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Published byKenny Powe Modified over 2 years ago

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The Spine

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Forces during Lifting What type of forces? zCompressive zShear

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

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

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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

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

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

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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

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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

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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

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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?

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

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Abdominal Force (F A ) Abdominal Pressure Abdominal Force P A = 10 -4 [43 - 0.36 H ][M L5/S1 ] 1.8 H = hip angle

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

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

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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

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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)

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Compressive Forces on L5/S1

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Horizontal F torso F muscle = F Comp F shear (Sacral Cutting Plane) 90 0 - F Load

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Sacral Cutting Plane

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Compressive Forces on L5/S1 Pelvic

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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)

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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)

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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

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

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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

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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 = ?

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Example: Forces on L5/S1 F A = ? P A = 10 -4 [43 - 0.36 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 = 22500 Ncm = 225 Nm

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Example: Forces on L5/S1 F A = ? P A = 10 -4 [43 - 0.36 H ][M L5/S1 ] 1.8 H = 90 0 M L5/S1 =225 Nm P A = 10 -4 [43 - 0.36(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.

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Example: Forces on L5/S1 F A = ? P A = 10 -4 [43 - 0.36(90)][225 ] 1.8 P A = 18.2 mmHg P A = 0.24 N/cm 2 * 1 N/cm 2 = 75 mmHg

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Example: Forces on L5/S1 P A = 0.24 N/cm 2 F A = (0.24 N/cm 2 )(465 cm 2 ) F A = 111.6 N

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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 = 111.6 N

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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 + 13500 Ncm - 1451 Ncm 5 cm F m = 4210 N (946 lbs.)

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Example: Forces on L5/S1 F comp = cos F torso wt. + cos F load - F A + F m = 40 0 +

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Example: Forces on L5/S1 Knee angle = 90 0 Torso angle = 60 0 = 40 0 + = 12 0 = 52 0 Pelvic

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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) – 111.6 N + 4210 N F comp = 277 N + 277 N – 111.6 N + 4210 N F comp = 4652.5 N

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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 = 354.6 N + 354.6 N F shear = 709.2 N

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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?

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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)?

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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.

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Consider This?

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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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