176_HW6

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C176

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Mechanical Engineering

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Oct 30, 2023

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1 ME C176/BIOE C119 H OMEWORK 6 1. A sagittal slice from an elderly human vertebral body is shown in the Figure 1 , with the plastic (UHMWPE, elastic modulus E PL ) component of a “total disc replacement” attached on top. The endplate has been removed during surgery and you can assume that the plastic implant is supported solely by the underlying trabecular bone. Vertical loading (point force P ) from the articulating metal component (not shown) occurs at the mid-point. Assume we can model this structure as a beam on an elastic foundation, in which the plastic implant is the beam of thickness d and length L and the trabecular bone is a uniform elastic foundation of effective height H . (i) Derive the relationship between k (the “foundation modulus”) and E TB (the Young’s modulus of the trabecular bone) in terms of the relevant dimensions. Assume that the bone and implant have a uniform depth b into the page. (ii) In the context of BOEF, explain the concept of “rigid” versus “flexible” behaviors and distinguish them from “compliant” and “stiff” behaviors. Recall: (iii) If H = 40 mm, b = 30 mm, L = 50 mm, E UHMWPE =1000 MPa, and E TB =500 MPa, calculate the minimum value of plastic thickness required to ensure there is no lift-off in the trabecular bone foundation, for this central loading condition. (iv) For part (iii), how would this minimum value of plastic thickness change for an old osteoporotic individual vs. a young active individual. Discuss briefly any implications for design of this type of implant. 2. A finite-length composite beam is supported by an elastic foundation ( Figure 2 ). The beam is 50 mm long and 5 mm wide and is made of a 2-mm-thick layer of CoCr alloy with a 6- mm-thick layer of polyethylene bonded on top (see cross section in Figure 1). The polyethylene has a modulus of 1.0 GPa and the CoCr has a modulus of 210 GPa. The beam is supported by cancellous bone having a Young’s modulus 500 MPa. Recall that k ¹ E for the foundation. A 1000 N load is applied at the center of the beam. Determine the maximum stress in the beam and plot a graph of the strain in the bone (elastic foundation) for each point along the length of the beam. The foundation strain is the foundation deflection normalized by the “effective l = k 4 EI 4 Figure 1

2 height” h* of the foundation. Although the value of h* is ambiguous, for the purposes of this problem assume it is 40 mm. For force P, lambda, foundation stiffness k, length L, and distances a and b, the equation for deflection v(x) of a finite beam, from page 208, is: v=(P*lambda/k)*1/((sinh(lambda*L))^2-(sin(lambda*L))^2)*(A*B+C*(E+F)) where: A=2*cosh(lambda*x)*cos(lambda*x); B=sinh(lambda*L)*cos(lambda*a)*cosh(lambda*b)- sin(lambda*L)*cosh(lambda*a)*cos(lambda*b); C=cosh(lambda*x)*sin(lambda*x)+sinh(lambda*x)*cos(lambda*x); E=sinh(lambda*L)*(sin(lambda*a)*cosh(lambda*b)-cos(lambda*a)*sinh(lambda*b)); F=sin(lambda*L)*(sinh(lambda*a)*cos(lambda*b)-cosh(lambda*a)*sin(lambda*b)); The moment M(x) within a finite beam, from page 209, is: M=P/(2*lambda)*1/((sinh(lambda*L))^2-(sin(lambda*L))^2)*(A*B+C*(E+F)); where: A=2*sinh(lambda*x)*sin(lambda*x); B=sinh(lambda*L)*cos(lambda*a)*cosh(lambda*b)- sin(lambda*L)*cosh(lambda*a)*cos(lambda*b); C=cosh(lambda*x)*sin(lambda*x)-sinh(lambda*x)*cos(lambda*x); E=sinh(lambda*L)*(sin(lambda*a)*cosh(lambda*b)-cos(lambda*a)*sinh(lambda*b)); F=sin(lambda*L)*(sinh(lambda*a)*cos(lambda*b)-cosh(lambda*a)*sin(lambda*b)); Figure 2. Finite-length beam on elastic foundation with centered loading (left) and composite cross-section (right).

176_HW6

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