ELASTICALLY COUPLED BEAM APPARATUS

cageysyndicateUrban and Civil

Nov 15, 2013 (3 years and 6 months ago)

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Factory & Works:

A
darsh Nagar

New Hardwar Road

Roorkee


247 667

Phone:

01332


273721

INSTRUCTIONAL MANUAL

OF

ELASTICALLY COUPLED
BEAM APPARATUS




ENGINEERING MODELS & EQUIPMENT

Regd. Office:

Mahavir Jain Market

7


Civil Lines, Post Box No.


13

Roorkee


247 667

Phone:

01332


275639, 273121 (Resi)

Fax:

01332


271621

E
-
mail:

enggmod@nde.vsnl.net.in
,
enggmodels@sancharnet.in


Visit us at:

www.enggmod.com

By:


2

ELASTICALLY COUPLED BEAM APPARATUS



CONTENTS:







Page No.


1.0

Theory








03


2.0

Objective








08


3.0

Apparatus








08


4.0

Suggested Experimental work





08


5.0

Results & Discussions






0
8


6.0

Sample Data Sheet







09


7.0

Precautions








09









3

ELASTI
CALLY COUPLED BEAM APPARATUS


1.0

THEORY:






















Let R
1
, R
2
, R
3

be the internal forces in the three suspension rods AD, BE and CF
respectively.

Lengths AD=BE=CF=L

Lengths AB=BC=DE=EF=L

Moment of
inertia of the cross section of the beam ABC and DEF = I

Young’s Modulus of elasticity of the material of the beam ABC and DEF= E

Modulus of elasticity of the material of rods AD, BE and CF =E
1
, E
2
,

E
3

respectively.

Cross sectional area of rods AD, BE and
CF= A
1
, A
2
, A
3

respectively.


Condition 1


When support B exists, the beam ABC becomes in
-
operative. The central deflection
at point E, of beam DEF due to load W at G and the upward force R
2

at E, relative to
the deflected positions of points D and F is gi
ven by

EI
L
R
EI
WL
y
6
96
11
3
2
3










(1)































































































A

B

C

R
1

R
2

R
3

A
1

A
2

A
3

E
1

E
2

E
3

D

E

F

L/2

L/2

L

W

L

D

F

E

Y
1

Y
2

Y
3

F
1

Y

2
3
1
Y
Y


D
1

E
1


4

The case is equivalent to simply supported beam supported at D and F subjected
to loads W and R
2

a
t G and E respectively. Now it is required to find the
deflection at the point E. the deflection shall be obtained by area moment theorem
for two different loading.

EI
WL
L
L
EI
WL
L
L
EI
WL
L
DD
3
16
5
2
3
3
1
2
8
3
2
3
2
1
2
3
2
8
3
2
2
1
'
































































































































































































































D

G

E

W


Y

F

L/2

L/2

L

Fig. (a)

B.M. D.

EI
WL
8
3

Fig. (b)

D

D'

E

F

E'


E"

W

L/2

3/2

L

Fig. (c)

D

D"

E
1


E
1
"

R
2

F

EI
L
R
2
2

B.M.D
.

Fig. (d)

Fig. (e)

R
1

R
2

R
3

E


5

EI
WL
L
L
EI
WL
E
E
EI
WL
DD
EE
24
3
4
1
2
1
"
'
32
5
2
'
'
3
3











Downwar
d deflection due to W



EI
WL
EI
WL
E
E
EE
EE
y
3
3
1
96
11
24
1
32
5
"
'
'
"













Also,

EI
L
R
DD
EE
EI
L
R
L
L
EI
L
R
DD
4
"
2
1
'
2
2
2
2
1
"
3
2
1
3
2
2









EI
L
R
L
L
EI
L
R
E
E
12
3
2
2
1
"
'
3
2
2
1
1








Upward deflection due to R
2

1
1
1
1
2
"
'
'
"
E
E
EE
EE
y





EI
L
R
EI
L
R
6
)
12
1
4
1
(
3
2
3
2







Net deflection y at point E=
EI
L
R
EI
WL
y
y
6
96
11
3
2
3
2
1







Let

elongation of suspension rod

AD =Y
1

elongation of suspension rod

BE =Y
2

elongation of suspension rod

CF =Y
3


then total deflection of point E relative to its original position is

EI
L
R
EI
WL
Y
Y
Y
6
96
11
2
3
2
3
3
1
2










(2)


Now, we know that
3
3
3
3
2
2
2
2
1
1
1
1
,
,
E
A
L
R
Y
and
E
A
L
R
Y
E
A
L
R
Y




Substituting the values of Y
1
, Y
2
, and Y
3

in Eq. (2)



6

EI
L
R
EI
WL
E
A
L
R
E
A
L
R
E
A
L
R
6
96
11
2
1
3
2
3
3
3
3
1
1
1
2
2
2















(3)

Also,
W
R
R
R



3
2
1







(4)

Taking moment of all the forces about F, we get

0
.
2
3
2
.
2
1









L
R
L
W
L
R

2
4
3
2
1
R
W
R










(5)

Now solving
Eq. (3), (4) and (5) we get the values of R
1
, R
2

and R
3.

From Eq. (3)
be substituting the values of R
1
' R
2

and R
3

from Eq. (4) and (5).


)
6
(
3
2
4
12
11
3
2
3
12
11
8
8
1
8
3
96
11
6
4
4
6
96
11
)
2
4
3
(
)
2
4
3
(
2
1
3
3
2
1
3
3
1
2
3
3
1
1
3
3
3
1
1
3
3
3
3
1
1
2
2
2
2
3
3
3
2
2
1
1
2
2
2
2



























































EI
L
K
K
K
EI
L
K
K
W
R
E
A
L
E
A
L
EI
L
W
E
A
WL
E
A
WL
EI
WL
EI
L
E
A
L
E
A
L
E
A
L
R
EI
L
R
EI
WL
E
A
L
R
W
R
W
E
A
L
R
W
E
A
L
R




Where,
3
3
3
2
2
2
1
1
1
,
,
E
A
L
K
E
A
L
K
E
A
L
K























EI
L
K
K
K
EI
L
K
K
W
W
R
3
3
2
1
3
3
1
1
3
2
4
12
11
3
4
4
3

























EI
L
K
K
K
K
EI
L
K
K
K
EI
L
K
K
K
W
3
2
4
12
11
3
3
2
3
12
3
4
3
3
2
1
3
2
1
1
3
3
2
1


7




























EI
L
K
K
K
EI
L
K
K
W
3
3
2
1
3
3
2
3
2
4
12
13
2
12
4






(7)

and R
3

= W
-

R
1

-

R
2






































EI
L
K
K
K
EI
L
K
K
W
EI
L
K
K
K
EI
L
K
K
W
W
3
3
2
1
3
3
1
3
3
2
1
3
3
2
3
2
4
12
11
3
2
3
2
4
12
13
2
12
4




































EI
L
K
K
K
EI
L
K
K
EI
L
K
K
EI
L
K
K
K
W
3
3
2
1
3
3
1
3
3
2
3
3
2
1
3
2
4
6
11
2
6
12
13
2
12
3
8
4
16
4
4























EI
L
K
K
K
EI
L
K
K
W
3
3
2
1
3
2
1
3
2
4
4
1
4
2
4





(8)

Condition II


When the upper end of the central suspension rod is attached to the centre of a
similar elastic beam.

T
his is achieved by removing the support at B, then beam AC will also deflect
due to the load R
2

applied at its center. Here the total deflection of point E relative
to its original position as given by Eq. (3) is equal to the elongation of member
BE+ the c
entral deflection of beam ABC and Eq. (3) will modified to


EI
L
R
EI
WL
E
A
L
R
E
A
L
R
EI
L
R
E
A
L
R
6
96
11
2
1
6
3
2
3
3
3
3
1
1
1
3
2
2
2
2

















(3A)

Now solution of Eqs. (3A), (4) and (5) will yield the values of R
1
, R
2

and R
3.























EI
L
K
K
K
EI
L
K
K
W
R
3
3
2
1
3
3
1
2
3
4
4
12
11
3
2






(9)



8























EI
L
K
K
K
EI
L
K
K
W
R
3
3
2
1
3
3
2
1
3
4
4
12
37
2
12
4






(10)

and





















EI
L
K
K
K
EI
L
K
K
W
R
3
3
2
1
2
1
3
3
4
4
12
5
4
2
4





(11)

Here,
AE
L
K

for any spring is the extension of the spring per unit load.


2.0

OBJECTIVE:


To calculate experimentally and theoretically the loads in the three suspension
rods supporting an elastic beam with a concentrate
d load hung midway between
two of the suspension rods under two conditions.


1.

When the suspension rods are attached at their upper ends to rigid supports.

2.

When upper end of the central suspension rod is attached to the centre of a similar
elastic beam.



3.0

APPARATUS:


Apparatus consists of a three parallel bar suspension system with elastic beam at
their upper and lower ends. The upper ends of the two outer suspension rods are
tied to a vertical wooden board while central suspension rod may be tied to the
c
enter of another elastic beam supported at two outer ends only.


4.0

SUGGESTED EXPERIMENTAL WORK:


Step
1
:

Tighten the screws at the top of the beam ABC for making the supports rigid to
achieve Condition I. Load the beam DEF at the quarter point by 200gms to
initialize the system. Now apply the loads in steps of 1kg up to 4.0kg and measure
the extension of the springs for knowing the force in the member.

Step
2
:

Now release the middle screw so that the top beam ABC also becomes operative
to achieve Condition II
. Load the lower beam in the step of 1kg.up to 4kg and
measure extension in the spring for calculating the forces.


5.0

RESULTS AND DISCUSSION:



Compare the theoretical and experimental values of forces in various members in
both cases.



9

6.0

SAMPLE DATA SHEET:


T
able
-
1

Observed value of reactions



Load

(kg)

Reading at

R
1

(Reaction)

R
2

(Reaction)

R
3

(Reaction)

Condition I









Condition II








Table
-
2

Comparison of Results


Applied Load

R
1

R
2

R
3

Remarks

Condition I






Observed

Calculated

Condition
II






Observed

Calculated


7.0

PRECAUTIONS:


(a)

Increase the load on the spring gradually while finding the value of K of
individual spring.

(b)

Load the lower beam without any jerk.