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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.

The Spine

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## 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

### 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

### Determining External Forces

• 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

### What next?

ML5/S1 = 0

Which of these variables do we know?

### Abdominal Force (FA)

Abdominal Pressure  Abdominal Force

PA = 10-4[43 - 0.36H][ML5/S1]1.8

H = hip angle

### Abdominal Force (FA)

FA = PA (465 cm2)

465 cm2 = average diaphragm surface area

### Erector Spinae Force (Fm)

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

E

### 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)

### Compressive Forces on L5/S1

Horizontal

900 - 

Fmuscle = FComp

Fshear

(Sacral Cutting Plane)

Ftorso

Pelvic

### 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)

### 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)

### 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#)

* 1 lbs. = 4.45 N

### Example: Forces on L5/S1

Given:

Hip angle = 900

Knee angle = 900

Torso angle = 600

b = 20 cm

h = 30 cm

### Example: Forces on L5/S1

ML5/S1 = 0

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

E

### Example: Forces on L5/S1

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

E

Ftorso = 450 N

b = 20 cm

h = 30 cm

D = 13 cm

E = 5 cm

FA = ?

### 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

### 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.

### 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

### Example: Forces on L5/S1

PA = 0.24 N/cm2

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

FA = 111.6 N

### Example: Forces on L5/S1

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

E

Ftorso = 450 N

b = 20 cm

h = 30 cm

D = 13 cm

E = 5 cm

FA = 111.6 N

### 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.)

### Example: Forces on L5/S1

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

 = 400 + 

### Example: Forces on L5/S1

Knee angle = 900

Torso angle = 600

• = 400 + 

• = 120

 = 520

Pelvic

### 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

### 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

### 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?

### Questions?

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

• Compressive forces (explain theoretically and mathematically)?

• Shear forces (explain theoretically and mathematically)?

### 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.

### 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).