Reinforced Concrete Design IIIReinforced Concrete Design IIIReinforced Concrete Design IIIReinforced Concrete Design III

Dr. Nader Okasha

L t 2Lecture 2Design of hollow block and ribbed slabs

PART I

One way ribbed slabs

٢

Ribbed slabs consist of regularly spaced ribs monolithically built

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 Temporary form

Figure [1] Hollow block floor Figure [2] Moulded floor

٣

Ribbed and hollow block slabs

The main advantage of using hollow blocks is the reduction inThe main advantage of using hollow blocks is the reduction inweight by removing the part of the concrete below the neutral axis.Additional advantages are:

1- Ease of construction.2 H ll bl k k it ibl t h th ili hi h i2- Hollow blocks make it possible to have smooth ceiling which isoften required for architectural considerations.3- Provides good sound and temperature insulation properties.3 Provides good sound and temperature insulation properties.

Hollow block floors proved economic for spans of more than 5 mwith light or moderate live loads, such as hospitals, offices orresidential buildings. They are not suitable for structures havingh li l d h h kiheavy live loads such as warehouses or parking garages.

٤

One-way v.s two-way ribbed slabs

If the ribs are provided in one direction only the slab is classified asIf the ribs are provided in one direction only, the slab is classified asbeing one-way, regardless of the ratio of longer to shorter paneldimensions. It is classified as two-way if the ribs are provided in twodirections. One way spans typically span in the shorter direction.One way ribbed slabs may be used for spans up to 6 - 6.5 m.

٥ Two-way slabOne-way slab

One-way ribbed (joist) slab

٦

Arrangements of ribbed slabs

٧

Arrangements of ribbed slabs

٨

Arrangements of ribbed slabs

٩

Arrangements of ribbed slabs

١٠

Arrangements of ribbed slabs

١١

a. Topping slab:

Key components of ribbed slabsACI 8.13.6.1a. Topping slab:

Topping slab thickness (t) is not to be less than 1/12 the cleardistance (lc) between ribs, nor less than 50 mm

C 8. 3.6.

12cl

t⎧⎪≥ ⎨⎪

and should satisfy for a unit strip:lc

Slab thickness (t)

50 mm⎪⎩

d s ou d s s y o u s p: ( )

2u cw lt ≥

Shrinkage reinforcement is provided in the topping slab in both

1240 cf ′Φ

Shrinkage reinforcement is provided in the topping slab in bothdirections in a mesh form.

١٢

b. Regularly spaced ribs:

Key components of ribbed slabsb. Regularly spaced ribs:Minimum dimensions:Ribs are not to be less than 100 mm in width, and a depth of not

th 3 5 ti th i i b idth d l imore than 3.5 times the minimum web width and clear spacingbetween ribs is not to exceed 750 mm.

ACI 8.13.2lc ≤ 750 mm ACI 8.13.3

h ≤ 3.5 bw

bw ≥ 100

١٣

Shear strength:

Key components of ribbed slabsACI 8.13.8g

Shear strength provided by rib concrete Vc may be taken 10% greaterthan those for beams.

Flexural strength:Flexural strength:Ribs are designed as rectangular beams in the regions of negativemoment at the supports and as T-shaped beams in the regions ofpositive moments between the supports.

ff i fl id h b i k h lf h di b ibbe

Effective flange width be is taken as half the distance between ribs,center-to-center.

١٤

Key components of ribbed slabsc. Hollow blocks:Hollow blocks are made of lightweight concrete or other

y p

g glightweight materials. The most common concrete hollow blocksizes 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,they impose an additional weight on the slab. In some cases, blocksy p g ,made of polystyrene, which is 1/15 of the weight of concreteblocks, are used.

To avoid shear failures, the blocks are terminated near the supportand replaced by solid parts. Solid parts are made under partitionsp y p p pand concentrated walls.

To avoid cracking due to shrinkage in top concrete flange theTo avoid cracking due to shrinkage in top concrete flange, theconcrete blocks should be watered prior to concrete placing.

١٥

Cross (distribution) ribs

Transversal ribs or cross ribs are added to one-way hollow blockTransversal ribs or cross ribs are added to one way hollow blockfloors for better distribution of the applied loads. They also help indistributing the concentrated loads due to walls in the transversedirection. The bottom reinforcement is taken as the reinforcement inthe main ribs, and the top reinforcement should be taken at least ½ ofth b tt i f t C ib ll 10 idthe bottom reinforcement. Cross ribs are usually 10 cm wide.

Arrangement of regularly spaced cross rib according to EgyptianArrangement of regularly spaced cross rib according to Egyptiancode:

١٦

Cross (distribution) ribs

No cross ribsNo cross ribs One cross rib Three cross ribs

١٧

Arrangement of hollow blocks and width of hidden beams

The number of blocks in each direction must be specified on theThe number of blocks in each direction must be specified on theconstruction drawings. Thus, the layout of the blocks must bepositioned so that enough solid parts are present near the supportingbeams. The normal width of solid part ranges between 0.8-2.0 m forfloors with hidden beams and ranges between 0.2-0.5 m for floors

ith j t d bwith projected beams.

The number of blocks (having sizes of 40 × 25 cm in plan) and theThe number of blocks (having sizes of 40 25 cm in plan) and thewidth of the beams must satisfy:

250 100LIn the rib direction (mm):

P di l t ib di ti ( )

1 1250 100c crL n n= × + ×

400 ( 1)L n b n= × + ×

١٨

Perpendicular to rib direction (mm): 2 2 2400 ( 1)c wL n b n= × + × −

Arrangement of hollow blocks and width of hidden beams

1 1250 100c crL n n= × + ×=width of main ribwb

2 2 2400 ( 1)c wL n b n= × + × −

١٩

Minimum thickness of one way slabs ACI Table 9.5(a)

Minimum Cover ACI 7.7.1

a - Concrete exposed to earth or weatherfor Φ<16mm------40 mm and for Φ>16mm----- 50 mm

b - Concrete not exposed to earth or weatherfor Φ<32mm------20 mm, otherwise ------ 40 mm ٢٠

Loads Assigned to Slabs 1 2 D L 1 6 L Lwu=1.2 D.L + 1.6 L.L

a- Dead Load (D L) :a- Dead Load (D.L) : 1- Weight of slab covering materials2- Equivalent partition weightq p g3- Own weight of slab

b Li L d (L L)b- Live Load (L.L)

٢١

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 = 0.315 kN/m2plaster (1.5cm thick) 0.015 21 0.315 kN/m

tiles 2 5 cmtiles cement mortar

sand

2.5 cm2.5 cm

5 cm

slab

plaster

slab

1.5 cm

٢٢

plaster

2-Equivalent partition weightThis load is usually taken as the weight of all walls (weight of 1m spanThis 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 blocksA bl k i h hi k 10 i h 10 kA block with thickness 10cm weighs 10 kg 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 =Load / 1m surface for 20 cm block 12.5 × 20 +2×30=310 kg/m2 = 3.1 kN/m2

Weight of 1m span of wall with height 3m:

20 cm

٢٣

Weight of 1m span of wall with height 3m:For 10 cm block wt. = 1.85 kN/m2 × 3 = 5.6 kN/mFor 20 cm block wt. = 3.1 kN/m2 × 3 = 9.3 kN/m

3- Own weight of slab

Find the total ultimate load per rib for the ribbed slab shown:

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.

٢٤

Solution3- Own weight of slab

• 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= 2 25 + 1 6 = 3 85 kN/m2• Total slab own weight 2.25 + 1.6 3.85 kN/m

L d p ibLoad per rib

Total dead load= 3.85 + 2.315 + 0.75 = 6.915 kN/m2

٢٥

Ultimate load = 1.2(6.915) + 1.6(2) = 11.5 kN/m2

Ultimate load per rib = 11.5 × 0.5 = 5.75 kN/m

Minimum live Load values on slabsType of Use Uniform Live Load

kN/m2

Residential 2

b- Live Load (L.L)

ResidentialResidential balconies

23

Computer use 5Offices 2( )

It depends on the function for which the floor is constructed.

WarehousesLight storageHeavy Storage

612

SchoolsSchoolsClassrooms 2

Librariesrooms 3roomsStack rooms

36

Hospitals 2Assembly Halls

Fixed seatingMovable seating

2.55

Garages (cars) 2.5StoresStores

Retailwholesale

45

Exit facilities 5

٢٦

ManufacturingLightHeavy

46

Loads Assigned to BeamsBeams are usually designed to carry the following loads:Beams are usually designed to carry the following loads:

- Their own weight- Weights of partitions applied directly on them

Fl l d- Floor loads

L

SS S2S1

٢٧

Shrinkage Reinforcement RatioShrinkage Reinforcement RatioAccording to ACI Code and for fy =420 MPa

hbA ×⇒ 0018000180ρ

ACI 7.12.2.1

where, b = width of strip, and h = slab thickness

hbA shrinkagesshrinkage ×=⇒= 0018.00018.0 ,ρ

Mi i R i f t R ti f M i R i f tMinimum Reinforcement Ratio for Main Reinforcement

0 0018s min s shrinkageA A . b h≥ = × ACI 10.5.4s ,min s ,shrinkage ACI 10.5.4

Check shear capacity of the section

c w 1 1 (1.1)0.17 u cV . V f ' b d≤ Φ = ΦOtherwise enlarge depth of slab

٢٨

Spacing of Reinforcement Barsa- Flexural Reinforcement BarsFlexural reinforcement is to be spaced not farther than three times the slab thickness (hs), nor farther apart than 45 cm, center-to-center.

3 h⎧max

3 hS smaller of

45cms⎧

≤ ⎨⎩ ACI 10.5.4

b- Shrinkage Reinforcement BarsShrinkage reinforcement is to be spaced not farther than five times the slabShrinkage reinforcement is to be spaced not farther than five times the slabthickness, nor farther apart than 45 cm, center-to-center.

5 hll s⎧

⎨ ACI 7 12 2 2maxS smaller of45cm

s⎧≤ ⎨

⎩ACI 7.12.2.2

٢٩

Summary of one-way ribbed slab design procedureSummary of one way ribbed slab design procedure

1. The direction of ribs is chosen.2. Determine h, and select the hollow block size, bw and t3. Provide shrinkage reinforcement for the topping slab in bothdirections.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 momentregions and rectangular beams in the regions of negative moment.8. Neat sketches showing arrangement of ribs and details of thereinforcement are to be prepared.p p

٣٠

Example 1

Determine the arrangementDetermine the arrangementof blocks and width ofhidden beams for the planpshown. The blocks usedhave the size of 40 × 20 cmin plan. The live load is 4kN/m2.

٣١

Solution

٣٢Note that the width of hollow blocks in Gaza is 250 mm NOT 200 mm

Solution

٣٣

Solution

٣٤

Solution

٣٥

Example 2

Design a one-way ribbed slab to cover a 3 8 m x 10 m panel shown in theDesign a one way ribbed slab to cover a 3.8 m x 10 m panel, shown in thefigure below. The covering materials weigh 2.25 kN/m2, equivalentpartition load is equal to 0.75 kN/m2, and the live load is 2 kN/m2.p q ,Use fc’=25 MPa, fy=420MPa

m3.

8

10 m

٣٦

Solution1. The direction of ribs is chosen:Ribs are arranged in the short direction as shown in the figure

3.8

m

5 0 m 5 0 m

3

5.0 m 5.0 m2. Determine h, and select the hollow block size, bw and t:From ACI Table 9.5(a), hmin = 380/16 = 23.75cm use h = 24 cm.L t idth f b b 10Let width of web, bw =10 cmUse hollow blocks of size 40 cm × 25 cm × 17 cm (weight=0.17 kN)Topping slab thickness = 24 – 17 = 7cm > lc/12 =40/12= 3.3cm > 5cm OKpp g cFor 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

٣٧

2 28 9 400 16 OK0 9 1240 251240

u c

c

w l . ( )t mm( . )f

≥ = =′Φ

3. Provide shrinkage reinforcement for the topping slab in both directions:

Solutiong pp g

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 m31.0 m

Volume of hollow blocks in 1m2

= 8 × 0.4 × 0.25 × 0.17 = 0.136 m3

Net concrete volume in 1m2

= 0 24 0 136 = 0 104 m3 m

0.05 m

= 0.24- 0.136 = 0.104 m3

Weight of concrete in 1m2

= 0.104 × 25 = 2.6 kN/m2

Weight of hollow blocks in 1m2

1.0

0.25

m

g= 8 × 0.17= 1.36 kN/m2

Total dead load /m2

= 2.25 + 0.75 + 2.6 + 1.36

0.4 m 0.4 m0.1 m4

m

7 cm

= 7.0 kN/m2 0.24

٣٨

Solutionwu=1.2(7)+1.6(2)=11.6 kN/m2wu 1.2(7) 1.6(2) 11.6 kN/mwu/m of rib =11.6x0.5= 5.8 kN/m of rib5. 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

u,max1.1Φ 1.1 0.75 0.17 25 100 208 14400 N = 14.4 V 11cV kN kN= × × × × × = > =

Though shear reinforcement is not required, 4 φ 6 mm stirrups per meterrun are to be used to carry the bottom flexural reinforcement.

٣٩

7. Design flexural reinforcement for the ribs:

Solutiong

There is only positive moments over the simply supported beam, and thesection of maximum positive moment is to be designed as a T-sectionAssume that a<70mm and Φ=0 90→Rectangular section with b = b =500mmAssume that a<70mm and Φ=0.90→Rectangular section with b = be =500mm

247

50

kN.m

6

2

0.85 25 2 10 5 10ρ 1 1420 0 9 0 85 25 500 208

.⎛ ⎞× × ×= − −⎜ ⎟⎜ ⎟× × × ×⎝ ⎠

10

24

As105

k

2

420 0 9 0.85 25 500 208

0 0013ρ 0 0013 500 208 135

.

.A b d mm

⎜ ⎟× × × ×⎝ ⎠=

= = × × =

Use 2φ10mm (As,sup= 157 mm2)ρ 0 0013 500 208 135s eA b d . mm= = × × =

The assumption is right

s y

c e

A f 157 420a 6.2mm 70mm0.85f 'b 0.85 25 500

×= = = <

× ×The assumption is right

٤٠

SolutionCheck As min

cs,min w w

y y

0 25 f ' 1.4A max b d b df f

.;

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

s,min

2 2s,min s,supA 70mm A 157 mm OK= < =

Check Φ=0 9 (ductility of the section)

a 6.2c 7 3 mmβ 0 85

.= = =

Check Φ=0.9 (ductility of the section)

1

t

β 0.85d c 208 7 3ε 0.003 0.003

c 7.3.− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

tε 0 083 0 005 Tension controlled 0 9. . . OK⎝ ⎠ ⎝ ⎠

= >> ⇒ ⇒ Φ =

٤١

Solution8. Neat sketches showing arrangement of ribs and details of the reinforcement are to beprepared

3.8

m

1Φ10

m

1Φ10

m

1Φ10

m

1Φ10

m

AA

5.0 m 5.0 mΦ6mm meshΦ6mm stirrups

7cm

17cm

Φ6mm mesh@20 cm

Φ6mm stirrups@25 cm

24cm17cm

40 cm 10102Φ10mm 2Φ10mm

S i A ASection A-A

٤٢

PART II

Two way ribbed slabs

٤٣

Method of analysis

i iR Lφ γ≥ ∑i

٤٤

Method of coefficients

S

1.0m wSls www +=

Lwlw wβ=

sw wα=

wllw wβ

Rectangularity ratio:

Case

L L 0.76L 0.76LLL L 0.87L0.87L

٤٥

r S S S 0.87S0.87SS 0.76S 0.76SS

Method of coefficients

ECP 203 load coefficients LL< 5kN/m2

2.01.91.81.71.61.51.41.31.21.11.0r

0.850.800.750.700.650.600.550.500.450.400.35α

0 080 090 110 120 140 160 180 210 250 290 35β

Marcus load coefficients LL ≥ 5kN/m2

0.080.090.110.120.140.160.180.210.250.290.35β

2.01.91.81.71.61.51.41.31.21.11.0r

0.7570.7300.6990.6630.6230.5770.5260.4700.4110.3550.292α

0.0470.0560.0670.0790.0950.1140.1370.1650.1980.2400.292β

٤٦

Minimum slab thickness:

To avoid the complexity of calculating α for a two way ribbed slabTo avoid the complexity of calculating α for a two way ribbed slab,one of three equations of the ACI 318-89 which provides an upperbound for the deflection control of the slab thickness can be used forsimplicity.

max

(800 /1.4)36000

n yl fh

+=

36000

٤٧

Design of beams in two way ribbed slabs:

The loads acting on beams have tributary areas which are boundedThe 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

٤٨S L

Design of beams in two way ribbed slabs:

L

S

qu(S/2)

L

ong beam

S

qu(S/2)

mShort beam

L

An equivalent uniformly distributed load can be established for a beam in a two

SShort beam

q yway system. For a triangular load distribution, the equivalent shear force coefficient Cs is

٤٩

equal to 0.5 and the equivalent bending moment coefficient Cb is equal to 0.67.

Design of beams in two way ribbed slabs:q (S/2)

LonS

qu(S/2)

L

ng beam

qu(S/2)

SShort beam

L

For a trapezoidal load distribution, Cs and Cb are given in the following table.

S

Shear and moment equivalent load coefficients for trapezoidal load distribution2.01.91.81.71.61.51.41.31.21.11.0r =L/S

٥٠

0.7500.7370.7220.7060.6880.6670.6430.6150.5830.5450.500Cs0.9170.9080.8970.8850.8700.8520.8300.8030.7690.7250.667Cb

Design of beams in two way ribbed slabs:

Original distribution Equivalent distribution for shear

Equivalent distribution for moment

wu 0.67 wu0.5 wu

S S S

wu Cs wuCb wu

L L L

٥١

Summary of two-way ribbed slab design procedure

1. Evaluate overall slab thickness and key ribbed slab components.2 Determine the total factored load on the slab2. Determine the total factored load on the slab.3. Determine load distributions in the two principal directions.4 Determine the shear force and bending moments4. Determine the shear force and bending moments.5. Check web width for beam shear.6 D i ib i f6. Design rib reinforcement.7. Design drawing.

٥٢

Example 3pDesign the two-way ribbed slab shown in the figure below. The coveringmaterials weigh 1.5 kN/m2, equivalent partition load is equal to 0.75kN/ 2 t h ll bl k 40 25 17 i di ikN/m2, concrete hollow blocks are 40cm×25cm×17cm in dimension,each 17 kg in weight and the live load is 4 kN/m2. All beams are 30 cmwide. Use fc’=30 MPa, fy=420 MPa.

8.0 m

8.0 m

٥٣ 8.0 m 8.0 m

S l tiSolution:

1- Evaluate overall slab thickness and key ribbed slab components:

The largest slab thickness is given by:

800 30 770nl cm= − =

The largest slab thickness is given by:

max

(800 /1.4)36000

n yl fh

+=

max(770)(800 420 /1.4) 23.5

36000h cm+

= =

٥٤Take width of rib = 12 cm.

S l tiSolution:

2- Determine the total factored load on the slab:e e e e o c o ed o d o e s b

Total volume (hatched) = 0.52 × 0.62 × 0.23 = 0.074152 m3

V l f h ll bl k 2[0 4 0 25 0 17] 0 034 3Volume of hollow blocks = 2[0.4 × 0.25× 0.17] = 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/m2Weight of concrete /m 1.0038 /[(0.52)(0.62)] 3.11 kN/m

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

٥٥

S l tiSolution:

3- Determine load distributions in the two principal directions: 0 52 3 Determine load distributions in the two principal directions:

8 1.08

LrS

= = =

0.52 m

w =w =0 35(14 1) =4 92 kN/m2

8 m

w1=w2=0.35(14.1) =4.92 kN/m2

Load per rib in Direction 1:w /m of rib =4 92x0 62= 3 06 kN/m of rib

8 m 0.62 mwu/m of rib =4.92x0.62= 3.06 kN/m of rib

Load per rib in Direction 2:w /m of rib =4 92x0 52= 2 56 kN/m of rib 8 8

m

wu/m of rib 4.92x0.52 2.56 kN/m of rib 8 m 8 m

2

٥٦ 0.4

0.5

1Direction

S l tiSolution:

4- Determine the shear force and bending moments:e e e e s e o ce d be d g o e s

Using the ACI 8.3 coefficients

3 06 kN/m3.06 kN/m

Direction 1

Maximum factored shear force = 1.15wuln/2 = 1.15(3.06) (7.7/2) = 13.5 kN Maximum factored negative moment = wuln

2/9 = 3.06(7.7)2/9 = 20.2 kN.mM i f t d iti t l 2/14 3 06(7 7)2/14 12 9 kNMaximum factored positive moment = wuln

2/14 = 3.06(7.7)2/14 = 12.9 kN.m

2.56 kN/mDirection 2

Maximum factored shear force = 1 15w l /2 = 1 15(2 56) (7 7/2) = 11 4 kN

٥٧

Maximum factored shear force = 1.15wuln/2 = 1.15(2.56) (7.7/2) = 11.4 kN Maximum factored negative moment = wuln

2/9 = 2.56(7.7)2/9 = 16.9 kN.mMaximum factored positive moment = wuln

2/14 = 2.56(7.7)2/14 = 10.9 kN.m

S l tiSolution:

5- Check web width for beam shear:5 C ec web w d o be s e

Reinforcement is closest to the outside surface of concrete.Direction 1

Effective depth d = 23 – 2 – 0.60 – 0.8 = 19.6 cm, assuming φ16 mmreinforcing bars and φ6 mm stirrups.

1 1Φ 1 1 0 75 0 17 30 120 196 18067 NV

u,max

1.1Φ 1.1 0.75 0.17 30 120 196 18067 N = 18.1 V 13 5

cVkN . kN

= × × × × × =

> =

Direction 2Effective depth d = 23 – 2 – 0.60 – 1.6 - 0.8 = 18.0 cm, assuming φ16 mmreinforcing bars and φ6 mm stirrups.

Direction 2

u,max

1.1Φ 1.1 0.75 0.17 30 120 180 16593 N = 16.6 V 11 4

cVkN . kN

= × × × × × =

> =

٥٨

All though shear reinforcement is not required, 4 φ 6 mm stirrups per meterrun are used to carry the bottom flexural reinforcement.

S l tiSolution:

6- Design rib reinforcement:6 es g b e o ce e

Direction 1

Positive moment Negative moment

Mu = 12.9 kN.m

Use mm 1φ 10 and mm 1 φ 12

Mu = 20.2 kN.m

Use mm 2φ 16 in each rib

٥٩

Use φ 0 a d φreinforcing bars in each rib.

Use mm 2φ 16 in each rib.

S l tiSolution:

6- Design rib reinforcement:6 es g b e o ce e

Direction 2

Positive moment Negative moment

Mu = 10.9 kN.m

Use mm 1φ 10 and mm 1 φ 12

Mu = 16.9 kN.m

Use mm 2φ 16 in each rib

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Use φ 0 a d φreinforcing bars in each rib.

Use mm 2φ 16 in each rib.

S l tiSolution:

7- Design drawing:7 es g d w g

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S l tiSolution:

7- Design drawing:7 es g d w g

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