6.3 Computation of reaction forces
6.3.1 Equilibrium revisited
- The equations of motion are three dimensional
- Three translational degrees of freedom \[\begin{split} \sum F_x =& \; 0 \\ \sum F_y =& \; 0 \\ \sum F_z =& \; 0 \\ \end{split}\]
- Three rotational degrees of freedom \[\begin{split} \sum M_x =& \; 0 \\ \sum M_y =& \; 0 \\ \sum M_z =& \; 0 \\ \end{split}\]
6.3.2 Force diagrams, statically equivalent forces, and “free body diagrams”
Draw structural bodies in equilibrium, ie properly constrained or with balanced forces
To compute reaction forces \(R\) and \(B\), use statically equivalent forces (ie place the weight of the forearm at its center of gravity)
6.3.3 Force diagrams, statically equivalent forces, and “free body diagrams”
Complete calculations using the equilibrium equations
Equillibrium: \[\begin{split} \sum F_y =& \; 0 \\ \sum M =& \; 0 \\ \end{split}\]
6.3.4 Force diagrams, statically equivalent forces, and “free body diagrams”
- Equillibrium: \[\begin{array} \sum F_y =& \; +B - G - W -R = 0 \\ \sum M_R =& \; +(0 \, {\mathrm cm}) (R) + (3 \, {\mathrm cm}) (B) - (15 \, {\mathrm cm}) (G) - (30 \, {\mathrm cm}) (W) = 0 \\ \end{array}\]
6.3.5 Force diagrams, statically equivalent forces, and “free body diagrams”
Use the moment equation first: \[\begin{split} (0\, {\mathrm cm})(R) + (3\, {\mathrm cm})(B) - (15\, {\mathrm cm})(G) - (30\, {\mathrm cm})(W) =& \; 0 \\ \end{split}\] \[\begin{split} (3\, {\mathrm cm})(B) =& \; (15\, {\mathrm cm})(G) + (30\, {\mathrm cm})(W) \\ \end{split}\]
Use known values \[\begin{split} G =&\; 15 \, \mathrm{N}\\ W =&\; 20 \, \mathrm{N}\\ \end{split}\]
Therefore: \[\begin{split} B = \frac{(15 \, {\mathrm cm})(15 \mathrm{N}) + (30\, {\mathrm cm})(20 \mathrm{N})}{3\, {\mathrm cm}} = 275 \, \mathrm{N} \end{split}\]
6.3.6 Force diagrams, statically equivalent forces, and “free body diagrams”
Use the force equation next: \[\begin{split} +B - G - W -R = 0 \\ R = +B - G - W \\ \end{split}\]
Use known values \[\begin{split} G =&\; 15 \, \mathrm{N}\\ W =&\; 20 \, \mathrm{N}\\ B =&\; 275 \, \mathrm{N}\\ \end{split}\]
Therefore: \[\begin{split} R = 275 \, \mathrm{N} - 20 \, \mathrm{N} - 15 \, \mathrm{N}= 240 \, \mathrm{N}\\ \end{split}\]
6.3.7 The problem of redundancy (a mathematical problem)
Problem: human body has “redundant” and/or opposing muscle groups which make it difficult or impossible to compute the individual muscle loads (without making assumptions)
For example:
- Agonist/Antagonist muscles can be used to hold a
fixed location despite external load
- ie flex your bicep/triceps
- Hold position with minimal or maximal force!
- This allows the body to have broad function and also to compensate for injury/fatigue/etc
- Agonist/Antagonist muscles can be used to hold a
fixed location despite external load
This “redundancy” makes for an indeterminate set of equations (more unknowns than equations)
6.3.8 Additional examples of static analysis
- Compute the elbow joint loads due to lifting a box
- Compute the tibio-talar resultant force and moment in the ankle joint when standing
6.3.9 Indeterminance
- Forces from:
- Glenoid contact
- Deltoid
- Pectoralis major
- Supraspinatus
- Infraspinatus
- Teres minor/major
- Subscapularis
- Latissimus dorsi
- Biceps brachii
- Triceps brachii
- Ligaments/capsule
- …
- We have 6 equations and \(>\) 6 unknowns… must make assumptions!
6.3.10 Example–rigid link analysis of body segments
- Three body segments – \(6 \times 3 = 18\) equations (9 equations for planar model)
- If our intent is to calculate the inter-segmental resultant forces, we can accomplish this goal without regard to internal muscle forces (known external forces)
- Calculation of muscle forces requires significant additional work and assumption
- Stress analysis would require calculation of the muscle forces
6.3.11
6.3.12
6.3.13 The Joint Force Distribution Problem
Rigid body mechanics often insufficient for desired purpose
- ie a first step that gives input loads to a higher fidelity model
Thus, we must distribute the inter-segmental resultant forces to the muscles and other soft tissues
Additional information about the neuromusculoskeletal system helps
- Use information about the load carrying elements to supplement
the number of available equations
- ie Assume relationships between muscle forces
- Specify performance criterion and assume an optimum is found by
the body
- Minimize energy used
- Minimize a contact force
- Both methods can be employed
- Use information about the load carrying elements to supplement
the number of available equations
6.3.14 Auxiliary conditions
- We can assume auxiliary conditions to enable the calculation
- Force reduction
- Assume several muscles have zero force (one muscle dominates)
- Assume several muscles act as one at a coincident point of action
- Assume muscle exertions are proportional to each other
- Assume muscle exertions are proportional to cross section
- Muscle scaling – 0.2MPa max achievable stress
- Force reduction
- We can use external measurements
- EMG Measurements (Electromyogram)
- Electrical impulses of the muscles indicative of contraction effort
- Use “constitutive” descriptions of tissues. (ie solve the
statically indeterminate mechanics problem)
- Almost always measured on separate tissue, assumed accurate
- EMG Measurements (Electromyogram)
- These methods add equations but also add questions
- ie what is the muscle cross sectional area
6.3.15 Optimization Technique
- This method attempts to find the “best” solution among an infinite number of possible solutions
- Assumes that the body chooses the best solution neurologically
- ie minimize contact forces (through sensation – pain, etc)
- Must write an objective function
- \(f(\vec{x}) = f(x_1, x_2, x_3,...)\)
- \(\vec{x}\) is the vector of unknown variables
- Minimize wrt a set of constraint equations (\(h(x), g(x)\) – equality
and inequality constraints)
- equilibrium, constitutive, force limits, etc
- Challenges:
- Is the objective function “correct”?
- Is it appropriate to assume that the body finds the optimum?