Presentation on theme: "Lab 6: Torsion test (AISI 1018 Steel, cold drawn )"— Presentation transcript:
1 Lab 6: Torsion test (AISI 1018 Steel, cold drawn ) To study the linearly elastic behavior of metallic material under torsion and to determine the shear modulus of elasticity, G, and Poisson’s Ratio, , for metals using torsional stress-strain relationships.To study the complete behavior of metallic materials under torsion and to determine qualitatively the relationship between torsional load and angle of twist for a full range of strains till failure.To determine whether the metallic materials fail in tension, compression, or shear when it is subjected to pure shear.
2 Hooke’s law for shearing stress and strain For any homogeneous isotropic material, the initial portion of the shearing stress-strain diagram is a straight line.Values of shearing stress that do not exceed the proportional limit in shear,txy = G gxytyz = G gyztzx = G gzxxyzt = shearing stress, (psi or N/m2)g = shearing strain, (angle of radians)G = Modulus of rigidity or shear modulus (psi or N/m2)yytyxtyz0.5gzy0.5gxytxytzyxz0.5gxy0.5gzy
3 t = G * g g = r x (f / L) gmax = R x ( f / L ) ; r = rmax = R Determine qualitatively the relationship between torsional load and angle of twist for a full range of strains till failure.T’External T’ = Internal TdFrT = f(r dF)dFdFdFT = fr (t dA)Shearing straindAt = dF/dAT = fr ((r / R) t max dA)T = (t max / R) fr2 dA)T = (t max / R) ( J )tmax = T R / JTL x g = r x fg = r x (f / L)gmax = R x ( f / L ); r = rmax = RG * g = (r / R) G gmaxt = (r / R) t maxt = G * gg/gmax = r / Rg = ( r / R ) x gmaxHooke ’s law in shear
4 Experiment Assumption t = G * gg = t / Ggmax = tmax / Gtmax = T R/ Jgmax = T r / G Jgmax = R x ( f / L )TR x ( f / L ) = T R / G J( f / L ) = T / G Jf = T L / G J to be provef / L = q = angle of twist per unit length according to the test setupT = G ( J * q ); J = p d4 / 32?
5 t (N/m2) T t = T’ t = G * g g T T * R J (ue) = g x 10-6 = shear strain Determine the shear modulus of elasticity, G, and Poisson’s Ratio, , using torsional stress-strain relationships.T * RJTt =; T (N-m)t (N/m2)J = p d4 / 32 ; (m4)T’R = d / 2Circular shafts subjected to torsionGShear stressTt maxta = T * r / Jtmax = T * R / Jrt = G * gg(ue) = g x 10-6 = shear strainEx (ue) = 5 x 10-6
6 Determine the Poisson’s Ratio, , using torsional stress-strain relationships. Poisson ratio (Siméon Denis Poisson ( ) denoted by the ,“nu”The elongation produced by an axial tensile force (P) in the direction of the force (x axis) is accompanied by a contraction in any transverse direction (y and z axis). All materials considered will be assumed to be both homogeneous and isotropic, their mechanical properties will be assumed independent of both position and direction.The strain must have the same value for any transverse direction (y and z axis),sz = sy= 0 ; ey = ez K0 = - sx / E ; sx = P/A = - (lateral strain / axial strain) = -ey / ex = -ez / ex ; ex = sx / ERelationship between E and Gy =EP’- 1x2GPzEs = 200 GPa
7 Experiment Shear stress (T r / J ) The Tinius Olsen torsion testing machineDiameter of specimen……..(19)……….mm = …19… x 10-3 mMeasure Torque T (N-m) load cellMeasure angle of twist per unit length q, (degree)(Note 360 degree = 2p rad )Measure Shear strain g measured by Strain Gage on surface of stressed element.Es = 200 GPa
8 Failure of material under torsion Ductile materials fail in shear when subjected to torsion. Specimens will break along a plane perpendicular to its longitudinal axisBrittle materials generally weak in tension than in shear. When subjected to torsion a specimen tents to break along surfaces which are perpendicular to the direction in which tension is maximum.F’F
9 Analysis of dataPlot the experimented torsional shear stress and shear strain diagram from the first phase of torsion test .Use a spreadsheet to plot the torsional shear stress-shear strain curve from the computer file.This curve extends only up to a stress level below the proportional limit. Determine the experimental shear modulus of elasticity from this curve “G”.Then in conjunction with the Young’s modulus, or modulus of elasticity E, from a reference value, calculate the Poisson’s ratio.Calculate the theoretical torque from experimental q and compare to the experimental torque results.
10 t = G * g = T = G ( J q ) E 2G - 1 Experiment Theory q q * 2p/360 T =E2G- 1T = G ( J q )ExperimentTheoryApplied TorqueTAngle of Twistq% Compareq * 2p/360T(N-m)T(N-m)ExpJ qrad(N-m)(degree)(rad)Theory
11 DiscussionPlot the experimentally obtained torsional shear stress , and shear strain , curves on two graphs, one for the first phase (linearly elastic behavior) and another one for second phase (elastic and plastic behavior) of torsion tests, respectively. Use the data given to you by the instructor.Based on the shear modulus of elasticity G, and the Poisson’s ratio, , obtained from both tests, compare their values with the authoritative tests as given in the text book and other resources. Compare and discuss about G and value with reference.Describe the behavior of this material as it responds to increasing applied torque. Pay special attention to the region above the yield where linear elastic theory no longer applies.Sketch and describe the appearance of the failed bar, and discuss the mode of failure (ductile failure or brittle failure under torsion).
12 Conclusion t = G * g T = G ( J q ) Two equations are proven in this experiment. What are they?Is the Hooke’s law for shearing stress and strain valid?According to your data and analysis, What evidences do you use to support your conclusion?What is the failure modes of ductile and brittle materials when they are subjected to pure shear? Did this material fail in tension, compression or shear? What observations bring you to that conclusion?