TUGAS 3 MAINTENANCE & ORGANISASI DEPARTEMEN PERAWATAN DALAM SUATU INDUSTRI

TUGAS 3

MAINTENANCE & ORGANISASI DEPARTEMEN PERAWATAN DALAM SUATU INDUSTRI

Nama : Eko Adi Prasetyanto

Kelas : 3ic09

NPM : 23414453

UNIVERSITAS GUNADARMA

FAKULTAS TEKNOLOGI INDUSTRI

JURUSAN TEKNIK MESIN

2017

1.1   Maintenance

Maintenance adalah sebuah aktifitas dimana bertujuan untuk memperbaiki serta melakukan perawatan pada sebuah mesin dan sebagainya. Contoh pada sebuah mobil yang kita miliki bisa melakukan fungsinya dengan benar dan maksimal kita semestinya harus tau cara mengoperasikannya, prinsip kerja dari mesin tersebut, karena jika pengoperasian tidak sesuai maka akan menimbulkan kerusakan pada alat tersebut. Setelah tau cara mengoperasikan dan prinsip kerjanya, barulah kita mengetahui bagaimana cara perawatannya, jenis perawatan seperti apa yang harus kita berikan dan kapan saja kita melakukan perawatan (Maintenance) pada mesin tersebut.

Pada dasarnya Maintenance atau Perawatan Mesin/Peralatan kerja memerlukan beberapa kegiatan seperti dibawah ini :
– Kegiatan Pemeriksaan/Pengecekan
– Kegiatan Meminyaki (Lubrication)
– Kegiatan Perbaikan/Reparasi pada kerusakan (Repairing)
– Kegiatan Penggantian Suku Cadang (Spare Part) 

1.2   Jenis Jenis Maintenance

1.       Breakdown Maintenance (Perawatan saat terjadi Kerusakan)

Breakdown Maintenance adalah perawatan yang dilakukan ketika sudah terjadi kerusakan pada mesin atau peralatan kerja sehingga Mesin tersebut tidak dapat beroperasi secara normal atau terhentinya operasional secara total dalam kondisi mendadak.

2.      Preventive Maintenance (Perawatan Pencegahan)

Preventive Maintenance atau kadang disebut juga Preventative Maintenance adalah jenis Maintenance yang dilakukan untuk mencegah terjadinya kerusakan pada mesin selama operasi berlangsung. Preventive dibagi menjadi 2 :

. Periodic Maintenance (Perawatan berkala)

. Predictive Maintenance

3.      Corrective Maintenance

Corrective Maintenance adalah Perawatan yang dilakukan dengan cara mengidentifikasi penyebab kerusakan dan kemudian memperbaikinya sehingga Mesin atau peralatan Produksi dapat beroperasi normal kembali.

1.3    ORGANISASI DEPARTEMEN PERAWATAN DALAM SUATU INDUSTRI

Secara garis besar pengertian manajemen pemeliharaan yaitu pengorganisasian operasi pemeliharaan untuk memberikan performansi mengenai peralatan produksi dan fasilitas industri. Dasar pemikiran yang sehat dan logis adalah suatu persyaratan terbaik dalam mengorganisasikan pemeliharaan. Pengorganisasian ini mencakup penerapan dari metode manajemen dan memerlukan perhatian yang sistematis.

Hal ini merupakan pekerjaan yang harus dipertimbangkan secara sungguhsungguh dalam mengatur perlengkapan. Dimana perlengkapan itu merupakan peralatan, material, tenaga kerja, biaya, teknik atau tata cara yang diterapkan serta waktu pelaksanaannya. Dengan mengetahui tujuan dan sistem manajemen yang diterapkan, maka akan dapat mengatasi masalah, megambil tindakan serta mengerti dengan jelas permasalahan yang sedang dihadapi. 

Pada Maintenance perorganisasian departemen perawatan haruslah dibuat agar terstruktur dengan baik dengan masing masing job yang sudah tertata. Contoh pada organisasi departemen di atas adalah bahwa dibawah Manager Pabrik ada suatu sub seperti (Perawatan Mekanik ; Perawatan Listrik ; Perawatan Sipil ) di kepalai oleh Kepala Departemen Perawatan, dimana bias dibilang pusat.

Perawatan Mekanik

Dalam perawatan mekanik juga masih di bagi menjadi beberapa tugas di antaranya :

a.       Pengawas perawatan pabrik yang dimana tugas dari perawatan pabrik adalah perawatan terhadap perlatan, pemeriksaan, serta service yang sudah termasuk pelumasan, serta bila ada alat baru mereka yang menginstalasi

b.      Pengawas bengkel perbaikan dimana tugasnya yaitu mengawasi setiap bengkel yang ada di pabrik tersebut

c.       Pengawas pembangkit tenaga dimana tugasnya yaitu melakukan pengawasan terhadap perlengkapan mekanik pada pembangkit

Perawatan Listrik

Dalam perawatan mekanik juga masih di bagi menjadi beberapa tugas di antaranya :

a.       Pengawas Telepon dimana tugasnya adalah mengawasi baik instalasi maupun perawatan pada telepon internal

b.      Pengawas pembangkit tenaga (Listrik) dimana tugasnya ialah mengawasi melakukan perawatan dan perlengkapan kelistrikan pada pembangkit tenaga

c.       Pengawas Peralatan Listrik dimana tugasnya ialah melakukan pengawasan pada penerangan, perbaikan mesin pabrik (bagian kelistrikan)

Perawatan civil

Dimana diantaranya masih terbagi menjadi beberapa bagian dalam membagi tugas diantaranya:

a.       Pengawas peralatan gedung yang tugasnya ialah melakukan pengawasan serta mempersiapkan peralatan gedung

b.      Pengawasa peralatan jalan tugasnya ialah melakukan pengawasan terhadap perawatan jalan di suatu gedung atau pabrik

c.       Pengawasan perawatan saluran air dan sanitari tugasnya ialah mengawasi bagaimana kendala dan perawatan saluran air dalam suatu pabrik agar tidak terjadi hal hal yang tidak di inginkan

http://maintenance-group.blogspot.co.id/2010/09/manajemen-pemeliharaan.html

http://ilmumanajemenindustri.com/jenis-maintenance-perawatan-mesin-peralatan-kerja/

Maintenance Costing Tugas bulan ke 2

Maintenance Costing / Biaya Pemeliharaan

1.                  MAINTENANCE LABOR COST ESTIMATION

The cost of labor is an important component of the maintenance cost. Labor costs are made up of payroll information that is usually obtained from labor distribution reports prepared by the accounting department. The information is required for four key areas:

1. Total number of hours worked annually on a per-employee basis

2. Hourly cost of employee benefits on a per-employee basis

3. Ratio of cost of annual benefits to yearly wages

4. Base pay rates per hour by labor grade

·         The cost per employee is expressed by :

C_em = LR(1 + BR)TAH……………………………………..Eq(1.1)

Where :

C_em = cost per employee,

LR = hourly labor rate,

BR = benefit ratio,

TAH = total number of annual hours.

·         The total labor cost is given by :

TLC = C_emN……………………………………………….Eq(1.2)

Where :

TLC = total labor cost,

N = number of employees

Example :

Assume that in a maintenance organization we have the following data:

TAH = 2000 h

 BR = 0.2

LR = $15 per hour

N = 40

Calculating the total labor cost associated with the maintenance activity by

inserting the above values into Eqs. (1.1) and (1.2), we get :

C_em = 15 × (1 + 0.2) × 2,000 = $36,000

And

TLC = 36,000 × 40 = $1,440,000

Thus, the total labor cost associated with the maintenance activity will be $1.44 million.

2.                  CORRECTIVE M AINTENANCE LABOR COST ESTIMATION

In this case corrective maintenance labor cost is estimated when the item/system mean time between failures (MTBF) and mean time to repair (MTTR) are known. Consequently, the annual labor cost of corrective maintenance is expressed by :

.........……..Eq(2.1)

Where :

CM_al = annual corrective maintenance labor cost,

LCH = corrective maintenance labor cost per hour,

SOH = annual scheduled operating hours.

 Example :

A system is scheduled to operate for 2000 hours per year. The system’s MTBF and MTTR are 400 h and 20 h, respectively. Calculate the annual labor cost of corrective maintenance if the maintenance labor cost is $20 per hour. By substituting the specified data into Eq. (2.1), we get :

The annual labor cost of corrective maintenance is $2000.

3.                  EQUIPMENT OWNERSHIP CYCLE MAINTENANCE COST  ESTIMATION

This section is concerned with estimating the maintenance cost of the entire ownership cycle of equipment, so that its present value can be added to acquisition cost to obtain equipment life cycle cost. The following two formulas are often used to find present value of a sum of money. They can equally be used to obtain present values of equipment ownership cycle maintenance costs :

FORMULA I

This formula is used to estimate the present value of a single amount of money after k periods and is expressed by :

       

....……….Eq(3.1)

Where :

PV = present value,

AM = single amount,

k = number of interest or conversion periods (normally taken as years),

i = interest rate per period.

FORMULA II

This formula is concerned with obtaining the present value of equal amounts of, say, maintenance costs occurring at the end of each of k conversion periods (usually years). The present value is given by :

...………Eq(3.2)

Where :

            MC = maintenance cost occurring at the end of each (interest) conversion period.

Example :

A maintenance department considers procuring an engineering system. Two manufacturers are bidding to provide the system and their corrective maintenance costrelated data are given in Table 3.1. Determine which of the two systems will be less costly with respect to present

values of corrective maintenance and by how much :

Tabel 3.1 Maintenance Cost-Related Data

Manufacturer A System

The annual expected cost, MC, of a corrective maintenance action is :

MCa = 1000 × 2.5 = $2500

The present value, PV_A, of the engineering system life cycle corrective maintenance

cost using Eq. (3.2) and the specified and calculated values is :

Manufacturer B System

The annual expected cost, MC, of a corrective maintenance action is :

MCb = 1400 × 2 = $2800

The present value, PV_B, of the engineering system life cycle corrective maintenance

cost using Eq. (3.2) and the specified and calculated values is :

Manufacturer A’s engineering system will be less costly with respect to present

values of corrective maintenance by $2,260.81.

4.                  ACTIVITY EXPECTED DURATION TIME ESTIMATION

 The PERT scheme calls for three estimates of activity duration time using the following

formula to calculate the final time:

……………Eq(4.1)

Where :

            Ta = activity expected duration time,

OT = optimistic or minimum time an activity will require for completion,

PT = pessimistic or maximum time an activity will require for completion,

MT = most likely time an activity will require for completion. This is the time

          used for CPM activities.

Example 3.1

Assume that we have the following time estimates to accomplish an activity:

OT = 55 days

PT = 80 days

MT = 60 days

Calculate the activity expected duration time. Substituting the given data into Eq. (4.1), we get :

The expected duration time for the activity is 62.5 days.

5.                  MAINTAINABILITY MEASURES AND FUNCTIONS

Various measures are used in maintainability analysis: for example, mean time to repair

(MTTR), mean preventive maintenance time, and mean maintenance downtime. Maintainability

functions are used to predict the probability that a repair, starting at time t= 0, will be completed in a time. Some maintainability measures and functions are presented below.

MEAN TIME TO R EPAIR

Mean time to repair (MTTR) is probably the most widely used maintainability measure.

It measures the elapsed time required to perform a given maintenance activity. MTTR

is expressed by :

…………..Eq(5.1)

Where :

            k  number of units or parts,

li  failure rate of unit/part, for i  1, 2, 3, k

CMT_i corrective maintenance/repair time required to repair unit/part i, for i  1, 2, 3.….,k

Usually, times to repair follow exponential, lognormal, and normal probability

distributions.

Example :

MAIN TIME BETWEEN FAILURE

Mean time between failure  (MTBF) is probably the most time between the occurring of failure in maintainability measure.

It measures the elapsed time required to perform a given maintenance activity.

EXAMPLE :

Data historical of machine in industry :

a.       Hour ( 0 to 200) = up-time

b.      Hour ( 200 to 280 ) = down-time

c.       Hour ( 280 to 420 ) = up-time

d.      Hour ( 420 to 490) = down-time

e.       Hour ( 490 to 590 ) = up-time

f.       Hour ( 590 to 650 ) = down-time

g.      Hour (650 to 780 ) = up-time

To calculate of :

·         the mean time to between failure (MTBF ) and mean downtime (MDT) ?

MTBF ( Main time between Failure ) = 

MDT ( Main down time )                   =    

Tugas 2 - Perawatan Mesin

Tanggal : 4 April 2017              

Materi : ( Maintenance Costing ) / Biaya Pemeliharaan                3IC09

NB : 1. Tugas dikirm ke Blog dan Student site + Lampirkan Materi Maintenance Costing nya !!

         2. Dikirim ke email matkul perawatan mesin : tekper2017@gmail.com

Kerjakanlah soal – soal maintenance costing berikut ini !!

1.      To Calculate the mean time to repair (MTTR ) ?

a.       Eight Subsytem : a, b, c, d, e, f, g, h form an electronic system.

b.      The constants failure rates of these subsystems are :

la = 0.002 , lb = 0.004 , lc = 0.006 , ld = 0.008 , le = 0.010 , lf = 0.012 , lg = 0.014 , lh = 0.016 failures per hour.

c.       The corresponding estimate corrective maintenance time are :

Ta = 2 hour, Tb =  3 hour, Tc = 4 hour, Td = 5 hour, Te = 6 hour, Tf = 7 hour, Tg = 8 hour, Th = 9 hour.

2.      Data historical of machine in industry :

h.      Hour ( 0 to 200) = up-time

i.        Hour ( 200 to 210 ) = down-time

j.        Hour ( 210 to 400 ) = up-time

k.      Hour ( 400 to 415) = down-time

l.        Hour ( 415 to 590 ) = up-time

m.    Hour ( 590 to 600 ) = down-time

n.      Hour (600 to 750 ) = up-time

To calculate of :

·         the mean time to between failure (MTBF ) and mean downtime (MDT) ?

·          the annual labor cost of corrective maintenance (CM_al) ?

Whereas : SOH = 3500 hour, MTTR = 30 h, LCH = $ 40.

3.      Maintenance Cost-Related Data :

Description	Manufacture A system	Manufacture B system	Manufacture C System
Expected Life	15 years	15 years	15 years
Expected cost of  a corrective maintenance action	$2000	$2500	$3500
Annual failure rate	3 failures per year	3.5 failures per year	2.5 failures per year
Annual interest rate	4 %	4 %	4 %

Determine which of the three system will be less costly with present value of corrective maintenance ( Find the value of (PV)?

NB : Using the Formula II.

4.      Assume that we have the following time estimates to accomplish an activity :

OT = 85 days

PT = 105 days

MT = 90 days

To Calculate the activity expected duration time  ( T_a )?

           

5.      Assume that in maintenance organization we have the following data :

TAH = 2500 hour

BR = 0.2

LR = $25

N  = 50

To Calculate the cost per employee (C_em)and The total labor cost ( TLC )?

           

My Answer To The Question Of Maintenance Costing:

Carilah MTTR ( MAIN TIME BETWEEN FAILURE )

1.      Diketahui ada 8 sub system yang membentuk elektronik system dengan tingkat konstanta kegagalan sub system ini :

la = 0.002 , lb = 0.004 , lc = 0.006 , ld = 0.008 , le = 0.010 , lf = 0.012 , lg = 0.014 , lh = 0.016 failures per hour.

Dengan estimasi waktunya:

Ta = 2 hour, Tb =  3 hour, Tc = 4 hour, Td = 5 hour, Te = 6 hour, Tf = 7 hour, Tg = 8 hour, Th = 9 hour

            JAWAB :

           

  

                       

      2.  Diketahui :

         Data historical of machine in industry :

a.      Hour ( 0 to 200) = up-time

b.      Hour ( 200 to 210 ) = down-time

c.      Hour ( 210 to 400 ) = up-time

d.        Hour ( 400 to 415) = down-time

e.       Hour ( 415 to 590 ) = up-time

f.        Hour ( 590 to 600 ) = down-time

g.      Hour (600 to 750 ) = up-time

Ditanya :

·         the mean time to between failure (MTBF ) and mean downtime (MDT) ?

·          the annual labor cost of corrective maintenance (CM_al) ?

Whereas : SOH = 3500 hour, MTTR = 30 h, LCH = $ 40.

Jawab :

          3.      Maintenance Cost-Related Data

            Ditanya : Menemukan nilai biaya perbandingan 1 perusahaan ( PV) :

            Jawab :

            Langkah pertama cari

                     

   

    4.      Diketahui :

OT = 85 days

PT = 105 days

MT = 90 days

Ditanya : Berapa lama membuat suatu produk ( T_a )?

Jawab :

5. Diketahui :

TAH = 2500 hour

BR = 0.2

LR = $25

N  = 50

Ditanya: To Calculate the cost per employee (C_em)and The total labor cost ( TLC ) atau gaji karyawan dan total biaya karyawan?

Jawab :

a.       Mencari gaji karyawan (personal) / C_em

                 =          25 (1 + 0.2) 2500

                                                                      =          $75,000

      

                                                                               =   $75,000 x 50

                                                                               =    $3,750,000

Thus, the total labor cost associated with the maintenance activity will be $3,750,000million.