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Determining Yield Shear Stress and Modulus of Rigidity for Specimen of Brass and Mild Steel - Lab Report Example

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"Determining Yield Shear Stress and Modulus of Rigidity for Specimen of Brass and Mild Steel" paper contains an experiment that there is a linear relationship between torque and rotation. It can be deduced from this experiment, as well that steel has a higher mechanical strength compared to brass. …
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Determining Yield Shear Stress and Modulus of Rigidity for Specimen of Brass and Mild Steel
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Extract of sample "Determining Yield Shear Stress and Modulus of Rigidity for Specimen of Brass and Mild Steel"

? EN1048 Engineering Applications In this experiment the modulus of rigidity and yield shear stress are determined for specimens of mild steel and brass of circular cross section. In this case, the shear property was used to measure the maximum torque measured at the notional strength, rigidity and the shear strength by ?y= yield shear, stress. Generally, as far as this experiment was concerned, three theory equations were applied. These were (1) , (2) And (3) Introduction Most components used in the current world, are characterized by a shear stress often induced in them when a torque is applied (Hearn, 1997). Fenner and other (1984) indicated that the modulus of rigidity is given by the ratio of stress to shear strain often represented mathematically as ?/?. As widely cited, modulus of rigidity is essential in measuring the material’s shear stiffness and it is known to be analogous to Young’s modulus for any elastic behaviour in compression or tension (Benham, Crawford & Armstrong, 1996). Given that a ductile material component useful in providing shear, an individual need to note that the maximum shear stress is safety, which is less than the materials yield shear stress (?y). However, a brittle material the criterion of failure is based upon its tensile stress (Gere &Timoshenko, 1984). Research indicate that the theory of torsion often give the torque in form of rotation as (1) The equation 1 above is often derived from Hooke’s law and it is also valid for the shear stresses though somewhat lower than shearing proportional limit (Case, Chilver & Ross, 1993). In a solid circular section, it has been outlined mathematically that the polar second moment of the area of cross section is given by equation (2) (Ryder, 1969) (2) This implies that from (1), it can be shown that the maximum shear stress, that occurs at r = d/2, is provided by equation (3) (3) This experiment deals with determining yield shear stress along with the modulus of rigidity for the specimen of brass and mild steel of a circular cross section. Apparatus In this experimental set up, there were various sets of apparatus used. Among them were the torsion machine, a torsion bar, a gearbox, a torque meter, two protractors, a counter, and a hand-wheel. This way, a torsion machine was used for carrying out a test involving a specimen that was accommodated within the socket at each end. The 60:1 reduction gearbox served the purpose of manually applying the angular displacement at the left- hand side. The torsion bar was essential in this experiment given that it was used for measuring the torque when at right-side with a torque meter being used for reading digital data. The two protractors listed above were vital in this experiment since they were used for purposes of measuring the rotation of the specimen. They were placed at the gearbox, in which case, one was placed at the input hand wheel along side a 6 degrees scale, while the second gearbox was placed at the specimen shaft at 360 degrees scale. The counter, on the other hand, was used in providing the number of revolutions each of which might be zeroed. Last but not least, the hand-wheel placed at the torque meter assembly was used in returning the right-hand of the given specimen back to the original position. Procedure It is worth enlisting that this lab exercise involved 2 experiments: experiment 1 and experiment 2. This implies that there were specific procedures for specific experimental set up. In experiment 1, which involved determining the modulus of rigidity G for brass and mild steel, the procedure for the experiment involved the following steps. 1. The diameter of the each of the cylindrical portion of the specimen was measured. 2. Using the level, the torque meter hand-wheel was adjusted in ensuring that the torsion bar, as well as the deflection bar is maintained close to and as parallel as possible. 3. The dial gauge was then set to zero through rotating the outer bezel. 4. The specimen was then accommodated in the right hand hexagonal socket followed by accommodating the left hand socket. 5. The specimen was twisted with an increment of one degree while ensuring the dial gauge, which is at the torque meter, is brought to zero. These reading were recorded on the torque meter. 6. The rotation was increased incrementally by one degree while taking the readings. This process was continued until a point where five degrees rotation was applied. 7. While ensuring that the dial gauge is zeroed through adjusting of the hand-wheel before recording the readings, the specimen were untwisted while hand-wheeling to zero loading. 8. The procedure was repeated using other material test. 9. the modulus of rigidity for the specimen was determined using a given equation followed by plotting a graph for the measurements. In experiment 2, which was based upon the determination of yield shear stress (?y) for brass and mild steel, the procedure is as follow. 1. The diameter of each of the cylindrical portion specimen was measured. 2. A bold line was then marked along the specimen’s length. This made it possible to find the axial distribution of the permanent shear distribution deformation through observing the fractured specimen. 3. A specimen was then inserted into a torsion machine just like it had been done before. 4. The protractor scales, as well as the gauge scales were brought to zero. The measurements for this experiment were captured in degrees with the help of the rev. Counter and protractor scales. 5. The specimen was ultimately twisted until it failed to record the digital and rotation data at some suitable increment. This was done while ensuring that the dial gauge is returned to zero through adjusting the hand-wheel. 6. A graph was plotted for the torque versus rotation while determining from the graph the applied torque at yield point (shearing proportional limit. Using equation 3, a shear stress was used to estimate the shear stress (?max) at which the material begins to yield in shear. Results The results were recorded in tables 1, 2, and 3 shown below Table 3: Reading for torgue meter for specimen 2 Rotation (degrees) Rotation (Radians) Torgue (NM) (loading) Torque (NM) (unoading) 1 0.017 0.321 0 2 0.035 0.592 0.354 3 0.052 1.031 0.845 4 0.070 1.659 1.521 5 0.087 2.180 2.180 Table 1 Given that the Modulus of rigidity for the specimen is given by J= ?d4/32 Thus substituting the values from the table into the equation, it can be shown that ?6.4610-3) 4/32 = 1.7110-3 m4 ? = T/? L/J =32.46 0.1/I = 22.5 Gpa Modulus of rigidity for specimen J= ?d4/32 = ?6.4610-3) 4/32= 1.49 10-10 m4 G = T/? L/J = 20 = 13.4 Gpa. Table 4: Reading for maximum torgue specimen 1 Rod rotation in degrees Torgue (NM) 3 1.521 6 4.563 9 8.112 12 11.83 15 14.619 18 16.291 21 17.06 24 18.455 27 18.421 30 18.59 54 18.928 78 18.097 102 18.097 126 19.097 150 19.266 162 Breaks Ґmax = ?y = 16T/?d3= 16 (19.097)/ ?(64 10-3)3 = 360.08mpa Table 5: Reading for maximum torgue specimen 2 Rod rotation in degrees Torgue (NM) 3 0.524 6 1.944 9 3.718 12 5.571 15 7.098 18 8.264 21 8.928 24 9.430 27 9.633 30 9.971 33 10.056 57 10.732 105 11.661 153 12.168 201 12.506 249 12.844 345 13.182 393 13.520 421 13.520 465 14.027 1038 Breaks Ґmax = ?y = 16T/?d3= 16 (13.520)/ ?(62 10-3)3 = 283.4 ?pa. Discussion Basing on the graphs of torgue against Rotation shown above, it is clear that steel has a higher mechanical strength (31.56 Mpa) compared to brass (27.33). Steel can, therefore, be said to be somewhat harder than brass and it can easily be bend. Steel does not either break easily. Moreover, it was observed that the axial distortion of the permanent shear strain of specimen is normal to the longitudinal axis and it does produce the fracture surface along the plane of maximum shear stress. This implies that the failure characteristics indicate that the material is ductile. According to Pytel and Singer (1987) the yield shear asily. stress and the modulus of rigidity (G) are some of the material properties known to be independent of the dimensions of the specimen. In this experimental set up, sources of errors might be from a tortion machine following poor calibration and perhaps from no parallax method. Conclusion It is clear from this experiment that there is a linear relationship between torque and rotation. It can be deduced from this experiment, as well that steel has a higher mechanical strength compared to brass. It is also more rigid that brass given that its value for modulus of rigidity is higher than that of brass. References Hearn, E.J., 1997. Mechanics of Materials, Vol. 1, Butterworth-Heinemann. New York: Oxford Publishers. Pytel, A., &Singer, F.L., 1987. Strength of Material. Ney York: Harper and Row. Fenner, R.T., 1984. Mechanics of Solids. Blackwell. London. Johns and Sons. Benham, P.P., Crawford, R.J., & Armstrong, C.G., 1996. Mechanics of Engineering Materials, London: Longman. Gere, J.M., &Timoshenko, S.P.,1984. Mechanics of Materials. New York: Brooks-Cole. Ryder, G.H., 1969. Strength of Materials. New York: Macmillan. Case, J., Chilver, A.H, & Ross, C.T.F., 1993. Strength of Materials and Structures. London: Arnold. Young, W.C., 1989. Roark’s Formulae for Stress and Strain, London: McGraw Hill. Read More
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