MACHINE DESIGN: PAST BOARD EXAM PROBLEMS CHAPTER 10 - CLUTCHES

00:46:35
https://www.youtube.com/watch?v=pKlrCXJOk54

Résumé

TLDRIl video riguarda il design di macchine, focalizzandosi sugli ingranaggi, in particolare sulle frizioni. Spiega in dettaglio come le frizioni collegano o scollegano due alberi rotanti, discutendo i tipi di frizioni come quelle a disco o a cono e i metodi di calcolo come il metodo di pressione uniforme e quello di usura uniforme. Vengono esaminati problemi pratici e reali, utilizzando vari parametri come pressione, diametro e velocità per calcolare la capacità della frizione di trasmettere potenza. Alla fine, l'autore analizza diversi problemi, guidando passo passo attraverso calcoli di coppia e potenza.

A retenir

  • 🔧 La frizione collega o scollega alberi rotanti.
  • 🔍 Due tipi comuni: a disco e a cono.
  • 📏 Utilizza diametri per calcolare la coppia.
  • 🔄 La coppia è la forza moltiplicata per il raggio.
  • 📈 Calcola la capacità in base a potenza e velocità.
  • 🔗 Il metodo di pressione uniforme utilizza formule specifiche.
  • 🔒 Considera la pressione massima per la sicurezza.
  • 🧮 Il numero di superfici di attrito influisce sul risultato.
  • 🌟 Problemi reali possono utilizzare vari metodi di calcolo.
  • 💡 Ogni calcolo fornisce informazioni chiave per il design.

Chronologie

  • 00:00:00 - 00:05:00

    视频介绍了离合器及其在机械设计中的作用，强调了离合器的基本功能：连接和断开传动轴。主要讨论了盘式离合器和锥形离合器两种类型，其中盘式离合器的功能通过均匀压力法和均匀磨损法进行数值分析。

  • 00:05:00 - 00:10:00

    视频深入介绍了盘式离合器的扭矩计算，通过实际力、摩擦系数以及大直径和小直径的关系来计算。强调在考试中这些直径经常是已知的，并介绍了均匀压力法的计算公式。

  • 00:10:00 - 00:15:00

    盘式离合器传递的扭矩详细计算过程，展示了摩擦表面数量的计算，以及实际力的计算方法。这部分对考试中的公式记忆提供策略，并详细解释了相关力学公式。

  • 00:15:00 - 00:20:00

    视频继续介绍锥形离合器的特点和计算方法，包括通过直径之差计算的面宽。并且，探讨了使用均匀磨损法的情况，特别是如何在已知平均直径的情况下应用计算。

  • 00:20:00 - 00:25:00

    通过实例展示了典型的考试题，例子中涉及多片离合器的扭矩和功率计算。使用均匀压力法，不仅计算了扭矩，还进一步计算了功率。这部分为观众在真实考试中提供实战指南。

  • 00:25:00 - 00:30:00

    通过进一步的实例，解释了锥形离合器在均匀压力下的计算，包括面角、实际力及相关扭矩的计算步骤。这部分帮助观众理解如何在实际应用中操作布局和参数计算。

  • 00:30:00 - 00:35:00

    第三个例子展示了如何在已知功率和转速的情况下，通过均匀磨损法计算实际所需的力。这部分反复强调了不同方法间切换的策略，当给定平均直径时，应优先使用均匀磨损法。

  • 00:35:00 - 00:40:00

    通过不同行例，解答了关于盘式离合器的多个样题，解释了如何从给定条件推算离合器所需的具体参数，如实际力和功率。此外，这部分还帮助观众掌握了用于推算的实际计算技巧。

  • 00:40:00 - 00:46:35

    视频总结了离合器计算实例，通过逐步分析过去问题，帮助观众更好理解考试题目中的变式。鼓励观众运用学到的公式和方法，提升机械设计领域的实际技能。

Afficher plus

Carte mentale

Mind Map

Vidéo Q&R

  • Qual è la funzione principale degli ingranaggi?

    Gli ingranaggi collegano o scollegano due alberi rotanti.

  • Quali sono i tipi comuni di ingranaggi?

    Disco o a piatto e cono sono tipi comuni di ingranaggi.

  • Cosa fa la frizione a cono in un veicolo?

    Trasmette il 30% di tutta la potenza dall'albero motore all'albero condotto.

  • Come aiuta la frizione nel trasferimento di coppia?

    Clutch aiuta a gestire il trasferimento di coppia tra due alberi.

  • È importante sapere la pressione o la dimensione di un diametro specifico per calcolare la forza?

    Sì, è fondamentale per calcolare la capacità della frizione di gestire la potenza.

  • Che metodo si usa per calcolare ingranaggi con diametri medi dati?

    Utilizza il metodo di pressione uniforme per problemi con diametri medi dati.

  • Come si considera il numero di superfici di attrito?

    Viene moltiplicato il numero di superfici di attrito per calcolare la forza totale.

  • Di cosa hai bisogno per determinare la capacità della frizione di trasmettere potenza?

    Dalla velocità dell'albero e dalla coppia trasmessa.

Voir plus de résumés vidéo

Accédez instantanément à des résumés vidéo gratuits sur YouTube grâce à l'IA !
Sous-titres
en
Défilement automatique:
  • 00:00:02
    hi guys or welcome to drive again so in
  • 00:00:05
    this video we're going to uh discuss and
  • 00:00:07
    continue our machine design journey
  • 00:00:10
    and the topic for today is chapter 10
  • 00:00:12
    club chess and this is a pass board exam
  • 00:00:16
    problem so
  • 00:00:18
    illegal and let us start answering
  • 00:00:19
    password take some questions
  • 00:00:24
    and okay so
  • 00:00:31
    let us have a recap or review of
  • 00:00:33
    formulas for clutches
  • 00:00:35
    but first let us define the clutches
  • 00:00:37
    first so a clutch
  • 00:00:39
    is a mechanical device that engages
  • 00:00:42
    and engages power transmission
  • 00:00:45
    especially
  • 00:00:46
    from a drive shop
  • 00:00:49
    or driving shop to a driven shop in the
  • 00:00:51
    simplest application clutch just connect
  • 00:00:53
    and disconnect two rotating shops so the
  • 00:01:00
    definition of clutches or highlighted in
  • 00:01:02
    the words is the word
  • 00:01:04
    uh
  • 00:01:05
    connect and disconnect or engage yes
  • 00:01:08
    and this engages
  • 00:01:10
    so
  • 00:01:15
    so there are two types of clutches
  • 00:01:19
    didn't discuss during our machine design
  • 00:01:21
    days in college or support exam and that
  • 00:01:24
    was
  • 00:01:25
    disk or plate clutch
  • 00:01:27
    yan yanyukura
  • 00:01:30
    and of course the cone clutch so there
  • 00:01:32
    are many types of clutches
  • 00:01:34
    no
  • 00:01:35
    hydraulic electrical so in software
  • 00:01:41
    [Music]
  • 00:01:44
    these are the two useful types that
  • 00:01:45
    didn't discuss
  • 00:01:47
    okay so
  • 00:01:49
    this part
  • 00:01:52
    is
  • 00:02:03
    so you have your rpm here and so
  • 00:02:07
    new pair of clutches
  • 00:02:10
    okay so that was your clutch
  • 00:02:13
    that uh
  • 00:02:14
    engage and disengage power transmission
  • 00:02:17
    especially from a drive shaft
  • 00:02:19
    to a drip bench up so let us assume this
  • 00:02:21
    is the drive shaft
  • 00:02:24
    and this is the dream bench
  • 00:02:27
    this is the driving shop and this is the
  • 00:02:29
    driven shop so boston made the long shop
  • 00:02:32
    knocking on egg and disconnect none
  • 00:02:34
    clutches nato
  • 00:02:48
    my spring
  • 00:03:02
    so that was represented by alpha and
  • 00:03:04
    will be discussed later and champion
  • 00:03:09
    which is the so-called face width
  • 00:03:16
    is so that was represented by b so that
  • 00:03:20
    was space
  • 00:03:23
    with
  • 00:03:25
    okay
  • 00:03:27
    so that was the basic parts of the
  • 00:03:29
    clutch assembly
  • 00:03:33
    [Music]
  • 00:03:36
    connect and disconnect to rotating
  • 00:03:38
    shafts so these two types of clutches uh
  • 00:03:42
    [Music]
  • 00:03:44
    was numerically functioned by using two
  • 00:03:49
    methods of formula that was the uniform
  • 00:03:51
    pressure method
  • 00:03:56
    and the uniform where method so magita
  • 00:03:58
    magikita natanzama susun slide
  • 00:04:01
    okay
  • 00:04:03
    so this is
  • 00:04:04
    our first type of clutch which is the
  • 00:04:06
    disk
  • 00:04:07
    plate clutch
  • 00:04:09
    again you have the two rotating shafts
  • 00:04:11
    the driving shaft and the driven shaft
  • 00:04:55
    by clutch of course this is
  • 00:04:59
    rotating
  • 00:05:02
    and of course my torque
  • 00:05:04
    so torque has a formula of torque is
  • 00:05:06
    equal to f which is the actual force
  • 00:05:09
    multiplied by f which is the coefficient
  • 00:05:12
    of friction between contacting surfaces
  • 00:05:14
    and of course d and the small d was the
  • 00:05:18
    small diameter and the big diameter so
  • 00:05:22
    engineer hb san public small diameter
  • 00:05:27
    big diameter so some aboard exam problem
  • 00:05:30
    pass boards
  • 00:05:31
    yes
  • 00:05:32
    on the problem the small and big
  • 00:05:34
    diameter are already stated but for this
  • 00:05:36
    presentation i will show
  • 00:05:38
    that this
  • 00:05:40
    so report my line
  • 00:05:45
    so this is
  • 00:05:47
    the
  • 00:05:48
    large diameter
  • 00:05:49
    and of course this is the
  • 00:05:52
    small diameter so such a funny made the
  • 00:05:55
    drawing pair in the union shop mismo
  • 00:05:57
    it was the clutch
  • 00:06:01
    so that was the capital d
  • 00:06:03
    the big diameter and this the the inside
  • 00:06:06
    or the small diameter so atona magnum
  • 00:06:10
    space with which is atom
  • 00:06:13
    of course
  • 00:06:15
    our pace with
  • 00:06:16
    obviously
  • 00:06:18
    that was the difference of the two
  • 00:06:20
    diameter the
  • 00:06:22
    capital d and the small d or the
  • 00:06:25
    outside or the inside diameter the
  • 00:06:27
    difference of the outside inside
  • 00:06:28
    diameter divided by two
  • 00:06:31
    or
  • 00:06:32
    b plus d naught small plus b is equal to
  • 00:06:36
    d
  • 00:06:36
    so
  • 00:06:38
    tamana mayan d
  • 00:06:39
    and transpose
  • 00:06:41
    so okay so that was the torque
  • 00:06:42
    transmitted by clutch
  • 00:06:45
    torque is equal to
  • 00:06:47
    f
  • 00:06:48
    or actual force times the coefficient of
  • 00:06:50
    friction times one third times d cubed
  • 00:06:53
    minus d cubed normally
  • 00:06:56
    over d squared minus d squared of course
  • 00:06:58
    lagging now union capital d
  • 00:07:15
    so i'm pre friction radius i atom 130 q
  • 00:07:18
    minus d cube over t squared minus
  • 00:07:22
    d squared
  • 00:07:25
    and of course multiplied by nfs which is
  • 00:07:28
    the number of friction surfaces
  • 00:07:30
    so marine time formula and a number of
  • 00:07:32
    friction surfaces is equal to number of
  • 00:07:34
    plates
  • 00:07:35
    minus one case
  • 00:07:41
    emergency
  • 00:07:44
    [Music]
  • 00:07:53
    number of friction surfaces is equal to
  • 00:07:55
    number of the plates minus one common
  • 00:07:57
    number of plates i decided
  • 00:08:00
    a number of friction surpluses i this is
  • 00:08:03
    cassette by pair and villa again has a
  • 00:08:05
    picture that was only a one pair
  • 00:08:09
    of
  • 00:08:13
    that was that will become three and
  • 00:08:15
    shampre
  • 00:08:16
    pairs
  • 00:08:18
    nfs is equal to np minus one so the
  • 00:08:21
    actual force temperature lagging actual
  • 00:08:23
    force needed to engage the clutch
  • 00:08:26
    represented by capital f f is equal to p
  • 00:08:29
    what is that p small p that was the
  • 00:08:31
    pressure developed at the surface of the
  • 00:08:33
    disc or plate times pi over for d
  • 00:08:35
    squared minus d squared so in order for
  • 00:08:38
    us to memorize this easily this formula
  • 00:08:42
    just
  • 00:08:43
    remember the
  • 00:08:46
    pressure is equal to force over area and
  • 00:08:48
    just assume that this is a hollow
  • 00:08:50
    circular
  • 00:08:52
    tube so that will become f is equal to
  • 00:08:54
    pi over four
  • 00:08:56
    d squared minus d squared alumni area
  • 00:08:58
    now hollow is five now hollow circular
  • 00:09:01
    troop is pi over four times d squared
  • 00:09:03
    minus d squared so
  • 00:09:05
    cross multiply that will become f is
  • 00:09:07
    equal to p
  • 00:09:08
    times pi
  • 00:09:09
    over four d squared minus this way so
  • 00:09:14
    uniform pressure methods
  • 00:09:22
    the numerator cubing depends in the
  • 00:09:24
    denominator square modeling difference
  • 00:09:26
    times one third that was your friction
  • 00:09:28
    radius and of course aluminum material
  • 00:09:30
    torque is equal to force
  • 00:09:32
    times radius so just multiply it by
  • 00:09:36
    this f or coefficient friction and this
  • 00:09:39
    r is the friction radius so fabulous
  • 00:09:43
    nasa bracket and of course multiply by
  • 00:09:46
    friction straight basis
  • 00:09:49
    okay so we have also the space weed here
  • 00:09:52
    no no discus
  • 00:09:54
    that is the difference of the diameters
  • 00:09:57
    divided by two
  • 00:10:00
    of course using uniform where method
  • 00:10:03
    so torque is equal to
  • 00:10:05
    actual force times coefficient of
  • 00:10:07
    friction times
  • 00:10:08
    d
  • 00:10:10
    uh sum of the two diameters divided by
  • 00:10:12
    four times and so the linked and then
  • 00:10:14
    uniform where
  • 00:10:16
    method but guys this is the trick here
  • 00:10:20
    when i will use the uniform pressure at
  • 00:10:22
    the uniform where method kappa lamps
  • 00:10:26
    method
  • 00:10:27
    and use the upn perro kapagnaki takayo
  • 00:10:33
    diameter or average diameters of
  • 00:10:35
    problematic use the uniform where method
  • 00:10:38
    y because we can
  • 00:10:41
    rewrite this formula by
  • 00:10:44
    actual force times coefficient of
  • 00:10:46
    friction
  • 00:10:48
    times one half times the average of
  • 00:10:52
    the two diameters
  • 00:10:55
    as one half times one half is one fourth
  • 00:10:57
    times d plus d
  • 00:11:04
    [Music]
  • 00:11:27
    plus d over 2 because d plus d
  • 00:11:29
    over 2 is the mean diameter or average
  • 00:11:32
    diameter so that was the torque
  • 00:11:34
    transmitted by clutch
  • 00:11:36
    and simply this is your friction reduce
  • 00:11:40
    friction reduce
  • 00:11:43
    and of course my torque my actual force
  • 00:11:46
    density uniform wear method
  • 00:11:48
    that was f
  • 00:11:49
    is equal to pi
  • 00:11:52
    times p max
  • 00:11:54
    or maximum pressure developed
  • 00:11:58
    at the surface of the disk or plate
  • 00:11:59
    times d over two times the difference of
  • 00:12:02
    the two diameters or paramasma
  • 00:12:07
    tanda and that was pi times p or
  • 00:12:10
    pressure times the small d or
  • 00:12:12
    multiplied by the
  • 00:12:15
    half of the
  • 00:12:17
    difference of the two diameter so anita
  • 00:12:20
    mandeling tandem
  • 00:12:25
    f is equal to pi times maximum pressure
  • 00:12:28
    times d over two times d minus d
  • 00:12:31
    okay
  • 00:12:33
    of course the average phone clutch
  • 00:12:35
    and as you can see guys
  • 00:12:46
    as i said earlier on the first slide
  • 00:12:50
    with a guy but i'm going to
  • 00:13:05
    d squared minus d squared
  • 00:13:08
    among eddy divide mean torque by sine
  • 00:13:10
    alpha so none jump ring force
  • 00:13:13
    actual force in coefficient of friction
  • 00:13:14
    at the number of friction surfaces
  • 00:13:18
    okay
  • 00:13:20
    and chambra's actual force
  • 00:13:23
    divided
  • 00:13:27
    and of course p pressure coefficient of
  • 00:13:30
    friction in f and this is alpha is our
  • 00:13:34
    cone angle means
  • 00:13:37
    for computing but
  • 00:13:39
    in the long diameters no problem oh man
  • 00:13:42
    and of course b is equal to face with so
  • 00:13:45
    d minus the over too long
  • 00:14:01
    to d minus d over 2 sine alpha
  • 00:14:04
    now using
  • 00:14:05
    uh
  • 00:14:06
    trigonometric identities
  • 00:14:10
    clutch
  • 00:14:11
    phase weight is equal to
  • 00:14:13
    the half of the deepers of the diameter
  • 00:14:15
    so just divide it by sine alpha
  • 00:14:18
    and chambra is a uniform where method
  • 00:14:20
    same parents
  • 00:14:23
    disc clutch
  • 00:14:27
    present
  • 00:14:28
    f actual force coefficient of friction
  • 00:14:31
    of course the two diameters and the
  • 00:14:32
    number of friction story places and
  • 00:14:35
    chambray eggshell force nothing
  • 00:14:38
    needed to engage engage the clutch
  • 00:14:42
    uniform where method than this clutch
  • 00:14:44
    so no need to divide it by sine alpha
  • 00:14:46
    it's not divided
  • 00:14:48
    and again shampoo and genuine pressure
  • 00:14:50
    develop the surface of the distribute
  • 00:14:52
    volume e max f is the coefficient of
  • 00:14:55
    friction and of course we have the cone
  • 00:14:57
    angle and the face width
  • 00:15:04
    so number of frictions are processed so
  • 00:15:05
    for a multiple disc clutch as i said
  • 00:15:07
    earlier multiply the equation of torque
  • 00:15:09
    by the number of pairs of friction
  • 00:15:11
    surfaces nfs so nfs is equal to np minus
  • 00:15:14
    1 and this is the number
  • 00:15:17
    of plates
  • 00:15:23
    description
  • 00:15:24
    and of course the friction power is
  • 00:15:26
    equal to
  • 00:15:27
    t times n over k so
  • 00:15:30
    torque times angular speed
  • 00:15:33
    for sharp speed divided by the constant
  • 00:15:35
    so alumni
  • 00:15:41
    first
  • 00:15:42
    question so a local utility
  • 00:15:46
    vehicle production company have
  • 00:15:47
    considered to use in one of its designs
  • 00:15:50
    a multi-disc
  • 00:15:52
    clutch consisting of a nine steel disc
  • 00:15:55
    and eight bronze discs with effective
  • 00:15:58
    outside the inside diameter of 220 and
  • 00:16:00
    160 millimeters respectively
  • 00:16:03
    if the shaft turns at 1 500 rpm and an
  • 00:16:07
    actual force of 600 newton is applied
  • 00:16:10
    find the horsepower capacity considering
  • 00:16:13
    a coefficient of friction of 0.25 so
  • 00:16:17
    meeting method
  • 00:16:25
    uniform pressure method but the question
  • 00:16:28
    is looking for the fine
  • 00:16:30
    the horsepower capacity
  • 00:16:33
    of course power nanometer
  • 00:16:42
    so we have a formula of torque for
  • 00:16:43
    unicorn pressure method of
  • 00:16:46
    uh actual force times coefficient of
  • 00:16:48
    friction times the
  • 00:16:50
    friction radius so you say nothing to my
  • 00:16:53
    even
  • 00:16:54
    actual force
  • 00:16:56
    so marine time actual force
  • 00:17:00
    actual force of 600 newtons
  • 00:17:17
    coefficient of friction is
  • 00:17:19
    0.25
  • 00:17:22
    of course the
  • 00:17:24
    the outside diameter
  • 00:17:26
    and 220
  • 00:17:36
    so how about our nfs
  • 00:17:40
    so on nfs not then we have a total of
  • 00:17:43
    nine steel disk and eight brands this
  • 00:17:46
    meaning we have a total of
  • 00:17:50
    uh rather
  • 00:17:53
    we have a total of
  • 00:17:55
    number of plates not an i nine plus
  • 00:17:57
    eight which is 17 minus 1 we have
  • 00:18:01
    16 pairs of
  • 00:18:04
    number of friction surfaces
  • 00:18:11
    of course
  • 00:18:18
    okay so everything is given here
  • 00:18:22
    of course in order for us to get the
  • 00:18:23
    power
  • 00:18:26
    so
  • 00:18:27
    powers equal to torque times and
  • 00:18:28
    multiplied by the constant
  • 00:18:31
    so
  • 00:18:33
    during our ninth nine chapters in the
  • 00:18:37
    nine chapters
  • 00:18:42
    so let us not review this anymore
  • 00:18:45
    of course uh cal q torque is equal to
  • 00:18:49
    600 newton detonal unit analysis guys
  • 00:18:53
    times
  • 00:18:54
    0.25 times
  • 00:18:57
    1 3
  • 00:19:00
    2
  • 00:19:01
    20 cube
  • 00:19:02
    minus 160 cube
  • 00:19:07
    divided by
  • 00:19:09
    220 squared minus 160 squared
  • 00:19:13
    times 16
  • 00:19:15
    so that will be
  • 00:19:16
    so guys i'm gigging formula unit is m
  • 00:19:20
    i say m
  • 00:19:22
    over m squared is equal to m
  • 00:19:25
    and of course that was the friction
  • 00:19:26
    radius
  • 00:19:28
    m m to the first degree long
  • 00:19:30
    unit so
  • 00:19:31
    i'll keep a 600 times 0.25
  • 00:19:34
    times 220 cubed minus 2 1 160 cubed
  • 00:19:40
    divided by 2 times square minus 160
  • 00:19:42
    squared divided 3
  • 00:19:45
    times 16 so that was
  • 00:19:49
    229 thousand
  • 00:19:52
    894.74
  • 00:19:55
    newton mn
  • 00:19:57
    so in order for us to get the power
  • 00:20:00
    our torque should be in newton mm which
  • 00:20:02
    is correct
  • 00:20:04
    our n should be in rpm and constant up
  • 00:20:09
    9.549 times 10 raised to 6 in order for
  • 00:20:13
    us to get a power up
  • 00:20:15
    kilowatts and just
  • 00:20:17
    convert it later paramugging horsepower
  • 00:20:19
    because horsepower you know not at nasa
  • 00:20:21
    choices so maritime torque in newton mm
  • 00:20:26
    speed
  • 00:20:27
    which is
  • 00:20:29
    the shaft turns down at 0.5 rpm so that
  • 00:20:33
    was one
  • 00:20:34
    thousand five hundred rpm so the power
  • 00:20:37
    is that we're gonna small i capital p
  • 00:20:40
    so that was two
  • 00:20:41
    two nine eight nine four point seventy
  • 00:20:44
    four
  • 00:20:45
    newton mm times one thousand
  • 00:20:48
    five hundred all over
  • 00:20:51
    nine point five four nine times ten
  • 00:20:53
    raised to six
  • 00:20:55
    and the power
  • 00:20:57
    is a
  • 00:20:59
    torque
  • 00:21:05
    so that was uh
  • 00:21:07
    36
  • 00:21:10
    point 11
  • 00:21:12
    kilowatts
  • 00:21:19
    and of course we have the power in
  • 00:21:21
    kilowatts just convert it on
  • 00:21:24
    hp which is in every hp we have 0.746
  • 00:21:28
    kilowatts one horsepower
  • 00:21:30
    the power is equal to
  • 00:21:35
    and
  • 00:21:35
    48.41
  • 00:21:37
    horsepower
  • 00:21:41
    so the answer is
  • 00:22:16
    which is the answer is 8.41 horsepower
  • 00:22:19
    or
  • 00:22:20
    letter a
  • 00:22:24
    okay so that was the first sample
  • 00:22:26
    problem digitizer number two
  • 00:22:29
    determine the power capacity of cone
  • 00:22:32
    clutch
  • 00:22:33
    under uniform pressures
  • 00:22:36
    under uniform pressure pedophone clutch
  • 00:22:38
    pixel
  • 00:22:44
    with the following operating data major
  • 00:22:46
    diameter of 250 mm minor diameter of 200
  • 00:22:50
    mm length of conical elements in contact
  • 00:22:53
    functional being length of electronical
  • 00:22:55
    elements in contact that was the phase
  • 00:22:57
    width
  • 00:22:58
    rotational speed 870 rpm coefficient of
  • 00:23:01
    friction 2.3 and allowable pressure of
  • 00:23:04
    70 kilo pascal so all most of the
  • 00:23:06
    details
  • 00:23:13
    the question again is looking for the
  • 00:23:15
    power capacity so the power capacity
  • 00:23:18
    again is equal to torque multiplied by
  • 00:23:20
    the speed divided
  • 00:23:38
    [Music]
  • 00:23:51
    capital b which is the 250 our minor
  • 00:23:53
    diameter is 200
  • 00:23:56
    length of conical elements is
  • 00:23:59
    125 mm
  • 00:24:02
    coefficient friction data point 3 and
  • 00:24:04
    allowable pressure of 70 kilo pascal so
  • 00:24:06
    we are looking for this actual force and
  • 00:24:09
    this angle because
  • 00:24:15
    face with which is
  • 00:24:32
    elements or the phase width is
  • 00:24:35
    d minus d so that was 250
  • 00:24:38
    minus 200 divided by
  • 00:24:41
    2 sine alpha
  • 00:24:43
    so our b is 125.
  • 00:24:48
    so direct submulans are calculatorship
  • 00:24:50
    so
  • 00:24:52
    and we're going to have iconic our same
  • 00:24:54
    angle up 250 minus 200 by 2 dividing 125
  • 00:25:01
    and you have an 11 point
  • 00:25:04
    54 degrees of angle
  • 00:25:07
    so check make natal and currently we're
  • 00:25:09
    looking only now for the actual force
  • 00:25:12
    which has this formula f is equal to
  • 00:25:16
    uh
  • 00:25:17
    pressure times pi over 4 times d squared
  • 00:25:19
    minus d squared or
  • 00:25:22
    just
  • 00:25:22
    pressure times an area of alloy circular
  • 00:25:33
    so again actual force is
  • 00:25:37
    look guys the unit even here 70
  • 00:25:40
    kilopascals and we all know that
  • 00:25:42
    millimeter young given
  • 00:25:45
    some diameters and we all know that
  • 00:25:46
    kilopascal is kilo newton per meter
  • 00:25:49
    squared so from the atomic converter
  • 00:25:51
    into meter that's converting kilonewtons
  • 00:25:53
    a newton
  • 00:25:54
    kilopascal
  • 00:26:00
    we all know that
  • 00:26:02
    one thousand
  • 00:26:04
    kilopascal is equal to one
  • 00:26:07
    mega
  • 00:26:08
    pascal
  • 00:26:11
    divide 1000 so 70 divided 1000 that was
  • 00:26:14
    0.07
  • 00:26:17
    mega pascals
  • 00:26:20
    that was
  • 00:26:21
    0.07 mega pascals or newton times mm
  • 00:26:25
    square newton per mm square times pi
  • 00:26:27
    over 4
  • 00:26:29
    times 250 squared
  • 00:26:32
    minus 200 squared so that was mm squared
  • 00:26:36
    so one cell cancel and we have a newton
  • 00:26:39
    here
  • 00:26:40
    so that was times 0.25 pi times 250
  • 00:26:43
    squared minus 200 squared and the answer
  • 00:26:46
    is one two three seven
  • 00:26:50
    newton so check maintenance f
  • 00:26:52
    and we're going to
  • 00:26:54
    substitute it directly so that was one
  • 00:26:57
    two three seven newtons
  • 00:27:00
    times
  • 00:27:03
    point
  • 00:27:04
    three
  • 00:27:06
    divided by sine alpha
  • 00:27:08
    sine
  • 00:27:09
    11.54 degrees
  • 00:27:13
    times one-third
  • 00:27:16
    or times the friction rate u so that was
  • 00:27:18
    250 cubed
  • 00:27:20
    minus 200 cubed
  • 00:27:23
    divided by 250 squared
  • 00:27:27
    minus 200 squared
  • 00:27:31
    so sanji media mean amboses
  • 00:27:34
    so one two three seven times point three
  • 00:27:38
    divided by sine 11.54
  • 00:27:41
    times 250 cubed minus 200 cubed times
  • 00:27:44
    250 squared minus 200 squared times one
  • 00:27:48
    third
  • 00:27:49
    so that was 209
  • 00:27:53
    1548
  • 00:27:56
    65 newton this is m gaianina
  • 00:28:01
    newton
  • 00:28:02
    mm so far back
  • 00:28:04
    for the power cause
  • 00:28:06
    in order for us to have a power in
  • 00:28:07
    kilowatts
  • 00:28:08
    choices kilowatts
  • 00:28:10
    we need to have a torque of newton mm
  • 00:28:13
    and rpm
  • 00:28:15
    of speed
  • 00:28:17
    so
  • 00:28:18
    according to power so power is 209
  • 00:28:22
    548.65
  • 00:28:25
    so a long number of friction surprises
  • 00:28:34
    rotational speed of 870 rpm
  • 00:28:39
    so this is rpm newton mm
  • 00:28:42
    and of course the constant is nine point
  • 00:28:45
    five point nine times ten raised to six
  • 00:28:47
    in order for us to have a power in
  • 00:28:48
    kilowatts
  • 00:28:50
    so times eight hundred and seventy
  • 00:28:52
    divided by nine point five four nine
  • 00:28:55
    divided ten raised to six so the answer
  • 00:28:57
    is 19.1
  • 00:29:00
    kilowatts so the answer is closest to
  • 00:29:05
    guys
  • 00:29:10
    is
  • 00:29:24
    19.2 kilowatts
  • 00:29:27
    so the answer is
  • 00:29:29
    okay so we have already two items so how
  • 00:29:32
    about this one third item
  • 00:29:35
    account clutch is to transmit 30
  • 00:29:37
    kilowatts at one 250 rpm the phase angle
  • 00:29:40
    of a cone clutch is 12.5 degrees
  • 00:29:42
    and the mean diameter is to be 355 mm
  • 00:29:47
    the coefficient of prediction between
  • 00:29:48
    the contacting surface is 0.2 and the
  • 00:29:50
    maximum normal pressure is 830
  • 00:29:53
    83 kilopascal rather determine the
  • 00:29:55
    actual force in kilonewton required to
  • 00:29:57
    transmit the power so
  • 00:29:59
    in an
  • 00:30:00
    word which is
  • 00:30:02
    the word mean diameter
  • 00:30:04
    meaning matic use the uniform where
  • 00:30:08
    method okay so the formula
  • 00:30:12
    for the torque
  • 00:30:16
    is determine the actual force
  • 00:30:18
    of course
  • 00:30:34
    a we are looking for the actual force
  • 00:30:39
    no
  • 00:30:40
    and the pad given in torque in order for
  • 00:30:41
    us to get this actual force so
  • 00:30:45
    nothing
  • 00:30:45
    on torque so because the first sentence
  • 00:30:48
    account clutch is to transmit
  • 00:30:51
    30 kilowatts so we have this formula
  • 00:30:54
    so
  • 00:30:56
    30 kilowatts is equal to torque the
  • 00:30:59
    newton m
  • 00:31:01
    and n should be in rpm which is on sub
  • 00:31:03
    b1 250rpm
  • 00:31:08
    and of course a constant of
  • 00:31:11
    9.549 times 10 raised to 6 in order for
  • 00:31:14
    us to get the torque and of course the
  • 00:31:16
    answer is 30 times 9.549 times 10 raised
  • 00:31:20
    to 6 divided by 1 to 50. so that was 229
  • 00:31:26
    1776
  • 00:31:29
    newton mm submarine italian
  • 00:31:44
    newton
  • 00:31:47
    two two nine one seven six
  • 00:31:50
    newton mm is equal to f for actual force
  • 00:31:54
    multiplied by the coefficient friction
  • 00:31:56
    and subject and energy
  • 00:31:58
    the coefficient of prediction between
  • 00:32:00
    the contact length so every place is
  • 00:32:01
    point two
  • 00:32:04
    and assign
  • 00:32:06
    face angle up
  • 00:32:09
    the face angular point clutch is twelve
  • 00:32:10
    point five degrees
  • 00:32:15
    and of course
  • 00:32:17
    i'm given is a mean diameter
  • 00:32:20
    355 mm if it's been probably nothing but
  • 00:32:23
    evaluating d plus the over four
  • 00:32:26
    into one half times
  • 00:32:28
    d plus d
  • 00:32:29
    over two or
  • 00:32:31
    one half times mean diameter
  • 00:32:34
    so that was one half times mean diameter
  • 00:32:37
    so that was 355 mm
  • 00:32:40
    so newton mm 2 mm answer mm expect that
  • 00:32:43
    f in newton
  • 00:32:45
    so that was
  • 00:32:46
    229176 times sine of 12.5 degrees
  • 00:32:50
    times 2 divided 0.2 divided 355 mm the
  • 00:32:54
    answer is
  • 00:32:55
    one point
  • 00:32:58
    uh
  • 00:33:01
    the answer is one three nine seven
  • 00:33:05
    one three nine seven point
  • 00:33:08
    twenty
  • 00:33:08
    six newtons
  • 00:33:11
    okay wait let me go on
  • 00:33:16
    just to be sure
  • 00:33:27
    okay one three seven okay one three the
  • 00:33:30
    answer is one three nine seven point
  • 00:33:31
    twenty six and of course
  • 00:33:34
    one thousand newton is equal to one kilo
  • 00:33:36
    newton the answer should be in kilo
  • 00:33:39
    newton
  • 00:33:42
    so just divide it by one thousand that
  • 00:33:44
    was
  • 00:33:45
    1.397
  • 00:33:48
    kilo newton
  • 00:33:50
    and of course that was closest to 1.4
  • 00:33:53
    kilo newton so the answer is n
  • 00:33:55
    the answer is letter a
  • 00:33:59
    g
  • 00:34:00
    that's which letter a so that long
  • 00:34:02
    letter a no no
  • 00:34:04
    no no no
  • 00:34:32
    okay
  • 00:34:34
    so again last item per tree on questions
  • 00:34:37
    or that long question element
  • 00:34:40
    so this is a password exam problem
  • 00:34:43
    uh pajero chip wagon's engine develop 40
  • 00:34:46
    kilowatts at one to rpm with a single
  • 00:34:49
    plate clutch having two pairs of
  • 00:34:51
    friction surfaces transmitting the power
  • 00:34:54
    consider the coefficient of friction to
  • 00:34:56
    be 0.3 and the mean diameter of the
  • 00:34:59
    discs to be 200 mm first question of
  • 00:35:02
    this number for item determine the
  • 00:35:04
    actual force required to engage the
  • 00:35:06
    clutch
  • 00:35:07
    and to
  • 00:35:08
    transmit the power
  • 00:35:11
    okay so
  • 00:35:13
    you mean diameter
  • 00:35:15
    use
  • 00:35:16
    immediately
  • 00:35:18
    the uniform wear
  • 00:35:20
    method
  • 00:35:21
    to determine the actual force required
  • 00:35:23
    to engage the clutch and to transmit the
  • 00:35:26
    power
  • 00:35:29
    okay so how we're going to do this one
  • 00:35:32
    so maritime formula torque view ewm
  • 00:35:36
    and this is not as this is not a cone
  • 00:35:38
    clutch this is uh
  • 00:35:40
    uh
  • 00:35:42
    you call that a disc or plate clutch
  • 00:35:44
    because i'm giving us angle
  • 00:35:46
    so this is a flat
  • 00:35:49
    now so our torque
  • 00:35:51
    the economy formula and torque is uh
  • 00:35:55
    actual force times coefficient of
  • 00:35:57
    friction times
  • 00:35:59
    one port over i one four times d plus d
  • 00:36:02
    or kappa given mean diameter that was f
  • 00:36:05
    times coefficient of friction times
  • 00:36:08
    d one half
  • 00:36:10
    times the
  • 00:36:13
    mean
  • 00:36:14
    diameter
  • 00:36:15
    okay so in explaining the sungalenian
  • 00:36:20
    so in order for us to get this actual
  • 00:36:22
    force we should get
  • 00:36:23
    the coefficient friction and this
  • 00:36:25
    outside inside diameter and of course
  • 00:36:27
    the torque but the thing is
  • 00:36:31
    smaller inside casing
  • 00:36:34
    diameters
  • 00:36:37
    of course
  • 00:36:40
    consider the efficient friction to be
  • 00:36:42
    0.3
  • 00:36:43
    so how are we going to get this torque
  • 00:36:45
    so maritime 40 kilowatts at one to rpm
  • 00:36:48
    and we all know that
  • 00:36:50
    power is equal to torque
  • 00:36:52
    times speed over k so nothing a
  • 00:36:55
    rectangle
  • 00:36:56
    that
  • 00:36:58
    signals
  • 00:36:59
    torque is equal to
  • 00:37:01
    power times constant divided by rpm so
  • 00:37:04
    that was 40
  • 00:37:06
    kilowatts and constant nothing is
  • 00:37:09
    9.54910
  • 00:37:10
    raised to 6 and of course our speed here
  • 00:37:13
    should be in rpm per pack and given is
  • 00:37:16
    1200 rpm so
  • 00:37:21
    uh let us expect the power
  • 00:37:24
    oh no no no the torque
  • 00:37:27
    let us expect that torque is in
  • 00:37:30
    newton mm so that was
  • 00:37:33
    40 kilowatts times 9.549 times trains 10
  • 00:37:37
    raised to 6.
  • 00:37:39
    divided by 1 2
  • 00:37:41
    so that was 318 thousand
  • 00:37:45
    300 newton mm
  • 00:37:50
    actual force times 0.3
  • 00:37:53
    times
  • 00:37:54
    mean diameter of 200
  • 00:37:57
    of course d by 2 at the end
  • 00:37:59
    so our actual force should be in
  • 00:38:03
    newton so dividing the nations of 1000
  • 00:38:06
    so that was 318 300 times 2
  • 00:38:10
    divided by 0.3 divided by 200
  • 00:38:14
    comma 1 40 plus e
  • 00:38:29
    [Music]
  • 00:38:31
    of the disk is to be
  • 00:38:34
    200 mm
  • 00:38:38
    some of the guys with that
  • 00:38:44
    one up hundred
  • 00:38:48
    four is equal to ah okay
  • 00:38:51
    so i forget the number of prediction
  • 00:38:53
    surpasses
  • 00:38:55
    sorry guys
  • 00:38:57
    so anxiety the number of friction
  • 00:38:59
    surfaces is equal to
  • 00:39:03
    uh with a single plate clutch having two
  • 00:39:05
    pairs
  • 00:39:07
    of friction
  • 00:39:08
    surfaces
  • 00:39:11
    somatic indeed minus one because
  • 00:39:17
    not two pairs of clutch but two pairs of
  • 00:39:19
    friction surprises meaning
  • 00:39:25
    a number of frequencies should be three
  • 00:39:27
    minus one can again two percent
  • 00:39:33
    nfs
  • 00:39:37
    nfs
  • 00:39:39
    and seduna
  • 00:39:42
    okay so
  • 00:39:43
    three hundred eighteen thousand three
  • 00:39:44
    hundred
  • 00:39:45
    times two divided point three divided by
  • 00:39:48
    two hundred and divide two so that was
  • 00:39:50
    five three or five newtons
  • 00:39:54
    or divide mulan
  • 00:39:56
    1000 that was 5.305
  • 00:40:00
    kilo newton and the closest answer is of
  • 00:40:03
    course
  • 00:40:04
    obviously letter a again
  • 00:40:08
    okay the hotel is the second item
  • 00:40:10
    determine the inside diameter of the
  • 00:40:11
    disk if the lining pressure is limited
  • 00:40:14
    to 200 kilo pascal given a 200 kilo
  • 00:40:18
    pascal um lining pressure and you know
  • 00:40:20
    nothing in the converter
  • 00:40:23
    and we all know that one thousand kilo
  • 00:40:25
    pascal is equal to one mega pascal just
  • 00:40:27
    divided by one thousand paramaging mega
  • 00:40:30
    pascal
  • 00:40:35
    time
  • 00:40:43
    because it is also equivalent to 0.2
  • 00:40:45
    newton
  • 00:40:46
    per square millimeter
  • 00:40:49
    okay so that i mean the inside diameter
  • 00:40:50
    of the discs
  • 00:40:53
    lung so we may get the inside diameter
  • 00:40:56
    and the outside diameter again
  • 00:41:00
    so it's
  • 00:41:01
    the equation
  • 00:41:03
    i think
  • 00:41:06
    so determine the inside here and
  • 00:41:08
    determine the outside here
  • 00:41:14
    so
  • 00:41:16
    so i answered nothing
  • 00:41:25
    newton and we all know that formula for
  • 00:41:28
    the uniform where method for the actual
  • 00:41:30
    force is equal to
  • 00:41:32
    uh maximum pressure
  • 00:41:36
    times uh
  • 00:41:38
    pi over two
  • 00:41:39
    times small d times the difference of
  • 00:41:42
    two diameters
  • 00:41:45
    and chambray marine time five thousand
  • 00:41:47
    three hundred 305 newtons is equal to
  • 00:41:50
    0.2 mega pascal
  • 00:41:52
    or newton per mm squared times pi over 2
  • 00:41:56
    times small d
  • 00:41:58
    times the difference up
  • 00:42:00
    small diameter so this is
  • 00:42:03
    an uh unit of power one and unit of
  • 00:42:06
    power one so this tan them is the unit
  • 00:42:09
    of power two so that will become mm
  • 00:42:11
    squared so canceling mm square against
  • 00:42:13
    the newton mixer being time in
  • 00:42:16
    expect that the answer should be in
  • 00:42:17
    millimeters
  • 00:42:19
    of choices
  • 00:42:20
    so marina time second uh first equation
  • 00:42:34
    so that will become
  • 00:42:36
    5305
  • 00:42:38
    divided by 0.2 times 2 divided by
  • 00:42:42
    so that was 16 thousand
  • 00:42:46
    eight hundred and eighty-six
  • 00:42:48
    is equal to d
  • 00:42:50
    times d minus d
  • 00:42:53
    so are we going to get
  • 00:42:56
    the two values which is for
  • 00:42:59
    outside the inside diameter and remember
  • 00:43:02
    guys that maram patang is angbala we all
  • 00:43:04
    know that the
  • 00:43:06
    mean diameter
  • 00:43:08
    or the average of two diameters is 200
  • 00:43:12
    so meaning
  • 00:43:13
    d plus small d is equal to 400
  • 00:43:17
    and we may uh
  • 00:43:18
    transpose this one so that will become
  • 00:43:22
    uh
  • 00:43:23
    the ah capital
  • 00:43:25
    [Music]
  • 00:43:26
    is a substitute so that was 400 minus
  • 00:43:29
    small diameter spraying nothing
  • 00:43:31
    so that was 16 886
  • 00:43:35
    it's equal to d
  • 00:43:36
    times d
  • 00:43:42
    400 minus 2d
  • 00:43:44
    support 100 minus d minus d so that was
  • 00:43:47
    400 minus 2d so i ship sold or
  • 00:43:51
    you may use the mode setup no
  • 00:44:00
    so that was 16886 is equal to
  • 00:44:03
    x times 400 minus 2x let us
  • 00:44:07
    substitute the x with this small
  • 00:44:09
    diameter d
  • 00:44:11
    so ships or magnetic values
  • 00:44:14
    given choices from 139 yen but us so let
  • 00:44:18
    us say 150.
  • 00:44:31
    involves a second degree or third degree
  • 00:44:34
    just look
  • 00:44:50
    so the small d or the inside diameter is
  • 00:44:53
    139.46
  • 00:44:56
    mm which is obviously letter a nanoman
  • 00:45:00
    and of course
  • 00:45:02
    maritime
  • 00:45:05
    d is equal to 400 minus the small d or
  • 00:45:08
    400 minus 139.46
  • 00:45:12
    so that was
  • 00:45:14
    260.54
  • 00:45:40
    that's enough for today for this clutch
  • 00:45:43
    uh review
  • 00:45:54
    [Music]
  • 00:46:12
    lang
  • 00:46:14
    do like my facebook page derive nasa
  • 00:46:18
    description and of course please
  • 00:46:19
    subscribe to my channel for more updates
  • 00:46:22
    and for more
  • 00:46:24
    engineering videos especially on this
  • 00:46:26
    machine design
  • 00:46:27
    and guys good luck and god bless and
  • 00:46:29
    grab you bye
Tags
  • macchine
  • ingranaggi
  • frizione
  • potenza
  • trasmissione
  • metodo di pressione uniforme
  • metodo di usura uniforme
  • progettazione
  • calcoli
  • coppia