EAA 204 – LIGHT STRUCTURE
LIGHT STRUCTURE EXPERIMENT
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
LIGHT STRUCTURE EXPERIMENT
1.
U1
-
Combined Bending and Torsion Experiment
2.
U2
-
Shear Force and Bending Moment Experiment
3.
U3
-
Beam Bending Experiment
4.
U4
-
Unsymmetrical Loaded Cantilever Experiment
5.
U5
-
Tensile Test
1
EAA 204 – LIGHT STRUCTURE
S1- COMBINED BENDING AND TORSION EXPERIMENT
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
OBJECTIVES
1. To study the behaviour of structural member subjected to bending and torsion or
combination or both load system.
2. To determine the critical level of the combined action of bending and torsional loading
on structural member.
2
EAA 204 – LIGHT STRUCTURE
INTRODUCTION AND THEORY
A structural member will experience deformation when it is loaded. Different loading
system will cause different types of stresses to develop in the member. A structural member
is in torsion if subjected to twisting couples or torques and is in bending if subjected to
couples acting in the longitudinal plane.
In general, the design of engineering structure components is not straight forward because
it may involve the combination of few stresses in those components. For example, the
bridge pier is subjected to stresses due to bending and torsion. Few theories have been
established to assess the strength of the material subjected to the combination of bending
and torsional stresses but it is difficult to prove by experiment which theory will gives the
most accurate result. Hence, most of the code of practice will give empirical value from few
experiments as a guide to design.
In order to get the critical bending and torsion ratio for a material is a serial complicated
experiment. Although the apparatus used in this`experiment has been equipped with seven
bending and torsion ratio which can be applied, accurate result for the critical bending and
torsion ratio cannot be achieved due to the following reasons:1. For each bending and torsion ratio, the experiment specimen must consist of the same
material. This condition is difficult to achive because the material of the specimen is
not 100% uniform although it comes from the same sample.
2. To overcome problem (1) above, the same specimen may be used for each experiment
in the series. By this method, the uniformity of the specimen can be maintained.
However, due to the reason that each experiment will be conducted until the material
achieves elastic failure stage, the material has undergone through strain hardening. This
means that for the subsequent experiments, the results obtained will be influenced by
strain hardening.
APPARATUS
1.
2.
3.
4.
5.
Circular Plate
Load Hanger and Set of Weights (Weight of Load Hanger is 500g)
Dial Gauge
Specimens (4 specimens to be tested at angle 0 0, 15 0, 75 0 and 900)
Spirit Level
3
EAA 204 – LIGHT STRUCTURE
PROCEDURE
Each group will be given four aluminium specimen. These specimen will be tested by using
four different angle which consist of 0o,15o, 75o and 90o. Each group needs only to compare
their result from this two experiment and determine bending and torsion ratio which is
more critical.
1. Place the first specimen in the apparatus by clamping it to the fixed support. Make sure
that the distance of the sholuder of the necking section from surface of the fixed
support is about 3mm. This will give the distance between two support surfaces of
approximately 25.4mm. Fasten the circular plate and lock the fixed supports.
2. Then, place the dial gauge opposite to the point where the load is to be applied. For
example, if the specimen is to be tested at 0o angle, place the dial gauge at the location
directly opposite. Tighten the screw to fix the location of the dial gauge. Adjust the dial
gauge to the reading of zero.
3. Place the load hanger (weighing 500g) as the first load at 00 angle and record dial gauge
reading. Dial gauge reading have to be multiplied with 0.01mm in order to get
deflection value in the unit of mm.
4. Add weights with the increment of 200g load until the specimen achieves the state of
elastic failure. Record the reading of dial gauge for each addition of load. This can be
observed when the change in dial gauge reading starts to become not uniform for each
addition of load.
5. Repeat the experiment for the second, third & forth specimen at angle of 15o, 75o and
90o respectively.
RESULT
1. Plot load (g) versus deflection (mm) diagram for all four experiment on the same graph.
2. Obtain the slope of each of the diagram. This will give the stiffness of the material for
the corresponding bending to torsion ratio.
3. Compare and comment on the values of stiffness obtained.
4
EAA 204 – LIGHT STRUCTURE
S1- COMBINED BENDING AND TORSION EXPERIMENT FORM
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
Name:
_____________________________________________________________________
Matric No.: _______________________
Group: _______________________
Date of Experiment: _________________________
RESULTS:
Load
(g)
Torsion Angle 0°
Dial
Gauge
Reading
Deflection
(mm)
Torsion Angle 15°
Dial
Gauge
Reading
Deflection
(mm)
5
Torsion Angle 75°
Dial
Gauge
Reading
Deflection
(mm)
Torsion Angle 90°
Dial
Gauge
Reading
Deflection
(mm)
EAA 204 – LIGHT STRUCTURE
Graph: Load versus Deflection diagram for all four experiment
Calculation: Stiffness of material for each experiment
DISCUSSION AND CONCLUSION:
6
EAA 204 – LIGHT STRUCTURE
S2 - SHEAR FORCE AND BENDING MOMENT EXPERIMENT
7
EAA 204 – LIGHT STRUCTURE
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
OBJECTIVES
Part 1: Shear Force Experiment
a) To understand the action of shear force on a beam.
b) To show that the shear force at a cut section of a beam is equal to the algebraic sum of
the forces acting to the left or right of the section.
Part 2: Bending Moment Experiment
a) To understand the action of bending moment on a beam.
b) To show that the bending moment at a section of a loaded beam is equal to the
algebraic sum of the moment to the left or right of the section.
INTRODUCTION
A length of material supported at two points in a such way that it will carry loads is
called a beam. The loading perpendicular to its longitudinal axis causes bending and in
most cases transverse sharing. In the simplest example the loads and supporting reactions
act in a vertical plane containing the longitudinal axis, and the beam has a rectangular cross
section.
The loads and support reactions are the external forces acting on the beam and
they must be in equilibrium. But in order to study the strength of the beam, it is necessary
to know how these external forces affect it. As the theory appendix shows, the
mathematical method is to assume the beam is cut into two parts by a transverse section
and then to examine the equilibrium of each part. To maintain equilibrium, it is evident
that certain forces must be introduced at the cut, and when the cut is not there these same
forces exist internally in the material of the beam.
8
EAA 204 – LIGHT STRUCTURE
Figure 1
In this experiment, a horizontal beam has been actually cut into two parts (A) and (B)
by a vertical cross section (Figure 1), and is then held together with springs (or spring
balances) which must produce a system of forces equivalent to those which would exist
internally in the beam at that section if it had not been cut. Since the forces in part (A)
acting on part (B) must be equal and opposite to those in (B) acting on (A), it follows that
the same values will be obtained by working to the right or left of section plane. Given a
horizontal beam with vertical loading the internal forces will be:1.
2.
for vertical equilibrium a shearing force in the section plane.
for equilibrium of moments a moment of resistance due to compression in the
top half of the beam section and tension in the bottom half.
In the experimental beam the second system of forces is replaced by compression at a
hinge in the beam and tension in the under slung spring (or spring balance). The vertical
restraint is provided by the half housing at the end of part (A) which fits on the ball
bearings pinned to the mating end of part (B).
THEORY
SHEAR FORCE AND BENDING MOMENT
9
EAA 204 – LIGHT STRUCTURE
P
(A)
(B)
RA
RB
Figure 2
Consider abeam resting on supports at A and B and supporting a load P (Figure 2).
If the beam is cut by a vertical section XX, then for the beam to remain in equilibrium
each part must be in equilibrium. Neglecting the self weight of the beam any extra forces
acting on part (A) to preserve equilibrium must be transmitted from part (B) across the
section XX, and vice versa. Also the action of a part (A) on part (B) must be equal and
opposite to that of (B) and (A).
As the load is solely vertical and in the plane of the beam there cannot be
horizontal reactions, so equilibrium yields two conditions:1. Vertical equilibrium
2. Equilibrium of moments
In the first place these are used for the beam as a whole to evaluate the reactions at A and
B.
SHEAR FORCE
X
P
10
EAA 204 – LIGHT STRUCTURE
+ve
Qx
(A)
(B)
+ve
RA
RB
X
Figure 3 - Shear Force
For vertical equilibrium of part (A) there must be a shearing force, Qx acting as shown and
equal to –RA.
For vertical equilibrium of part (B) the shearing force, Qx evidently acts as shown and has
the value of RB – P.
To clarify the results a sign convention must be used. The normal one is shown, which
leads to the –ve values above.
BENDING MOMENT
P
QX
Mx
QX
l
Mx
+ ve
+ ve
(A)
(B)
A
B
a
b
RA
RB
Figure 4 – Bending Moment
For equilibrium of moments take an axis as shown in the section XX to eliminate Qx.
Then using the sign convention given for part (A) of the beam:M x  RA .a
And for part (B) of the beam
M x  RA .b  P(b  )
It can easily been proved that these have the same value by substituting for RB and P  as
follows:-
11
EAA 204 – LIGHT STRUCTURE
Mx  b( P  RB )  P
 b.RA  RA (a  b)
 RA .a
Example :P1
P2
P3
A
B
1
2
3
RA
a
b
RB
L
Shearing force
Qx   RA  P1
or
RB  P3  P2
Bending Moment
M x  RA .a  P1 (a   1 ) or RB .b  P3 ( 3  a)  P2 ( 2  a)
Numerical Example
5N
A
200
X
10N
X
B
12
EAA 204 – LIGHT STRUCTURE
500
300
1000
Reactions. Take moment about A
1000RB – (500 x 10)- (200 x 5)=0
RB= 6(N)
By vertical equilibrium
RA=15-6 = 9 (N)
Section XX (to the right)
Shear force = 6 – 10 = -4(N)
Bending moment = (700 x 6) – (200 x 10)= 2200 (Nmm)
APPARATUS
Part 1: Shear Force Experiment
1. A pair of simple supports
2. Special beam with cut section
3. A set of weights
4. A load hanger
Part 2: Bending Moment Experiment
1.
2.
3.
4.
5.
Aluminum base support
Beam with section connected by a pin
A pair of simple support
A set of weights
A load hanger
13
EAA 204 – LIGHT STRUCTURE
Shear Force Apparatus
Bending Moment Apparatus
PROCEDURE
14
EAA 204 – LIGHT STRUCTURE
Part 1a: SHEAR FORCE
1. Place the beam on the supports so that the cut section is located at 600mm from the
left support and 300mm from the right support.
2. Place the load hanger at the specified location at = 100mm.
3. Level both beam (A) and (B) by using both of force gauge . Record the initial force
gauge reading as R1.
4. Place the load (F = 25N) on the load hanger at  1 and level the beam (A) and (B) as
step 3.
5. Record the force gauge reading as R2. The difference between R1 and R2 represents the
shear force 25N at the cut section.
6. Repeat step 2 to 5 by using different length such as  2= 250mm,  3=300mm and
 4=400mm at load hanger (B).
Part 1b: Shear Force
Repeat steps as stated in Part 1a but place the load hanger at the specified location (=
300mm) and apply a different set of loading (F = 25N, 30N, 35N and 40N).
Part 2: Bending Moment
1. Place the beam on the supports so that the cut section is located at 600mm from the
left support and 300mm from the right support.
2. Hang the load hanger at the middle of beam (B).
3. Level both beam (A) and (B) by using both of force gauge . Record the initial force
gauge reading as R1.
4. Place the load (F = 5N) on the load hanger at beam (B). Again, level both beam (A)
and (B) and record the reading as R2.
5. The difference between R1 and R2 represents the shear force 25N at the cut section.
6. Repeat step 2 to 5 by using different load such as 10N, 15N , 20N and 25N at load
hanger (B).
RESULT
Part 1 – Shear Force
1. For each load case, calculate the theoretical and experimental shear force values.
Compare these values and determine the percentage error.
2. Plot theoretical and experimental shear force values against distant, 
3. Plot theoretical and experimental shear force values against loading
Part 2: Bending Moment
1. For each load case, calculate the theoretical and experimental bending moment
values. Compare these values and determine the percentage error.
2. Plot theoretical and experimental bending moment values against loading
DISCUSSION AND CONCLUSION
15
EAA 204 – LIGHT STRUCTURE
1. What happen to the shear force value when the load is applied at different locations on
the beam?
2. What happen to the shear force when the different load is applied at the same location
in the beam?
3. What happen to the magnitude of bending moment when the applied load is
increased?
4. Comment on the accuracy of the experiment. Does the experimental data verify the
theoretical data?
S2 - SHEAR FORCE AND BEND ING MOMENT EXPERIMEN T FORM
16
EAA 204 – LIGHT STRUCTURE
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
Name:
_____________________________________________________________________
Matric No.: _______________________
Group: _______________________
Date of Experiment: _________________________
RESULTS:
Part 1a: Shear Force
Dist.

(mm)
Shear Force
(Theory),
F/L
(N)
A
B
Force Gauge
Reading, (N)
R1
R2
Shear Force
(Experiment),
R2- R1 (N)
C
D
E
Error
%
Error
F = E-B
G=(G/B)100
100
200
300
400
Part 1b: Shear Force
Load,
F
(N)
Shear Force
(Theory),
F/L
(N)
A
B
Force Gauge
Reading, (N)
R1
R2
Shear Force
(Experiment),
R2- R1 (N)
C
D
E
Error
%
Error
F = E-B
G=(F/B)100
25
30
35
40
Part 2: Bending Moment
Load,
F
(N)
Bending
Moment
(Theory)
0.1F
(Nm)
Force Gauge Reading, (N)
R1
R2
R3
= R2- R1
17
Bending
Moment
(Experiment)
R3  0.175
(Nm)
Error
%
Error
EAA 204 – LIGHT STRUCTURE
A
B
C
D
E
F
G = F-B
0
5
10
15
20
25
GRAPHS:
1. Theoretical and experimental shear force values against distant, 
2. Theoretical and experimental shear force values against loading
3. Theoretical and experimental bending moment values against loading
CALCULATIONS:
DISCUSSION AND CONCLUSION:
18
H=(F/B)100
EAA 204 – LIGHT STRUCTURE
S3 – BEAM BENDING EXPERIMENT
19
EAA 204 – LIGHT STRUCTURE
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
OBJECTIVE
To verify the theory of pure bending.
APPARATUS
1.
2.
SM104 MK III Beam Apparatus
Two load cells, three dial gauges, two weight hangers, beams (1 steel, 1 brass and 1
aluminum) and set of weights
INTRODUCTION
When a beam is loaded, the loads acting on the beam cause the beam to bend, there by
deforming its axis into a curve. The defection of the beam at any point along its axis is the
displacement of that point from its original position, measured in the vertical direction.
THEORY
When a beam is loaded, it bends in the plane of the applied moment. Theory of pure
bending of beam states that stress distribution within the beam and curvature of the beam
are related by:-
20
EAA 204 – LIGHT STRUCTURE
M 
E
 
I
y R
where M
-------------------- Eq. (1)
is the bending moment
is the second moment of area of the beam section (moment of inertia)
is the bending stress at distance y from the neutral axis
is the distance from the neutral axis
is the modulus of elasticity
is the radius of curvature
I

y
E
R
It can also be shown that the curvature of a beam 1/R is given, to a close approximation, by
the second derivative of the deflection. If z is the deflection of the beam at distance x from
a chosen origin then:
d 2z 1 M
 
dx 2 R EI
-------------------- Eq. (2)
The objective of this experiment is to show that equation (2) is valid.
Using equation (2) it can be shown that the deflection of a beam subjected to direct loading
can always be expressed in the form:
za
WL3
EI
-------------------- Eq. (3)
where :-
z
a
W
L
E
R
is the deflection
is a constant whose value depends upon the type of loading and supports
is the load acting on the beam
is the span
is the modulus of elasticity
is the radius of curvature
Consider a deflected beam which is subjected to loads due to weights as shown in Figure 1.
L
Beam specimen
h1
W
h2
a
h3
W
a
b
b
L/4
dial gauge
L/4
L/4
21
L/4
EAA 204 – LIGHT STRUCTURE
Figure 1
h
a
a
2R-h
Figure 2
The radius of curvature is determined by the Sagital method. Referring to Figure 2, the
radius of curvature is obtained from the relationship:
h(2R - h)
=
i.e.
2hR - h2
a2
=
a2
-------------------- Eq. (4)
Since h is relatives small compared to R, h2 will become a very small value and can
therefore be neglected. Therefore
2hR
or
=
a2
1 2h

R a2
-------------------- Eq. (5)
Also from equation (2) and Figure 1,
1 M

R EI
and
-------------------- Eq. (6)
M
=
Wb
-------------------- Eq. (7)
Substituting equation (6) and (7) into equation (5) yields:-
22
EAA 204 – LIGHT STRUCTURE
h
Wa 2b
2 EI
-------------------- Eq. (8)
By plotting h against W using equation (8), a straight line graph is obtained and the
gradient is:-
a 2b
2 EI
Thus when b = 300mm as used in the experiment, the gradient is
150a 2
EI
-------------------- Eq. (9)
Therefore for constant b, a graph of 150a2 against gradient should be linear and its gradient
will give a measure of the flexural rigidity (EI) of the beam.
PROCEDURE
1. Measure the length, thickness and width of the beam. Determine the mid-span of
the beam.
2. Choose the suitable reading on the upper scale of the apparatus for the mid-span of
the beam. (One of the 10cm markers is most convenient).
3. Set up one of the load cells so that it is 1/4 –span to the left of the marker chosen in
step 2. (Do not forget to take account of any offset in the position cursor).
4. Set up the second load cell 1/4-span to the right of the mid-span reading. Lock the
knife edge.
5. Place the beam in position with 1/4 -span overhang at either end.
6. Place one hanger near the left-hand end of the beam so that the load is applied on
the centre-line of the beam. Place the second hanger at the right-hand end of the
beam so that the hangers are the same distance from the supports. (See Figure 1).
7. Place one dial gauge at mid-span and the other two equidistant on either side of it
as shown in Figure 1. Adjust the three dial gauges to read zero.
23
EAA 204 – LIGHT STRUCTURE
8. Apply equal loads to the two hangers.
9. Gently tap the beam near its mid-span and take the readings of the dial gauges, h1,
h2 and h3.
10. Increase the load and repeat steps 8 and 9 for at least five values of W.
11. Record the readings of the dial gauges in the table as shown in the following section.
RESULT
1.
2.
Plot (h2-y) against load, W for test 1, 2 and 3. Obtain the gradient of each graph.
Plot 150a2 versus the gradient of (h2-y) versus load, W. Compare the experiment
and theoretical values of the flexural rigidity of the beam.
CONCLUSION
Using the results, verify the theory of pure bending.
S3 – BEAM BENDING EXPERIMENT FORM
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
Name:
_____________________________________________________________________
24
EAA 204 – LIGHT STRUCTURE
Matric No.: _______________________
Group: _______________________
Date of Experiment: _________________________
RESULTS:
Length
(mm)
Thickness
(mm)
Width
(mm)
I (mm4)
(Moment of
Inertia)
E (N/mm2)
(Modulus of
Elasticity)
Aluminum
Steel
Brass
Specimen:
Aluminum
W
(N)
b
(mm)
a
(mm)
h1
(mm)
h2
(mm)
h3
(mm)
y
h  h 
3
1
2
(mm)
Test 1
5
10
15
20
25
300
300
300
300
300
100
100
100
100
100
Test 2
5
10
15
20
25
300
300
300
300
300
150
150
150
150
150
5
300
200
25
h2-y
(mm)
EAA 204 – LIGHT STRUCTURE
Test 3
Specimen:
Steel
10
15
20
25
300
300
300
300
200
200
200
200
W
(N)
b
(mm)
a
(mm)
h1
(mm)
h2
(mm)
h3
(mm)
y
h  h 
3
1
2
(mm)
Test 1
5
10
15
20
25
300
300
300
300
300
100
100
100
100
100
Test 2
5
10
15
20
25
300
300
300
300
300
150
150
150
150
150
26
h2-y
(mm)
EAA 204 – LIGHT STRUCTURE
Test 3
5
10
15
20
25
300
300
300
300
300
200
200
200
200
200
Specimen:
Brass
W
(N)
b
(mm)
a
(mm)
h1
(mm)
h2
(mm)
h3
(mm)
y
h  h 
3
1
2
(mm)
Test 1
5
10
15
20
25
300
300
300
300
300
100
100
100
100
100
Test 2
5
10
15
20
25
300
300
300
300
300
150
150
150
150
150
5
10
15
20
300
300
300
300
200
200
200
200
Test 3
27
h2-y
(mm)
EAA 204 – LIGHT STRUCTURE
25
300
200
GRAPHS:
1. (h2 – y) against load (W) for tests 1, 2 and 3 for each specimen (in a single graph).
2. 150a2 against the gradient of the above graphs (in a single graph).
CALCULATION:
DISCUSSION AND CONCLUSION:
28
EAA 204 – LIGHT STRUCTURE
S4 – UNSYMMETRICALLY LOADED CANTILEVER EXPERI MENT
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
29
EAA 204 – LIGHT STRUCTURE
OBJECTIVES
1. To determine the deflection of a cantilever specimen
subjected to unsymmetrical load at different angles
2. To compare the theoretical and experimental values for
deflection and flexibility of the cantilever specimen.
INTRODUCTION AND THEORY
When an elastic structural member in equilibrium is
subjected to a load system, deformation will occur in that
structural member. The point where the load is applied will
displace. In general, the deformation can be expressed
using the following expression:
30
EAA 204 – LIGHT STRUCTURE
δ
=
f
×
P
..........Eq.
(1)
where
δ
f
P
=
=
=
Deflection
Flexibility of Structure
Applied Load System
For a cantilever structure, when a load is applied at the
free end, the corresponding deflection at the free end of
the structure is given by the formula:-
 
PL3
3EI
..........Eq.
(2)
where
δ
L
E
P
I
=
=
=
=
=
Deflection at the end (mm)
Length of the specimen (mm)
Young Modulus (N/mm²)
Applied Load (N)
Moment of Inertia (mm4)
Lets consider the following figure. A load P is applied at
the end of cantilever specimen with an angle θ, from y axis
x
d
b
90o y
31
EAA 204 – LIGHT STRUCTURE
Py = P sin θ
θ
P
Px = P cos θ
o
0
X
The load component in x axis direction is
Px
=
P cos θ
..........Eq.
(3)
The load component in y axis direction is
Py
=
P sin θ
..........Eq.
(4)
Therefore, deflection in x axis direction is
x 
Px L3
3EI yy
..........Eq.
(5)
And deflection in y-axis direction is
y 
Py L3
..........Eq.
3EI xx
(6)
APPARATUS
1. Cat. No. 151 Unsymmetrically Loaded Cantilever (Norwood
Instruments Ltd.) with Specimen and Two Dial Indicator
Gauges
2. Load Cord Hook and Load Hanger
3. Set of Loading Weights
4. Measuring Tape
32
EAA 204 – LIGHT STRUCTURE
PROCEDURE
1. Measure the size of the specimen section (width and
thickness) and the length between the end caps and enter
the measured data thereon in the given table. The length
of the specimen is measured without discarding the
specimen from the apparatus.
2. Place the hook wire ‘S’ in the hole. Set the top ring at
00 and zero the dial gauges. Dial Gauge 1 is the dial
gauge that touches the thickness of the specimen and Dial
Gauge 2 is the one that touches the width of the
specimen.
3. Place the load hanger on the hook wire ‘S’ and apply a
load (5N) on it. The weight of the load hanger (1N) is
not negligible and must be considered in the total
applied load. Record the readings of both dial gauges.
4. Rotate the top ring to different angles at 150 intervals
around the specimen until 1800 with a constant load.
Record the readings of both dial gauges for every angle.
5. Repeat steps 1 to 4 with a constant load of 10N (total
applied load = 11N).
RESULT
1. Plot the test results on a four quadrant axis, using the
gauge readings. Then on the same axes, plot the calculated
values of δy and δx using Eq. (5) and Eq.(6). Mark the
points in a different manner to differentiate between the
experimental results and calculated values.
2. For each of the data points where you have a calculated
δx and δy, calculate the ratio of δx/Px and δy/Py. Plot
δy/Py versus δx/Px.
DISCUSSION AND CONCLUSION
33
EAA 204 – LIGHT STRUCTURE
From the plot,
1. Compare two sets of results. How well do they agree?
2. Where do the maximum and minimum values of deflection
occur for the x and y direction?
From the plot of δy/Py against δx/Px, what do you notice
about the shape of the plot?
S4 – UNSYMMETRICALLY LOADED CANTILEVER EXPERI MENT
FORM
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
Name:
_____________________________________________________________________
34
EAA 204 – LIGHT STRUCTURE
Matric No.: _______________________
Group: _______________________
Date of Experiment: _________________________
RESULTS:
Specimen: Aluminum
1
2
Length, l
(mm)
Width, b
(mm)
Thickness,
d (mm)
Constant Load
Weight of Load Hanger
Total Applied Load
= 5 N ; 10 N
= 1 N ; 1 N
= 6 N ; 11 N
Dial Gauge 1 Reading without Loading = 0
Dial Gauge 2 Reading without Loading = 0
Value of Ixx and Iyy:
bd 3
=
I xx 
12
________ mm4
b3d
=
12
________ mm4
I yy 
a)
Calculation Sheet
Load = 5N
35
3
Average
EAA 204 – LIGHT STRUCTURE
Angle
θ
(o)
Dial
Dial
Px =
Gauge 1
Gauge 2
Pcosθ
δx
δy
(x0.01mm) (x0.01mm)
0
15
30
45
60
75
90
105
120
135
150
165
180
Load = 10N
36
Py =
Psinθ
δx
δy
(Theory) (Theory)
(mm)
(mm)
EAA 204 – LIGHT STRUCTURE
Angle
θ
(o)
Dial
Dial
Px =
Gauge 1
Gauge 2
Pcosθ
δx
δy
(x0.01mm) (x0.01mm)
Py =
Psinθ
δx
δy
(Theory) (Theory)
(mm)
(mm)
0
15
30
45
60
75
90
105
120
135
150
165
180
b)
Ratio of δx/Px and δy/Py for a constant load of 5N
(Theory)
37
EAA 204 – LIGHT STRUCTURE
δx (mm)
δy / Px (mm/N)
δy (mm)
δy / Py (mm/N)
c)Ratio of δx/Px and δy/Py for a constant load of 5N
(Experiment)
38
EAA 204 – LIGHT STRUCTURE
δx (mm)
δx/Px (mm/N)
δy (mm)
39
δy/Py (mm/N)
EAA 204 – LIGHT STRUCTURE
c) Ratio of δx/Px and δy/Py for a constant load of 10N (Theory)
δx (mm)
δy / Px (mm/N)
δy (mm)
40
δy / Py (mm/N)
EAA 204 – LIGHT STRUCTURE
e)Ratio of δx/Px and δy/Py for a constant load of 10N
(Experiment)
δx (mm)
Graphs:
δx/Px (mm/N)
δy (mm)
δy/Py (mm/N)
1. Deflection in x (δx)and y (δx) direction versus
angle θ
2. δy/Py against δx/Px (for a constant load of 5N
and 10N (theory and experiment)
DISCUSSION AND CONCLUSION:
41
EAA 204 – LIGHT STRUCTURE
42
EAA 204 – LIGHT STRUCTURE
S5 – TENSILE TEST
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
43
EAA 204 – LIGHT STRUCTURE
OBJECTIVE
To study the behaviour and properties mechanical of mild
steel specimen subjected to tensile load.
APPARATUS
1.
2.
3.
4.
5.
Tensile Testing Machine, SM100
Extensometer, SM100d
Pump, SM100k
Digital Load Gauge, E102
Tensile Test Specimen
INTRODUCTION
The understanding of the behavior and mechanical properties
of the material used in the design of machines and
structures is essential. The way to determine how materials
behave when they are subjected to loads is to perform
experiments in the laboratory. The usual procedure is to
place small specimens of the material in testing machines,
apply
the
loads,
and
then
measure
the
resulting
deformations (such as changes in length and changes in
diameter).
44
EAA 204 – LIGHT STRUCTURE
THEORY
Tensile test results are normally converted to
stresses and strains and plotted as stress-strain diagram.
This diagram which represents the relation between stress
and
strain
for
a
given
material
is
an
important
characteristic of the material.
The axial stress  in a test specimen is calculated by
dividing the axial load P by the cross-sectional area A,
 = P/A
The loading applied to the specimen will cause
deformation in the specimen. The deformation per unit
length of the specimen is known as strain (). Therefore,
 = changes in length/initial length
The value of the strain is normally very small and it is
usually expressed in percentage.
Mild steel is one of the most widely and is found in
buildings,
bridges,
cranes,
ships,
towers,
vehicles
and
many other types of constructions. A stress strain diagram
for
a
typical
structural
steel
in
tension
is
shown
in
Figure 1. Strains are plotted on the horizontal exis and
stresses on vertical axis.
The diagram begins with a straight line from the
origin O to point A, which means that the relationship
between stress and strain in this initial region is not
only linear but also proportionality between stress and
strain no longer exists; hence the stress at A is called
the proportional limit. For low carbon steels, this limit
is in the range 210 to 350 kPa but high strength steels can
have proportional limits of more than 550 kPa. The slope of
the straight line from O to A is called the modulus of
elasticity.
45
EAA 204 – LIGHT STRUCTURE
Figure 1 : Stress- strain diagram for a typical structural
steel in tension (not to scale)
With an increase in stress beyond the proportional
limit, the strain begins to increase more rapidly for each
increment in stress. Consequently, the stress- strain curve
has a smaller and smaller slope, until at point B the curve
becomes horizontal. Beginning at this point, considerable
elongation of the test specimen occurs with no noticeable
increase in the tensile force (from B to C). This
phenomenon is known as yielding of the material, and point
B is called the yield point. The corresponding stress is
known as the yield stress of the steel. In the region from
B to C,the material becomes perfectly plastic, which means
that it deforms without an increase in the applied load.
The elongation of a mild steel specimen in the perfectly
plastic, which means that it deforms without an increase in
the applied load. The elongation of a mild steel specimen
in the perfectly plastic region is typically 10 to 15 times
the elongation that occurs in the linear region (between
the onset of loading and proportional limit). The presence
of very strains in the plastic region ( and beyond) is the
reason for not plotting this diagram to scale.
After undergoing the large strains that occur during
yielding in the region BC, the steel begins to strain
harden. During strain hardening, the material undergoes
changes in its crystaline structure, resulting in increased
resistance
of
the
material
to
further
deformation.
Elongation of the test specimen in this region requires an
increase in the tensile load, and therefore the stressstrain diagram has a positive slope from C to D. The load
46
EAA 204 – LIGHT STRUCTURE
eventually reaches its maximum value, and the corresponding
stress (at D point) is called the ultimate stress. Further
stretching of the bar is actually accompanied by a
reduction in the load, and fracture finally occurs at a
point such as E in Figure 1.
The yield stress and ultimate stress of a material are
also called the yield strength and ultimate strength,
respectively. Strength is a general term that refers to the
capacity of a structure to resist loads. For instance, the
yield strength of a beam is the magnitude of the load
required to cause yielding in the beam, and the ultimate
strength of a
truss is the maximum load it can support,
that is, the failure load. However, when conducting a
tension test of a particular material, we define load
carrying capacity by the stresses in the specimen rather
than by the total loads acting on the specimen. As a
result, the strength of a material is usually started as a
stress.
When a test specimen is stretched, lateral
contraction occurs, as previously mentioned. The resulting
decrease in cross-sectional area is too small to have a
noticeable
effect
on
the
calculated
values
of
the
stresses`up to about point C in figure 1, but beyond that
point the reduction in the area begins to alter the shape
of the curve. In the vicinity of the ultimate stress, the
reduction in area of the bar becomes clearly visible and a
pronounced necking of the bar occurs (see Figure 2). If the
actual cross-sectional area at the narrow part of the neck
is used to calculate the stress, the true stress-strain
curve ( the dashed line CE’ in Figure 1) is obtained. The
total load the bar can carry does indeed diminish after the
ultimate stress is reached ( as shown by curve DE), but
this reduction is due to the decrease in area of the bar
and not to a loss in strength of the material itself. In
reality, the material withstands an increase in true stress
up in failure (point E’). Because most structures are
expected to function at stresses below the proportional
limit, the conventional stress-strain curve OABCDE, wich is
based upon the original cross sectional area of the
specimen and is easy to determine, provides satisfactory
information for use in engineering design.
47
EAA 204 – LIGHT STRUCTURE
Figure 2 : Necking of a mild steel bar in tension
The
diagram
of
Figure
1
shows
the
general
characteristics of the stress-strain curve for mild steel,
but its proportions are not realistic because, as already
mentioned, the strain that occurs from B to C may be more
than 10 times the strain occuring from O to A. Furthermore,
the strains from C to E are many times greater than those
from B to C. The correct relationships are portrayed in
Figure 3, which shows a stress-strain diagram for mild
steel drawn to scale.
In this figure, the strains from zero point to point A
are so small in comparison to the strains from A to point E
that they cannot be seen, and the initial part of the
diagram appears to be a vertical line.
48
EAA 204 – LIGHT STRUCTURE
Figure 3 : Stress –strain diagram for atypical structural
steel in tension (not to scale)
The presence of a clearly defined yield point followed
by lage plastic strains is an important characteristic of
structural steel that is sometimes utilized in practical
design. Metal such as mild steel that undergo large
permanent strains before faiure are classified as ductile.
For instance, ductility is the property that enables a bar
of steel to be bent into a circular arch or drawn into a
wire without breaking. A desirable feature of ductile
materials is that visible distortions occur if the loads
become too large, thus providing an opportunity to take
remedial action before an actual fracture occurs. Also,
materials exhibiting ductile behavior are capable of
absorbing large amounts of strain energy prior to fracture.
When material such as aluminium does not have an
obvious yield point and yet undergoes large strains after
the proportional limit is exceeded, an arbitiary yield
stress may be determined by the offset method. A straight
line is drawn on the stress- strain diagram parallel to the
innitial linear part of the curve ( Figure 4) but offset by
some standard strain, such as 0.002 (or 0.2%).
49
EAA 204 – LIGHT STRUCTURE
Figure 4 : Arbitiary yield stress determined by the offset
method
The intersection of the offset line and the stressstrain curve (point A in the Figure 4) define the yield
stress is determined by an arbitrary rule and is not an
inherent physical property of the material, it should be
distinguished from a true yield stress by refering to it as
the offset yield stress. For a material such as aluminium,
the offset yield stress is slightly above the proportinal
limit. In the case of structural steel, with its abrupt
transition from linear region to the region of plastic
stretching, the offset stress is essentially the same as
both the yield stress and the proportional limit.
The ductility of a material in tension can be
characterized by its elongation and by the reduction in
area at the cross section where fracture occurs. The
percent elongation is defined as follows:Percent elongation =
in
which
Lo is
the
L1  Lo
(100)
Lo
original
gage
length
and
L1
is
the
distance between the gage marks at fracture. Because the
elongation is not uniform over the length of the specimen
50
EAA 204 – LIGHT STRUCTURE
but is concentrated in the region of necking, the percent
elongation depends upon the gage length. Therefore, when
the
percent
of
elongation
depends
upon
the
gage
length
should always be given. For a 2 in. Gage length, steel may
have an elongation in the range from 3% to 40%, depending
on composition; in the case of structural steel, values 20%
to
30%
varies
are
from
common.
1%
to
The
45%,
elongation
depending
of
upon
aluminium
alloys
composition
and
treatment.
The percent reduction in area measures the amount of
necking that occurs and is defined as follows:-
Percent reduction in area =
A0  A1
(100)
A1
in which Ao is the original cross sectional area at fracture
section. For ductile steels, the reduction is about 50%.
Materials that fail in tension at relatively low
values of strain are classified as brittle. Examples are
concrete, stone, cast iron, glass, ceramics, and a variety
of metalic alloys. Brittle materials fail with only little
elogation after the proportional limit ( the stress at
point in Figure 5) is exceeded. Furthermore, the reduction
in area is insignificant, and so the nominal fracture
stress (point B) is the as the true ultimate sress. Highcarbon steels have very high yield stresses – over 700Mpa
in some cases – but they behave in a brittle manner and
fracture occurs at an elongation of only a few percent.
51
EAA 204 – LIGHT STRUCTURE
Figure 5
material
:
Typical
stress-
strain
diagram
for
brittle
showing the proporional limit (point A) and
fracture stress (point B)
52
EAA 204 – LIGHT STRUCTURE
PROCEDURE
1. Measure the diameter of the specimen at few locations in
the region of extensometer and calculate the crosssectional area.
2. Mark a gauge length on the specimen. These are the points
where the extensometer arms are attached to the specimen.
Install the test specimen in between the two large grips
of the testing machine and extensometer to the specimen.
3. Load the specimen in tension slowly and uniformly. Record
the value of elongation at every 2 kN load increment in
elastic regi1on. When reaching yield point, more reading
should be taken.
4. Uninstall the extensometer once the yield stress is
reached. Install the digital load gauge at ‘Peak Hold’.
Increase the load slowly and uniformly until the specimen
fracture. Record the ultimate load and fracture load.
RESULT
1. Calculate stresses and strains and plot stress-strain
diagram. ( = extensometer reading/(3 x 50)).
2. Determine the value for modulus of elasticity, yield
stress and ultimate tensile stress.
3. Calculate the percentage of reduction in cross-sectional
area and in length.
DISCUSSION AND CONCLUSION
4. From the test results and stress-strain diagram, discuss
the characteristic of the mild steel specimen as shown in
the experiment. Make conclusion about the modulus of
elasticity, yield stress and ultimate tensile stress.
53
EAA 204 – LIGHT STRUCTURE
S5 – TENSILE TEST FORM
SCHOOL OF CIVIL ENGINEERING
UNIVERSITI SAINS MALAYSIA
ENGINEERING CAMPUS
Name:
_____________________________________________________________________
Matric No.: _______________________
Group: _______________________
Date of Experiment: _________________________
RESULTS:
Before Rupture
Diameter (mm)
Length, Lo (mm)
Cross-Sectional Area, Ao
(mm2)
54
After Rupture
EAA 204 – LIGHT STRUCTURE
Mechanical Properties of Mild Steel Specimen
Yield Stress
Tensile Strength
Modulus of Elasticity
% Elongation 
Total Length
x 100
Initial Length
% Reduction in Cross-sectional Area
Cross - Sectional Area At Rupture

100
Initial Cross - Sectional Area
Load F
(kN)
Stress, σ = F/A0
(x 106 Nm-2)
Elongation, ΔL
(x 10-2 mm)
55
Strain,
ε = ΔL/L0 x 10-4
EAA 204 – LIGHT STRUCTURE
CALCULATION:
Graph: Stress versus Strain
56
EAA 204 – LIGHT STRUCTURE
DISCUSSION AND CONCLUSION:
57