Characteristics of Lumbosacral Joint Stress under the Static Load Homework from Upright to Dorsiflexion Position

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1 Journal of TUS Vol.27 No ~337 Doctor Forum L 5/S 1 L 5/S cm L 5/S 1 L 5/S 1 L 5/S 1 G A Characteristics of Lumbosacral Joint Stress under the Static Load Homework from Upright to Dorsiflexion Position LUO Jiong School of PE Southwest University Chongqing China Abstract: According to related material of the clinical biomechanical this research choose L5/S1 joint center for research emphases and mechanical model was established by changing human body posture which shows that when the normal human was in an balance with upright and relax position the isometric contraction down about wall muscular of before and after spinal can not play an "soft pillar" role to L5/S1 disc center. Spinal wall muscle strength and joint center stress of L5/S1 are impacted not only by pulling Angle of wall muscle but also are more influence by the arm of force of front wall muscle. when arm is smaller less than 5.0cm before and after wall muscular strength and joint stress of L5/S1 have declined following pulling Angle increasing and there is exist a best position which make justly muscular wall strength impose soft pillar effect on the L5/S1 joint center. Appropriate using of chest and abdominal pressures also can not only ensure body balance but also obtain "soft pillar" effect on the L5/S1 joint and at the same time spinal muscular wall is needed simply smaller force.in other factor unchanged the flexor a condition of human spine shall have much more before and after wall strength and stress of L5/S1 places than upright position which hints people to avoid proneness to bend over while daily work especially when the people proneness to move things which more should reduce heavy torque as far as possible. Key words: static homework lumbosacral joint stress characteristics soft pillar CT MRI X CT MRI [1-4] L 5 /S L 5 /S L 5 /S 1 L 5 /S 1 L 5 /L 4 85%~95% [4-6] L 5 /S 1 lumbosacral disc L 5 /S

2 334 Journal of TUS Vol.27 No L 5 /S 1 64% [2-3] W 1 = N CT MRI [2 3] [2-3] L 5 /S 1 X CT MRI P m1 Y d 1 =10 cm 3 L 5 /S 1 P m2 L 5 /S 1 d d cm [7] 30~ cm 7.0 cm % [9] d cm L 5 /S 1 L 5 /S W=0 F A =0 P m1 P m2 θ 1 θ 2 0 W 1 b L 5 /S m θ=90 O L 5 /S 1 P m1 =0 P m2 F A L 5 /S 1 P m2 = N=0.35W 1 F c =508 N=1.35W 1 L 5 /S 1 F c Y L 5 /S 1 O Y X θ 1 θ 2 F S [8 9] F A L 5 /S 1 W 1 W W 1 W L 5 /S 1 O b h d 1 d 2 1 ΣF X =0 圯 F s +P m1 sin θ1 +(W 1 +W)cosθ-P m2 sinθ 2 =0 1 ΣF Y =0 圯 F c -P m2 cosθ 2 +P m1 cosθ 1 +F A -(W 1 +W)sinθ=0 2 ΣM=0 圯 F A d+p m1 d 1 cosθ 1 +P m2 d 2 cosθ 2 -W 1 b-wh=0 3 2 Figure 2 Body lumbosacral joint Stress when it is balanced upright regardless of load conditions Figure 1 1 chart of lumbosacral joint Stress analysis 0.35 W 1 L 5 /S W 1 L 5 /S 1 O P m1 P m2 =0 L 5 /S 1 2 Q=60 Kg

3 Journal of TUS Vol.27 No cm [11] W=0 FA= θ 1 θ 2 P m1 P m2 F c θ1 θ2 P m1 P m2 θ θ F c N P m1 P n2 θ 1 θ 2 0 P m W N 1.33 W 1 F c θ 1 >θ 2 [10] θ 1 15 θ 2 10 L 5 /S 1 F c =0 14 θ θ 2 11 d 2 d cm d P m1 P m2 F c d 2 θ 1 θ 2 3 d 2 P m1 P m2 L 5 /S 1 4 d 2 P m1 P m2 L 5 /S 1 5 d 2 P m1 P m2 Fc θ 1 Fc θ 2 L 5 /S 1 Fc d 2 Figure 3 When d 2 was identified as a Figure 4 When d 2 was identified as a Figure 5 When d 2 was identified as a certain value the change curve the Pulling certain value the change curve the Pulling certain value the change curve the Pulling force of P m1 and P m2 and L 5 /S 1 stress Fc force of P m1 and P m2 and L 5 /S 1 stress Fc force of P m1 and P m2 and L 5 /S 1 stress Fc respectively with θ 1 respectively with θ 2 respectively with d 2 θ 1 =15 θ 2 =10 d cm P m1 P m2 F c d 2 P m1 P m2 F c d 2 P m1 P m2 F c d 2 =5.5 cm P m1 = N=1.26W 1 P m2 = N=1.88W 1 L 5 / [13] S 1 F c = N=-2.07W 1 400~600 N W W W N d=5.5 cm d cm F c N d cm F c N [12] N d cm θ1 θ2 d W=0 F A 0 3 Lumbar discherniation [14] LDH ZHONG [15] MORGAN [16]

4 336 Journal of TUS Vol.27 No P A 90 mmhg [14] CHAFFIN [17] F m1 sin15 -F m2 sin10 =0 F c +F m2 cos10 +F m1 cos15 +F A -W 1 =0 4 5 F m1 0.1cos15 +F A 0.07-F m cos10 -W =0 6 y W 1 b=0.25 m d CHAFFIN [18] 15 cm P m1 = F A P m2 = F A 7 8 F A d=7 cm d= W=0 F A 0 P m1 =0 45 P m2 F s +W 1 cos45 =0 F c +P m2 +Fa-W 1 sin45 = F c =9.73 F A F A F A 0.15+P m W =0 12 P m1 P m2 ZHONG [15] MORGAN [16] P m2 = F A 13 F c = F A 14 F A 106 N P m1 P m2 0 F c = N W 1 P m2 = F c 15 0 P m2 =1 710N F c =1 976 N P m F c 9 F c F A FA 2.59 F A =50 N 100 N 150 N 465 P m2 F c cm 2 P A 90 mmhg [17-18] W 1 F A =550 N P m1 = N=-5.20 W 1 P m2 = N=-7.75 W 1 F c = N=12.19 W F c W P m2 F c F A P m2 F c L 5 /S 1 KRAG [18] L 5 /S 1 P m1 P m2 L 5 /S 1 P m1 P m2 3 L 5 /S 1 1 L 5 /S 1 2 L 5 /S 1 L 5 /S 1 L 5 /S 1 L 5 /S L 5 /S 1 2 d 2 d cm L 5 /S 1

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