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1.INTRODUCTION1.OBJECTIVE1.THEORY1.1Torsion of Circular Elastic Bars1.Bending of Circular Elastic Bars.Combined Bending and Torsion of Circular Elastic Bars4.EXPERIMENTAL PROCEDURE64.1Apparatus and Equipment64.Procedure84..1Variation of Torques Versus Torsional Strain 4..Variation of Bending Moment Versus Bending Strain 5.RESULTS105.1Variation of Torques Versus Torsional Strain 105.Variation of Bending Moment Versus Bending Strain 116..DISCUSSIONS147.CONCLUSIONS168.REFERENCES16.APPENDIX and LOG SHEETS (i) AbstractThe circular shaft is mainly use to transmit torque in rotary machines and acts as a beam in the bending theory. The aim of this experiment is to verify the simple bending and torsion theories. Torsion and bending of circular elastic bars were carried out with different weights and equal weights respectively. The readings for strain measurement are then obtained from the strain gauge. 1. IntroductionThe 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 of the shaft is governed by the theory of circular. 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 the combined torsional and bending stress.. Theory .1 Torsion of Circular Elastic BarsTo 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 follows1 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 g 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 t is proportional to shear strain g.In the elastic case, on the basis of the previous assumptions, and at any give section a relationship can be written as follows,(tmax/c)¨°A rdA = T, where tmax and c are constants.(1)However, ¨°A rdA, 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, tmax = Tc/J, ()which 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,gmax = tmax/G = Tc/(JG),()where T is the applied torque, c is the radius of the circular shaft, J = pd4/, 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 BarsThe 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¦Ê = 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 Hooke¡¯s law for uniaxial stress applies.¦ max = E¦Åmax (6)Substitute (5) into (6), the maximum bending strain can be derived as¦Åmax = Mc / ( EI ) (7)M = applied bending momentE = 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 BarsThe 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. Below are the general expressions for normal stress and shear stress, respectively on any plane located by the angle q and caused by a known system of stresses, in order to establish the Mohr¡¯s circle. Using a basic trigonometric relation (cos¦È + sin¦È = 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,two principle stressesMaximum shear stressAs for the expressions for shear and normal strain to establish the Mohr¡¯s circle. The Equations are as below. Using a basic trigonometric relation (cos¦È + sin¦È = 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 ¦ÅAvg, and has a radius R equal to the maximum shear strain, as shown in the figure below,Two principle strainsMaximum shear strainNoteCorresponding variable in the transformation equation for the plane stress and plane strain4. Experimental Procedures4.1 Apparatus and Equipments(a) Solid Circular shaft (b) Strain Gauge(c) (i) Strain Rosette(ii) Loading frame(iii) Weights(Iv) Vernier Caliper 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, which is firmly fixed to the end of the shaft. A series of notches on the loading frame allow for variation of moment arm 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 ¨C Wa (counterclockwise +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 atrain rosette, with measurements of e0, e+45 and e-45. The strain measurement e0 directly measures the bending strain and the strain measurements e+45 and e-45 together can be used to measure the torsional shear strain with the use of Mohr¡¯s circle. The torsional strain can be obtained as, et = (e+45 - e-45)/ = g/(8)4. ProceduresThere 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 torques versus torsional strain 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 e0, e+45, and e-45.Neither W1 nor W should exceed 5kg.Once you have reached the end, reverse the direction of shift W where, the weights are keep 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 e0, e+45, and e-45 in the table provided. The total weight of W1 + W should not exceed 16 kg, in order not to damage the specimen. 5. Experimental and Theoretical Results Table 5.1 Variation of torques versus the torsional strain W1 = 0, W = (+0.1)x.8N,Radius of shaft c = 0.008m, L = 0.168m (measured)Bending Moment M = (W1+W)L, Torque T = W1a1 ¨C Wa,E = 00 GPa I = 1/4pr4 = .7x10-G = 75 GPaJ = 1/pr4 = 7.45x10-Distance a increasesExperimentalTheorya/me0e+45e-45g = e+45-e-45Me0 = Mc/EITg = Tc/GJ0.0457185.1557.-1.-1.0.065105.1557.-1.0-8.0.085455.1557.-.5-7.40.10617075.1557.-.1-46.50.1614-4475.1557.-.74-55.60.1464-7565.1557.-4.6-64.80.1665-14665.1557.-4.7-7.80.186457-18755.1557.-5.58-8.Table 5. Variation of torques versus the tortional strain Distance a decreasesExperimentalTheorya/me0e+45e-45g = e+45-e-45Me0 = Mc/EITg = Tc/GJ0.18660-14745.1557.-5.58-8.0.1665-1165.1557.-4.7-7.80.14550-16665.1557.-4.6-64.80.1584-55.1557.-.74-55.60.1057545.1557.-.1-46.50.0856575.1557.-.5-7.40.0655855.1557.-1.0-8.0.045145.1557.-1.-1.Table 5. Variation of bending moment versus bending positionW1 = W increases, a1 = a = outmost position From measurement, radius of shaft c = 0.008m, L = 0.168m,Bending Moment M = (W1 + W)L,Torque T = 0 E = 00 GPa, I = .7x10-G = 75 Gpa J = 7.45x10-ExperimentalTheoryWe0e+45e-45g = e+45-e-45e0 = Mc/EIg = Tc/GJ00000001.1x.844118-541.40.1x.8700-1078.00.1x.811545-1114.704.1x.8155464-1151.0 Graph of Experimental e0 against Theoretical e0 and Experiment g against Theoretical g Table 5.1 and Table 5.Graph of Experimental e0 against Theoretical e0 and Experiment g against Theoretical gTable 5.6. Discussions6.1 What are the likely experimental errors during the experiment. 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 caliberated before used to minimize errors.- Readings should be taken only when the weights are in equilibrium condition or stable condition.6. Prove equation (14) by formulae, as well as Mohr¡¯s circle.By formulaeSince 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 direction. ¦Å¦È = (¦Åx + ¦Åy ) / + [(¦Åx ¨C ¦Åy) cos¦È] / + [(¦Ãxy) sin¦È] / At ¦È = +450¦Å45 = (¦Åx + ¦Åy ) / + ¦Ãxy / (1) ¦È = -450¦Å-45 = (¦Åx + ¦Åy ) / - ¦Ãxy / ()(1) ¨C ()¦Ãxy = ¦Å45 - ¦Å-45 (proven)By Mohr¡¯s circle¦Å¦Ã / Y(¦Åy, + ¦Ã /)¦Åx(¦Åy, - ¦Ã / ) ¦Åmin ¦Åava ¦ÅmaxNote ¦Åava = (¦Åx + ¦Åy ) / R = ¡Ì{[(¦Åx + ¦Åy ) / ] + (¦Ã / )}From diagram¦Åmin = ¦Å-45¦Åmax = ¦Å45Therefore, ¦Åava = (¦Å45 - ¦Å-45) / (¦Å45 - ¦Å-45) / = ¦Ã /(proven) 8. Conclusion This experiment discusses the bending and torque transmitted to the shaft when load is applied to it. The main objective was met successfully. Firstly, we have found that the magnitude of the experimental torsional shear strain g and bending strain e0 was similar to the magnitude of the theoretical torsional shear strain g bending strain e0. And by plotting the graphs of experimental e0 and g against theoretical e0 and g we obtain linear graphs where e0 is vertical straight line in the positive region while g increasing linearly in the negative region. The negative region is due to the direction of the Torque, T we calculated using W1a1- Wa taking the counter-clockwise to be positive. By drawing the Mohr¡¯s circle, the maximum shear strain and maximum shear stress can be proven at the location of the experimented circular shaft. In conclusion, it is seen that the theory torsion and bending have been verified and the understanding of combined torsional and bending stress . References .1. Engineering Mechanics of Solids. Egor P. Popov. Prentice Hall, Englewood Cliffs, New Jersey, USA.. 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