Design of Ribbed Slabs with Polystyrene (EPS)

Polystyrene slab

Ribbed slabs incorporate voids to the soffits of slabs or replace voids with lighter materials. Generally, where the span of a solid slab is very large such that the thickness of the slab exceeds 200mm, they are considered to be uneconomical. Apart from being uneconomical, the cumulative weight of the slab leads to very large structural memebers which in-turn impacts significantly on the foundation. Thus, ribbed slabs are one of the closest effective and possible solutions available in such scenarios.

Ribbed slabs are designed ignoring the volume of concrete below the neutral axis of the slab. This is based on the assumption that the volume of concrete below the neutral axis is almost useless and all tensile stresses are fully resisted by the reinforcement only. Hence removing the concrete below the neutral axis or replacing them with lighter materials reduces the dead loads of slabs and increase the efficiency of the concrete section.

Several light-weight materials have been incorporated into the voids of ribbed slabs. This includes Expanded Polystyrene (EPS) Clay bricks, Terra-Cotta Blocks, amongst others. EPS is primarily advantageous in terms of cost, excellent fire resistance, sustainability benefit, improvement in terms of energy efficiency and it’s generally lighter. On the flip side, it is less stiff when compared to the clay bricks and blocks. However, the assumption is that these materials do not contribute to the structural strength of the slab, hence stiffness is not an issue.

Figure 1: Expanded Polystyrene (EPS)

The analysis and design of ribbed slabs with EPS is straightforward and handled in the same way as one-way slabs. This has already been covered in previous posts, however, we’re going to repeat the process in this post. The basic difference is that the weight of the polystyrene is calculated and included as an action on the slab.

See: Designing a Troughed Floor-Worked Example

Worked Example

The figure shown below is the structural layout of a ribbed slab, for an office building,  the voids are to be replaced with high-density EPS. Design the flexural and shear reinforcement required in the ribs and verify deflection. Assuming the slab will be constructed from C30/37 concrete, 460Mpa bars and EPS with a density of 18kg/m3.

Worked example on Ribbed Slabs with EPS
Figure 1: Structural Layout of Ribbed Slab
Worked Example Section through Ribbed Slabs with EPS
Figure 2: Section through Ribs
Actions

Preliminary sizing of the ribbed slab has been carried out based on the recommendation of span: depth ratio in the Concrete Centre manual “Economic concrete element

Permanent Actions:

The permanent actions includes the weight of toppings, rib, finishes and EPS

topping\quad =0.075\times 25\times 0.60\quad =1.125kN/m/rib
ribs\quad =0.15\times 0.225\times 25=\quad 0.84kN/m/rib
finishes\&services\quad @1.5kN/{ m }^{ 2 }\quad =1.5\times 0.6\quad =1.13kN/m/rib
EPS\quad @18{ kN/m }^{ 3 }\quad =0.45\times 0.225\times \frac { 18 }{ { 10 }^{ 3 } } \times 9.81 =0.018kN/m/rib
permanent\quad actions\quad { g }_{ k }=3.11kN/m/rib
Variable Actions
floor\quad imposed\quad load\quad @2.5kN/{ m }^{ 2 }\quad =2.5\times 0.6=1.5kN/m/rib
demountable\quad partitions\quad @2.5kN/{ m }^{ 2 }=0.5\times 0.6=0.3kN/m/rib
variable\quad actions\quad { q }_{ k }=1.8kN/m/rib
Design Actions

By inspection the permanent actions are less than 4.5 times the variable actions. Thus equation 6.10b of BS EN 1990 is more critical

{ w }_{ d }=1.35{ \xi g }_{ k }+1.5{ q }_{ k }
=\left( 1.35\times 0.925\times 3.11 \right) +\left( 1.5\times 1.8 \right) \quad =6.58kN/m
Structural Analysis

The slab is a one way continuous ribbed slab with equal span and by inspection all conditions required for carrying out analysis using simple coefficients are met. Therefore, the simplified coefficients for one way slab will be used to analyze the slab.

At midspan of (1-2) and (2-3)

{ M }_{ Ed }=0.075{ w }_{ d }{ l }^{ 2 }=0.075\times 6.58\times 7{ .5 }^{ 2 }=27.76kN/rib

At Interior Support 2

{ M }_{ Ed }=0.086{ w }_{ d }{ l }^{ 2 }=0.086\times 6.58\times 7{ .5 }^{ 2 }=31.84kN/rib

The maximum shear force occurs at the interior panel. Thus, this will be used to design the shear reinforcement.

{ V }_{ max }=0.6{ w }_{ d }l=0.6\times 6.58\times 7.5=29.6kN
Flexural Design
Midspan (1-2) & (2-3)
{ M }_{ Ed }=27.76kN.m

Assuming a cover of 25mm, longitudinal reinforcement of 12mm and links of 8mm. Effective depth:

d=h-\left( c+\frac { \phi }{ 2 } +links \right) =300-(25+12/2+8)=261mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 27.76\times { 10 }^{ 6 } }{ 600\times { 261 }^{ 2 }\times 30 } =0.023
z=d\left[ 0.5+\sqrt { 0.25-0.822k } \right] \le 0.95d
=0.95d\quad =0.95\times 261\quad =247.95{ mm }

By inspection, the neutral axis is the section flange, therefore the section can be designed as a rectangular section.

{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 27.76\times { 10 }^{ 6 } }{ 0.87\times 460\times 247.95 } =279.8{ mm }^{ 2 }

Try 3T12mm bars @ bottom of ribs (Asprov = 339mm2/rib)

Interior Support 2
{ M }_{ Ed }=31.84kN.m
d=h-\left( c+\frac { \phi }{ 2 } +links \right) =300-(25+12/2+8)=261mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 31.84\times { 10 }^{ 6 } }{ 600\times { 261 }^{ 2 }\times 30 } =0.026
z=d\left[ 0.5+\sqrt { 0.25-0.822k } \right] \le 0.95d
=0.95d\quad =0.95\times 261\quad =247.95{ mm }
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 31.84\times { 10 }^{ 6 } }{ 0.87\times 460\times 247.95 } =320.87{ mm }^{ 2 }

Try 3T12mm bars @ Top of ribs (Asprov = 339mm2/rib)

Verify minimum area of steel
{ A }_{ s,min }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } bd\ge 0.0013bd
=0.26\frac { 2.9 }{ 460 } \times 150\times 261\ge 0.0013\times 150\times 261
=64.18{ mm }^{ 2 }<339{ mm }^{ 2 }\quad o.k

Therefore, adopt bars provided from flexural design.

Shear Design

The design shear force is taken at a distance d from the interior support, hence:

{ V }_{ Ed }=29.6-\left( 0.261\times 6.58 \right) \quad =27.88kN
Concrete Shear Resistance
\rho =\frac { { A }_{ s } }{ bd } =\frac { 339 }{ 150\times 261 } =0.0087
k=1+\sqrt { \frac { 200 }{ d } } =1+\sqrt { \frac { 200 }{ 261 } } =1.88
{ V }_{ Rd,c }=0.12k\left( 100\rho { f }_{ ck } \right) ^{ 1/3 }bd\ge 0.035{ k }^{ 3/2 }\sqrt { { f }_{ ck } } bd
=0.12\times 1.88\left( 100\times 0.0087\times 30 \right) ^{ 1/3 }\times 150\times 261\\ \ge 0.035\times { 1.88 }^{ 3/2 }\sqrt { 25 } \times 150\times 261
=26.21kN

Since the design shear force is greater than concrete shear strength, shear reinforcement is required and should be provided.

Shear Reinforcement
\theta =\quad 0.5{ sin }^{ -1 }\left[ \frac { 5.56{ V }_{ Ed } }{ { b }_{ w }d\left( 1-\frac { { f }_{ ck } }{ 250 } \right) { f }_{ ck } } \right]
=0.5{ sin }^{ -1 }\left[ \frac { 5.56\times 27.88\times { 10 }^{ 3 } }{ 150\times 261\left( 1-\frac { 30 }{ 250 } \right) 30 } \right] =4.31^{ \circ }
cot\theta \quad =14.33\quad take\quad cot\theta =2.5
z=0.9d\quad =0.9\times 261=234.9mm
\frac { { A }_{ s } }{ { s }_{ v } } \ge \frac { { V }_{ Ed } }{ z{ cot\theta f }_{ ywd } } =\frac { 27.88\times { 10 }^{ 3 } }{ 234.9\times 2.5\times \frac { 410 }{ 1.15 } } =0.13
min.\quad spacing\quad =0.75d\quad =0.75\times 261=195.75mm

Use T8-175mm @ Centres (Asprov/Sv- 0.57)

Deflection Verification

Deflection is verified by comparing the actual span: effective depth ratio with the limiting span: effective depth ratio.

Limiting Span-depth Ratio
{ \left[ \frac { l }{ d } \right] }_{ limit }=N\cdot K\cdot F1\cdot F2\cdot F3
\rho =\frac { { A }_{ s,req } }{ { A }_{ c } } =\frac { 279.8 }{ \left( 600\times 75 \right) +150\left( 261-75 \right) } =0.0038
{ \rho }_{ o }={ 10 }^{ -3 }\sqrt { { f }_{ ck } } ={ 10 }^{ -3 }\sqrt { 30 } =0.0055
N=\left[ 11+\frac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2\sqrt { { f }_{ ck } } { \left( \frac { { \rho }_{ o } }{ \rho } \right) }^{ 3/2 } \right] =25.5
\frac { { b }_{ eff } }{ b } =\frac { 600 }{ 150 } =4>3\quad ;\quad F1\quad =0.8
F2\quad =\frac { 7 }{ { l }_{ eff } } =\frac { 7 }{ 7.5 } =0.933\quad ;\quad K=1.3\quad (end\quad spans)
F3=\frac { 310 }{ { \sigma }_{ s } } \le 1.5
{ \sigma }_{ s }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { g }_{ k }+{ \varphi }_{ 2 }{ q }_{ k } }{ { w }_{ d } } \right] \left( \frac { { A }_{ s,req } }{ { A }_{ s,pro } } \right)
=\frac { 460 }{ 1.15 } \left[ \frac { 3.11+0.3(1.8) }{ 6.58 } \right] \frac { 279.8 }{ 339 } =183.1Mpa
F3=\frac { 310 }{ 183.1 } =1.69\le 1.5
\left[ \frac { l }{ d } \right] _{ limit }=25.5\times 1.3\times 0.8\times 0.933\times 1.5\quad =37.11
Actual Span-Depth Ratio
{ \left[ \frac { l }{ d } \right] }_{ Actual }=\frac { span }{ effective\quad depth }= \frac { 7500 }{ 261 } =28.74

Since the actual span-effective depth ratio is less tha the limiting span-effective depth ratio, deflection is deemed satisfactory.

Slab Topping

Provide wire mesh or light reinforcement in the topping to avoid cracks and shrinkage.

Thank you for viewing this post! please share.

Omotoriogun Victor
About Omotoriogun Victor 66 Articles
A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices

1 Comment

  1. Hi there! I know this is kinda off topic but I was wondering which blog platform
    are you using for this site? I’m getting sick and tired of WordPress because I’ve had problems with hackers and I’m looking at alternatives for another platform.

    I would be awesome if you could point me in the direction of
    a good platform.

Leave a Reply

Your email address will not be published.


*