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Sunday, April 8, 2012

DEFLECTION OF CANTILEVER BEAMS

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TABLE OF CONTENTS


Page


1. INTRODUCTION


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. OBJECTIVES





. THEORY


.1 Torsion of Circular Elastic Bars


. Bending of Circular Elastic Bars


. Combined Bending and Torsion of Circular Elastic Bars 4





4. EXPERIMENTAL PROCEDURE 5


4.1 Apparatus and Equipment 5


4. Procedure 6


4..1 Variation of Torques Versus Torsional Strain


4.. Variation of Bending Moment Versus Bending Strain








5.. DISCUSSIONS 1 7





6. EXPERIMENTAL / THEORETICAL RESULTS 1





7. DISCUSSIONS 1





8. CONCLUSION 14





. REFERENCE 14





10. APPENDIX / LOG SHEETS


1. Introduction


The circular shaft is one of the most common mechanical engineering components. It is mainly used to transmit torque in rotary machines. Sometimes the same shaft is also subjected to bending due to loading transverse to the shaft, in which case the shaft acts as a beam in the bending theory.


The torsion theory is accurate when stresses remain elastic and the solution provided (the well-known torsion formula) is known to be ‘exact’. The bending of the shaft as beam is governed by the bending theory, which also requires the materials of the shaft to be loaded within the elastic range and the deflection be small. When both torsion and bending are applied to the shaft, the total stresses in the shaft can be decomposed into two parts, one is produced by bending, and the other produced by torsion. The total stress obeys the Principle of Superposition for elastic problem.


. Objectives


.1 To review and verify the torsion theory with its limitation.


. To review and verify the bending theory with its limitation.


. To understand Mohr’s Circle and stress transformation.


. Theory


.1 Torsion of Circular Elastic Bars


To establish a relation between the internal torque and the stress it sets up in members with circular solid cross sections, it is necessary to make two assumptions. These, in addition to the homogeneity of the material, are as follows


1) A plane section of material perpendicular to the axis of a circular member remains plane after the torques are applied, i.e. no warpage or distortion of parallel planes are normal to the axis of a member takes place.


For small deformations it is assumed that parallel planes perpendicular to the axis remain a constant distant apart. This is not true if deformations are large. However, since the usual deformations are very small, stresses not considered here are negligible.


) In a circular member subjected to torque, shear strains ᠊ vary linearly from the central axis reaching maximum at the periphery (the outside surface).


) If attention is confined to the linearly elastic material, Hook’s law applies, and follows that shear stress  is proportional to shear strain ᠊.


In the elastic case, on the basis of the previous assumptions, and at any give section a relationship can be written as follows,


(max/c)᠘A dA = T, where max and c are constants. (1)


However, ᠘A dA, the polar moment of inertia of a cross-sectional area, is also a constant for a particular cross-sectional area.


It will be designated by J in this case.


By using the symbol J for the polar moment of inertia of a circular area;


Equation 1 can written more compactly as,


max = Tc/J, ()


the above is the well-known torsion formula for circular shafts that expresses maximum shear stress in terms of resisting torque and the dimensions of a member. And otherwise it is express as follows,





᠊max = max/G = Tc/(JG), ()


where T is the applied torque, c is the radius of the circular shaft, J = ᠋d4/,


d is the diameter of the circular shaft, G = 75 GPa is the shear modulus of the steel shaft used in the present experiment.


. Bending of Circular Elastic Bars


The deformation of member caused by the bending moment M is measured by the curvature of the neutral surface. The curvature in the elastic range can thus be derived as


6 = M / (EI) (4)


In the case of pure bending, as long as the stresses remain in the elastic range, the neutral axis passes through the centroid of the cross section. It is being noted that I is the moment of inertia of the cross section with respect to the centroidal axis, which is perpendicular to the plane of the couple M.


Hence, maximum bending stress can be expressed as


max = Mc / I (5)


Since, in the elastic range, the normal stresses remain below the yield strength in the shaft, there will be permanent deformation, and Hook’s law for uni-axial stress applies.


max = Emax (6)


Substitute (5) into (6), the maximum bending strain can be derived as


max = Mc / ( EI ) (7)


M = applied bending moment


E = Young’s modulus of the shaft material used in the present experiment (00GPa)


I = Moment of inertia (Appendix)


c = Distance between the outer most surface and the neutral axis


. Combined Bending and Torsion of Circular Elastic Bars


The normal stress and shear stress at the top or bottom point of the shaft are calculated from the Equations () and (5). The normal strain and the shear strain at the top or the bottom point of the shaft are calculated from the Equations () and (7). As the directions of stresses and strains due to torsion and bending are different, the combined stress and strain will have to be calculated through the stress and strain transformation. The Mohr’s circle for stresses and strain can be conveniently constructed to perform such calculation.





4. Experimental Procedures


4.1 Apparatus and Equipments


Figure 1 Schematic set-up of the experiment.


A solid circular shaft is clamped at one end (A), and subjected to two dead weights W1 and W at the other end (B) --- Figure 1. The two dead weights are hung on a loading frame that is welded to the end of the shaft. A series of notches on the loading frame allow for variation of the arm of moment created by each dead weight about the longitudinal axis of the shaft.


The magnitude of the torque and the bending moment created by the two dead weights are


T = W1a1 � Wa (counter clockwise = +ve), (6)


M = (W1 + W) L, (7)


where, L is the distance between point C and the loading point B. The bending moment is for a point C on the shaft.


The applied torque is zero when W1a1 = Wa.


The strains at the top surface of the shaft are measured by a rectangular strain rosette, with measurements of 0, +45 and -45. The strain measurement 0 directly measures the bending strain and the strain measurements +45 and -45 together can be used to measure the torsion shear strain with the use of Mohr’s circle. The torsion strain can be obtained as,


᠊ = +45 - -45 (8)


4. Procedures


There are altogether three readings of strains, two weights, and two distances to be recorded at each load step.


The experiments are divided into two parts.


4..1 Variation of torsion strain versus the torque


Fixed weight W1 at a certain weight (W1 = 0kg) and position, use a certain W (.1kg) and shift the hanging position of W from one end of the bar to the other end. At each successive position of W, record the position of W, a, and the strains 0, +45, and -45.


Neither W1 nor W should exceed 5kg.


Once you have reached the end, reverse the direction of shift W where, the weights are kept the same, back from the above end to the original end. At each successive position record the unloading path.


4.. Variation of bending moment versus the bending strain


Fix the magnitudes and position of both weights such that W1 = W and a1 = a. This arrangement leads to no torque produced on the shaft. Record the strains readings and a1 = a. Apply successive weights in increment of 1kg to both weights, such that no torque is caused by the simultaneous increment of the weight. At each step, record the position and the strains in 0, +45, and -45 in the table provided. The total weight of W1 + W should not exceed 16 kg, in order not to damage the specimen.


5. Discussions 1


5.1 What are the likely experimental errors during the experiment?


Experimental errors are likely to be caused by





1. Inaccurate weights. An examination of the weights showed that some of the weights were chipped. This would affect the bending moment and thus the experiment.


Weights should be well-maintained and given to students.


. The shaft was assumed to be perfectly elastic, but this may not be true because former students may have exceeded the 5kg limit previously.


A new shaft should have been used.


. The weights were loaded in such a way that the centers of gravity were not in lines with those of the loading frames.


Instructions should have been passed down to inform students to take note of the proper loading procedure.


4. The bolt that tightened the shaft to the fixed end might have loosened during loading.


The bolt should have been tightened before the experiment. A clamp could have used to hold the bolt in place.


5. The shaft was assumed to be level with the ground, but there was no equipment to check this.


Provide a device (spirit leveler) to check for this point.


6. Parallax error may have occurred during measurements of lengths and placing of the strain gauge parallel to the shaft axis.


Measurements have to be taken a few times to ensure accuracy.


7. The strain rosette was noticed to have non-zero readings after the weights were unloaded. This may be due to residual stresses in the shaft.


Time should have been given for the shaft to return to its original state before taking the next reading.


5. Prove equation (8) by formulae, as well as Mohr’s circle.


Theory


Below are the general expressions for normal stress and shear stress, respectively on any plane located by the angle ᠋ and caused by a known system of stresses, in order to establish the Mohr’s circle.








Using a basic trigonometric relation (cos + sin = 1) to combine the two above equations we have,





This is the equation of a circle, plotted on a graph where the abscissa is the normal stress and the ordinate is the shear stress. This is easier to see if we interpret x and y as being the two principle stresses, and @xy as being the maximum shear stress. Then we can define the average stress, avg, and a radius R (which is just equal to the maximum shear stress),





The circle equation above now takes on a more familiar form,





The circle is centered at the average stress value, and has a radius R equal to the maximum shear stress as shown in the figure below,


The expressions for shear and normal strain to establish the Mohr’s circle are shown as





Using a basic trigonometric relation (cos + sin = 1) to combine the above two formulas we have,





This equation is an equation for a circle. To make this more apparent, we can rewrite it as,





where,





The circle is centered at the average strain value Avg, and has a radius R equal to the maximum shear strain, as shown in the figure below,


Note


Corresponding variable in the transformation equation for the plane stress and plane strain


By formulae


Since the strain rosette is mounted on the surface of the shaft, where the shaft is in plane strain, we can use the transformation equation for plane strain to calculate the strains in various directions.


 = (x + y ) / + [(x � y) cos] / + [(/xy) sin] /


At  = +450


45 = (x + y ) / + /xy / (1)


At  = -450


-45 = (x + y ) / - /xy / ()


(1) � ()


/xy = 45 - -45 (proven)


By Mohr’s circle


/ /





Y(y, + / /)





x(y, - / / )


min


ava


max


Note


ava = (x + y ) /


R = ͦ{[(x + y ) / ] + (/ / )}


From diagram


min = -45


max = 45


Therefore,


ava = (45 - -45) /


(45 - -45) / = / / (proven)





6. Experimental and Theoretical Results





Table 6.1 Variation of torsion strain versus the torque


W1 = 0, W = (+0.1)x .8N,


Radius of shaft c = 0.008m, L = 80mm (measured)


Bending Moment M = (W1+W) L,


Torque T = W1a1 � Wa,


E = 00 GPa I = 1/64᠋r4 = .17x10-


G = 75 GPa J = 1/᠋r4 = 6.44x10-


Distance a increases (W = kg x .8 m/s )


Experimental Theory


a/m 0 +45 -45 ᠊ = +45--45 M 0 = Mc/EI T ᠊ = Tc/GJ


0.040 67 16 17 5.488 68.5 0.784 1.00


0.060 67 5 14 1 5.488 68.5 1.176 1.0


0.080 67 8 11 7 5.488 68.5 1.568 6.00


0.100 67 41 8 5.488 68.5 1.60 .50


0.10 67 45 5 40 5.488 68.5 .6 .00


0.140 67 48 1 47 5.488 68.5 .744 45.50


Table 6. Variation of bending moment versus bending strain


W1 = W increases, a1 = a = outmost position


From measurement, radius of shaft c = 8mm, L = 80mm


Bending Moment M = (W1 + W)L,


Torque T = 0


E = 00 GPa, I = 1/64᠋r4 = .17x10-


G = 75 Gpa J = 1/᠋r4 = 6.44x10-


a1 = a = 100mm


Experimental Theory


W1=W 0 +45 -45 ᠊ = +45--45 0 = Mc/EI ᠊ = Tc/GJ


0 0 0 0 0 0 0


1 66 5 0 68.1 0


1 4 4 7 16.47 0


18 75 65 10 04.71 0


4 6 86 7 7.5 0


5 0 1 110 1 41.1 0


7. Discussions


7.1 What if the strain gage is not perfectly aligned with the axis of the shaft (See figure below)? Calculate the effect on the strain reading if the deviation is degrees. Compute your data with the theory. What angle, if not degrees, will account for the discrepancy?


7. What if the strain gauge is not perfectly on the top or the bottom of the shaft? (See figure below) What is the effect of a degrees off-centeredness? If your data does not correlate with the theory well, what angle of off-centeredness can account for this discrepancy?


7. Inspect and measure the actual strain gauge bonding and the respective deviation as discussed in (6.1) and (6.).


7.4 What other factors you can think of that has influenced your results. Investigate your hypothesis.


No sensor is perfect. Therefore, we can analyze a measurement system in terms of desired, interfering and modifying inputs. Here we do the same for the specific case of the strain gauge.


1. Desired input Strain


. Interfering input Change of resistance through-


• Vibrations


• Differential thermal expansion


• Electromagnetic interference (EMI), creep, etc.


. Modifying input Changes in gauge factor through-


• Temperature


• Misalignment between gauge and strain. (measurement system rather than gauge error)


• Ageing, corrosion, brittle, bond deterioration, etc.


Experimental errors are likely to be caused by





- Theoretically, the strain gauge should be perfectly on top of the shaft. It should be parallel to the shaft axis. When working on the experiment, to place the strain gauge parallel to the shaft axis is difficult therefore a slight tilt or displacement will result in the experimental error causing inaccurate results obtained.


- Vibration caused by surroundings and humans at the workplace as well as vibration from placing the weight may also caused inaccurate readings.


- Errors in data logger.


Ways to improve in the errors


- Ensure that the data logger to be calibrated before used to minimize errors.


- Readings should be taken only when the weights are in equilibrium condition or stable condition.


8. Conclusion


Both the torsion and bending theories were verified in this experiment.


However, there were some deviations from the theoretical results due to experimental errors especially this point - the strain rosette was noticed to have non-zero readings after the weights were unloaded. This may be due to residual stresses in the shaft, or the wrong choice of measuring equipment was chosen. Time should have been given for the shaft to return to its original state before taking the next reading, or more accurate and stable equipment should have been used.


Also, the choice of plotting the experimental results of t against the theoretical results did not show the deviation of the experimental results. A better method would have been to plot both the theoretical and experimental results of t against a.


The concept of torsion and bending stress was also better understood using this experiment.


. References


.1. Engineering Mechanics of Solids. Egor P. Popov. Prentice Hall, Englewood Cliffs, New Jersey, USA.





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