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Trapezoidal loads shear - moment diagram
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Ronan
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PostPosted: Tue Jun 28, 2011 6:42 am    Post subject: Trapezoidal loads shear - moment diagram Reply with quote

Hi all,

I'm experiencing a difficulty understanding how the trapezoidal loads are distributed and how to shear moment diagrams are drawn for structural members subjected to trapezoidal loading.

For example below example is frame member subject to trapeze load, due to the fact that this memeber is extracted from 3D frame building it has initial shear and  moment at ends resulted from fixed - end moments(equivalent nodal loads) loading.

So my hand-calculations is as below and want to compare them with software calulated values.
Vshear = 150 - q*x2/(2*A);
Mmoment=150*x -q*x3/(2*3*A);

A = triangular horizontal leg-length of trapezoidal load. (A=100cm , span = 400cm; q=1kg/cm )

the above formulas produces the consistent results for boundary conditions and same with software results but in-between values are wrong.

For example if x  = 30cm then Vshear = 145.5 kg where as software finds 142.5 kg.

http://i55.tinypic.com/1zdcfop.jpg

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Any comments will be appreciated.
Regards,
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vikram.jeet
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PostPosted: Tue Jun 28, 2011 12:27 pm    Post subject: Trapezoidal loads shear - moment diagram Reply with quote

Generally trapezoidal and triangular loadings on slabs can be converted into
equivalent u.d loadings for BM as well as for Shear

As we all know (nothing new)

Triangular loading on shorter side beams of slab
equivalent UDL
For BM computations w (equivalent) = w*Lx /3
For SF computations w (equivalent) = w*Lx /4

Trapezoidal loading on Longer side beams of slab  
equivalent UDL
For BM computations w (equivalent) =( 3-m^2)*w*Lx /6
For SF computations w (equivalent) = ( 2-m)*w*Lx /4

where Lx/Ly=m
Lx= short span of slab

From the reaction computed using equivalent udl, the BM
diagram can be drawn for intermediate values using actual loading
of trapezium

However BM at midspan will be =w(equivalent) *L^2/8
L may be Lx or Ly for that direction

best regds

vikramjeet






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Ronan
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PostPosted: Tue Jun 28, 2011 1:19 pm    Post subject: Reply with quote

Your comments will really be appreciated!!!
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suresh_sharma
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PostPosted: Wed Jun 29, 2011 2:32 am    Post subject: Reply with quote

Remove moment from the support. Most likely you will get the same shear from manual calculation that you got from the software. your formula is based on as if there is no moment at the support.

Please also advise the software that you have used for modeling the beam and obtaining the forces.
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vikram.jeet
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PostPosted: Wed Jun 29, 2011 4:53 am    Post subject: Trapezoidal loads shear - moment diagram Reply with quote

Er Ronan

Value of calculated SF of 145.5 Kg is OK

Software must be dividing the triangular area into small(tiny) rectangular areas
i.e adopting simpson's rule to calculate area and thus small difference
in computer value from actual one.

SF due to end moments is zero since both the moments are equal and
hogging

best regds

vikramjeet
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suresh_sharma
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PostPosted: Wed Jun 29, 2011 8:30 am    Post subject: Reply with quote

Dear Vikramjeet,

What you have said is correct for designing purpose because the loading suggested by you and practised by most of the designers before the advent of software for analysis and design is almost OK and acceptable but definitely approximate. But Ronan's query is based on exact calculation and distribution of load not the approximate one.
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sspawar
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PostPosted: Wed Jun 29, 2011 11:36 am    Post subject: Reply with quote

I agree with Er. Suresh , - Software will consider fixed end moments and forces also.
Remove moment from the support. Most likely you will get the same shear from manual calculation that you got from the software. your formula is based on as if there is no moment at the support.
Regards

suresh_sharma wrote:
Remove moment from the support. Most likely you will get the same shear from manual calculation that you got from the software. your formula is based on as if there is no moment at the support.

Please also advise the software that you have used for modeling the beam and obtaining the forces.
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vikram.jeet
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PostPosted: Wed Jun 29, 2011 12:04 pm    Post subject: Trapezoidal loads shear - moment diagram Reply with quote

Er Ronan

Value of calculated SF of 145.5 Kg is OK

Software must be dividing the triangular area into small(tiny) rectangular areas
i.e adopting simpson's rule to calculate area and thus small difference
in computer value from actual one.

SF due to end moments is zero since both the moments are equal and
hogging

best regds

vikramjeet
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vikram.jeet
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PostPosted: Wed Jun 29, 2011 12:20 pm    Post subject: Trapezoidal loads shear - moment diagram Reply with quote

Dear Er Suresh Sharma ji

Yes this was practiced during manual design era.

I agree with you that this is equivalent ud loading and derived from the fact
that simply supported midspan Mo due to Triangular/Tapezoidal(as the case may be)
will produce same BM as that produced by equivalent uD load.
This may be exact for working out mid span mo in a simply supported beams .
However in frames, the FEM calculated from this loading may have approximation
within acceptable limits.

best regds

vikramjeet



Dear Vikramjeet,

What you have said is correct for designing purpose because the loading suggested by you and practised by most of the designers before the advent of software for analysis and design is almost OK and acceptable but definitely approximate. But Ronan's query is based on exact calculation and distribution of load not the approximate one
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sankar
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PostPosted: Thu Jun 30, 2011 8:39 am    Post subject: Trapezoidal loads shear - moment diagram Reply with quote

The equivalent loading concept equate mid-span moment, however, total loading in the span increased approximately by 25%.


Regards,
Sankar Chakraborty

--Original Message--
From: vikram.jeet [mailto:forum@sefindia.org]
Sent: Thursday, June 30, 2011 11:39 AM
To: general@sefindia.org
Subject: [SEFI] Re: Trapezoidal loads shear - moment diagram

Dear Er Suresh Sharma ji

Yes this was practiced during manual design era.

I agree with you that this is equivalent ud loading and derived from the fact
that simply supported midspan Mo due to Triangular/Tapezoidal(as the case may be)
will produce same BM as that produced by equivalent uD load.
This may be exact for working out mid span mo in a simply supported beams .
However in frames, the FEM calculated from this loading may have approximation
within acceptable limits.

best regds

vikramjeet



Dear Vikramjeet,

What you have said is correct for designing purpose because the loading suggested by you and practised by most of the designers before the advent of software for analysis and design is almost OK and acceptable but definitely approximate. But Ronan's query is based on exact calculation and distribution of load not the approximate one

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