Designing a Troughed Floor -Worked Example

Introduction

In the design of reinforced concrete element, the tensile stress in the concrete element is assumed to be fully carried by the reinforcement. This is because the tensile strength of concrete is negligible and considered inefficient in resisting tensile stresses. It, therefore, follows that the volume of concrete in the tensile zone is ineffective in resisting tensile stresses. Thus, by introducing voids to the soffits of slabs or replacing them with lighter materials reduces the dead weight of the slab and increases the efficiency of the concrete section in bending. This is the principle of the ribbed floor system.

The ribbed floor system is very useful in long-span floors up to 12m, it provides a lighter and stiffer slab than an equivalent flat slab, reducing the extent of foundations. They also provide a very good form where slab vibration is an issue, such as laboratories and hospitals

A troughed floor is almost not different from the ribbed floor except that it combines the advantages of a ribbed floor and those of level soffits. These lightweight floors provide longer spans than one-way solid or flat slabs. They create level soffits and the provision of holes causes little or no problem in the ribbed area. To provide a level soffit the depth of the beams are kept equal to that of the ribs, as a result, the depths of the section depends on the width of the beam used, which tend to be very wide and heavily reinforced in order to control deflection.

Geometry

The width of ribs is governed by cover and bar spacing. The spacing between ribs is typically between 600, 750, 900, and 1200mm depending on the span. With an average width of 150mm, the self-weight of the slab can be estimated using the following expression

self\quad weight\quad =\frac { 25 }{ y } \left\{ r\left[ \frac { s }{ c } \left( w+k \right) +t \right] +h\left( u+v \right) \right\}

The terms required for estimating the self weight of the floor from above expression are defined in figure 1.

Figure 1: Terms for estimating a troughed floor self weight

Analysis and Design

The analysis of a trough slab is undifferent from that of a continuous one-way slab. Therefore, coefficients for one way slabs are also applicable and may be used to obtain the moment and shear forces provided the conditions set-out for their use are satisfied. i.e.

  • The area of each bay exceed 30m2
  • The ratio of the characteristic imposed loads to dead load doesn’t exceed 1.25
  • The characteristic imposed load doesn’t exceed 5kN/m2
  • Spans are approximately equal ( generally within 0.85 of the maximum span)

Alternatively, the slab can be analyzed using any chosen method of elastic analysis assuming simple supports.

The ribs behave like beams and therefore the rules defined for beams are followed. The section is designed as a solid section at the support while at the span as a flanged section. Light reinforcement or fabric mesh based on the minimum area of steel is to be provided in the topping to avoid cracking. However, if the spacing between ribs exceeds 900mm, the toppings must be designed for moment and shears assuming one-way action between ribs.

Worked Example

A troughed floor is required for a shopping mall. The layout of the floor is shown in figure 2. Preliminary sizing of the structural components of the floor have been carried out and is indicated on the figure. The floor is to constructed from C25/30 concrete and steel bars having a yield strength of 500mpa.

Figure 2a: Troughed Slab Layout
Figure 2a: Section through Ribs

Guidance on how to size a trough slab can be found in the post on preliminary sizing of structural elements or the concrete Centre publication: Economic concrete frame elements.

Actions

The sizes of the troughed floor components has been carried out, therefore as usual we proceed with the actions on the floor.

Permanent Actions
a.\quad self\quad weight\quad =\frac { 25 }{ y } \left\{ r\left[ \frac { s }{ c } \left( w+k \right) +t \right] +h\left( u+v \right) \right\}
\quad =\frac { 25 }{ 9.6 } \left\{ 8.1\left[ \frac { 0.35 }{ 0.75 } \left( 0.15+0.0167 \right) +0.1 \right] +0.45(0.75+0.75 \right\}
\quad =5.50kN/{ m }^{ 2 }
b.\quad finishes\quad \& \quad services\quad =1.70kN/{ m }^{ 2 }
Permanent\quad Actions\quad { g }_{ k }=5.5+1.70 =7.2kN/{ m }^{ 2 }
Variable Actions
a.\quad Imposed\quad load(shopping\quad areas)=4kN/{ m }^{ 2 }
b.\quad movable\quad partitions\quad say\quad =1kN/{ m }^{ 2 }
Varaible\quad Actions\quad { q }_{ k }=4+1=5.0kN/{ m }^{ 2 }
Design Value of Actions

By inspection, the permanent actions are less than 4.5 times the variable actions , therefore equation 6.13 can be used.

{ n }_{ s }=1.35{ \xi g }_{ k }+1.5{ q }_{ k }
=(1.35\times 0.925\times 7.2)+(1.5\times 5)=16.5kN/{ m }^{ 2 }

Analysis of Slab

A trough slab is basically a one way slab and by inspection the conditions necessary for applying coefficients of one way continuous slab in analysis are met. Therefore, we can use coefficient to analyze this slab.

Fig 3: Analysis of Trough Slab
F={n}_{s}l=16.5\times9.6=158.4kN/{ m }^{ 2 }
{ c }_{ nom }={ c }_{ min }+{ \triangle c }_{ dev }
{ c }_{ nom }=20+5=25mm

Flexural Design

The slab have been analyzed and the cover to reinforcement determine, the next step is to provide longitudinal reinforcement in the ribs.

End Supports A & C
{ M }_{ Ed }=60.83\times 0.75=45.62kN.m/rib
assuming\quad 8mm\quad links\quad and\quad 16mm\quad bars
d=h-({ c }_{ nom }+links+\phi /2)=450-(25+8+16/2)=409mm
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 45.62\times { 10 }^{ 6 } }{ 750\times { 409 }^{ 2 }\times 25 } =0.015
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 409=388.55
{ A }_{ s }=\frac { 45.62\times { 10 }^{ 6 } }{ 0.87\times 500\times 388.55 } =269.9{ mm }^{ 2 }/rib
Try\quad 2H16/rib\quad ({ A }_{ s,prov }=402{ mm }^{ 2 })
End Spans A-B & B-C
{ M }_{ Ed }=114.05\times 0.75=85.54kN/rib
assuming\quad 8mm\quad links\quad and\quad 20mm\quad bars
d=h-({ c }_{ nom }+links+\phi /2)=450-(25+8+20/2)=407mm
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 85.54\times { 10 }^{ 6 } }{ 750\times { 407 }^{ 2 }\times 25 } =0.028
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 407=386.65
verify\quad neutral\quad axis
x=2.5(d-z)=2.5(407-386.65)=50.88mm
50.88<100mm\quad therefore\quad design\quad as\quad rectangular\quad beam
{ A }_{ s }=\frac { 85.54\times { 10 }^{ 6 } }{ 0.87\times 500\times 386.65 } =508.6{ mm }^{ 2 }/rib
Try\quad 2H20/rib\quad ({ A }_{ s,prov }=628{ mm }^{ 2 })
Interior Support B
{ M }_{ Ed }=130.75\times 0.75=98.06kN/rib;\quad d=407mm
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 98.06\times { 10 }^{ 6 } }{ 750\times { 407 }^{ 2 }\times 25 } =0.0316
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 407=386.65
{ A }_{ s }=\frac { 98.06\times { 10 }^{ 6 } }{ 0.87\times 500\times 386.65 } =583.02{ mm }^{ 2 }/rib
Try\quad 2H20/rib\quad ({ A }_{ s,prov }=628{ mm }^{ 2 })

Shear Design

Haven determined the longitudinal bars, the next step is to determine the shear reinforcement in the ribs. The critical section for shear is taken at d from the supports and the width of the rib is taken at the middle of reinforcements.

Shear is more critical at the interior support, therefore will be used to size the shear reinforcement in the rib

Support B
{ V }_{ Ed }=(95.04-0.407\times 16.5)\times =88.41kN/m
=88.41\times 0.75=66.30kN/rib
{ V }_{ Rd,c }=0.12k(100\rho { f }_{ ck })^{ 1/3 }{ b }_{ w }d\ge{V}_{Rd,min}
{ b }_{ w }=150+2(43tan10)=159.6mm
k=1+\sqrt { \frac { 200 }{ d } } =1+\sqrt { \frac { 200 }{ 407 } } =1.7<2
\rho =\frac { { A }_{ s } }{ { b }_{ w }d } =\frac { 314 }{ 159.6\times 407 } 0.0048\quad (assuming\quad 1H20)
{ V }_{ Rd,c }=0.12\times 1.7(100\times 0.0048\times 25)^{ 1/3 }\times 159.6\times 407
=30.40kN
{ V }_{ Rd,min }=0.035{ k }^{ 3/2 }\sqrt { { f }_{ ck } } { b }_{ w }d
{ V }_{ Rd,min }=0.035\times { 1.7 }^{ 3/2 }\sqrt { 25 } \times 159.6\times 407
=25.2kN
{ V }_{ Ed }>{ V }_{ Rd }\quad (66.30kN>30.40kN)
Therefore\quad shear\quad reinforcement\quad is\quad required
Shear Reinforcement
\theta =0.5{ sin }^{ -1 }\left( \frac { 5.56{ V }_{ Ed } }{ { b }_{ w }d(1-{ f }_{ ck }/250){ f }_{ ck } } \right)
\theta =0.5{ sin }^{ -1 }\left( \frac { 5.56\times 66.30\times { 10 }^{ 3 } }{ 159.6\times 407(1-25/250)25 } \right) ={ 7.6 }^{ \circ }
cot\theta =7.5>2.5\quad therefore\quad cot\theta =2.5
\frac { { A }_{ sv } }{ { s }_{ v } } \ge \frac { { V }_{ Ed } }{ z{ f }_{ ywd }cot\theta } \quad z=0.9d=0.9\times 407=366.3mm
\frac { { A }_{ sv } }{ { s }_{ v } } \ge \frac { 66.30\times { 10 }^{ 3 } }{ 366.3\times 500\times 2.5 } =0.15
max\quad spacing\quad =0.75d=0.75\times 366.3=301.5mm
Use\quad H8\quad @\quad 250mm\quad c/c\quad (0.34)

Deflection Verification

limiting (L/d) ratio
\left[ \frac { l }{ d } \right] _{ limit }=N\cdot k\cdot F1\cdot F2\cdot F3
\rho =\frac { { A }_{ s } }{ ch+{ b }_{ w }(d-h) }
=\frac { 508.8 }{ (750\times 100)+159.6(407-100) } =0.0041
\\ { \rho }_{ o }={ 10 }^{ -3 }\sqrt { { f }_{ ck } } ={ 10 }^{ -3 }\sqrt { 25 } =0.005\quad since\quad \rho <{ \rho }_{ o }
N=\left[ 11+\frac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2{ f }_{ ck }\left( \frac { { \rho }_{ o } }{ \rho } -1 \right) ^{ 3/2 } \right]
N=\left[ 11+\frac { 1.5\sqrt { 25 } \times 0.005 }{ 0.0041 } +3.2\sqrt { 25 } \left( \frac { 0.005 }{ 0.0041 } -1 \right) ^{ 3/2 } \right]
=21.80
k=1.3(end\quad span)
F1\quad =0.82\quad F2=\frac { 7.0 }{ l } =\frac { 7.0 }{ 9.6 } =0.73
F3=\frac { 310 }{ { \sigma }_{ s } } \le 1.5
{ \sigma }_{ s }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { g }_{ k }+{ q }_{ k } }{ { n }_{ s } } \right] \frac { { A }_{ s,req } }{ { A }_{ s,pro } } \cdot \frac { 1 }{ \delta }
=\frac { 500 }{ 1.15 } \left[ \frac { 7.2+0.6(5) }{ 16.5 } \right] \frac { 508.6 }{ 628 } \cdot 0.93=202.5mpa
F3=\frac { 310 }{ 202.5 } =1.5
\left[ \frac { l }{ d } \right] _{ limit }=21.80\times 1.3\times 0.82\times 0.73\times 1.5=25.44
Actual (L/d) ratio
\left[ \frac { l }{ d } \right] _{ Actual }=\frac { span }{ effective\quad depth } =\frac { 9600 }{ 407 } =23.58

Therefore since the actual l/d ratio is less than the limiting l/d ratio, deflection is o.k

Minimum Areas Verification

{ A }_{ smin }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } { b }_{ t }d\ge 0.0013{ b }_{ t }d
{ f }_{ ctm }=0.3{ { f^{ 0.66 } }_{ ck } }=0.3\times { 25 }^{ 0.66 }=2.56mpa
=0.26\frac { 2.56 }{ 500 } \cdot { 10 }^{ 3 }\times 100\ge 0.0013\cdot { 10 }^{ 3 }\times 100=133.12{ mm }^{ 2 }
Use\quad A142\quad Mesh\quad in\quad Topping

The figure shown below is the design summary of the trough slab. Detailing of the slab is done by combining figure 3 with the layout in line with best practice.

Figure 4: Summary of Design

To conclude this design example, we are going to design the band beam on gridlines B-B.

BAND BEAM ON GRIDLINE B-B

The size of the band beams are indicated on the floor layout (figure 1). The band beam is 450mm deep and 1500 wide.

Action on Beam

Permanent Actions
a.\quad load\quad from\quad troughed\quad floor\quad =9.6\times 7.2\times 1.1=76kN/m
b.\quad self\quad weight\quad of\quad beam\quad =\quad 8.75kN/m
Permanent\quad Action\quad { G }_{ k }=76+8.75=84.75kN/m
Variable Actions
{ Q }_{ k }=9.6\times 5\times 1.1=45.4kN/m
Design Values of Actions
{ w }_{ max }=1.35\xi { G }_{ k }+1.5{ Q }_{ k }
=(1.35\times 0.925\times 84.75)+(1.5\times 45.4)=174kN/m
{ w }_{ min }=1.35\xi { G }_{ k }
=(1.35\times 0.925\times 84.75)=105.83kN/m

Analysis of Beam

Figure 5: Analysis of Band Beam

We can’t apply coefficient in analyzing this beam because the spans are largely unequal. Analysis of a sub-frame must, therefore, be carried out to determine the design moments and shears. Only the moment and shear diagrams are presented in this post. The procedure for analyzing frames have been covered in a previous post, kindly review it if in doubt. How to Analyse Element in Frames

Figure 6: Bending Moment & Shera Force Diagrams for Beam

Flexural Design

Support 1 (same as 5)
{ M }_{ Ed }=394kNm
assuming\quad 8mm\quad links\quad \& \quad 20mm\quad bars\quad
d=h-(c+links+\phi /2)=450-(30+8+20/2)=390mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 394\times { 10 }^{ 6 } }{ 1500\times { 390 }^{ 2 }\times 25 } =0.070<0.168
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
z=0.93d=0.93\times 390=359.91mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 394\times { 10 }^{ 6 } }{ 0.87\times 410\times 359.91 } =2497.2{ mm }^{ 2 }
Try\quad 9H20’s\quad ({ A }_{ s,prov }=2826{ mm }^{ 2 })
Span 1-2 (same as 4-5)
{ M }_{ Ed }=313kNm\quad \quad d=390mm
b={ b }_{ w }+{ b }_{ eff,1 }+{ b }_{ eff,2 }
{ b }_{ eff,1 }={ b }_{ eff,2 }=0.2b+0.1{ l }_{ o }\le 0.2{ l }_{ o }\le b
b=\left( \frac { 9600-750-750 }{ 2 } \right) =4050mm
{ l }_{ o }=0.85l=0.85\times 6000=5100mm
{ b }_{ eff,1 }=(0.2\times 4050)+(0.1\times 5100)\le 0.2(5100)\le 4050\\ { b }_{ eff,1 }={ b }_{ eff,2 }=1020mm
b=1500+1020+1020=3540mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 313\times { 10 }^{ 6 } }{ 3540\times { 390 }^{ 2 }\times 25 } =0.024<0.168
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
z=0.935=0.95\times 390=370.5mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 394\times { 10 }^{ 6 } }{ 0.87\times 410\times 370.51 } =1942.1mm
Try\quad 8H20’s\quad ({ A }_{ s,prov }=2512m{ m })^{ 2 }
Support 2 (same as 4)
{ M }_{ Ed }=769kNm\quad
assuming\quad 8mm\quad links\quad and\quad 25mm\quad bars
d=450-(30+8+25/2)=387.5mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 769\times { 10 }^{ 6 } }{ 1500\times { 387.5 }^{ 2 }\times 25 } =0.137<0.168
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d\\ z=0.86=0.86\times 387.5=333.25mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 769\times { 10 }^{ 6 } }{ 0.87\times 410\times 333.25 } =5304.77mm^{ 2 }
Try\quad 12H25’s\quad ({ A }_{ s,prov }=5892{ mm }^{ 2 })
Span 2-3 (Same as 3-4)
{ M }_{ Ed }=456kNm\quad \quad d=387.5mm
b={ b }_{ w }+{ b }_{ eff,1 }+{ b }_{ eff,2 }
{ b }_{ eff,1 }={ b }_{ eff,2 }=0.2b+0.1{ l }_{ o }\le 0.2{ l }_{ o }\le b
b=\left( \frac { 9600-750-750 }{ 2 } \right) =4050mm
{ l }_{ o }=0.7l=0.7\times 7500=5250mm
{ b }_{ eff,1 }=(0.2\times 4050)+(0.1\times 5250)\le 0.2(5250)\le 4050
1335\le 1050\le 2050
b=1500+1050+1050=3600mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 456\times { 10 }^{ 6 } }{ 3600\times { 387.5 }^{ 2 }\times 25 } =0.031<0.168
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95=0.95\times 387.5=368.13mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 456\times { 10 }^{ 6 } }{ 0.87\times 410\times 368.13 } =2847.61mm
Try\quad 10H20\quad ({ A }_{ s,prov }=3240{ mm }^{ 2 })
Support 3
{ M }_{ Ed }=836.5kNm\quad d=387.5mm
k=\frac { { M }_{ Ed } }{ { bd }^{ 2 }{ f }_{ ck } } =\frac { 836.5\times { 10 }^{ 6 } }{ 1500\times { 387.5 }^{ 2 }\times 25 } =0.149<0.168
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
z=0.84d=0.84\times 387.5=325.5mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 836.5\times { 10 }^{ 6 } }{ 0.87\times 500\times 325.5 } =5907.8mm^{ 2 }
Try\quad 14H25’s\quad ({ A }_{ s,prov }=6874{ mm }^{ 2 })

Shear Design

The shear force is critical at support C.

{ V }_{ Ed }=661-(0.3875\times 174)=593.6kN
{ V }_{ Rd,c }=0.12k(100\rho { f }_{ ck })^{ 1/3 }{ b }_{ w }d
k=1+\sqrt { \frac { 200 }{ d } } =1+\sqrt { \frac { 200 }{ 387.5 } } =1.72
\rho =\frac { { A }_{ s } }{ b_{ w }d } =\frac { 3240 }{ 1500\times 387.5 } =0.0055
{ V }_{ Rd,c }\quad =0.12\times 1.72(100\times 0.0055\times 25)^{ 1/3 }\cdot 1500\times 387.5
=287.41kN
{ V }_{ Rd,min }=0.035{ k }^{ 2/3 }\sqrt { { f }_{ ck } } { b }_{ w }d
=0.035\times { 1.72 }^{ 2/3 }\sqrt { 25 } \times 1500\times 387.5\\ =229.5kN
{ V }_{ Ed }>{ V }_{ Rd }\quad (593.6>287.4kN)
Therefore\quad shear\quad reinforcement\quad is\quad required
Shear reinforcement
\theta =0.5{ sin }^{ -1 }\left( \frac { 5.56{ V }_{ Ed } }{ { b }_{ w }d(1-{ f }_{ ck }/250){ f }_{ ck } } \right)
\theta =0.5{ sin }^{ -1 }\left( \frac { 5.56\times 593.4\times { 10 }^{ 3 } }{ 1500\times 387.5(1-25/250)25 } \right) ={ 7.31 }^{ \circ }
cot\theta =7.8>2.5\quad therefore\quad cot\theta =2.5
\frac { { A }_{ sv } }{ { s }_{ v } } \ge \frac { { V }_{ Ed } }{ z{ f }_{ ywd }cot\theta } \quad z=0.9d=0.9\times 387.5=348.75mm
\frac { { A }_{ sv } }{ { s }_{ v } } \ge \frac { 594.4\times { 10 }^{ 3 } }{ 348.75\times 500\times 2.5 } =1.35
max\quad spacing\quad =0.75d=0.75\times 387.5=290.6mm
\frac { { A }_{ sw,min } }{ { s }_{ v } } =\frac { 0.08\sqrt { { f }_{ ck } } { b }_{ w } }{ { f }_{ yk } } =\frac { 0.08\sqrt { 25 } \times 1500 }{ 500 } =1.2
Use\quad 6H8-175mm\quad (1.72)

This concludes the design of the band beam as well as the floor, clearly, deflection is not critical in this beam and doesn’t need to be checked. The detail drawing for the band beam can be obtained easily from figure 7. When detailing the beam, minimum areas of steel and bar spacing must be checked in line with the code requirements and best practice.

Figure 7: Summary of Band Beam Design

THANKYOU FOR YOU TIME!!!

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

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