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Materi # 8 - BIOMEKANIKA

Sritomo W.Soebroto, Arif Rahman dan Adhitya Sudiarno Lab. & Perancangan Sistem Kerja

Institut Teknologi Sepuluh Nopember Surabaya

Sritomo W. Soebroto, et.al.

Biomekanika & Manual Handling 1

Outline Pertemuan

Biomekanika : arti dan aplikasinya

Analisa Mekanika Tubuh

Perhitungan Dasar Biomekanika

Studi Kasus Biomekanika

Manual Handling & Back Safety

Sritomo W. Soebroto, et.al.

Biomekanika & Manual Handling 2

MENGANGKAT, MENDORONG, DENGAN POSISI SETEGAK

MUNGKIN UNTUK MEMINIMALKAN BEBAN INTERNAL

23/10/2013 Biomekanika 3

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 dari ilmu fisika (khususnya mekanika) dan teknik, dengan berdasar pada biologi dan juga pengetahuan lingkungan kerja.

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

Kinematika : pergerakan/ motion pada segmen segmen tubuh. Kinetik : efek dari gaya dan momen yang terjadi pada tubuh. Mekanika digunakan sebagai penyusun konsep, analisa, dan desain dalam sistem biologi makhluk 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 melakukan kerja dipengaruhi faktor manusia, sikap kerja dan jenis pekerjaan


Biomekanika Biomekanika :

Biostatik : bagian daari biomekanika umum yang hanya menganalisa bagian tubuh dalam keadaan diam maupun bergerak pada garis lurus dengan kecepatan seragam (uniform)

Biodinamik : berkaitan dengan gerakan-gerakan tubuh taanpa mempertimbangkan gaya yang terjadi (kinematik) dan gaya yang disebabkan gaya yang bekerja dalam tubuh ( kinetik)

Occupational biomekanika : Bagian dari mekanik terapan yang mempelajari interaaksi fisik antara pekerja dengan mesin, material, dan peralatan dengan tujuan untuk meminumkan keluhan pada sistem kerangka otott agar produktivitas kerja dapat meningkat (Chaffin & Anderson,1984)


occupational biomechanics


Occupational biomechanics adalah sub disiplin dalam kerangka besar biomekanika yang mempelajari hubungan antara pekerja dengan alat kerja, workstation, mesin, dan material untuk meningkatkan performa dengan meminimalisasi terjadinya cidera musculoskeletal.

Sehingga studi utama tentang Occupational biomechanics berkaitan dengan masalah musculoskeletal. Sistem musculoskeletal terdiri atas tulang, otot, dan jaringan penghubung (ligament, tendon, fascia, dan cartilage). Fungsi utama sistem tersebut adalah mendukung dan melindungi tubuh dan bagian bagian tubuh, menjaga postur tubuh, produce pergerakan tubuh, serta menghasilkan panas dan mempertahankan suhu tubuh.

why occupational biomechanics ???

Jeffress (1999) indicated that approximately 650,000 workers every year suffer serious injuries and illnesses caused by overexertion, repetition, and other types of physical stress. Such injuries cost U.S. business between $15 to $20 billion dollars a year in workman compensation. According to US Department of labor, back injuries accounted for nearly 20% of all injuries and illnesses in the work place. In the UK, similar numbers appear with 27% of all reported accidents involving manual 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

10/23/2013 Sritomo W.S & Dyah SD 10

Body Segments Group 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


Biomechanics of the back


LIFTING AND BACK

STRESS

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 (1) 30o

Angle of forehand from horizontal (2) 30o

Angle of upper arm from horizontal ( 3) 80o

Angle of trunk from horizontal ( 4) 45o

Body weight (mass) 70 kg


Calculation

Work = m.g.h.f m = 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

1 = Angle of the hand relative to horizontal 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 1 = 0


Fx = Fx w = 0

Fy = Fy w - Wo - WH = 0

Mw = Mw - (Wo +WH) . SL1 . Cos 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 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

2 = Angle of the lower arm relative to horizontal = 300

for this examples

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

2 = 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.. 2. SL2 . Cos 2 - Fyw . SL2 . Cos 2 - Fxw . SL2 . Sin 2 = 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 3 = Angle of the lower arm relative to horizontal = 80

0 for this examples

SL3 = Measured length from the elbow to shoulder SL3 = 0.30 m 3 = 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. 3. SL3 . Cos 3 - Fye . SL3 . Cos 3 - Fxe . SL3 . Sin 3 = 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 4 = Angle of the trunk relative to horizontal = 45

0 for this examples

SL4 = Measured length for L5/S1 to shoulder SL4 = 0.36 m 4 = Location of c.m. as a portion of SL from L5/S1: 4 = 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. 4. SL4 . Cos 4 - Fys . SL4 . Cos 4 - Fxe . SL4 . Sin 4 = 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.m F = 167.5 N.m = 4187 N 0. 04 m

Where : F = Muscle force required in erector spinae to maintain static equilibrium d = 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
