fem slab

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AAPPPPLLIICCAATTIIOONN OOFF PPLLAATTEESS && SSHHEELLLLSS

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Table of ContentsPart A

1. INTRODUCTION TO PLATE AND SHELL ELEMENT 22. ASSIGNMENT 1 SIMPLY SUPPORTED ONE-WAY SLABS 63. ASSIGNMENT 2 TWO-WAY SLABS 114. ASSIGNMENT 3 CANTILEVER BEAM 155. ASSIGNMENT 4 WATER TANKS 17

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global or local directions.

Linearly varying pressure on element surface in local directions.

Temperature load due to uniform increase or decrease of temperature.

Temperature load due to difference in temperature between top and

bottom surfaces of the element.

The distinguishing features of this finite element are:

Displacement compatibility between the plane stress component of one

element and the plate bending component of an adjacent element which is

at an angle to the first is achieved by the elements. This compatibility

requirement is usually ignored in most flat shell/plate elements.

The out of plane rotational stiffness from the plane stress portion of each

element is usefully incorporated and not treated as a dummy as is usually

done in most commonly available commercial software.

Despite the incorporation of the rotational stiffness mentioned previously,

the elements satisfy the patch test absolutely.

These elements are available as triangles and quadrilaterals, with corner

nodes only, with each node having six degrees of freedom.

These elements are the simplest forms of flat shell/plate elements

possible with corner nodes only and six degrees of freedom per node. Yet

solutions to sample problems converge rapidly to accurate answers even

with a large mesh size.

These elements may be connected to plane/space frame members with

full displacement compatibility. No additional restraints/releases are

required.

Out of plane shear strain energy is incorporated in the formulation of the

plate bending component. As a result, the elements respond to Poisson

boundary conditions which are considered to be more accurate than the

customary Kirchoff boundary conditions.

The plate bending portion can handle thick and thin plates, thus extending

the usefulness of the plate elements into a multiplicity of problems. In

addition, the thickness of the plate is taken into consideration in

calculating the out of plane shear.

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The plane stress triangle behaves almost on par with the well known linear

stress triangle. The triangles of most similar flat shell elements incorporate

the constant stress triangle which has very slow rates of convergence.

Thus the triangular shell element is very useful in problems with double

curvature where the quadrilateral element may not be suitable.

Stress retrieval at nodes and at any point within the element.

Plate Element Local Coordinate System

The orientation of local coordinates is determined as follows:

The vector pointing from I to J is defined to be parallel to the local x- axis.

The cross-product of vectors IJ and IK defines a vector parallel to the local

z-axis, i.e., z = IJ x IK.

The cross-product of vectors z and x defines a vector parallel to the local

y- axis, i.e., y = z x x.

The origin of the axes is at the center (average) of the 4 joint locations (3

joint locations for a triangle).

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Results:

Elementsizemxm

No. ofElements

BMkN.m

%error

inBM

Thicknessm

1.00 18 10.20-

10.29 0.150.50 72 10.93 -2.93 0.150.30 200 11.40 1.32 0.150.25 288 11.50 2.17 0.150.20 450 11.60 3.02 0.150.15 800 11.60 3.02 0.150.10 1800 11.60 3.02 0.15

Convergence

-12.00

-10.00

-8.00

-6.00

-4.00

-2.00

0.00

2.00

4.00

18 72 200 288 450 800 1800

No. of eleme nts

%Error

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Exact solution = w*l^2/8 = 11.25 kN.m

Conclusion :

From the above results we can conclude that the bending moments in the

slab converges to exact solution by mesh refinement .

For the element size 0.5m x 0.5m we get minimum +positive error and for

element size 0.3m x 0.3m we get minimum negative error.

Refinement of mesh further does help, and the negative error goes on

increasing till the element size of 0.1m x 0.1m.

PART 2Here we will model the supporting edge with beam elements instead of pinned

supports of different sizes and get the results of bending moments for both slab

and beam and check with exact classical solution.

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Results:

Beamsizemmxmm

BM inslabkN.m

BM inbeamkN.m

%errorin BM

300x300 13.40 40.88 -65.12300x450 12.00 55.45 -21.73300x600 11.61 60.81 -11.00300x750 11.46 63.00 -7.14300x900 11.39 64.03 -5.42

300x1200 11.32 64.88 -4.04

Convergence

-70.00

-60.00

-50.00

-40.00

-30.00

-20.00

-10.00

0.00

300x300 300x450 300x600 300x750 300x900 300x1200

Size of Beam

%Error

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Conclusion :

From the above results it can be concluded that, increasing the size ofedge beams the result converges to exact solution.

Hence relative stiffness of beams and slab is governing criteria forconvergence of the bending moments for both, beam and slab.

If the sizes of supporting beams are lesser than required, there is no slab-beam effect and act together as slab elements with just edged havingmore stiffness.

Again if the sizes of edge beams are different in sizes, the results arehigher for stiffer beam and carry more bending moments than the onewhich is comparatively weaker.

300x900

BM = 69.5kN.m

300x600

BM = 55.6kN.m

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3. AASSSSIIGGNNMMEENNTT 22 TTWWOO--WWAAYY SSLLAABBSS

Problem Statement :

1. Comparative study of two-way slabs bending moments using IS-456guidelines (Table 26) and finite element analysis.

2. Convergence of results for slab with line loads.3. Effect on panel under consideration due to change in size of adjacent

panel.

Data :

Slab Panel : 5m x 6m

Loading : 10kN/m2

Slab : 150mm

Boundary condition : One short edge continuous (case 8)

For the given problem we will start the meshing of panel from 1m x 1m and use

mesh refinement for convergence.

Analysis will be done using staad-prov8i software.

Results

Elementsizemxm

No. ofElements

BM MX+ve

BM MY+ve

BM MX-ve

BM MY-ve

% errorBM

X+ve

% errorBM

Y+ve

% errorBM Y-

ve

1.00 30 11.20 10.20 0.00 12.10 -31.70 -5.39 -17.77

0.50 120 11.30 10.30 0.00 17.40 -30.53 -4.37 18.10

0.25 480 11.40 10.40 0.00 20.90 -29.39 -3.37 31.82

0.20 750 11.50 10.40 0.00 21.70 -28.26 -3.37 34.33

0.10 3000 11.50 10.45 0.00 23.20 -28.26 -2.87 38.58

IS-456 14.75 10.75 0.00 14.25

Convergence

-60.00

-40.00

-20.00

0.00

20.00

40.00

60.00

0 500 1000 1500 2000 2500 3000 3500

No. of Elements

%Erro MX+ve

My +ve

MY -ve

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Conclusion :

For the given case, the mesh refinement converges the positive X and Y

moments for size 0.25m x 0.25m.

From the results it can be concluded that the positive moments from IS-

456 table is higher than results from finite elements analysis after

convergence.

For negative Y moments the results crosses the exact solution for element

size between 1.0m x 1.0m to 0.5m x 0.5m.

So for negative moments further mesh refinement add to positive error

and the results of finite element analysis are more than the one calculated

from IS-456, which is exactly opposite of the case with positive moments.

Also on changing the size of adjacent panel by half of the original, there is

decrease in negative moments at support and increase in mid span

moments.

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4. AASSSSIIGGNNMMEENNTT 33 CCAANNTTIILLEEVVEERR BBEEAAMM

Problem Statement :

Convergence of results for cantilever beam with concentrated load at end.

Data :

Beam span : 3m

Loading : 10kN at free end

Size : 300x600mm

For the given problem we will start meshing size from 0.2m x 0.2m and furtherrefinement will be done to get the convergence.

Results :

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Element sizemxm

No. ofElements

BMkN.m

%errorBM

0.2mx0.2m 45 25.8 -16.500.15mx0.15m 80 27.4 -9.550.10mx0.1m 180 28.6 -4.730..05mx0.5m 720 29.511 -1.66

Convergence

-20.00

-15.00

-10.00

-5.00

0.00

0 100 200 300 400 500 600 700 800

No. of Elements

%Erro

BM

Conclusion :

For the given problem we can conclude that the result for bendingmoments converges to exact solution for the plate size of 0.05m x 0.05m.

Also the results vary for different width of beam, here we tried to convergethe results for beam size of 300x600mm

The results for width other than 300mm, the bending moments aredifferent for same number of elements.

Hence unlike classical solution, where we dont take the effect of width incalculation of bending moments, in finite element analysis the width of

beam influences the bending moments.

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5. AASSSSIIGGNNMMEENNTT 44 WWAATTEERR TTAANNKKSS

Problem Statement :

Comparison of .moments, shear and hoop forces for water tanks using FEMand IS-3370.a) Rectangular tanksb) Cylindrical tanks

Data :

a) Rectangular tanks

Tank size : 3.75m x 5.0m hieght

Loading : 50kN/m

Wall: 150mm

For the given problem we will use the meshing of panel 0.25m x 0.25m as we

have concluded this size from above examples

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Vertical Table 1

y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)

b/H x/H MY staad %error MY staad %error MY staad %error

0.75 0 0.00 0.49 100.00 0.00 0.08 100.00 0.00 -0.36 100.00

1/4 1.25 1.65 24.24 0.00 0.42 100.00 -2.50 -1.84 -35.87

1/2 6.25 4.80 -30.21 2.50 1.68 -48.81 -5.00 -3.59 -39.31

3/4 11.25 10.00 -12.50 6.25 4.75 -31.58 -6.25 -3.90 -60.46

1 0.00 3.55 100.00 0.00 2.12 100.00 0.00 -0.72 100.00

Horizontal Table 1


b/H x/H MX staad %error MX staad %error MX staad %error

0.75 0 0.00 1.02 100.00 0.00 0.15 100.00 0.00 -1.23 100.00

1/4 7.50 6.68 -12.28 2.50 1.06 -135.85 -15.00 -10.91 -37.45

1/2 13.75 12.90 -6.59 3.75 2.48 -51.27 -27.50 -21.53 -27.76

3/4 13.75 14.02 1.93 6.25 3.93 -59.03 -31.25 -24.42 -27.951 0.00 2.54 100.00 0.00 1.09 100.00 0.00 -3.52 100.00

Vertical Table 2



0.75 0 0.00 -0.15 100.00 0.00 -0.11 100.00 0.00 -0.59 100.00

1/4 1.25 1.00 -25.00 0.00 -0.27 100.00 -3.75 -2.21 -69.68

1/2 6.25 4.34 -44.11 2.50 1.42 -75.56 -5.00 -3.64 -37.36

3/4 12.50 14.02 10.83 7.50 4.71 -59.40 -6.25 -3.89 -60.59

1 0.00 3.55 100.00 0.00 2.12 100.00 0.00 -0.72 100.00

Horizontal Table 2y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)


0.75 0 6.25 5.35 -16.82 1.25 0.75 -66.67 -10.00 -6.42 -55.69

1/4 10.00 8.11 -23.30 2.50 0.89 -179.64 -16.25 -12.86 -26.38

1/2 13.75 13.13 -4.72 5.00 2.39 -109.21 -27.50 -21.76 -26.36

3/4 15.00 9.89 -51.67 5.00 3.91 -27.91 -32.50 -24.45 -32.92

1 0.00 2.54 100.00 0.00 1.09 100.00 0.00 -3.52 100.00

Vertical Table 3



0.75 0 0.00 -0.12 100.00 0.00 -0.09 100.00 0.00 -0.61 100.00

1/4 1.25 0.91 -36.91 0.00 -0.05 100.00 -2.50 -2.51 0.40

1/2 6.25 5.13 -21.83 2.50 2.00 -25.00 -3.75 -3.26 -15.03

3/4 8.75 7.76 -12.76 3.75 4.00 6.25 -3.75 -2.77 -35.38

1 -30.00 -20.67 -45.14 -18.75 -11.91 -57.43 0.00 -1.07 100.00

Horizontal Table 3



0.75 0 5.00 5.37 6.82 1.25 0.75 -66.67 -8.75 -6.57 -33.18

1/4 10.00 7.87 -27.10 2.50 0.97 -157.73 -13.75 -12.54 -9.65

1/2 12.50 11.74 -6.48 3.75 2.47 -51.82 -21.25 -19.79 -7.39

3/4 8.75 9.72 9.98 3.75 3.19 -17.55 -16.25 -17.81 8.75

1 -6.25 -3.18 -96.36 -3.75 -1.85 -103.14 0.00 -1.07 100.00

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Conclusion :

For the water tank wall panels, with the given panel ratio and wallthickness the results are having 50% error

This error may be due to the fact that IS-3370 coefficients does not takeinto effect the thickness of wall panel of tank apart from the width andheight.

b) Cylindrical tanks

Tank size : 5m (Height) x 6m (Radius)

Loading : 50kN/m

Wall: 150mm

Boundary conditions : Top of tank Free and bottom hinged

IS 3370 : Table 12,13 and 14.

Results

Hoop Tension IS 3370 Moment My IS 3370

7.95 -0.03 -0.01 0.00

37.05 29.38 -0.02 0.00

66.60 59.10 -0.06 0.00

97.65 89.75 -0.13 -0.13

131.40 122.55 -0.18 -0.13

168.60 159.47 -0.04 -0.12

206.10 197.78 0.56 0.03

230.40 228.12 1.89 1.53229.35 225.07 2.78 3.29

119.85 153.30 3.66 4.17

42.75 0.00 1.57 0.00

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Vertical Moments

0

1

2

3

4

5

6

-1.00 0.00 1.00 2.00 3.00 4.00 5.00

kNm

mt.

IS 3370

Staad

Hoop Tension

0

1

2

3

4

5

6

-100.00 0.00 100.00 200.00 300.00

kN

mt.

IS 3370

Staad

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Conclusion:

For the given case, the results of staad pro and IS 3370 coefficients

are nearly same with negligible, 1% error in case of hoop tension and

about 13% error in moments.

fem slab

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