Deflection of Beams Problem | Macaulay's Method | simply supported beam | GATE

00:19:38
https://www.youtube.com/watch?v=Qt-lls7jV5I

Ringkasan

TLDRThe video lecture covers how to employ Macaulay's Method to calculate deflection in a simply supported beam with a practical example. It begins by explaining Macaulay's Method, which involves using double integration of the bending moment equation EI d²y/dx². The lecture discusses determining the bending moment at any section in systematic order, where the slope and deflection are found through integrating this bending moment equation. The example considers a beam of 6 meters length with two point loads (48 kN and 45 kN) placed 1 meter and 3 meters, respectively, from the left support. The lecture details constructing a free body diagram, calculating reaction forces at supports using equilibrium concepts, and applying Macaulay's Method to find the deflection under each load. It further discusses the partitioning of sections and finding boundary conditions to compute integration constants. Finally, it calculates the deflection at specified points, emphasizing that a negative sign in results indicates deflection below the baseline. The video closes with calculated deflections at points D and C.

Takeaways

  • 📚 Macaulay's Method uses double integration to determine beam deflection.
  • 🧮 Bending moment is modeled by EI d²y/dx² equation.
  • 🔄 Integrate bending moment to get slope and defect of beam.
  • 📏 Example: 6m beam with 48kN and 45kN point loads.
  • 🔗 Reaction forces require equilibrium calculations at supports.
  • 🚫 Negative deflection indicates displacement below the beam's baseline.
  • ✍️ Solve integration constants using boundary conditions.
  • 🖊️ Calculate deflections at specific sections using modified equations.
  • 🔁 Different beam sections require adjusting mathematical approach.
  • 🔨 Practical use of structural analysis for engineering solutions.

Garis waktu

  • 00:00:00 - 00:05:00

    In this lecture, the speaker explains Macaulay's method, a technique for determining deflection in a simply supported beam using double integration of the bending moment. The method involves determining the bending moment at any section x, which may be at the start or the end of the beam. By integrating the bending moment equation, one can find the slope, and by integrating the slope, the deflection of the beam is derived. The procedure is introduced with an example problem involving a beam with point loads, using equilibrium concepts to calculate reaction forces at supports.

  • 00:05:00 - 00:10:00

    The speaker continues by describing the procedure of applying Macaulay's method. A section x is considered in the beam, and distances of forces relative to this section are calculated. The moment about section x is determined using forces and distances, integrated to find the slope and deflection. The integration introduces constants, requiring boundary conditions for solution. The boundary conditions are zero deflections at the supports, allowing determination of integration constants. These constants are substituted into an equation that calculates deflection at any beam point.

  • 00:10:00 - 00:19:38

    Finally, the deflection at specific points on the beam is determined using the derived equations and substituting appropriate values for x, resulting in deflection values at points D and C. These calculations include conversion of units for consistency with given Young's modulus and moment of inertia values. The deflections are computed as negatives because they occur below the baseline of the beam. The example demonstrates Macaulay's method through practical calculation steps and provides specific deflection results for given loads and beam conditions.

Peta Pikiran

Mind Map

Pertanyaan yang Sering Diajukan

  • What is Macaulay's Method?

    Macaulay's Method is a technique used in structural analysis to determine deflection in beams, utilizing double integration of bending moment equations.

  • What is the bending moment equation used in Macaulay's Method?

    The bending moment in Macaulay's Method is given by the equation EI d²y/dx².

  • What does the double integration in Macaulay's Method provide?

    Double integration of the bending moment equation first provides the slope, then the deflection of the beam.

  • How does the example problem define the beam and loads?

    The problem involves a simply supported beam of 6 meters with two point loads of 48 kN and 45 kN placed at 1 meter and 3 meters from the left support.

  • What is the moment of inertia given in the example?

    The moment of inertia is given as 85 x 10^6 mm⁴.

  • What are the boundary conditions used in the solution?

    The boundary conditions are x=0, y=0 at the right support, and x=6 m, y=0 at the left support, as the beam is simply supported at both ends.

  • What are the deflections calculated at points C and D?

    The deflection under load at point D is -16.7 mm, and at point C, it is -9.02 mm.

  • What is the value of Young's Modulus used in the problem?

    Young's Modulus is given as 2 x 10^5 N/mm².

  • What is the significance of the negative sign in the deflection calculation?

    The negative sign indicates the deflection occurs below the baseline of the beam.

  • How are the reaction forces at the supports determined?

    Reaction forces at the supports are determined using equilibrium equations of forces and moments in the beam system.

Lihat lebih banyak ringkasan video

Dapatkan akses instan ke ringkasan video YouTube gratis yang didukung oleh AI!
Teks
en
Gulir Otomatis:
  • 00:00:00
    in this video lecture i am going to
  • 00:00:02
    explain
  • 00:00:03
    how to use macaulay's method to
  • 00:00:05
    determine the deflection in a
  • 00:00:07
    simply supported beam with an example
  • 00:00:09
    problem before moving on to solving
  • 00:00:11
    the problem let me give an overview
  • 00:00:14
    about
  • 00:00:14
    what is macaulay's method and how it
  • 00:00:16
    works it's a simple
  • 00:00:18
    method which works based on the double
  • 00:00:20
    integration concept
  • 00:00:21
    in which the bending moment is given by
  • 00:00:25
    the equation
  • 00:00:26
    e i d square y by d x square in this
  • 00:00:30
    method bending moment at any section x
  • 00:00:34
    is determined in a systematic order
  • 00:00:37
    the section x should be taken either in
  • 00:00:40
    the first
  • 00:00:41
    or the last portion of the beam the
  • 00:00:45
    slope
  • 00:00:45
    dy by dx is obtained by
  • 00:00:48
    integrating the bending moment equation
  • 00:00:51
    that means the integrating the bending
  • 00:00:53
    moment equation
  • 00:00:54
    we can get the slope value
  • 00:00:59
    again by integrating this slope equation
  • 00:01:02
    we can get the deflection of b so
  • 00:01:05
    again if you integrate this equation
  • 00:01:09
    we are going to get the deflection value
  • 00:01:13
    that is why we say that it works based
  • 00:01:16
    on the double integration concept
  • 00:01:19
    now let us solve the problem so if we
  • 00:01:21
    read the problem
  • 00:01:22
    a beam of length 6 meter is simply
  • 00:01:26
    supported at its ends
  • 00:01:27
    and carries two point loads of 48 kilo
  • 00:01:31
    newton
  • 00:01:32
    and 45 kilo newton at a distance of
  • 00:01:35
    one meter and three meter respectively
  • 00:01:38
    from
  • 00:01:38
    left support determine the deflection
  • 00:01:41
    under each load consider the young's
  • 00:01:44
    modulus as
  • 00:01:45
    2 multiplied by 10 to the power 5 newton
  • 00:01:48
    per mm square
  • 00:01:50
    and the moment of inertia value was
  • 00:01:53
    85 multiplied by 10 to the power 6
  • 00:01:56
    millimeter power
  • 00:01:57
    4 first let us construct the given
  • 00:02:00
    system we have 2 point load 48 kilo
  • 00:02:04
    newton and 40 kilo newton
  • 00:02:06
    which are located at a distance of 1
  • 00:02:08
    meter and
  • 00:02:09
    3 meter from the left support the entire
  • 00:02:13
    length of the beam is
  • 00:02:14
    6 meter it is simply supported
  • 00:02:17
    to construct a free body diagram we must
  • 00:02:20
    know the value of reaction force at a
  • 00:02:23
    and reaction
  • 00:02:24
    force at b at a we have a reaction force
  • 00:02:27
    which is r
  • 00:02:28
    a and at b we have a reaction force
  • 00:02:31
    rb now let us find out the reaction
  • 00:02:35
    forces at
  • 00:02:36
    a and b for that we need to apply the
  • 00:02:39
    equilibrium concept
  • 00:02:40
    that is sigma fy is equal to 0
  • 00:02:44
    that means the net forces which are
  • 00:02:46
    acting in the vertical direction
  • 00:02:48
    is equal to zero so r
  • 00:02:52
    a plus rb that means the net forces
  • 00:02:55
    which are acting in the upward direction
  • 00:02:57
    is equal to 48 plus 40 and these two are
  • 00:03:00
    the point loads which are acting in the
  • 00:03:01
    downwards direction r a plus rb
  • 00:03:04
    is equal to 88
  • 00:03:07
    and then we will apply the next
  • 00:03:09
    equilibrium condition
  • 00:03:10
    that is moment about any point is equal
  • 00:03:13
    to zero here we take moment about
  • 00:03:15
    a is equal to 0 so the first
  • 00:03:19
    force which creates the moment about a
  • 00:03:21
    is rb
  • 00:03:22
    rb multiplied by the distance between
  • 00:03:26
    a b which is 6 meter it creates a
  • 00:03:29
    counter clockwise moment it creates a
  • 00:03:32
    counter clockwise moment about a
  • 00:03:34
    so we put positive sign here the next
  • 00:03:37
    moment is created by this point load 40
  • 00:03:40
    kilo newton
  • 00:03:41
    above point a it creates
  • 00:03:45
    clockwise moment about a so we put
  • 00:03:48
    minus sign here 40 multiplied by
  • 00:03:51
    the distance between a and d which is 3
  • 00:03:54
    so 40 multiplied by 3
  • 00:03:56
    the next moment is created by this point
  • 00:03:59
    load 48 kilo newton
  • 00:04:01
    above 0.8 which is 48 multiplied by
  • 00:04:04
    1 meter is equal to 0
  • 00:04:07
    by solving this equation we can find
  • 00:04:10
    the value of rb which is equal to 28
  • 00:04:14
    kilo newton after substituting this
  • 00:04:18
    rb value in this equation we can get the
  • 00:04:21
    value of
  • 00:04:21
    r a which is equal to 60 kilo newton
  • 00:04:25
    now we have calculated the reaction
  • 00:04:28
    forces at a
  • 00:04:29
    and b now the free body diagram is
  • 00:04:32
    complete now let us see how to apply the
  • 00:04:35
    mccollux method to find out the
  • 00:04:37
    deflection
  • 00:04:38
    in the simply supported beam
  • 00:04:41
    so this is the free body diagram and
  • 00:04:43
    here
  • 00:04:45
    the position of c and d are referred
  • 00:04:48
    from the
  • 00:04:49
    right support b so this distance ac is
  • 00:04:54
    1 meter the entire beam length is 6
  • 00:04:56
    meter so
  • 00:04:57
    this length that is c to b measures 5
  • 00:05:00
    meter
  • 00:05:01
    similarly a to d it is given as 3 meter
  • 00:05:05
    so the remaining length is 3 meter that
  • 00:05:07
    is d 2 b
  • 00:05:10
    now we are going to consider a section
  • 00:05:13
    x in the first part of the beam that is
  • 00:05:17
    the section ac which is located at a
  • 00:05:21
    distance of
  • 00:05:22
    x from the right support p
  • 00:05:26
    it is the first step in mccallis method
  • 00:05:29
    now
  • 00:05:29
    we have to find out the distance of
  • 00:05:32
    force
  • 00:05:33
    at c and d with respect to this section
  • 00:05:36
    x
  • 00:05:36
    so this distance is going to be x minus
  • 00:05:40
    3 meter because the total distance is x
  • 00:05:43
    and this distance is 3 meter so it is
  • 00:05:45
    going to be x minus
  • 00:05:46
    3 meter and this distance is
  • 00:05:49
    x minus 5 meter now let us calculate the
  • 00:05:54
    moment
  • 00:05:54
    about this section x so according to
  • 00:05:58
    mccollum's method we represent the
  • 00:06:00
    pending moment as e i d square y by d x
  • 00:06:03
    square
  • 00:06:04
    which is equal to the moment about
  • 00:06:07
    gex so the moment about x is
  • 00:06:10
    found by using the moments which are
  • 00:06:13
    created by the reaction force
  • 00:06:15
    and these two point loads first
  • 00:06:18
    this moment is created by this force
  • 00:06:21
    multiplied by this distance
  • 00:06:23
    so it creates a counter clockwise moment
  • 00:06:25
    so we put positive here
  • 00:06:26
    so 28 multiplied by the distance x
  • 00:06:30
    and the next moment is created by this
  • 00:06:33
    force that is 40 kilo newton
  • 00:06:35
    so it is going to be 40 multiplied by x
  • 00:06:38
    minus 3
  • 00:06:39
    minus because this force is going to
  • 00:06:42
    create a
  • 00:06:43
    clockwise moment about this section x so
  • 00:06:45
    it is going to be minus
  • 00:06:47
    and 48 multiplied by x minus 5
  • 00:06:51
    so this is the bending moment about x
  • 00:06:53
    and which is equal to
  • 00:06:55
    e i d square y by d x square
  • 00:06:58
    so we have considered some partition
  • 00:07:01
    here
  • 00:07:02
    to separate the section in the
  • 00:07:06
    beam so this 28 x refers
  • 00:07:09
    this section we have considered this
  • 00:07:11
    section as one
  • 00:07:14
    and the next section is cd
  • 00:07:18
    so it is going to be the second section
  • 00:07:22
    and the next one is this ac so this is
  • 00:07:25
    going to be the third section
  • 00:07:27
    and the corresponding moment value is
  • 00:07:31
    given here so this representation will
  • 00:07:34
    help us to find out the
  • 00:07:36
    deflection at individual sections
  • 00:07:40
    now we are going to integrate this
  • 00:07:42
    equation with respect to k x
  • 00:07:46
    to get the value of slope so
  • 00:07:49
    by integrating this one we are going to
  • 00:07:52
    get this value that is e
  • 00:07:53
    i d y by d x is equal to 28
  • 00:07:57
    x square by 2 plus c 1 this integration
  • 00:08:00
    constant
  • 00:08:02
    minus 40 x minus 3 whole power
  • 00:08:06
    2 divided by 2 minus 48
  • 00:08:10
    x minus 5 whole power 2 divided by
  • 00:08:13
    2 so here also we can consider this
  • 00:08:17
    partition that is 1 2 and 3
  • 00:08:21
    now let us simplify the equation it is
  • 00:08:24
    14 and it is
  • 00:08:28
    20 and it is
  • 00:08:31
    24. now let us write this simplified
  • 00:08:34
    equation that is 14x square
  • 00:08:37
    plus c1 minus john t
  • 00:08:40
    x minus 3 whole power 2 minus 24
  • 00:08:44
    x minus 5 whole power 2. here also
  • 00:08:47
    we consider this partition 1 2 3 to
  • 00:08:51
    represent the individual section in the
  • 00:08:52
    beam so this equation is
  • 00:08:55
    considered as equation number one
  • 00:08:59
    now we are going to integrate this
  • 00:09:01
    equation to get the
  • 00:09:03
    deflection equation so by integrating
  • 00:09:06
    the double equation
  • 00:09:07
    we get e i y
  • 00:09:10
    which is equal to 14 x cube divided by 3
  • 00:09:13
    plus c 1 x
  • 00:09:15
    plus c 2 so it is the second integration
  • 00:09:18
    so we get the another constant which is
  • 00:09:21
    called as c2 minus 20 x minus 3
  • 00:09:25
    the whole power 3 divided by 3 minus
  • 00:09:28
    24 x minus 5 the whole power 3 divided
  • 00:09:31
    by
  • 00:09:32
    3 let us simplify this equation
  • 00:09:36
    which is 1 this is
  • 00:09:39
    8 so the other time we will keep
  • 00:09:43
    as it is okay so it is going to be 14
  • 00:09:45
    divided by 3 x cube
  • 00:09:47
    plus c 1 x plus c2 minus 20 divided by 3
  • 00:09:53
    multiplied by x minus 3 whole power 3
  • 00:09:55
    minus
  • 00:09:56
    8 x minus 5 the whole power 3
  • 00:09:59
    here also we have to consider this
  • 00:10:02
    partition
  • 00:10:03
    and this equation is referred as
  • 00:10:06
    equation number
  • 00:10:07
    two so we are going to use this equation
  • 00:10:10
    to calculate
  • 00:10:11
    the deflection at individual section and
  • 00:10:13
    for that we need to calculate
  • 00:10:15
    these two integration constants that is
  • 00:10:17
    c1
  • 00:10:18
    and c2 to find out the value of c1 c2
  • 00:10:23
    we need to apply the boundary condition
  • 00:10:25
    now we are going to see
  • 00:10:26
    how to find the boundary conditions
  • 00:10:29
    which are required to find out the c1
  • 00:10:31
    and c2
  • 00:10:32
    for that we need to consider this being
  • 00:10:35
    here if i put x is equal to 0
  • 00:10:39
    i am going to be here at the right
  • 00:10:40
    support if i put x is equal to 0
  • 00:10:43
    so when x is equal to 0 the deflection
  • 00:10:45
    value is equal to 0
  • 00:10:47
    the reason is at point b
  • 00:10:51
    we have a simply support so when x is
  • 00:10:55
    equal to 0 there is no deflection at
  • 00:10:56
    this point
  • 00:10:58
    since it is simply supported there is no
  • 00:11:01
    motion along
  • 00:11:02
    y direction so the deflection value is
  • 00:11:04
    going to be zero
  • 00:11:05
    similarly if i consider this point
  • 00:11:09
    the x is equal to 6 meter i will be
  • 00:11:12
    getting this point
  • 00:11:14
    so when x is equal to 6 meter again the
  • 00:11:16
    y value is going to be 0 because
  • 00:11:19
    this end a is also simply supported so
  • 00:11:23
    when x is equal to 6 meter y is equal to
  • 00:11:25
    0. so these two are all the boundary
  • 00:11:26
    conditions the first one is
  • 00:11:29
    x equal to zero y is equal to zero this
  • 00:11:32
    is boundary condition one
  • 00:11:34
    and the second boundary condition is
  • 00:11:37
    x is equal to six meter and y is equal
  • 00:11:39
    to zero
  • 00:11:41
    we are going to apply these boundary
  • 00:11:43
    conditions to find out dc1
  • 00:11:45
    and c2 that is x is equal to 0
  • 00:11:48
    and y is equal to 0 and the second one
  • 00:11:51
    is when x is equal to 6 y is equal to
  • 00:11:54
    0 first let us apply the boundary
  • 00:11:56
    condition
  • 00:11:58
    1 in equation number
  • 00:12:02
    2 so by substituting x is equal to 0
  • 00:12:06
    and y is equal to 0 this term becomes 0
  • 00:12:08
    and this term
  • 00:12:09
    also 0 this term also 0
  • 00:12:13
    and c2 is there and this
  • 00:12:16
    tam is going to be 20 by 3 0 minus 3 the
  • 00:12:20
    whole power 3 minus 8
  • 00:12:21
    0 minus 5 the all power 3.
  • 00:12:24
    now we are going to find out whether
  • 00:12:27
    these two parts are needed or not
  • 00:12:29
    so we have to be very careful about that
  • 00:12:32
    because
  • 00:12:33
    if any negative value is present
  • 00:12:36
    within this bracket we should not
  • 00:12:38
    consider that section
  • 00:12:41
    the reason is when x is equal to 0 it is
  • 00:12:44
    corresponding to the first part of the
  • 00:12:46
    equation because here in the diagram you
  • 00:12:49
    can see that
  • 00:12:50
    when x is equal to 0 it lies in the
  • 00:12:54
    point b so it represent the first
  • 00:12:56
    section
  • 00:12:57
    so we need to consider only the
  • 00:13:00
    first portion so the remaining two
  • 00:13:03
    portion
  • 00:13:04
    we should not consider or otherwise we
  • 00:13:07
    can remember like this
  • 00:13:08
    if you get any negative value within
  • 00:13:10
    this bracket
  • 00:13:12
    we should not consider that part so the
  • 00:13:15
    second part
  • 00:13:17
    and the third part should not be
  • 00:13:18
    considered because x is equal to 0
  • 00:13:21
    which is only corresponding to this
  • 00:13:25
    first part of the equation so we should
  • 00:13:26
    not consider these two
  • 00:13:28
    now by simplifying this equation we get
  • 00:13:31
    c2 is equal to 0
  • 00:13:35
    and the next boundary condition is when
  • 00:13:37
    x is equal to 6 meter
  • 00:13:39
    y is 0 we are going to substitute this
  • 00:13:42
    boundary condition in equation number 2
  • 00:13:44
    and we get the y value is 0
  • 00:13:47
    and x value is substituted as 6
  • 00:13:51
    so here we need to consider the all the
  • 00:13:53
    part of the equation the reason is
  • 00:13:55
    within the bracket we don't get any
  • 00:13:57
    negative value
  • 00:13:58
    and technically when x is equal to 6
  • 00:14:01
    meter
  • 00:14:02
    it represents the entire length so we
  • 00:14:04
    need to consider the
  • 00:14:05
    entire part of the equation first part
  • 00:14:08
    second part as well as the third part of
  • 00:14:10
    the equation
  • 00:14:12
    so we need to consider the entire part
  • 00:14:13
    of the equation when x is equal to 6
  • 00:14:16
    meter
  • 00:14:16
    so by simplifying this equation we got
  • 00:14:19
    c1
  • 00:14:20
    as minus 136.60
  • 00:14:24
    now we have calculated the constants
  • 00:14:27
    c1 and c2 let us substitute
  • 00:14:30
    those values in equation number two
  • 00:14:33
    c2 is zero so we don't consider that tam
  • 00:14:36
    so here also we consider that three
  • 00:14:39
    partition
  • 00:14:40
    one two and three which represents
  • 00:14:44
    different section of the beam this
  • 00:14:48
    equation we represent that as
  • 00:14:49
    equation number three now the deflection
  • 00:14:53
    equation is ready to calculate the
  • 00:14:56
    deflection at
  • 00:14:57
    any point on the beam now we are going
  • 00:15:00
    to calculate the deflection at
  • 00:15:02
    d and deflection at c in order to find
  • 00:15:05
    out the deflection at d
  • 00:15:06
    we need to substitute x is equal to 3
  • 00:15:09
    meter
  • 00:15:11
    so in the equation number 3 we need to
  • 00:15:13
    substitute x is equal to 3 meter
  • 00:15:16
    to find out the deflection at this point
  • 00:15:19
    d
  • 00:15:20
    and if you want to find out the
  • 00:15:21
    deflection at c in equation number 3 we
  • 00:15:24
    need to substitute
  • 00:15:25
    x is equal to 5 meter first let us find
  • 00:15:28
    out the reflection at d
  • 00:15:30
    to find out the deflection we need to
  • 00:15:31
    substitute x is equal to 3 meter
  • 00:15:33
    in the equation number 3 so by
  • 00:15:36
    substituting this
  • 00:15:37
    value we got e i y which is equal to 14
  • 00:15:40
    by 3
  • 00:15:41
    into 3 power 3 minus 136.67 multiplied
  • 00:15:45
    by 3
  • 00:15:46
    here we have applied the first part of
  • 00:15:49
    the equation the reason is
  • 00:15:51
    x is equal to 3 meter when x is equal to
  • 00:15:54
    3 meter it lies
  • 00:15:55
    in the first part of the beam so we need
  • 00:15:57
    to consider the first
  • 00:15:59
    part of the equation so we have
  • 00:16:00
    considered the first part of the
  • 00:16:02
    equation
  • 00:16:03
    as x equal to 3 meter lies in the first
  • 00:16:05
    part of the
  • 00:16:06
    beam even if you substitute this value 3
  • 00:16:09
    here it becomes 0 and this dam
  • 00:16:11
    becomes negative within this bracket so
  • 00:16:14
    we need to ignore these two
  • 00:16:16
    terms by solving this terms
  • 00:16:19
    we can get minus 2 eight four kilo
  • 00:16:22
    newton
  • 00:16:22
    meter cube since we have integrated this
  • 00:16:25
    dam
  • 00:16:26
    two times we got this dam as meter cube
  • 00:16:29
    so the entire system represents
  • 00:16:33
    in kilo newton meter cube we need to
  • 00:16:35
    convert this dam into newton
  • 00:16:38
    mm power 3 the reason is the x modulus
  • 00:16:41
    value
  • 00:16:42
    is given in newton per millimeter square
  • 00:16:45
    and i value is given in millimeter power
  • 00:16:48
    4. so we need to convert this dam into
  • 00:16:50
    newton millimeter power 4
  • 00:16:53
    so which is equal to minus 284 into 10
  • 00:16:55
    to the power 12
  • 00:16:57
    newton millimeter power 3 this dam
  • 00:17:01
    is obtained by converting kilo newton
  • 00:17:04
    into
  • 00:17:04
    newton that is into 20 to the power 3
  • 00:17:07
    multiplied by converting this meter into
  • 00:17:10
    millimeter so
  • 00:17:11
    10 power 3 the whole power 3 so
  • 00:17:15
    totally we got 10 to the power 12 after
  • 00:17:17
    converting this
  • 00:17:19
    unit into newton meter power 3
  • 00:17:22
    we can substitute this value of e
  • 00:17:26
    and i here so we got the y value as
  • 00:17:29
    minus 16.7 mm this minus sign
  • 00:17:33
    indicates the deflection happens below
  • 00:17:36
    the baseline
  • 00:17:37
    after substituting x is equal to 3 meter
  • 00:17:42
    we got that deflection at point d as
  • 00:17:45
    16.7
  • 00:17:46
    mm now let us calculate the deflection
  • 00:17:49
    at c
  • 00:17:51
    so if you want to find out the
  • 00:17:52
    deflection at c we need to substitute x
  • 00:17:54
    is equal to 5 meter in the equation 3
  • 00:17:56
    and here we have to consider only the
  • 00:17:59
    first two part of the equation
  • 00:18:02
    because when you find out the deflection
  • 00:18:05
    at c
  • 00:18:06
    the x is going to be 5 meter
  • 00:18:11
    it lies in the second part of the bin so
  • 00:18:14
    we need to consider the first
  • 00:18:16
    two part of the beam to find out the
  • 00:18:18
    deflection at
  • 00:18:19
    c so e i y which is equal to 14 by 3
  • 00:18:24
    multiplied by 5 power 3 minus 136.67
  • 00:18:28
    multiplied by 5 and this is the second
  • 00:18:31
    term minus 20 divided by 3
  • 00:18:33
    5 minus 3 the whole power 3. even if you
  • 00:18:36
    apply the
  • 00:18:37
    3 here this third term becomes negative
  • 00:18:40
    so we have to
  • 00:18:42
    ignore this term so
  • 00:18:45
    by substituting the x in the
  • 00:18:49
    equation number three we got the values
  • 00:18:51
    minus 153.35 kilo newton meter cube so
  • 00:18:56
    we need to convert this kilo newton
  • 00:18:57
    meter cube into newton mm power 3
  • 00:19:01
    so that we can apply the value of e and
  • 00:19:04
    i
  • 00:19:04
    here so y is going to be minus
  • 00:19:08
    153.3 into 10 to the power 12 divided by
  • 00:19:12
    2 multiplied by 10 to the power 5
  • 00:19:15
    multiplied by 85 into 10 to the power 6
  • 00:19:18
    so the y value is minus 9.02
  • 00:19:22
    mm here also we got the negative sign
  • 00:19:24
    because
  • 00:19:25
    the deflection happened below the
  • 00:19:27
    baseline the deflection at c
  • 00:19:29
    is 9.02 mm
  • 00:19:34
    thank you for watching
Tags
  • Macaulay's Method
  • Beam Deflection
  • Structural Analysis
  • Double Integration
  • Bending Moment
  • Equilibrium Conditions
  • Example Problem
  • Reaction Forces
  • Simply Supported Beam
  • Deflection Calculation