Materi # 8a - Biomekanika

BIOMEKANIKA

Sritomo Wignjosoebroto,Dyah SDLaboratorium Ergonomi dan Perancangan Sistem Kerja

Jurusan Teknik Industri - Institut Teknologi Sepuluh NopemberSURABAYA

11/23/2010 Sritomo W.S & Dyah SD 1

MENGANGKAT, MENDORONG, DENGAN POSISI SETEGAK

MUNGKIN UNTUK MEMINIMALKAN BEBAN INTERNAL

23/11/2010 Biomekanika 2

BEBAN INTERNAL TULANG BELAKANG SANGAT

DIPENGARUHI OLEH POSISI MENGANGKAT


TEKANAN PERUT IKUT BERPERAN DALAM MENGURANGI

BEBAN INTERNAL PIRINGAN TULANG BELAKANG


Biomekanika“Bidang ilmu yang memadukan antara bidang ilmu biologi dan mekanika”

• Biomekanika adalah ilmu pengetahuan yang merupakan kombinasi dariilmu fisika (khususnya mekanika) dan teknik, dengan berdasar pada biologidan juga pengetahuan lingkungan kerja.

• Biomekanika menggunakan hukum – hukum fisika, mekanika teknik,biologi, dan prinsip fisiologi untuk menggambarkan kinematika dan kinetikyang terjadi pada anggota tubuh manusia.

• Kinematika : pergerakan/ motion pada segmen – segmen tubuh. Kinetik :efek dari gaya dan momen yang terjadi pada tubuh. Mekanika digunakansebagai penyusun konsep, analisa, dan desain dalam sistem biologimakhluk hidup.

• Biomekanika dari gerakan manusia adalah ilmu yang menyelidiki,menggambarkan dan menganalisis gerakan-gerakan manusia (Winter,1990)

• Pada dasarnya mempelajari dan menganalisis batas-batas kekuatan,ketahan, kecepatan, dan ketelitian yang dimiliki manusia dalam melakukankerja dipengaruhi faktor manusia, sikap kerja dan jenis pekerjaan


Biomekanika• Biomekanika :

– Biostatik : bagian daari biomekanika umum yanghanya menganalisa bagian tubuh dalam keadaandiam maupun bergerak pada garis lurus dengankecepatan seragam (uniform)

– Biodinamik : berkaitan dengan gerakan-gerakantubuh taanpa mempertimbangkan gaya yangterjadi (kinematik) dan gaya yang disebabkangaya yang bekerja dalam tubuh ( kinetik)

– Occupational biomekanika : Bagian darimekanik terapan yang mempelajari interaaksi fisikantara pekerja dengan mesin, material, danperalatan dengan tujuan untuk meminumkankeluhan pada sistem kerangka otott agarproduktivitas kerja dapat meningkat (Chaffin &Anderson,1984)


occupational biomechanics


Occupational biomechanics adalah sub – disiplindalam kerangka besar biomekanika yangmempelajari hubungan antara pekerja dengan alatkerja, workstation, mesin, dan material untukmeningkatkan performa dengan meminimalisasiterjadinya cidera musculoskeletal.

Sehingga studi utama tentang Occupationalbiomechanics berkaitan dengan masalahmusculoskeletal.

Sistem musculoskeletal terdiri atas tulang, otot,dan jaringan penghubung (ligament, tendon, fascia,dan cartilage). Fungsi utama sistem tersebut adalahmendukung dan melindungi tubuh dan bagian –bagian tubuh, menjaga postur tubuh, producepergerakan tubuh, serta menghasilkan panas danmempertahankan suhu tubuh.

why occupational biomechanics ???

Jeffress (1999) indicated that approximately650,000 workers every year suffer seriousinjuries and illnesses caused by overexertion,repetition, and other types of physical stress.Such injuries cost U.S. business between $15to $20 billion dollars a year in workmancompensation. According to US Departmentof labor, back injuries accounted for nearly20% of all injuries and illnesses in the workplace. In the UK, similar numbers appearwith 27% of all reported accidents involvingmanual handling.


Analisa Mekanik

Tiga jenis gaya yang bekerja di dalam tubuh manusia (Winter,1990) :

• Gaya Gravitasi:

Gaya yang melalui pusat massa dari segmen tubuh manusia dengan arah ke bawah

• Gaya Reaksi

Gaya yang terjadi akibat beban pada segmen tubuh atau berat segmen tubuh itu sendiri

• Gaya Otot

Gaya yang terjadi pada bagian sendi,baik akibat gesekan sendi atau akibat gaya pada otot yang melekat pada sendi.Gaya ini menggambarkan besarnya momen otot


Body SegmentsGroup body segments as a percentage of

Individual body segments mass as a percentage of

Total body mass ( %)

Group segment mass (%)

Total body mass (%)

Head and neck 8.4 Head

Neck

73.8

26.2

6.2

2.2

Torso (trunk) 50.0 Thorax

Lumbar

Pelvis

43.8

29.4

26.8

21.9

14.7

13.4

Each arm (total) 5.1 Upper arm

Forearm

Hand

54.9

33.3

11.8

2.8

1.7

0.6

Each leg (total) 15.7 Thigh

Lower leg (shank)

Foot

63.78

27.4

8.9

10.0

4.3

1.4


Contoh Perhitungan Dasar Biomekanik

• Mis. P = 10 N, W= 20 N

• F = (13*20) + (30*10) = 112 N Gaya reaksi : J = 112 – 20 -10 = 82 N

5

11/23/2010 Sritomo W.S & Dyah SD11

Biomechanics of the back


LIFTING AND BACK

STRESS

Cont.


Cont.


Biomechanical -- Static analysis

Example :

A male worker pick a container off a conveyor (located 35 cm above the floor). The container has a mass of 15 kg. This task is performed 360 times pershift


The measurement

Distance from wrist to center of mass (c.m.) of hand ( SL1)

0.07 m

Distance from wrist to elbow ( SL2) 0.28 m

Distance from elbow to shoulder ( SL3) 0.3 m

Distance from shoulder to L5/S1 disk (SL4) 0.36 m

Angle of hand from horizontal (θ1) 30o

Angle of forehand from horizontal (θ2) 30o

Angle of upper arm from horizontal ( θ3) 80o

Angle of trunk from horizontal ( θ4) 45o

Body weight (mass) 70 kg


Calculation

Work = m.g.h.fm = mass of the load in kilograms (kg)

g = gravitational constant ( 9.8 m. s -2 )

h = height of lift in meters (m)

F = frequency (number of lift per shift)

Work = (15 kg/lift). (9.8 m. s -2 ).(0.65 m – 0.35 m). (360 lifts/shift)

Work = 15876 J per shift


For the hand segment• Wo = Force due to the weight of the external load = m.g

= 15 kg . 9.8 m. s -2 = 147 N

• WH = Force due to the weight of the hand = m H. g

= ( 0.006 . 70 kg ). (9.8 m. s -2 ) = 4.1 N

• Mw = Resultant moment at the wrist to maintain static equilibrium

• Fxw = Resultant force in x- direction at the wrist to maintain static equilibrium

• Fyw = Resultant force in y- direction at the wrist to maintain static equilibrium

• θ 1 = Angle of the hand relative to horizontal θ 1 = 300 for this examples

• SL1 = Measured length from wrist to c.m. of hand ( at handles of box) SL1 = 0.07 m for this examples

• ∑ Fx = Fx w = 0

• ∑ Fy = Fy w - Wo - WH = 0

• ∑ Mw = Mw - (Wo +WH) . SL1 . Cos θ 1 = 0


∑ Fx = Fx w = 0

∑ Fy = Fy w - Wo - WH = 0

∑ Mw = Mw - (Wo +WH) . SL1 . Cos θ 1

= 0

Thus For each wrist :Fx w = 0

Fy w = Wo + WH = ( 147 N)/2 + 4.1 N = 77.6 N

Mw = (Wo +WH) . SL1 . Cos θ 1

=( 77.6 N) . (0.07 m ). (cos 300 )

= 4.7 N.m


For the lower arm segment• WLA = Force due to the weight of the lower arm = mLA . g

= ( 0.017 . 70 kg ). (9.8 m. s -2 ) = 11.7 N

• Mw = 4.7 N.m

• Fxw = 0

• Fyw = 77.6 N

• θ2 = Angle of the lower arm relative to horizontal = 300

for this examples

• SL2 = Measured length from wrist to elbow SL2 = 0.28 m

• λ2 = Location of c.m. as a portion of SL from elbow = 0.43 ( or 43 %)

• Me = Resultant moment at the elbow to maintain static equilibrium

• Fxe = Resultant force in x- direction at the wrist to maintain static equilibrium

• Fye = Resultant force in y- direction at the wrist to maintain static equilibrium

• ∑ Fx = - Fx w + Fxe = 0

• ∑ Fy = - Fy w - WLA + Fye = 0

• ∑ Me = Me - Mw - WLA.. Λ2. SL2 . Cos θ2 - Fyw . SL2 . Cos θ2 - Fxw . SL2 . Sin θ2 = 0


Thus For each elbow:Fx e = 0

Fy e = 77.6 N + 11.7 N = 89.3 N

Me = 4.7 N.m + 11.7 N . 0.43 . 0.28 m. 0.866 + 77.6 N .0.28 m . 0.866

= 24.7 N.m


For the upper-arm segment

• WUA = Force due to the weight of the trunk = mUA . g• = ( 0.028. 70 kg ). (9.8 m. s-2 ) = 19.2 N• Me = 24.7 N.m• Fxe = 0• Fye = 89.3 N• θ3 = Angle of the lower arm relative to horizontal = 800 for this

examples• SL3 = Measured length from the elbow to shoulder SL3 = 0.30 m• λ3 = Location of c.m. as a portion of SL from shoulder = 0.436 ( or 43.6

%)• Ms = Resultant moment at the shoulder to maintain static equilibrium• Fxs = Resultant force in x- direction at the shoulder to maintain static

equilibrium• Fys = Resultant force in y- direction at the shoulder to maintain static

equilibrium

• ∑ Fx = - Fxe + Fxs = 0• ∑ Fy = - Fye - WUA + Fys = 0• ∑ Me = Ms - Me - WUA. λ3. SL3 . Cos θ3 - Fye . SL3 . Cos θ3 - Fxe . SL3 . Sin θ3 = 0


Thus For each shoulder:

Fxs = 0

Fys = Fye + WUA = 89.3 N + 19.2 N = 108.5 N

Ms = 24.7 N.m + 19.2 N . 0.436 . 0.30 m. 0.174 + 89.3 N .0.3 m . 0.174

= 29.8 N.m


For the trunk segment

• WT = Force due to the weight of the trunk = mT . g• = ( 0.45. 70 kg ). (9.8 m. s -2 ) = 308.7 N• Ms = 29.8 N.m for each shoulder = 59.6 N.m for both shoulders• Fxs = 0• Fys = 108.5 for each shoulder = 217.0 N for both shoulders• θ4 = Angle of the trunk relative to horizontal = 450 for this

examples• SL4 = Measured length for L5/S1 to shoulder SL4 = 0.36 m• λ4 = Location of c.m. as a portion of SL from L5/S1: λ4 = 0.67 (

estimated)• Mt = Resultant moment at L5/S1 to maintain static equilibrium• Fxt = Resultant force in x- direction at L5/S1 to maintain static

equilibrium• Fyt = Resultant force in y- direction at L5/S1 to maintain static

equilibrium

• ∑ Fx = - Fxs + Fxt = 0• ∑ Fy = - Fye - WT + Fyt = 0• ∑ Me = Mt - Ms - WT. λ4. SL4 . Cos θ4 - Fys . SL4 . Cos θ4 - Fxe . SL4 . Sin θ4 = 0


Thus For each trunk:

Fxt = 0

Fyt = Fys + WT = 217.0 N + 308.7 N = 525.7 N

Me = 59.6 N.m + 308.7 N . 0.67 . 0.36 m. 0.707 + 217.0 N .0.36 m . 0.707

= 167.5 N.m


If the erector spinae muscle group is assumed the only muscle group in the back active to counter the moment L5/S1 and the moment arm of the erector spinae muscle group is known, the muscle force necessary in the erector spinae muscle group to maintain static equilibrium can be estimated

If the moment arm of the erector spinae muscle group is 0.04 m ( from L5/S1) we can determine the muscle force required by :

F. d = 167.5 N.mF = 167.5 N.m = 4187 N

0. 04 mWhere : F = Muscle force required in erector spinae to

maintain static equilibriumd = Moment arm length of erector spinae muscle

group ( from L5/S1)


• If we wish to examine the compressive and shear forces acting on the disk L5/S1) we can use the calculations above. Since the trunk is bend 450 angle , the vertical force can be resolved into equal compressive and shear components. The vertical force, other than that exerted by the erector spinae muscle group, is the sum of the weights of the load (box) the arms and trunk, therefore :

Fv = Total vertical force acting upon L5/S1 disk

= WO + WH + WLA + WUA + WT

= 147 + 2(4.1) + 2 (11.7) + 2(19.2) + 308.7= 525.7 N


The vertical force due to weight of the box, arms and trunk is resolved into its compressive (Fvc) and shear (Fvs) components :

Fvc = 525.7 . Cos 450 = 371.7 N

Fvs = 525.7 . Sin 450 = 371.7 N

The total compressive ( Fc) dan shear ( Fs) forces are found as follows :

Fc = 371.7 N + 4187 N = 4558.7 N

Fs = 371.7 N


Materi # 8a - Biomekanika

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