Drexel University

Drexel University
School of Biomedical Engineering, Science & Health Systems

Homework
Biomaterials Module
Assigned: 4 €“ 18 €“ 13
Due: 4 €“ 25 €“ 13 in class or Bb Learn(late submissions not accepted)
Reading

Textbook: Enderle, J., Blanchard, S.M. &Bronzino, J. (2006). INTRODUCTION TO BIOMEDICAL ENGINEEERING, 3rd EDITION. Academic Press. ISBN-13: 978-0-12-374979-6 (IBME)

Chapter 4 €“ Biomechanics: Sections 4.3 (2nd and 3rdEds)
Chapter 5 €“ Biomaterials: Sections 5.1 €“ 5.2 (3rd Ed), 6.1 €“ 6.2 (2nd Ed)

Biomechanics Module HomeworkProblem

Solve the Example Problem 4.9 (3rd Ed) or 4.8 (2nd Ed) for orthopedic nail-plate used to fix an intertrochanteric fracture. The hip applies an external force of 400 N at angle of 20o on the nail-plate during static standing as shown in the fig. 4.16 (3rd Ed) or 4.13 (2nd Ed.) below. The nail-plateextends axially from the greater trochanter through the femoral neck to the femoral head. It is rectangular stainless steel with cross-sectional dimensions of 10 mm (width) by 5 mm (height), and6 cm in length. It is well-fixed with screws onto the outside of the femoral shaft and isfriction fit into the trochanteric head. Find the i) forces, ii) moments, iii) stresses, and iv)strainsthat will develop in the nail-plate for the following cases:
A) The external hip force is 400 N at an angle of 20o(this is the book specification).
B) The external hip force is 400 N at an angle of 0o.
C) If the external hip force rises to 800 N at an angle of 20o, chart in a table (like the table below)whether the following materials will be adequate to meet the load demands. If not, discuss how it will fail.
a. Wrought stainless steel, cold worked,
b. Aluminum, 2024-T4,
c. UHMWPE,
d. Zirconia, Yttria stabilized,
e. Wrought Co-Cr-Mo alloy.

ONLY SOLVE PART C!

Part C. Table of Material Fitness for the Nail-Plate
Material Stress Loading Strain Loading Shear Stress Loading Elastic Modulus Any other parameters of your choice
A Can material A meet this demand Can material A meet this demand Can material A meet this demand Can material A meet this demand Can material A meet this demand
B Can material B meet this demand Etc. Etc. Etc. Etc.
C Can material C meet this demand Etc. Etc. Etc. Etc.
D Can material D meet this demand Etc. Etc. Etc. Etc.
E Can material E meet this demand Etc. Etc. Etc. Etc.
Fx=800 cos(20) = 751.75 N
Fy= 800 sin(20) = 273.62 N
Ma= 273.62* 0.06 = 16.42 NM
Stress = Fx/A = 751.75 / (0.005 * 0.01) = 1.50 MPa
Strain = F/EA = 751.75 / (180*10-9 (0.005* 0.01)) = 83.53*10-6

Material Stress Loading
(MPa) Strain Loading Shear Stress Loading Elastic Modulus
(Gpa) Stress yield
(MPa) Any other parameters of your choice
A 1.50
Meet demand 7.52×10-5
Meet demand 8.21
Meet demand 200
Meet demand 690 Can material A meet this demand
B 1.50
Meet demand 2.06 x10-4
Meet demand 8.21
Meet demand 73
Meet demand 303 Etc.
C 1.50
Meet demand 0.03
Meet demand 8.21
Meet demand 0.5
Meet demand 14 Etc.
D 1.50
Meet demand 7.52×10-5
Meet demand 8.21
Meet demand 200
Meet demand 250 Etc.
E 1.50
Meet demand 7.16×10-5
Meet demand 8.21
Meet demand 210
Meet demand 490 Etc.

All materials meet the demand since all forces are below the yield stress.