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# The Spine - PowerPoint PPT Presentation

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

What type of forces?

• Compressive

• Shear

External Forces

• Body (torso) weight due to gravity

• Weight of load due to gravity

Internal Forces

• Muscular Forces

• Abdominal Pressure

• Calculate the external forces – why?

• To determine internal forces – how?

• System in static equilibrium  sum of forces = 0

• Finternal = Fexternal or Finternal - Fexternal = 0

• First, determine the external moment about L5/S1, why?

• Force = Moment/MA

• Sum of ML5/S1 = 0

• Sum of external moments - sum of internal moments= 0

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

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

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

ML5/S1 = 0

Ftorso wt.b + Floadh – FAD – FME = 0

Which of these variables do we know?

Knowns vs. Unknowns

Ftorso wt.b + Floadh – FAD – FME = 0

Abdominal Pressure  Abdominal Force

PA = 10-4[43 - 0.36H][ML5/S1]1.8

H = hip angle

FA = PA (465 cm2)

465 cm2 = average diaphragm surface area

Knowns vs. Unknowns

Ftorso wt.b + Floadh – FAD – FME = 0

Ftorso wt.b + Floadh – FAD – FME = 0

Fm= Ftorso wt.b + Floadh - D(FA)

E

 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)

Horizontal

900 - 

Fmuscle = FComp

Fshear

(Sacral Cutting Plane)

Ftorso

Pelvic

 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)

 Fshear = 0

sin  Ftorso wt. + sin  Fload - Fshear = 0

Fshear = sin  Ftorso wt. + sin  Fload

 = sacral cutting plane (vertical orientation of the sacrum)

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

Given:

Hip angle = 900

Knee angle = 900

Torso angle = 600

b = 20 cm

h = 30 cm

ML5/S1 = 0

Ftorso wt.b + Floadh – FAD – FME = 0

Fm= Ftorso wt.b + Floadh - D(FA)

E

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 = ?

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

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.

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

PA = 0.24 N/cm2

FA = (0.24 N/cm2)(465 cm2)

FA = 111.6 N

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

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.)

Fcomp = cos  Ftorso wt. + cos  Fload - FA + Fm

 = 400 + 

Knee angle = 900

Torso angle = 600

• = 400 + 

• = 120

 = 520

Pelvic

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

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

• What component is the largest contributor to compressive forces on L5/S1?

• What component is the largest contributor to shear forces on the spine?

3. As the sacral cutting angle increases, what happens to:

• Compressive forces (explain theoretically and mathematically)?

• Shear forces (explain theoretically and mathematically)?

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

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).