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Design for impact load

 
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Padmanabhan G
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PostPosted: Wed Oct 13, 2010 8:34 am    Post subject: Design for impact load Reply with quote

Dear Sir,

Can any one suggest how to design RCC strucutres particulary floor for impact load from obects that are dropped from certain height.

With regards
Padmanabhan.G
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sspawar
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PostPosted: Wed Oct 13, 2010 10:11 pm    Post subject: Reply with quote

Dear ER. P'NABHAN,

Please refer books on Machine foundations. Somecode List is pasted below If you have access to these codes you can refer the same.
IS 1499 : 1977 Method for Charpy Impact Test (U-notch) for Metals2003
IS 1598 : 1977 Method for Izod Impact Test of Metals2003
IS 1757 : 1988 Method for charpy impact test (V notch) for metallic material2003
IS 2974 : Part 2 : 1980 Code of practice for design and construction of machine foundations: Part 2 Foundations for impact type machine (hammer foundations)2008
IS 2974 : Part 5 : 1987 Code of practice for design and construction of machine:foundations Part 5 Foundations for impact machines other than hammers (forging and stamping press, pig breakers, drop crusher and jolter)2008


Regards


Last edited by sspawar on Fri Oct 15, 2010 3:58 pm; edited 1 time in total
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vikram.jeet
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PostPosted: Thu Oct 14, 2010 4:52 am    Post subject: Design for impact load Reply with quote

Impact Load on a floor:

Theoretically speaking,the impact loading is infinite on a rigid(non yielding)
body since the acceleration at time of impact is infinite due to instantaneous
change of velocity from value v to zero.But every elastic material
deforms and effect of impact may reduce .

Normally the Impact loading is taken 2 times the load dropped.
Based on contact area with floor, the load intensity could be worked out
and shall be within permissible bearing stress of floor material

Floors subjected to frequent impacts may be provided with ironite flooring
to reduce damage .Fibre-reinforced concerte could also be thought of.

best regards

vikramjeet


Dear Sir,

Can any one suggest how to design RCC strucutres particulary floor for impact load from obects that are dropped from certain height.

With regards
Padmanabhan.G

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vikram.jeet
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PostPosted: Thu Oct 14, 2010 5:34 am    Post subject: Design for impact load Reply with quote

Impact loading on various structures:

In continuation to earlier post,

Also refer IS 875(part-2) for Impact loading on various structures
like Lifts & hoists and machinery, crane girders

IRC -6 for impact loading on Concrete and steel Bridges

vikramjeet
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Yogesh.Pisal
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PostPosted: Thu Oct 14, 2010 6:20 am    Post subject: Reply with quote

Dear Er Padmanabhan.G,

Here you need to firstly decide philosophy and then method of design. You can think of following points.

1. Decide the load path

e.g. If some object is dropping on RC slab then load path can be one way or two way depending on type of floor you are having.

In my opinion it is better to have one way bending (you can detail your slab accordingly).

Load path: Slab to Beams, Beams to Columns, Columns to foundation

2. Analysis and design of slab:

If magnitude of load is small then you can go for static design. As per strain energy theory you will end up with twice BM value in impact load as compared to static load

If load is huge then you have to go for dynamic analysis - Here you may need to go for displacement based design. You can use numerical integration for calculating response.

Refer: Introduction to structural Dynamics by Biggs or Dyanmics of Structures by AK Chopra.

3. Susequently consider the similar design for beams and columns

Regards,
Yogesh Pisal
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lele_raj
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PostPosted: Thu Oct 14, 2010 10:06 am    Post subject: Design for impact load Reply with quote

...Also, for cyclic impact load (e.g. crushers, screens etc.) pl. don't forget to perform a dynamic analysis. It's preferred that at least first two, if not three, natural frequencies of the supporting structure don't match with the excitation frequency.


Best regards,

Rajendra (Raj) Lele


P Please consider the environment before printing this e-mail or any attachment/s.



From: vikram.jeet <forum@sefindia.org>
To: general@sefindia.org
Sent: Thu, 14 October, 2010 5:27:23 PM
Subject: [SEFI] Re: Design for impact load

     Impact loading on various structures:

In continuation to earlier post,

Also refer IS 875(part-2) for Impact loading on various structures
like Lifts & hoists and machinery, crane girders

IRC -6 for impact loading on Concrete and steel Bridges

vikramjeet
-- ญญ

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M_N_NISAR
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Joined: 29 May 2010
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Location: Hyderabad, Andhra pradesh,India

PostPosted: Thu Oct 14, 2010 6:46 pm    Post subject: lift design loads Reply with quote

respected sefiance
i am just confused about the impact loading due to lift in a building
with reference to the given answers i want some basic and general concept ..some of my friends suggested me to double the dead loads on account of impact loading ..does in a lift horizontal loads my also occurs ??? and what will be the resultant ???
please suggest me
thanking you
m.n.nisar
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sspawar
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PostPosted: Thu Oct 14, 2010 10:58 pm    Post subject: Reply with quote

Dear Frnds,

Below article defines well regarding - Impact loading also summerise
why Limp =2L

----------------------------------------------------------------------------------------------------------
Generally when the strength of machine elements are considered it is assumed that the loading is static or applied gradually.  This loading condition is often not the case, the loading may be cyclic requiring assessment for fatigue.   Fatigue    or it may involve impact or suddenly applied loads.   When loads are applied suddenly and when the loads are applied as impact loads the resulting transient stresses (and deformations) induced in the machine elements are much higher than if the loads are applied gradually.    This effect is shown in the diagram below



It is normal practice to design machines such that impact loads are eliminated or reduced by inclusion of shock absorbers.  Inclusion of low cost, mass produced, shock absorbers can virtually eliminate the increased stresses and deformations resulting from impact loads.

Most ductile materials have strength properties which are a function of the loading speed.   The more rapid the loading the higher the tensile and ultimate strengths of the materials.   .  Two standard tests, the Charpy and Izod, measure the impact energy (the energy required to fracture a test piece under an impact load), also called the notch toughness.

The detailed assessment of the strength of machine elements under impact loading regimes involves use of advanced techniques including Finite Element Analysis.   Impact loads result in shock waves propogating through the elements with possible serious consequences.  It is possible to complete a relatively simply stress evaluation for suddenly applied and impact loads by using the principle of conservation of energy and conditional that the materials considered are operating within their elastic regions.

The equations are most accurate for relatively heavy impacting masses moving at low velocities on impact.

Notation


  • A = Area (m2)
  • E = Modulus of Elasticity (N/m2)
  • h = Drop distance (m)
  • k =Stiffness (N/m)
  • M = Mass-moving body (kg)
  • M1 = Mass of static bar or beam (kg)
  • K = Factor to allow for energy loss at impact
  • l = length of bar (l)
  • tp = time period (s)
  • v = velocity (m/s)
  • W = Weight - moving body(N)
  • V = Velocity (m/s)
  • w = Specific Weight (kg/m3)
  • σ = stress (N/m2)
  • σ stat = stress resulting from static load(N/m2)
  • δ = deflection (m)
  • δ stat = deflection resulting from static load(m)
  • μ = Ratio Moving Mass/Stationary bar
  • β = Constant = A Sqrt(wEg/W) - see text



    • Linear Impact deflections and stresses (gravity loads)

      Important note: The notes below represent a very simple view of the loading condition and do not consider more real case involving shock waves being propagated through the loaded member or the moving mass

      Consider a loading regime as shown below with a ring of Mass M(kg) with weight W= M.g(N) being dropped through a distance h onto a collar supported by a vertical bar which behaves as a spring with a stiffness of k (N/m).

      The support bar has a length l (m), an Area A (m2 ) With a modulus of elasticity E (N/m2 )


      In practice the weight would impact onto the support which would elastically deform until all of the potential energy has been absorbed.  The support would then contract initiating damped oscillations until the system assumes a stable static position.  The equations below determine the initial maximum deformation which provides the most highly stressed condition.



      In accordance with conservation of energy the potential energy of the weight is converted to elastic strain energy.


      This may be expressed as a quadratic equation


      This is solved for the maximum deflection
      δ
      max as follows


      The weight applied gradually would result in a deflection
      δ
      st thus


      Note:
      The stiffness k = Force /Deflection = F / δ :    E = stress /strain = σ /e = (F /A) /( δ /l)    Therefore k = EA/l

      Substituting this into the equation for δ max results in


      This can be expressed as


      The resultant maximum force is simply
      Pmax = k δ max



      and the resultant maximum stress = σ max = Pmax/ A

      This may be expressed as  
                                                               

      For the calculation of the stress due to a suddenly applied load with h = 0


      σ max = 2 σ st





      Linear impact deflections and stresses (kinetic impacts )


      Important note: The notes below represent a very simple view of the loading condition and do not consider more real case involving shock waves being propagated through the loaded member or the moving mass

      Impact loads based pimarily on kinetic energy e.g horizontal impacts are treated slightly differently.  For these applications the kinetic energy is converted into stored energy due to elasticity of the resisting element.

      Consider a Mass M(kg) with a velocity of v impacting on a collar which is supported by a bar with a stiffness of K (N/m) - Ignoring gravitational forces.

      The kinetic energy of the mass Mv2 /2 is transformed into stored energy in the support.


      The resulting equation is


      The resultant maximum deflection equals..


      Noting that the static deflection = Wl/AE this equation can be written


      The equivalent maximum stress =


      Noting that the static deflection = Wl/AE this equation can be written





      Beams




      Using similar principles as expressed above it can be easily proved using principles of conservation of energy that.


      and


      If the impact is horizontal instead of vertical . The resulting deformation and stress resulting from the impact are


      Note: The above approximate relationships can been applied generally to most structural systems subject to distortion with the elastic range when subject to impact loading




      Energy losses on impact


      The above equations are very approximate and include many assumptions.  A very important assumption is that all of the energy ( based on h or v2) is used up in producing the same distortion as would result from static loading.  In reality, some kinetic energy is lost in internal friction.  Account can be taken for these losses by multiplying h or v2 by an appropriate factor K.  This factor is derived from the Mass of the moving body (M) and the mass of the beam or bar (M1).  The factor is different for different loading systems ass follows.

      In the equations above h or v2 would be replaced by K.h or K v2
      The K factor is nearest unity when M is large compared to M1 . As an example for the axial impact
      if M1/M is say 0,1 then K = 0,95..
      if M1/M is say 10 then K = 0,15..
      This is illustrated in the figure below


      1) A moving mass M striking axially one end of a bar of mass M1. The other end of the bar being fixed...


      2) A moving mass M striking traversely the end of a beam of mass M1...


      3) A moving mass M striking traversely the center of a beam with simple supported ends of mass M1....


      4) A moving mass M striking traversely the center of a beam with fixed ends of mass M1. ..





      Impact stresses considering propagation of shock waves- Unsupported bar




      When a impact force is suddenly applied to and elastic body , a wave of stress is propogated traveling through the body with a velocity..


      w = weight /unit volume (kg/m2), v = velocity (m/s)

      The unsupported bar subject to the longitudonal impact from a rigid body with velocity v experiences a wave of compressive stress of intensity
      σ
      .


      If the mass of the moving body is very large compared to the mass of the bar the wave of compression bounces back from the far end of the bar as a wave of tension and returns to the struck end after a time period .
      t p = 2L / V


      If the mass of the moving body is very large compared to the bar so that it can be considered infinite then after breaking contact the moving bar will move away from the impacting mass with a velocity of vb= 2v.   The moving bar will be stress free.

      If the mass of the impinging body is
      μ
      time the mass of the bar then the bar will move away with an average velocity of


      The moving bar is left vibrating with a stress intensity of













      Impacts considering propagation of shock waves- bar with one end supported




      For the case of a bar with one end fixed , the wave of compressive stress resulting from the impact on the unsupported end is reflected back unchanged from the supported (fixed end) and combines with the advancing waves to produce a maximum stress approximately equal to..



      μ
      = Mass of Moving Body/Mass of Bar



      -------------------------------------------------------
      please refer These Threads also -
      http://www.roymech.co.uk/Useful_Tables/Fatigue/Mechanics_Impact.html
      http://www.pdhcenter.com/courses/s164/s164content.pdf
      http://people.virginia.edu/~lz2n/mse209/Chapter8.pdf

      ------------------------------------------

      One thing I want to add - Ironite or any treatment against wear and tear of top surface can never be associate with Impact resistance.
      Wear and tear or abration of an substance or object is measured by its hardness.
      Impact resistance is measured by its toughness or stiffness.
      Fatigue or durability is measured by its soundness.
      Comprressive strength is measured by its crushing value.

      Regards
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      vikram.jeet
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      Joined: 26 Jan 2003
      Posts: 2212

      PostPosted: Fri Oct 15, 2010 5:43 am    Post subject: Design for impact load Reply with quote

      Dear Er MN Nisar,

      As per IS 875(part-2) , Imact loading can be taken without any reservations/confusions

      For frames supporting Lifts & Hoists - - - - - - - - - - - - - - - - - - - - - - - 100%
      For foundations,footings and piers supporting - - - - - - - - - - -- - - - 40%
      Lifts and hoisting apparatus

      Generally in cranes , horizontal loads (in transverse as well as longitudinal  
      direction ) occur due to braking/tractive efforts on crane movement which is  
      horizontal . But for Lifts which are hoisted on pulley- drum through ropes
      and move within vertical members (guide rails) and movement is only vertical, the  
      question of horizontal loading does not arise under normal conditions(i.e. non seismic)

      Lifts are also provided with shock/Impact absorber( Buffers) in Lift pit to safeguard  
      in event of lift failure

      best regards

      vikramjeet

      respected sefiance
      i am just confused about the impact loading due to lift in a building
      with reference to the given answers i want some basic and general concept ..some of my friends suggested me to double the dead loads on account of impact loading ..does in a lift horizontal loads my also occurs ??? and what will be the resultant ???
      please suggest me
      thanking you
      m.n.nisar

      -- ญญ

      Posted via Email
      Back to top
      View user's profile Send private message
      Heena
      SEFI Member
      SEFI Member


      Joined: 10 Sep 2019
      Posts: 2

      PostPosted: Tue Sep 10, 2019 7:58 am    Post subject: Reply with quote

      sspawar wrote:
      Dear Frnds,

      Below article defines well regarding - Impact loading also summerise
      why Limp =2L

      ----------------------------------------------------------------------------------------------------------
      Generally when the strength of machine elements are considered it is assumed that the loading is static or applied gradually.  This loading condition is often not the case, the loading may be cyclic requiring assessment for fatigue.   Fatigue    or it may involve impact or suddenly applied loads.   When loads are applied suddenly and when the loads are applied as impact loads the resulting transient stresses (and deformations) induced in the Design of Machine Elements are much higher than if the loads are applied gradually.    This effect is shown in the diagram below



      It is normal practice to design machines such that impact loads are eliminated or reduced by inclusion of shock absorbers.  Inclusion of low cost, mass produced, shock absorbers can virtually eliminate the increased stresses and deformations resulting from impact loads.

      Most ductile materials have strength properties which are a function of the loading speed.   The more rapid the loading the higher the tensile and ultimate strengths of the materials.   .  Two standard tests, the Charpy and Izod, measure the impact energy (the energy required to fracture a test piece under an impact load), also called the notch toughness.

      The detailed assessment of the strength of machine elements under impact loading regimes involves use of advanced techniques including Finite Element Analysis.   Impact loads result in shock waves propogating through the elements with possible serious consequences.  It is possible to complete a relatively simply stress evaluation for suddenly applied and impact loads by using the principle of conservation of energy and conditional that the materials considered are operating within their elastic regions.

      The equations are most accurate for relatively heavy impacting masses moving at low velocities on impact.

      Notation


      • A = Area (m2)
      • E = Modulus of Elasticity (N/m2)
      • h = Drop distance (m)
      • k =Stiffness (N/m)
      • M = Mass-moving body (kg)
      • M1 = Mass of static bar or beam (kg)
      • K = Factor to allow for energy loss at impact
      • l = length of bar (l)
      • tp = time period (s)
      • v = velocity (m/s)
      • W = Weight - moving body(N)
    • V = Velocity (m/s)
    • w = Specific Weight (kg/m3)
    • σ = stress (N/m2)
    • σ stat = stress resulting from static load(N/m2)
    • δ = deflection (m)
    • δ stat = deflection resulting from static load(m)
    • μ = Ratio Moving Mass/Stationary bar
    • β = Constant = A Sqrt(wEg/W) - see text



      • Linear Impact deflections and stresses (gravity loads)

        Important note: The notes below represent a very simple view of the loading condition and do not consider more real case involving shock waves being propagated through the loaded member or the moving mass

        Consider a loading regime as shown below with a ring of Mass M(kg) with weight W= M.g(N) being dropped through a distance h onto a collar supported by a vertical bar which behaves as a spring with a stiffness of k (N/m).

        The support bar has a length l (m), an Area A (m2 ) With a modulus of elasticity E (N/m2 )


        In practice the weight would impact onto the support which would elastically deform until all of the potential energy has been absorbed.  The support would then contract initiating damped oscillations until the system assumes a stable static position.  The equations below determine the initial maximum deformation which provides the most highly stressed condition.



        In accordance with conservation of energy the potential energy of the weight is converted to elastic strain energy.


        This may be expressed as a quadratic equation


        This is solved for the maximum deflection
        δ
        max as follows


        The weight applied gradually would result in a deflection
        δ
        st thus


        Note:
        The stiffness k = Force /Deflection = F / δ :    E = stress /strain = σ /e = (F /A) /( δ /l)    Therefore k = EA/l

        Substituting this into the equation for δ max results in


        This can be expressed as


        The resultant maximum force is simply
        Pmax = k δ max



        and the resultant maximum stress = σ max = Pmax/ A

        This may be expressed as  
                                                                 

        For the calculation of the stress due to a suddenly applied load with h = 0


        σ max = 2 σ st





        Linear impact deflections and stresses (kinetic impacts )


        Important note: The notes below represent a very simple view of the loading condition and do not consider more real case involving shock waves being propagated through the loaded member or the moving mass

        Impact loads based pimarily on kinetic energy e.g horizontal impacts are treated slightly differently.  For these applications the kinetic energy is converted into stored energy due to elasticity of the resisting element.

        Consider a Mass M(kg) with a velocity of v impacting on a collar which is supported by a bar with a stiffness of K (N/m) - Ignoring gravitational forces.

        The kinetic energy of the mass Mv2 /2 is transformed into stored energy in the support.


        The resulting equation is


        The resultant maximum deflection equals..


        Noting that the static deflection = Wl/AE this equation can be written


        The equivalent maximum stress =


        Noting that the static deflection = Wl/AE this equation can be written





        Beams




        Using similar principles as expressed above it can be easily proved using principles of conservation of energy that.


        and


        If the impact is horizontal instead of vertical . The resulting deformation and stress resulting from the impact are


        Note: The above approximate relationships can been applied generally to most structural systems subject to distortion with the elastic range when subject to impact loading




        Energy losses on impact


        The above equations are very approximate and include many assumptions.  A very important assumption is that all of the energy ( based on h or v2) is used up in producing the same distortion as would result from static loading.  In reality, some kinetic energy is lost in internal friction.  Account can be taken for these losses by multiplying h or v2 by an appropriate factor K.  This factor is derived from the Mass of the moving body (M) and the mass of the beam or bar (M1).  The factor is different for different loading systems ass follows.

        In the equations above h or v2 would be replaced by K.h or K v2
        The K factor is nearest unity when M is large compared to M1 . As an example for the axial impact
        if M1/M is say 0,1 then K = 0,95..
        if M1/M is say 10 then K = 0,15..
        This is illustrated in the figure below


        1) A moving mass M striking axially one end of a bar of mass M1. The other end of the bar being fixed...


        2) A moving mass M striking traversely the end of a beam of mass M1...


        3) A moving mass M striking traversely the center of a beam with simple supported ends of mass M1....


        4) A moving mass M striking traversely the center of a beam with fixed ends of mass M1. ..





        Impact stresses considering propagation of shock waves- Unsupported bar




        When a impact force is suddenly applied to and elastic body , a wave of stress is propogated traveling through the body with a velocity..


        w = weight /unit volume (kg/m2), v = velocity (m/s)

        The unsupported bar subject to the longitudonal impact from a rigid body with velocity v experiences a wave of compressive stress of intensity
        σ
        .


        If the mass of the moving body is very large compared to the mass of the bar the wave of compression bounces back from the far end of the bar as a wave of tension and returns to the struck end after a time period .
        t p = 2L / V


        If the mass of the moving body is very large compared to the bar so that it can be considered infinite then after breaking contact the moving bar will move away from the impacting mass with a velocity of vb= 2v.   The moving bar will be stress free.

        If the mass of the impinging body is
        μ
        time the mass of the bar then the bar will move away with an average velocity of


        The moving bar is left vibrating with a stress intensity of













        Impacts considering propagation of shock waves- bar with one end supported




        For the case of a bar with one end fixed , the wave of compressive stress resulting from the impact on the unsupported end is reflected back unchanged from the supported (fixed end) and combines with the advancing waves to produce a maximum stress approximately equal to..



        μ
        = Mass of Moving Body/Mass of Bar



        -------------------------------------------------------
        please refer These Threads also -
        http://www.roymech.co.uk/Useful_Tables/Fatigue/Mechanics_Impact.html
        http://www.pdhcenter.com/courses/s164/s164content.pdf
        http://people.virginia.edu/~lz2n/mse209/Chapter8.pdf

        ------------------------------------------

        One thing I want to add - Ironite or any treatment against wear and tear of top surface can never be associate with Impact resistance.
        Wear and tear or abration of an substance or object is measured by its hardness.
        Impact resistance is measured by its toughness or stiffness.
        Fatigue or durability is measured by its soundness.
        Comprressive strength is measured by its crushing value.

        Regards



        Hello sspawar,
        Thank you for sharing this valuableinformation.
        Will you please share more information about Design of Machine Elements?
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