Ben Franklin thought that the North American wild turkey, a brightly plumed, cun

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Ben Franklin thought that the North American wild turkey, a brightly plumed, cunning bird of flight at that time, should become the national bird. Today this bird is bred to profound stupidity (source of the 1970's slang expression for a person doing something stupid, a "turkey") and to oversized proportions, fit only for the dinner plate. Americans consume near 20 lbs of turkey per person per year. Amongst the most favored part of this bird is the "drumstick" which is all "dark meat" and skin, large enough to be an entire meal. The bone in the drumstick is the tibiotarsus. Its response to torsional loading which is equivalent to loading the material in pure shear is to be analyzed in this Assignment. While this is certainly not the way this bone is typically loaded, but just as with an axial test, this test does permit determination of one of the elastic moduli of this material as well as its fracture strength. It should be noted that it is assumed here that the mechanical properties of bone are isotropic that is, they are identical for loadings along any arbitrary direction in the bone. We represent the tibiotarsus bone as a circular, tubular specimen with inner and outer radii given by ri and ro respectively. The loading in torsion about its longitudinal axis is specified by the applied moment (or torque) T. The specimen deforms by twisting which is specified by the angle of twist, , about the central axis. Shear stresses tau are developed in the specimen which vary linearly from zero at the central axis to their maximum value tau max at the outer surface. tau max = Tro/J where: J = pi/2 ( ) (units of [L4]) (1) Here J is the polar moment of inertia of the bone. While the tibiotarsus is not really a circular tube, for this study we shall approximate it so. From geometric considerations the angle of twist of a bone specimen of length L is directly related to the maximum shear strain gamma max at the surface of the specimen. Using the constiutive material "law" tau = G gamma, one obtains = L/ro gamma max = L/ro tau max/G = TL/JG which becomes: T = JG/L (2) It is clear that knowing the dimensions of the bone specimen, that the shear modulus G for the material can be determined from torque/angle-of-twist (T vs. ) data. Data from one such test which is described in detail on the WEB sitea is shown in the figure at right. Fig. 1 - Torsion/twist data from a turkey tibiotarsus specimen. Also evident from the torque/angle-of-twist data is the maximum torque, e.g. Tmax that the specimen was able to sustain. This can be used with Eq. (1) to determine the ultimate shear strength tau ult of the sample. If a specimen is torsionally loaded until the material completely fails, then the failure surface visible on the test specimen provides evidence as to the operative failure mode. Brittle materials tested either in tension or torsion typically fail suddenly by fracture with little apparent yielding. In a tensile test the fracture occurs when the axial, normal stress (e.g. Axial Load/Area) reaches the material's ultimate stress, sigmault. The resulting fracture surface is a flat surface normal to the loading axis. In a torsion test it is the identically same ultimate stress at which the fracture of the specimen occurs. One can easily show that for a specimen loaded in pure shear, the maximum tensile stress is on planes which are oriented at 45degree to the center axis of the specimen. And also, for that case, sigmault = taurlt. A picture of the failed turkey bone is shown at right. You can see that the fracture surface appears to be helical and oriented at approximately 45degree to the specimen axis. Fig. 2 - A fractured turkey tibiotarsus. And the tensile stress which resulted in the fracture of this bone during the torsion test is identical to the tensile stress that would be needed to fracture this material in a simple tensile test. A "virtual lab" detailing the sample preparation, torsional testing and data reduction can be found on the WEB site1. Table I - The dimensions of the test specimen shown were as follows: Specimen gage length, L : 116.4 mm Approximate outer radius, ro : 6.01mm Approximate inner radius, ri: 4.72 mm Tasks: Compute the polar moment of inertia of the specimen. Determine the shear modulus G of the turkey tibitarsus. Use at least five to ten points from the data shown in Fig. 1 and perform a least-squares analysis to determine this modulus. Determine the maximum torque Tult which the tibitarsus withstood. And from this compute the ultimate shear stress tau ult and the maximum tensile strength sigma ult of this bone.

Explanation / Answer

1. Polar moment of inertia J=*(ro4-ri4)/2 J = *(6.014-4.724)/2 J = 1270 mm4 or 1270*10-12m4 2.Take one point T = 8N.m = 0.3 so G = T*L/J G= 8*0.1164/(0.3*1270*10-12 ) G = 2.444 GPa 3.Tult = 12 N.m ult = Tro/J ult = 12*0.00601/1270*10-12 ult = 56.79 MPa ult = 56.79 MPa

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