report on impact analysis of a material

1. Discussion and data analysis.
Question 1: The mechanism of the impact testing machine
The general methods of testing materials are:
1. Tensile testing; this method involves material being stretched or bend at minimal strain
rate. As a result this method will give a figure of the material’s ability to overcome
slowly or static force applied to give yield strength. The material under this test tends to
be more ductile.
2. Impact testing; the material is usually fractured at high strain rate. It helps to give
magnitude of a material’s sharp impact which indicates its toughness. The material
being tested with this method displays brittle characteristic.
This test gives good method for examining relative toughness of a given material. It
demonstrates how material’s toughness changes with temperature. When using other tests,
such a demonstration is never clear.
A pendulum at a given angle is released enabling it to accelerate downwards. This enables it to
change its potential energy to kinetic energy. When it strikes test specimen, part of the kinetic
energy is absorbed by the specimen causing it to break. The pendulum will then continue its
oscillation with less kinetic energy which is changed to potential energy as it raises towards its
starting point. By finding the difference in height between the points where potential energy is
maximum and where kinetic energy is maximum, the amount of energy used to fracture the
specimen can be determined. The following figure shows pendulum with a point mass m on a
massless armof length r.

2. The initial pendulum energy is 1mgh , where  1 1 sin( 90)h r    .
Final pendulum potential energy is 2mgh , where 2 (1 sin( 90))h r    .
Difference in energy (potential energy) = (sin( 90) sin( 90))mgr    
When reaction R is measured at a distance x with pivot as the datum, it becomes easy to
determine value of mgr while the pendulum is held horizontally.

3. Question 2.
Calculating difference in potential energy of aluminum:
Let Initial potential energy be denoted by E1,
Final potential energy be denoted by E2, 1 (1 sin( 90))E mgr   
1
1
(1 sin( 90))
E 2.34(1 sin(99 90))
E mgr   
    = 2.71 J
2
2
(1 sin( 0))
E 2.34(1 sin(72 90))
E mgr   
    = 1.62 J
Difference in potential energy E ,
1 2E E E  
= 2.72 J-1.62 J
= 1.09 J
Since the difference in potential energy in equal to the amount of energy absorbed during
fracture then this theory is valid.

4. Question 3.
From this bar chart, it is found that mild steel has the highest amount of energy absorbed
before fracture. This indicates that mild steel (ER&)s-2) is more tough compared to copper,
aluminum HDPE and PVC.
0
0.5
1
1.5
2
2.5
3
Aluminium (5356) Copper Mild Steel (ER70s-2) HDPE PVC