Dr. Nader Okasha
ec ure Design of hollow block and ribbed slabs
PART I One way ribbed slabs
Ribbed and hollow block slabs
with a toping slab. The voids between the ribs may be either light material such as hollow blocks [figure 1] or it may be left unfilled [figure 2]. Topping slab
Rib
Hollow block
Figure [1] Hollow block floor
Temporary form Figure [2] Moulded floor
Ribbed and hollow block slabs
weight by removing the part of the concrete below the neutral axis. Additional advantages are: 1- Ease of construction. - o ow oc s ma e poss e o ave smoo often required for architectural considerations. -
ce ng w c
s
.
Hollow block floors proved economic for spans of more than 5 m with light or moderate live loads, such as hospitals, offices or residential buildings. They are not suitable for structures having eavy ve oa s suc as ware ouses or par ng garages.
One-way v.s two-way ribbed slabs
, being one-way, regardless of the ratio of longer to shorter panel dimensions. It is classified as two-way if the ribs are provided in two directions. One way spans typically span in the shorter direction. One way ribbed slabs may be used for spans up to 6 - 6.5 m.
One-way slab
Two-way slab
One-way ribbed (joist) slab
Arrangements of ribbed slabs
Arrangements of ribbed slabs
Arrangements of ribbed slabs
Arrangements of ribbed slabs
Arrangements of ribbed slabs
Key components of ribbed slabs
ACI 8.13.6.1 Topping slab thickness (t ) is not to be less than 1/12 the clear distance (l c ) between ribs, nor less than 50 mm a. To
in slab:
⎧ lc ⎪ t ≥ ⎨12 ⎩50 mm and should satisf for a unit stri : 2
t ≥
w u lc
Φ1240 f c ′
directions in a mesh form.
l c
Slab thickness (t)
Key components of ribbed slabs b. Re ularl s aced ribs: Minimum dimensions:
Ribs are not to be less than 100 mm in width, and a depth of not more an . mes e m n mum we w an c ear spac ng between ribs is not to exceed 750 mm. ACI 8.13.2 ACI 8.13.3 l ≤ 750 mm c
h ≤ 3.5 bw
bw ≥ 100
Key components of ribbed slabs
ACI 8.13.8 Shear strength provided by rib concrete Vc may be taken 10% greater than those for beams. Shear stren th:
Ribs are designed as rectangular beams in the regions of negative moment at the supports and as T-shaped beams in the regions of positive moments between the supports. Effective flange width be is taken as half the distance between ribs, center-to-center. b e
Key components of ribbed slabs c. Hollow blocks:
Hollow blocks are made of lightweight concrete or other lightweight materials. The most common concrete hollow block sizes are 40 × 25 cm in plan and heights of 14, 17, 20, and 24 cm. Hollow blocks do not contribute to the strength of the slab. In fact, the im ose an additional wei ht on the slab. In some cases, blocks made of polystyrene, which is 1/15 of the weight of concrete blocks, are used. To avoid shear failures, the blocks are terminated near the support and re laced b solid arts. Solid arts are made under artitions and concentrated walls. o avo crac ng ue o s r n age n op concre e ange, concrete blocks should be watered prior to concrete placing.
e
Cross (distribution) ribs
floors for better distribution of the applied loads. They also help in distributing the concentrated loads due to walls in the transverse direction. The bottom reinforcement is taken as the reinforcement in the main ribs, and the top reinforcement should be taken at least ½ of e o om re n orcemen . ross r s are usua y cm w e.
code:
Cross (distribution) ribs
One cross rib
Three cross ribs
Arrangement of hollow blocks and width of hidden beams
construction drawings. Thus, the layout of the blocks must be positioned so that enough solid parts are present near the supporting beams. The normal width of solid part ranges between 0.8-2.0 m for floors with hidden beams and ranges between 0.2-0.5 m for floors w pro ec e eams. ×
width of the beams must satisfy: In the rib direction (mm): erpen cu ar o r
rec on mm :
c1
c2
=
× n1 + 2
× ncr w
2
−
Arrangement of hollow blocks and width of hidden beams bw =width of main rib
Lc 1 = 250 × n1 + 100 × ncr Lc 2 = 400 × n 2 + bw × ( n 2 − 1)
Minimum thickness of one way slabs
Minimum Cover
ACI Table 9.5(a)
ACI 7.7.1
a - Concrete exposed to earth or weather for Φ<16mm------40 mm and for Φ>16mm----- 50 mm b - Concrete not exposed to earth or weather for Φ<32mm------20 mm, otherwise ------ 40 mm
Loads Assigned to Slabs w u= .
. + .
.
. 1- Weight of slab covering materials 2- E uivalent artition wei ht 3- Own weight of slab -
ve
oa
.
a- Dea Dead d Load Load (D. (D.L) L)
1- Weight of slab sl ab covering materials, total =2.315 kN/m2
tiles (2.5cm thick) =0.025×23 = 0.575 kN/m2 cement mortar (2.5cm thick) =0.025×21 = 0. 0.525 kN/m2 sand (5.0cm thick) =0.05×18 = 0.9 kN/m 2 2 = = ×
cement mortar sand
. 2.5 cm 5 cm
1.5 cm
2-Equivalent partition weight
of wall × total spans of all walls) carried by the slab divided by the floor area and treated as a dead load rather than a live load. To calculate the weight of 1m span of wall: Each 1m2 surface of wall contains 12.5 blocks A block with with thickness thickness 10cm weighs weighs 10 kg A block block with thickness 20cm weighs 20 kg
Each face of 1m2 surface has 30kg plaster Load / 1m2 surface surface for 10 10 cm bloc block k= 12.5 × 10 +2×30=185 kg/m2 = 1.85 kN/m2 2
= 12.5 × 20 +2×30=310 kg/m2 = 3.1 kN/m2
20 cm
For 10 cm block wt. = 1.85 kN/m2 × 3 = 5.6 kN/m For 20 cm block wt. = 3.1 kN/m2 × 3 = 9.3 kN/m
3- Own wei weight ght of slab slab
Example Find the total ultimate load er rib for the ribbed slab shown: Assume depth of slab = 25 cm (20cm block +5cm toping slab) Hollow Hollow block blockss are 40 cm × 25 cm × 20 cm in in dimen dimensio sion n Assume ribs have 10 cm width of web Assume equivalent partition load = 0.75 kN/m 2 Consider Consider live load load = 2 kN/m2.
3- Own weight of slab
Solution •
Total volume (hatched) = 0.5 × 0.25 × 0.25 = 0.03125 m3
•
Volume of one hollow block = 0.4 × 0.20 × 0.25 = 0.02 m3
•
Net concrete volume = 0.03125 - 0.02 = 0.01125 m 3
•
Weight of concrete = 0.01125 × 25= 0.28125 kN
•
Weight of concrete /m2 = 0.28125 /[(0.5)(0.25)] = 2.25 kN/m 2
•
Weight of hollow blocks /m2 = 0.2/[(0.5)(0.25)] = 1.6 kN/m 2
•
= .
. = .
2
oa per r Total dead load= 3.85 + 2.315 + 0.75 = 6.915 kN/m 2 Ultimate load = 1.2(6.915) + 1.6(2) = 11.5 kN/m 2 Ultimate load per rib = 11.5
×
0.5 = 5.75 kN/m
Minimum live Load values on slabs Type of Use
Uniform Live Load kN/m 2
b- Live Load L.L
It depends on the function for which the floor is constructed.
Residential balconies Computer use Offices Warehouses
3 5 2
Light storage
6
Heavy Storage
12
Classrooms Libraries
2
rooms
3
Stack rooms Hospitals Assembly Halls
6 2
Fixed seating
Movable seating Garages (cars) Stores
2.5 5 2.5
Retail
4
wholesale Exit facilities Manufacturing
5 5
Light
4
Heavy
6
Loads Assigned to Beams - Their own weight - Weights of partitions applied directly on them - F oor oa s
L
1
2
Shrinka e Reinforcement Ratio According to ACI Code and for f y =420 MPa shrinkage
= .
s , shrinkage
ACI 7.12.2.1
= .
where, b = width of strip, and h = slab thickness
n mum
A
,
e n orcemen
≥A
,
a o or
an
e n orcemen
= 0.0018 b × h
Check shear capacity of the section
V u ≤ 1 .1Φ V c = (1.1)0.17Φ f c ' b w d Otherwise enlarge depth of slab
Spacing of Reinforcement Bars a- Flexural Reinforcement Bars
Flexural reinforcement is to be spaced not farther than three times the slab thickness (hs), nor farther apart than 45 cm, center-to-center. Smax ≤ smaller of ⎨
s
⎩45cm
ACI 10.5.4
b- Shrinkage Reinforcement Bars
Shrinka e reinforcement is to be s aced not farther than five times the slab thickness, nor farther apart than 45 cm, center-to-center. Smax ≤ smaller of
5h
⎩45cm
CI 7.12.2.2
-
1. The direction of ribs is chosen. 2. Determine h , and select the hollow block size, b w and t 3. Provide shrinkage reinforcement for the topping slab in both directions. 4. The factored load on each of the ribs is computed. 5. The shear force and bending moment diagrams are drawn. 6. The strength of the web in shear is checked. 7. Design the ribs as T-section shaped beams in the positive moment regions and rectangular beams in the regions of negative moment. 8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared.
Example 1 of blocks and width of hidden beams for the
lan
shown. The blocks used have the size of 40 × 20 cm in plan. The live load is 4 kN/m2.
Solution
Note that the width of hollow blocks in Gaza is 250 mm NOT 200 mm
Solution
Solution
Solution
Example 2 -
.
,
figure below. The covering materials weigh 2.25 kN/m 2, equivalent artition load is e ual to 0.75 kN/m2 and the live load is 2 kN/m2. Use f c’=25 MPa, f y=420MPa
m 8 . 3
10 m
Solution 1. The direction of ribs is chosen:
Ribs are arranged in the short direction as shown in the figure
m 8 .
.
.
2. Determine h , and select the hollow block size, b w and t :
From ACI Table 9.5(a), h min = 380/16 = 23.75cm use h = 24 cm. e w o we , w = cm Use hollow blocks of size 40 cm × 25 cm × 17 cm (weight=0.17 kN) To in slab thickness = 24 – 17 = 7cm > l /12 =40/12= 3.3cm > 5cm OK For a unit strip of topping slab: wu=[1.2(0.07 × 25 + 0.75 + 2.25) + 1.6(2)] ×1m = 8.9 kN/m = 8.9 N/mm t ≥
w u l c2
Φ1240 f c ′
=
8.9( 400 ) 2 ( 0.9 )1240 25
= 16mm OK
Solution 3. Provide shrinka e reinforcement for the to
in slab in both directions:
Area of shrinkage reinforcement, As=0.0018(1000)70=126 mm2 Use 5 Φ 6 mm/m in both directions. 4. The factored load on each of the ribs is to be computed:
Total volume (in 1m2 surface) = 1.0 × 1.0 × 0.24 = 0.24 m3 Volume of hollow blocks in 1m2 = 8 × 0.4 × 0.25 × 0.17 = 0.136 m3 Net concrete volume in 1m2 . - . . Weight of concrete in 1m2 = 0.104 × 25 = 2.6 kN/m2 Wei ht of hollow blocks in 1m2 = 8 × 0.17= 1.36 kN/m2 Total dead load /m2 = 2.25 + 0.75 + 2.6 + 1.36 = 7.0 kN/m2
1.0 m
0.05 m m 0 . 1
m 5 2 . 0
0.4 m m 2 . 0
0.1 m
0.4 m
m c 7
Solution w =1.2 7 +1.6 2 =11.6 kN/m2 wu/m of rib =11.6x0.5= 5.8 kN/m of rib 5. Critical shear forces and bending moments are determined (simply supported beam):
Maximum factored shear force = w ul/2 = 5.8 (3.8/2) = 11 kN Maximum factored bending moment = w ul2/8 = 5.8 (3.8)2/8 = 10.5 kN.m 6. Check rib strength for beam shear:
Effective depth d = 24–2–0.6–0.6 =20.8 cm, assuming φ 12mm reinforcing bars and Φ 6 mm stirru s. 1.1ΦV c = 1.1× 0.75 × 0.17 × 25 × 100 × 208 = 14400 N = 14.4 kN > Vu,max = 11kN
Though shear reinforcement is not required, 4 φ 6 mm stirrups per meter run are to be used to carry the bottom flexural reinforcement.
Solution 7. Desi n flexural reinforcement for the ribs:
There is only positive moments over the simply supported beam, and the section of maximum positive moment is to be designed as a T-section . ρ=
0.85 × 25 ⎛
⎜1 − 1 −
e
2 ×10.5 × 106 .
2
.
= 0.0013 s
2
.
e
Use 2φ 10mm (As,sup= 157 mm2) a=
A s f y 0.85f c 'b e
=
157 × 420 0.85 × 25 × 500
= 6.2 mm < 70mm
⎞ ⎟
50
m . N k 5 0 1
7 24
As 10
Solution Check A ,
⎧⎪ 0.25 f c' ⎫⎪ 1.4 A s,min = max ⎨ bw d ; bw d ⎬ f y ⎪⎩ f y ⎪⎭ A s,min = 70 mm 2 < A s,sup = 157 mm 2 OK .
c=
a 1
=
6.2 .
= 7 .3 mm
⎛d−c⎞ ⎛ 208 − 7 .3 ⎞ 0.003 0.003 = ⎟ ⎜ ⎟ c 7.3
εt = ⎜
ε t = 0.083 >> 0.005 ⇒ Tension controlled ⇒ Φ = 0 .9 OK
Solution 8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared
m 8 . 3
m m 0 0 1 1 Φ Φ 1 1
A
A
m m 0 0 1 1 Φ Φ 1 1
5.0 m
5.0 m Φ6mm mesh @20 cm
@25 cm
7cm 24cm
2Φ10mm
10
40 cm Sect on A-A
10
2Φ10mm
PART II Two way ribbed slabs
Method of analysis
φ R ≥
∑ γ L i
i
i
Method of coefficients w
w = ws + wl
S
1.0m S
w s = α w
=
L l
Rectangularity ratio: Case
r
L
L
0.76 L
L
S
S
S
S
L
L
0.76S 0.87S
0.87 L 0.76 L 0.87 L S
0.87S 0.76S
Method of coefficients ECP 203 load coefficients LL< 5kN/m2 r
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
.
.
.
.
.
.
.
.
.
.
.
Marcus load coefficients LL ≥ 5kN/m2 r
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
0.292
0.355
0.411
0.470
0.526
0.577
0.623
0.663
0.699
0.730
0.757
0.292
0.240
0.198
0.165
0.137
0.114
0.095
0.079
0.067
0.056
0.047
Minimum slab thickness:
, one of three equations of the ACI 318-89 which provides an upper bound for the deflection control of the slab thickness can be used for simplicity.
h max =
l n (800 + f y / 1.4)
Design of beams in two way ribbed slabs:
by 45-degree lines drawn from the corners of the panels and the centerlines of the adjacent panels parallel to the long sides. S
L
S
L
Design of beams in two way ribbed slabs: qu(S/2)
qu(S/2)
L
L o n g b e a
L S
An e uivalent uniforml distributed load can be established for a beam in a two way system. For a triangular load distribution, the equivalent shear force coefficient C s is equal to 0.5 and the equivalent bending moment coefficient C b is equal to 0.67.
Design of beams in two way ribbed slabs: u
L o g b e a m
S L
qu(S/2)
L Short beam
For a trapezoidal load distribution, C s and C b are given in the following table. Shear and moment equivalent load coefficients for trapezoidal load distribution r =L/S
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
C
0.500
0.545
0.583
0.615
0.643
0.667
0.688
0.706
0.722
0.737
0.750
C
0.667
0.725
0.769
0.803
0.830
0.852
0.870
0.885
0.897
0.908
0.917
s
b
Design of beams in two way ribbed slabs: Original distribution
Equivalent distribution for shear
wu
0.5 wu
S
S
wu
L
Cs wu
L
Equivalent distribution for moment 0.67 wu
S
Cb wu
L
Summary of two-way ribbed slab design procedure
1. Evaluate overall slab thickness and key ribbed slab components. .
.
3. Determine load distributions in the two principal directions. .
e erm ne
e s ear orce an
en ng momen s.
5. Check web width for beam shear. .
es gn r
re n orcement.
7. Design drawing.
Exam le 3 Design the two-way ribbed slab shown in the figure below. The covering materials weigh 1.5 kN/m2, equivalent partition load is equal to 0.75 m , concre e o ow oc s are cm× cm× cm n mens on, each 17 kg in weight and the live load is 4 kN/m2. All beams are 30 cm wide. Use f c’=30 MPa, f y=420 MPa.
8.0 m
8.0 m
8.0 m
8.0 m
o ut on: 1- Evaluate overall slab thickness and key ribbed slab components: l n = 800 − 30 = 770cm
h max = h max =
l n (800 + f y / 1.4)
36000 (770)(800 + 420 / 1.4) 36000
Take width of rib = 12 cm.
= 23.5cm
o ut on: 2- Determine the total factored load on the slab:
Total volume (hatched) = 0.52 o ume o
o ow
oc s =
×
0.62 × 0.23 = 0.074152 m3
.
×
.
.
×
= .
m
Net concrete volume = 0.074152 - 0.034 = 0.040152 m3 Weight of concrete = 0.040152 2
= .
×
25= 1.0038 kN .
.
= .
2
Weight of hollow blocks /m 2 = 0.17(2)/[(0.52)(0.62)] = 1.05 kN/m 2 Total dead load= 3.11+1.05 + 1.5 + 0.75 = 6.41 kN/m 2 Ultimate load = 1.2(6.41) + 1.6(4) = 14.1 kN/m 2
o ut on: -
.
r =
w1=w2= .
.
L
m
8
= =1.0
S 8
= .
8 m
m 8
Load per rib in Direction 1: = . . = . u
0.62 m
Load per rib in Direction 2: = . . = . u
8m
8m
2
Direction
0 . 5
1
0.4
o ut on: 4- Determine the shear force and bendin moments:
Using the ACI 8.3 coefficients . Direction 1
Maximum factored shear force = 1.15w uln/2 = 1.15(3.06) (7.7/2) = 13.5 kN Maximum factored negative moment = wuln2/9 = 3.06(7.7)2/9 = 20.2 kN.m ax mum ac ore pos ve momen = wu n = . . = . .m Direction 2
2.56 kN/m
ax mum ac ore s ear orce = . wu n = . . . = . Maximum factored negative moment = w uln2/9 = 2.56(7.7)2/9 = 16.9 kN.m Maximum factored positive moment = w uln2/14 = 2.56(7.7)2/14 = 10.9 kN.m
o ut on: 5- Check web width for beam shear: Direction 1
Reinforcement is closest to the outside surface of concrete.
Effective depth d = 23 – 2 – 0.60 – 0.8 = 19.6 cm, assuming φ16 mm reinforcing bars and φ6 mm stirrups.
.
c
= . × .
× .
×
×
×
= = 18.1 kN > Vu,max = 13.5 kN
Effective depth d = 23 – 2 – 0.60 – 1.6 - 0.8 = 18.0 cm, assuming φ16 mm reinforcing bars and φ6 mm stirrups.
1.1ΦV c = 1.1× 0.75 × 0.17 × 30 ×120 ×180 = 16593 N = 16.6 kN > Vu,max = 11.4 kN All though shear reinforcement is not required, 4 φ 6 mm stirrups per meter run are used to carry the bottom flexural reinforcement.
o ut on: 6- Desi n rib reinforcement: Direction 1
Positive moment Mu = 12.9 kN.m Use mm 1 10 and mm 1 12 reinforcing bars in each rib.
Negative moment Mu = 20.2 kN.m .
o ut on: 6- Desi n rib reinforcement: Direction 2
Positive moment Mu = 10.9 kN.m Use mm 1 10 and mm 1 12 reinforcing bars in each rib.
Negative moment Mu = 16.9 kN.m .
o ut on: 7- Desi n drawin :
