6834 Lecture Notes - Lecture 8: Ground Reaction Force, Lumbar Vertebrae, Compressive Stress
21 Jun 2018
School
Department
Course
Professor
BIOMECHANICS 1/G MECHANICS OF BIOLOGICAL SYSTEMS 1
TASKS – SUGGESTED ANSWERS
TASK 1 (RELATES TO STRESSES INFORMATION)
Think of 2 activities or actions that relates to each stress type (compression, tension
and shear). It can look at whole body or bones or muscles etc.
Examples
Compression – humerus when doing a push-up / ground reaction force when
walking / spine compression in standing
Tensile – Humerus when doing a chin up, Eccentric muscle action / ACL in
knee valgus
Shear – force in lumbar vertebrae in a Deadlift
TASK 2 (RELATES TO STRESSES INFORMATION)
How much compressive stress is on the L1, L2 vertebral disc of 625 N women, given
that approximately 45% of her body weight is supported by the disc when she stands
in the anatomical position. (assume the disc is oriented horizontally and has a
surface area of 20 cm2)
Solution: (remove before publishing)
What we know:
Force = (625N) * 0.45
Area = 20 cm2
Formula: Stress = Force / Area
Solution:
Stress = (625N * 0.45) / 20 cm2
Stress = 281.25 / 20cm2
Stress = 14 N/cm2
TASK 3 (RELATES TO STRAIN AND STRESS STRAIN CURVES INFORMATION
Aim of this task is to use the definitions and the stress strain curve to interpret
Go to this website and create a stress/strain curve for the below materials.
https://mechanicalc.com/calculators/material-stress-strain-curve/
1
find more resources at oneclass.com
find more resources at oneclass.com
Document Summary
Think of 2 activities or actions that relates to each stress type (compression, tension and shear). It can look at whole body or bones or muscles etc. Compression humerus when doing a push-up / ground reaction force when walking / spine compression in standing. Tensile humerus when doing a chin up, eccentric muscle action / acl in knee valgus. Shear force in lumbar vertebrae in a deadlift. Stress = (625n * 0. 45) / 20 cm2. Task 3 (relates to strain and stress strain curves information. Aim of this task is to use the definitions and the stress strain curve to interpret. Go to this website and create a stress/strain curve for the below materials. https://mechanicalc. com/calculators/material-stress-strain-curve/ The data is derived from the following reference. Take a print screen of the figure and paste it below so you can refer to it after: Yield strength 3770. 98 psi (26 mpa) Ultimate strength (fracture point) 4641. 21(32 mpa)
