Torsion Test - ME 220 MECHANICS OF MATERIALS LAB Torsion...

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ME 220 MECHANICS OF MATERIALS LABTorsion LabTest: October 11th, 2007Submitted: October 18th, 2007Matthew OlsenSection 3
SummaryLike the tension test previously, the torsion test is another method of obtaining the stress-strainrelationship of a material. However, while the tension test provides the tensile strain and tensile stress, thetorsion test provides the shear stress and shear strain directly, which makes this test the preferred methodin obtaining said data.Along with the shear data, the modules of resilience and rigidity will also beobtained.To perform this test, a metal rod was fastened to a torsion machine. A troptometer was then usedto measure the angle of rotation. Then torsion was applied until the material fractures. The torsionreadings as well as the angular deformation were collected and used to calculate the stress, strain, andmodules.As expected, the steel performed as a ductile material. The material exhibited elastic behavior upto 2600 in-lbs, giving very little deformation, measuring an angle of 0.069 radians, before yielding tookover. At the 2800 in-lb mark, the specimen gave in to plastic deformation, jumping from the 0.069 radiansto 1.11 radians. At the point of fracture, the material had completed 4.84 revolutions with a maximumtorsion of 5800 in-lbs. The stress and strain at this point were 55819.13 psi and 2.53 in/in respectively.The proportional limit stress and strain were 31767.4 psi and 0.00576 in/in respectively.The cast iron behaved much as it had when put in tension. The material showed very little elasticdeformation, fracturing suddenly. While the steel fracture was clean with visible twist, the cast ironfracture showed a sharp 45 degree angle rotated surface, with many small pieces that flew away.Graphs and data of the experiment can be seen following this lab.
ResultsFrom the experiment, we received the torque and arc length of rotation. To calculate the angle ofrotation, divide the arc length by the radius of rotation, as follows:== ..+ .= .θSR0 24 in0 3735 in 4 448819 in0 048985 radiusCalculating the shear stress requires two different equations. For the linear portion, use thefollowing equation:==.=.τr2Tπr321800 in∙lbπ0 37352in21992 81 psiFollowing the proportional limit, a different equation applies:=+τr12πr3θdTdθ 3T=..-.- .+12π0 37353 in1 106241 radians2800 2600in∙lb1 106241 0 069395radians 32800in∙lb=.26310 08 psiTo calculate the shear strain, apply the following equation:== ...= .γrrθL0 03735 in1 106241 radians4 5 in0 091818 ininThe data table and graphs of torque versus rotation and shear stress versus shear strain. Theproportional limit, as seen in the shear stress-strain graph, is 2600 in-lbs.The shear modulus of rupturecan be calculated as follows:==.=.τu3Tmax2πr335800 in∙lb2π0 3735 in353149 3 psiThe modulus of rigidity is the slope of the elastic region, which is 5,816,168 psi. The modulus ofresilience is calculated as the area under the elastic region as follows:==..=.Modulus of Resilience12γplτpl120 00576 inin31767 4 psi37 79 psi

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Term
Fall
Professor
Ding
Tags
Arc Length, Shear Stress, Strain, Stress, Torsion, maximum shear stress, Matthew Olsen

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