Biomechanics Agenda Review Biomechanical modeling Review: - PowerPoint PPT Presentation

Biomechanics  Agenda  Review  Biomechanical modeling

Review: Skeletomuscular Levers  First-class: Fulcrum is  Second-class: Fulcrum  Third-class: Fulcrum is between the two loads, is at the end ; Force is at the end ; Force is which is good for fine exerted through a exerted through a positional control . longer moment arm shorter moment arm than the resistance. than the resistance 2

Review: In-Class Exercise ∑ F V = 0: R ELBOW –W FOREAR AND HAND -W LOAD =0 R ELBOW - 15.8N – 49N = 0 R ELBOW = 64.8N ∑ F H =0: N/A ∑ M A = 0: A M E – M FOREAR AND HAND - M LOAD =0 M E – 0.172m*15.8N - 0.355m*49N=0 M E =20.1NM

Link-Segment Model  Basic for calculations of joint reaction forces and net muscle moments throughout the body.

Link-Segment Model  Modeling components  Anthropometric data o Segment length o Segment weight o Moment of inertia  Posture data  External load: hand load  Internal load o Segment weight o Muscle contraction force

Link-Segment Model  Basic for calculations of joint reaction forces and net muscle moments throughout the body. Weight

Link-Segment Model  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 o I=mr 2  Shape of the body does not change  Constant segment length W BW

Link-Segment Model  Low Back Pressure, F M ? L: Low back W BW : Upper body weight (internal load ) θ F M F: Hand load (external load ) D M F M : Low back muscle force (internal load ) L W BW M M - M BW - M F ΣM L = 0 : =0 F M × D M - W BW × D BW - F × D F = 0 D BW F M = ( W BW × D BW + F × D F )/ D M D F F

Link-Segment Model  Low Back Pressure, Joint compression and shear forces? F M D M F M F Compression θ θ θ L L L F Shear W BW W BW F D BW F Compression = F M + (W BW + F) * cosθ D F F F Shear = (W BW + F) * sinθ

In-class Exercise Upper body weight, W BW = 300N Hand load, F = 100N F M Trunk flexion angle, θ = 30 degree D M D BW = 0.25 m, D F = 0.5 m, D M = 0.05 m Please calculate the following variables: θ 1. Muscle force at the low back joint, F M 2. Compression forces at the low back joint, F Compression 3. Shear forces at the low back joint, F Shear W BW D BW D F F

In-class Exercise 1. Muscle force at the low back joint, F M L: Low back F M D M ΣM L = 0 : θ M M - M BW - M F =0 L F M × D M - W BW × D BW - F × D F = 0 W BW F M = ( W BW × D BW + F × D F )/ D M F M = (300 × 0.25+ 100 × 0.5)/ 0.05=2500 N D BW D F F

In-class Exercise 2. Compression forces at the low back joint, F Compression F M F Compression = F M + (W BW + F) * cosθ D M 3 F Compression = 2500 + (300 + 100) × 2 =2846.4 N θ L W BW 3. Shear forces at the low back joint, F Shear F Shear = (W BW + F) * sinθ D BW 1 F Shear = (300 + 100) × 2 = 200 N D F F