Biomechanics of Lifting and. Lower Back Pain: part 2

Transcription

1 Biomechanics of Lifting and Outline Lower Back Pain: part 2 Spinal stability Shear forces S.N. Robinovitch Effect of abdominal pressure on lifting mechanics Cantilever model of lifting

2 Forces on the lumbar spine Moment due to applied load Erector Spinae Force Disc shear force (perpendicular to long axis of vertebrae) Disc compressive force (parallel to long axis of vertebrae)

3 Definition of stability Engineering definition of stability: system is in a state of stable equilibrium if, for all possible small displacements from equilibrium, restoring force arise which accelerate the system back toward the equilibrium position Clinical definition of spinal instability: loss of the spine s ability to maintain its patterns of displacement under physiologic loads

4 Effect of co-contraction on spinal stability Spine with ligaments but no muscle will buckle under 90 N force Co-contracted muscles act like cables to stabilize the spine. Increasing the force or stiffness in both cables: increases the load-carrying capacity of the spine Increases ability to withstand perturbations (surprise loads) from both directions Reduces risk for buckling

5 Shear Forces Shear forces act parallel to the vertebral end plate and promote relative sliding between vertebrae Shear forces at the L4-L5 arise from (1) weight of the HAT, (2) hand forces (if any), and (3) forces in muscles and ligaments that connect to the spine If the erector spinae line of action is parallel to the long axis of the vertebrae, it does not contribute to disc shear force As the lumbar spine becomes fully flexed, the contribution of ligaments to the supportive moment increases.while this increases spinal stability, it also increases disc shear forces.

6 Shear force convention By convention, positive shear (or anterior shear ) indicates a tendency for L4 to move forward on L5, as when the trunk is flexed forward (Figure A). Negative (or posterior ) shear indicates a tendency for L4 to move backward on L5, as when a person pushes forward with their hands (Figure B). A B

7 Shear force affects injury risk In the Ontario Back Pain Study, injured workers had workloads that involved: Higher peak hand forces Higher peak L4-L5 shear forces Higher cumulative moments (time integrated) Higher peak trunk velocities Probability Peak Reaction Shear (N) Norman et al, 1998

8 Chaffin s Cantilever Low-back Model of Lifting Toppling moments due to HAT weight and hand force are balanced by supporting moments from erector spinae and abdominal pressure Includes abdominal pressure, and allows that long axis of L5/S1 may be different than long axis of torso. Related reading: Chaffin and Andersson, Occupational Biomechanics, Chapter 6: Section shear axis comp axis

9 Governing equations: cantilever model Step 1. v " M L5/S1 = 0 gives : b*(mg) HAT + h(mg) load # D F A Let v M L5/S1 ( ) # E( F m ) = 0. ( ) external = b *(mg) HAT + h(mg) load and v M L5/S1 ( ) internal = #D F A ( ) # E F m ( ). Use E = 0.05 m and D = 0.11 m. Step 3. Let F A = P A * A where A is the diaghram area, assumed equal to 465 cm 2. Note :1 mm Hg = N/cm 2 =133 Pa Step 2. Define abdominal pressure P A (in mm Hg) as v depending on hip flexion and M L5/S1 v P A =10 #4 [ 43 # 0.36 *( 180 #$ H )] ( ) external : ( ) 1.8 ( M L5/S1 ) external where $ H is the included hip angle (knee - hip - shoulder).

10 Abdominal pressure Abdominal pressure (P A ) is developed through contraction of the diaphragm and abdominal wall muscles. Abdominal pressure is higher in fast than slow lifts. The internal force (F A ) created by the abdominal pressure is estimated using the following two assumptions (Morris et. al., 1961) average diaphragm area (A) of 465 cm 2 a line of action parallel to the compressive forces on the lumbar spine

11 Disc axes for compression and shear to calculate disc compression and shear force, the plane of L5/S1 must be determined. spinal curvature will cause each intervertebral joint to have unique coordinate axes the longitudinal axis of L5/S1 will differ from the angle T of the torso

12 Sacral joint rotation Angle (!) between the plane of L5/S1 and the horizontal is assumed to depend on posture as follows:! = 40 o + " Where " depends on the included knee angle K and the torso angle T as follows: " = T K TK T K 2 Alternatively,! can be estimated from spinal curvature For erect posture,!! 0 and!! 40 o

13 Sacral joint rotation (") scales with torso and knee angle When the torso flexes beyond deg, pelvis rotates forward (cw) at a rate of 2 deg for each 3 deg of forward torso flexion (T) When the knee flexes beyond 45 deg, pelvis rotates backward (ccw) at a rate of 1 deg for each 3 deg of knee flexion (K)

14 Calculation of compression and shear force " F = 0 : comp cos# ( mg) HAT + cos# ( mg) load $F A + F M $ F C = 0 (Eqn 6.51) " F = 0 : shear sin# mg ( ) HAT + sin# mg ( ) load $F S = 0 (Eqn 6.52)

15 Assumptions in the Cantilever Model As discussed in Ch. 53 & 54, assumptions in this model include: 1.2D analysis is valid 2.static (vs. dynamic) analysis is valid 3.ligament forces are negligible 4.single equivalent muscle for erector spinae 5.assumptions regarding muscle force direction and moment arm 6.assumptions regarding abdominal pressure and surface area 7.assumptions regarding orientation (rotation) of vertebral joints