One way joist slab (by Prof. A. Charif) Typical joist (rib)
Vertical section
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). bf hf hw bw
S Void or hollow block (Hourdis)
Joists are supported by beams, which are supported by girders or columns. ACI / SBC Conditions on joist dimensions: Web width: bw 100 mm
Web thickness: hw 3.5bw
S / 12 Flange thickness: h f 50 mm
Spacing: S 800 mm
The flange width is then: b f bw S 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.
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 and determine 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. Dead w jD (SDL c h jf ) x b jf c b jw h jw b Shjw The ultimate joist load is:
Live w jL LL x b jf
w ju 1.4w jD 1.7w jL
(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 that Vc Vu with Vc increased 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
One way joist slab example: 8.2 m
8.1 m
E 4.0 m D 4.0 m
Joist Data (mm)
C
50
250
4.0 m
120
B
500
120
4.0 m A 1
2
3
Typical joist (rib)
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 b 12 kN / m 3 Concrete: f c' 25 MPa c 24 kN / m3
Steel: f y 420 MPa
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/m2 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 wwall 14.4 kN / m
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) for hmin
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
L / 20
L / 24
L / 28
L / 10
L / 16
L / 18.5
L / 21
L/8
Solid oneway slab Beams or ribs
Spans 1 and 4: One end continuous
Spans 2 and 3: Both ends continuous Thus hmin 216 .22 mm
hmin
L 4000 216.22 mm 18.5 18.5
hmin
L 4000 190.48 mm 21 21
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 dimensions
Web width: bw 120 mm 100 mm
Web thickness: hw 250 mm 3.5bw 3.5 x120 420 mm
S / 12 500 / 12 41.67 mm Flange thickness: h f 50 mm 50 mm
Spacing: S 500 mm 800 mm
All dimension conditions are satisfied. The flange width is then: b f bw S 500 120 620 mm b)
Loading:
Dead: w jD (SDL c h jf ) x b jf c b jw h jw b Shjw w jD (1.5 24 x 0.05) 0.62 24 x 0.12 x 0.25 12 x 0.5 x 0.25 3.894 kN / m
Live
w jL LL x b jf 3 x 0.62 1.86 kN / m
The ultimate joist load is:
w ju 1.4w jD 1.7w jL 8.614 kN / m
Step 3: Flexural analysis All conditions of ACI/SBC coefficient method are satisfied. So
l Vu C v wu n 2
M u Cm wu (l n ) 2
ln is the clear length wu is the factored uniform load
l n 4.0
0.3 0.3 3.7 m 2 2
for all spans
For shear force, span positive moment and external negative moment, ln is the clear length of the span For internal negative moment, ln is 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:
RC-SLAB1 software output is:
Step 4: Flexural RC design Steel depth
d h cover
db ds 2
Cover = 20 mm
Assume bar diameter db = 12 mm and stirrup diameter ds = 8 mm Thus d 300 20
12 8 266 mm 2
RC design for internal negative moment Mu = 11.79 kN.m We find As = 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:
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 use a smaller diameter of 10 mm The spacing is S
Thus
(we consider 1 m strip)
Ab = 78.5 mm2
bAb 1000 x 78.5 872.2 mm As 90
Maximum spacing for shrinkage steel in slabs according to SBC / ACI is:
S max Min(4h , 300 mm) Min(4 x 50 , 300) 200 mm
we thus use 10 @200 mm
Step 6: Shear check Ultimate shear force Vu Cv w ju
Ln 3.7 1.15 x 8.614 18.33 kN 2 2
For joists, the nominal concrete shear strength Vc is increased by 10 %. Thus
Vc 1.1
f c' 6
bw d 1.1
25 120 x 266 29260 N 29.26 kN 6
Vc 0.75Vc 21.945 kN Vu 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 w 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
S
The section is b x hf = 1000 x 50 mm The ultimate uniform load is obtained from slab loading:
w 1.4SDL c h f 1.7 LL x1m 1.41.5 24 x 0.05 1.7 x 3 x1m 8.88 kN / m The maximum ultimate at fixed ends is: M u
w S 2 8.88 x 0.5 2 0.185 kN.m 12 12
As the member is un-reinforced, the nominal capacity must consider concrete tension strength, which is defined by SBC as: t 0.7 f c' 3.5 MPa (for ACI, it is t 0.5 f c' ).
The nominal moment for a rectangular section with maximum stress equal to tension strength is: Mn t
bh f
2
6
t
1000 x 50 2 3.5 1458333 N .mm 1.458 kN.m 6
Strength reduction factor for plain concrete is 0.65 Therefore M n 0.65 x1.458 0.948 kN.m
t
M n M u The flange is thus OK
Step 8: Detailing Standard execution plans conforming to ACI / SBC provisions for beams and ribs. Top steel
Max (Ln1/3 ,Ln2/3)
Max (Ln2/3 ,Ln3/3)
Ln1 /4
Ln1
Min. 150 mm
Bottom steel
Ln2
Ln3
Beam loading (uniform in kN/m) 8.2 m
8.1 m
E 4.0 m D 4.0 m C 4.0 m B 4.0 m A 1
2
3
Load transferred by joists to the beam according to its tributary width lt as 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: ltn lt bb The dead load includes possible wall loading Dead wbD
w jD b jf
ltn c bb hb SDL x bb wwall
Live wbL LL x lt
4 4 For internal beams (B, C, D) the tributary widths are: lt 4.0 m 2 2
ltn 4.0 0.3 3.7 m
For beam B without wall loading: wbD
3.894 3.7 24 x 0.3 x 0.6 1.5 x0.3 28.008 kN / m 0.62
The ultimate load is
wbL 3 x4 12 kN / m
wbu 59.61kN / m
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. Analysis and design of beams is performed using the same steps as in one way solid slabs.
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 Tsection or a rectangular section.
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.
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
PD
w jD b jf
At c i bwi hwi lti i wwall,i lti
Live PL LL x At
For beams / girders inside the tributary area, the total web self weight and total wall load is considered ( i 1) . For beams / girders on the border of the tributary area, only half is considered ( i 0.5) . lti is the member length inside the tributary area (clear length for beams and full length for girders).
