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- 1. Shoulder & Elbow BiomechanicsForces, Moments & Joint Reaction Forces Lennard Funk
- 2. Force• any inﬂuence that causes a free body to undergo an acceleration• = Mass x Acceleration• Newton’s second law
- 3. Moment (of force) /• Turning Force• = Force x Distance • = F.r
- 4. Moment = Force (F) x Distance (d)
- 5. Equilibrium = Sum of the moments
- 6. Joint Reaction Force• not the same as joint forces• "the equal and opposite forces that exist between adjacent bones at a joint caused by the weight and inertial forces of the two segments."
- 7. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.
- 8. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.
- 9. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.
- 10. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.
- 11. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.The sum of the moments (∑M) should be in equilibrium i.e. ∑M=0
- 12. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.The sum of the moments (∑M) should be in equilibrium i.e. ∑M=0i.e the clockwise moments must be equal to the anticlockwise moments.
- 13. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.The sum of the moments (∑M) should be in equilibrium i.e. ∑M=0i.e the clockwise moments must be equal to the anticlockwise moments.(30cm x .05BW) = (M x 3cm)
- 14. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.The sum of the moments (∑M) should be in equilibrium i.e. ∑M=0i.e the clockwise moments must be equal to the anticlockwise moments.(30cm x .05BW) = (M x 3cm)M = (30cm x .05BW) / 3cm
- 15. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.The sum of the moments (∑M) should be in equilibrium i.e. ∑M=0i.e the clockwise moments must be equal to the anticlockwise moments.(30cm x .05BW) = (M x 3cm)M = (30cm x .05BW) / 3cmProblem: Assuming BW is 70Kg and Gravity is 10m/s, calculate the abductionforce of deltoid (M):
- 16. The arm is in 90 degrees of abduction and it is assumed that only deltoid isactive.The force produced by deltoid muscle (M) acts 3cm from the centre ofrotation of the shoulder joint.The force produced by the weight of the arm is 0.05 times body weight (BW)and acts 30cm from the joint centre.The reaction force of the joint (J) is equal to M.The sum of the moments (∑M) should be in equilibrium i.e. ∑M=0i.e the clockwise moments must be equal to the anticlockwise moments.(30cm x .05BW) = (M x 3cm)M = (30cm x .05BW) / 3cmProblem: Assuming BW is 70Kg and Gravity is 10m/s, calculate the abductionforce of deltoid (M):............Next slide for answer ..........
- 17. Calculations:M = (0.3m x 0.05 (70Kg x 10m/s) / 0.03mTips:- Units should be converted to metres (m)- BW should be a Force (N), which is mass x acceleration = 70Kg x10m/s
- 18. Answer: M = 350NCalculations:M = (0.3m x 0.05 (70Kg x 10m/s) / 0.03mTips:- Units should be converted to metres (m)- BW should be a Force (N), which is mass x acceleration = 70Kg x10m/s
- 19. In this example the person is holding ball in the outstretched hand.
- 20. In this example the person is holding ball in the outstretched hand.∑M=0
- 21. In this example the person is holding ball in the outstretched hand.∑M=0[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0
- 22. In this example the person is holding ball in the outstretched hand.∑M=0[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0M = [(30cm x .05BW) + (60cm x .025BW)] / 3cm
- 23. In this example the person is holding ball in the outstretched hand.∑M=0[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0M = [(30cm x .05BW) + (60cm x .025BW)] / 3cmProblem: Assuming BW is 70Kg and Gravity is 10m/s, calculate theabduction force of deltoid (M):
- 24. Answer: M = 700N
- 25. In this example the person has their elbow bent.∑M=0[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0M = (15cm x .05BW) / 3cmProblem: Assuming BW is 70Kg and Gravity is 10m/s, calculate theabduction force of deltoid (M):................ Next slide for answer .......
- 26. Answer: M = 175N
- 27. In the next two slides you can see how Moments arerelevant in clinical practice.With a cuff deﬁcient shoulder, arm elevation (ﬂexion &abduction) are dependent entirely on the deltoidmuscle.By bending their elbow, the person reduces themoment of their arm (by shortening the distance the ofthe force from the shoulder). This reduces the Forcerequired by deltoid to elevate the arm. M = 350N M = 175N
- 28. Joint Reaction Force in practice•The joint reaction force (JRF) is a useful measure of joint stability•An imbalance in the force couples of the rotator cuff tendons alters the position, direction and magnitude of the JRF (Parsons et al. J Orthop Res, 2001)•It is also important in the process of
- 29. JRF JRF Inclination Activity of Daily Living %BW AngleAbduction 75 degrees without weight 85 33 Bergmann 45 degrees with 19.4 N weight 88 30 45 degrees without weightFlexion 51 33 Magnitude JRF 120 degrees without weight 124 34 Inclination Angle 90 degrees with 19.4 N weight 129 34 90 degrees without weight 77 33Extension, supine position, elbowflexed, 118 N resistance at elbow 82 32Lifting 13.6 N Coffeepot, straight arm 103 32Nailing 15 cm above head 88 40Steering automobile Slow, 7 Nm, both hands 42 32 Fast, 7 Nm, both hands 40 23 Fast, 7 Nm, one hand 109 19 Slow, 12 Nm, both hands 110 21 Steering wheel fixed, both hands 152 32Walking with 2 crutches, full support 118 21Lifting 96.8 N laterally 14 26Putting 24.2 N into shelf, 60 cm in front 69 32Combing hair Typical effort 64 47 Maximum effort 98 41 Average 31.3 Standard Deviation 7.2
- 30. Polar Reference Frame for designing a humeral head replacement "Y""Superior" """"""""Y""Superior """"""""""X""Anterior Z ""X""Anterior" Right Arm Joint Force Joint Force Inclination Angle
- 31. Elbow Biomechanics:In the example below, assume Force W is 20N and Force Pis 10N, calculate the force required by biceps to keep the 10N 20N
- 32. Resources1. Richards - Biomechanics in Clinic & Research, Churchill Livingstone2. Norden & Frankel - Biomechanics of the Musculoskeletal System3. Wikipedia.org4. Orthoteers.com

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