BMES 345 CH03 Problem Set 20220927

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BMES 345 PROBLEM SET – CHAPTER 3. LINEAR ELASTICITY Contents Problem 3.1: Linear Elasticity ....................................................................................................................................... 3 Problem 3.2: 1-D Hooke’s Law ...................................................................................................................................... 4 Problem 3.3: Linear Elasticity ....................................................................................................................................... 5 Problem 3.4: 1-D Hooke’s Law ...................................................................................................................................... 6 Problem 3.5: Linear Elasticity ....................................................................................................................................... 7 Problem 3.6: Linear Elasticity ....................................................................................................................................... 8 Problem 3.7: 3-D Hooke’s Law ...................................................................................................................................... 9 Problem 3.8: 3-D Hooke’s Law .................................................................................................................................... 10 Problem 3.9: 3-D Hooke’s Law .................................................................................................................................... 11 Problem 3.10: 3-D Hooke’s Law .................................................................................................................................. 12 Problem 3.11: Linear Elasticity ................................................................................................................................... 13 Problem 3.12: 3-D Hooke’s Law .................................................................................................................................. 14 Problem 3.13: Hooke’s Law in Shear ......................................................................................................................... 15 Problem 3.14: 3-D Hooke’s Law .................................................................................................................................. 16 Problem 3.15: 3-D Hooke’s Law .................................................................................................................................. 17 Problem 3.16: 1-D Hooke’s Law .................................................................................................................................. 18 [Solution] Problem 3.1 ................................................................................................................................................... 19 [Solution] Problem 3.2 ................................................................................................................................................... 20 [Solution] Problem 3.3 ................................................................................................................................................... 21 [Solution] Problem 3.4 ................................................................................................................................................... 22 [Solution] Problem 3.5 ................................................................................................................................................... 23 [Solution] Problem 3.6 ................................................................................................................................................... 25 [Solution] Problem 3.7 ................................................................................................................................................... 26 [Solution] Problem 3.8 ................................................................................................................................................... 28 [Solution] Problem 3.9 ................................................................................................................................................... 30 [Solution] Problem 3.10 ................................................................................................................................................ 33 [Solution] Problem 3.11 ................................................................................................................................................ 35 [Solution] Problem 3.12 ................................................................................................................................................ 36 [Solution] Problem 3.13 ................................................................................................................................................ 38 [Solution] Problem 3.14 ................................................................................................................................................ 40 1

[Solution] Problem 3.15 ................................................................................................................................................ 41 [Solution] Problem 3.16 ................................................................................................................................................ 43 2

Problem 3.1: Linear Elasticity Consider 4 bars of different dimensions, made of the same homogeneous, isotropic linear elastic material (note that the pictures below represent the undeformed dimensions of the bars): Each bar is subjected to the same tensile axial normal strain . Answer the following questions: a) Which bar(s) will experience the highest tensile normal force? b) Which bar(s) will experience the largest change in length? c) Which bar(s) will experience the largest tensile normal stress? d) Which bar(s) will experience the largest lateral normal strain? 3

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Problem 3.2: 1-D Hooke’s Law The following data were collected from compression testing of cortical bone. The cortical bone sample was a cylinder, with a height of 2 mm and a diameter of 6 mm. If the elastic modulus of this cortical bone sample is E = 19 GPa, fill in all of the blanks of the following table: Sample Height (mm) Strain Force (N) Stress (MPa) 2.00 0.000 0.0 0.0 1.99 0.010 13430.3 950.0 4

Problem 3.3: Linear Elasticity A sample of cardiac muscle is subjected to tensile testing. Based on this testing, the properties of the muscle (assuming it is a linear elastic material) are determined to be E = 33 kPa and ν = 0.35. Using this information, complete the missing entries in the table: Sample Length (mm) Sample Width (mm) Force (N) Stress (kPa) Strain 6.00 2.0000 0 0 0 6.05 1.9942 0.00055 0.275 6.10 1.9883 0.00110 0.0167 6.15 1.9825 0.825 0.0250 6.20 0.00220 1.100 0.0333 1.9708 0.00275 1.375 0.0417 5

Problem 3.4: 1-D Hooke’s Law Consider a portion of the femoral artery under tension: The femoral artery segment is subjected to tensile loads F = 0.17 N at each end of the vessel, and the resulting elongation of the vessel is δ = 1 mm. a) What is the elastic modulus of the femoral artery? b) If the tensile load is increased to F = 0.55 N, what is the resulting elongation? 6

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Problem 3.5: Linear Elasticity Consider the following drawing of a person’s flexed arm supporting a weight: In this diagram, d 1 is the distance to the center of mass of the forearm, d 2 is the distance to the center of mass of the ball, and d 3 is the distance to the insertion point of the biceps tendon (all measured from the elbow joint). In this example, the biceps tendon inserts at an angle of 85 o (1.484 radians) to the horizontal—nearly vertical, but not quite. Assume that the biceps is the only muscle doing appreciable work in this scenario, the weight (in this case an ostrich egg) has a mass of 1.4 kg, and the forearm has a mass of 1.1 kg. The biceps tendon has a circular cross section with diameter 4.5 mm. Using MRI, we are able to measure that the axial strain in the tendon is 0.009 and the lateral strain is 0.0027 under these loading conditions. Stating all assumptions, answer the following questions: a) Draw a properly labeled free body diagram for the forearm. Write (but do not solve) the equations of equilibrium necessary to solve this system. b) What is the stress in the biceps tendon? Is this a tensile, compressive, or shear stress? c) What is the shear modulus of the biceps tendon? 7

Problem 3.6: Linear Elasticity During moderate running the tendon experiences an axial tensile load of 4900 N. Let us assume that the Achilles tendon is 6 cm long and has an elliptical cross-section with major and minor axes of 13.3 and 5.6 mm, respectively. Stating all assumptions, determine the following: Assumptions: uniaxial loading, prismatic bar, isotropic, homogenous linear elastic material a) Normal stress in the tendon Area of ellipse  A = πab (where a = half of major axis, b = half of minor axis) Stress  σ = F A = 4900 N π ∗ 6.65 . ∗ 2.8 mm = 83.8 MPa b) Normal strain in the tendon Young’s Modulus  E = σ ε  Strain ε = σ E E (Achilles tendon) ~ 870 MPa ε = 83.8 MPa 870 MPa = 0.096 c) Change in length of the tendon Strain  ℇ = ∆L L 0  ∆ L = ℇ ∗ L 0 = 0.096 * 6 cm = 0.576 cm = 5.76 mm d) If the Poisson’s ratio of the Achilles tendon is ν = 0.38, what is the length of the major axis during the described tensile load? υ = − ε lateral ε axial  ε lateral = v ∗ ε axis = 0.38 *0.096 = -0.03648 ε lateral = ∆ L L 0  ∆ L 13.3 mm = -0.03648  ∆ L = -0.03648*13.3mm = -0.49 L deform = 13.3 − 0.49 = 12.81 mm 8

Problem 3.7: 3-D Hooke’s Law The distal biceps tendon connects the biceps muscle to the radius in the elbow joint. Ruptures of the biceps tendons represent about 3% of all biceps injuries, so there is interest in understanding the properties of these tendons to better prevent injuries and improve repair techniques. Measurements of the distal biceps tendon yielded the following mean dimensions (assuming the cross-section is approximately rectangular):  Length = 63.0 mm  Width = 6.0 mm  Thickness = 3.0 mm From cadaveric studies of the human distal biceps tendon, it was determined that the tendon elastic modulus E = 556 MPa. For the purposes of this problem, we will assume that the Poisson’s ratio ν = 0.3. Based on the provided information, answer the following questions (remember to state assumptions): a) During a peak tensile axial force of 300 N, what are the 3-D stress components in the distal biceps tendon? Please identify all stress components, even if they are zero. b) During a peak tensile axial force of 300 N, what are the 3-D strain components in the distal biceps tendon? Please identify all strain components, even if they are zero. c) What are the deformed dimensions of the distal biceps tendon during that 300 N load? 9

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Problem 3.8: 3-D Hooke’s Law During normal motion, the medial collateral ligament (MCL) is subjected to a tensile axial load of 60 N. Assume the MCL is a prismatic bar with length 10 cm, cross-sectional area 11 mm 2 , an elastic modulus E = 275 MPa, and a Poisson’s ratio ν = 0.4. Remember to state all assumptions. a) What are the normal and shear strains in the x-, y-, and z-directions in the MCL under this load? b) What is the deformed length of the ligament during normal motion? c) What is the cross-sectional area of the ligament under the aforementioned tensile load? 10

Problem 3.9: 3-D Hooke’s Law Most biomechanical studies of ligaments rely on human cadaveric or animal tissue to determine ligament mechanical properties. However, cadaveric tissue and animal models are beset with limitations and caveats, so there has been interest in developing non-invasive, MRI-based techniques to measure ligament biomechanics in situ . Using these imaging approaches, researchers are able to determine that the medial collateral ligament (MCL) has the following undeformed dimensions (assuming a rectangular cross-section):  Length = 112.1 mm  Width = 32.1 mm  Thickness = 2.1 mm Using free body approaches, the researchers are able to determine that the MCL experiences a peak force of 1,670 N (tension) during valgus loading of the knee. Via MRI, the deformed dimensions of the MCL are:  Length = 120.46 mm  Thickness = 2.05 mm Based on the provided information, answer the following questions (remember to state assumptions): a) What are the 3-D stresses in the MCL? Please identify all stress components, even if they are zero. b) What is the elastic modulus of the MCL? c) What is the Poisson’s ratio of the MCL? d) What are the 3-D strains in the MCL? Please identify all strain components, even if they are zero. e) It is hypothesized that ligament fibroblasts (the cells within ligaments) remodel ligament properties to maintain a desired strain. If the tension force increases to 2,200 N, and the ligament fibroblasts want to maintain the same strains as you calculated in part (d), describe one change the cells could make to the ligament’s geometric, material, or structural properties to achieve this outcome. Be concise, specific, and quantitative in your description. 11

Problem 3.10: 3-D Hooke’s Law Researchers at the Institute of Sports Medicine in Copenhagen used ultrasound imaging and biomechanical analysis of knee motion to determine the properties of human patellar tendons in vivo (Hansen et al., Clinical Biomechanics 2006). Using their results, as well as other literature values, we can construct a simple model of the patellar tendon under tension using the following parameters:  E = 1.09 GPa  Poisson’s ratio ν = 0.49  Length of the patellar tendon = 41 mm  Cross-sectional dimensions (assumed elliptical): semi-major axis a = 13 mm, semi-minor axis b = 4 mm (hint: area of an ellipse = πab ) Consider a situation where the patellar tendon is subjected to a tensile force F = 3100 N. Using the above parameters, answer the following questions (remember to state assumptions and show your work): a) What is the normal stress in the patellar tendon? b) What is the deformed length of the patellar tendon? c) What is the deformed cross-sectional area of the patellar tendon? 12

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Problem 3.11: Linear Elasticity The lining of blood vessels, called the endothelium, is made up of cells called endothelial cells. These cells are constantly subjected to shear forces generated by the fluid flow in the circulatory system. Consider an endothelial cell in a coronary artery, as pictured (side view): The cell has a surface area of 280 µm 2 , a height of 2 µm and a length of 30 µm. The shear stress applied to the cell surface by blood flow is  = 55 dynes/cm 2 . Previous studies have shown that an endothelial cell has an approximate shear modulus G = 210 Pa. Answer the following questions (note—1 N = 10 5 dynes): a) What is the shear force applied to the surface of the endothelial cell, in nanonewtons (nN)? b) Assuming that the bottom surface of the cell is strongly adhered to the basement membrane, how much does the top surface displace (in µm)? c) What is the approximate elastic modulus E of the endothelial cell? 13

Problem 3.12: 3-D Hooke’s Law The epithelial cells lining the renal tubules are exposed to fluid flow that applies shear stress to the cells: Grabias BM, Konstantopoulus K. The physical basis of renal fibrosis: effects of altered hydrodynamic forces on kidney homeostasis. Am J Physiol Renal Physiol (2014). We can model a single renal epithelial cell exposed to fluid shear as follows: The cell has a luminal surface area (in contact with the fluid) of 99 µm 2 , a height of 15 µm and a length of 10 µm. The force applied to the cell surface is V = 4.26 x 10 -10 N and the resulting displacement of the cell’s top surface is measured (via microscopic image analysis) to be 0.18 µm in the direction of fluid flow. Assume that the cell can be modeled as a rectangular prism (constant width), the basal surface is anchored perfectly to the basement membrane, and the cell is exposed to pure shear. Also assume that only the top surface (not the sides) are in contact with flowing fluid and therefore the applied force. Answer the following questions (state all additional assumptions): a) What are the 3-D stresses in the epithelial cell? Please identify all stress components, even if they are zero. b) What are the 3-D strains in the epithelial cell? Please identify all strain components, even if they are zero. c) What is the cell’s shear modulus G ? 14

Problem 3.13: Hooke’s Law in Shear During loading of the knee, the femur translates relative to the tibia, applying shear forces to the articular cartilage that surfaces the bones. Let us consider Scenario A (pictured), where the articular cartilage of the tibia experiences pure shear in response to a shear stress τ = 29 kPa. Assuming that we can treat articular cartilage as an isotropic, homogeneous, linear elastic material, answer the following questions: a) What is the resulting shear strain γ ? b) What is the shear modulus G of articular cartilage? c) One potential complication in understanding the mechanics of articular cartilage is that it contains chondrocytes (cartilage cells) that have significantly different properties than the surrounding tissue. If we now consider Scenario B, how would this affect the assumptions that we made regarding articular cartilage? 15

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Problem 3.14: 3-D Hooke’s Law Let us consider a bone screw used in a plate and screw fixation to treat a bone fracture. The screw is subjected to the forces as pictured: What are the stresses and strains in the screw along the cross-sectional plane ( A = 12.6 mm 2 ) that lies between the plate and the bone (marked by the dotted line)? The screw is made of Ti-6Al-7Nb, which has properties E = 105 GPa, G = 39 GPa, and ν = 0.36. Remember to state all assumptions. 16

Problem 3.15: 3-D Hooke’s Law During loading of the knee, the femur translates relative to the tibia, simultaneously applying both normal and shear stresses to the articular cartilage that surfaces the bones. Consider the pictured situation, where the articular cartilage of the tibia experiences the following stresses (all other stress components are zero): σ x = 100 kPa σ y =− 300 kPa τ xy = 50 kPa We know from a previous experiment that the elastic modulus of tibial articular cartilage is E = 5 MPa and the Poisson’s ratio is ν = 0.4. We are also able to observe the deformation of an initially square material element within the cartilage (shown above). Answer the following questions (remember to show your work): a) Please state your assumptions. - Prismatic bar, uniaxial loading, homogenous, isotropic linear elastic materials b) What are the 3-D strain components in the articular cartilage during loading? Make sure to identify all components, even if they are zero. 5 MPa  5000 kPa ε x = σ x E − v ( σ y E + σ z E )= 100 kPa 5000 kPa − 0.4 ( − 300 kPa 5000 kPa + 0 ) = 0.044 ε y = σ y E − v ( σ x E + σ z E ) = − 300 kPa 5000 kPa − 0.4 ( 100 kPa 5000 kPa + 0 ) = -0.068 ε z = σ z E − v ( σ x E + σ y E ) = 0 − 0.4 ( 100 kPa 5000 kPa + − 300 5000 kPa ) = 0.016 γ xy = τ xy G = 2 ( 1 + v ) E τ xy = 2 ( 1 + 0.4 ) 5000 kPa ( 50 kPa ) = 0.028 γ x z = 0 17

γ yz = 0 c) What are the dimensions b, h, and Δx for the deformed material element? b = b 0 ( 1 + ε x ) = ( 0.75 mm ) ( 1 + 0.044 mm ) = 0.783 mm h = h 0 ( 1 + ε y ) = ( 0.75 mm ) ( 1 − 0.068 ) = 0.699 mm γ xy = tan ( γ ¿¿ xy )= ∆ x h →∆ x = γ xy h = ( 0.028 ) ( 0.699 ) = 0.020 mm ¿ 18

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Problem 3.16: 1-D Hooke’s Law Titanium alloys are frequently used in total joint replacements. The provided Excel file (ch3_ti6al7nb) contains force-displacement data obtained from a tensile test of a specimen made from Ti-6Al-7Nb: 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 Force (N) Displacement (mm) The specimen had an initial length L = 50 mm, and cross-sectional dimensions w = 12.5 mm and t = 2 mm (following guidelines from ASTM E8/E8M-21, Standard Test Methods for Tension Testing of Metallic Materials ). From the force-displacement data, determine the elastic modulus E for Ti-6Al- 7Nb. Assumptions: homogenous material, prismatic bar, uniaxial loading and linear elastic material We know: Stress  σ = Eε Therefore: elastic modulus  E = σ ε Using the excel file, we calculate the strain and stress from the initial time to the time in which deformation begins to occur. Recall: Strain  ℇ = ∆L L 0 Stress  σ = F A @ time = 1 s ℇ = 0.025 mm 50 mm = 0.00025 σ = 787.25 N 12.5 mm ∗ 2 mm = 31.49 MPa 19

We do these calculations until deformation begins to occur; around time = 20 s (eye-balled). We then plot the stress vs strain, and the slope would be the elastic modulus for this specimen. Based on the stress vs strain plot above, the slope is 109376 MPa (~109 GPa) which is the elastic modulus for this specimen. 20

[Solution] Problem 3.1 Consider 4 bars of different dimensions, made of the same homogeneous, isotropic linear elastic material (note that the pictures below represent the undeformed dimensions of the bars): Each bar is subjected to the same tensile axial normal strain . Answer the following questions: a) Which bar(s) will experience the highest tensile normal force? Since all the bars are made of the same linear elastic material, Hooke’s law tells us that for a given normal strain, we will get the same normal stress in every bar. Thus, the bar that will experience the highest tensile force will be the one with the largest cross-sectional area, which is bar D . b) Which bar(s) will experience the largest change in length? Since all the bars are experiencing the same axial normal strain, the longest bars will have the largest change in length. Thus, bars C and D will experience the largest change in length. c) Which bar(s) will experience the largest tensile normal stress? Since all the bars are made of the same linear elastic material, Hooke’s law tells us that for a given normal strain, we will get the same normal stress in every bar . d) Which bar(s) will experience the largest lateral normal strain? Since the bars are all made of the same linear elastic material, that means they will have the same Poisson’s ratio. Since each bar is subjected to the same axial normal strain, the Poisson effect dictates that every bar will experience the same lateral normal strain. 21

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[Solution] Problem 3.2 The following data were collected from compression testing of cortical bone. The cortical bone sample was a cylinder, with a height of 2 mm and a diameter of 6 mm. If the elastic modulus of this cortical bone sample is E = 19 GPa, fill in all of the blanks of the following table: Sample Height (mm) Strain Force (N) Stress (MPa) 2.00 0.000 0.0 0.0 1.99 0.005 2686.1 95.0 1.98 0.010 5372.1 190.0 1.95 0.025 13430.3 475.0 1.9 0.050 26860.6 950.0 To fill out this table, you need to assume that cortical bone behaves like a linear elastic material, and then use the following relationships: A = π r 2 = π ( 3 mm ) 2 = 28.27 mm 2 E = 19 GPa σ = F A ε = δ L = h − h 0 h 0 σ = Eε 22

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[Solution] Problem 3.3 A sample of cardiac muscle is subjected to tensile testing. Based on this testing, the properties of the muscle (assuming it is a linear elastic material) are determined to be E = 33 kPa and ν = 0.35. Using this information, complete the missing entries in the table: Sample Length (mm) Sample Width (mm) Force (N) Stress (kPa) Strain 6.00 2.0000 0 0 0 6.05 1.9942 0.00055 0.275 0.0083 6.10 1.9883 0.00110 0.55 0.0167 6.15 1.9825 0.00165 0.825 0.0250 6.20 1.9766 0.00220 1.100 0.0333 6.25 1.9708 0.00275 1.375 0.0417 Row 2 ε = δ L = 6.05 mm − 6 mm 6 mm = 0.0083 Alternatively, you can use the stress and the Young’s modulus to find the strain. Row 3 σ = Eε = ( 33 kPa ) ( 0.0167 ) = 0.551 kPa Row 4 We need to find the cross-sectional area of the sample to find the force in row 4 (we will use row 2’s data): σ = F A → A = F σ = 0.00055 N 275 Pa = 0.000002 m 2 σ = F A →F = σA = ( 825 Pa ) ( 0.000002 m 2 ) = 0.00165 N Row 5 We need to use the Poisson’s ratio to find the new width of the sample. ν = − ε ' ε →ε ' =− νε =− ( 0.35 ) ( 0.0333 ) =− 0.011655 ε ' = ∆w w →∆ w = wε ' = ( 2 mm ) ( − 0 . 011655 ) =− 0 . 02331 mm 2 mm − 0.02332 mm = 1.9767 mm Row 6 Here we use the strain and stretch ratio to find the deformed length. 23

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L = λL 0 = ( 1 + ε ) L 0 = ( 1 + 0.0417 ) ( 6 mm ) = 6.25 mm [Solution] Problem 3.4 Consider a portion of the femoral artery under tension: The femoral artery segment is subjected to tensile loads F = 0.17 N at each end of the vessel, and the resulting elongation of the vessel is δ = 1 mm. a) What is the elastic modulus of the femoral artery? δ = FL EA → E = FL Aδ A = π ( r 2 2 − r 1 2 ) = π [ ( 9 mm ) 2 −( 8.3 mm ) 2 ] = 38.04 mm 2 E = ( 0.17 N )( 100 mm ) ( 38.04 mm 2 )( 1 mm ) = 0.447 MPa = 447 kPa b) If the tensile load is increased to F = 0.55 N, what is the resulting elongation? Since we are treating the artery as a linear elastic material, the change in force should cause a proportional change in elongation of the artery (assuming the artery remains linear elastic at this higher stress / strain level). 0 . 55 N 0 . 17 N = δ 1 mm →δ = 3 . 24 mm 24

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[Solution] Problem 3.5 Consider the following drawing of a person’s flexed arm supporting a weight: In this diagram, d 1 is the distance to the center of mass of the forearm, d 2 is the distance to the center of mass of the ball, and d 3 is the distance to the insertion point of the biceps tendon (all measured from the elbow joint). In this example, the biceps tendon inserts at an angle of 85 o (1.484 radians) to the horizontal—nearly vertical, but not quite. Assume that the biceps is the only muscle doing appreciable work in this scenario, the weight (in this case an ostrich egg) has a mass of 1.4 kg, and the forearm has a mass of 1.1 kg. The biceps tendon has a circular cross section with diameter 4.5 mm. Using MRI, we are able to measure that the axial strain in the tendon is 0.009 and the lateral strain is 0.0027 under these loading conditions. Stating all assumptions, answer the following questions: a) Draw a properly labeled free body diagram for the forearm. Write (but do not solve) the equations of equilibrium necessary to solve this system. 26

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The equations of equilibrium for this system are two balance equations for force and one balance equation for a moment about a point (we will use the elbow): ∑ F x = 0 = F jx − F biceps cos1 . 484 ∑ F y = 0 = F biceps sin 1 . 484 − F jy − W arm − W egg ∑ M elbow = 0 = ( F biceps sin 1 . 484 ) d 3 − ( W arm ) d 1 − ( W egg ) d 2 b) What is the stress in the biceps tendon? Is this a tensile, compressive, or shear stress? To find the stress in the biceps tendon, we need to solve for the biceps force. We can do this using the moment balance equation alone. F biceps = W arm d 1 + W egg d 2 ( sin 1 . 484 ) d 3 = ( 1 . 1 kg ) ( 9 . 81 m s − 2 ) ( 15 cm ) +( 1 . 4 kg )( 9 . 81 m s − 2 )( 35 cm ) ( 0 . 9962 )( 4 cm ) F biceps = 161 . 3 N Now we assume the biceps tendon is under uniaxial load and has a constant cross-section to find the stress in the tendon: σ = F A = 161.3 N π ( 2.25 mm ) 2 = 10.14 MPa This stress tends to stretch or elongate the tendon, so it is a tensile normal stress. c) What is the shear modulus of the biceps tendon? If we assume the tendon is a linear elastic material, we can relate the stress to the strain using Hooke’s law: σ = Eε→ E = σ ε = 10.14 MPa 0.009 = 1126.7 MPa To find the shear modulus from the elastic modulus, we need to first find the Poisson’s ratio: ν = − ε ' ε = 0.0027 0.009 = 0.3 G = E 2 ( 1 + ν ) = 1126.7 MPa 2 ( 1 + 0.3 ) = 433.3 MPa 27

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[Solution] Problem 3.6 During moderate running the tendon experiences an axial tensile load of 4900 N. Let us assume that the Achilles tendon is 6 cm long, and has an elliptical cross-section with major and minor axes of 13.3 and 5.6 mm, respectively. Stating all assumptions, determine the following: a) Normal stress in the tendon We will assume that the tendon can be modeled as a prismatic bar with an elliptical (as described). The normal stress is given by: σ = F A = 4900 N ( 0.25 ) π ( 13.3 mm )( 5.6 mm ) = 83.8 MPa b) Normal strain in the tendon Here we assume that the tendon behaves like a linear elastic (isotropic, homogeneous) material and thus we can use 1-D Hooke’s law. Using that assumption, we need to either estimate or look up the Young’s modulus of a human Achilles tendon. Depending on your source, you will likely find values that range from 400 MPa to 1.1 GPa. Using results from Lichtwark and Wilson (J Exp Biol 2005, http://jeb.biologists.org/content/208/24/4715 ), we will use a Young’s modulus E = 0.87 GPa: ε = σ E = 83.8 MPa 0.87 GPa = 83.8 MPa 870 MPa = 0.0963 = 9.63% This is a great example of precisely why we measure the mechanical properties of materials. It allows us to determine how much stress or strain a material will experience when loaded, and thus lets us model that behavior ahead of time. c) Change in length of the tendon Using the normal strain calculated in part (b), we can find the change in length of the tendon: δ = Lε = ( 6 cm ) ( 0.0963 ) = 0.578 cm = 5.78 mm d) If the Poisson’s ratio of the Achilles tendon is ν = 0.38, what is the length of the major axis during the described tensile load? Recall that the Poisson’s ratio relates the lateral strain to the axial strain. We can determine the lateral strain, and then use the definition of normal strain to determine the new length of the major axis of the tendon. ν = − ε ' ε →ε ' =− νε =− ( 0.38 ) ( 0.0963 ) =− 0.0366 ε ' = δ L →δ = Lε ' = ( 13 . 3 mm ) ( − 0 . 0366 ) =− 0 . 487 mm 28

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L + δ = 13.3 mm − 0.487 mm = 12.813 mm [Solution] Problem 3.7 The distal biceps tendon connects the biceps muscle to the radius in the elbow joint. Ruptures of the biceps tendons represent about 3% of all biceps injuries, so there is interest in understanding the properties of these tendons to better prevent injuries and improve repair techniques. Measurements of the distal biceps tendon yielded the following mean dimensions (assuming the cross-section is approximately rectangular):  Length = 63.0 mm  Width = 6.0 mm  Thickness = 3.0 mm From cadaveric studies of the human distal biceps tendon, it was determined that the tendon elastic modulus E = 556 MPa. For the purposes of this problem, we will assume that the Poisson’s ratio ν = 0.3. Based on the provided information, answer the following questions (remember to state assumptions): a) During a peak tensile axial force of 300 N, what are the 3-D stress components in the distal biceps tendon? Please identify all stress components, even if they are zero. Key assumptions 1. Isotropic, homogeneous, linear elastic material 2. Prismatic bar 3. Uniaxial loading With these assumptions in place, we can determine the stress components. Since this is a uniaxial loading scenario, all the stress components except one normal stress are zero. We will define the axial direction as the x-direction (you can define any of the coordinate directions as the axial direction): σ x = F A = 300 N ( 6 mm )( 3 mm ) = 300 N 18 mm 2 = 16 . 67 MPa σ y = σ z = τ xy = τ xz = τ yz = 0 b) During a peak tensile axial force of 300 N, what are the 3-D strain components in the distal biceps tendon? Please identify all strain components, even if they are zero. Because it is uniaxial loading and there is only one non-zero stress component (the normal stress in the axial direction), we immediately know that all shear strain components are zero: γ xy = γ xz = γ yz = 0 We can then use 3-D Hooke’s law to find the 3 normal strain components: ε x = 1 E [ σ x − ν ( σ y + σ z ) ] = 16 . 67 MPa 556 MPa = 0 . 030 29

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ε y = 1 E [ σ y − ν ( σ x + σ z ) ] =− ( 0 . 3 ) 16 . 67 MPa 556 MPa =− 0 . 009 ε z = 1 E [ σ z − ν ( σ x + σ y ) ] =− ( 0 . 3 ) 16 . 67 MPa 556 MPa =− 0 . 009 c) What are the deformed dimensions of the distal biceps tendon during that 300 N load? We can use the strain components from part (b) and the original dimensions to find the deformed dimensions of the tendon during loading: ε x = ∆ L L →∆ L = ( 0 . 030 ) ( 63 . 0 mm ) = 1 . 89 mm L ' = L + ∆ L = 64 . 89 mm Since the normal strains in the y- and z-directions are the same, it does not matter which direction is matched up with which dimension: ε y = ∆w w →∆w = ( − 0 . 009 ) ( 6 . 0 mm ) =− 0 . 054 mm w ' = w + ∆w = 5 . 946 mm ε z = ∆t t →∆t = ( − 0 . 009 ) ( 3 . 0 mm ) =− 0 . 027 mm t ' = t + ∆t = 2 . 973 mm 30

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[Solution] Problem 3.8 During normal motion, the medial collateral ligament (MCL) is subjected to a tensile axial load of 60 N. Assume the MCL is a prismatic bar with length 10 cm, cross-sectional area 11 mm 2 , an elastic modulus E = 275 MPa, and a Poisson’s ratio ν = 0.4. Remember to state all assumptions. a) What are the normal and shear strains in the x-, y-, and z-directions in the MCL under this load? Assumptions:  Linear elastic material  Isotropic material  Homogeneous material  (given) Prismatic bar  Uniaxial load We can use our version of 3-D Hooke’s law, noting that all stress components are zero except for a single normal stress (we will set the long axis of the MCL in the x-direction): σ x = F A = 60 N 11 mm 2 = 5 . 455 MPa ε x = 1 E [ σ x − ν ( σ y + σ z ) ] = 5 . 455 MPa 275 MPa = 0 . 0198 ε y = 1 E [ σ y − ν ( σ x + σ z ) ] = − ( 0 . 4 ) ( 5 . 455 MPa ) 275 MPa =− 0 . 0079 ε z = 1 E [ σ z − ν ( σ x + σ y ) ] = − ( 0 . 4 ) ( 5 . 455 MPa ) 275 MPa =− 0 . 0079 γ xy = τ xy G = 0 γ xz = τ xz G = 0 γ yz = τ yz G = 0 b) What is the deformed length of the ligament during normal motion? We use the x-normal strain and our definition of normal strain to find the change in length: = → = = . = . 𝜺 𝜹𝑳 𝜹 𝜺𝑳 𝟎 𝟎𝟏𝟗𝟖𝟏𝟎 𝒄𝒎 𝟎 𝟏𝟗𝟖 𝒄𝒎 We then add this change in length to the original length (since the MCL is under tension) to obtain a deformed length of 10.198 cm. c) What is the cross-sectional area of the ligament under the aforementioned tensile load? We can describe the cross-sectional area as the integral of all the differential area: 31

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A = ∫ dA = ∫ dydz →dA = dydz During loading, the differential dimensions dy and dz will change according to the y- and z- normal strains, i.e., the lateral normal strain (due to the Poisson effect): dA ' = dy ' dz ' = ( 1 + ε y ) dy ( 1 + ε z ) dz = dydz ( 1 + ε ' ) 2 = dA ( 1 + ε ' ) 2 ∫ dA ' =( 1 + ε ' ) 2 ∫ dA = ( 1 − 0 . 00792 ) 2 ( 11 mm 2 ) = 10 . 826 mm 2 32

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[Solution] Problem 3.9 Most biomechanical studies of ligaments rely on human cadaveric or animal tissue to determine ligament mechanical properties. However, cadaveric tissue and animal models are beset with limitations and caveats, so there has been interest in developing non-invasive, MRI-based techniques to measure ligament biomechanics in situ . Using these imaging approaches, researchers are able to determine that the medial collateral ligament (MCL) has the following undeformed dimensions (assuming a rectangular cross-section):  Length = 112.1 mm  Width = 32.1 mm  Thickness = 2.1 mm Using free body approaches, the researchers are able to determine that the MCL experiences a peak force of 1,670 N (tension) during valgus loading of the knee. Via MRI, the deformed dimensions of the MCL are:  Length = 120.46 mm  Thickness = 2.05 mm Based on the provided information, answer the following questions (remember to state assumptions): a) What are the 3-D stresses in the MCL? Please identify all stress components, even if they are zero. Assumptions: 1. Uniaxial tension along the length, defined as the x-direction 2. MCL is a prismatic bar Since it is uniaxial tension, all stress components in the x-, y-, and z-directions are zero except for the normal stress in the x-direction. We calculate that using equation for normal stress: σ x = F A = 1670 N ( 32 . 1 mm )( 2 . 1 mm ) = 1670 N 67 . 41 mm 2 = 24 . 77 MPa σ x = 24 . 8 MPa σ y = σ z = 0 τ xy = τ xz = τ yz = 0 b) What is the elastic modulus of the MCL? Assumptions: 1. MCL is made of an isotropic, homogeneous, linear elastic material We can directly calculate the elastic modulus of the MCL from 1-D Hooke’s law: 33

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σ x = E ε x → E = σ x ε x Therefore, we need to find the normal strain in the x-direction. We can use the undeformed and deformed length of the MCL to find this value: ε = δ L = 120.46 mm − 112.1 mm 112.1 mm = 8.36 mm 112.1 mm = 0.0746 E = σ x ε x = 24.77 MPa 0.0746 = 332 MPa c) What is the Poisson’s ratio of the MCL? We can determine the Poisson’s ratio by calculating the transverse normal strain, e.g., the strain in the direction of the thickness of the MCL: ε = δ L = 2.05 mm − 2.1 mm 2.1 mm = − 0.05 mm 2.1 mm =− 0.0238 ν = − ε ' ε = − ( − 0.0238 ) ( 0.0746 ) = 0.319 d) What are the 3-D strains in the MCL? Please identify all strain components, even if they are zero. Since we assumed the MCL is an isotropic, homogeneous, linear elastic material in part (b), we can use Hooke’s law in 3-D to find the strains. Since the shear stress components are all zero, the shear strain components are also all zero. ε x = 1 E [ σ x − ν ( σ y + σ z ) ] = 24 . 77 MPa 332 MPa = 0 . 0746 ε y = 1 E [ σ y − ν ( σ x + σ z ) ] = − ( 0 . 319 ) ( 24 . 77 MPa ) 332 MPa =− 0 . 0238 ε z = 1 E [ σ z − ν ( σ x + σ y ) ] = − ( 0 . 319 ) ( 24 . 77 MPa ) 332 MPa =− 0 . 0238 γ xy = γ xz = γ yz = 0 Alternatively, we already calculated the normal strain in the x-direction in part (b). In part (c), we calculated the transverse normal strain, which is the same in both the y- and z-directions because this is a uniaxial loading situation. e) It is hypothesized that ligament fibroblasts (the cells within ligaments) remodel ligament properties to maintain a desired strain. If the tension force increases to 2,200 N, and the ligament fibroblasts want to maintain the same strains as you calculated in part (d), describe one change the cells could make to the ligament’s geometric, material, or structural properties to achieve this outcome. Be concise, specific, and quantitative in your description. 34

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To maintain the same strain in response to a 2,200 N uniaxial tensile force (an increase of 31.7%), one of the following changes would need to occur:  The elastic modulus of the ligament could increase commensurate with the increased stress (due to the increased force), resulting in no net change in the normal strain. The elastic modulus would need to increase to 437 MPa.  The cross-sectional area of the ligament could increase to match the increased force, resulting in no change in normal stress and therefore no change in normal strain (assuming the elastic modulus is unchanged). The cross-sectional area would need to increase to 88.8 mm 2 (up from 67.4 mm 2 ).  The stiffness of the ligament could increase (by changes in the cross-sectional area and elastic modulus in combination) commensurate with the increased force, resulting in no net change in the elongation of the ligament and therefore the normal strain. The ligament’s current stiffness is (1670 N / 8.36 mm) = 199.8 N/mm. To accommodate the new force, the ligament’s stiffness would need to increase to (2200 N / 8.36 mm) = 263.2 N/mm (a 31.7% increase). 35

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[Solution] Problem 3.10 Researchers at the Institute of Sports Medicine in Copenhagen used ultrasound imaging and biomechanical analysis of knee motion to determine the properties of human patellar tendons in vivo (Hansen et al., Clinical Biomechanics 2006). Using their results, as well as other literature values, we can construct a simple model of the patellar tendon under tension using the following parameters:  E = 1.09 GPa  Poisson’s ratio ν = 0.49  Length of the patellar tendon = 41 mm  Cross-sectional dimensions (assumed elliptical): semi-major axis a = 13 mm, semi-minor axis b = 4 mm (hint: area of an ellipse = πab ) Consider a situation where the patellar tendon is subjected to a tensile force F = 3100 N. Using the above parameters, answer the following questions (remember to state assumptions and show your work): a) What is the normal stress in the patellar tendon? We use our definition of normal stress during axial loading and the cross-sectional area of the tendon: A = πab = π ( 13 mm ) ( 4 mm ) = 163.4 mm 2 σ = F A = 3100 N 163.4 mm 2 = 18.97 MPa = 1.897 x 10 7 Pa b) What is the deformed length of the patellar tendon? First we need to determine the normal strain that occurs during loading. Here we use the normal stress (from part a) and assume that the patellar tendon behaves according to Hooke’s law: σ = Eε ε = σ E = 18.97 MPa 1090 MPa = 0.0174 We can then determine the deformed length by multiplying the original length by the stretch ratio: L = L 0 λ = L 0 ( 1 + ε ) = ( 41 mm ) ( 1 + 0.0174 ) = 41.71 mm c) What is the deformed cross-sectional area of the patellar tendon? Here we realize that the lateral dimensions (semi-major and semi-minor axes) will contract as the patellar tendon is placed under tension, due to the Poisson effect. We can use the 36

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Poisson’s ratio to determine the new dimensions and then calculate the cross-sectional area. We find the lateral strain: ν = − ε ' ε ε ' =− νε =− ( 0 . 49 ) ( 0 . 0174 ) =− 0 . 00853 We then use the lateral strain to find the deformed cross-sectional dimensions, in a manner similar to part (b): a = a 0 λ = a 0 ( 1 + ε ' ) b = b 0 λ = b 0 ( 1 + ε ' ) A = πab = π a 0 b 0 ( 1 + ε ' ) 2 = π ( 13 mm ) ( 4 mm ) ( 1 − 0.00853 ) 2 = 160.6 mm 2 37

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[Solution] Problem 3.11 The lining of blood vessels, called the endothelium, is made up of cells called endothelial cells. These cells are constantly subjected to shear forces generated by the fluid flow in the circulatory system. Consider an endothelial cell in a coronary artery, as pictured (side view): The cell has a surface area of 280 µm 2 , a height of 2 µm and a length of 30 µm. The shear stress applied to the cell surface by blood flow is  = 55 dynes/cm 2 . Previous studies have shown that an endothelial cell has an approximate shear modulus G = 210 Pa. Answer the following questions (note—1 N = 10 5 dynes): a) What is the shear force applied to the surface of the endothelial cell, in nanonewtons (nN)? The shear force is equal to the shear stress times the area over which the shear stress is acting, in this case the top surface of the cell: V = τA = (55 dyne cm -2 )(280 µm 2 ) = (55 dyne cm -2 )(280 x 10 -8 cm 2 ) = 0.000154 dynes = 1.54 nN b) Assuming that the bottom surface of the cell is strongly adhered to the basement membrane, how much does the top surface displace (in µm)? We can use Hooke’s law for shear (if we assume the cell is made of an isotropic, homogeneous, linear elastic solid) to determine the shear strain of the endothelial cell, and then using the height of the cell, we can determine the displacement of the cell’s upper surface: τ = Gγ → γ = τ/G = (55 dyne cm -2 )/(210 Pa) = 5.5 Pa / 210 Pa = 0.0262 γ ≈ tan γ = x / h → x = γh = (0.0262)(2 µm) = 0.0524 µm = 52.4 nm c) What is the approximate elastic modulus E of the endothelial cell? This requires an estimate of the Poisson’s ratio of the cell (or looking it up). If we take ν = 0.3, we can use the following equation relating the shear modulus to the elastic modulus of an isotropic, homogeneous, linear elastic solid: G = E 2 ( 1 + ν ) → E = 2 G ( 1 + ν ) = 2 ( 210 Pa ) ( 1 + 0.3 ) = 546 Pa We can reasonably assume a Poisson’s ratio in the range of 0.2 to 0.5 for the endothelial cell. 38

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[Solution] Problem 3.12 The epithelial cells lining the renal tubules are exposed to fluid flow that applies shear stress to the cells: Grabias BM, Konstantopoulus K. The physical basis of renal fibrosis: effects of altered hydrodynamic forces on kidney homeostasis. Am J Physiol Renal Physiol (2014). We can model a single renal epithelial cell exposed to fluid shear as follows: The cell has a luminal surface area (in contact with the fluid) of 99 µm 2 , a height of 15 µm and a length of 10 µm. The force applied to the cell surface is V = 4.26 x 10 -10 N and the resulting displacement of the cell’s top surface is measured (via microscopic image analysis) to be 0.18 µm in the direction of fluid flow. Assume that the cell can be modeled as a rectangular prism (constant width), the basal surface is anchored perfectly to the basement membrane, and the cell is exposed to pure shear. Also assume that only the top surface (not the sides) are in contact with flowing fluid and therefore the applied force. Answer the following questions (state all additional assumptions): a) What are the 3-D stresses in the epithelial cell? Please identify all stress components, even if they are zero. Key assumptions: 1. Isotropic, homogeneous, linear elastic material 2. Loading results in pure shear With these assumptions in place, we observe that there is a tangential (shear) force acting in one direction, so this is a case of pure shear with only one non-zero shear stress component (we will arbitrarily pick the xy component; it does not matter which one you choose); all other 39

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stress components are zero. We can calculate the non-zero shear stress by using the shear force V and the area upon which it is acting, i.e., the luminal surface area: τ xy = V A = 4 . 26 x 10 − 10 N 99 x 10 − 12 m 2 = 4 . 30 Pa σ x = σ y = σ z = τ xz = τ yz = 0 b) What are the 3-D strains in the epithelial cell? Please identify all strain components, even if they are zero. Because it is pure shear and there is only one non-zero shear stress component, we immediately know that all normal strain components are zero. Similarly, since shear stresses do not interact in pure shear of an isotropic, linear elastic material, 2 of the shear strain components are also zero: ε x = ε y = ε z = γ xz = γ yz = 0 To find the one non-zero shear strain component (the xy component), we use our definition of shear strain, the displacement of the top surface of the cell, and the height of the cell: tan γ xy = 0.18 μm 15 μm We can either solve for the shear strain using the equation above, or realizing the strain is small here, use a small angle approximation for the tangent: tan γ xy ≅ γ xy →γ xy = 0.18 μm 15 μm = 0.012 c) What is the cell’s shear modulus G ? Using the shear stress from part (a), the shear strain from part (b), and shear equations for 3-D Hooke’s law, we can find the shear modulus: γ xy = τ xy G →G = 4 . 30 Pa 0 . 012 = 358 Pa 40

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[Solution] Problem 3.13 During loading of the knee, the femur translates relative to the tibia, applying shear forces to the articular cartilage that surfaces the bones. Let us consider Scenario A (pictured), where the articular cartilage of the tibia experiences pure shear in response to a shear stress τ = 29 kPa. Assuming that we can treat articular cartilage as an isotropic, homogeneous, linear elastic material, answer the following questions: a) What is the resulting shear strain γ ? We can calculate the shear strain experienced by the articular cartilage in Scenario A using the provided geometry: γ ≈ tan γ = 4 μm 50 μm = 0.08 Alternatively, you can forego the approximation and calculate the shear strain exactly: tan γ = 0.08 γ = tan − 1 ( 0.08 ) = 0.0798 b) What is the shear modulus G of articular cartilage? Here we use the assumption that cartilage is an isotropic, homogeneous, linear elastic material and invoke Hooke’s law in shear: τ = Gγ G = τ γ = 29 kPa 0.08 = 362.5 kPa 41

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c) One potential complication in understanding the mechanics of articular cartilage is that it contains chondrocytes (cartilage cells) that have significantly different properties than the surrounding tissue. If we now consider Scenario B, how would this affect the assumptions that we made regarding articular cartilage? We assumed that cartilage is homogeneous; however, if there are cells in the tissue that take up a non-negligible fraction of the tissue, and have significantly different properties, we can no longer assume that cartilage properties do not vary with position. Furthermore, if a cell is not isotropic or linear elastic (could be viscoelastic or nonlinear elastic), their presence could impact our assumptions of an isotropic, linear elastic material. 42

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[Solution] Problem 3.14 Let us consider a bone screw used in a plate and screw fixation to treat a bone fracture. The screw is subjected to the forces as pictured: What are the stresses and strains in the screw along the cross-sectional plane ( A = 12.6 mm 2 ) that lies between the plate and the bone (marked by the dotted line)? The screw is made of Ti-6Al-7Nb, which has properties E = 105 GPa, G = 39 GPa, and ν = 0.36. Remember to state all assumptions. If we assume that Ti-6Al-7Nb behaves like an isotropic, homogeneous, linear elastic material, we can use 3-D Hooke’s law. We start by observing that, for a linear elastic material, the stress at the identified location is simply the superposition of the conditions causing pure shear (i.e., the shear force V ) and the conditions causing the normal stress (i.e., the uniaxial force F ). In this case, I will define the force F as acting in the y-direction, so that gives normal stresses: σ y = F A = 50 N 12 . 6 mm 2 = 3 . 97 MPa We observe that the other two normal stress components are zero . Similarly, all the shear stress components are zero except for one (we will say the xy component): τ average = τ yx = V A = 100 N 12 . 6 mm 2 = 7 . 94 MPa With these two being the only stress components, we can superpose them and input them into 3-D Hooke’s law to find the strains: ε x = 1 E [ σ x − ν ( σ y + σ z ) ] = −( 0 . 36 )( 3 . 97 MPa ) 105 , 000 MPa =− 1 . 36 x 10 − 5 ε y = 1 E [ σ y − ν ( σ x + σ z ) ] = 3 . 97 MPa 105 , 000 MPa = 3 . 78 x 10 − 5 ε z = 1 E [ σ z − ν ( σ x + σ y ) ] = −( 0 . 36 )( 3 . 97 MPa ) 105 , 000 MPa =− 1 . 36 x 10 − 5 γ xy = τ xy G = 7 . 94 MPa 39 , 000 GPa = 0 . 000204 γ xz = τ xz G = 0 γ yz = τ yz G = 0 43

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[Solution] Problem 3.15 During loading of the knee, the femur translates relative to the tibia, simultaneously applying both normal and shear stresses to the articular cartilage that surfaces the bones. Consider the pictured situation, where the articular cartilage of the tibia experiences the following stresses (all other stress components are zero): σ x = 100 kPa σ y =− 300 k Pa τ xy = 50 kPa We know from a previous experiment that the elastic modulus of tibial articular cartilage is E = 5 MPa and the Poisson’s ratio is ν = 0.4. We are also able to observe the deformation of an initially square material element within the cartilage (shown above). Answer the following questions (remember to show your work): a) Please state your assumptions. Assumptions 1. Linear elastic material 2. Isotropic material 3. Homogeneous material b) What are the 3-D strain components in the articular cartilage during loading? Make sure to identify all components, even if they are zero. Because of our assumptions, and given that we are provided the stress components, we can find the strain components using 3-D Hooke ’s law: 45

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ε x = 1 E [ σ x − ν ( σ y + σ z ) ] = 100 kPa 5000 kPa − ( 0 . 4 ) ( − 300 kPa ) 5000 kPa = 0 . 044 ε y = 1 E [ σ y − ν ( σ x + σ z ) ] = − 300 kPa 5000 kPa − ( 0 . 4 ) ( 100 kPa ) 5000 kPa =− 0 . 068 ε z = 1 E [ σ z − ν ( σ x + σ y ) ] = ( − 0 . 4 ) ( 100 kPa − 300 kPa ) 5000 kPa = 0 . 016 γ xy = τ xy G = 2 ( 1 + ν ) E τ xy = 2 ( 1 + 0 . 4 ) 5000 kPa ( 50 kPa ) = 0 . 028 γ xz = τ xz G = 0 γ yz = τ yz G = 0 c) What are the dimensions b, h, and Δx for the deformed material element? We can directly apply the strain components we found in part b to find the deformed dimensions of the material element. The dimensions b and h depend on the original dimensions (0.75 mm x 0.75 mm) and the x- and y- normal strain components: b = b 0 ( 1 + ε x ) = ( 0.75 mm ) ( 1 + 0.044 ) = 0.783 mm h = h 0 ( 1 + ε y ) = ( 0.75 mm ) ( 1 − 0.068 ) = 0.699 mm To find Δx , we need to use the xy-shear strain. Since we assumed linear elasticity, we also effectively assumed small deformations (or small angles), so we can use approximations for the tangent of the shear strain angle (alternatively you do not need to use small angles and can use the tangent to find the displacement, but the difference will be very small): γ xy ≅ tan γ xy = ∆ x h ∆ x = γ xy h = ( 0.028 ) ( 0.699 mm ) = 0.020 mm 46

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[Solution] Problem 3.16 Titanium alloys are frequently used in total joint replacements. The provided Excel file (ch3_ti6al7nb) contains force-displacement data obtained from a tensile test of a specimen made from Ti-6Al-7Nb: 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 Force (N) Displacement (mm) The specimen had an initial length L = 50 mm, and cross-sectional dimensions w = 12.5 mm and t = 2 mm (following guidelines from ASTM E8/E8M-21, Standard Test Methods for Tension Testing of Metallic Materials ). From the force-displacement data, determine the elastic modulus E for Ti-6Al- 7Nb. Assuming that our specimen is a prismatic bar undergoing uniaxial tension (which are good assumptions given that this is a uniaxial tension test of a specially fabricated Ti-6Al-7Nb sample), and that we can treat Ti-6Al-7Nb as a homogeneous, linear elastic material, we can compute the normal stress and normal strain from the provided force and displacement data: σ = F A = F ( 12.5 mm )( 2 mm ) ε = ∆ L L 0 = ∆ L 50 mm The result is the normal stress – normal strain curve. To find the elastic modulus, we find the slope of the stress-strain curve. In this case, we would want to identify the linear region of the curve, which appears to extend to about ε = 0.005. If we do a linear regression of that region, we obtain the elastic modulus E : 47

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y = 109376x - 0.2255 R² = 0.9994 -100 0 100 200 300 400 500 600 700 800 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010 Stress (MPa) Strain In this case, the slope = 109,376 MPa, or E = 109 GPa. We can also see that the data beyond ε = 0.005 begin to deviate significantly from the linear trendline, so selecting an appropriate linear region is very important. 48

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BMES 345 CH03 Problem Set 20220927

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