April 10, 2018 UNIVERSITY OF RHODE ISLAND Department of Electrical, Computer and Biomedical Engineering BME 207 Introduction to Biomechanics Spring 2018 Homework 9 Prolem 1 The intertrochanteric nail from Homework 4 is used to repair a hip fracture. The stainless steel nail and the femur (cortical one) have these properties: stainless one steel units tension compression shear all loads E 180 GPa 15 17 G 77 GPa 3.3 ν 0.28 0.41 σ yield 502 MPa 80 135 σ fail 860 MPa 127 184 78 The unloaded nail is 6 cm long with a rectangular cross-section, 10 mm wide 5 mm high. At the proximal end, the nail stailizes the femoral head and neck (caput femoris and collum femoris); the distal end is fixed to a vertical support which is attached to the femoral shaft (corpus femoris). The hip applies a 400 N force when the patient stands still. Compute the: A. average normal stress (along the axial direction) in the nail; B. average shear stress at the proximal tip of the nail; C. the dimensions of the nail when loaded. D. Sketch the shear force V and ending moment M along the length of the nail. 400 N 20 Compute the: E. maximum flexural stress and its axial location along the nail (call this location X); F. maximum shear stress at X. G. Sketch the flexural and shear stresses on the eam profile at X. H. Compute and sketch the net axial stress at X and identify the neutral axis. I. At X, sketch material elements located at y = +h/2, y = 0, and y = h/2. J. At X, compute the principal stresses, the angle of the principal stress planes, the maximum shear stress, and the angle of the maximum shear stress planes at each of the y locations listed aove. K. If these stresses were applied to an unroken, healthy femur (with no nail), would the femur deform elastically, plastically, or fail? Why? 6 cm - 1 -
Answers A. σ = 7.52 MPa (compressive) B. τ = 2.74 MPa C. L X = 5.99975 cm, L Y = 10.0 + 117 10 6 cm, L Y = 5.0 + 58.5 10 6 cm E. σ flex = 196.8 MPa (at y = +h/2), σ flex = 196.8 MPa (at y = h/2) F. τ max = 4.1 MPa H. σ net = 189.3 MPa (at y = +h/2) and 204.3 MPa (at y = h/2) J. at y = +h/2: σ 1 = 189.3 MPa, σ 2 = 0 MPa, θ P = 0, τ max = 94.6 MPa, θ S = 45-2 -
Prolem 2 A cantilever eam supports a vertical load on the free end. The cross-sectional area is 5.5 cm 2. The cross-sectional face can e one of the five geometries elow. If the ending moment at some particular location is M = 250 N-cm, which geometry provides the lowest maximum flexural stress at the same location? Recall the flexure formula is σ x = My/I, where y is measured from the neutral axis. rectangle h = 2.500 cm = 2.200 cm h I = h3 12 circle = 1.323 cm I = π 4 r4 o annulus = 1.500 cm r i = 0.707 cm r i I = π 4 ( r 4 o r 4 i ) I-eam h = 5.000 cm = 5.000 cm H = 4.416 cm B = 4.416 cm h H B/2 I = h3 BH 3 12 H-eam h = 5.000 cm = 5.000 cm H = 4.416 cm B = 4.416 cm B/2 I = (h H) 3 + H ( B) 3 12 H h - 3 -
Prolem 3 A different cantilever eam is eing tested as part of the design for a prosthetic tiia. The cross section can e one of the four geometries elow. Which cross section provides the lowest maximum shear stress when the load is pure torsion? For a general purpose prosthetic tiia, ending stresses and shear stresses oth need to e considered. Which cross section provides the lowest maximum flexural (ending) stress? circle = 1.323 cm I = π 4 r4 o J = π 2 r4 o annulus = 1.500 cm r i = 0.707 cm r i I = π ( ) r 4 4 o ri 4 J = π ( ) r 4 2 o ri 4 a elliptical a = 1.500 cm ar = 1.167 cm I = π 4 a3 J = π 4 a ( a 2 + 2) τ max = 4V 3A area = A = πa elliptical tue a o = 2.150 cm a i = 1.600 cm o = 1.582 cm i = 1.032 cm a o a i o i I = π 4 J = π 4 τ max = 4V 3A ( a 3 o o a 3 i i ) [ ( ) ( )] ao o a 2 o + 2 o ai i a 2 i + 2 i area = A = π (a o o a i i ) - 4 -
Prolem 4 A pure ending load acts on a cantilever eam with a rectangular cross section. The two plots elow show the shear stress profile at some particular location on the eam s length. = 10 cm + h 2 y, m τ(y) = 7500 750000 y 2 Pa same h = 20 cm τ(y) h 2 0 + h 2 y, m h 2 A. What is the shear force V that generates the shear stress? Hint: Integrate (in the y direction) the product of the shear stress and the differential area over the cross-sectional face of the eam: 000 111 h dy so the differential area = da = dy, then: V = τ da = +h/2 τ dy h/2 B. What is the average shear stress? C. What is the maximum shear stress? Compute this two ways: using the τ max equation in the textook, and using the function τ(y) aove. Answers A. V = 100 N B. τ = 5000 Pa C. τ max = 7500 Pa - 5 -
Prolem 5 A downhill skier hits a fixed rock at a distance d = 90 cm anterior to the centerline of the tiia. The rock generates a force F = 16.67 N perpendicular to the long axis of the ski. The minimum cross-section of the tiia is modeled as an elliptical tue, with an outer major axis of 20 mm and a minor axis of 16 mm. The inner major axis is 18 mm and the minor axis is 14.5 mm. Since the ski-rock impact happens quickly, we can assume the shear failure stress is the same as the shear ultimate stress, which is approximately 60 MPa. Is the tiia likely to fracture? Regardless of the previous answer, if the tiia does fracture, what will the crack pattern look like? y d x - 6 -