1. Abstract—In this lab, students learned how to use charpy
impact testing to test brass andmarble. Using LabVIEW, data was
collectedfrom these test. Using the data, students were responsible
for calculating the impact energy due to the collision. In order for
proper analysis, non-conservative forces must be accounted for in
the calculations. Brass was the stronger material as the work
required to cause fracture was higher for brass than for marble.
Index Terms—charpy impact testing, brass, marble, work
I. INTRODUCTION
HE purpose of this lab is to test brass and marble using
charpy impact testing.A charpy impact testeris simply an
impactor mounted as a simple pendulum with the purpose of
creating an impact force on the test specimen. The impactor has
a position sensor that creates a voltage that is proportional to
the angular position relative to the ground that allows for
accurate data collection. Impact testing allows for materials to
be tested using dynamic loading. Dynamic loading is different
from static loading in that in dynamic loading, the applied force
is not constant. In order to calculate the force, two strain gages
are mounted in a ½ bridge configuration wired to a Tacuna
strain gage amplifier whose output is wired to a DAC. With the
data obtained in LabVIEW, students were responsible for
calculating the windage and friction losses from the pendulum
swinging, the absorbed impact energy of both specimens, the
work done during the impact of each specimen, and the elastic
and plastic portions of energy for brass.
II. PROCEDURE
Hang down test
In order to properly measure position, a hang down test was
performed. In the hang down test,the impactor is placed at the
position of lowest potential energy and data is recorded using
LabVIEW. From the data, students found the average position
and a scaling factor. The scaling is found using the following
formula.
𝑆𝑐𝑎𝑙𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟 =
360
𝑉𝑠𝑒𝑛𝑠𝑜𝑟
(1)
Once the scaling factor is found, it is multiplied by the tared
voltage reading which is found by using the following formulas:
∆𝑉𝑔 =
𝑉𝑎𝑚𝑝
1100
(2)
𝐶𝑎𝑙𝑖𝑏𝑟𝑎𝑡𝑒𝑑 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = 𝑆𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∗ ∆𝑉𝑔 (3)
This value yields the calibrated initial position in degrees.
Calculating windage and friction
In this lab, specimens testing is done via dynamic loading. In
order to perform this properly, the windage and friction losses
have to be accounted for. This is done by displacing the
pendulum from its lowest potential energy position and
releasing it to allow it to freely oscillate for at least two cycles.
The difference between the initial position and the first zero
velocity location after release represents the energy losses due
to non-conservative forces. Students must account for these
losses in order to have accurate and meaningful data. Using the
data in the swing down file, students found a new calibrated
position by subtracting the average position found in the hang
down test by the position in the swing down file. This value is
then multiplied by the scaling factor to get the calibrated
position is degrees. Using the following formula, the totallosses
are found.
𝑀𝑔( 𝐻𝑖 − 𝐻𝑓 ) = 𝑙𝑜𝑠𝑠𝑒𝑠 (4)
Where 𝐻𝑖 and 𝐻𝑓 are the initial and final heights found using
the following:
𝐻𝑖 = 𝐿 𝑔 − 𝐿 𝑔 cos 𝜃𝑖 (5)
𝐻𝑓 = 𝐿 𝑔 − 𝐿 𝑔 cos 𝜃𝑓 (6)
Where 𝜃𝑖 and 𝜃𝑓 are maximum and minimum theta values
from the theta vs. time plot for the swing down test located in
the results section. M is the mass of the impactor (2.3 kg) and
Lab 5: Charpy Impact Testing of Brass and
Marble
Ballingham, Ryland
Section 3236 3/25/2016
T
2. <Section####_Lab#> Double Click to Edit 2
2
g is acceleration due to gravity.
Specimen testing
Once the windage and friction losses are accounted for,
material testing can be performed. The test of both materials
were done sequentially with brass being the first material brass
being tested. Before testing, the test specimen is mounted in
the charpy impact tester. Once the specimen is mounted, the
impactor is brought back to approximately 90° and released.
Data is recorded of each test and uploaded to canvas foranalysis
in order to calculate the energy absorbed by the specimens, the
work done by the impactor on the specimens during impact and
the elastic and plastic portions of energy absorption for each
specimen. The figure below shows a more detailed
representation on how testing is performed.
Fig. 1. Simplified charpy impact test diagram
Fig. 2. Detailed view of bar dimensions and components
III. RESULTS
Fig. 3. Theta vs. time plot for the swing down test
Fig. 4. Theta vs.time plot forthe brass test
3. <Section####_Lab#> Double Click to Edit 3
3
Fig. 5. Theta vs.time plot forthe marble test
Fig. 6. Impact force vs arc length for brass specimen
Fig. 7. First peak for brass specimen
Fig. 7. Impact force vs arc length for marble specimen
Fig. 8. First peak for marble specimen
IV. DISCUSSION
1. The average initial position from the hang down test was
found 2.66 V. From (1) the scaling was found to be 72.85.From
(2) ∆𝑉𝑔 was found to be 0.00245.This value was multiplied the
scale factor to get an initial position of 0.179°. The maximu m
and minimum angles were found to be 89.5° and -88.1°. By (3)
and (4), 𝐻𝑖 was found to be 0.704 m and 𝐻𝑓 was found to be
0.687 m. Using (4) the losses were found to be 0.384 J.It makes
sense that the losses would be relatively small since there isn’t
a lot of locations on the impact tester for frictional losses to
occur.
2. The corrected strain energy absorption numbers (SEA) are
found from the following formula:
𝑆𝐸𝐴 = 𝑀𝑔(𝐻𝑖 − 𝐻𝑓) − 𝑙𝑜𝑠𝑠𝑒𝑠 (7)
Finding 𝐻𝑖 and 𝐻𝑓 is done the same way when finding the
losses.For the marble the maximum and minimum angles were
found to be 81.37° and 59.03°. Using appropriate geometry, 𝐻𝑖
and 𝐻𝑓 were found to be 0.603 m and 0.345 m respectively.
Using (7) the energy absorbed after losses for the marble was
found to be 5.44 J.
For brass, using the same process as above the energy
absorbed after losses was found to be 5.33 J.
3. The work done on the brass specimen is found by
numerically integrating the first peak in arc length vs impact
force curve using the trapezoidal method in Excel. In order to
calculate the impact force, the strain of each specimen must be
found. Strain is found using the following formula:
𝜀 =
2∆𝑉𝑔
𝑉𝑠 𝐺𝑓
(8)
Where ∆𝑉𝑔 is found from (2), 𝐺𝑓 is 2.1 and 𝑉𝑠 is given in the
data file as strain 𝑉𝑒. After the strain is found, the moment of
4. <Section####_Lab#> Double Click to Edit 4
4
inertia of the rectangular tube cross section is found. The
moment of inertia (I) is found to be 8.628 x 10−9
𝑚4
. Now the
force due to the rotation of the tester can be found. This found
using the following formula:
𝐹 =
𝐼𝑙𝜖𝐸
𝑐𝑎𝑣
(9)
Where l is the total length of the bar and impactor which is
0.88 m, E is the modulus of elasticity of the impactor (205 GPa)
[1], c is the distance to the neutral axis which is 0.0127 m, a is
the distance fromthe center of mass to the sample which is 0.17
m and v is the distance to the strain gage from the pivot point
which is 0.255 m. Once this force is found the impact force can
be found using the following formula:
𝐹𝑖𝑚 =
𝐹𝑏
𝑙
(10)
Where b is the length from the pivot point of the testerto the
center of mass of the assembly and l is the distance from the
pivot point of the assembly to the test specimen. After this the
arc length can be found from the following equation:
𝑑 = 𝑙𝜃 (11)
Once the arc length is found, the arc length vs the impact
force can be plotted. Like stated earlier, the work done is found
by numerically integrating the first peak of each curve using the
trapezoidal method in Excel. Doing this yields 4.02 J for the
work done during impact on the brass specimen and 1.48 J for
the marble specimen [3].
4. Theoretically, the change in potential energy should equal
the work done during impact. This only applies to a collision
that is perfectly elastic doesn’t happen in the real world. Since
this collision isn’t perfectly elastic, there will be a small
difference in the two values of change in potential energy and
work done.The difference between the two values is a result of
energy loss due to windage and friction.
5. The elastic energy (𝜀 𝑒𝑙) for the brass specimen is
calculated using the following formula:
𝜀 𝑒𝑙 =
𝑃2
𝐿3
96𝐸𝐼
(12)
Where P is the impact force, L is the length of the specimen,
E is the modulus o elasticity of the specimen and I is the
moment of inertia of the specimen. The elastic portion of the
energy occurs before the yield point on the stress vs. strain
curve for brass. Once the brass specimen passes its yield limit
of 135 MPa, the plastic portion of energy begins to occur.
Marble doesn’t have an elastic portion of energy since it is a
brittle material. The value obtained for the brass specimen for
the elastic energy was found to be 0.229 J. The plastic energy
(𝜀 𝑝𝑙) portion is found by using the following formula:
𝜀 𝑝𝑙 = 𝑆𝐸𝐴 − 𝜀 𝑒𝑙 (13)
Doing this yields a plastic portion of energy of 5.61 J for the
brass specimen.
6. Uncertainties were calculated the formulas found in the
appendix. The results are found in table II in the appendix.
V. CONCLUSION
This lab shows that the rate at which a force is applied affects
material properties. Once data is collected, students were
responsible for finding the initial and final heights to find the
change in potentialenergy (energy absorbed)ofeach specimen,
the work done during impact of each specimen and the elastic
and plastic portions of energy for the brass specimen. The brass
specimen is the strongerofthe two specimens as the work done
on the brass to cause fracture is higher than the work done to
fracture the marble specimen. This makes sense since the
ductile material (brass) is going to have a higher strain energy
thus requiring a higher value of work (energy) to break it.
APPENDIX
TABLE I
ENERGY VALUES
Energy Values (J) Brass Marble
Energy absorbed 5.71 5.82
Losses 0.384 0.384
Energy absorbed after
losses
5.33 5.44
Work done duringimpact 4.02 1.48
Elastic strain energy 0.229 -
Plastic portion of strain
energy
5.99 -
Uncertainty equations used:
𝑈𝜃 = √(
𝜕𝜃
𝜕𝑉 𝑝
)
2
(𝑈𝑉 𝑝)2
+ (
𝜕𝜃
𝜕𝑉 𝑠
)
2
( 𝑈𝑉 𝑠
)2 (14)
𝑈∆𝑉𝑔 = √(
𝜕 𝑉𝑔
𝜕𝑉𝑎𝑚𝑝
)
2
(U𝑉𝑔)2
(15)
𝑈 𝜀 = √(
𝜕𝜀
𝜕∆𝑉𝑔
)
2
(𝑈∆𝑉𝑔
)2 +(
𝜕𝜀
𝜕 𝑉𝑠
)
2
(𝑈 𝐼)2 + (
𝜕𝑃
𝜕𝐿
)
2
(𝑈 𝐿)2 + (
𝜕𝑃
𝜕𝑐
)
2
(𝑈 𝑐)2 (16)
𝑈 𝐻 = √(
𝜕𝐻
𝜕𝐵
)
2
(𝑈 𝐵)2
+ (
𝜕𝐻
𝜕𝜃
)
2
(𝑈 𝜃)2
(17)
