One Way Continous Slab-Worked Example

One way slabs are the most basic form of slab and beam construction, generally employed for utilitarian purposes in offices, retail developments, warehouses, stores and similar buildings. They are very economical for small to moderate spans of between 4-6m.

This post presents a practical example of a continuous one-way slab. It is an extension of the previous post on designing a concrete slab to Eurocode. Kindly review that post before assimilating the contents of this post.

Worked Example

The figure shown below is the floor layout of a shopping mall building. The chosen scheme for this structure consists of a solid one way slabs and deep beams. Preliminary sizing of the slabs has been carried out and is indicated on the drawing. The slab is to be designed from C20/25 concrete and reinforcement steel bars with a yield strength of 410mpa.

Actions

The slab thickness has been established already, therefore, we can start by determining the design value of the actions on the slab.

Permanent Actions

a.\quad self\quad weight\quad of\quad slab\quad =0.175\times 25=4.35kN/{ m }^{ 2 }

b.\quad finsishes\quad \& \quad services\quad say\quad =1.5kN/{ m }^{ 2 }

c.\quad partion\quad allowance\quad say\quad =1kN/m^{ 2 }

Permanent\quad Actions\quad { g }_{ k }\quad =6.875kN/{ m }^{ 2 }

Variable Actions

The floor imposed load is taken from the U.K National Annex to part 1-1 of Actions on Structures BS-EN 1991-1-1.

a.\quad floor\quad imposed\quad load(shopping\quad areas)\quad =4kN/{ m }^{ 2 }

b.\quad movable\quad partions\quad =1kN/{ m }^{ 2 }

Variable\quad Actions\quad { q }_{ k }=5kN/{ 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 6.875)+(1.5\times 5)=16.1kN/{ m }^{ 2 }

Analysis of Slab

The slab is a one-way continuous slab and by inspection the conditions for applying coefficient in the analysis are satisfied. since:

Area of each bay = (4.8×27)=129.6m²>30m²
The ratio of the variable actions to permanent actions =5/6.875=0.72<1.25
Imposed load =5kN/m²
The spans are all equal (4.8m)

F={ n }_{ s }l=16.1\times 4.8=77.28kN


	END SUPPORTS A&E	END SPANS A-B & D-E	FIRSR INTERIOR SUPPORTS B & D	INTERIOR SPANS B-C & C-D	INTERIOR SUPPORT C
Moment (kN.m)	0.04FL=14.83	0.075FL=27.82	0.086FL=31.90	0.063FL=23.37	0.063FL=23.37
Shear (kN)	0.46F=35.55	-	0.6F=46.34	-	0.5F=38.64

Cover

{ c }_{ nom }={ c }_{ min }+{ \triangle c }_{ dev }

{ c }_{ min }=15+10\quad =25mm

Flexural Design

Haven analyzed and determine the cover to the slab, the next step is to provide reinforcement in the slab.

End Support A & E

{ M }_{ Ed }=14.83kN.m

assuming\quad 12mm\quad bars

d=h-({ c }_{ nom }+\phi /2)\quad =175-(25+12/2)=144mm

k=\frac { M }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 14.83\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times 144^{ 2 }\times 20 } =0.0358

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.95d=0.95\times 144=136.8mm

{ A }_{ s }=\frac { M }{ 0.87{ f }_{ yk }z } =\frac { 14.83\times { 10 }^{ 6 } }{ 0.87\times 410\times 136.8 } =303.91{ mm }^{ 2 }

Try\quad Y12-250mm\quad c/c\quad ({ A }_{ s,prov }=452{ mm }^{ 2 })

End Span A-B (Same as D-E)

{ M }_{ Ed }=27.82kN.m\quad ;\quad d=144mm

k=\frac { M }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 27.82\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times 144^{ 2 }\times 20 } =0.0671

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.94d=0.94\times 144=135.36mm

{ A }_{ s }=\frac { M }{ 0.87{ f }_{ yk }z } =\frac { 27.82\times { 10 }^{ 6 } }{ 0.87\times 410\times 135.36 } =576.19{ mm }^{ 2 }

Try\quad Y12-175mm\quad c/c\quad ({ A }_{ s,prov }=645{ mm }^{ 2 })

First Interior Supports B & D

{ M }_{ Ed }=31.90kN.m\quad ;\quad d=144mm

k=\frac { M }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 31.90\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times 144^{ 2 }\times 20 } =0.077

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.93d=0.93\times 144=133.92mm

{ A }_{ s }=\frac { M }{ 0.87{ f }_{ yk }z } =\frac { 31.90\times { 10 }^{ 6 } }{ 0.87\times 410\times 133.92 } =667.8{ mm }^{ 2 }

Try\quad Y12-150mm\quad c/c\quad ({ A }_{ s,prov }=753{ mm }^{ 2 })

Interior Spans B-C & C-D (Same as Interior Support C)

{ M }_{ Ed }=23.37kN.m\quad ;\quad d=144mm

k=\frac { M }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 23.37\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times 144^{ 2 }\times 20 } =0.056

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.95d=0.95\times 144=135.36mm

{ A }_{ s }=\frac { M }{ 0.87{ f }_{ yk }z } =\frac { 23.37\times { 10 }^{ 6 } }{ 0.87\times 410\times 135.36 } =484.02{ mm }^{ 2 }

Try\quad Y12-225mm\quad c/c\quad ({ A }_{ s,prov }=502{ mm }^{ 2 })

Now that the reinforcement have been provided in all sections of the slab, the next step will be to verify deflection.

Deflection Verification

By observation, deflection is more critical in the end spans i.e. span A-B and D-E. Therefore the verification will be carried out only for the end spans.

Limiting L/d ratio

\left[ \frac { l }{ d } \right] _{ limit }=N\cdot k\cdot F1\cdot F2\cdot F3

\rho =\frac { { A }_{ s,req } }{ bd } =\frac { 576.19 }{ { 10 }^{ 3 }\times 144 } =0.004

{ \rho }_{ o }={ 10 }^{ -3 }{ f }_{ ck }={ 10 }^{ -3 }\times 20\quad =0.0045 ;\quad i.e\quad \rho <{ \rho }_{ o }

N=\left[ 11+\frac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2\sqrt { { f }_{ ck } } \left( \frac { { \rho }_{ o } }{ \rho } -1 \right) ^{ 3/2 } \right]

=\left[ 11+\frac { 1.5\sqrt { 20 } \times 0.0045 }{ 0.004 } +3.2\sqrt { 20 } \left( \frac { 0.0045 }{ 0.004 } -1 \right) ^{ 3/2 } \right]

N=19.18

k=1.3(end\quad span)\quad ;\quad F1=1.0\quad ;\quad F2=1.0

F3=\frac { 310 }{ { \sigma }_{ s } } \le 1.5

{ \sigma }_{ s }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { g }_{ k }+{ \psi q }_{ k } }{ { n }_{ s } } \right] \left( \frac { { A }_{ s,req } }{ { A }_{ s,pro } } \right) \cdot \frac { 1 }{ \delta }

\\ =\frac { 410 }{ 1.15 } \left[ \frac { 6.875+0.6(5) }{ 16.1 } \right] \frac { 576.19 }{ 645 } \cdot 1=195.35mpa

F3=\frac { 310 }{ 195.35 } =1.58 \le1.5; =1.5

\left[ \frac { l }{ d } \right] _{ limit }=19.18\times 1.3\times 1.0\times 1.0\times 1.5=37.40

Actual L/d ratio

\left[ \frac { l }{ d } \right] _{ Actual }=\frac { span }{ effective\quad depth } =\frac { 4800 }{ 144 } =33.33

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

Detailing Checks

Minimum Area of Steel

{ A }_{ s,min }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } bd\le 0.0013bd

{ f }_{ ctm }=0.3{ { f }^{ 2/3}_{ ck } }=\quad 2.21mpa

=0.26\times \frac { 2.21 }{ 410 } \times { 10 }^{ 3 }\times 144\ge 0.0013\times { 10 }^{ 3 }\times 144

=201.81{ mm }^{ 2 }\quad o.k

Maximum Spacing

{ \sigma }_{ s }=195.35mpa\quad ;\quad spacing\quad 275mm

Crack Control

As the slab is less than 200mm all measures set-out to prevent cracking are not necessary

Finally, the design summarized above in figure 3 together with the general arrangement drawing illustrated in figure1 is combined to detail the slab to normal best practice.

THANKYOU FOR READING!!!

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