RC 2 2015 16 Chapter 2 Example 3

Reinforced concrete structures II – Ribbed slab Example 2
Example 2.3[shear design, ribbed slab]

Design the ribbed slab system in example 2 for shear using the same material properties and loading.
Solution
A. Rib shear design
Step 1. Material property
Concrete fcu=25Mpa fck=20Mpa
fcd=11.33Mpa fctk=1.5 Mpa
fctm=2.2 Mpa fctd=1.0 Mpa
Steel fyk=300 Mpa fyd=260.87Mpa
𝜀 yd=1.3%0 Es= 200 Gpa
Step 2. Loading on Ribs

Gd= 4.17 KN/m
Qd = 2.4 KN/m
Step 3. Analysis
- Shear envelope diagram
Step 4. Concrete shear capacity

1
𝑉𝑅𝑑,𝑐 = [𝐶𝑅𝑑,𝑐 . 𝐾. (100𝜌1 . 𝑓𝑐𝑘 )3 + 𝐾1 𝜎𝑐𝑝 ] 𝑏𝑤 . 𝑑 > (𝑉𝑚𝑖𝑛 + 𝐾1 . 𝜎𝑐𝑝 )𝑏𝑤 . 𝑑
1
Where:
0.18 0.18
 𝐶𝑅𝑑,𝑐 = = = 0.12
𝛾𝑐 1.5
200
 𝐾 =1+ √ ≤ 2.0 𝑑 = 233
𝑑
K=1.92
𝐴𝑠
 𝜌1 = < 0.02
𝑏𝑤 .𝑑
226.19
𝜌1 = { = 0.0121, 𝑓𝑜𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ 2∅12
80𝑥233
 𝑓𝑐𝑘 = 20 𝑀𝑝𝑎
 K1= 0.15
𝑁𝐸𝐷
 𝜎𝑐𝑝 = < 0.2𝑓𝑐𝑑 = 0 … … … (𝑁𝐸𝐷 = 0)
𝐴𝑐
3 1
 2
𝑉𝑚𝑖𝑛 = 0.035. 𝐾 2 . 𝑓𝑐𝑘 = 0.416
Therefore
𝐹𝑜𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ 2∅12 ⟹ 𝑉𝑅𝑑,𝑐 = 12.43 𝐾𝑁 > 7.75 𝐾𝑁
Step 5: Diagonal compression check of concrete

𝛼𝑐𝑤 .𝑏𝑤 .𝑍.𝑣.𝑓𝑐𝑑
𝑉𝑅𝑑,𝑚𝑎𝑥 =
(cot 𝜃+tan 𝜃)
𝐴𝑠𝑤
𝑉𝑅𝑑,𝑠 = . 𝑍. 𝑓𝑦𝑤𝑑 . cot 𝜃
𝑆
𝑉𝑅𝑑,𝑚𝑎𝑥
𝑉𝑅𝑑 (minimum of) = {
𝑉𝑅𝑑,𝑠
𝑤ℎ𝑒𝑟𝑒
𝛼𝑐𝑤 = 1 𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑛𝑜𝑛 − 𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑑 𝑚𝑒𝑚𝑏𝑒𝑟
𝑙𝑖𝑚𝑖𝑡 𝜃 ⟹ 1 ≤ cot 𝜃 ≤ 2.5

𝑇𝑎𝑘𝑒 … . . cot 𝜃 = 2.5
𝑏𝑤 = 80 𝑚𝑚
2
𝑓𝑐𝑑 = 11.33 𝑀𝑝𝑎
Z=0.9xd = 209.7
1𝑥80𝑥209.7𝑥0.6𝑥11.33
⟹ 𝑉𝑅𝑑,𝑚𝑎𝑥 = 1
= 39.325 𝐾𝑁
(2.5 + )
2.5
Check this value with values form the shear envelop

⟹ 𝑉𝑅𝑑,𝑚𝑎𝑥 > 𝑉𝐸𝐷 … … . . 𝑎𝑡 𝑎𝑙𝑙 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛𝑠, 𝒕𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝑶𝒌!
Step 6. Calculate the required shear reinforcement

𝐴𝑠𝑤 . 𝑓𝑦𝑑 1
≤ . 𝛼𝑐 . 𝑣. 𝑓𝑐𝑑
𝑏𝑤 . 𝑆 2
𝑆𝑚𝑎𝑥 = 0.75. 𝑑. (1 + cot 𝛼), 𝛼 = 900
𝑆𝑚𝑎𝑥 = 0.75d
0.08. √𝑓𝑐𝑘
𝜌𝑚𝑖𝑛 = = 0.00119
𝑓𝑦𝑘
𝜋
𝐴𝑠𝑤 2𝑥(∅62 )
4
𝑆= = = 593.99𝑚𝑚
𝑏𝑤 . 𝜌𝑤 . sin 𝛼 80𝑥0.00119𝑥1
- Check this value with Smax
Required: shear reinforcement
𝐴𝑠𝑤 𝑉𝑒𝑑 2 ∗ 𝜋 ∗ 62
= 𝐴𝑠𝑤 = = 56.54𝑚𝑚2
𝑆 0.78 ∗ 𝑑 ∗ 𝑓𝑦𝑘 ∗ 𝑐𝑜𝑡𝜃 4
𝐴𝑠𝑤 ∗ 0.78 ∗ 𝑑 ∗ 𝑓𝑦𝑘 ∗ 𝑐𝑜𝑡𝜃
𝑆=
𝑉𝑒𝑑
VEd– d from the face of the columns but since the loads are small just take the values at the centre of the
columns.
𝜃- since VRd,max>VEd lets take the conservative value
𝜃 = 22° 𝑐𝑜𝑡𝜃 = 2.5
3
56.54 ∗ 0.78 ∗ 233 ∗ 300 ∗ 2.5 7706.68

𝑆= = 𝐾𝑁𝑚𝑚
𝑉𝑒𝑑 𝑉𝑒𝑑
Scal
span Location VEd(KN) (mm) Sprovid
near A 10.67 722 ф 6 C/C 175 mm
AB
near B -16.34 471 ф 6 C/C 175 mm
near B 13.825 557.44 ф 6 C/C 175 mm
BC
near C 13.825 557.44 ф 6 C/C 175 mm
near C 16.34 471.64 ф 6 C/C 175 mm
CD
near D 10.67 722.27 ф 6 C/C 175 mm
Smin = 594 mm
Smax = 0.75 *d = 0.75 * 233 = 174.75 → 175mm
Provide ф 6 C/C 175 mm throughout
 Additional torsion force on the longitudinal reinforcement.
∆𝐹𝑡𝑑 = 0.5 ∗ VEd ∗ cotθ take VEd = 16.34 (the largest value)
∆𝐹𝑡𝑑 = 0.5 ∗ 16.34 ∗ cot(45) = 𝟐𝟎. 𝟒𝟐𝟓𝐊𝐍

Note: this has to be added to the Flexural computation ( on the tension reinforcements.)
Step 1: Shear envelope
 A and D
o bw = 300, D= 300mm, d=259
o ф8 stirrup, ф 16 rebar
 B and C
o bw = 600, D= 300mm, d=257
4
o ф8 stirrup, ф 20 rebar
Step 2: concrete shear capacity

1
𝑉𝑅𝑑, 𝑐 = [𝐶𝑅𝑑, 𝑐 ∗ 𝑘 ∗ (log 𝑓𝑐𝑘) 3 + K1 ∗ ρσcp] ∗ bco ∗ d
> [Vmin + K1 ∗ σcp] ∗ bco ∗ d
𝐶𝑅𝑑, 𝑐 = 0.12
200 200
K=1+√ =1+√ ≤ 2.0
𝑑 233
𝐾 = 1.926
𝐴𝑠
𝜌= < 0.02
𝑏𝑤 ∗ 𝑑
For A and D:
Support: 6ф16
1206.37
𝜌= = 0.0155
300∗259
Span: 4ф16, 𝜌 = 0.01035
For B and C:
Support: 10ф20 , 𝜌 = 0.02
Span: 4ф16, 𝜌 = 0.0122
𝑓𝑐𝑘 = 20𝑀𝑝𝑎
𝜎𝑐𝑝 = 0
3 1
𝑉𝑚𝑖𝑛 = 0.035 ∗ 𝑘 2 ∗ 𝑓𝑐𝑘 2 = 0.416
 A and D
5
o Span - VRD,c = 56.412 KN > 32.3 KN

o Support - VRD,c = 49.22 KN > 32.2 KN
 B and C
o Span - VRD,c = 121.88 KN > 32.3 KN
o Support - VRD,c = 103.36 KN > 32.3 KN
Step 3: Diagonal compression check of concrete

𝛼𝑐𝑤 ∗ 𝑏𝑤 ∗ 𝑓 ∗ 𝑣 ∗ 𝑓𝑐𝑑
𝑉𝑅𝑑,𝑚𝑎𝑥 =
cot 𝜃 + tan 𝜃
 Girder A and D
 𝛼𝑐𝑤 = 1
 𝑉 = 0.6
 𝑏𝑤 = 300𝑚𝑚
 𝑓𝑐𝑑 = 11.33𝑀𝑝𝑎
 cot 𝜃 = 2.5
 𝑧 = 0.9 ∗ 𝑑 = 0.9 ∗ 239 = 233.1
1 ∗ 0.6 ∗ 300 ∗ 233.1 ∗ 11.33
𝑉𝑅𝑑 𝑚𝑎𝑥 = 1
2.5 ∗
2.5
𝑉𝑅𝑑 𝑚𝑎𝑥 = 163.92𝐾𝑁
 𝑉𝑅𝑑 𝑚𝑎𝑥 > 𝑉𝑅𝑑…………𝑶𝑲
 Girder A and D
 𝛼𝑐𝑤 = 1
 𝑉 = 0.6
 𝑏𝑤 = 600𝑚𝑚
 𝑓𝑐𝑑 = 11.33𝑀𝑝𝑎
 cot 𝜃 = 2.5
 𝑧 = 0.9 ∗ 𝑑 = 0.9 ∗ 257 = 231.3
1 ∗ 0.6 ∗ 600 ∗ 231.3 ∗ 11.33
𝑉𝑅𝑑 𝑚𝑎𝑥 = 1
2.5 ∗
2.5
𝑉𝑅𝑑 𝑚𝑎𝑥 = 325.32𝐾𝑁
 𝑉𝑅𝑑 𝑚𝑎𝑥 > 𝑉𝑅𝑑…………𝑶𝑲
Step 4: Shear reinforcement required

𝐴𝑠𝑤,𝑚𝑎𝑥 ∗ 𝑓𝑦𝑑 1
< ∗ 𝛼𝑐 ∗ 𝑉𝑦𝑐
𝑏𝑤 ∗ 𝑠 2
𝐴𝑠𝑤 𝑉𝐸𝑑
=
𝑠 0.78 ∗ 5 ∗ 𝑓𝑦𝑑 ∗ cot 𝜃
6
𝜋 ∗ 82
𝐴𝑠𝑤 = 2 ∗ = 100.53
4
0.78 ∗ 𝑑 ∗ 𝑓𝑦𝑘 ∗ 𝐴𝑠𝑤 ∗ 𝑐𝑜𝑡𝜃
𝑆=
𝑉𝑒𝑑
𝑆𝑚𝑎𝑥 = 0.75 ∗ 𝑑 ∗ (1 + cot 𝜃)
𝑆𝑚𝑎𝑥 = 194.25 𝑚𝑚 𝑓𝑜𝑟 𝐴 𝑎𝑛𝑑 𝐷
𝑆𝑚𝑎𝑥 = 192.74 𝑚𝑚 𝑓𝑜𝑟 𝐵 𝑎𝑛𝑑 𝐶
𝐴𝑠𝑤
𝑆𝑚𝑖𝑛 = = 281.59 𝑓𝑜𝑟 𝐴𝑛𝑑 𝐷
𝑏𝑤 ∗ 𝜌𝑤 ∗ sin 𝑥
𝑆𝑚𝑖𝑛 = 140.79 𝑚𝑚 𝑓𝑜𝑟 𝐵 𝑎𝑛𝑑 𝐶
Span location VEd(KN) d Scal (mm) S provided

A and D near 1 39.67 259 767.92 ф 8 C/C 560 mm
near 2 66.6 259 456.92 ф 8 C/C 450 mm
near 2 66.6 259 456.92 ф 8 C/C 450 mm
near 3 39.67 259 767.92 ф 8 C/C 560 mm
B and C near 1 107.24 257 281.86 ф 8 C/C 280 mm
near 2 180.74 257 167.24 ф 8 C/C 190 mm
near 2 180.74 257 167.24 ф 8 C/C 190 mm
near 3 107.24 257 281.86 ф 8 C/C 280 mm
Step 7. Detailing
7

RC 2 2015 16 Chapter 2 Example 3

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Reinforced concrete structures II – Ribbed slab Example 2

Example 2.3[shear design, ribbed slab]

fcd=11.33Mpa fctk=1.5 Mpa

fctm=2.2 Mpa fctd=1.0 Mpa

Steel fyk=300 Mpa fyd=260.87Mpa

𝜀 yd=1.3%0 Es= 200 Gpa

Step 2. Loading on Ribs

Step 4. Concrete shear capacity

𝐹𝑜𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ 2∅12 ⟹ 𝑉𝑅𝑑,𝑐 = 12.43 𝐾𝑁 > 7.75 𝐾𝑁

Step 5: Diagonal compression check of concrete

𝑙𝑖𝑚𝑖𝑡 𝜃 ⟹ 1 ≤ cot 𝜃 ≤ 2.5

𝑓𝑐𝑑 = 11.33 𝑀𝑝𝑎

Check this value with values form the shear envelop

Step 6. Calculate the required shear reinforcement

𝑆𝑚𝑎𝑥 = 0.75. 𝑑. (1 + cot 𝛼), 𝛼 = 900

- Check this value with Smax

Required: shear reinforcement

𝜃- since VRd,max>VEd lets take the conservative value

𝜃 = 22° 𝑐𝑜𝑡𝜃 = 2.5

56.54 ∗ 0.78 ∗ 233 ∗ 300 ∗ 2.5 7706.68

Smax = 0.75 *d = 0.75 * 233 = 174.75 → 175mm

Provide ф 6 C/C 175 mm throughout

 Additional torsion force on the longitudinal reinforcement.

∆𝐹𝑡𝑑 = 0.5 ∗ 16.34 ∗ cot(45) = 𝟐𝟎. 𝟒𝟐𝟓𝐊𝐍

Step 1: Shear envelope

Step 2: concrete shear capacity

Span: 4ф16, 𝜌 = 0.01035

Support: 10ф20 , 𝜌 = 0.02

Span: 4ф16, 𝜌 = 0.0122

o Span - VRD,c = 56.412 KN > 32.3 KN

Step 3: Diagonal compression check of concrete

 𝑉𝑅𝑑 𝑚𝑎𝑥 > 𝑉𝑅𝑑…………𝑶𝑲

Step 4: Shear reinforcement required

Span location VEd(KN) d Scal (mm) S provided

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Smax = 0.75 d = 0.75 233 = 174.75 → 175mm