# Cantilevers, one way slabs

The structural frame
The construction
The reinforcement I
The reinforcement II
Quantity/Cost estimation
Detailing drawings
Introduction >

Wind and Seismic Forces >
Structural model and Analysis
Slabs
Seismic behavour of frames
Appendix A
Appendix B
Appendix C
Appendix D
Introduction >
Modelling slabs

Materials
To be continued >
Introduction

## Static analysis

Static analysis of cantilevers (as well as one-way slabs) takes into account that loads are applied on 1.00 m wide strips.

## General

One-way slabs are those supported on two opposite edges.
If a one-way slab is supported on more than two edges and its aspect ratio, i.e. the ratio of the larger to the smaller theoretical span, is greater than 2.0 (such as slab s3 in the same figure), it is considered as one-way slab in the principal direction while taking into account the secondary stresses in rest edges.

## Static analysis

Continuous one-way slabs are analysed considering a frame of continuous member s of rectangular shape cross-section, having width equal to 1.00 m and height equal to the thickness of the slab. The strip loads comprise self-weight, dead and live loads.

Analysis is performed:

a) approximately, by applying all design loads p=1.35g+1.50q (when live load is relatively small)

## Example: Three-span continuous slab

Bending moment diagram

The three slabs (previous figure) have L1=4.50 m, h1=180 mm, g1=10.0 kN/m2, q1=2.0 kN/m2, L2=4.00 m, h2=140 mm, g2=5.0 kN/m2, q2=2.0 kN/m2, L3=4.00 m, h3=140 mm, g3=5.0 kN/m2, q3=2.0 kN/m2, where loads g include self-weight. Perform static analysis considering global loading in ultimate limit state.

Design load in each slab is equal to pi=γg×gi+γq×qi=1.35×gi+1.50×qi, thus on 1.00 m wide strip, it is:

p1=1.35×10.0+1.50×2.0=16.5 kN/m

p2=p3=1.35×5.0+1.50×2.0=9.75 kN/m

The three-span continuous slab will be solved through Cross method.

Fundamental design span moments (table b3)

M10=-p1×L12/8=-16.5×4.502/8=-41.8 kNm

M12=M21=-p2×L22/12=-9.75×4.002/12=-13.0 kNm

M23=-p3×L32/8=-9.75×4.002/8=-19.5 kNm

Moments of inertia I

I01=Ic=1.0×0.183/12=4.86×10-4 m4

I12=I23=1.0×0.143/12=2.29×10-4 m4=0.47Ic

Stiffness factors k, distribution indices υ

V01=16.5×4.50/2-22.6/4.50=32.1 kN

V10=-16.5×4.50/2-22.6/4.50=-42.1 kN

V12=9.75×4.00/2+(-13.9+22.6)/4.00=21.7 kN

V21=-9.75×4.00/2+(-13.9+22.6)/4.00=-17.3 kN

V23=9.75×4.00/2+13.9/4.00=23.0 kN

V32=-9.75×4.00/2+13.9/4.00=-16.0 kN

maxM01=32.12/(2×16.5)=31.2 kNm

maxM12=21.72/(2×9.75)-22.6=1.5 kNm

maxM23=16.02/(2×9.75)=13.1 kNm

## The elastic line of the three slabs

Front view of the elastic line (from pi-FES with active module\SLABS) (project <B_451>)

## Effect of live load on the static analysis of one-way slabs Accurate analysis method (2)

Analysis with maximum design live loads qd,i=0.35gi+1.50qi

## Effect of live load on the static analysis of one-way slabs Accurate analysis method (3)

Analysis with maximum design live loads qd,i=0.35gi+1.50qi

Minimum support moments

## Effect of live load on the static analysis of one-way slabs Accurate analysis method (4)

Analysis with maximum design live loads qd,i=0.35gi+1.50qi

## Example

The continuous slab shown in the figure, of span length L=5. 00 m and of thickness h=160 mm, is subjected to covering load ge=1.0 kN/m2 and live load q=5.0 kN/m2. Concrete class C50/60. Calculate the shear forces and bending moments envelopes for the three slabs, in ultimate limit state.

The design dead load for each slab is gd=1.00·5.0=5.0 kN/m and the total design load is pd=γg·g+γq·q=1.35·5.0+1.50·5.0=14.25 kN/m.

Manual calculations:   I=(b·h3)/12=(1.0·0.163)/12=341·10-6 m4

Modulus of elasticity for concrete C50/60 is equal to E=37.3 GPa.

E·I=37.3·109N/m2·341·10-6m4=12.719·106 N·m2

For I10=I12=I23=Ic, stiffness factors k distribution indices υ are:

Due to frame symmetry:  και

## Analysis using the software (project <<B_453-1>)

Shear force envelope for q=5.0 kN/m2

Bending moment envelope for q=5.0 kN/m2

Deflection envelope for q=5.0 kN/m2

Shear force diagram for p=14.25 kN/m

Bending moment diagram for p=14.25 kN/m

Deflection diagram for pd=14.25 kN/m

Comparing the results of analysis with global loading and unfavourable loadings, relatively small differences regarding shear forces are noticed, while significant differences regarding span bending moments and deflections are mainly registered at the mid-span.

## Effect of live load on the static analysis of one-way slabs Simplified method for envelope estimation

 2 spans 3 spans 4 spans Infinite spans

Unfavourable loadings may not be considered in static analysis of slabs of common buildings, provided that the support and span bending moments are multiplied by an incrementation factor according to the following.

• ·In case of one way slabs with almost equal spans and load ratio "dead" / "live" = gd/qd = 1.0g / (0.35g + 1.50q) 1.0 or gd/pd = 1.0g / (1.35g+1.50q) 0.50 (valid for most cases of loads used for resintential design), the incrementation factors of bending moments may be taken from the above table.

• ·In case of two-way slabs having almost equal spans, similar increments are valid although reduced by 50%.

• ·Regarding shear forces, the average increment for all support types is approximately 10%.