* f,x- " 6 378/.99x : *7389*.4
F&$ .: Second loadin+ rate.
$t
t ( 0 )=61,1289
t
F ( 0)=−5252,57
$t
t ( 1)= 61,6294 t F ( 1 )=−7798,64
Loading Rate=[ F ( 1)− F ( 0 )]/[ t ( 1 )−t ( 0 )]=[−7798,64 −(−5252,57 )]/[ 61,6294 −611289 ]=−5087,1546
¿ graph this rate egual ¿=−5091,9
2" .3 L)3&( R)-$
9
Time's* f,x- " 6 378;.2 /x : 4//34*.8/
F&$ 4: T*ird loadin+ rate.
$t t ( 0 ) = 120,1865 % at
t ( 1)=120,687
$t F ( 0)=−505,706 % at F ( 1 )=−3050,6
Loadingrate =[ F ( 1 )− F ( 0 )]/[ t ( 1 )−t ( 0 )]=[−3050,6 −(−505,706 )]/[ 120,687 −120,1865 ]=−5084,703
¿ graph this rate egual ¿=−5092,4
8
." E%),-& '3%, ,$(3 %)3&(
a" ial strain f,x- " ;** 3*4.58 x : 7.;8
F&$ 5, ial stress 7 aial strain 'or second loadin+.
Stress =
force ( F ) area( A )
, A t t ( 0 )=60,128 F =−131,868 A = π ∗50∗50=7853,9816
F −131,868 =−0,017 Stress = = A 7853,9816 Strain = Deformation ( ΔL )/ original length ( L)
LVDT ( 1 ) + LVDT ( 2 ) Deformation= ,Soatt ( 60,128 ) , 2
ΔL =(0,002 +(−0,005 ))/ 2=−0,00172
L=1300 mm,Strain =
ΔL −0,00172 ,= =−1,323∗10−5 L 130
¿ also , slope of this graph= Elastic m odulus( E2)= 23353 /7
for second loading
4" E%),-& '3%, -7&3 %)3&(
a" ial strain f,x- " ;* 278.4/ x : 7.*2
F&$ 6, ial stress 7 aial strain 'or t*ird loadin+.
At t ( 120,1865 ) , F =−505,706 A = π ∗ 50∗50 =7853,9816
F −505,706 Stress = = =−0,06 A 7853,9816 Strain = Deformation ( ΔL )/ original length ( L)
LVDT ( 1 ) + LVDT ( 2 ) Deformation = ,Soatt ( 120,1865 ) , 2
ΔL =
0,002+ (−0,007 ) 2
=−2,5∗10−3
//
−3
L=130 mm,Strain =
ΔL −2,5∗10 =−1,923∗10−5 ,= 130 L
¿ also , slopeof this graph= Elastic m odulus( E3)= 23410
for thirdloading
5" P&,,(, )-& ,$(3 %)3&(
a"ial strain
f,x- " 6 7.; /x 6 7
F&$ 9, transverse strain8 aial strain second loadin+
Transverse strain= Lateral deformation / Original laterallength
Lateraldeformat i on =(0,44 / 1)∗(Value of OD ) At t ( 60,12893 )= OD=0,00029825305488 So , Lateral deformation=0,44 / 0,0014343=1475,258
then, Tranverse strain =lateral deformatin / original length=1475,258 /100 ( originallength )=14,753 a!ial strain has alread" calculated at figure 4
#oisson$ sratio =−( tranverse strain / a!ialstrain ) , ¿ slope of Figure 7 Transverse strain/ a!ial strain=−0,209
/;
#oison$ s ratio ( v2 ) =−(−0,209 )=0,209 for second loading
6" P&,,(, )-& -7&3 %)3&(
a"ial strain
f,x- " 6 7.;/x 6 7
F&$ 8, transverse strain 8aial strain 'or t*ird loadin+
Lateraldeformat i on =(0,44 / 1)∗(Value of OD ) At t ( 120,1865 )=OD =0,00044 So , Lateral deformation=0,44 / 0,00044 =1000
then, Tranverse strain =lateral deformatin / original length=1000 / 100 ( originallength )=10 a!ial straanhas alread" calculated at figure 5 third loading
#oisson$ sratio =−( tranverse strain / a!ialstrain ) , ¿ slope of Figure 8 Transverse strain / a!ial strain=−0,210 #oison $ s ratio ( v 3)=−(−0,210)= 0,210 for third loading
/*
9" R$,%-, L)3&( R)-$
oadin+ Rate 'or Second oadin+ 9 85-(161 :s81 oadin+ Rate 'or T*ird oadin+ 9 85-(264 :s81 E%),-& M3%,
;lastic 0od!l!s 'or Second oadin+ ;2 9 23353 0Pa ;lastic 0od!l!s 'or T*ird oadin+ ;3 9 2341- 0Pa vera+e ;lastic 0od!l!s ;av+ 9 23353 < 2341- = 2 9 233%2 0Pa P&,,(, R)-&
Poisson>s Ratio 'or Second oadin+ v;- " 7%;78 Poisson>s Ratio 'or T*ird oadin+ v*- " 7%;/7
$verae Poisson>s Ratio vav 9 -621- < -62-( = 2 " 7%;/7 8" D&,,,&( R$,%-,
&n this experiment we found out how to calculate both the Elastic Modulus and the Poisson’s Ratio of concrete usin their relation with stress and strain. The Elastic Modulus showed us the sti1ness of concrete while the Poisson’s Ratio showed us the tendency of concrete to expand#contract in a lateral direction when a compressive#tensile force is applied to it lonitudinally.
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owever% we only considered the second two readins and nelected the ?rst one because it may contain unwanted errors by the experimenter or the machine and provide us with inaccurate results. Moreover% we used two +@ATs in this experiment and averaed their results. This was done to avoid any possible errors and acBuire results as accurate as possible. The oriinal Poisson’s Ratio for concrete ,vconcrete - varies between 7%/3 to 7%;3 and the value determined from this experiment i.e. v concrete " 7%;/7 lies well inside this rane which aCrms the accuracy of this experiment. Moreover% the oriinal value for the ;lastic 0od!l!s 'or concrete ;concrete varies beteen 14--- 0Pa to 41--- 0Pa and t*e val!e ac?!ired 'rom t*is eperiment i.e. ; concrete 9 233%2 0Pa lies entirely inside t*is ran+e *ic* con'irms t*e acc!racy o' o!r res!lts.
For an isotropic material% its Dul= Modulus ,- and (hear Modulus ,- can be calculated by usin the followin formulas " E # * ,/ 6 ;v- and " E # ; ,/ : v- respectively. (ince% concrete is an isotropic material and the Elastic Modulus ,E- as well as the Poisson’s Ratio ,v- are =nown% both the Dul= Modulus ,- and the (hear Modulus ,- can be calculated. .ul$ Modulus '/*
" E # * ,/ 6 ;v " ;**9; # * ,/ G ; ,7%;/7- " /*2*5 MPa S0ear Modulus '*
" E # ; ,/ : v " ;**9; # ; ,/ : ,7%;/7- " 844; MPa
/3
Finally% if we endeavour to calculate the Elastic Modulus from the force G lonitudinal deformation raph for this specimen% we can determine its value because both the dimensions for this specimen and relation of Elastic Modulus with stress and strain are =nown. 2- Conclusion
ence% this experiment showed us the relation of Elastic Modulus and Poisson’s Ratio with stress and strain.
References!
/. ErdoIan% Turhan J et al. ,;7/7-. Introduction to Material Science for Civil Engineers. ,/st ed.-. $n=ara) METH Press.
;. http)##www.enineerintoolbox.com#concrete6properties6dK/;;*.html *. http)##www.eduresourcecollection.com#civilKsmK(trains.php 2. http)##www.efunda.com#formulae#solidKmechanics#matKmechanics#elasticK constantsKK=.cfm 3. https)##en.wi=ipedia.or#wi=i#HniversalKtestinKmachine
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