1.0 TITLE: Torsion of Bars
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ABSTRACT AND OBJECTIVES
(a) To investigate the relationship between torque, T and the angle of twist, Φ of a circular cross section specimen in terms of the formula:
Φ = TL / GJ
Where, T = torque L = length of rod G = modulus of rigidity of material J = Πd4 / 32 D = diameter of rod
(b) To determine the Modulus of rigidity (G) of materials
In this experiment, the material rod will be clamped first, then the load is apply to the rod via a cord. The length of the shaft is either set 400 mm for the first test followed by 200 mm or vice versa. The rotation scale is set with the pointer pointing 0o of twist. Load is apply to the rod by putting it on the hanger. Finally read and record the twist of angle for each loading. Diameter of the rod needs to be measure and record at the end of the experiment. From the result we obtained, the relationship between torque, T and the angle of twist, Φ of a circular cross section specimen or material is proved to be directly proportional to each another and twist Φ, also proportional to the length of the shaft, L. And the modulus of rigidity, G for:
Aluminium alloy = 2532.62 Nmm-2 Steel = 4545.97 Nmm-2 Nylon = 104.81 Nmm-2
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3.0 APPARATUS
A base frame has a clamp at one end and a ball bearing in a housing (Plummer block) at the other. A short shaft in the bearing has a three jaw chuck facing the clamp and a torsion head at the outward side. A hangar cord is wound round the torsion head with an effective diameter of 75 mm. Specimens in the form of lengths of rod are gripped by the fixed clamp and the rotating chuck 450 mm away. An arc shaped scale of degrees is mounted on a base which can be moved along the length of the specimen. A pointer on a spring steel strip registers the rotation of the specimen when a load is applied to the hanger cord.
4.0
INTRODUCTION
A torsion bar is a flexible spring that can be moved about its axis via twisting. Torsion bars are designed and based on the amount of torque used in the twisting of the spring, the angle of the twist, the overall dimensions of the torsion bar and what materials the torsion bar is made from. The most common place to find a torsion bar is in the suspension of a car or truck, in machines used for production or in other precision devices. The flexibility of the spring is the main reason that a torsion bar is used. If a more rigid structure were used such as a steel rod were used too much load bearing pressure would be placed on the both the wheels and the under body of the vehicle. A torsion bar works by resisting the torque placed on it. When one end of the torsion bar is affixed to an object that cannot be moved, the other end of the bar is twisted, thus causing torque to build up. When this happens, the torsion bar is resistant to the torque and will quickly go back to its starting position once the torque is removed. In general, the object that cannot be moved is usually a frame. If there is not any force applied to the torsion bar, it will stay at the same position until force is applied. An example the other end of the bar will be contacted to a control arm. The control arm moves in a fixed manner on the frame and this creates the twisting movement on the bar. This, in turn, supplies the torque needed to make a spring
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Torsion is the twisting of an object due to an applied torque. In circular sections, the
resultant shearing stress is perpendicular to the radius. For solid or hollow shafts of uniform circular cross-section and constant wall thickness, the torsion relations are:
•
R is the outer radius of the shaft i.e. m, ft.
•
τ is the maximum shear stress at the outer surface.
•
φ is the angle of twist in radians.
•
T is the torque (N·m or ft·lbf ).
•
ℓ is the length of the object the torque is being applied to or over.
•
G is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa), lbf/in 2 (psi), or lbf/ft2.
•
J is the torsion constant for the section. It is identical to the polar moment of inertia for a round shaft or concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and 3
for thin walled tubes of arbitrary shape the shear flow ap proximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and finite element analysis (FEA) may be the best method. •
the product GJ is called the torsional rigidity.
The shear stress at a point within a shaft is:
Where: •
r is the distance from the center of rotation
Note that the highest shear stress is at the point where the radius is maximum, the surface of the shaft. High stresses at the surface may be compounded by stress concentrations such as rough spots. Thus, shafts for use in high torsion are polished to a fine surface finish to reduce the maximum stress in the shaft and increase its service life. The angle of twist can be found by using:
5.0 THEORY
When analysis the stresses and deformation that takes place in circular shafts, an important property of circular shafts is demonstrated: When a circular shaft is subjected to torsion, every cross section rotates as a rigid slab. This property will enable to determine the distribution of shearing strains in a circular shaft and to conclude that the shearing strain varies linearly with the distance from the axis of the shaft. Review the derivation and interpretation of the theory of torsion of circular shafts. Start by looking at a small section of length dx of a circular shaft under torsion. During 4
twisting, one end of the shaft will rotate about the longitudinal axis with respect to the other end. The magnitude of this rotation is measured in terms of the angle in radians by which one end rotates relative to the other. This is called the ‘Angle of Twist’.
Considering the case of a shaft of length, L and of uniform cross section of radius, c subjected to a torque, T at its free end. (Figure 1.1)
Figure 1.1
The angle of twist, Φ and the maximum shearing strain, γmax are related as follows:
γmax = cΦ / L
In the elastic range, the shearing strain and shear, τ is related by Hooke’s Law:
γmax = τmax / G = Tc / JG
Equating the right-hand members of equation (1.3) and (1.4), and solving for the angle of twist, we write:
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Φ = TL/JG
Note that all the relations here are based solely on the geometry of the circular shaft. Hence they are valid for any type of material. This is not so in what follows, the calculation of stresses based on linear elastic material behavior.
6.0 PROCEDURE
1. The aluminium alloy rod was clamped in position and the load hanger is put on the cord. 2. The rotation scale and pointer was set 400 mm from the fixed clamp and the pointer is put to zero. 3. A load of 40 N by 5 N increments was added and the twist of the specimen for each increment is recorded in a table. 4. After the load was removed, the rotation scale and pointer is moved to 200 mm from the clamp and the above procedure is repeated. 5. The diameter of the rod was measured and recorded.
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6. Now, the specimen was changed for the steel rod and the procedure is repeated for the 400 mm length only. 7. Finally, the Nylon rod was clamped in position and the twist over 400mm and 200mm is measured when a load up to 40 N by increments of 5 N is applied to the torsion head. The load was removed and whether full elastic recovery had occurred is noted.
7.0 RESULT
7.1 Experiment Data
1. Test 1
Material = Aluminium Alloy
Diameter = 6.26 mm Twist of rod
Hanger Load (N) 5
Torque (Nmm) 15.65
Over 400 (0) 1.0
Over 200 mm (0) 0.5
10
31.30
2.0
1.0
15
46.95
3.0
1.5
20
62.60
4.0
2.0
25
78.25
5.0
2.5
7
30
93.90
6.0
3.0
35
109.55
7.0
3.5
40
125.20
8.0
4.0
Table 1
2. Test 2
Material = Steel
Diameter = 5.03 mm
Hanger Load ( N )
Torque ( Nmm )
Twist over 400 mm ( o )
0
0.0
0.0
5
12.58
1.0
10
25.15
2.0
15
37.73
3.0
20
50.30
4.0
25
62.88
5.0
30
75.45
6.0
35
88.03
7.0
40
100.60
8.0
Table 2 3. Test 3
Material= Nylon
Diameter = 6.51 mm Twist of rod
Hanger Load (N) 1
Torque (Nmm) 3.26
2
6.51
8.0
3.0
3
9.77
13.0
5.0
4
13.02
19.0
7.0
5
16.28
25.0
10.0
0
0.00
0.0
0.0
Over 400 mm (o) 4.0
Table 3
7.2
EXAMPLE OF CALCULATION
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Over 200 mm (o) 1.0
ø (rad) = ø (o) x (π / 180) = 1 x (π / 180) = 0.0175 rad Radius, r = d/2 =6.26/2 =3.13 mm Torque, T = F x r = 10 x 3.13 = 31.30 Nmm Key words; Ø = angle of twist F = force r = radius d = diameter
8.0
DISCUSSION
1) Material : Aluminum Aloy
2) Material : Steel
3) Material : Nylon
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Calculations
The value of angle of twist can be defined from this equation:-
Ø = TL / GJ
while;
From the graph above the value of L / GJ is slope of graph,m but instead of degree the angle must be change to unit of radians to compare with units below:
G
=modulus of rigidity of material, N / m2
T
=
torque, N.m
L
=
length of rod, m
Ø
=
angle of twist, radian polar moment of area = πd4 / 32, m4
J= d
Slope, m = L / GJ
=
diameter of rod, m
so
G = L / mJ
Calculation of G for: 1) Aluminium
J = πd4 / 32 = π x (6.26)4 / 32 = 150.76 mm4
a) For length,L = 400mm , m = 0.0011 G = L/mJ = 400 / (0.0011 x 150.76) = 2412.02 N/mm2 10
b) For length,L = 200m, m = 0.0005 G = L/mJ = 200/ (0.0005 x 150.76) = 2653.22 N/mm 2
Overall, G = [(2412.02 + 2653.22) ] / 2 = 2532.62 N/mm2
2) Steel
J = πd4 / 32 = π (5.03)4 / 32 = 62.85 mm4
a) For length,L = 400mm , m = 0.0014 G = L/mJ = 400 / (0.0014 x 62.85) = 4545.97 N/mm2 3) Nylon
J = πd4 / 32 = π x (6.51)4 / 32 = 176.33 mm4
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a) For length,L = 400m , m = 0.0255 G = L/mJ = 400/ (0.0255 x 176.33) = 88.96 N/mm 2
b) For length,L = 200m, m = 0.0094 G = L/mJ = 200 / (0.0094 x 176.33) = 120.66 N/mm 2
Overall, G = [(88.96 + 120.66)] / 2 = 104.81 N/mm2
a. To what extent does the experiment verify the torque / twist formula?
From the data that we get, the graph plotted shows that the angle of twist against torque is increasing linearly for all types of specimens, but it is not directly proportional as mention in theory. From the formula given: Φ α T and Φ α L Φ = TL / GJ where Φ = angle of twist (radians) T = torque L = length of the shaft G = modulus of rigidity (constant) Angle of twist is directly proportional to torque in theory. It can be conclude that the experiment does not fulfill the torque/twist formula in theory.
b. Were all the graphs (angle of twist against torque) linear, thus exhibiting purely elastic behavior?
Yes, all the graphs are linear, does exhibiting purely elastic behaviour. A relation will be derived between the angle of twist of a circular shaft and the torque exerted on the shaft. 12
The entire shaft will be assumed to remain elastic. But in the elastic range the yield is not exceeded anywhere in the shaft, Hooke’s Law apply and we have ymax= τmaxG As long as the yield stress of the material is not exceeded the point obtained by plotted Φ against T will fall in a straight line. The slope of this line represents the quantity JG/L, from which the modulus of rigidity G may be computed.
c. Handbook values for G (modulus of rigidity) are 26200 N / mm2 for aluminum alloy and about 79000 N /mm2 for steel. Comment on the experimental comparisons.
Specimen Steel Aluminum
G theoretical
G expectation
2 N/mm
2 N/mm
79000 26200
4545.97 2532.62
For steel: % error compare to handbook value = [(79000 – 4545.97) / 79000] x 100% = 94.25%
For aluminum: % error compare to handbook value = [(26200 – 2532.62) / 26200] x 100% = 90.33%
There are many of reasons why we got the different value compared to the theoretical value. Firstly, it depends on the temperature of the specimen while the experiment has been conducted. It may also have an error in taking diameter of the typical torsion of bars-test specimen. Beside, specimen malformation may probably occur.
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Secondly, environment factor plays an important role in order to get the best result in this experiment. In this case, we did this experiment in the room temperature which is 27oc. This temperature may yield different result compared with the theoretical value. Other than that, air resistance also affected our result. Although this is only a small factor, it would give error and consequently the result was not accurately.
Thirdly, the loads applied were not consistent. For example, 5N’s block + 5N’s block ≠ 10N’s block for the loading block. This factor gives a big effect to our experiment. As a result, we didn’t get the linear straight line for the angle of twist versus torque graphs. To solve this matter, we made some interpolation so that the graph accurate with the theoretical value
There were some errors occurred during the experiment was carried out. Below were the errors that occurred: (a) When reading the value, there were some parallax error and the apparatus also might have zero error. (b) When reading the value angle of twist, there were some zero-error occur where the pointer of the angle measurement apparatus is not pointing zero value before loading added. (c) Old material such as rusty steel maybe causing reasonable error.
Recommendations
Below are some recommendations that we suggest to improve the experiment (a) Reduce the parallax error and zero error. (b) It is better using new materials to run the experiment.
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9.0
CONCLUSION
After analyzed data and discussion about the experiment result, we can conclude that there are relationship between torque and angle of twist and obey the equation below:-
Φ = TL / GJ
This equation is a convenient method to determine the modulus of rigidity (G) of any materials. From the experiment value of G for specimens;
Specimen
Value of G (modulus of rigidity) N/mm 2
Steel
4545.97
Aluminum
2532.62
Nylon
104.81
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10. REFERENCE
http://en.wikipedia.org/wiki/Torsion_%28mechanics%29 www.attension.com/encyclopedia/torsion+bar+experiment en.wikipedia.org/wiki/cavendish_experiment http://www.ehow.com/how-does_5459141_torsion-bar-works.html http://instruct1.cit.cornell.edu/courses/virtual_lab/chalktalks/theory/1.shtml http://instruct1.cit.cornell.edu/courses/virtual_lab/chalktalks/theory/2.shtml
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