Download - Example 2.2 [Ribbed slab design] - · PDF fileExample 2.2 [Ribbed slab design] A typical floor system of a lecture hall is to be designed as a ribbed slab. ... -Effective width computation

Reinforced concrete structures II – Ribbed slab Example 2

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Example 2.2 [Ribbed slab design]

A typical floor system of a lecture hall is to be designed as a ribbed slab. The joists which are spaced at

400mm are supported by girders. The overall depth of the slab without finishing materials is 300mm.

Imposed load of 1.5KN/m2 for partition and fixture is considered in the design. In addition, the floor has

a floor finish material of 3cm marble over a 2cm cement screed and it ha 2cm plastering as ceiling. Take

the unit weight of ribbed block to be 2KN/m2.

Use: C 20/25

S – 300

Class 1 works

a) Analyze the ribbed slab system, considering the effects of loading pattern

b) Design the ribbed slab system


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Solution:

Step 1:Material property

Concrete:

𝑓𝑐𝑡𝑘, 0.05 = 1.5𝑀𝑝𝑎

𝑓𝑐𝑡𝑚 = 2.2𝑀𝑝𝑎

ɤ𝑐 = 1.5

𝑓𝑐𝑘 = 20𝑀𝑝𝑎, 𝑓𝑐𝑢 = 25𝑀𝑝𝑎

𝑓𝑐𝑑 = 0.85 ∗ 20

1.5= 11.33𝑀𝑝𝑎

Rebar

𝑓𝑦𝑘 = 300𝑀𝑝𝑎

𝑓𝑦𝑑 =𝑓𝑦𝑘

1.15= 260.87𝑀𝑝𝑎

ɛ𝑦𝑑 =𝑓𝑦𝑑

𝐸𝑠=

260.87

200= 1.74‰

Step 2: Verify if the general requirements for Rib slab are met using Euro Code 2

1. The centers of the ribs should not exceed 1.5 m:

- This is satisfied, as the center-to-center spacing between the ribs is 400mm.

2. The depth of ribs excluding topping should not exceed four times their average width.

- Also satisfied as 80 x 4 > 240 mm.

3. The minimum rib width should be determined by consideration of cover, bar spacing

and fire resistance

- BS 8110 code - recommends 125 mm,

- Assume for this example the conditions are satisfied hence assume requirement

satisfied.

4. The thickness of structural topping or flange should not be less than 50mm or one tenth

of the clear distance between ribs.

- 60 mm satisfies this requirement.

Step 3: Loading

Dead load:

Joist→ 0.2 * 0.08 * 25 = 0.4

Topping→ 0.4 * 0.06 * 25 = 0.6

Floor finish → 0.4 * 0.03 * 27 = 0.32

Cement Screed → 0.4 * 0.02 * 23 = 0.184


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Plastering → 0.4 * 0.02 * 23 = 0.184

Partition and fittings → 0.4*1.5 = 0.6

Ribbed block → 0.4 * 2 = 0.8

Gk = 3.092 KN/m

Live load:

Qk = 4KN/m2 * 0.4 = 1.6 KN/m

Design load:

𝐺𝑑 = 1.35 ∗ 𝐺𝑘 = 1.35 ∗ 3.092 = 4.174𝐾𝑁/𝑚

𝑄𝑑 = 1.5 ∗ 𝑄𝑘 = 1.5 ∗ 1.6 = 2.4𝐾𝑁/𝑚

Step 4: Analysis (for Ribs)

i) Full design load

ii) Maximum support moment [at B and C]


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iii) For maximum span moment [ at span AB and CD]

iv) Maximum span moment [ at BC]


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v) Only dead load acting

- Moment envelope diagram for the rib

- Maximum reaction envelope

- Minimum reaction envelope


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Step 5.Loading on Girders

- Assume Width of girders A , D ………….W=300mm

B, C ………….W=600mm

For all girders …...D=300mm

- Note: the section should be checked for serviceability

Self-weight: A & D …………= 0.3 x 0.3 x 25 = 2.25 KN. m

B&C …………= 0.6 x 0.3 x 25 = 4.5 KN. m

Design loads: A & D …………Gd = 1.35 x2.25 = 3.04 KN. m

B & C …….……Gd= 1.35 x4.5 = 6.08 KN. M

Step 6. Analysis of Girders

i. For Girder on axis “A” and “D”

- To get to maximum support moment [at 2]

From the maximum reaction of the Ribs divided by the rib spacing at A and D 10.67

0.4=10.67 KN.m

Total load W=26.68 + 3.04 = 29.72 KN.m


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- To get maximum span moment on girder A & D at 12 & 23

From the maximum reaction of the Ribs divided by the rib spacing A & D

10.67

0.4=10.67 KN.m

From the minimum reaction of the Ribs divided by the rib spacing 6.24

0.4=15.52 KN.m

- Moment envelop for girder A and D


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ii. For Girder on axis “B” and “C”

- Loading: Self-weight = 6.08 KN.m

Reactions from the ribs (divided by the rib spacing) 29.8

0.4= 74.5 𝐾𝑁. 𝑚and

18.467

0.4= 46.15 𝐾𝑁. 𝑚

- To get maximum support moment [at “2”]

- To get maximum span moment [at “12” or “23”]


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- Moment envelop diagram for girders on axis B and C

Step 7.Loading on the Beam …. Axis 1, 2 and 3

- Self-weight width= 200 mm

Depth= 300 mm

N.B: cross section should be checked for serviceability.

Since there are columns at the intersection of the beams and girders, the beams will only support

their own loads.

DL = 0.2 x 0.3 x 25 = 1.5 KN/m

Gd= 1.35 x 1.5 = 2.025 KN/m

- Beam analysis


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Step 8. Design

1. Rib design

Cross section at span

ℎ𝑓 = 60 𝑚𝑚

𝑏𝑤 = 80 𝑚𝑚

ℎ = 260 𝑚𝑚

𝑇𝑎𝑘𝑒 𝑐𝑜𝑣𝑒𝑟 15 𝑚𝑚

𝑑 = 260 − 15 − 6 −12

2= 233 𝑚𝑚

- Effective width computation

𝑏𝑒𝑓𝑓,𝑖 = 0.2𝑏𝑖 + 0.1𝑙𝑜 ≤ 0.2𝑙𝑜

I. For end span(sagging moment)

𝑙𝑜 = 0.85𝑙1

𝑙𝑜 = 0.85 ∗ 4000 = 3400𝑚𝑚

𝑏1 = 𝑏2 = 160 𝑚𝑚

𝑏𝑒𝑓𝑓 1 = 𝑏𝑒𝑓𝑓 2 = 372 < 680 < 𝑏1


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𝑏𝑒𝑓𝑓 = ∑ 𝑏𝑒𝑓𝑓,𝑖 + 𝑏𝑤 ≤ 𝑏

𝑏𝑒𝑓𝑓 = 824 ≤ 400 𝑵𝑶𝑻 𝑶𝑲

𝒃𝒆𝒇𝒇 = 𝟒𝟎𝟎 𝒎𝒎

II. For interior sagging moment (+ve)

𝑙𝑜 = 0.7𝑙2

𝑙𝑜 = 0.7 ∗ 4000 = 2800𝑚𝑚

𝑏1 = 𝑏2 = 160 𝑚𝑚

𝑏𝑒𝑓𝑓 1 = 𝑏𝑒𝑓𝑓 2 = 312 < 560 < 𝑏1


𝑏𝑒𝑓𝑓 = 704 ≤ 400 𝑵𝑶𝑻 𝑶𝑲

𝒃𝒆𝒇𝒇 = 𝟒𝟎𝟎 𝒎𝒎

III. For support hogging moment (-ve)

𝑙𝑜 = 0.15(𝑙1 + 𝑙2)

𝑙𝑜 = 1200𝑚𝑚

𝑏1 = 𝑏2 = 160 𝑚𝑚

𝑏𝑒𝑓𝑓 1 = 𝑏𝑒𝑓𝑓 1 = 152 < 240 < 𝑏1


𝑏𝑒𝑓𝑓 = 384 ≤ 400 𝑶𝑲

𝒃𝒆𝒇𝒇 = 𝟑𝟖𝟒 𝒎𝒎

Note: However since it is a negative moment the width of the compression zone will be, b= 80mm

- Design of the T-section

A. Positive span moment AB and CD


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𝑀𝑠𝑑 = 9.506 𝐾𝑁𝑚 𝑏𝑒𝑓𝑓 = 400 𝑚𝑚 𝑑 = 233𝑚𝑚 𝑓𝑐𝑑 = 11.33 𝑚𝑝𝑎 𝑓𝑦𝑑 = 260.87 𝑚𝑝𝑎

µ𝑠𝑑 =𝑀𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

9.506 ∗ 106𝑁𝑚𝑚

11.33 ∗ 400 ∗ 2332= 0.0386

𝜇𝑠𝑑 < 𝜇𝑠𝑑,𝑙𝑖𝑚 = 0.295 𝑺𝒊𝒏𝒈𝒍𝒚 𝒓𝒆𝒊𝒏𝒇𝒐𝒓𝒄𝒆𝒅

𝐾𝑥 = 0.055 𝑋 = 𝐾𝑥𝑑 = 12.815 𝑚𝑚 < ℎ𝑓 𝒅𝒆𝒔𝒊𝒈𝒏 𝒂𝒔 𝒂 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒔𝒆𝒄𝒕𝒊𝒐𝒏

𝐾𝑧 = 0.975 𝑍 = 𝐾𝑧𝑑 = 227.175𝑚𝑚

𝐴𝑠 =𝑀𝑠𝑑

𝑓𝑦𝑑𝑍=

9.506 ∗ 106𝑁𝑚𝑚

260.87 ∗ 227.175= 160.40 𝑚𝑚2

𝐴𝑠𝑚𝑖𝑛 =0.26𝑓𝑐𝑡𝑚

𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 𝑑 = 233𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎 𝑓𝑦𝑘

= 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 35.54 𝑚𝑚2 < 𝐴𝑠 𝑶𝑲!

𝑢𝑠𝑖𝑛𝑔 ∅ 12 𝑎𝑠 = 113.1𝑚𝑚2 𝑛 =𝐴𝑠

𝑎𝑠= 1.418 𝒖𝒔𝒆 𝟐∅𝟏𝟐𝒃𝒐𝒕𝒕𝒐𝒎 𝒃𝒂𝒓𝒔

B. Negative moment on the rib support B and C

𝑀𝑠𝑑 = 11.512 𝐾𝑁𝑚 𝑏𝑤 = 80 𝑚𝑚 𝑑 = 233𝑚𝑚 𝑓𝑐𝑑 = 11.33 𝑚𝑝𝑎 𝑓𝑦𝑑 = 260.87 𝑚𝑝𝑎



11.512 ∗ 106𝑁𝑚𝑚

11.33 ∗ 80 ∗ 2332= 0.2339

𝜇𝑠𝑑 < 𝜇𝑠𝑑,𝑙𝑖𝑚 = 0.295 𝑆𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑

𝐾𝑧 = 0.88 𝑍 = 𝐾𝑧𝑑 = 205.04 𝑚𝑚


𝑓𝑦𝑑𝑍=

11.512 ∗ 106𝑁𝑚𝑚

260.87 ∗ 205.04= 215.22𝑚𝑚2


𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 𝑑 = 233𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎

𝑓𝑦𝑘 = 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 35.54 𝑚𝑚2 < 𝐴𝑠 𝑶𝑲!


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𝑎𝑠= 1.9029 𝒖𝒔𝒆 𝟐∅𝟏𝟐 𝒃𝒂𝒓𝒔 𝒂𝒕 𝒕𝒉𝒆 𝒕𝒐𝒑

C. Span moment between B and C

𝑀𝑠𝑑 = 4.63 𝐾𝑁𝑚 𝑏𝑒𝑓𝑓 = 400 𝑚𝑚 𝑑 = 233𝑚𝑚 𝑓𝑐𝑑 = 11.33 𝑚𝑝𝑎 𝑓𝑦𝑑 = 260.87 𝑚𝑝𝑎



4.63 ∗ 106𝑁𝑚𝑚

11.33 ∗ 400 ∗ 2332= 0.0188


𝐾𝑥 = 0.07 𝑋 = 𝐾𝑥𝑑 = 16.31 𝑚𝑚 < ℎ𝑓 𝒅𝒆𝒔𝒊𝒈𝒏 𝒂𝒔 𝒂 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒔𝒆𝒄𝒕𝒊𝒐𝒏.

𝐾𝑧 = 0.985 𝑍 = 𝐾𝑧𝑑 = 229.505𝑚𝑚


𝑓𝑦𝑑𝑍=

4.63 ∗ 106𝑁𝑚𝑚

260.87 ∗ 229.505= 77.33𝑚𝑚2


𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 𝑑 = 233𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎 𝑓𝑦𝑘 = 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 35.54 𝑚𝑚2 < 𝐴𝑠𝑶𝑲!


𝑎𝑠= 0.6837 𝒖𝒔𝒆 𝟐∅𝟏𝟐 𝒃𝒐𝒕𝒕𝒐𝒎 𝒃𝒂𝒓𝒔

Note: For the shear design of the ribs, refer to Example 2.3 for ribbed slabs.

2. Girder design

a. Girder at A and D

- Positive span moment

𝑀𝑠𝑑 = 37.728 𝐾𝑁𝑚 𝑏𝑤 = 300 𝑚𝑚 𝐷 = 300 𝑚𝑚 𝑓𝑐𝑑 = 11.33 𝑚𝑝𝑎 𝑓𝑦𝑑 = 260.87 𝑚𝑝𝑎

𝑑 = 300 − 25 − 8 −16

2= 259 𝑚𝑚



37.728 ∗ 106𝑁𝑚𝑚

11.33 ∗ 300 ∗ 2592= 0.165


𝐾𝑧 = 0.907 𝑍 = 𝐾𝑧𝑑 = 234.91 𝑚𝑚


𝑓𝑦𝑑𝑍=

37.728 ∗ 106𝑁𝑚𝑚

260.87 ∗ 234.91= 615.525𝑚𝑚2


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𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 𝑑 = 259 𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎 𝑓𝑦𝑘 = 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 148.148𝑚𝑚2 < 𝐴𝑠𝑶𝑲!

𝑢𝑠𝑖𝑛𝑔 ∅ 16𝑎𝑠 = 200.96 𝑚𝑚2 𝑛 =𝐴𝑠

𝑎𝑠= 3.0629 𝒖𝒔𝒆 𝟒∅𝟏𝟔 𝒃𝒐𝒕𝒕𝒐𝒎 𝒃𝒂𝒓𝒔

- Negative support moment


𝑑 = 300 − 25 − 8 −16

2= 259 𝑚𝑚



59.442 ∗ 106𝑁𝑚𝑚

11.33 ∗ 300 ∗ 2592= 0.260


𝐾𝑧 = 0.841 𝑍 = 𝐾𝑧𝑑 = 217.819 𝑚𝑚


𝑓𝑦𝑑𝑍=

59.442 ∗ 106𝑁𝑚𝑚

260.87 ∗ 217.819= 1046.1𝑚𝑚2


𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 𝑑 = 259 𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎 𝑓𝑦𝑘 = 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 148.148 𝑚𝑚2 < 𝐴𝑠 𝑶𝑲!

𝑢𝑠𝑖𝑛𝑔 ∅ 16 𝑎𝑠 = 200.96 𝑚𝑚2 𝑛 =𝐴𝑠

𝑎𝑠= 3.0629 𝒖𝒔𝒆 𝟔∅𝟏𝟔 𝒃𝒐𝒕𝒕𝒐𝒎 𝒃𝒂𝒓𝒔

b. Girder at B and C

Positive span moment


𝑑 = 300 − 25 − 8 −20

2= 257 𝑚𝑚



101.59 ∗ 106𝑁𝑚𝑚

11.33 ∗ 600 ∗ 2572= 0.226


𝐾𝑧 = 0.867 𝑍 = 𝐾𝑧𝑑 = 222.819 𝑚𝑚


𝑓𝑦𝑑𝑍=

101.59 ∗ 106𝑁𝑚𝑚

260.87 ∗ 222.819= 1747.73𝑚𝑚2


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𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 𝑑 = 259 𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎 𝑓𝑦𝑘

= 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 148.148 𝑚𝑚2 < 𝐴𝑠𝑶𝑲!

𝑢𝑠𝑖𝑛𝑔 ∅ 20𝑎𝑠 = 314𝑚𝑚2 𝑛 =𝐴𝑠

𝑎𝑠= 5.566 𝒖𝒔𝒆 𝟔∅𝟐𝟎 𝒃𝒐𝒕𝒕𝒐𝒎 𝒃𝒂𝒓𝒔

Negative support moment


𝑑 = 300 − 25 − 8 −20

2= 257 𝑚𝑚



161.16 ∗ 106𝑁𝑚𝑚

11.33 ∗ 300 ∗ 2572= 0.358

𝜇𝑠𝑑 > 𝜇𝑠𝑑,𝑙𝑖𝑚 = 0.295 𝐷𝑜𝑢𝑏𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑒𝑐𝑡𝑖𝑜𝑛

𝐾𝑧,𝑙𝑖𝑚 = 0.814

𝑀𝑠𝑑,𝑙𝑖𝑚 = µ𝑠𝑑,𝑙𝑖𝑚𝑓𝑐𝑑𝑏𝑑2 = 0.295 ∗ 11.33 ∗ 600 ∗ 2572 = 132.455 𝐾𝑁𝑚

𝑍 = 𝐾𝑧,𝑙𝑖𝑚 ∗ 𝑑 = 0.814 ∗ 257 = 209.19 𝑚𝑚

𝐴𝑠1 =𝑀𝑠𝑑,𝑙𝑖𝑚

𝑍𝑓𝑦𝑑+

𝑀𝑠𝑑,𝑠 − 𝑀𝑠𝑑,𝑙𝑖𝑚

𝑓𝑦𝑑(𝑑 − 𝑑2)=

132.455 ∗ 106

260.87 ∗ 209.19+

(161.16 − 132.455) ∗ 106

260.87 ∗ (257 − 35)

= 2915.617𝑚𝑚2

𝒖𝒔𝒆 𝟏𝟎 ⌀𝟐𝟎

Compression reinforcement design

- Check if the reinforcement has yielded

𝑑2

𝑑=

35

257= 0.14𝜀𝑠2 = 2.6‰ (𝑟𝑒𝑎𝑑 𝑓𝑟𝑜𝑚 𝑐ℎ𝑎𝑟𝑡)

𝜀𝑠2 = 2.6‰ > 𝜀𝑦𝑑 𝑢𝑠𝑒 𝑓𝑦𝑑 = 260.87

- Calculate the stress in the concrete at the level of compression reinforcement to avoid

double counting of area.

𝜀𝑐𝑠2 = 2.6‰ ≥ 2‰ , Therefore, we take

𝜀𝑐 = 3.5‰ 𝑎𝑛𝑑 𝜎𝑐𝑑,𝑠2 = 11.33 𝑚𝑝𝑎


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𝐴𝑠2=

1

(𝜎𝑠2−𝜎𝑐𝑑,𝑠2)(

𝑀𝑠𝑑𝑠−𝑀𝑠𝑑,𝑙𝑖𝑚𝑑−𝑑2

)=

1

(260.87 − 11.33)((161.16 − 132.455) ∗ 106

(257 − 35)= 518.160𝑚𝑚2

𝒖𝒔𝒆 𝟐⌀𝟐𝟎

3. Beams Design

i) Negative support moment

𝑀𝑠𝑑 = 3.491𝐾𝑁𝑚 𝑏𝑤 = 200 𝑚𝑚 𝐷 = 300 𝑚𝑚 𝑓𝑐𝑑 = 11.33 𝑚𝑝𝑎 𝑓𝑦𝑑 = 260.87 𝑚𝑝𝑎

𝑑 = 300 − 25 − 8 −12

2= 261 𝑚𝑚



3.491 ∗ 106𝑁𝑚𝑚

11.33 ∗ 200 ∗ 2612= 0.0226


𝐾𝑧 = 0.985 𝑍 = 𝐾𝑧𝑑 = 257.08 𝑚𝑚


𝑓𝑦𝑑𝑍=

3.491 ∗ 106𝑁𝑚𝑚

260.87 ∗ 257.08= 52.04𝑚𝑚2


𝑓𝑦𝑘𝑏𝑡𝑑 𝑤ℎ𝑒𝑟𝑒 𝑏𝑡 = 𝑏𝑤 = 200 𝑑 = 261 𝑚𝑚 𝑓𝑐𝑡𝑚 = 2.2 𝑚𝑝𝑎 𝑓𝑦𝑘 = 300 𝑚𝑝𝑎

𝐴𝑠𝑚𝑖𝑛 = 99.52𝑚𝑚2 > 𝐴𝑠𝑁𝑜𝑡 𝑶𝑲!

𝑢𝑠𝑖𝑛𝑔 ∅ 12𝑎𝑠 = 113.04𝑚𝑚2 𝑛 =𝐴𝑠,𝑚𝑖𝑛

𝑎𝑠= 0.88 𝒖𝒔𝒆 𝟐∅𝟏𝟐 𝒃𝒐𝒕𝒕𝒐𝒎 𝒃𝒂𝒓𝒔

𝑼𝒔𝒆 𝟐∅𝟏𝟐 𝒃𝒐𝒕𝒕𝒐𝒎 𝒂𝒏𝒅 𝒕𝒐𝒑 𝒃𝒂𝒓 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒍𝒆𝒏𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒎.

Step 8. Transverse reinforcement

Secondary reinforcement is required for temperature and shrinkage.

𝐴𝑠2 = 20% 𝐴𝑠,𝑚𝑖𝑛

𝐴𝑠2 = 0.12% 𝐴𝑡𝑜𝑝𝑝𝑖𝑛𝑔

𝑆𝑝𝑎𝑐𝑖𝑛𝑔 𝑆 =𝑏𝑎𝑠

𝐴𝑠

𝒖𝒔𝒆 ∅ 𝟖 𝒄𝒄⁄ 𝟐𝟎𝟎 𝒎𝒎


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Step 9. Detailing


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