Flexural Young’s Modulus of Aluminium

Introduction:
We are currently producing about 800 million tonnes of steel worldwide for a variety of uses and
about 40% of this production originates from recycled steel. Steel is possibly the most versatile
construction material - it is strong in tension and in compression, it is easily formable and has
uniform quality. The inclusion of steel in concrete to produce reinforced concrete has extended its
versatility.
Objective:
The purpose of this laboratory is to use a 1-point beam bending test in order to determine the
flexural stiffness of different metallic materials; steel, copper, brass and aluminium, specific
objectives are:
 Awareness of safety in the laboratory
 Understand the different metallic materials behaviour
 Perform an experimental test to determine the flexural behaviour of the metallic materials
 To use the standard beam bending equations to back calculate the metallic flexural Young’s
Modulus.
 Analyse and interpret data against typical values for steel
 Enhance ability to work in teams and communicate
 To gain experience of the process of experimental materials testing
 To gain experience of working in a laboratory environment
Procedure:
Figure 1 shows the experimental setup. Essentially a particular metallic beam sample has been
placed on two supports and is to be loaded in the mid-span as shown in Figure 2.

Questions:
(1) Draw the beam’s cross-section below and, without disturbing the beam, measure the
dimensions and thickness of the specimen.
Steel Brass Aluminium
b=20.8mm b=20.8mm b=20.8mm
d=4mm d=4mm d=4mm
L=510mm L=510mm L=510mm
(2) What is the distance of the neutral axis from the beam’s upper surface?
Since the object is uniform about the x axis and y axis. So, the neutral axis will be passing through the
centre of gravity of the object. So, the distance of the neutral axis from the beam’s upper surface is d/2. And
that will be 2 mm.

Since the section of the all the material are same the distance between neutral axis and upper surface is also
same for the all section.
(3) Determine the 2nd moment of area of the cross-section along the beam’s neutral axis?
Here beam’s neutral axis is at the geometric centroid of the cross-section so,
The equation for finding the 2nd moment of inertia along the beam’s neutral axis is
I = b d3
12
Now put the value of cross section in this equation.
I =(20.8)(4)3
12
I = 1331.2
12
I =110.933 mm4
Since the cross section of the all the material are same 2nd moment of inertia is also same.
(4) Carefully add weights to the beam at the places indicated on the beams upper surface. After
each increment record the number indicated on the dial gauge and convert to displacement
as appropriate in Table 1.
Table 1: Experimental Data Steel
Increment Number Weight Added Per Position
(N)
Displacement (mm)
1 100 g 5 mm
2 500 g 46 mm
3 1000 g 90 mm
4 1900 g 177 mm
Table 2: Experimental Data Brass
Increment Number Weight Added Per Position (N) Displacement (mm)
1 100 g 1 mm
2 500 g 28 mm
3 1000 g 66 mm
4 1900 g 126 mm
Table 3: Experimental Data Aluminium
Increment Number Weight Added Per Position
(N)
Displacement (mm)
1 100 g 10 mm
2 500 g 62 mm
3 1000 g 130 mm
4 1900 g 246 mm