Aim
To understand the deflection of beam
Objective
1. Learn basic working of beam
2. Perform theoretical calculation for deflection of beam
3. Perform series of experiment with different material, shapes and load for beam
4. Compare the result and Discuss the finding
Introduction
Load supporting element in any structure that is installed in horizontal direction and take load perpendicular to its length is called a beam. Based on the way it is installed in structure, beam can be classified in to four main types. First simply supported beam which have supports at its both ends, second is overhanging beam which have length extended over its supports, third is overhanging beam which have more than two supports and fourth is cantilever beam which fixed only at one end and other end is free. There are three basic types of supports available for beams which includes fixed ends support, pined connection support and roller end support. From these three types any two types can be used to support single beam.
When load is applied on beam, its produce bending moment in beam by apply shear force across its cross section area. Beam reacts against this bending moment and shear force according to its moment of inertia. Moment of inertia of beam is its shape dependent property which shows the beam ability to resist bending moment applied on it. Another factor which help beam to resist bending is modulus of elasticity of beam. It’s the material related property of beam which defines the strain produce when a certain load is applied on it. Moment of inertia of beam and modulus of elasticity of beam material when combine define the beam reaction against the bending moment due to applied load. Another factor which affects the working of beam is its length. Greater the length of beam greater will be the bending moment produce due to external load.
Experimental Procedure
1. Selection one material from the four different material available and after that select the shape on which experiment will be conducted
2. Note down the materials Modulus of elasticity and moment of inertia of the selected shape
3. Select the number of weights and magnitude of each weight.
4. Fix the position of application of load
5. Use all the above information to find the reaction of beam theoretically
6. Use the information available to form the moment equation of beam
7. Use the moment equation of beam in Macaulay theory to drive the equation of slope and deflection
8. Use Equation of deflection to find the deflection of beam due to load at certain positions
9. Place the selected beam on apparatus and attach weight hanger and deflection measuring instrument
10. Apply load on beam by placing weight on hangers
11. Note the deflection of beam on selected points
12. Compare the theoretical and experimental results
13. Repeat the procedure for four different beams
Moment of inertia
I=1/12*b*h^3
I=1/12*19*〖6.4〗^3
I=415 mm^4
Modulus of elasticity of Aluminium
E=6.9 GPa=6.9*〖10〗^4 N/mm^2
So
E*I=6.9*〖10〗^4*415=2863*〖10〗^4 N/mm^2
Reactions of support
∑▒〖Fy=0〗
R1+R2=10+10
R1=20R2
∑▒〖M1=0〗
F1*250+F2*450=R2*length of beam
10*250+10*450=R2*700
R2=(10*250+10*450)/700
R2=10 N
Using
R1=20R2
R1=2010=10 N
Moment Equation
M=R1*xF1(x250)F2(x450)
According to Macaulay Theory
EI* (d^2 y)/(dx^2 )=M
So
EI* (d^2 y)/(dx^2 )= R1*xF1(x250)F2(x450)
Integrate
EI* dy/dx=〖10*x〗^2/2(10*(x250)^2)/2(10〖*(x450)〗^2)/2+A
Integrate
EI*y=(10*x^3)/6(10〖*(x250)〗^3)/6(10〖*(x450)〗^3)/6+B*x+A
Put initial boundary conditions of x=y=0
EI*0=(10*0^3)/6(10〖*(0250)〗^3)/6(10〖*(0450)〗^3)/6+B*0+A
As per Macaulay Theory ignore the negative terms inside brackets
A=0
Put initial boundary conditions of x=700 and y=0
EI*0=(10*〖700〗^3)/6(10〖*(700250)〗^3)/6(10〖*(700450)〗^3)/6+B*700+0
B*700= (10*〖700〗^3)/6+(10〖*(450)〗^3)/6+(10〖*(250)〗^3)/6
B=(571666666.7+151875000+26041666.67)/700
B= 519642.85
So equation of deflection is
EI*y=(10*x^3)/6(10〖*(x250)〗^3)/6(10〖*(x450)〗^3)/6519642.85*x
Beam 3
Moment of inertia
I=1/12*b*h^3
I=1/12*19*〖3.2〗^3
I=52 mm^4
Modulus of elasticity of Aluminium
E=10.5 GPa=10.5*〖10〗^4 N/mm^2
So
E*I=10.5*〖10〗^4*52=546*〖10〗^4 N/mm^2
Reactions of support
∑▒〖Fy=0〗
R1+R2=2
R1=2R2
∑▒〖M1=0〗
F1*250+F2*450=R2*length of beam
2*350=R2*700
R2=(2*350)/700
R2=1 N
Using
R1=2R2
R1=21=1 N
Moment Equation
M=R1*xF1(x350)
According to Macaulay Theory
EI* (d^2 y)/(dx^2 )=M
So
EI* (d^2 y)/(dx^2 )= R1*xF1(x350)
Integrate
EI* dy/dx=〖1*x〗^2/2(2*(x350)^2)/2 A
Integrate
EI*y=(1*x^3)/6(2〖*(x350)〗^3)/6+B*x+A
Put initial boundary conditions of x=y=0
EI*0=(1*0^3)/6(2〖*(0350)〗^3)/6+B*0+A
As per Macaulay Theory ignore the negative terms inside brackets
A=0
Put initial boundary conditions of x=700 and y=0
EI*0=(1*〖700〗^3)/6(2〖*(700250)〗^3)/6+B*700+0
B*700= (1*〖700〗^3)/6+(2〖*(350)〗^3)/6
B=(571666666.7+14291666.67)/700
B= 61250
So equation of deflection is
EI*y=(1*x^3)/6(2〖*(x350)〗^3)/661250*x
Results
beam 1


material

Aluminum


position of left support


500mm


position of right support


1200mm


load 1 (N)


10 N


position of load 1


250mm


load 2


X


position of load 2


X


Distance along length

EXP Deflection in mm

Calculation deflection mm


X

y

y


500

0.03

0.00


600

1.41

1.54


700

2.57

2.90


800

3.21

3.93


900

3.14

3.93


1000

2.37

2.90


1100

1.28

1.54


1200

0.02

0.00

beam 2


material

Aluminum


position of left support


500mm


position of right support


1200mm


load 1 (N)

10N


position of load 1


750mm


load 2

10N


position of load 2


950mm


Distance along length

Deflection in mm

Calculation deflection mm


X

y



500

0.03

0.00


600

2.31

1.76


700

4.26

3.16


800

5.45

3.88


900

5.51

3.73


1000

4.4

2.72


1100

2.44

1.01


1200

0.01

0.00

beam 3


material

Brass


position of left support


500mm


position of right support


1200mm


load 1 (N)

22 N


position of load 1


850mm


load 2 (n)

X


position of load 2


X


Distance along length

EXP Deflection in mm

Calculation deflection mm


X

Y



500mm

0.01

0.00


600mm

4.75

4.58


700mm

8.65

8.40


800mm

11.11

10.71


900mm

11.27

10.71


1000mm

9.02

8.40


1100mm

4.86

4.58


1200mm

0.02

0.00

Beam 4


material

steel


position of left support


500mm


position of right support


1200mm


load 1 (N)

1.16 N


position of load 1


750mm


load 2 (n)

1.16 N


position of load 2


950mm


Distance along length

EXP Deflection in mm

Calculation deflection mm


X

Y



500

0

0.00


600

5.43

4.61


700

9.95

8.30


800

12.82

10.17


900

12.84

9.78


1000

10.3

7.12


1100

5.6

2.64


1200

0.02

0.00

Discussion
From the experimental and
theoretical calculation made above it can be concluded that theoretical values
are always less than the experimental values. This is because theoretical
values are made with ideal cases ignoring many facts of real life, like damaged
apparatus, human error and human/machine limitations. Other facts which can be
concluded, the increase in modulus of elasticity of beam decrease deflection
and similar to that increase in moment of inertia of beam decreases the
deflection. Different material act differently during experiment.
Conclusion
The aim of this lab work
to study the deflection of beam has been completed successfully and four
different experiments have been conducted on three different materials with
four different shapes. At the end of this lab work it can be concluded that
increase in modulus of elasticity and moment of inertia decreases the
deflection where increase in number of loads, magnitude of load and distance of
load from ends increase the deflection of beam. Each beam show deflection based
on its modulus of elasticity and moment of inertia. It also can be concluded
that experimental values of deflection are always greater than calculated
values.
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