Different Methods of Slab Analysis

Abstract

This article briefly explains the different methods of slab analysis and design. These methods include simple Eurocode 2 method, yield line method, Hilleborg strip method and Finite element method. The main objective of this article is to analyze the slab using different methods and compare their results. The report shows a brief introduction of all these method with their history, limitations, advantages and disadvantages. The report also shows all the necessary calculations involved in the analysis and design of slab. At the end the report compares all these methods with their results

Introduction

Slab is one of the main components of structure system. It is flexural member resisting tension and compression at a time. Thickness of the slab should be enough to resist the shear stresses. So Normally Shear reinforcement is not provided in the slab. Usually, slab takes area load perpendicular to its plane. But sometimes it has also designed for concentrated loads. For example, a heavy machinery is installed at the center of slab. Since slab is the most important structure member so proper analysis and design of slab is very important. There are different types of slab depending on materials, size, loading criteria. Few of them are flat slab, waffle slab, one-way ribbed slab, two-way slab supported by beams etc. There are different methods available for the design of slab such as moment coefficient method, yield line method, strip method, direct design method and finite element method. Application of specific method is based on a type and scope of the slab. For example, Rectangular slab may be designed by any of the method, but triangular slab is not designed by moment coefficient method. Finite element method is based on computer program because of number of algebraic equations while other methods may be done using hand calculations. Some of these methods are based on elastic theories while few methods calculate the collapse load (Inelastic)

Literature Review/Methodology

Yield Line Theory

Conventional Slab system is designed based on moment coefficients and these coefficients are calculated based on elastic calculations. Whilst in strength method of design, concrete’s strength is taken up to ultimate capacity which is calculated based on inelastic calculation. In this way, a contradiction arises in the analysis and design of structure elements. Limit or plastic designed eliminates this elastic and inelastic difference and perform ultimate analysis of structure member.

Conventional slab design has a lot of restrictions: for example, slab should be rectangular, square, having uniform distributed load and no opening inside the slab. But in real life, slab may have curved shape, triangular shape, concentrated load or large opening. So, yield line method is used is a type of plastic analysis which perform inelastic analysis of slab and also used for complex shapes like triangular, circular etc.

When load is applied on a slab, it resists this load up to its nominal strength. But when load is further increasing plastic hinges (Inelastic rotation) occurs at a point of maximum bending moment. Then the load transfer into other region of slab and form a pattern of crack, called yield lines. This method is based on upper bound theorem which follows virtual work method. According to this to make a structure safe, summation of all the external work due to loading is equal to the internal work due to moment and rotation should be equal.

This method is applicable on slab having:

o?? Under reinforced section

o?? Class B or C steel (Ductile)

o?? x/d should not increase 0.25 for C50/60 concrete

o?? Slab with complex shapes

Disadvantages of Yield line analysis

o?? Required experience to predict actual collapse mechanism

o?? Does not show the behavior of slab under service load because all calculations are done on the basis of ultimate load.

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Hilleborg Strip method

This method is based on lower bound theorem which states that if we found the stress distribution in a structure that satisfies the equilibrium without disturbing the yield line, then the structure is safe against the external load.?

This method was presented by Hillerborg in early 1950s and further studied by wood and Armer in 1960s which is currently formed. This method assumes that at the failure point, external load is carried out by bending separately, either by X and Y axis, without causing any torsion or twisting moment. This method is easily applied to any slab but difficult for the slab that are resting directly on column. For this Hilleborg introduced new method called advanced strip method.

Assumptions

It is also assumed that distribution of reinforcement is known in both directions.

This method is used as tool to recheck the design or it is used as iteration method unless satisfactory result may not come.

In this method, first slab is divided into different moment fields that satisfy the equilibrium, then reinforcement is calculated on each moment field.

Advantages of Strip method

The results that are found using this method are on safer side, which is practically prefer.

It is design method, so amount of reinforcement can be calculated using this method.

It can be applied to any slab having complex geometry or slab with large opening.

Finite Element Analysis of Slab

FEA is one of the most power and versatile computer programs used for the analysis of two- or three-dimensional structure system that use the concept of ODE (ordinary differential equation) and PDE (Partial differential equation). FEA use displacement as an unknown to solve the multiple matrixes. In this method slab is divided into a number of elements that has finite size. Each element is connected with node. The divided element may be rectangular, square, triangle etc. In this way divided element made mesh. Using different algebraic equation setting the displacement as an unknown, each divided element is analyzed and generate a whole member result. The smaller the size of divided element, the more accurate are the results.

Advantages of FEA

o?? It can easily apply on the slab having complex geometry

o?? It can easily apply on the slab having large opening

o?? Give estimate of displacement of complex slab

o?? Easily apply on the slab having un-usual loading like concentrated loading

o?? As it is computer-based program, so it takes less time

Disadvantages of FEA

o?? Older version of software takes much time to set the model

o?? Redistribution of moments are not easily attained

o?? Difficult to understood for new users

o?? Human error may occur while modeling the slab and this error is not easily checked.

o?? This method needs Engineering judgments to predict the behavior of concrete while designing through this method

It was developed in 1951. As this method requires huge mathematical calculations hence it was limited in academic before the advancement in computer program. First, FEA was applied to a plate structure by R J Melosh in 1961

Design and calculation

Provision of Eurocode 2

This method is a simplest way of design the slab. This method designs the slab on the provision of moment coefficients which depends on the aspect ratio of the slab and boundary conditions. Moments are calculated using table given by Eurocode 2. Reinforcement and area of steel is calculated using these moments

Limitations

o?? Moments are calculated based on elastic theories.

o?? Only apply on the slab with simple geometry like Rectangle or square.

o?? Slab having no concentrated loads

o?? Slab having no opening

Figure 1 Plan of Slab

Assume the size of slab with opening = 6m x 5m

Slab with concentrated loads = 9m x 5m

Assume = fck = 30 N/mm2

fyk = 500 N/mm2

Load calculations

Dead load of slab = 24 x 0.2 = 4.8 KN/m2

Assume other permanent loads = 3KN/m2

Total Permanent load = 4.8 + 3 = 7.8 KN/m2

Assume live load = 3KN/m2

Ultimate load = 1.35(7.8) + 1.5(3)

?????????????????????? = 15.03 KN/m2

Slab = 6m x 5m

Ly = 6m, Lx = 5m

Ly/Lx = 6/5 = 1.2

Figure 2 Moment coefficients in two-way slab

Negative moment = 0.078 x 15.03 x 52 =29.30 KN.m

Positive moment (5m) = 0.059 x 15.03 x 52 =22.16 KN.m

Positive moment (6m) = 0.059 x 15.03 x 62 =31.92 KN.m

Bending moment M/bd2fck

.d = 200 – 20 = 180mm

K = (29.30 10^6)/1000180^2*30 = 0.030< 0.132

Z/d= [0.5 + (0.25 – 0.030/0.9) ^1/2 = 0.96 > 0.95

Z= 0.95 x 180 = 171mm

As = (M)/ 0.87*fyk*z

???? = (29.30 10^6)/0.87500*171

???? = 393.9mm2/m

Provide H10 bar at 180mm center to center

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Bending moment = 22.16 KN.m

.d = 200 – 20 = 180mm

K = (22.16 10^6)/1000180^2*30 = 0.022< 0.132

Z/d= [0.5 + (0.25 – 0.030/0.9) ^1/2 = 0.97 > 0.95

Z= 0.95 x 180 = 171mm

As = M/ 0.87*fyk*z

???? = (22.16 10^6)/0.87500*171

???? = 297.90mm2/m

Provide H10 bar at 250mm center to center

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Bending moment = 31.92 KN.m

.d = 200 – 20 = 180mm

K = (31.92 10^6)/1000180^2*30 = 0.032< 0.132

Z/d= [0.5 + (0.25 – 0.030/0.9) ^1/2 = 0.96 > 0.95

Z= 0.95 x 180 = 171mm

As = M/ 0.87*fyk*z

???? = (31.92 10^6)/0.87500*171

???? = 429.11mm2/m

Provide H10 bar at 180mm center to center

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Slab = 9m x 5m

Ly = 9m, Lx = 5m

Ly/Lx = 9/5 = 1.8

Negative moment = 0.087 x 15.03 x 52 =32.69 KN.m

Positive moment (5m) = 0.065x 15.03 x 52 =24.42 KN.m

Positive moment (9m) = 0.065 x 15.03 x 92 =79.13 KN.m

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Bending moment = 32.69 KN.m

.d = 200 – 20 = 180mm

K = (32.69 10^6)/1000180^2*30 = 0.033< 0.132

Z/d= [0.5 + (0.25 – 0.030/0.9) ^1/2 = 0.96 > 0.95

Z= 0.95 x 180 = 171mm

As = M/ 0.87*fyk*z

???? = (32.69 10^6)/0.87500*171

???? = 439.47mm2/m

Provide H10 bar at 170mm center to center

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Bending moment = 79.13 KN.m

.d = 200 – 20 = 180mm

K = (79.13 10^6)/1000180^2*30 = 0.0814< 0.132

Z/d= [0.5 + (0.25 – 0.0814/0.9) ^1/2 = 0.96 > 0.95

Z= 0.89 x 180 = 160.2mm

As = M/ 0.87*fyk*z

???? = (79.13 10^6)/0.87500*160.2

???? = 1135.5mm2/m

Provide H12 bar at 100mm center to center

Yield Line Theory

Figure 5 Yield line mechanism

M’ = support moment

M = field moment

Assume = support moment = field moment

According to virtual work method

External work done = Internal work done

External work = Collapse load x Area x Displacement

Dead load of slab = 4.8 KN/m2

Other load = 6KN/m2

Total load Wu = 10.8 KN/m2

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For region A

10.8 x 5 x 5 x 1/3 Δ max = 90 Δ max

300 x 1/5 x 0.5 = 30 Δ max x 2 = 60 Δ max

For region B

10.8 x 4.5 x 5 x 1/2 Δ max = 108 Δ max

400 x 1/4 x 0.5 = 50 Δ max

Total external load = 90 + 60 + 108 + 50 = 308 Δ max

Internal Work done = moment x rotation x length of projected yield line

Region 1 = m’ x 1/5 x 5 = m’

??????????????? m x 1/5 x 5 = m

Region 2 = m x 1/2.925 x 5 = 1.71m

Region 3 = m x 1/2.08 x 5 = 2.40m

Region 4 & 5 = 2 x m x 1/2.5 x 4 = 3.2m

As assumed m = m’

Total Internal work done = 9.31m

Equating the virtual work equation Δ max = 1

308 = 9.31m

Moment = 33.08 KN.m

Slab with opening

Figure 6 Yield line mechanism of slab with opening

6m – 1.3 (1.2) = 5.1m

External work done

For Region A

10.8 x 5.1 x 5 x 1/3 Δ max = 91.8 Δ max

For region B

10.8 x 0.9 x 5 x 1/3 Δ max = 16.2 Δ max

Total external load = 108 Δ max

Internal Work done = moment x rotation x length of projected yield line

Region 1 = mx 1/5.1 x 5 = 0.98m’

Region 2 = m x 1/1.9 x 5.1 = 2.68m

Region 3 = m x 1/1.9 x 5.1 = 2.68m

Region 4 = m x 1/0.9 x 5 = 5.55m’

As assumed m = m’

Total Internal work done = 11.89m

Equating the virtual work equation Δ max = 1

108 = 11.89m

Moment = 9.08 KN.m/m

Design

Bending moment = 33.08 KN.m

.d = 200 – 20 = 180mm

K = (33.08 10^6)/1000180^2*30 = 0.0814< 0.132

Z/d= [0.5 + (0.25 – 0.0814/0.9) ^1/2 = 0.96 > 0.95

Z= 0.95 x 180 = 171mm

As = M/ 0.87*fyk*z

???? = (33.08 10^6)/0.87500*171

???? = 445mm2/m

Provide H10 bar at 175mm center to center

Bending moment = 9.08 KN.m

.d = 200 – 20 = 180mm

K = (9.08 10^6)/1000180^2*30 = 0.009< 0.132

Z/d= [0.5 + (0.25 – 0.0814/0.9) ^1/2 = 0.98 > 0.95

Z= 0.95 x 180 = 171mm

As = M/ 0.87*fyk*z

???? = (9.08 10^6)/0.87500*171

???? = 123mm2/m

Provide H10 bar at 350mm center to center

Hilleborg Strip method

Figure 7 Distribution of stresses

Wu = 10.8 KN/m2

Strip 1

Right Moment = 5.4 x 1.25 x 1.25/2 = 4.21 KN-m

Left moment = 300 x 0.5 + 5.4 x 1.25 x 1.25/2 = 154.21 KN-m

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Strip 2

Support moment = 10.8 x 1.25 x 1.25/2

???????????????????????? = 8.4375 KN-m

Vertical strip 3

Support moment = 5.4 x 1.25 x 1.25/2

???????????????????????? = 4.21875 KN-m

Vertical Strip 4

Max moment = Wl2/8

?????????????????? = 10.8 x 52 /8

??????????????????? = 33.75 KN-m/m

Bending moment = 33.75 KN.m

.d = 200 – 20 = 180mm

K = (33.75 10^6)/1000180^2*30 = 0.0814< 0.132

Z/d= [0.5 + (0.25 – 0.0814/0.9) ^1/2 = 0.96 > 0.95

Z= 0.95 x 180 = 171mm

As = M/ 0.87*fyk*z

???? = (33.75 10^6)/0.87500*171

???? = 455mm2/m

Provide H10 bar at 170mm center to center

Slab with opening

Wu = 10.8 KN/m2

Design for 1m strip

Strip 1

Moment left = 5.4 x 1.25 x 1.25/22 = 4.21875 KN-m

Strip 2

Moment left = 10.8 x 1.25 x 1.25/22 = 8.43 KN-m

Strip 3 vertical

10.8 x 52 /8 = 33.75 KN-m/m

Horizontal Strip (Right side)

W (0.5 x 0.25 + 1.75) + w (0.5 x 0.25) + 154.21 = 10.8 x 1.25 x 1.25/2 + 0.5

2w = -139.031 KN/m

W = 69.5 KN/m

Moment = 69.5 x 0.5 x 0.5/2 = 8.6875 KN-m

Comparison of different methods with results

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? Conventional slab system is designed on the basis of moment coefficients and these moment coefficients are calculated using linear elastic theories. This method is applying on the slab supported by beams, wall or column, but the panel should be square or rectangular. Load should be uniform while Yield line analysis, strip analysis and finite element analysis is applied on each type of slab, either it has complex geometry or concentrated load.

The main difference between yield line analysis and strip analysis is that results of strip analysis are always on safer side, while yield line analysis’s result may be on unsafe side. Strip analysis is a tool for the design of slab by which amount and distribution of tensile steel is calculated whilst yield line is a tool to predict the collapse load of a given structure.

As these both methods are limit-based method, so neither of them provides any information regarding cracking in a structure under serviceable condition.

As per economy is concerned, as result of strip analysis are always on safer side provides more reinforcement as compared to yield line analysis. Hence, slab design by yield line analysis is more economical as strip method. Strip method is better to use for slab having large opening because yield line often led to uniform bar spacing.

Compared to finite element method with others, FEA is a computer-based program. It has a lot of deep understanding to set the limit of computer program. Although the results obtained through this method is more accurate but if limits are not set properly it may lead very irrational results.

Results of this report using strip analysis and yield line analysis are quite closer except one strip. Because of Eurocode 2 method’s limitations results are very different. Results obtained through FEA has little bit difference du e to support condition apply in computer program.

References

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o?? Baumann, R.A. and Weisgerber, F.E., 1983. Yield-line analysis of slabs-on-grade.?Journal of Structural Engineering,?109(7), pp.1553-1568.

o?? Park, R. and Gamble, W.L., 1999.?Reinforced concrete slabs. John Wiley & Sons.

o?? Hillerborg, A., 2012.?Strip method design handbook. CRC Press.

o?? Jofriet, J.C. and McNeice, G.M., 1971. Finite element analysis of reinforced concrete slabs.?Journal of the structural division,?97(3), pp.785-806.

Written by

Engineer Fahad Yaqoob

MSc, BSc Civil Engineering

Northwestern Polytechnic University

[email protected]

[email protected]

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