Suspended Hinged Bar Two Different Material Wires

7/27/2019 Suspended Hinged Bar Two Different Material Wires

1/6

SUSPENDED HINGED BAR BY USING TWO DIFFERENT MATERIAL

WIRES

To determine the forces(f1&f2) and the ratio of forces(f1/f2) and carried by

different wire in a system of a hinged bar suspended by a steel and brass wireexperimentally and compare with the theoretical values.

Statically Indeterminate

A Statically determinate forces system is one in which the values of all the

external forces acting on the body can be determined by the equations of static

equilibrium alone. The forces acting on a body can not be determined by the

equations of equilibrium. Such a case of force system is said to be statically

indeterminate.

Experiment is aimed at finding these unknown forces using simple dial gauges.

A hinged flat suspended by two wires of different materials is an indeterminate

system. It will have one unknown force in each wire and a vertical unknown

reaction at the hinged support. In this case there are only two effective

equations of equilibrium. There are no horizontal forces and hence the system

is indeterminate of order one. To determine the unknown forces this method

is used, which considers the deformation of the system. The procedure to be

followed in analyzing an indeterminate system is to first to write all equations

of static equilibrium that pertain to the system and then supplement these

equations with other equations based upon the deformation of the structure.

Equations involving deformation will be written so that total number of

equations from both statics and deformations is equal to the number of unknown

forces involved.

APPARATUS REQUIRED:

Hinged bar suspended by Steel and Brass wires of equal length and diameter,Dial gauges (2 Nos.), weights and pan loading frame.

PROCEDURE:

The hinged flat in suspended using two wires of equal diameter, one of steel and

the other of brass. The flat is suspended horizontally. Two dial gauges are

placed below the point of attachment of each wire. Weights are placed in the

pan in steps of 0.5 to 1 Kg. Dial gage readings are noted and tabulated. The

first weight of 0.2 kg makes the wires initially straight. Hence the difference inthe dial gauge readings between 0.5 to 1 kg is taken for the calculation of load

1


2/6

in the wires. Load on each wire is also calculated using theory from Strength

of Materials compared with the values obtained from the dial gage readings.

The ratio of the loads on each wire also calculated.

f1l1 a1 E11 -------- f 1 = -------- 1

a1E1 l1

f2l2 a2 E22 = -------- f 2 = -------- 2

a2 E2 l2

l1 = l2= l a1 = a2 = a

E1= steel (youngs modulus) E2 = Brass (Youngs modulus)

S. No. P in kg 1(STEEL) 2(BRASS)

1 0.5

2 1

1(BRASS) 2(STEEL)

3 0.5

4 1

The ratio of forces in the Steel and Brass wire is determined experimentally and

theoretically and found agree with near by value. The individual forces on each

wire obtained experimentally are higher than the theoretical values due to the

slip of the wire, which also contributes to the dial gage readings.

The wires may be interchanged and repeat the experiment.

2


3/6

f1 Position Steel f2 Position Brass

f1 E1

---- = 2----- .(1)f2 E2

3/2 W 2f1 = f2 ..(2)

Substituting Eqn.2 in eqn. (1) : E1 = 200 GPa for steel; E2 = 100 GPa for Brass

f1 200

------------------- = 2 --------

3 100---- W 2f12

f1

----------- = 4

3W-4f1

-----------

2

2f1

--------- =8

3W-4f1

2f1 = 12W 16f1

18f1=12W

12W

f1 = -----------9

2 W

f1= ------ ----------------------------(3)

3

3


4/6

Substitute eqn (3) in eqn (2)

3W

------- - 2f1= f2

2

3W - 4W = f2

---- -----

2 3

9W - 8W

------------- = f2

6

W

f2 = ----------- ---------------(4)6

Actual brass youngs modulus is 95.1 10-3

Due to commercial ingredients at the time of manufacturing we have to take

100 x 10-3 Gpa

Interchange the wire position

f1 Position Brass f2 Position Steel

f1l f2 l

---- = 2 -----

aE1 aE2

f1 E1---- = 2 ----- .(1)

f2 E2

3----- W 2f1 = f2 .. (2)

2

Substituting in eqn. (1) for f2 ; E1 = 100 for brass GPa; E2 = 200 for steel GPa

f1 100

------------------- = 2 -------- = 1

3 200

---- w 2f12

4


5/6

3

f1 = ------ (W 2f1)1

2

3 3 1 W

3f1 = ----- w = ----- ----- = ------

2 2 3 2

w w

f1= ----- & from eqn. (2) f2 = ----

2 2

EXPERIMENTAL CALCULATION:

l1 = l2 = l= 0.017m d1= d2 = 1mm a1 = a2 = (/4) d2 (dia of wire)

E1 = MS GPa 200 for steel E2 = (for Brass) 100GPa

S. No. P in kg 1(STEEL) 2(BRASS)

1 0.5 414 426

2 1 424 438

1(BRASS) 2(STEEL)

3 0.5 455 432

4 1 458 480

1 =414 and 2 = 426

A1E1 (0.001)2 100 1

f1 = -------- 1 = -------------------------------N For brass

l1 4 l1

(0.001)2 100 426

f1= ------------------------------ = 1.97 N

40.017

5


6/6

A2E2 (0.001)2 200 2

f2 = -------- 2 = ------------------------------ For steel

l2 40.017

2W 0.51

(0.001)2 200 414 f1= ------ = ----------9.81 =1.635Nf2= ---------------------------- =3.82 N 3 3

40.001

W 0.5

f2 = ----- = ------ 9.81 =0.8175N

1.97 6 6

f1 / f2 = ------------ = 0.57 0.8175

3.82 f2= ----------- = 0.50N Ratio force

Ratio of force 1.635

From theory P1 = P/2 and P2 = P/2

(From the Experiment Ratio of forces must be to one)

CONCLUSION:

Hence the ratio of forces from the experiment less than one, condition was

proved. Similarly need to calculate the other possible combination of given

wires.

6

Documents

Suspended Hinged Bar Two Different Material Wires