SLIDE 1 Agenda
- Review
- Biomechanical modeling
Biomechanics
SLIDE 2 First-class: Fulcrum is between the two loads, which is good for fine positional control.
Review: Skeletomuscular Levers
2
Second-class: Fulcrum
is at the end; Force is exerted through a longer moment arm than the resistance.
Third-class: Fulcrum is
at the end; Force is exerted through a shorter moment arm than the resistance
SLIDE 3
Review: In-Class Exercise
∑ FV = 0: RELBOW –WFOREAR AND HAND-WLOAD =0 RELBOW -15.8N – 49N = 0 RELBOW= 64.8N ∑ FH =0: N/A ∑ MA = 0: ME – MFOREAR AND HAND-MLOAD =0 ME – 0.172m*15.8N-0.355m*49N=0 ME =20.1NM A
SLIDE 4
Basic for calculations of joint reaction forces and net muscle moments throughout the body.
Link-Segment Model
SLIDE 5 Modeling components
- Anthropometric data
- Segment length
- Segment weight
- Moment of inertia
- Posture data
- External load: hand load
- Internal load
- Segment weight
- Muscle contraction force
Link-Segment Model
SLIDE 6
Basic for calculations of joint reaction forces and net muscle moments throughout the body.
Link-Segment Model
Weight
SLIDE 7 Assumptions
- A fixed point mass, located at its center of mass.
- Joints are represented by simple hinge joints (not free to
translate, but free to rotate)
- Constant moment of inertia
- I=mr2
Shape of the body does not change
Link-Segment Model
WBW
SLIDE 8 Low Back Pressure, FM?
Link-Segment Model
F WBW DBW DF L: Low back FM
DM
L
F: Hand load (external load) WBW : Upper body weight (internal load) FM : Low back muscle force (internal load) ΣML = 0: MM
=0 FM = (WBW× DBW + F× DF)/ DM FM × DM - WBW× DBW - F× DF = 0
θ
SLIDE 9
Low Back Pressure, Joint compression and shear forces?
Link-Segment Model
FCompression FShear
L θ FCompression = FM + (WBW + F) * cosθ L θ FM WBW F FShear = (WBW + F) * sinθ F WBW DBW DF FM
DM
L θ
SLIDE 10
In-class Exercise
Upper body weight, WBW = 300N Hand load, F = 100N Trunk flexion angle, θ = 30 degree DBW = 0.25 m, DF = 0.5 m, DM = 0.05 m Please calculate the following variables:
1. Muscle force at the low back joint, FM 2. Compression forces at the low back joint, FCompression 3. Shear forces at the low back joint, FShear
F WBW DBW DF FM
DM
θ
SLIDE 11 In-class Exercise
1. Muscle force at the low back joint, FM
F WBW DBW DF FM
DM
θ L: Low back L
ΣML = 0: MM
=0 FM = (WBW× DBW + F× DF)/ DM FM × DM - WBW× DBW - F× DF = 0 FM = (300 × 0.25+ 100 × 0.5)/ 0.05=2500 N
SLIDE 12 In-class Exercise
F WBW DBW DF FM
DM
θ L FCompression = FM + (WBW + F) * cosθ FShear = (WBW + F) * sinθ
- 2. Compression forces at the low back joint, FCompression
- 3. Shear forces at the low back joint, FShear
FCompression = 2500 + (300 + 100) ×
3 2 =2846.4 N
FShear = (300 + 100) ×
1 2 = 200 N
