Reinforced Concrete Design III
Dr. Nader Okasha
Lecture L t 2 Design of hollow block and ribbed slabs
PART I One way ribbed slabs
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Ribbed and hollow block slabs Ribbed slabs consist of regularly spaced ribs monolithically built 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
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Ribbed and hollow block slabs The main advantage of using hollow blocks is the reduction in weight by removing the part of the concrete below the neutral axis. Additional advantages are: 1- Ease of construction. 2 Hollow 2H ll bl k make blocks k it possible ibl to t have h smooth th ceiling ili which hi h is i often required for architectural considerations. 33- Provides good sound and temperature insulation properties. 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 h heavy li loads live l d suchh as warehouses h or parking ki garages. ٤
One-way v.s two-way ribbed slabs If the ribs are provided in one direction only, only the slab is classified as 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.
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One-way slab
Two-way slab
One-way ribbed (joist) slab
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Arrangements of ribbed slabs
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Arrangements of ribbed slabs
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Arrangements of ribbed slabs
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Arrangements of ribbed slabs
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Arrangements of ribbed slabs
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Key components of ribbed slabs ACI C 8. 8.13.6.1 3.6. Topping slab thickness (t) is not to be less than 1/12 the clear distance (lc) between ribs, nor less than 50 mm a. Topping slab:
⎧ lc ⎪ t ≥ ⎨12 ⎪⎩50 mm
andd sshould ou d ssatisfy s y for o a uunit sstrip: p: t≥
lc Slab thickness ((t))
w u l c2 Φ1240 f c ′
Shrinkage reinforcement is provided in the topping slab in both directions in a mesh form. ١٢
Key components of ribbed slabs b. Regularly spaced ribs: Minimum dimensions:
Ribs are not to be less than 100 mm in width, and a depth of not more than th 3.5 3 5 times ti th minimum the i i webb width idth andd clear l spacing i 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
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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 strength: g
Flexural strength:
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 ff i flange fl width id h be is i taken k as half h lf the h distance di b between ribs, ib center-to-center. b e
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Keyy components p of ribbed slabs c. Hollow blocks: Hollow blocks are made of lightweight g g 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, theyy impose p an additional weight g 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 replaced p byy solid pparts. Solid pparts are made under ppartitions and concentrated walls.
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To avoid cracking due to shrinkage in top concrete flange, flange the concrete blocks should be watered prior to concrete placing.
Cross (distribution) ribs Transversal ribs or cross ribs are added to one one-way way hollow block 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 th bottom the b tt reinforcement. i f t Cross C ribs ib are usually ll 10 cm wide. id Arrangement of regularly spaced cross rib according to Egyptian code:
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Cross (distribution) ribs
No cross ribs
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One cross rib
Three cross ribs
Arrangement of hollow blocks and width of hidden beams The number of blocks in each direction must be specified on the 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 with ith projected j t d beams. b The number of blocks (having sizes of 40 × 25 cm in plan) and the width of the beams must satisfy:
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In the rib direction (mm):
Lc 1 = 250 × n1 + 100 × ncr
P Perpendicular di l to t rib ib direction di ti (mm): ( )
Lc 2 = 400 × n 2 + bw × (n 2 − 1)
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)
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Minimum thickness of one way slabs
Minimum Cover
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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 wu=1.2 1 2 D.L D L + 1.6 1 6 L.L LL
a- Dead Load (D (D.L) L) : 1- Weight of slab covering materials 2- Equivalent q ppartition weight g 3- Own weight of slab
b Live bLi L Load d (L (L.L) L)
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a- Dead Load (D.L) ( ) 1- Weight of slab 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 = 0.525 kN/m2 sand (5.0cm thick) =0.05×18 = 0.9 kN/m2 plaster (1.5cm thick) =0.015×21 0.015 21 = 0.315 kN/m2
tiles cement mortar sand
22.55 cm 2.5 cm 5 cm
slab
plaster ٢٢
1.5 cm
2-Equivalent partition weight
This load is usually taken as the weight of all walls (weight of 1m span 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 bl blockk with i h thickness hi k 10 weighs 10cm i h 10 kg k A block with thickness 20cm weighs 20 kg Each face of 1m2 surface has 30kg plaster Load / 1m2 surface for 10 cm block = 12.5 × 10 +2×30=185 kg/m2 = 1.85 kN/m2 Load / 1m2 surface for 20 cm block = 12.5 × 20 +2×30=310 kg/m2 = 3.1 kN/m2
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20 cm
Weight of 1m span of wall with height 3m: 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 weight of slab
Example Find the total ultimate load per rib for the ribbed slab shown: Assume depth of slab = 25 cm (20cm block +5cm toping slab) Hollow blocks are 40 cm × 25 cm × 20 cm in dimension Assume ribs have 10 cm width of web Assume equivalent partition load = 0.75 kN/m2 Consider live load = 2 kN/m2.
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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 m3
•
Weight of concrete = 0.01125 × 25= 0.28125 kN
•
Weight of concrete /m2 = 0.28125 /[(0.5)(0.25)] = 2.25 kN/m2
•
Weight of hollow blocks /m2 = 0.2/[(0.5)(0.25)] = 1.6 kN/m2
•
Total slab own weight weight= 2.25 2 25 + 1.6 1 6 = 3.85 3 85 kN/m2
L dp Load per rib ib Total dead load= 3.85 + 2.315 + 0.75 = 6.915 kN/m2 Ultimate load = 1.2(6.915) + 1.6(2) = 11.5 kN/m2 ٢٥
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/m2
b- Live Load ((L.L))
It depends on the function for which the floor is constructed.
Residential
2
Residential balconies Computer use Offices Warehouses
3 5 2
6
Light storage
Heavy Storage Schools
12
Classrooms Libraries
2
rooms
3
Stack rooms Hospitals Assembly Halls
6 2
Fixed seating
Movable seating Garages (cars) Stores
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2.5 5 2.5
Retail
4
wholesale Exit facilities Manufacturing
5 5
Light
4
Heavy
6
Loads Assigned to Beams Beams are usually designed to carry the following loads: - Their own weight - Weights of partitions applied directly on them - Floor Fl loads l d
L
S1 ٢٧
S2
Shrinkage Reinforcement Ratio According to ACI Code and for fy =420 MPa
ACI 7.12.2.1
ρ shrinkage = 0.0018 ⇒ As , shrinkage = 0.0018 b × h where, b = width of strip, and h = slab thickness
Mi i Minimum Reinforcement R i f t Ratio R ti for f Main M i Reinforcement R i f t
A s ,min min ≥ A s ,shrinkage shrinkage = 0.0018 b × h
ACI 10.5.4
Check shear capacity of the section
V u ≤ 1.1Φ V c = (1.1)0.17Φ f c ' b wd Otherwise enlarge depth of slab ٢٨
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. ⎧ 3 hs Smax ≤ smaller of ⎨ ACI 10.5.4 ⎩45cm b- Shrinkage Reinforcement Bars Shrinkage reinforcement is to be spaced not farther than five times the slab thickness, nor farther apart than 45 cm, center-to-center. Smax
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⎧ 5 hs ≤ smaller ll of ⎨ ⎩45cm
ACI 7.12.2.2 7 12 2 2
Summary of one one-way way ribbed slab design procedure 1. The direction of ribs is chosen. 2. Determine h, and select the hollow block size, bw 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 p p reinforcement are to be prepared. ٣٠
Example 1 Determine the arrangement of blocks and width of p hidden beams for the plan shown. The blocks used have the size of 40 × 20 cm in plan. The live load is 4 kN/m2.
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Solution
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Note that the width of hollow blocks in Gaza is 250 mm NOT 200 mm
Solution
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Solution
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Solution
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Example 2 Design a one one-way way ribbed slab to cover a 3.8 3 8 m x 10 m panel, panel shown in the figure below. The covering materials weigh 2.25 kN/m2, equivalent ppartition load is equal q to 0.75 kN/m2, and the live load is 2 kN/m2.
3.8 m
Use fc’=25 MPa, fy=420MPa
10 m
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Solution 1. The direction of ribs is chosen:
3.8 m 3
Ribs are arranged in the short direction as shown in the figure
50m 5.0
50m 5.0
2. Determine h, and select the hollow block size, bw and t:
From ACI Table 9.5(a), hmin = 380/16 = 23.75cm Î use h = 24 cm. L t width Let idth off web, b bw =10 10 cm Use hollow blocks of size 40 cm × 25 cm × 17 cm (weight=0.17 kN) Topping pp g slab thickness = 24 – 17 = 7cm > lc/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 w u l c2
8.9( 400 ) 2 t≥ = = 16mm OK ( 0.9 )1240 25 Φ1240 f c ′
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Solution 3. Provide shrinkage g reinforcement for the topping pp g 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:
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1.0 m
0.4 m
0.1 m
0.4 m
7 cm
0.25 m
1.0 m
0.05 m
0.244 m
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 = 0.240 24 0.136 0 136 = 0.104 0 104 m3 Weight of concrete in 1m2 = 0.104 × 25 = 2.6 kN/m2 Weight g 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
Solution wu=1.2(7)+1.6(2)=11.6 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 = wul/2 = 5.8 (3.8/2) = 11 kN Maximum factored bending moment = wul2/8 = 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 φ12mm reinforcing bars and Φ 6 mm stirrups. p 1.1ΦV c = 1.1× 0.75 × 0.17 × 25 × 100 × 208 = 14400 N = 14.4 kN > Vu,max = 11 kN
Though shear reinforcement is not required, 4 φ 6 mm stirrups per meter run are to be used to carry the bottom flexural reinforcement.
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Solution 7. Design g 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 Assume that a<70mm and Φ=0.90→Rectangular Φ=0 90→Rectangular section with b = be =500mm
As = ρ be d = 0.0013 × 500 × 208 = 135 mm
2
Use 2φ10mm (As,sup= 157 mm2) As f y
157 × 420 = 6.2 mm < 70mm 0.85f c 'b e 0.85 × 25 × 500 The assumption is right a=
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=
⎞ ⎟ ⎟ ⎠
50
105 k kN.m
0.85 × 25 ⎛ 2 ×10.5 ×106 ρ= ⎜1 − 1 − ⎜ 420 ⎝ 0.9 × 0.85 0 85 × 25 × 500 × 2082 = 0.0013
7 24
As 10
Solution Check As,min s min
⎧⎪ 0.25 f c ' ⎫⎪ 1.4 A s,min = max ⎨ bw d ; bw d ⎬ fy ⎪⎩ f y ⎭⎪ A s,min = 70 mm 2 < A s,sup = 157 mm 2 OK
Check Φ=0.9 Φ=0 9 (ductility of the section)
c=
a 6.2 = = 7 .3 mm β1 0 0.85 85
⎛d−c⎞ ⎛ 208 − 7 .3 ⎞ εt = ⎜ 0.003 = ⎟ ⎜ ⎟ 0.003 7.3 ⎠ ⎝ c ⎠ ⎝ ε t = 0.083 >> 0 .005 ⇒ Tension controlled ⇒ Φ = 0 .9 OK
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Solution
A
1Φ10 m
A
1Φ10 m
1Φ10 m
1Φ10 m
3.8 m
8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared
5.0 m
5.0 m Φ6mm mesh @20 cm
Φ6mm stirrups @25 cm
7cm 24cm 17cm 2Φ10mm
10
40 cm S i A-A Section AA
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10
2Φ10mm
PART II Two way ribbed slabs
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Method of analysis
φ R ≥ ∑ γ i Li i
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Method of coefficients 1.0m
wS
w = ws + wl
S
w s = αw
w l = βw
wl
L
Rectangularity ratio: Case
r ٤٥
L S
L S
0.76L S
L S
L L 0.87L 0.76L 0.87L 0.76S 0.87S 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
α
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
β
0 35 0.35
0 29 0.29
0 25 0.25
0 21 0.21
0 18 0.18
0 16 0.16
0 14 0.14
0 12 0.12
0 11 0.11
0 09 0.09
0 08 0.08
Marcus load coefficients LL ≥ 5kN/m2
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r
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
α
0.292
0.355
0.411
0.470
0.526
0.577
0.623
0.663
0.699
0.730
0.757
β
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: To avoid the complexity of calculating α for a two way ribbed slab, slab 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 =
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l n (800 + f y /1.4) 36000
Design of beams in two way ribbed slabs: The loads acting on beams have tributary areas which are bounded 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
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S
L
Design of beams in two way ribbed slabs: qu(S/2)
qu(S/2)
L
Long beam m
S
L Short beam S
An equivalent q uniformlyy distributed load can be established for a beam in a two way system. For a triangular load distribution, the equivalent shear force coefficient Cs is equal to 0.5 and the equivalent bending moment coefficient Cb is equal to 0.67. ٤٩
Design of beams in two way ribbed slabs: qu(S/2)
Lon ng beam
S L
qu(S/2)
L Short beam S
For a trapezoidal load distribution, Cs and Cb are given in the following table. Shear and moment equivalent load coefficients for trapezoidal load distribution
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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
Cs
0.500
0.545
0.583
0.615
0.643
0.667
0.688
0.706
0.722
0.737
0.750
Cb
0.667
0.725
0.769
0.803
0.830
0.852
0.870
0.885
0.897
0.908
0.917
Design of beams in two way ribbed slabs: Original distribution
wu
0.5 wu
S
S
wu
L
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Equivalent distribution for shear
Cs wu
L
Equivalent distribution for moment 0.67 wu
S
Cb wu
L
Summary of two-way ribbed slab design procedure 1. 2 2. 3. 4 4. 5. 6 6. 7.
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Evaluate overall slab thickness and key ribbed slab components. Determine the total factored load on the slab. slab Determine load distributions in the two principal directions. Determine the shear force and bending moments. moments Check web width for beam shear. D i rib Design ib reinforcement. i f Design drawing.
Example p 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 kN/ 2, concrete kN/m t hollow h ll bl k are 40cm×25cm×17cm blocks 40 25 17 i dimension, in di i each 17 kg in weight and the live load is 4 kN/m2. All beams are 30 cm wide. Use fc’=30 MPa, fy=420 MPa.
8.0 m
8.0 m
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8.0 m
8.0 m
S l ti Solution: 1- Evaluate overall slab thickness and key ribbed slab components: l n = 800 − 30 = 770cm The largest slab thickness is given by:
h max = h max
l n (800 + f y /1.4)
36000 (770)(800 + 420 /1.4) = = 23.5cm 36000
Take width of rib = 12 cm. ٥٤
S l ti Solution: 2- Determine ee e thee total o factored c o ed load o d on o thee slab: s b
Total volume (hatched) = 0.52 × 0.62 × 0.23 = 0.074152 m3 V l Volume off hhollow ll blocks bl k = 2[0.4 2[0 4 × 0.25 0 25× 0.17] 0 17] = 0.034 0 034 m3 Net concrete volume = 0.074152 - 0.034 = 0.040152 m3 Weight of concrete = 0.040152 × 25= 1.0038 kN Weight of concrete /m2 = 1.0038 /[(0.52)(0.62)] = 3.11 kN/m2 Weight of hollow blocks /m2 = 0.17(2)/[(0.52)(0.62)] = 1.05 kN/m2 Total dead load= 3.11+1.05 + 1.5 + 0.75 = 6.41 kN/m2 Ultimate load = 1.2(6.41) + 1.6(4) = 14.1 kN/m2
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S l ti Solution: 33- Determine load distributions in the two principal directions:
r=
0 52 m 0.52
L 8 = =1.0 S 8 8m
w1=w2=0.35(14.1) =0 35(14 1) =4.92 =4 92 kN/m2 8m
Load per rib in Direction 1: wu/m of rib =4.92x0.62= =4 92x0 62= 3.06 3 06 kN/m of rib
0.62 m
Load per rib in Direction 2: wu/m of rib =44.92x0.52 92x0 52= 2.56 2 56 kN/m of rib
8m
8m
2 0.5
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Direction
1
0.4
S l ti Solution: 4- Determine ee e thee shear s e force o ce and d bending be d g moments: o e s
Using the ACI 8.3 coefficients 3 06 kN/m 3.06
Direction 1
Maximum factored shear force = 1.15wuln/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 M i Maximum ffactored t d positive iti momentt = wuln2/14 = 3.06(7.7) 3 06(7 7)2/14 = 12.9 12 9 kN.m kN Direction 2
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2.56 kN/m
Maximum factored shear force = 1.15w 1 15wuln/2 = 1.15(2.56) 1 15(2 56) (7 (7.7/2) 7/2) = 11.4 11 4 kN Maximum factored negative moment = wuln2/9 = 2.56(7.7)2/9 = 16.9 kN.m Maximum factored positive moment = wuln2/14 = 2.56(7.7)2/14 = 10.9 kN.m
S l ti Solution: 55- C Check ec web width w d for o be beam sshear: e 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.
1 1× 0.75 0 75 × 0.17 0 17 × 30 × 120 × 196 = 18067 N 1 1ΦV c = 1.1 1.1Φ = 18.1 kN > Vu,max = 13.5 kN Direction 2 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
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All though shear reinforcement is not required, 4 φ 6 mm stirrups per meter run are used to carry the bottom flexural reinforcement.
S l ti Solution: 66- Design es g rib b reinforcement: e o ce e Direction 1
Positive moment Mu = 12.9 kN.m Use mm 1φ φ 100 aandd mm 1 φ 12 reinforcing bars in each rib. ٥٩
Negative moment Mu = 20.2 kN.m Use mm 2φ 16 in each rib. rib
S l ti Solution: 66- Design es g rib b reinforcement: e o ce e Direction 2
Positive moment Mu = 10.9 kN.m Use mm 1φ φ 100 aandd mm 1 φ 12 reinforcing bars in each rib. ٦٠
Negative moment Mu = 16.9 kN.m Use mm 2φ 16 in each rib. rib
S l ti Solution: 77- Design es g drawing: d w g
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S l ti Solution: 77- Design es g drawing: d w g
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