INTRODUCTION
TITLE
Deflection of a Simple Supported Beam
OBJECTIVE :
1) To observe the deflection of a simple supported beam with variable loads.
2) To find the relationship between the deflection of a simple supported beam and the
variable length of the beam.
Introduction :
A beam is a length of material supported at its two ends, in such a way so as to bear loads. The load perpendicular to its longitudinal axis will result in bending and, in most cases, transverse shearing. In the simplest of situations, the beam is taken to have a rectangular cross-section and the loads and supporting reactions act in the vertical plane containing the longitudinal axis. The loads and the reactions at the supports are considered external forces and they must be in equilibrium for the entire beam to be in equilibrium. To study the strength of the beam, it is necessary to know how these external forces affect it. As in the theory appendix, the mathematical approach is to assume that an imaginary transverse section cuts the beam into two parts, and then to examine the equilibrium of each part. To maintain the equilibrium, certain forces must be introduced at the point of cut. When the cut is not present, these forces must continue to exist internally in the material of the beam. Before the slope or the displacement at a point on a beam (or shaft) is determined, it is often helpful to sketch the deflected shape of the beam when it is loaded, in order to “visualize” any computed result and thereby partially check these results. The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve. For most beams the elastic curve can be sketched without much difficulty. When doing so, however, it is necessary to know how the slope or displacement is restricted at various types of supports. In general, supports that resist a force, such as a fixed wall, restrict displacement, and those that resist a moment, such as a fixed wall, restrict rotation or slope. Due to the nature characteristic of the materials, when a force acting on a long beam, the force will cause the beam to bend. If the force is acting downwards, the moments at the two ends will act upwards and same for the opposite way.
Theory :
In this experiment, we will examine the deflection of a simply beam subjected to an increasing point load. We will also vary the beam length by changing the distance between the supports. This means we can find out the relationship between the deflection and the length of the beam.
From the figure 1, we have:
1 = - ( (1)
← y
If the material is homogeneous and behaves in a linear-elastic manner, then ( = ( / E. Also, since the flexure formula applies, ( = - My/L. Combining these equations and substituting into equation (1), we have,
1 = M (2)
( EI
Where, ( = the radius of curvature at a specific point on the elastic curve (1/( is
Referred to as the curvature)
M = the internal moment in the beam at the point where ( is to be determined
E = the material’s modulus of elasticity
I = the beam’s moment of inertia computed about the neutral axis
The elastic curve for a beam can be expressed mathematically as v = f(x). To obtain this equation, we must first represent the curvature (1/p) in terms of v and x. In most calculate books, it is shown that this relationship is
1 = d2v/dx2 (3)
( [1 + (dv/dx)2]3/2
substituting equation (2) into equation (3), we get
d2v/dx2 = M (4)
[1 + (dv/dx)2]3/2 EI
The above equation gives a non-linear second-order differential equation. Its solution, which is called the elastic, gives the exact shape of the elastic curve, assuming, of course, that beam deflections occur only due to bending.
In order to solve most of the deflection problems, equation (4) can be modified by written as :
d2v = M (5)
dx2 EI
P
A B
P/2 P/2
Figure 2 : The apparatus in this experiment
x
M
P/2
Figure 3 : Segment AB
+ (M = 0 , M – Px/2 = 0
M = Px/2
From equation 5, EI d2v/dx2 = M
= Px/2
EI dv/dx = Px2 / 4 + C1
EIv = Px3 / 12 + C1x + C2 (6)
The constant of integration are obtained by applying the boundary condition v = 0 at x = 0,
and the symmetry condition that dv/dx = 0 at x = L/2. This leads to
C1 = PL2 / 16 and C2 = 0
Substitutes inside equation (6),
Hence, EI v = (Px3 / 12 + PL2x / 16
The maximum deflection happened at x = L/2, we have
EIv = P(L/2)3 / 12 ( PL2(L/2) / 16
EIv = PL3 / 96 ( PL3 / 32
EIv = ( PL3 / 48 (7)
Modifying equation (7) that y-axis is positive downwards, and P = W in this experiment,
EIv = WL3 / 48
v = WL3 (8)
48 EI
where W = Load (N);
L = Distance between the two support (m);
E = Young’s modulus for cantilever material (Nm-2);
I = Second moment of area of the cantilever (m4)
In this experiment,
I = bd3 / 12
where, b = width (m)
d = depth (m)
Apparatus :
1. TQ Deflection of Beams
2. Cantilevers
3. Aluminium beam
4. Steel beam
5. Brass beam
6. Vernier gauge
7. Knife-edge
8. Digital dial test indicator
9. Variable loads
Procedure :
Part 1
Figure 3: Simply supported beam set-up and
schematic (fixed beam with variable load)
1. A vernier gauge is used to measure the width and the depth of the mild steel test beam. The values are recorded next to the results table and they are used to calculate the second moment of the area, I by using the formula given in the theory above.
2. Any clamps from the backboard are removed. The length L is set to 400 mm, and the beam is set up as shown in figure 3.
3. The digital dial test indicator is slide into position on the beam and is locked by using the thumbnut at the rear.
4. A knife-edge hanger is slide to the position shown. The frame is tapped lightly. The digital dial test indicator is set to zero by using the ‘origin’ button.
5. Masses are applied to the knife-edge hanger in the increments from 0 to 500 g.
6. The frame is tapped lightly each time, and the digital dial test indicator readings are recorded for each increment of mass. Theoretical deflection is computed too.
7. All the readings are recorded into a table.
8. A graph of Deflection versus Applied Mass for a simply supported beam is plotted according to the readings from the tables.
9. The procedures of 1 – 8 are repeated by substituted the brass test beam for mild steel
test beam.
Part 2
Figure 4 : Simply supported beam set-up and
schematic (fixed beam load with variable length)
1. The beam is set up with the length (that is, distance between knife-edge supports) at 200 mm. The digital dial test indicator and load hanger is ensured that still central to the beam, as shown in figure 4.
2. The frame is tapped lightly and the digital dial test indicator is set to be zero using the ‘origin’ button. A 500 g mass is applied and the deflections are recorded.
3. The procedure is repeated for each increment of beam length. All the readings are recorded into a table.
4. A graph Deflection versus length, L3 is plotted.
5. The procedures of 1 – 4 are repeated by substituted the brass test beam for mild steel
test beam.
RESULT:
Part 1
Steel
Modulus of Elasticity, E = 207 GNm-2
Width b = (18.9+19.0+18.9) / ( mm
= 18.93 mm
Depth d = ((.5+(.4+(.5) / 3 mm
= (.47 mm
I = m4
= 6.59 x 10-11 m4
|Mass (g) |Actual deflection |Actual deflection |Actual deflection |Theoretical deflection (mm) |
| |(mm) |(mm) |(mm) | |
| |1 |2 |Average | |
|0 |0 |0 |0 |0 |
|100 |0.08 |0.10 |0.090 |0.096 |
|200 |0.21 |0.22 |0.215 |0.191 |
|300 |0.34 |0.34 |0.340 |0.288 |
|400 |0.46 |0.46 |0.460 |0.384 |
|500 |0.59 |0.58 |0.585 |0.479 |
Steel
From the graph deflection versus applied mass,
Gradient, kS = (0.585 – 0.340) / (500 – 300)
= 1.225 x 10-3 mm/g
Brass
Modulus of Elasticity, E = 105 GNm-2
Width b = (19.2+19.2+19.2) / 3 mm
= 19.2 mm
Depth d = (3.2+3.3+3.3) / 3 mm
= 3.27 mm
I = m4
= 5.59x 10-11 m4
|Mass (g) |Actual deflection (mm) |Actual deflection (mm)|Actual deflection (mm)|Theoretical deflection (mm) |
| |1 |2 |Average | |
|0 |0 |0 |0 |0 |
|100 |0.26 |0.25 |0.255 |0.223 |
|200 |0.55 |0.47 |0.510 |0.446 |
|300 |0.79 |0.75 |0.770 |0.669 |
|400 |1.05 |1.05 |1.050 |0.891 |
|500 |1.30 |1.29 |1.295 |1.114 |
Brass
From the graph deflection versus applied mass,
Gradient, kBr = (1.295 – 0.510) / (500 – 200)
= 2.62 x 10-3 mm/g
Part 2
Steel
W = 0.5 ( 9.81
= 4.905 N
|Length,L (mm) |L3 (x10-3m3) |Actual |Theoretical Deflection, (mm) |
| | |Deflection, (mm) | |
|200 |8.00 |0.09 |0.06 |
|250 |15.63 |0.17 |0.12 |
|300 |27.00 |0.28 |0.20 |
|350 |42.88 |0.43 |0.32 |
|400 |64.00 |0.58 |0.48 |
|450 |91.13 |0.93 |0.68 |
|500 |125.00 |1.31 |0.94 |
Steel
From the graph deflection versus length of beam,
Gradient, kS = (0.93 – 0.17) / (450 – 250)
= 3.80 x 10-3
Brass
W = 0.5 ( 9.81
= 4.905 N
|Length,L (mm) |L3 (x10-3m3) |Actual Deflection, (mm) |Theoretical Deflection, (mm) |
|200 |8.00 |0.15 |0.14 |
|250 |15.63 |0.30 |0.27 |
|300 |27.00 |0.57 |0.47 |
|350 |42.88 |0.92 |0.75 |
|400 |64.00 |1.29 |1.11 |
|450 |91.13 |1.88 |1.59 |
|500 |125.00 |2.64 |2.18 |
Brass
From the graph deflection versus length of beam,
Gradient, kS = (1.88 – 0.30) / (450 – 250)
= 7.90 x 10-3
Discussion :
For part 1, from the results above, we can conclude that the deflection increased as the applied mass increased for all the beams (aluminium beam, brass beam and steel beam) . But, the values we gained from the experiment (actual deflection) were different from the theoretical calculations (theoretical deflection).
For part 2, there is logical that when the length of the beam is increasing, the deflection of the beam will increase too. It means that the longer length of the beam will cause the greater of the deflection of the beam.
The value for actual deflection and theoretical deflection are different by a small value. The factors that would cause the errors above are as below :
1. Temperature.
Temperature is one of the main factor that cause the errors because high temperature would cause the specimen to expand, and soften the specimen. In most of the factory, the theoretical Young’s modulus is obtained on 28(C experimentally.
2. Oxidation layer on the beams’ surface.
The second factor is the oxidation layer on the beams’ surface that might harden the specimens. For example, Aluminium oxide is harder than Aluminium. The layer might affect the deflection too. So the accuratecy of the experiment is not very high.
3. The level of the beams.
The third factor is the level of the beams. If the beam is imbalance. The reactions occurred on the two supports is different. So, the moment on the segment will be affected. Therefore, the equation can not applied under this circumstance.
4. The impurity of the cantilevers.
The forth factor is the impurity of the cantilevers that may cause the change of the Young’s modulus. The bigger or smaller substance, or holes inside the cantilever may soften or harden it. Then, the deflection may also affected too.
5. Length of the cantilevers.
The fifth factor is the length of the cantilever.If the length of the cantilever is too long and the the distance between the two supports is short, the end of the beam will react on the endwise because of the force applied by the weight of the beam. The extract forces might affect the deflection.
The apparatus can be modified to obtain a better results: -
1. The weight for each mass is too small. If there is a small error occurred on each mass, there will become a big error when they are gathered.
So, to solve the problem, larger weight of mass(100g @ 200g each) should be used.
2. A level should added to the backboard and the backboard is movable so that we can adjust the two supports so that they are at the same level before starting the experiment.
3. Two more supports should added to the backboard to balance the weight occurred on the endwise of the beam.
Examples
In real life, the study of deflection is very useful. In some cases, this type of bending is desirable. For examples, in constructing bridges and buildings. A beam is used as a supporter in building a bridge. The beams are bent upwards before it used to build bridges. The beam can support more loads on the curve surface. The deflection of beam is also used in weighing heavy loads such as heavy vehicles. The stress occurred on the beam is converted to weight. In sports, there are a lot of activities which using the characteristic of deflection. For examples, in diving. The diver uses the characteristic of elastic of the diving board to rebound. And in gymnasium, a participant uses the characteristic of deflection of the beam so that he can turn 360( for few rounds. In the design of vehicles, beams suspend the wheels of vehicles so that the wheels are movable up and down. In some cases, the deflection of beams is undesirable. For examples, in rotating beams such as the shaft of the engine. The deflection of the beam is highly deniable. The deflection of the beam may cause the beam rotates out of its axis and cause the rotation unstable.
Conclusion :
From the experiment, we can conclude that the deflection of aluminium , brass , and steel beams are proportional to the applied mass. But the deflection value for aluminium is biggest than the value for brass and steel. In the other words, the deflection increases as the applied mass increased.
Deflection, v ( M
where, M = mass
The deflection of beam is proportion to the power of three of length.
The deflection increases as the length increased.
Deflection, v ( L3
where, L = length
References :
1. Russell C. Hibbeler, (1997). ‘Mechanics of Materials’. Prentice Hall International, Inc.
2. William D Callister, JR.(1999). ‘Materials Science and Engineering an Introduction’, 4th edition. John Willey & Sons, Inc.
Graph 1 : Graph Deflection versus Mass Applied for aluminium
Graph 2 : Graph Deflection versus Applied Mass for brass.
Graph 3 : Graph Deflection versus Applied Mass for steel.
Graph 4 : Graph Deflection versus Beam Length for aluminium.
Graph 5 : Graph Deflection versus Beam Length for brass.
Graph 6 : Graph Deflection versus Beam Length for steel.
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