Link Segment Analysis Net Muscle Moments and Net Joint Forces

Link Segment Analysis Net Muscle Moments and Net Joint Forces

Back Moments So far we have focused on back moments with simple models that assumed we knew the location of the upper body centre of gravity. Dealing with the model segment by segment allows you to calculate moments about each joint.

Link Segment Models Link segment approaches are suitable when the moments and reactive forces at numerous joints are needed. Some studies have shown that lifting under certain conditions is often limited more by the stresses on the lumbar spine than by limitations of strength. Hence numerous muscle moments at various joints are often not needed.

Link Segment Models By looking at each segment in sequence we can work our way through the body and calculate muscle moment and NET joint forces at each joint. However, unless we know the insertion point and line of action of a suitable single muscle crossing a joint (or group such as erector spinae and quads) we cannot calculate contact boneon bone forces.

Static Equilibrium Equations F y = 0 F x = 0 M = 0

Sample Problem (50% male anthropometry) What flexor muscle moment is needed to hold the forearm/hand segments in the position shown? Taking moments about the elbow. Hence the system in question is the forearm and hand. Draw a diagram. To calculate the answer the first step is to calculate the moment arms from the elbow. Forearm = 10.9 cm [1] Hand = 25.3 + 9.2 = 34.5 cm [1] Moments: 0.109x1.2x-9.81 = -1.294 Nm 0.345x0.4-9x.81 = -1.354 Nm Total = -2.65 Nm [3] Therefore the elbow flexor muscle moment is +2.65 Nm

Additional Question If the forearm flexors insert 3 cm from the axis of rotation of the elbow, what is the muscle force and bone-on-bone force? Moment = Force x Distance 2.65 = F x 0.03 F = 88.333 N Looking at the free-body diagram again. F=0 88.33-11.772-3.924 + F R =0 F R = -72.64 N

Forces Acting on the Link Segment Model Gravitational forces. Ground reaction forces and other external forces. Muscle forces Joint reaction forces.

Calculate the moment at the elbow and shoulder joints. assume this is a static task use the anthropometry table (50%ile male) model the forearm and hand as one segment the force is distributed equally between the arms the force is being applied at the hands COM. Forearm angle from left horizontal = 0º 60 kg Shoulder angle from left horizontal = 30º Trunk angle from left horizontal = 60º

M 1 Ry 1 Load/2 mg H mg F Mass: Hand = 3.92 N Forearm = 11.77 N Upper arm = 20.60 N _Fy = 0 Ry 1 (294.3 + 3.92) 11.77 = 0 Ry 1 = 310 N _M = 0 M 1 [298.22 * (0.253 + 0.092)] [11.77 * 0.109] = 0 M 1 102.9 1.28= 0 M 1 = 104.18 Nm Length: Forearm = 25.3 cm Upper arm = 32.2 cm Distance to COM: Hand = 9.2 cm Forearm = 10.9 cm Upper arm = 14.0 cm M 2 -Ry 1 -M 1 mg UA _M = 0 M 2 104.18 - [310 * (0.322 * cos30º)] [20.60 * (0.14 * cos30º] = 0 M 2 104.18 86.45 2.5= 0 M 2 = 193.13 Nm

Muscle Strength Prediction

Muscle Moments As already discussed, we cannot calculate accurately the muscle forces from our models at most joints. However, we can calculate accurately the muscle moment required at these joints. Simple experiments can be conducted to assess what is the mean and standard deviations of strength across various joints, (i.e. max. joint torque M j + SD).

Prediction Tables and Equations Some tables exist that show the maximum strength a population can exert, but they tend to focus on a set angle. That is they will give you the maximum strength at a given angle. For example in the 4D WATBAK program they present data from Askew et. al. (1987) for 90 o elbow angle, and from Koski & McGill (1994) for 0 o shoulder angle (vertical) This is obviously of limited value.

Mean Isometric Joint Moments of Force (Nm) [strength values] Stobbe 82 Male %var. Female %var. Hip Flexion 232 17 168 38 Extension 251 31 128 41 Knee Flexion 118 4 69 32 Extension 267 3 123 33 Trunk Flexion 301 19 174 24 Extension 391 19 244 22 % var. = % variability = (S.D./mean) x 100

Strength Prediction Equations There is a large variation in strength as joint angle changes: Mj =ƒ _j The adjacent joint may also affect max. torque, particularly if there are 2-joint muscles present (i.e. muscles acting across both joints). A set of equations have been derived that estimate joint strength as a function of posture and gender. For example: S E = [336.29 + 1.544_ E - 0.0085_ 2 E - 0.5_ S ][G] This allows us to compare the demands of a particular task with population strength data.

S E = [336.29 + 1.544 E - 0.0085 E 2-0.5 S ][G] Joint Moment- Strength Mean Prediction Equations U of Michigan What is the predicted male mean strength for elbow flexion (equation above) if E = 140 o and s = 20 o? L5/S1 Hip H T K S Elbow Knee Shoulder E Ankle A

SE = [336.29 + 1.544 E - 0.0085 E 2-0.5 S ][G] SE = [336.29+ 1.544(140) - 0.0085(140 2 ) - 0.5(20)][0.1913] SE = [336.29 + 216.16-166.6-10][0.1913] SE = 71.9 Nm (this is the mean predicted strength) Coefficient of variation = 0.2458 Coeff. of variation = Standard deviation/mean S.D. = 0.2458 x 71.9 = 17.67 Nm S = 20 o Shoulder E = 140 o We must then decide what % of the population we want to accommodate. Lets say 95% 80 o Elbow

Kroemer Page 142-143 Modelling to the 5 th percentile strength will accommodate 95% of the population. From the Kroemer chapter: X (p) = m + ks [where: m = mean and S = S.D.] k-value for 5 th percentile = -1.64 So substituting: X (5) = 71.9 + 17.67(-1.64) X (5) = 71.9 28.98 X (5) = 42.9 Nm What would the strength of the 1 st percentile be (accommodating 99% of the population)?

z-value for 1 st percentile = -2.33 X (1) = 71.9 + 17.67(-2.33) X (1) = 71.9 41.17 X (1) = 30.7 Nm This shows that by accommodating 4% more of the population you have to accept a strength reduction of 28% (12.2/42.9 x 100%) For this reason we would tend to try to accommodate either 95%, 90% or even 75% of the population

Problem What is the maximum weight you would recommend for packaging that has to be lifted by a male workforce in both hands in the position opposite? To keep things simpler focus on the elbow and shoulder and ignore L5/S1. 80 o Shoulder 20 o 140 o Elbow

A Problem! Strength and anthropometry do not correlate. That is, the 10%tile male in terms of strength is not the 10%tile male in terms of stature, limb lengths and/or body mass. The solution in the absence of specific anthropometry is to develop your model around the 50%tile male, female or general population anthropometry. Use simple male 50%tile Kin201 Anthropometry for this problem

Solution Step 1: Determine percentage of male population to be accommodated (e.g. 10%). Step 2: Calculate maximum allowable muscle moment (strength) for average male. Step 3: Calculate maximum allowable muscle moment (strength) for 10%tile. Step 4: Construct link segment model where instead of the unknown being the muscle moment(s) the unknown is the weight of the load.

83.9 [0.2845] 83.9 19.39 83.9 - (19.39) = 59.1 Nm

Geometry Review your geometry if you can t follow this process! Shoulder 20 o 80 o 10 o 10 o 80 o 80 o Elbow 100 o 40 o 80 o

80 o For one arm

What about the female workforce? mass (kg) length (cm) distance to COM (cm) Hand 0.4-6.4 Elbow 1.0 24.1 10.4 Shoulder 1.7 27.8 12.1