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.
8. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
9. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation of the shoulder joint.
10. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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.
11. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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.
12. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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
13. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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
i.e the clockwise moments must be equal to the anticlockwise moments.
14. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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
i.e the clockwise moments must be equal to the anticlockwise moments.
(30cm x .05BW) = (M x 3cm)
15. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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
i.e the clockwise moments must be equal to the anticlockwise moments.
(30cm x .05BW) = (M x 3cm)
M = (30cm x .05BW) / 3cm
16. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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
i.e the clockwise moments must be equal to the anticlockwise moments.
(30cm x .05BW) = (M x 3cm)
M = (30cm x .05BW) / 3cm
Problem: Assuming BW is 70Kg and Gravity is 10m/s, calculate the abduction
force of deltoid (M):
17. The arm is in 90 degrees of abduction and it is assumed that only deltoid is
active.
The force produced by deltoid muscle (M) acts 3cm from the centre of
rotation 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
i.e the clockwise moments must be equal to the anticlockwise moments.
(30cm x .05BW) = (M x 3cm)
M = (30cm x .05BW) / 3cm
Problem: Assuming BW is 70Kg and Gravity is 10m/s, calculate the abduction
force of deltoid (M):
............Next slide for answer ..........
18. Calculations:
M = (0.3m x 0.05 (70Kg x 10m/s) / 0.03m
Tips:
- Units should be converted to metres (m)
- BW should be a Force (N), which is mass x acceleration = 70Kg x
10m/s
19. Answer: M = 350N
Calculations:
M = (0.3m x 0.05 (70Kg x 10m/s) / 0.03m
Tips:
- Units should be converted to metres (m)
- BW should be a Force (N), which is mass x acceleration = 70Kg x
10m/s
20.
21. In this example the person is holding ball in the outstretched hand.
22. In this example the person is holding ball in the outstretched hand.
∑M=0
23. In this example the person is holding ball in the outstretched hand.
∑M=0
[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0
24. In this example the person is holding ball in the outstretched hand.
∑M=0
[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0
M = [(30cm x .05BW) + (60cm x .025BW)] / 3cm
25. In this example the person is holding ball in the outstretched hand.
∑M=0
[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0
M = [(30cm x .05BW) + (60cm x .025BW)] / 3cm
Problem: Assuming BW is 70Kg and Gravity is 10m/s, calculate the
abduction force of deltoid (M):
28. In this example the person has their elbow bent.
∑M=0
[(30cm x .05BW) + (60cm x .025BW)] - (M x 3cm) = 0
M = (15cm x .05BW) / 3cm
Problem: Assuming BW is 70Kg and Gravity is 10m/s, calculate the
abduction force of deltoid (M):
................ Next slide for answer .......
31. In the next two slides you can see how Moments are
relevant in clinical practice.
With a cuff deficient shoulder, arm elevation (flexion &
abduction) are dependent entirely on the deltoid
muscle.
By bending their elbow, the person reduces the
moment of their arm (by shortening the distance the of
the force from the shoulder). This reduces the Force
required by deltoid to elevate the arm.
M = 350N M = 175N
32. 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
33. JRF JRF Inclination
Activity of Daily Living
%BW Angle
Abduction
75 degrees without weight 85 33
Bergmann
45 degrees with 19.4 N weight 88 30
45 degrees without weight
Flexion
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 33
Extension, supine position, elbow
flexed, 118 N resistance at elbow 82 32
Lifting 13.6 N Coffeepot, straight arm 103 32
Nailing 15 cm above head 88 40
Steering 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 32
Walking with 2 crutches, full support 118 21
Lifting 96.8 N laterally 14 26
Putting 24.2 N into shelf, 60 cm in front 69 32
Combing hair
Typical effort 64 47
Maximum effort 98 41
Average 31.3
Standard Deviation 7.2
34. 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
35. Elbow Biomechanics:
In the example below, assume Force W is 20N and Force P
is 10N, calculate the force required by biceps to keep the
10N
20N
36. Resources
1. Richards - Biomechanics in Clinic &
Research, Churchill Livingstone
2. Norden & Frankel - Biomechanics of the
Musculoskeletal System
3. Wikipedia.org
4. Orthoteers.com
