Module 1

Question 1

Why do we have Auxillary conditions/Optimisation techniques?

Answer 1

There is a joint force distribution problem by which rigid body mechanics alone is not enough to determine all internal forces.
This can be quantifiably described as statistically indeterminate problems whereby the number of equations does not equal the number unknowns.
This can also be interpreted as the muscle redundancy problem, implying that some muscles are not influential with respect to the problem we are dealing with and hence, do not need to be included.

Question 2

How do we deal with the join force distribution/muscle redundancy problem?

Answer 2

Auxillary conditions and Optimisation techniques

Question 3

Auxillary conditions

Answer 3

The addition of constraint conditions in order to get an equal number of equations and unknowns.

1. FORCE REDUCTION - define relationships among forces and combine muscle actions into single resultant or synchronous activity groups.
2. MUSCLE SCALING - the maximum stress that most muscles can sustain is 0.2MPa. Estimate crossectional area of muscle and find the force of the muscle using Stress = F/A and Fi = (Ai/A1)*F1. ASSUMING MUSCLE IS AT MAX STRESS
3. SOFT TISSUE FORCE-DEFORMATION RELATIONSHIPS - found from material testing

Question 4

Optimisation

Answer 4

Addition of performance criterion based on the idea that the physiological system performs in an optimal state.

1. Implement design variable (muscles, contact forces etc.)
2. Chosen such that they have an objective function (ie. minimising energy use, maximising range of motion/force)
3. Subject to constraints (ASSUMPTIONS) (ie. equillibrium, muscle and ligament forces are tensile, joint contact forces are compressive, joint friction is minimal (forces act perpendicular to surface)).

Question 5

Planes of motion

Answer 5

Sagittal - split left and right - flexion/extension
Frontal - split front to back - adduction/abduction
Transverse - splits top to bottom - rotation/pronation/supination.

Question 6

Statics

Answer 6

Sum of forces = 0

Sum of moments = 0

Question 7

Moment

Answer 7

Moment about O = F x D whereby D is the perpendicular distance from O to the line of action of F

Question 8

Free body diagrams

Answer 8

1. Simplify problem - what are u interested in? Add necessary reaction forces
2. Consider external forces (weight/gravity)
3. Add in muscle forces if interested
4. Add in joint contact forces if interested
5. Are any moments created?

Question 9

Relative velocity

Answer 9

1. VECTORS (i, j, k) - usually given a value you need to work backwards from.
2. Find position vectors of all points Ra/b
3. Find velocity vectors Va = Vb + Va/b.
   Va/b = W*Ra/b
   *From rest -> velocity = 0

Question 10

Relative acceleration

Answer 10

1. VECTORS (i, j, k) - usually given a value you need to work backwards from.
2. Find position vectors of all points Ra/b
3. Find acceleration vectors Aa = Ab + (Aa/b)n + (Aa/b)t.
   (Aa/b)n = W(WRa/b)
   (Aa/b)t = Alpha*Ra/b
   *From rest -> velocity = 0

Question 11

Mass moment of Inertia

Answer 11

Body’s resistance to being rotated

I = mr^2

Question 12

Area moment of Inertia

Answer 12

Body’s resistance to bending
I = Ar^2
(w.r.t x or y axis) -> Ix = Integral(x^2)dA or Iy = Integral(y^2)dA (replace dA w.r.t x/y)

Question 13

Polar moment of Inertia

Answer 13

Body’s resistance to torsion (twisting about z axis)
J = Integral(r^2)dA
(replace dA w.r.t r)

Question 14

Parallel axis theorem

Answer 14

Determines moment of inertia of an object about a new axis if M.O.I about com is know
I = Icom + d^2M (MMOI)
I = Icom + d^2A (AMOI)

Question 15

Radius of Gyration

Answer 15

Distance of the total moment of inertia of an object about an axis if it were a point mass.
I = d^2*(M or A)
–> d = sqrt(I/(M or A))