Agenda Review Biomechanical modeling

Biomechanics

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

Review: In-Class Exercise

∑ FV = 0: RELBOW –WFOREAR AND HAND-WLOAD =0RELBOW -15.8N – 49N = 0RELBOW= 64.8N

∑ FH =0: N/A

∑ MA = 0: ME – MFOREAR AND HAND-MLOAD =0ME – 0.172m*15.8N-0.355m*49N=0ME =20.1NM

A

Basic for calculations of joint reaction forces and netmuscle moments throughout the body.

Link-Segment Model

Modeling components Anthropometric datao Segment lengtho Segment weighto Moment of inertia

Posture data External load: hand load Internal loado Segment weighto Muscle contraction force

Link-Segment Model

Basic for calculations of joint reaction forces and netmuscle moments throughout the body.

Link-Segment Model

Weight

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 inertiao I=mr2

Shape of the body does not change Constant segment length

Link-Segment Model

WBW

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 - M BW - MF =0

FM = (WBW× DBW + F× DF)/ DM

FM × DM - WBW× DBW - F× DF = 0

θ

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) * sinθF

WBW

DBW

DF

FM DM

L

θ

In-class Exercise

Upper body weight, WBW = 300NHand load, F = 100NTrunk flexion angle, θ = 30 degreeDBW = 0.25 m, DF = 0.5 m, DM = 0.05 mPlease calculate the following variables:

1. Muscle force at the low back joint, FM2. Compression forces at the low back joint, FCompression3. Shear forces at the low back joint, FShear

F

WBW

DBW

DF

FM DM

θ

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 - M BW - MF =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

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) × 32

=2846.4 N

FShear = (300 + 100) × 12

= 200 N