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The Spine

The Spine. Forces during Lifting. What type of forces? Compressive Shear. What causes these forces?. External Forces Body (torso) weight due to gravity Weight of load due to gravity. What causes of these forces?. Internal Forces Muscular Forces Abdominal Pressure.

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The Spine

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  1. The Spine

  2. Forces during Lifting What type of forces? • Compressive • Shear

  3. What causes these forces? External Forces • Body (torso) weight due to gravity • Weight of load due to gravity

  4. What causes of these forces? Internal Forces • Muscular Forces • Abdominal Pressure

  5. How do you determine the magnitude of these forces? • Calculate the external forces – why? • To determine internal forces – how? • System in static equilibrium  sum of forces = 0 • Finternal = Fexternal or Finternal - Fexternal = 0

  6. Determining External Forces • First, determine the external moment about L5/S1, why? • Force = Moment/MA

  7. Moments about L5/S1 • Sum of ML5/S1 = 0 • Sum of external moments - sum of internal moments= 0

  8. Estimation of External Moment about L5/S1 ML5/S1 = Mtorso wt. + Mload ML5/S1 = Ftorso wt.b + Floadh 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 ML5/S1 = Merector spinae + Mabdominal pressure ML5/S1 = Ferector spinaeE + FabdominalD E = moment arm of erector spinae (5 cm) D = moment arm of abdominal force

  10. Moment arm of FA (D) • varies with sine of the torso angle • 7 cm for erect sitting • 15 cm when torso is 900 from vertical

  11. What next? ML5/S1 = 0 Ftorso wt.b + Floadh – FAD – FME = 0 Which of these variables do we know?

  12. Knowns vs. Unknowns Ftorso wt.b + Floadh – FAD – FME = 0

  13. Abdominal Force (FA) Abdominal Pressure  Abdominal Force PA = 10-4[43 - 0.36H][ML5/S1]1.8 H = hip angle

  14. Abdominal Force (FA) FA = PA (465 cm2) 465 cm2 = average diaphragm surface area

  15. Knowns vs. Unknowns Ftorso wt.b + Floadh – FAD – FME = 0

  16. Erector Spinae Force (Fm) Ftorso wt.b + Floadh – FAD – FME = 0 Fm= Ftorso wt.b + Floadh - D(FA) E

  17. Compressive Forces on L5/S1  Fcomp = 0 Fcomp = reactive force cos  Ftorso wt. + cos  Fload - FA + Fm - Fcomp = 0 Fcomp = cos  Ftorso wt. + cos  Fload - FA + Fm  = sacral cutting plane (vertical orientation of the sacrum)

  18. Compressive Forces on L5/S1

  19. Compressive Forces on L5/S1   Horizontal 900 -   Fmuscle = FComp Fshear (Sacral Cutting Plane) Ftorso FLoad

  20. Sacral Cutting Plane

  21. Compressive Forces on L5/S1  Pelvic

  22. Compressive Forces on L5/S1  Fcomp = 0 cos  Ftorso wt. + cos  Fload - FA + Fm - Fcomp = 0 Fcomp = cos  Ftorso wt. + cos  Fload - FA + Fm  = sacral cutting plane (vertical orientation of the sacrum)

  23. Shear Forces on L5/S1  Fshear = 0 sin  Ftorso wt. + sin  Fload - Fshear = 0 Fshear = sin  Ftorso wt. + sin  Fload  = 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 = 900 Knee angle = 900 Torso angle = 600 b = 20 cm h = 30 cm

  26. Example: Forces on L5/S1 ML5/S1 = 0 Ftorso wt.b + Floadh – FAD – FME = 0 Fm= Ftorso wt.b + Floadh - D(FA) E

  27. Example: Forces on L5/S1 Fm= Ftorso wt..b + Floadh - D(FA) E Ftorso = 450 N b = 20 cm Fload = 450 N h = 30 cm D = 13 cm E = 5 cm FA = ?

  28. Example: Forces on L5/S1 FA = ? PA = 10-4[43 - 0.36H][ML5/S1]1.8 ML5/S1 = Ftorso wtb + Floadh = (450 N)(20 cm) + (450 N)(30 cm) ML5/S1 = 22500 Ncm = 225 Nm

  29. Example: Forces on L5/S1 FA = ? PA = 10-4[43 - 0.36H][ML5/S1]1.8 H = 900 ML5/S1 =225 Nm PA = 10-4[43 - 0.36(60)][225 ]1.8 NOTE: the values for H and ML5/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 FA = ? PA = 10-4[43 - 0.36(90)][225 ]1.8 PA = 18.2 mmHg PA = 0.24 N/cm2 * 1 N/cm2 = 75 mmHg

  31. Example: Forces on L5/S1 PA = 0.24 N/cm2 FA = (0.24 N/cm2)(465 cm2) FA = 111.6 N

  32. Example: Forces on L5/S1 Fm= Ftorso wt.b + Floadh - D(FA) E Ftorso = 450 N b = 20 cm Fload = 450 N h = 30 cm D = 13 cm E = 5 cm FA = 111.6 N

  33. Example: Forces on L5/S1 Fm = (450 N)(20 cm) + (450 N)(30 cm) - (13 cm)(111.6 N) 5 cm Fm = 9000 Ncm + 13500 Ncm - 1451 Ncm 5 cm Fm = 4210 N (946 lbs.)

  34. Example: Forces on L5/S1 Fcomp = cos  Ftorso wt. + cos  Fload - FA + Fm  = 400 + 

  35. Example: Forces on L5/S1 Knee angle = 900 Torso angle = 600 • = 400 +  • = 120  = 520  Pelvic

  36. Compressive Forces on L5/S1 Fcomp = cos  Ftorso wt. + cos  Fload - FA + Fm Fcomp = (cos 520)(450 N) + (cos 520)(450 N) – 111.6 N + 4210 N Fcomp = 277 N + 277 N – 111.6 N + 4210 N Fcomp = 4652.5 N

  37. Shear Forces on L5/S1 Fshear = sin  Ftorso wt. + sin  Fload Fshear = (sin 520)(450 N) + (sin 520)(450 N) Fshear = 354.6 N + 354.6 N Fshear = 709.2 N

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

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