INSTRUCTIONAL MANUAL
OF
ELASTICALLY COUPLED
BEAM APPARATUS
By:
ENGINEERING MODELS & EQUIPMENT
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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
08
6.0
Sample Data Sheet
09
7.0
Precautions
09
3
ELASTICALLY COUPLED BEAM APPARATUS
1.0
THEORY:
A
B
R1
C
R3
R2
A1
A2
E1
A3
E2
D
W
L/2
L
E3
F
E
L/2
L
Let R1, R2, R3 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 =E1, E2, E3
respectively.
Cross sectional area of rods AD, BE and CF= A1, A2, A3 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 R2 at E, relative to
the deflected positions of points D and F is given by
11 WL3 R2 L3
y

(1)
96 EI
6 EI
D
Y1
E
F
Y1  Y3
2
Y2
Y3
F1
Y
D1
E1
4
The case is equivalent to simply supported beam supported at D and F subjected
to loads W and R2 at 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.
1 L 3 WL 2 L 1 3L 3 WL  L 1 3L 
DD '   
   

   
2 2 8 EI 3 2 2 2 8 EI  2 3 2 
5 WL3

16 EI
R1
R2
R3
E
G
D
F
Y
W
Fig. (a)
L/2
L/2
L
3 WL
8 EI
Fig. (b)
B.M. D.
D
F
E
Fig. (c)
E"
W
E'
L/2
3/2 L
D'
D"
E1’
Fig. (d)
R2
E1 "
D
Fig. (e)
F
E
R2 L
2 EI
B.M.D.
5
DD ' 5 WL3
EE ' 

2
32 EI
1 1 WL
L
E ' E"  
 L
2 4 EI
3
3
WL

24 EI
 Downward deflection due to W
y1  EE"  EE ' E ' E"
1 
 5
  
 32 24 
11 WL3

96 EI
Also,
R L3
1 R L
DD"   2  2 L  L  2
2 2 EI
2 EI
3
R L
1
 EE '1  DD"  2
2
4 EI
1 R2 L
L R2 L3
E '1 E"1  
 L 
2 2 EI
3 12 EI
Upward deflection due to R2
y 2  EE"1  EE'1 E'1 E"1
R L3 1 1
R L3
 2 (  ) 2
EI 4 12
6 EI

WL3
EI
Net deflection y at point E= y1  y 2 
Let
11 WL3 R2 L3

96 EI
6 EI
elongation of suspension rod AD =Y1
elongation of suspension rod BE =Y2
elongation of suspension rod CF =Y3
then total deflection of point E relative to its original position is
Y  Y 11 WL3 R2 L3
Y2  1 3 

2
96 EI
6 EI
Now, we know that Y1 
RL
R1 L
R L
, Y2  2 , and Y3 3
A1 E1
A2 E 2
A3 E3
Substituting the values of Y1, Y2, and Y3 in Eq. (2)
(2)
6
R3 L  11 WL3
R2 L
R 2 L3
1 R L
 
(3)
  1 

A2 E 2
2  A1 E1
A3 E 3  96 EI
6 EI
Also, R1  R2  R3  W
(4)
Taking moment of all the forces about F, we get
3 
R1 .2 L  W  L   R2 .L  0
2 
R
3
(5)
R1  W  2
4
2
Now solving Eq. (3), (4) and (5) we get the values of R1, R2 and R3. From Eq. (3)
be substituting the values of R1' R2 and R3 from Eq. (4) and (5).
R
R
3
 3

( W  2 ) L (W  R2  W  2 ) L 
R2 L 1  4
11 WL3 R2 L
2
4
2
 




A2 E 2 2 
A1 E1
A3 E 3
96
EI
6 EI



 L
L
L
L3  11 WL3 3 WL
1 WL
 R2 






8 A1 E1 8 A3 E 3
 A2 E 2 4 A1 E1 4 A3 E 3 6 EI  96 EI

W  11 L3
3L
L 




8 12 EI A1 E1 A3 E 3 

11 L3 
3
K

K

3
 1

12 EI 

R2 

2 L3 
K

4
K

K

2
3
 1

3 EI 

W
2
Where, K 1 
L
L
L
, K2 
, K3 
A1 E1
A2 E 2
A3 E 3

11 L3 
3
K

K

3
 1

12 EI 
3
W 
R1  W 
4
4 
2 L3 
K1  4K 2  K 3 

3 EI 


2 L3
11 L3 
3
K

12
K

3
K


3
K

3
K

K

2
3
1
1
2
 1

EI
12 EI 
W 

2 L
4

 K 1  4 K 2  K 3  K 3  3 EI 


(6)
7

13 L3 
12
K

2
K

2
3


12 EI 
W 
(7)

4 
2 L3 
K1  4K 2  K 3 

3 EI 

and R3 = W- R1 - R2


13 L3 
11 L3 
12
K

2
K

3
K

K

2
3
3


 1

12 EI 
12 EI 
W 
W 
W 

4 
2 L3  2 
2 L3 
K1  4K 2  K 3 

K1  4K 2  K 3 

3 EI 
3 EI 



8 L3
13 L3
11 L3 
2
4
K

16
K

4
K


12
K

2
K


6
K

2
K

2
3
3
1
3
 1

3 EI
12 EI
6 EI 
W 

4

2 L3 
K1  4K 2  K 3 

3 EI 


1 L3 
  2 K 1  4 K 2 

4 EI 
W 
(8)

4 
2 L3 
 K 1  4 K 2  K 3 

3 EI 

Condition II
When the upper end of the central suspension rod is attached to the centre of a
similar elastic beam.
This is achieved by removing the support at B, then beam AC will also deflect
due to the load R2 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 central deflection of beam ABC and Eq. (3) will modified to
R L  11 WL3 R2 L3
R2 L R2 L3 1  R1 L
(3A)

 
 3  

A2 E 2
6 EI 2  A1 E1 A3 E3  96 EI
6 EI
Now solution of Eqs. (3A), (4) and (5) will yield the values of R1, R2 and R3.

11 L3 
 3K 1  K 3 

12 EI 
W 
(9)
R2 
2 
4 L3 
 K 1  4 K 2  K 3 

3
EI


8

37 L3 
12 K 2  2 K 3 

12 EI 
W 
(10)
R1 
4 
4 L3 
 K 1  4 K 2  K 3 

3 EI 

5 L 

  2K1  4K 2 

W 
12 EI 
and R3 
(11)
4 
4 L3 
 K 1  4 K 2  K 3 

3 EI 

L
Here, K 
for any spring is the extension of the spring per unit load.
AE
2.0
OBJECTIVE:
To calculate experimentally and theoretically the loads in the three suspension
rods supporting an elastic beam with a concentrated load hung midway between
two of the suspension rods under two conditions.
1.
2.
When the suspension rods are attached at their upper ends to rigid supports.
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
center of another elastic beam supported at two outer ends only.
4.0
SUGGESTED EXPERIMENTAL WORK:
Step1: 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.
Step2: 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:
Table-1
Observed value of reactions
Load
(kg)
Reading at
R2 (Reaction)
R1 (Reaction)
R3 (Reaction)
Condition I
Condition II
Table-2
Comparison of Results
Applied Load
Condition I
Condition II
7.0
R1
R2
R3
Remarks
Observed
Calculated
Observed
Calculated
PRECAUTIONS:
(a)
(b)
Increase the load on the spring gradually while finding the value of K of
individual spring.
Load the lower beam without any jerk.