Lab report for the measurement of Young

Lab report for the measurement of Young’s modulus
Hanchen Nan------ID: 10118555
Hengyang Luo ----ID: 10119611
Jingzhi Liu---------ID: 10119993
Lib instructor: Yan Zhao
Date of experiment: March 8, 2011
Contents
1.Introduction (2)
a)Abstract (2)
b)Introduction (2)
c)Theory (2)
2.Procedure (4)
a)Measure the diameter of the text sample (4)
b)Install the sample (4)
c)Inspection (5)
d)Recording (5)
3.Result (6)
4.Analysis (9)
a)Relative error (9)
b)Error analysis (9)
c)How to reduce error (10)
5.Conclusion (11)
1.I ntroduction
a)Abstract
The goal of this experiment was to determine the Young’s modulus of different metals and determine metal types by measuring Young’s modulus of the metal.
b)Introduction
Measure the diameter d of the text samples with micrometer caliper. And calculate the cross-section area A. Stretch the wire by the force of 45N, and then decrease the force of 5N per time. Measure and record the change of length ∆L. Use these data to calculate out Y.
c)Theory
Figure 1 show the apparatus used in this experiment.
A uniform metal wire or bar length L and cross-sectional area s has an extension of ΔL along the le ngth direction of the force F. According to Hooke’s law, in the elastic limit the fraction, change rate of wire length ΔL/L and tensile stress F /A are proportional.
Namely: Y=F/A
∆L/L
(Equation 1)
Where: F: Tensile force
A: cross-section area
∆L/L: Fraction change of wire length
Figure 2 shows a schematic representation of an object under tension.
Figure 2: An object under tension
In the equation the ratio coefficient Y is the Young’s modulus of the solid. And the international unite of Young’s modulus is Nm-2.
In the experiment if the metal wire is d in diameter, the cross-sectional area
A=1
4πd2, take it into figure 1 then get the equation Y=4FL
πd2∆L
(Equation 2)
The equation 2 indicate that in the same length L diameter d and tensile force the metal wire, which has a high Young’s modulus, has a small elongation ∆L. Thus, the Young’s modulus expresses the stretching (or compression) deformation capacity of metal when it resistance forces.
The values of Young’s modulus for some common materials are listed in Table 1
Table 1 –Values of Young’s for some common materials
2.P rocedure
The experiment was conducted on March 8, 2011, 09:00a.m. – 11:00a.m.
a)Measure the diameter of the text sample
Measure the diameter of the text samples with micrometer caliper on five different locations.
b)Install the sample
Take off the dial indicator from the deformation instrument.
Loose the two screw of gauges fixation and spin right the hand wheel to make it close to the base of tensile deformation instrument. Then turn the hand wheel three circles anticlockwise.
Traverse the text sample wire though the left retaining clip of the text sample. And fasten the wire tightly by the right retaining clip. Make the dynamometer show the numerical reading of 10N by tensioning the wire rod and then screw down the left retaining clip with the special key. Turn the hand wheel slowly and make the dynamometer show 30N and 50N. Each time fasten the wire tightly.
Regulate the location of the dial indicator fixation, move it to the middle of the pillar.
Then fasten the dial indicator fixation tightly and install the dial indicator on the fixation. Adjust the dial plate and make the pointer locate in the middle of the range. Make sure that the dial indicator can stretch and shrink freely.
c)Inspection
Inspect weather the pointer of the dial indicator moves sequential.
Turn the hand wheel anticlockwise to make the indication of the ergometer increase from 5N to 50N, and record the indication of the dial indicator when the indications of ergometer are 10N and 50N. Then turn the hand wheel clockwise to make the indication of ergometer is 5N. And turn the hand wheel anticlockwise and record again.
The readings of two-times measurements should not much difference in the 2 scale marks.
d)Recording
Measure the length between the right and left retaining clip when the indication of ergometer was 10N. Record the indication of the dial indicator when the indication of the ergometer were 10N, 15N, 20N, 25N, 30N, 35N，40N, 45N. Then turn the force back to 5N, and repeat the measurement.
3. Result
Table 2: Measured indication of the dial indicator when the wire was
stretch by different tensile force.
According to the equation S=
4πd 2 and the diameters measured in the
experiment, the area of the brass wire was 610087.2-⨯2
m and the area of the iron wire was 610196.2-⨯2
m .
Table 3
2168
F/A
4566210
Figure 4
Curve graphs figure 3 and figure 4, which were reflecting the relation between
y = -22143x + 6E+07
R² = 0.9883
05000000
100000001500000020000000250000000
500
1000
1500
2000
2500
3000
unit unit
Brass
2m N L
L ∆y = -24154x + 6E+07
R² = 0.9868
05000000
100000001500000020000000250000000
500
1000
150020002500
unit unit
Iron
2m N L
L ∆L
L ∆
and , were made according to the experimental values in table 1 and
table 2. And in order to get an accurate curve graph, 610-was adopted as the
unit of x-coordinate. The gradient of regression line figure 1 is 22143, so the
experimental Yo ung’s modulus of brass is 2 2 3 Nm −2 and Y exp brass
=
2 2 4
3 Nm −2. Identically the Young’s modulus of iron can be calculated,
and Y exp iron =2 4 54 Nm −2.
N F L L ∆
4.Analysis
a)Relative error
According to the table 1 the theoretical Young’s modulus of brass is Nm−2and the theoretical Young’s modulus of iron is
Nm−2. Compare the data which were got in the experiment Y exp brass=2 2 43Nm−2and Y exp brass=2 2 43Nm−2. It seems that the data which were got in the experiment are having a big error with the theoretical data. But the data are all in the same order of magnitude. So the experiment was successful.
b)Error analysis
Systematic error: The lengths of wires were measured by ruler, and the precision of ruler was 1mm. So the completely accurate lengths cannot be measured by using a ruler. And the diameters of the wires were measured by a dial indicator. Though the precision was 0.001mm the completely accurate diameters cannot be measured. This kind of error was made by the limitation of measuring tools. And in the process of stretching wires there might have tiny slide between retaining clips and wires.
Accidental error: The error was unavoidable in data reading. By using a same measuring tool to measure a same thing different can have a different data reading. This kind of error was made by the person. And the result can also be influenced by the data processing method.
c)How to reduce error
In the process of experiment guarantee that there are no slip between wires and retaining clips. In order to decrease the error the average value of the data got by several measuring was used in the calculation. And EXCEL was used to deal the data. By using curve graphs to deal with data and the unit of
x-coordinate should be adopted appropriate. In this experiment there has a 2.6% error between the unit adopted 1 and− , and the unit adopted − was more accurate.
5.C onclusion
According to the gradient of the curve graphs, the Young’s modulus of brass and iron were got, and Y exp brass=2 2 43Nm−2and Y exp iron=2 4 54 Nm−2. And referring to the table 1, the theoretical Young’s modulus of brass is Nm−2and the theoretical Young’s modulus of iron is Nm−2.
Thought the values had a big different from the theoretical values, the numerical values were all in the same order of magnitude. So the experiment was successful.
The laminations of the measuring tools and the process of reading are both the reason of producing errors. In order to increase the precision of the value more accurate tools should be used. And use the average value to decrease the accident errors.。