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The Spine. Forces during Lifting What type of forces? zCompressive zShear.

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Presentation on theme: "The Spine. Forces during Lifting What type of forces? zCompressive zShear."— Presentation transcript:

1 The Spine

2 Forces during Lifting What type of forces? zCompressive zShear

3 What causes these forces? External Forces zBody (torso) weight due to gravity zWeight of load due to gravity

4 What causes of these forces? Internal Forces zMuscular Forces zAbdominal Pressure

5 How do you determine the magnitude of these forces? 1.Calculate the external forces – why? 2.To determine internal forces – how? 3.System in static equilibrium  sum of forces = 0 4.F internal = F external or F internal - F external = 0

6 Determining External Forces zFirst, determine the external moment about L5/S1, why? z Force = Moment/MA

7 Moments about L5/S1 zSum of M L5/S1 = 0 zSum of external moments - sum of internal moments= 0

8 Estimation of External Moment about L5/S1 M L5/S1 = M torso wt. + M load M L5/S1 = F torso wt. b + F load h b = distance from L5/S1 to COM of torso h = distance from L5/S1 to COM of load

9 Estimation of Internal Moment about L5/S1 M L5/S1 = M erector spinae + M abdominal pressure M L5/S1 = F erector spinae E + F abdominal D E = moment arm of erector spinae (5 cm) D = moment arm of abdominal force

10 Moment arm of F A (D) zvaries with sine of the torso angle z7 cm for erect sitting z15 cm when torso is 90 0 from vertical

11 What next?  M L5/S1 = 0 F torso wt. b + F load h – F A D – F M E = 0 Which of these variables do we know?

12 Knowns vs. Unknowns F torso wt. b + F load h – F A D – F M E = 0

13 Abdominal Force (F A ) Abdominal Pressure  Abdominal Force P A = [  H ][M L5/S1 ] 1.8  H = hip angle

14 Abdominal Force (F A ) F A = P A (465 cm 2 ) 465 cm 2 = average diaphragm surface area

15 Knowns vs. Unknowns F torso wt. b + F load h – F A D – F M E = 0

16 Erector Spinae Force (F m ) F torso wt. b + F load h – F A D – F M E = 0 F m = F torso wt. b + F load h - D(F A ) E

17 Compressive Forces on L5/S1  F comp = 0 F comp = reactive force cos  F torso wt. + cos  F load - F A + F m - F comp = 0 F comp = cos  F torso wt. + cos  F load - F A + F m  = sacral cutting plane (vertical orientation of the sacrum)

18 Compressive Forces on L5/S1

19 Horizontal F torso F muscle = F Comp  F shear (Sacral Cutting Plane)    F Load

20 Sacral Cutting Plane

21 Compressive Forces on L5/S1  Pelvic

22 Compressive Forces on L5/S1  F comp = 0 cos  F torso wt. + cos  F load - F A + F m - F comp = 0 F comp = cos  F torso wt. + cos  F load - F A + F m  = sacral cutting plane (vertical orientation of the sacrum)

23 Shear Forces on L5/S1  F shear = 0 sin  F torso wt. + sin  F load - F shear = 0 F shear = sin  F torso wt. + sin  F load  = sacral cutting plane (vertical orientation of the sacrum)

24 Example: Forces on L5/S1 Calculate the compressive & shear forces on the L5/S1 IV disk for a 200 lbs. UPS driver who has to lift a maximal load of 100 lbs. from the floor to waist level. Given: Torso weight: 450 newtons (100#) Load weight: 450 newtons (100#) * 1 lbs. = 4.45 N

25 Example: Forces on L5/S1 Given: Hip angle = 90 0 Knee angle = 90 0 Torso angle = 60 0 b = 20 cm h = 30 cm

26 Example: Forces on L5/S1  M L5/S1 = 0 F torso wt. b + F load h – F A D – F M E = 0 F m = F torso wt. b + F load h - D(F A ) E

27 Example: Forces on L5/S1 F m = F torso wt.. b + F load h - D(F A ) E F torso = 450 N b = 20 cm F load = 450 N h = 30 cm D = 13 cm E = 5 cm F A = ?

28 Example: Forces on L5/S1 F A = ? P A = [  H ][M L5/S1 ] 1.8 M L5/S1 = F torso wt b + F load h = (450 N)(20 cm) + (450 N)(30 cm) M L5/S1 = Ncm = 225 Nm

29 Example: Forces on L5/S1 F A = ? P A = [  H ][M L5/S1 ] 1.8  H = 90 0 M L5/S1 =225 Nm P A = [ (60)][225 ] 1.8 NOTE: the values for  H and M L5/S1 must be entered into the equation in degrees and Nm, respectively; however since this is a regression equation; units are NOT maintained as in typical algebraic equations.

30 Example: Forces on L5/S1 F A = ? P A = [ (90)][225 ] 1.8 P A = 18.2 mmHg P A = 0.24 N/cm 2 * 1 N/cm 2 = 75 mmHg

31 Example: Forces on L5/S1 P A = 0.24 N/cm 2 F A = (0.24 N/cm 2 )(465 cm 2 ) F A = N

32 Example: Forces on L5/S1 F m = F torso wt. b + F load h - D(F A ) E F torso = 450 N b = 20 cm F load = 450 N h = 30 cm D = 13 cm E = 5 cm F A = N

33 Example: Forces on L5/S1 F m = (450 N)(20 cm) + (450 N)(30 cm) - (13 cm)(111.6 N) 5 cm F m = 9000 Ncm Ncm Ncm 5 cm F m = 4210 N (946 lbs.)

34 Example: Forces on L5/S1 F comp = cos  F torso wt. + cos  F load - F A + F m  = 

35 Example: Forces on L5/S1 Knee angle = 90 0 Torso angle = 60 0  =   = 12 0  = 52 0  Pelvic

36 Compressive Forces on L5/S1 F comp = cos  F torso wt. + cos  F load - F A + F m F comp = (cos 52 0 )(450 N) + (cos 52 0 )(450 N) – N N F comp = 277 N N – N N F comp = N

37 Shear Forces on L5/S1 F shear = sin  F torso wt. + sin  F load F shear = (sin 52 0 )(450 N) + (sin 52 0 )(450 N) F shear = N N F shear = N

38 Questions? 1.What component is the largest contributor to compressive forces on L5/S1? 2.What component is the largest contributor to shear forces on the spine?

39 Questions? 3. As the sacral cutting angle increases, what happens to: A.Compressive forces (explain theoretically and mathematically)? B.Shear forces (explain theoretically and mathematically)?

40 Questions? 4. As the torso angle increases and the position of the lower extremities remains fixed, describe what happens with relative compressive and shear forces.

41 Consider This?

42 Questions? 5. Describe mathematically how a “deep squat” affects compressive and shear forces on the spine compared with an “erect” knee angle position [knees extended] (hint: refer to the pelvic rotation vs. torso axis graph).


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