Mech Lab 2

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Lab 2Torsion of Bars

Group Members (1a)

Delano Cantera

Paolo Ferrer

Zafar Javed

Jonathan van Leeuwen

Date performed

February 6, 2012, 2:30 PM

Date submitted

February 28, 2012, 9:45 AM

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Introduction & Background

Objective:

Investigate torsion angles from torsional loadings in different cross sections of brass

bars.

Background:

Torsion tests can be carried out on most materials, using standards specimens, to

determine mechanical properties such as modulus of elasticity in shear, yield shear

strength, ultimate shear strength, modulus of rupture in shear, and ductility. Torsion tests

can also be carried out on full-size parts (shafts, axles, and twist drills) and structures

(beams and frames) to determine their response to torsional loading. In torsion testing,

unlike tension testing and compression testing, large strains can be applied before plastic

deformation occurs, and complications due to friction between the test specimen and dies

do not arise. Torsion tests are most frequently carried out on prismatic bars of circular

cross section by applying a torsional moment about the longitudinal axis. [1]

Manufacturers change the torsion bar or key to adjust the ride height, usually to

compensate for heavier or lighter engines. While the ride height may be adjusted by

turning the adjuster bolts on the stock torsion key, rotating the stock key too far can bend

the adjusting bolt and (more importantly) place the shock piston outside its standard travel.

Over-rotating the torsion bars can also cause the suspension to hit the bump-stop

prematurely, causing a harsh ride. [2]

Experimental Methodology

Fig. 1

1. Universal Test Frame SE

2. Chuck, adjustable

3. Adjustable angle scale

4. Chuck-mounted test bar

5. Clamped-on pointer for angle scale

6. Additional test bars of differing cross-section

7. Wheel for introducing moment

8. Adjustable guide pulley

9. Hanger with weights

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To do the experiment we set up an apparatus that would apply a torsion load on a variety

of different bars of the same material but with different physical conditions such as its

shape or being slotted or tubular. The bars are to be held in place by a chuck, being fixed on

the left end and attached to a pulley on the right side which the torque will be transmitted

through. Weights strung from the pulley will apply a moment force called torque on the

rods creating shear stress on the bar. From this we will measure the angle of twist in thebar by the 2 attached pointers corresponding with a fixed angle scale on the apparatus.

Operation

The operations of the experiment are as followed:

Removal of previous test bar

- Release both pointers from the angle scale- Open the left chuck and move it to the left- Release the right chuck, take out the test bar, and remove the pointers

Fit the new test bar

- Slip pointers loosely onto desired test bar

- Insert test bar with pointers into right chuck, thread pointers between angle scalesand insert test bar into left chuck

- Firmly close both chucks to prevent test bar from slipping through under load- Set desired distance between angle scales- Adjust pointer in front of angle scale and secure in 0 degrees position

Experiment

- Select test bar and mount it as described above- Set torsion length L and adjust scale pointers to 0 degrees- Attach weight in 5N increments from 0 to 15N (increments or 1N from 0-3 for

slotted bar) recording the difference between the two torsion angles after applying

each new weight

- (NOTE: do not overload bars max 20N for each bar with the exception of the slottedbar which its max load should be 3N)

Results

Knowing we were applying increments of 5N forces on the pulley, with the equation M =

F*D, where F is the force applied, D is the radius of the pulley and M is the moment force

(torque) we can find that applied forces 1, 2, 3, 5, 10, and 15N with a pulley radius of

110mm create torques of 110, 220, 330, 550, 1100, and 1650N*mm respectively.

From our measurements we can determine the angle where = 2 - 1

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Table 1

Test bar 1: Round bar D=6mm

Load Force in N Torque in N*mm Angle measured in degrees

0 0 0

5 550 3-2 = 110 1100 8-6 = 2

15 1650 11-7 = 4

Test bar 2: Tube D=6 x 1 mm


0 0 0

5 550 3-2 = 1

10 1100 8 - 4 = 4

15 1650 11-5 = 6

Test bar 3: Square tube Dimensions: 6 x 6 x 1mm


0 0 0

5 550 5-2 = 3

10 1100 8-3 = 5

15 1650 13-6 = 7

Test bar 4: slotted tube D=6 x 1mm


0 0 0

1 110 14-8 = 62 220 28-16 = 12

3 330 44-21 = 23

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Analytical Methodology

Torsion is the twisting of an object due to an applied torque to the object, and the angle of

twist or torsion angle caused by this applied torque is calculated using the following

formula:

=

.

(Eq.1)

where

MT = torsion moment

L = the length of the object the torque is being applied to or over (200mm for the

four test bars used in this experiment)

G = shear modulus (for brass G=40000 N/mm2)

Ip = polar moment of inertia of area

Torsion Moment

The torsion moment (MT) applied to an object is calculated as follows:

MT = F. R (Eq.2)

where F = force applied

R = radius of the pulley

Calculations of torsion moment

Example Calculation:

Test Bar 1:

F = 10 Newton

R = 110 mm (the same for all test bar in this experiment)

MT = F. R = (10 N)(110mm) = 1100 N-mm

The calculated torsion moment for each loading in different test bars are tabulated in table

2.

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Table 2 (Torsion Moment)

TEST BAR 1Round Bar

TEST BAR 2Tube

TEST BAR 3Square Tube

TEST BAR 4Slotted Tube

Load

F in N

Torque

MT in N

mm

Load

F in N

Torque

MT in N

mm

Load

F in

N

Torque

MT in N

mm

Load

F in

N

Torque

MT in N

mm

0 0 0 0 0 0 0 0

5 550 5 550 5 550 1 110

10 1100 10 1100 10 1100 2 220

15 1650 15 1650 15 1650 3 330

Torsion angle calculations:

With the calculated values of moment of inertia and polar moment of inertia, equation one

can now be used to calculate the torsion angle for each different bar and loadings. Now we

will observe how the moment of inertia affects the torsion angle of the object.

Bar 1:

Sample Calculations:

MT = 550 Newton L= 200 mm

Ip= 127.23 mm4 G= 40000 N/mm2

=

.

=( )( )

( )

x

= 1.23o

Bar 2:



Ip= 102.10 mm4 G= 40000 N/mm2

=

.

=( )( )

( )

x

= 1.54o

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Bar 3:



Ip= 125 mm4 G= 40000 N/mm2

=

.

=( )( )

( )

x

= 1.26o

Bar 4:



Ip= 127.23 mm4 G= 40000 N/mm2

=

.

=( )( )

( )

x

= 6.13o

Other calculation results for the four test bars in different loadings are listed in table 3.

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Table 3

Test bar 1: Round bar D=6mm

Load F in N Torque MT N mm measured in calculated in

0 0 0 0

5 550 3-2 = 1 1.23

10 1100 8-6 = 2 2.48

15 1650 11-7 = 4 3.72

Test bar 2: Tube D=6 x 1 mm


0 0 0 0

5 550 3-2 = 1 1.54

10 1100 8-3 = 3 3.09

15 1650 11-5 = 4 4.63

Test bar 3: Square tub 6 x 6 x 1 mm


0 0 0 0

5 550 5-2 = 3 1.26

10 1100 8-3 = 5 2.52

15 1650 13-6 = 7 3.78

Test bar 4: Slotted tube D= 6 x 1mm


0 0 0 0

1 110 14-8 = 6 6.13

2 220 28-16 = 12 12.26

3 330 44-21 = 23 18.39

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Fig. 2

Fi vs MT (CHANGE Fi INTO SYMBOL)

Fig. 2

Explain a bit about the graph

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Comparison & Discussion

We notice that the hollow bars polar moment of inertia is less than the bar without the

hole; from here we analyzed that as the cross section of the bar decreases, and its polar

moment of inertia also decreases. We also found out that the polar moment of inertia

greatly decreases when the section is open, as observed from bar 4 compared to bar 2. Wealso notice that there is not much difference with the polar moment of inertia between the

round bar (bar 1) having the diameter equal to the length of the square tube (bar 3).

Polar Moment of Inertia

The polar moment of inertia (Ip) of a beams cross-sectional area measures the beamsability to resist torsion, and the larger the polar moment of inertia the beam has the less the

beam will twist. Analyzing Eq.1, we can conclude that having a bigger value of lpwill result

in a larger denominator value resulting into a smaller torsion angle.

The calculation of the polar moment of inertia of an object varies with the form of each

object since the cross sections for each object differs from the others. The calculations of

the Ip for the four test bars used are the following:

TABLE 4 polar moment of inertia [3]

Bar

1

Ip = (/32)*D4 = (/32)(1296mm4) Ip = 127.23 mm4

Bar

2

Ip = (/32)(D4 - d4)= (/32)(1040mm4) Ip = 102.10 mm4

Bar

3

Ip = a3 . t = (125 mm3)(1 mm) Ip = 125 mm4

Bar

4

Ip = *

( ) + () Ip = 5.14 mm4

We notice that the hollow bars polar moment of inertia is less than the bar without the

hole, from here we analyzed that as the cross section of the bar decreases, and its polar

moment of inertia also decreases. We also found out that the polar moment of inertia

greatly decreases when the section is open, as observed from bar 4 compared to bar 2. We

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also notice that there is not much difference with the polar moment of inertia between the

round bar (bar 1) having the diameter equal to the length of the square tube (bar 3). During

the experiment we observed that some factors may affect the overall results of the

experiment these sources of error are as follows.

- The test bars may have been under a load that was not being considered in ourcalculations, exerted from the chuck.

- While adjusting the weight used to exert the torsion on the system the scales or indication

could have been accidentally changed.

- Due to manufacturing error or inconsistencies the test bars may have not been of uniform

shape or to exact specifications as intended.

- The force of friction acting in between the pulley and the axel it is on would result in a

different torsion than expected.

Even with these conditions that could have affected the accuracy of our results our

experimental values were quite close to those calculated at the theoretical level. From

comparing these results and calculations we are quite confident that our data represents anaccurate portrayal of the effects torsional loading has on the angle of twist in a bar within

the constraints of real life conditions.

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Reference:[1]http://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnext

oid=90d64ef322e18110VgnVCM100000701e010aRCRD

[2]http://en.wikipedia.org/wiki/Torsion_bar_suspension

[3]Lab manual of MECH. 2222.
http://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnextoid=90d64ef322e18110VgnVCM100000701e010aRCRDhttp://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnextoid=90d64ef322e18110VgnVCM100000701e010aRCRDhttp://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnextoid=90d64ef322e18110VgnVCM100000701e010aRCRDhttp://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnextoid=90d64ef322e18110VgnVCM100000701e010aRCRDhttp://en.wikipedia.org/wiki/Torsion_bar_suspensionhttp://en.wikipedia.org/wiki/Torsion_bar_suspensionhttp://en.wikipedia.org/wiki/Torsion_bar_suspensionhttp://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnextoid=90d64ef322e18110VgnVCM100000701e010aRCRDhttp://www.asminternational.org/portal/site/www/AsmStore/ProductDetails/?vgnextoid=90d64ef322e18110VgnVCM100000701e010aRCRD

Mech Lab 2

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