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Aim 

To determine the effect of compressive axial load due to transverse load on deflection of the beam 


Objection

1. Experimentally determine the deflection in beam with vertical end
2. Experimentally determine the deflection in beam with oblique end
3. Theoretically determine the deflection in beam with vertical end
4. Theoretically determine the deflection in beam with oblique end


Introduction


Beam
A structural member who is designed to resist the bending stresses produce in it when an external load is applied on it.

Simply supported beam
Types of beam which has support available at its both ends is called simply support beam

Overhanging beam
Types of beam which has support available at its both ends but has some of its area extended over one of its ends is called overhanging beam

Cantilever beam
Type of beam which has support available only at one of its ends and other ends is free is called cantilever beam

I beam
Type of beam which has cross section shape like English alpha bet I is called I beam



Procedure


Vertical end condition
1. first step in this experiment is to fix the vertical end of the apparatus with the beam
2. After that check the apparatus for any damages or errors present in it. if there is any of them make sure they are removed before experiment starts
3. Third step is attached hanger to the beam at its central point. This is important to attach the hanger in the middle as the apparatus has deflection measuring arrangement only at that point
4. After that add weight of 5 N and check the reading of the beam deflection
5. Repeat the above step for different weight by adding 10 N each time and check the deflection reading for each of them
6. When loaded up to 100 N, start removing weight and check deflection for that point too.

Oblique end condition
1. first step in this experiment is to fix the oblique end of the apparatus with opposite end of the beam
2. After that check the apparatus for any damages or errors produce due to the vertical end condition test. if there is any of them make sure they are removed before oblique end condition experiment starts
3. Third step is attached hanger to the beam at its central point. This is important to attach the hanger in the same spot as done for the vertical end condition test in order to ckect the true effect of oblique end condition
4. After that add weight of 5 N and check the reading of the beam deflection
5. Repeat the above step for different weight by adding 10 N each time and check the deflection reading for each of them.
6. When loaded up to 100 N, start removing weight and check deflection for that point too.


Readings


Force on beam
Loaded deflection mm oblique
Unloaded deflection  mm oblique
Loaded deflection mm vertical
Unloaded deflection  mm vertical
Average deflection oblique
Average deflection mm vertical
5
1
1.5
0
1
1.25
0.5
15
4
3.5
3
4
3.75
3.5
25
6
5.5
5
5
5.75
5
35
8
7.5
7
7
7.75
7
45
11
10
10
10
10.5
10
55
13
12.5
12
12
12.75
12
65
16
15
14
15
15.5
14.5
75
18
17
17
17
17.5
17
85
21
20.5
20
20
20.75
20
95
23.5
23.5
23
23
23.5
23
105
27
26
26
26
26.5
26



 Calculation

Support Reactions

sum of forces in x direction is equa to zero

A-F+B=0

A=F-B

sum of moment is equal to zero

A*0.8=F*0.4

B=(15*0.4)/0.8=7.5 N

A=F-B

A=15-7.5=7.5 N

Shear force
Shear force Before B

Shear force  = A=7.5 N

Shear force After b

SShear force  = A– F=7.5-15= -7.5 N


Bending moment
At A

Bending Moment = A*L =7.5*0 = 0 Nm

At B

Bending Moment = F*L =1 5*0.4 =6 Nm

At C

Bending Moment = B*L =75*0 = 0 Nm


Beam Deflection
moment of inertia=1/12 a*b^3

moment of inertia=1/12*50*〖125〗^3

moment of inertia=8*〖10〗^(-9)  m^4

modulus of elasticity=-(Force*〖Length〗^3)/(48*deflection*moment of inertia)

modulus of elasticity=-(15*〖0.8〗^3)/(48*0.003*8*〖10〗^(-9) )=6666666667 Pa=6.67 GPa

The actual value of modulus of elasticity for aluminium is 68 GPa which is far too much as compared to the experimental value of modulus of elasticity of aluminium which is only 6.67 GPa. The difference between values of modulus of elasticity is may be due to faults in apparatus or due to the errors made by operator during experimentation and observation.

Compressive force 

Compressive load when load is 5 N

C S =(Moment*b/2)/(moment of inertia)

C S =(6*0.0125/2)/(8*〖10〗^(-9) )=4687500 Pa

C S=Force/Area

4687500=Force/(5*125)

F=2929.68N


Moment equation
M=nornmal load bending+axial compressive load bending

Moment=F*L/2+Fc*deflection


M/EI=(〖 d〗^2 y)/(dx^2 )  

EI (〖 d〗^2 y)/(dx^2 )  =F*L+Fc*deflection

Integrating

dy/dx  =1/EI(F*L^2/2+Fc*deflection*L+C_1

Integrating


y =1/EI(F*L^3/6+Fc*deflection*L^2/2+C_1*L+C_2

Put x=0 y= 0

C_2=0

Put x = 0.4 y = 0


0=15*〖0.4〗^3/6+〖0.4*C〗_1

C1 =0.4

Deflection at 0.4 meter due to load of 15 N

70*〖10〗^9*8*〖10〗^(-9)*y =15*〖0.4〗^3/6+0.14*0.4+2929*y 〖0.4〗^2/2

560y-234y =0.216

y =0.216/326
y=6.63*〖10〗^(-4)  m
y =0.66 mm 



Discussion

The graph of deflection made against load is shown below and according to that graph the deflection of oblique end condition is much more than that of vertical end condition. According to graph the initial loading which is from 0 to 15 N does not show much difference in deflection but as the graph moves from 15 to 100 N load the deflection of oblique end condition increase much more rapidly than that of the vertical end condition. So from this it can be concluded that the axial compressive load due to oblique end condition has considerable effect on deflection of beam.


Aim

Aim of this lab experiment is to examine the bending of different shapes of timber supporting structure

Objective

  1. Experimentally determine the deflection in timber beam
  2. Experimentally determine the deflection in timber plank
  3. Calculate the second moment of inertia for timber beam
  4. Calculate the second moment of inertia for timber plank
  5. Study the effect of shape on deflection and second moment of inertia
  6. Study the deflection of three flitch beams
  7. Compare the result of three flitch beams


Theory


Beam
Beam is the simplest, most common and most important supporting structure member of anybody like buildings, bridges and vehicles. Beams provide support to the structures by resisting against the forces which are applied on that structure. Forces applied on beam try to bend the beam and produce deflection in its shape. Deflection produce in beam depends on its shape, material and size. There are many different types of beam based on their end supports and shapes. 

Based on the end support beams can be classified as follow
Simply supported beam
Cantilever beam
Over hanging beam

Based on the shape beams can be classified as follow
Square beam
Rectangular beam
I beam

Timber
Timber is processes wood that’s shaped like a plank or beam and they are used in construction of different structures like houses, bridges and vehicles. 

Deflection
Deflection of beam is change in its shape or movement of beam from its neutral position due to the application of forces on it. Beams are fixed at their both or at least on end and when force is applied on beam at any position (other than the fix end) displace the beam from the initial position in the direction of force, this displacement of beam is called deflection. Deflection of simply supported beam can be calculated as follow

δ=-(FL^3)/48EI

In above equation 
δ is the reaction or deflection of beam
F is force applied to the beam
E the modulus of elasticity of material 
I is second moment of the inertia of the beam

Second moment of inertia
Second moment of inertia is a geometric property of anybody which depends on its shape and size and defines the resisting ability of body when an external force is applied on that body.
Second moment of inertia of any rectangle can be calculated as follow

I=1/12*B*D^3

In above equation 
I is Second moment of inertia
B and D are dimensions of body as follow



Risk Assessment

Bending of timber beam to check the deflection of beam involves applying load on timber beam. Load is applied in this experiment with the help of weights which are added on hander attached to the beam. Beam which is made of timber is purchased from market which can have defects like crakes which are invisible from outside. When load is applied to the beam crakes may lead to quick failure of beam much before it is expected. Sudden failure of beam can cause damage to students performing the experiment. This can be avoided by simply using the safety precautions like applying load slowly and gradually. 

Methodology

Beam of timber beam and plank was done by following the procedure mention below.

1. First step of this experiment if to setup the apparatus which involve placing the apparatus on horizontal place and attaching beam on it. Both ends of beam will be fixed on opposite ends of the apparatus (beam should be kept straight and horizontal).

2. Second step is to inspect the apparatus for any kind of error like zero error. Necessary work should be done in order to remove the present error before starting the experiment.

3. Third step is to add weight on beam hanger slowly and gradually. Starting from 0.5 KN increase the load 0.5 KN every time until beam brakes. 

4. Note the deflection of beam for every increase of 0.5 KN and fill the provided sheet

5. Repeat the steps 1 to 4 for plank and three different shapes of flitch beam

Raw Results


1.      Joist Plank


JOIST (PLANK)
B
169mm
D
44mm

Load (kN)
Deflection (mm)
Left
Right
0.5
7.61
7.47
1
7.95
7.81
1.5
7.97
7.87
2
8.19
8.35
2.5
8.24
8.42
3
8.77
9.04
3.5
11.57
11.93
4
12.98
12.91


2.      Joist Beam


JOIST (BEAM)
B
48.5mm
D
163mm

Load (kN)
Deflection (mm)
Left
Right
1
1.48
1.32
2
2.8
2.64
3
4.21
4.03
4
5.56
5.37
5
7.21
7.01
6
8.54
8.37
7
9.96
9.89
8
11.38
11.74
9
12.95
12.63
10
1.85
1.98
11
3.71
3.94


I=1/12*B*D^3
I=1/12*48.5*〖163〗^3
I=1.75*〖10〗^(-5)  m^4

3.      Flitch A



FLITCH A
B
36.5mm
D
100mm
T
7mm
Weight
13.2kg

Load (kN)
Deflection (mm)
Left
Right
0.5
0.6
0.6
1
1.15
1.16
1.5
1.75
1.76
2
2.21
2.23
2.5
2.76
2.78
3
3.31
3.35
3.5
3.82
3.85
4
4.36
4.4
4.5
4.87
4.92
5
5.44
5.54




4.      Fitch B



FLITCH B
B
36mm
D
100mm
T
2.5mm
Weight
12.4kg

Load (kN)
Deflection (mm)
Left
Right
0.5
0.67
0.67
1
1.26
1.27
1.5
1.85
1.87
2
2.53
2.54
2.5
3.13
3.15
3
3.74
3.77
3.5
4.32
4.34
4
4.53
4.98
4.5
5.09
5.54
5
5.65
6.1




5.      Fitch C



FLITCH C
D
107mm
B
36mm
T
3.5mm
X
250mm
Y
100mm
Z
25mm
Weight
7.65kg

Load (kN)
Deflection (mm)
Left
Right
0.5
1.28
1.35
1
3.08
3.04
1.5
4.38
4.39
2
6.02
6.04
2.5
7.6
7.56
3
8.74
8.68
3.5
9.89
9.84
4
11.26
11.19
4.5
12.41
12.33
5
18.33
18.15

Calculations
1.      Second moment of inertia of Joist Plank

JOIST (PLANK)
B
169mm
D
44mm


I=1/12*B*D^3
I=1/12*169*〖44〗^3
I=1.2*〖10〗^(-6)  m^4


2.      Second moment of inertia Joist Beam


JOIST (BEAM)
B
48.5mm
D
163mm

I=1/12*B*D^3
I=1/12*48.5*〖163〗^3
I=1.75*〖10〗^(-5)  m^4




Discussion

As mention in theory section that deflection of beam depends on the size and shape of beam, the graphs made between deflection and load for beam and plank experimentally proves that theory. According graph between the displacement and load for plank show very small increase in displacement when load is increase but after 3 KN the displacement increased rapidly from 9 to 13 mm whereas the displacement of beam increase very gradually when load applied on it increased. Comparing the deflection of both plank and beam shows that to reach the highest value of 13 mm deflection plank takes 4 KN of load whereas beam takes about 9 KN of load which shows that beam has much more resistance against deflection as compared to plank. This is due to the fact that the second moment of inertia of beam is much more than that of plank due to which it deflection is much less than that of plank.

Graph made between displacement and load for three different flitch shapes show that shapes and size does effect the deflection of flitch. According to graph all shapes starts from same point but as the load increase from 0 to 1 KN the deflection made by shape C is much more as compared to shape A and B. Deflection made by shape A and B is almost the same as there is also almost same. As the load increase further till 5 KN the displacement made by shape C reaches maximum of 18 mm whereas for shape A and B it is only 6 mm. this show that shape A and B has much more second moment of inertia as compared to shape C.

Conclusion

Experiment has been performed to find the effect of shape of timber on the bending of the timber and according to the results obtain it can be said that the deflection of timber does depends on the size and shape of the timber. Comparison between plank and beam shows that beam has much ore resistance against deflection as compared to plank and comparison between different flitch shapes show that shape and size does effect the deflection.

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