joist slab

7/29/2019 Joist Slab

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One way joist slab (by Prof. A. Charif)

Joists (Ribs) are closely spaced T-beams. The space between the beams may be left void or filled with

light hollow blocks called Hourdis. This type of floor is very popular and offers many advantages

(lighter, more economical, better isolation).

Joists are supported by beams, which are supported by girders or columns.

ACI / SBC Conditions on joist dimensions:

Web width: mmbw 100 Web thickness: ww bh 5.3

Flange thickness:

mm

S

hf 50

12/

Spacing: mmS 800

The flange width is then: Sbb wf

ACI and SBC codes specify that concrete shear strength may be increased by 10 % in joists.

Usually stirrups are not required in joists, but are used to hold longitudinal bars. It is therefore

recommended to consider stirrups when computing longitudinal steel depth.

Analysis and design of joist slabs is equivalent to analysis and design of joist as T-beams. Shrinkage

reinforcement must then be provided in the secondary direction.

Typical joist (rib)

Vertical section

bw S

bf

hw

hf

Void or hollow

block (Hourdis)


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Steps for the analysis and design of a joist slab:

(1) Thickness: Determine minimum thickness and:

If the thickness is unknown choose a value greater or equal to the minimum value

If the thickness is given, check that it is greater or equal to the minimum value

(2) Geometry and Loading: Check the joist dimensions anddetermine the dead and live uniform loading

on the joist (kN/m) using the given area loads (kN/m2) for live load and super imposed dead load as well

as the joist self weight. If hollow blocks (Hourdis) are present, their weight must be added to dead load.

DeadjwbjwjwcjfjfcjD ShhbbxhSDLw )( Live jfjL bxLLw

The ultimate joist load is: jLjDju www 7.14.1

(3) Flexural analysis: Determine the values of ultimate moments at major locations (exterior negative

moment, interior negative moment and positive span moment) using the coefficient method (if conditions

are satisfied) with appropriate clear lengths and moment coefficients.

(4) Flexural RC design: Perform RC design using standard methods starting with the maximum moment

value. Determine the required steel area and compare with code minimum steel area. Determine the bar

number and check bar spacing.(5) Shrinkage reinforcement:

Determine shrinkage (temperature) reinforcement and the corresponding spacing.

(6) Shear check: Perform shear check, that is, check thatuc VV with Vcincreased by 10%

If it is not checked, stirrups must be provided.

(7) Flange check: Part of the flange is un-reinforced. It must be checked as a plain concrete member.

(8) Detailing: Draw execution plans


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One way joist slab example:

The above figure shows a one-way joist slab with beams and girders (same floor as in one

way solid slab example).

Beams are in X-direction (perpendicular to slab strip) and girders are in Y-direction

(parallel to joists). Joists are in Y-direction.

The space between joists is filled by hollow blocks (hourdis) with a density 3/12 mkNb

Concrete: 3' /2425 mkNMPaf cc Steel: MPafy 420

All beams and girders have the same section 300 x 600 mm.

All columns have the same square section 300 x 300 mm.

Superimposed dead load SDL = 1.5 kN/m

2

Live load LL = 3.0 kN/m2

All external beams and girders as well as the internal beam C support a wall with a

uniform weight of mkNwwall /4.14

4.0 m

4.0 m

4.0 m

4.0 m

8.2 m 8.1 m

A

B

C

D

E

1 2 3

T ical oist rib

500120120

50250

Joist Data (mm)


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Solution of joist slab example:

The joist is modeled as a continuous beam with four equal spans

Step 1: Thickness use Table 9.5(a) forhmin

Table 9.5(a): Minimum thickness for beams (ribs) and one-way slabs

unless deflections are computed and checked

Simply

supported

One end

continuous

Both ends

continuous Cantilever

Solid one-

way slabL / 20 L / 24 L / 28 L / 10

Beams

or ribsL / 16 L / 18.5 L / 21 L / 8

Spans 1 and 4: One end continuous mmL

h 22.2165.18

4000

5.18min

Spans 2 and 3: Both ends continuous mmL

h 48.19021

4000

21min

Thus mmh 22.216min The total joist thickness is h = hf+ hw = 50 + 250 = 300 mm

The joist thickness is thus OK (No deflection check required)

Step 2: Geometry and Loading

a) Geometry: Check joist dimensionsWeb width: mmmmbw 100120 Web thickness: mmxbmmh ww 4201205.35.3250

Flange thickness:

mm

mmSmmhf

50

67.4112/50012/50 Spacing: mmmmS 800500

All dimension conditions are satisfied. The flange width is then: mmSbb wf 620120500

b) Loading: Dead: jwbjwjwcjfjfcjD ShhbbxhSDLw )( mkNxxxxxwjD /894.325.05.01225.012.02462.0)05.0245.1(

Live mkNxbxLLw jfjL /86.162.03

The ultimate joist load is: mkNwww jLjDju /614.87.14.1


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Step 3: Flexural analysis

All conditions of ACI/SBC coefficient method are satisfied.

So2)( numu lwCM

2

n

uvu

lwCV

lnis the clear length wu is the factored uniform load

mln 7.32

3.0

2

3.00.4 for all spans

For shear force, span positive moment and external negative moment, lnis the clear length of the span

For internal negative moment, lnis the average of clear lengths of the adjacent spans.

Cm and Cv are the moment and shear coefficients given by ACI tables. The moment coefficients and values

are:


6/13

RC-SLAB1 software output is:


7/13

Step 4: Flexural RC design

Steel depths

b dd

hd 2

cover Cover = 20 mm

Assume bar diameterdb = 12 mm and stirrup diameterds = 8 mm

Thus mmd 2668

2

1220300

RC design for internal negative moment Mu = 11.79 kN.m

We findAs = 121.88 mm2

requiring two 12 mm bars (we may use two 10 mm bars).

We also find that one 12 mm bar is sufficient for bottom reinforcement (for positive moment) and for

external top reinforcement (external negative moment).

RC-SLAB1 design output is:


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Step 5: Shrinkage reinforcement

As in one way solid slabs, shrinkage steel (in secondary slab direction) is equal to minimum steel.

Ashr= Asmin = 0.0018 bh = 0.0018 x 1000 x 50 = 90 mm2

(we consider 1 m strip)

We use a smaller diameter of 10 mm Thus Ab = 78.5 mm2

The spacing is mmx

A

bAS

s

b 2.87290

5.781000

Maximum spacing for shrinkage steel in slabs according to SBC / ACI is:

mmxMinmmhMinS 200)300,504()300,4(max we thus use 10 @200 mm

Step 6: Shear check

Ultimate shear force kNxLwCV njuvu 33.1827.3614.815.1

2

For joists, the nominal concrete shear strength Vc is increased by 10 %.

Thus kNNxdbf

V wc

c 26.29292602661206

251.1

61.1

'

ucc VkNVV 945.2175.0 So shear is OK. No stirrups required

Step 7: Flange check

The flange part between the webs must be checked as a plain concrete member. It is considered as fixed to

both webs with a length equal to spacing S. We consider a 1m strip.

We have a doubly fixed beam with length S= 500 mm = 0.5 m

The section is bxhf= 1000 x 50 mm

The ultimate uniform load is obtained from slab loading:

mkNmxxxmxLLhSDLw fc /88.8137.105.0245.14.117.14.1

The maximum ultimate at fixed ends is: mkNxSw

Mu .185.012

5.088.8

12

22

As the member is un-reinforced, the nominal capacity must consider concrete tension strength, which is

defined by SBC as: MPafct 5.37.0' (for ACI, it is '5.0 ct f ).

S

w


9/13

The nominal moment for a rectangular section with maximum stress equal to tension strength is:

mkNmmNxbh

Mf

tn .458.1.14583336

5010005.3

6

22

Strength reduction factor for plain concrete is 65.0

Therefore mkNxMn .948.0458.165.0 un MM The flange is thus OK

Step 8: Detailing

Standard execution plans conforming to ACI / SBC provisions for beams and ribs.

Ln1 Ln2 Ln3

Ln1/4

Max (Ln2/3 ,Ln3/3)Max (Ln1/3 ,Ln2/3)

Min. 150 mmBottom steel

Top steel

t

t


10/13

Beam loading (uniform in kN/m)

Load transferred by joists to the beam according to its tributary width ltas in one way solid slabs.

Area load (kN/m2) used for this purpose is equal to the joist load (kN/m) divided by the flange width.

In order to avoid duplication of the joist-beam joint weight, we must use the beam clear tributary width ltn.

It is obtained by subtracting the beam width: bttn bll

The dead load includes possible wall loading

Dead wallbbbctnjf

jD

bD wbxSDLhblb

ww Live tbL lxLLw

For internal beams (B, C, D) the tributary widths are: mlt 0.42

4

2

4

mltn 7.33.00.4

For beam B without wall loading:

mkNxxxwbD /008.283.05.16.03.0247.362.0

894.3 mkNxwbL /1243

The ultimate load is mkNwbu /61.59

The effective section of the beam is a T-section for internal beams and L-section for external beams.

However with small flange thicknesses, rectangular section is frequently considered.

Analysisand design of beams is performed using the same steps as in one way solid slabs.

4.0 m

4.0 m

4.0 m

4.0 m

8.2 m 8.1 m

A

B

C

D

E

1 2 3


11/13

Analysis and design of beam B:

The following figure is produced by RC-SLAB1 software. It performs various checks and gives the

analysis results and diagrams.

The next figure, also produced by RC-SLAB1 software, shows the flexural design results considering a T-

section or a rectangular section.


12/13

In theory the rectangular section and T-section designs give the same result for negative moments (flange

in tension) and different results for positive moments (flange in compression).

It can be observed that for negative external moment, rectangular and T-section designs give the same

result (four 16-mm bars). For the span positive moment, rectangular section design gives eight bars

whereas T-section design requires seven bars.

It is however worth noting that the two designs are also different for the internal negative moment. This in

fact is due to the required number of layers. The initial result of twelve bars is maintained for the T-section

because they fit in one single layer in the flange. For the rectangular section, the twelve bars require more

layers and successive design-checks are performed by RC-SLAB1 software. The final design requires

thirteen bars in three layers.


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Girder loading (uniform and concentrated)

The girder is subjected to uniform loading as well as concentrated forces transferred from supported

beams just as in the case of one way solid slab.

Concentrated forces on columns

The internal forces in the columns may be determined as in the case of one way solid slab, using the

tributary area concept. The area load is equal to the joist line load divided by the flange width.

The dead force includes area loading as well the self weight of the webs of all beams and girders in the

tributary area. It also includes possible wall loads.

Dead tiiwallitiwiwiictjf

jD

D lwlhbAb

wP , Live tL AxLLP

For beams / girders inside the tributary area, the total web self weight and total wall load is considered

( )1i . For beams / girders on the border of the tributary area, only half is considered ( )5.0i . lti is

the member length inside the tributary area (clear length for beams and full length for girders).

joist slab

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