BMEG 311 Homework 3

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West Virginia University *

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311

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Mechanical Engineering

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Apr 3, 2024

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BMEG 311 Biomaterials Homework 3 Due Thursday, September 14 th before 11:59PM EST Madison Paris 1. A force of 85 kN is applied to a bar with cross-sectional dimensions of 10mm x 22mm. Given the information above and the data for hypothetical materials in Table 1, show your work and answer the following questions: σ = 85000 N / 2.2 cm^2 = 38636.36 Pa / 100 = 386.36 MPa a. Which bar materials will plastically deform? Explain why. 386.36 > 330 386.36 < 410 386.36 > 150 Material A and material C will plastically deform because the calculated stress (386.36 MPa) exceeds both yield strengths of 330 MPa and 150 MPa. b. Which bar materials will experience necking? Explain why. 386.36 < 400 386.63 < 570 386.63 > 210 Material C will experience necking because the calculated stress (386.63 MPa) is greater than its ultimate tensile strength (210 MPa) Table 1. Data for Hypothetical Bar Materials Material Yield Strength (MPa) Young’s Modulus (GPa) Ultimate Tensile Strength (MPa) Strain at Fracture Fracture Strength (MPa) A 330 210 400 0.23 380 B 410 310 570 0.15 520 C 150 140 210 0.40 180

2. Before testing, a sample was 30 mm long; at fracture, the sample was extended to 53 mm long. The stress-strain curve is shown below. Show your work and determine the following properties of the material (if there are no calculations needed, describe how you determined the values): a. Ultimate Tensile Strength (UTS) Highest point on the stress-strain curve. b. Yield Strength Parallel offset line 0.2% positive strain from the linear portion. c. Fracture Strength Max strength of a material before it fractures, or the end point of the line. d. Young’s Modulus Slope of the linear portion of the line before its yield strength. e. % Elongation % E = ∆L / L0 = (53 - 30) / 30 = 0.76 = 76% f. For A-D, label each value on the graph

3. When a person is moving, the knee undergoes a range of forces during the cycle of the step. For a person walking, the maximum load the knee undergoes is 3900 N. The average length of articular cartilage is 3.50 mm and it covers the tibia with an average cross-sectional diameter of 60 mm. Load = 3900 N length = 3.5 mm diameter = 60 mm a. Using the stress-strain curve for cartilage below, determine the change in cartilage length when a person is walking. Stress = F/A = 3900 N/ 0.00283 m^2 = 1379342.9 Pa = 1.38 MPa E = (2.1-1.05)/ (0.1-0.05) = 21 MPa Strain = stress/E = 1.38 MPa / 21 MPa = 0.066 MPa ∆L = strain * L0 = 0.066 MPa * 0.0035 m = 0.000231 m = 0.231 mm b. If a person increases their walking speed, would the cartilage show signs of creep, stress relaxation or strain rate dependence? Briefly explain your answer and describe how creep, stress relaxation and strain rate dependence are different. When a person increases their walking speed, the primary response in cartilage is likely to be strain rate dependence due to the change in the rate at which it experiences deformation. Creep and stress relaxation can occur over more extended time periods, they are less likely to occur during a brief increase in walking speed. Creep is a time-dependent deformation that occurs when a material is put under a constant load or stress over an extended period. Stress relaxation is also time-dependent, but involves a constant strain while the stress or load decreases over time. Strain rate dependence refers to how a material's mechanical properties change in response to the rate at which it is deformed.

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4. When going to the gym, lifting free weights directly above your head is a mode of compressive loading. Assume a 30 kg bar with weights applies a 275 N force on the humerus bones. The Young’s Modulus of the long bone is 22 GPa, the Ultimate Compressive Strength is 180 MPa, and the Poisson’s ratio is 0.4. Knowing the initial length of the bone is 38 cm and the initial diameter is 3 cm, what is the change in the radius of the humerus bone? Show your work for all steps. *For compressive loading, the applied force is treated as negative, so it will be -275 N, but the equations are the same as shown in the slides besides this change. Diameter= 3 cm -> radius= 1.5 cm -> 0.015 m F= -275 N L= 0.38 m E= 22 GPa -> 22,000,000,000 Pa v= 0.4 A= pi * r^2= pi * (0.015 m)^2= 0.000707 m^2 Stress = F/A = -275 N / 0.000707 m^2 = -388967.47 Pa Strain = stress/E = -388967.47 Pa / 22,000,000,000 Pa = -1.768 x 10^-5 Pa εlat = - εlong * v = -1.768 x 10^-5 * 0.4 = -7.072 x 10^-6 ∆r = r0 * εlat = 0.015 m * 7.072 x 10^-6 = 1.061 x 10^-7 m