Orthopedic Mechanics, Forces and Moments

orthopedic mechanics

1. Department of Biomedical Engineering
Lecture 2
8/27/2025 BMEN 457/657-Fall 2025
BMNE 457/657: Orthopedic Biomechanics

2. Question:
Is the solution for the forces sensitive to the
choice of the pivot point in a system that is in
mechanical equilibrium?
8/27/2025 BMEN 457/657-Fall 2025

3. 8/27/2025 BMEN 457/657-Fall 2025
�
� �
�
�� ��
�
��
�
�
�
a b c
��
Σ�
�=�
��−�
��+�
��+�+�
�0=0
Σ�
�=0
Σቚ�
�±??????
=�
��±�−�
�(�±�)+�
��+�±�+�
�0±�=0
=Σ�
�±�(Σ�
�)=0
Σ�
�=0 as long as
(Σ�
�)=0

4. Forces on knee joint
8/27/2025 BMEN 457/657-Fall 2025
FindcontactforcesbetweenthePatellaandfemur
F
QuadricepsTendon≈F
PatellaTendon

5. Forces on knee joint
8/27/2025 BMEN 457/657-Fall 2025
55°
75°
�
�
??????
�
�
•Point A is the contact point between
the kneecap (Patella) and femur.
�
�
Σ�
�=0→??????
�=�
??????cos55°+�
??????cos(75°)
Σ�
�=0→�
�=−{�
??????[sin55°−sin(75°)]}
??????
�=1040.5�,�
�=+183.5�
�
??????
�
??????
Force equilibrium

6. When the moment is considered
8/27/2025 BMEN 457/657-Fall 2025
55°
75°
�
Moment equilibrium
�
�
�
??????
�
??????
Σ�
�=0
�
�
�
�=�∗�
??????[−sin55°+sin(75°)]
�
�
�=�∗183.5�
�
�
�+�∗�
??????sin55°−�∗�
??????sin(75°)=0

7. 8/27/2025 BMEN 457/657-Fall 2025
Quiz.FindFQ
(
Patellatendonforce)
Please note that our
interest in this problem is
estimating the force in the
patella-tibia connection
that is generated along the
tendon. There are other
forces in knee ligaments,
but they don’t produce
significant moments
around the z-axis (out-of-
plane axis)
Note: the symbol ⊥stands for “perpendicular”.
•r
⊥: perpendicular distance from F
Q to the knee joint.
•r’
⊥: perpendicular distance from W
legto the knee joint.
•r’’
⊥: perpendicular distance from W
childto the knee joint.

8. Which one is correct?
The Slido appmust be installed on every computer you’re
presenting from

9. Forces in upper leg
8/27/2025 BMEN 457/657-Fall 2025
??????=10
Σ�
�=0
�
??????∗??????
⊥−�
??????�??????∗??????
′
⊥−�
�ℎ????????????�∗??????
′′
⊥=0
�
??????∗2−40∗20−100∗38=0
�
??????=
40∗20+100∗38
2
=2300�
�=??????∗??????

10. Joint/Body Mechanics
•Whole body models used for gait
analysis
•Displacement markers used to
define bone positions
•Compute muscle forces for
rehabilitation/injury
•System of equations from free
body diagrams of each bone
•However, often more unknowns
than available equations
8/27/2025 BMEN 457/657-Fall 2025

11. Joint/Body Mechanics
•Max 6 equilibrium equations for
each bone
•ΣMx, ΣMy, ΣMz
•ΣFx, ΣFy, ΣFz
•Generally 6 contact forces at
joints plus numerous muscle
attachments (ie, moments)
•Statically indeterminate problem
Ft1
F
N
Ft2
Fx
Fy
Fz
8/27/2025 BMEN 457/657-Fall 2025

12. Statically indeterminate problem
8/27/2025 BMEN 457/657-Fall 2025
Indeterminate
Determinate

13. Joint/Body Mechanics
•Max 6 equilibrium equations for
each bone
•ΣMx, ΣMy, ΣMz
•ΣFx, ΣFy, ΣFz
•Generally 6 contact forces at
joints plus numerous muscle
attachments (i.e., moments)
•Statically indeterminate problem
•Can directly measure muscle
activity via electromyography
(EMG) for added constraints
8/27/2025 BMEN 457/657-Fall 2025

14. Body/Joint Mechanics
Lifting a Patient Forces on Spine
NPR 2/11/15
Source: Spine Research Institute at Ohio State University
8/27/2025 BMEN 457/657-Fall 2025

15. •Why bother?
Stress and Strain
F
F
L
δ
E: Young’s Modulus
A: Cross-sectional Area
k: Extensional rigidity
�=??????�=
��
�
�
F
δ
k
�=??????�Hooke’s Law:
F
δ
k
8/27/2025 BMEN 457/657-Fall 2025
δ
�
�
The same material yields a
different displacement applying
the same force.
Rigidity is shape-dependent.
F
δ
k
2
k
1
1
2
We need to find a non-shape-
dependent relationship between
force and displacement.

16. •Why bother?
Stress and Strain
F
F
L
δ
E: Young’s Modulus
A: Cross-sectional Area
k: Extensional rigidity
�=??????�=
��
�
�
Hooke’s Law:
F
δ
k
8/27/2025 BMEN 457/657-Fall 2025

17. •Why bother?
Stress and Strain
F
F
L
δ
E: Young’s Modulus
A: Cross-sectional Area
�=??????�=
��
�
�
Hooke’s Law:
F
δ
k
�=
�
�
??????=
�
�
??????=��
σ
ε
E
8/27/2025 BMEN 457/657-Fall 2025
Rearranging terms:
�
�
=�
�
�
Since:
We obtain the stress-strain
relationship, which is shape-
independent:

18. “Brain Teaser” …
8/27/2025 BMEN 457/657-Fall 2025
Question: Was Galileo correct?
[He hypothesized that the cross--‐sectional Geometry of long
bones would have To increase more quickly than length to
Support the increased weight of larger animals.]
L
A
r
�=�∗??????
2
The animal is scaled up but the
materials of the bone stay the
same –no change in stiffness.
�~?????? V~??????
3
W=????????????=��??????~??????
3
The stress that the bones need
to bear increases at a rate~??????
3
2.
σ=
�
�
~
??????
3
??????
2
~??????
3
2
??????~??????
3
2
r should increase
at the same rate
rate to keep up.
�~??????
2
The load supported by the
bones increases ~??????
3
while
the cross-section only ~??????
2
.

19. “Brain Teaser” …
8/27/2025 BMEN 457/657-Fall 2025
Question: Was Galileo correct?
10 different adult primate femurs
Elephant femur