Faculty of Architecture BDes
inArch
Design Studies
Constructing the Environment
DESA1102 Structures
LAB 2 REPORT
–
Elasticity
Date Group Name
SID
PART A1
–
DEFLECTION
OF DIFFERENT BEAMS
Three different beams were tested
for deflection, using a single point
load at midspan.
Considering the two timber beams, does the small one def
lect more than the large one?
The large beam contains 4 times as much material as the small one. Is the deflection reduced by a factor of
4, or more, or less?
Considering the equal

sized timber and steel beams, does the timber one deflect more than the s
teel one?
By approximately what factor?
The deflection formula has
E
on the bottom line. From your results, is
E
greater for timber or for steel?
The deflection formula has I on the bottom line. I is calculated for any rectangular section as
BD
3
/12
. I
f the
rectangular steel section were laid on its side, so that
D
became 6mm and
B
became 11mm, would its
deflection be more or less than it was on edge?
For a given span and a given load, what can you do to minimise deflection?
(circle answer)
Use a m
aterial with a
big / small
value of Modulus of Elasticity,
E
Use a section with a
big / small
value of Moment of Inertia,
I
Use a layout with
built

in / simple
support conditions, if possible.
Beam material
Timber
Timber
Steel
Cross section D x B
11 x 6
16 x 16
11 x 6
Dial gauge initial reading
Dial gauge final reading
Deflection
Which deflections are
greatest or least
?
PART A2
–
COMPARATIVE STIFFNESS OF MATERIALS
Four beams of different materials were
tested for deflection, using a single
point load at midspan.
Is the difference in deflection between the materials due to:
(circle a) or b))
a)
the different s
trength? or
b) the different stiffness of the materials?
PART B
–
YOUNG’S MODULUS AND STRENGTH OF COPPER WIRE
A copper wire was suspended from a bracket and loads applied to its lower end. The elongation was
measured with a vernier attached to a f
ixed wire. First, a set of readings was taken while loading and
unloading. Secondly a set of readings was taken while loading to failure.
To determine the stress in the wire, we divide the force by the cross

sectional area of the wire.
Diameter of the wi
re = mm Cross

sectional area A =
D
2
/4 = mm
2
To determine the strain, we divide the change in length by the original length of the wire.
Length of wire (from the support to the vernier) = mm
Deflection (mm)
Load (kg)
steel
alum
brass
0.5
1.0
Deflection (mm)
Load (kg)
perspex
0.1
0.2
x 9.81
/ = MPa
/ =
/ = MPa
Readings on copper wire
First loading
Unloading
Second loading
Load (kg)
Reading, R
R

Ro
Reading, R
R

Ro
Readi
ng, R
R

Ro
0.50
Ro =
0
Ro =
0
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
Finding the elastic properties of
copper
First, plot the results using the raw figures (kg for load and mm for extension). Use the first graph for the first
set of readings, choosing a scale for elongation that will fit in with the readings. Use the second graph for the
loading to failure.
Choose scales for both axes to fit in the results. Note that there may be an uneven portion at
the beginning of the graph, where kinks in the wire are straightened out.
0
Elongation (mm)
0
0.5
1.0
1.5
2.0
2.5
Load (kg)
Choose a straight section of the first graph to represent elastic behavio
ur. The Modulus of Elasticity for this
section is Change in Stress / Change in Strain. Do the calculations below:
Change in load (kg) x g (m/s
2
) / area of wire (mm
2
) = change in stress (N/mm
2
or MPa)
Change in elongation (mm)
/ original length (mm) = change in strain (ratio)
Change in stress (MPa) / change in strain (ratio) = Modulus of Elasticity (MPa)
From the second graph, pick the yield str
ess (where the elongation begins to increase rapidly), and the load at
failure (ultimate stress)
Yield stress = load x g / area = Ult. Stress = load x g / area =
El ongation (mm)
Load (kg)
PART C: COMPRESSIVE TESTS ON TIMBER
Three hardwood and three softwood samples were tested parallel to the grain and three hardwood samples
were tested perpendicular to the grain under compression until the samples failed.
Record the result
s for each test and make a sketch of the mode of failure. Calculate the mean results for
the three cases. Calculate the difference as a percentage of the mean softwood compressive strength.
Which is stronger in compression:
(circle one
)
a) softwood?
b) hardwood?
Which is stronger in compression:
(circle one)
a) parallel to the grain?
b) perpendicular to the grain?
When would you use timber in compression only?
So when using timber in compressi
on, what would you use?
What is the difference in the failure modes between the softwood and the hardwood samples?
Parallel to the grain
Perpendicular to the grain
Test No.
Softwood
Hardwood
Hardwood
1
Value
Mode of Failure
Value
Mode of Failure
Value
Mode of Failure
2
3
Mean
Value
Percentage
difference =
Mean /
Mean
Softwood x
100
What is the difference in the failure modes between the parallel and the perpendicular grain samples?
ADDITIONAL QUESTIONS to be answered
at the end of the lab session
1.
In Part B, what value did you get for the Modulus of Elasticity of copper? Given the values in the Lab 2
Notes , do you think this is a likely ‘ball park’ figure, or do you suspect there is an error?
2.
From your experi
ment, do you think copper is a ductile or brittle material?
3.
In measuring the elongation of the copper wire, and its diameter, how accurately could you measure?
Given the smallest division you could measure, would you be confident that the to
tal elongation was
within 10%, or 1% or 0.1% of the real value?
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