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1.0 INTRODUCTION Foundation is the substructure below ground that transfer and spread the load from a structureβs column and wall to the ground. Foundation is important for the overall stability of the structure. As earth under the foundation is the most variable of all the material that are considered in the design and engineering of an engineering structure, it is very critical that the safe bearing capacity of the soil must not be exceeded. If the safe bearing capacity of soil is exceeded, it can cause excessive settlement which results in damage to the building and water or gas pipelines. Therefore, it is necessary to conduct site investigation to survey the soil under a proposed structure to determine the soil properties in order to estimate its safe bearing pressure and calculate possible settlements of the structure. There are two types of foundation which includes shallow foundation (pad footing) and deep foundation (deep pile). This report only covers pad footing.
2.0 PROCEDURE Step 1. Determine the plan size of the footing (if not specified, skip to step 2 if size is specified) - Self-weight of the footing must be assumed first before starting if not given. - The size of the footing must be computed using serviceability limit, meaning that the loadings used are not factored using the following formula: Total axial load = 1.0 Gk + 1.0 Qk (in kN)
[Equation 1]
Gk = total axial load (permanent) = dead load from column + self-weight of footing Qk = total axial load (variable)
Required size of footing = Total axial load / safe bearing pressure on soil [Equation 2] -Usually the calculated required size of footing is slightly increased for safety.
Step 2. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing ππΈπ· = 1.35Gk + 1.5Qk Earth pressure =
[Equation 3] ππΈπ·
πππ§π ππ ππππ‘πππ
[Equation 4] 1
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column - If height and effective depth is specified, use the specified dimension. π
π
ππ ππ Shear force at the face of the column, ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
[Equation 5]
π’ is the perimeter of the column.
Step 4. Check for punching shear - Find the basic control perimeter of punching shear. [Equation 6]
Punching shear control perimeter = perimeter of column + 4ππ
- Find the area within the control perimeter. Shear force at the face of the column,
,
= 0.5 2
[0.6 (1 β 250)] 1.5
Area within control perimeter (square) = (π + 4π) β (4 β π)2π is the perimeter of the column.
2
[Equation 5] [Equation 7]
or
Area within control perimeter (rectangle) = (a + 4d) (b + 4d) - (4 β π)2π2
[Equation 6]
Figure 1 Area within control perimeter - Find punching shear force Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter) [Equation 8]
Punching shear stress =
ππΈπ· πππππππ‘ππ Γπ
[Equation 9] 2
- Find ππ π·,πΆ and ππ π·,πΆ πππ 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
[Equation 10]
200
Where k = (1 + β π1 =
) β€ 2.0 with d in mm
π
π΄π ,ππππ’ππππ ππ€ π 3
β€ 0.02 1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π)
[Equation 11]
- Assume π1 to be 0.02 as the area of steel required is not known yet. - If ππΈπ· <ππ π·,πΆ , h of the footing is adequate. - If ππΈπ· >ππ π·,πΆ , h of the footing has to be increased.
Step 5. Find the reinforcement steel bar required to resist bending. - The moment has to be found from the face of the column.
Column
Face of the column
Footing
Figure 2. Cross section of footing and column ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ K=
π πππ ππ€ π 2
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
where Kβ= 0.167
Z = d (0.5 + β0.25 β π΄π π πππ’ππππ =
(πππππ‘πππ β πππππ’ππ )
πΎ ) 1.134
where 0.82d < Z < 0.954d
π 0.87ππ¦π π§
Number of Steel Bars Required =
π΄π π΄πππ ππ π π‘πππ πππ
π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
3
Step 6. Repeat the check for punching shear (step 4) - The checking is done using the actual π1, where area of steel reinforcement required is found in step 5.
Step 7. Check the shear force at critical zone ( ππΈπ· ) - Shear critical zone is 1.0d away from face of column. 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 3
[Equation 12]
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π)
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [
[Equation 13] (πππππ‘πππ β πππππ’ππ ) 2
β π]
[Equation 14]
If ππΈπ· <ππ π·,πΆ , no shear reinforcement required.
4
3.0 DATA & RESULT
B
A 6.069m
C 6.069m
D 6.069m
1 3.069m
2 3.069m
3 Figure 3. Plan View
b =2000mm
h =2000mm
Figure 4. Section of Footing - πππ = 30π/ππ2; ππ¦π = 500π/ππ2 ; Diameter of main steel bar=16mm; Cover= 45mm - Assume that safe bearing pressure on soil = 200kN/π2 - It can be seen that there are four footings to be designed: 1. Footing A1, A3, D1, D3 2. Footing B1, B3, C1, C3 3. Footing A2 and D2 4. Footing B2 and C2 5
3.1 Design of Footing A1, A3, D1, D3 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 161.78kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 161.78 + π πππ π€πππβπ‘ ππ ππππ’ππ = 161.78 + (2.4 Γ 0.3 Γ 0.3)(25) = 161.78 + 5.4 = 167.18kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 167.18 (2000 Γ2000)Γ 10β6
= 41.80kN/ππ
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 200mm d = thickness - cover - β /2 = 200 - 45 - 16/2 = 147mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(147) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 931.39kN > 167.18kN β΄ ππ π·,πππ₯ is adequate.
6
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(147) = 3047.26mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 147)2 β (4 β π)(2 Γ 147)2 = 788544 β 74197.30 = 7.14 Γ 105 ππ2 = π. πππ ππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 41.80 Γ (2 Γ 2 β 0.714) = 41.80 Γ (3.345) = 137.35kN Punching shear stress =
137.35 Γ 103
3047.26 Γ147 = 0.307N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(2) (30) ](3047.26 Γ 147) = 242.88kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π) 1 3
= [0.12(2)(100 Γ 0.02 Γ 30) ](3047.26 Γ 147) = 420.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
200
k = (1 + β
π
) <2
π1 = 0.02
200
= (1 +β147 ) = 2.17 > 2 β΄k=2
ππ π·,πΆ πππππππ‘ππΓπ 420.88Γ103 3047.26 Γ147
= 0.94N/πππ > Punching shear stress = 0.307N/ππ2 β΄ h is adequate for punching shear.
7
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ (2β 0.3)
= 41.8 Γ 2 Γ
2
Γ
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 30.20kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 30.20 Γ106
(30)(2000)(147)2
= 0.023 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.023 1.134
where 0.82d < Z < 0.954d
)
= 0.979d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 30.20 Γ106 0.87(500)(0.954 Γ147)
= 495.05πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
495.05 π162 /4
= 2.46 β3 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 3 Γ π162 /4 = 603.19πππ
8
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(2) (30) ](3047.26 Γ 147) = 242.88kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π πππ.ππ
= (1 +β147 )
=
= 2.17 > 2
= 0.002
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(147)
1
= [0.12(2)(100 Γ 0.002 Γ 30)3 ](3047.26 Γ 147) = 195.35kN < ππ π·,πΆ πππ = 242.88kN
β΄ k =2
β΄ ππ π·,πΆ = 242.88kN > ππΈπ· = 137.35kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 41.8 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.147]
= 58.77kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(2)2 (30)2 ](2000 Γ 147) = 159.41kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(2)(100 Γ 0.002 Γ 30)3 ](2000 Γ 147) = 128.22kN < ππ π·,πΆ πππ = 159.41kN
β΄ ππ π·,πΆ = 159.41kN > ππΈπ· = 58.77kN β΄ No shear reinforcement is required.
9
3.2 Design of Footing B1, B3, C1, C3 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 291.63kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 291.63 + π πππ π€πππβπ‘ ππ ππππ’ππ = 291.63 + (2.4 Γ 0.3 Γ 0.3)(25) = 291.63 + 5.4 = 297.03kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 297.03 (2000 Γ2000)Γ 10β6
= 74.26kN/ππ < safe bearing pressure on soil = 200kN/π2 β΄ Earth pressure exerted is safe.
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 300mm d = thickness - cover - β /2 = 300 - 45 - 16/2 = 247mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(247) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 1564.99kN > 297.03kN β΄ ππ π·,πππ₯ is adequate.
10
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(247) = 4303.89mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 247)2 β (4 β π)(2 Γ 247)2 = 1658944 β 209482.3 = 1.45 Γ 106 ππ2 = π. ππππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 74.26 Γ (2 Γ 2 β 1.45) = 74.26 Γ (2.55) = 189.36kN Punching shear stress =
189.36 Γ 103
4303.89 Γ247 = 0.178N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
200
k = (1 + β
π
) <2
π1 = 0.02
200
= (1 +β247 ) = 1.9 < 2
1 3
= [0.12(1.9)(100 Γ 0.02 Γ 30) ](4303.89 Γ 247) = 948.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
ππ π·,πΆ πππππππ‘ππΓπ 948.88Γ103 4303.89 Γ247
= 0.89N/mππ > Punching shear stress = 0.178N/ππ2 β΄ h is adequate for punching shear.
11
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ (2β 0.3)
= 74.26 Γ 2 Γ
2
(πππππ‘πππ β πππππ’ππ )
(2β 0.3)
Γ
2
1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 53.65kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 53.65 Γ106
(30)(2000)(247)2
= 0.015 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.015 1.134
where 0.82d < Z < 0.954d
)
= 0.987d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 53.65 Γ106 0.87(500)(0.954 Γ247)
= 523.40πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
523.4 π162 /4
= 2.6 β3 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 3 Γ π162 /4 = 603.19πππ
12
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π πππ.ππ
= (1 +β247 )
=
= 1.9 < 2
= 0.001
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(247)
1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](4303.89 Γ 247) = 349.57kN < ππ π·,πΆ πππ = 533.72kN
β΄ ππ π·,πΆ = 533.72kN > ππΈπ· = 189.36kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 74.26 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.247]
= 89.56kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(1.9)2 (30)2 ](2000 Γ 247) = 248.02kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](2000 Γ 247) = 162.44kN < ππ π·,πΆ πππ = 248.02kN
β΄ ππ π·,πΆ = 248.02kN > ππΈπ· = 89.56kN β΄ No shear reinforcement is required.
13
3.3 Design of Footing A2 and D2 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 311.78kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 311.78 + π πππ π€πππβπ‘ ππ ππππ’ππ = 311.78 + (2.4 Γ 0.3 Γ 0.3)(25) = 311.78 + 5.4 = 317.18kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 317.18 (2000 Γ2000)Γ 10β6
= 79.30kN/ππ < safe bearing pressure on soil = 200kN/π2 β΄ Earth pressure exerted is safe.
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 300mm d = thickness - cover - β /2 = 300 - 45 - 16/2 = 247mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(247) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 1564.99kN > 317.18kN β΄ ππ π·,πππ₯ is adequate.
14
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(247) = 4303.89mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 247)2 β (4 β π)(2 Γ 247)2 = 1658944 β 209482.3 = 1.45 Γ 106 ππ2 = π. ππππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 79.3 Γ (2 Γ 2 β 1.45) = 79.3 Γ (2.55) = 202.22kN Punching shear stress =
202.22 Γ 103
4303.89 Γ247 = 0.19N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
200
k = (1 + β
1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
π
) <2
π1 = 0.02
200
= (1 +β247 ) = 1.9 < 2
1 3
= [0.12(1.9)(100 Γ 0.02 Γ 30) ](4303.89 Γ 247) = 948.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
ππ π·,πΆ πππππππ‘ππΓπ 948.88Γ103 4303.89 Γ247
= 0.89N/mππ > Punching shear stress = 0.19N/ππ2 β΄ h is adequate for punching shear.
15
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ (2β 0.3)
= 79.3 Γ 2 Γ
2
Γ
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 57.29kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 57.29 Γ106
(30)(2000)(247)2
= 0.016 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.016 1.134
where 0.82d < Z < 0.954d
)
= 0.986d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 57.29 Γ106 0.87(500)(0.954 Γ247)
= 558.91πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
558.91 π162 /4
= 2.78 β3 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 3 Γ π162 /4 = 603.19πππ
16
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π πππ.ππ
= (1 +β247 )
=
= 1.9 < 2
= 0.001
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(247)
1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](4303.89 Γ 247) = 349.57kN < ππ π·,πΆ πππ = 533.72kN
β΄ ππ π·,πΆ = 533.72kN > ππΈπ· = 202.22kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 79.3 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.247]
= 95.64kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(1.9)2 (30)2 ](2000 Γ 247) = 248.02kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](2000 Γ 247) = 162.44kN < ππ π·,πΆ πππ = 248.02kN
β΄ ππ π·,πΆ = 248.02kN > ππΈπ· = 95.64kN β΄ No shear reinforcement is required.
17
3.4 Design of Footing B2 and C2 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 559.63kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 559.63 + π πππ π€πππβπ‘ ππ ππππ’ππ = 559.63 + (2.4 Γ 0.3 Γ 0.3)(25) = 559.63 + 5.4 = 565.03kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 565.03 (2000 Γ2000)Γ 10β6
= 141.26kN/ππ < safe bearing pressure on soil = 200kN/π2 β΄ Earth pressure exerted is safe.
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 300mm d = thickness - cover - β /2 = 300 - 45 - 16/2 = 247mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(247) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 1564.99kN > 565.03kN β΄ ππ π·,πππ₯ is adequate.
18
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(247) = 4303.89mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 247)2 β (4 β π)(2 Γ 247)2 = 1658944 β 209482.3 = 1.45 Γ 106 ππ2 = π. ππππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 141.26 Γ (2 Γ 2 β 1.45) = 141.26 Γ (2.55) = 360.21kN Punching shear stress =
360.21 Γ 103
4303.89 Γ247 = 0.34N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
200
k = (1 + β
1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
π
) <2
π1 = 0.02
200
= (1 +β247 ) = 1.9 < 2
1 3
= [0.12(1.9)(100 Γ 0.02 Γ 30) ](4303.89 Γ 247) = 948.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
ππ π·,πΆ πππππππ‘ππΓπ 948.88Γ103 4303.89 Γ247
= 0.89N/mππ > Punching shear stress = 0.34N/ππ2 β΄ h is adequate for punching shear.
19
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ = 141.26 Γ 2 Γ
(2β 0.3) 2
Γ
(πππππ‘πππ β πππππ’ππ )
(2β 0.3) 2
1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 102.06kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 102.06 Γ106
(30)(2000)(247)2
= 0.028 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.028 1.134
where 0.82d < Z < 0.954d
)
= 0.975d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 102.06 Γ106 0.87(500)(0.954 Γ247)
= 995.68πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
995.68 π162 /4
= 4.95 β5 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 5 Γ π162 /4 = 1005.31πππ
20
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π ππππ.ππ
= (1 +β247 )
=
= 1.9 < 2
= 0.002
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(247)
1
= [0.12(1.9)(100 Γ 0.002 Γ 30)3 ](4303.89 Γ 247) = 440.43kN < ππ π·,πΆ πππ = 533.72kN
β΄ ππ π·,πΆ = 533.72kN > ππΈπ· = 360.21kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 141.26 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.247]
= 170.36kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(1.9)2 (30)2 ](2000 Γ 247) = 248.02kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(1.9)(100 Γ 0.002 Γ 30)3 ](2000 Γ 247) = 204.67kN < ππ π·,πΆ πππ = 248.02kN
β΄ ππ π·,πΆ = 248.02kN > ππΈπ· = 170.36kN β΄ No shear reinforcement is required.
4.0 CONCLUSION In this report, 4 footings were designed: 1. Footing A1, A3, D1, D3
2. Footing n B1, B3, C1, C3
3. Footing A2 and D2
4. Footing B2 and C2
The design of these footing is done using Eurocode 2. All the footing is singly reinforced.
21
2.0 PROCEDURE Step 1. Determine the plan size of the footing (if not specified, skip to step 2 if size is specified) - Self-weight of the footing must be assumed first before starting if not given. - The size of the footing must be computed using serviceability limit, meaning that the loadings used are not factored using the following formula: Total axial load = 1.0 Gk + 1.0 Qk (in kN)
[Equation 1]
Gk = total axial load (permanent) = dead load from column + self-weight of footing Qk = total axial load (variable)
Required size of footing = Total axial load / safe bearing pressure on soil [Equation 2] -Usually the calculated required size of footing is slightly increased for safety.
Step 2. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing ππΈπ· = 1.35Gk + 1.5Qk Earth pressure =
[Equation 3] ππΈπ·
πππ§π ππ ππππ‘πππ
[Equation 4] 1
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column - If height and effective depth is specified, use the specified dimension. π
π
ππ ππ Shear force at the face of the column, ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
[Equation 5]
π’ is the perimeter of the column.
Step 4. Check for punching shear - Find the basic control perimeter of punching shear. [Equation 6]
Punching shear control perimeter = perimeter of column + 4ππ
- Find the area within the control perimeter. Shear force at the face of the column,
,
= 0.5 2
[0.6 (1 β 250)] 1.5
Area within control perimeter (square) = (π + 4π) β (4 β π)2π is the perimeter of the column.
2
[Equation 5] [Equation 7]
or
Area within control perimeter (rectangle) = (a + 4d) (b + 4d) - (4 β π)2π2
[Equation 6]
Figure 1 Area within control perimeter - Find punching shear force Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter) [Equation 8]
Punching shear stress =
ππΈπ· πππππππ‘ππ Γπ
[Equation 9] 2
- Find ππ π·,πΆ and ππ π·,πΆ πππ 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
[Equation 10]
200
Where k = (1 + β π1 =
) β€ 2.0 with d in mm
π
π΄π ,ππππ’ππππ ππ€ π 3
β€ 0.02 1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π)
[Equation 11]
- Assume π1 to be 0.02 as the area of steel required is not known yet. - If ππΈπ· <ππ π·,πΆ , h of the footing is adequate. - If ππΈπ· >ππ π·,πΆ , h of the footing has to be increased.
Step 5. Find the reinforcement steel bar required to resist bending. - The moment has to be found from the face of the column.
Column
Face of the column
Footing
Figure 2. Cross section of footing and column ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ K=
π πππ ππ€ π 2
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
where Kβ= 0.167
Z = d (0.5 + β0.25 β π΄π π πππ’ππππ =
(πππππ‘πππ β πππππ’ππ )
πΎ ) 1.134
where 0.82d < Z < 0.954d
π 0.87ππ¦π π§
Number of Steel Bars Required =
π΄π π΄πππ ππ π π‘πππ πππ
π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
3
Step 6. Repeat the check for punching shear (step 4) - The checking is done using the actual π1, where area of steel reinforcement required is found in step 5.
Step 7. Check the shear force at critical zone ( ππΈπ· ) - Shear critical zone is 1.0d away from face of column. 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 3
[Equation 12]
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π)
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [
[Equation 13] (πππππ‘πππ β πππππ’ππ ) 2
β π]
[Equation 14]
If ππΈπ· <ππ π·,πΆ , no shear reinforcement required.
4
3.0 DATA & RESULT
B
A 6.069m
C 6.069m
D 6.069m
1 3.069m
2 3.069m
3 Figure 3. Plan View
b =2000mm
h =2000mm
Figure 4. Section of Footing - πππ = 30π/ππ2; ππ¦π = 500π/ππ2 ; Diameter of main steel bar=16mm; Cover= 45mm - Assume that safe bearing pressure on soil = 200kN/π2 - It can be seen that there are four footings to be designed: 1. Footing A1, A3, D1, D3 2. Footing B1, B3, C1, C3 3. Footing A2 and D2 4. Footing B2 and C2 5
3.1 Design of Footing A1, A3, D1, D3 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 161.78kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 161.78 + π πππ π€πππβπ‘ ππ ππππ’ππ = 161.78 + (2.4 Γ 0.3 Γ 0.3)(25) = 161.78 + 5.4 = 167.18kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 167.18 (2000 Γ2000)Γ 10β6
= 41.80kN/ππ
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 200mm d = thickness - cover - β /2 = 200 - 45 - 16/2 = 147mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(147) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 931.39kN > 167.18kN β΄ ππ π·,πππ₯ is adequate.
6
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(147) = 3047.26mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 147)2 β (4 β π)(2 Γ 147)2 = 788544 β 74197.30 = 7.14 Γ 105 ππ2 = π. πππ ππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 41.80 Γ (2 Γ 2 β 0.714) = 41.80 Γ (3.345) = 137.35kN Punching shear stress =
137.35 Γ 103
3047.26 Γ147 = 0.307N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(2) (30) ](3047.26 Γ 147) = 242.88kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π) 1 3
= [0.12(2)(100 Γ 0.02 Γ 30) ](3047.26 Γ 147) = 420.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
200
k = (1 + β
π
) <2
π1 = 0.02
200
= (1 +β147 ) = 2.17 > 2 β΄k=2
ππ π·,πΆ πππππππ‘ππΓπ 420.88Γ103 3047.26 Γ147
= 0.94N/πππ > Punching shear stress = 0.307N/ππ2 β΄ h is adequate for punching shear.
7
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ (2β 0.3)
= 41.8 Γ 2 Γ
2
Γ
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 30.20kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 30.20 Γ106
(30)(2000)(147)2
= 0.023 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.023 1.134
where 0.82d < Z < 0.954d
)
= 0.979d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 30.20 Γ106 0.87(500)(0.954 Γ147)
= 495.05πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
495.05 π162 /4
= 2.46 β3 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 3 Γ π162 /4 = 603.19πππ
8
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(2) (30) ](3047.26 Γ 147) = 242.88kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π πππ.ππ
= (1 +β147 )
=
= 2.17 > 2
= 0.002
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(147)
1
= [0.12(2)(100 Γ 0.002 Γ 30)3 ](3047.26 Γ 147) = 195.35kN < ππ π·,πΆ πππ = 242.88kN
β΄ k =2
β΄ ππ π·,πΆ = 242.88kN > ππΈπ· = 137.35kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 41.8 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.147]
= 58.77kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(2)2 (30)2 ](2000 Γ 147) = 159.41kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(2)(100 Γ 0.002 Γ 30)3 ](2000 Γ 147) = 128.22kN < ππ π·,πΆ πππ = 159.41kN
β΄ ππ π·,πΆ = 159.41kN > ππΈπ· = 58.77kN β΄ No shear reinforcement is required.
9
3.2 Design of Footing B1, B3, C1, C3 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 291.63kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 291.63 + π πππ π€πππβπ‘ ππ ππππ’ππ = 291.63 + (2.4 Γ 0.3 Γ 0.3)(25) = 291.63 + 5.4 = 297.03kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 297.03 (2000 Γ2000)Γ 10β6
= 74.26kN/ππ < safe bearing pressure on soil = 200kN/π2 β΄ Earth pressure exerted is safe.
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 300mm d = thickness - cover - β /2 = 300 - 45 - 16/2 = 247mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(247) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 1564.99kN > 297.03kN β΄ ππ π·,πππ₯ is adequate.
10
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(247) = 4303.89mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 247)2 β (4 β π)(2 Γ 247)2 = 1658944 β 209482.3 = 1.45 Γ 106 ππ2 = π. ππππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 74.26 Γ (2 Γ 2 β 1.45) = 74.26 Γ (2.55) = 189.36kN Punching shear stress =
189.36 Γ 103
4303.89 Γ247 = 0.178N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
200
k = (1 + β
π
) <2
π1 = 0.02
200
= (1 +β247 ) = 1.9 < 2
1 3
= [0.12(1.9)(100 Γ 0.02 Γ 30) ](4303.89 Γ 247) = 948.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
ππ π·,πΆ πππππππ‘ππΓπ 948.88Γ103 4303.89 Γ247
= 0.89N/mππ > Punching shear stress = 0.178N/ππ2 β΄ h is adequate for punching shear.
11
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ (2β 0.3)
= 74.26 Γ 2 Γ
2
(πππππ‘πππ β πππππ’ππ )
(2β 0.3)
Γ
2
1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 53.65kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 53.65 Γ106
(30)(2000)(247)2
= 0.015 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.015 1.134
where 0.82d < Z < 0.954d
)
= 0.987d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 53.65 Γ106 0.87(500)(0.954 Γ247)
= 523.40πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
523.4 π162 /4
= 2.6 β3 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 3 Γ π162 /4 = 603.19πππ
12
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π πππ.ππ
= (1 +β247 )
=
= 1.9 < 2
= 0.001
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(247)
1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](4303.89 Γ 247) = 349.57kN < ππ π·,πΆ πππ = 533.72kN
β΄ ππ π·,πΆ = 533.72kN > ππΈπ· = 189.36kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 74.26 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.247]
= 89.56kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(1.9)2 (30)2 ](2000 Γ 247) = 248.02kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](2000 Γ 247) = 162.44kN < ππ π·,πΆ πππ = 248.02kN
β΄ ππ π·,πΆ = 248.02kN > ππΈπ· = 89.56kN β΄ No shear reinforcement is required.
13
3.3 Design of Footing A2 and D2 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 311.78kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 311.78 + π πππ π€πππβπ‘ ππ ππππ’ππ = 311.78 + (2.4 Γ 0.3 Γ 0.3)(25) = 311.78 + 5.4 = 317.18kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 317.18 (2000 Γ2000)Γ 10β6
= 79.30kN/ππ < safe bearing pressure on soil = 200kN/π2 β΄ Earth pressure exerted is safe.
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 300mm d = thickness - cover - β /2 = 300 - 45 - 16/2 = 247mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(247) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 1564.99kN > 317.18kN β΄ ππ π·,πππ₯ is adequate.
14
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(247) = 4303.89mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 247)2 β (4 β π)(2 Γ 247)2 = 1658944 β 209482.3 = 1.45 Γ 106 ππ2 = π. ππππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 79.3 Γ (2 Γ 2 β 1.45) = 79.3 Γ (2.55) = 202.22kN Punching shear stress =
202.22 Γ 103
4303.89 Γ247 = 0.19N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
200
k = (1 + β
1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
π
) <2
π1 = 0.02
200
= (1 +β247 ) = 1.9 < 2
1 3
= [0.12(1.9)(100 Γ 0.02 Γ 30) ](4303.89 Γ 247) = 948.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
ππ π·,πΆ πππππππ‘ππΓπ 948.88Γ103 4303.89 Γ247
= 0.89N/mππ > Punching shear stress = 0.19N/ππ2 β΄ h is adequate for punching shear.
15
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ (2β 0.3)
= 79.3 Γ 2 Γ
2
Γ
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 57.29kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 57.29 Γ106
(30)(2000)(247)2
= 0.016 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.016 1.134
where 0.82d < Z < 0.954d
)
= 0.986d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 57.29 Γ106 0.87(500)(0.954 Γ247)
= 558.91πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
558.91 π162 /4
= 2.78 β3 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 3 Γ π162 /4 = 603.19πππ
16
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π πππ.ππ
= (1 +β247 )
=
= 1.9 < 2
= 0.001
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(247)
1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](4303.89 Γ 247) = 349.57kN < ππ π·,πΆ πππ = 533.72kN
β΄ ππ π·,πΆ = 533.72kN > ππΈπ· = 202.22kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 79.3 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.247]
= 95.64kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(1.9)2 (30)2 ](2000 Γ 247) = 248.02kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(1.9)(100 Γ 0.001 Γ 30)3 ](2000 Γ 247) = 162.44kN < ππ π·,πΆ πππ = 248.02kN
β΄ ππ π·,πΆ = 248.02kN > ππΈπ· = 95.64kN β΄ No shear reinforcement is required.
17
3.4 Design of Footing B2 and C2 Step 1. Compute the earth pressure (associated with the critical loading arrangement at ultimate limit state) exerted by the footing.
ππΈπ· = 559.63kN (found from progress report 5: Design of RC Column) ππΈπ·,π‘ππ‘ππ = 559.63 + π πππ π€πππβπ‘ ππ ππππ’ππ = 559.63 + (2.4 Γ 0.3 Γ 0.3)(25) = 559.63 + 5.4 = 565.03kN
Density of Reinforced Concrete = 25kN/π3
Assumptions: safe bearing pressure on soil = 200kN/π2
Earth pressure = =
ππΈπ· πππ§π ππ ππππ‘πππ 565.03 (2000 Γ2000)Γ 10β6
= 141.26kN/ππ < safe bearing pressure on soil = 200kN/π2 β΄ Earth pressure exerted is safe.
Step 3. Assume a height (h) and effective depth (d) taking account of compression anchorage requirements of starter bars and check the shear force at the face of the column Assume height of footing = 300mm d = thickness - cover - β /2 = 300 - 45 - 16/2 = 247mm π
π
ππ ππ ππ π·,πππ₯ = 0.5π’π [0.6 (1 β 250 )] 1.5
= 0.5(1200)(247) [0.6 (1 β
30 30 )] 250 1.5
Perimeter of the column, π’ = 4 Γ 300 = 1200mm
= 1564.99kN > 565.03kN β΄ ππ π·,πππ₯ is adequate.
18
Step 4. Check for punching shear Punching shear control perimeter = perimeter of column + 4ππ
= 4 Γ 300 + 4π(247) = 4303.89mm Area within control perimeter (square) = (π + 4π)2 β (4 β π)2π2 = (300 + 4 Γ 247)2 β (4 β π)(2 Γ 247)2 = 1658944 β 209482.3 = 1.45 Γ 106 ππ2 = π. ππππ Punching shear force, ππΈπ· = Earth pressure Γ (Area of footing β Area within perimeter)
= 141.26 Γ (2 Γ 2 β 1.45) = 141.26 Γ (2.55) = 360.21kN Punching shear stress =
360.21 Γ 103
4303.89 Γ247 = 0.34N/πππ
3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
200
k = (1 + β
1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](πππππππ‘ππ Γ π)
π
) <2
π1 = 0.02
200
= (1 +β247 ) = 1.9 < 2
1 3
= [0.12(1.9)(100 Γ 0.02 Γ 30) ](4303.89 Γ 247) = 948.88kN > ππ π·,πΆ πππ = 242.88kN πΎπ π·,πΆ =
=
ππ π·,πΆ πππππππ‘ππΓπ 948.88Γ103 4303.89 Γ247
= 0.89N/mππ > Punching shear stress = 0.34N/ππ2 β΄ h is adequate for punching shear.
19
Step 5. Find the reinforcement steel bar required to resist bending.
ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ = 141.26 Γ 2 Γ
(2β 0.3) 2
Γ
(πππππ‘πππ β πππππ’ππ )
(2β 0.3) 2
1
2
Γ
(πππππ‘πππ β πππππ’ππ ) 2
1
(2)
(2)
= 102.06kNm
K= =
π
where Kβ= 0.167
πππ ππ€ π 2 102.06 Γ106
(30)(2000)(247)2
= 0.028 β΄ section is singly reinforced.
Z = d (0.5 + β0.25 β = d (0.5 + β0.25 β
πΎ ) 1.134 0.028 1.134
where 0.82d < Z < 0.954d
)
= 0.975d >0.954d
β΄ Z = 0.954d
π΄π π πππ’ππππ =
=
π 0.87ππ¦π π§ 102.06 Γ106 0.87(500)(0.954 Γ247)
= 995.68πππ Number of Steel Bars Required =
=
π΄π π΄πππ ππ π π‘πππ πππ
995.68 π162 /4
= 4.95 β5 π΄π ππππ£ππππ = π Γ ππππ ππ π π‘πππ πππ
= 5 Γ π162 /4 = 1005.31πππ
20
Step 6. Repeat the check for punching shear 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](πππππππ‘ππ Γ π) 3 2
200
k = (1 + β
1 2
= [0.035(1.9) (30) ](4303.89 Γ 247) = 533.72kN
π
200
) <2
π1 =
π΄π ,ππππ£ππππ ππ€π ππππ.ππ
= (1 +β247 )
=
= 1.9 < 2
= 0.002
1 3
ππ π·,πΆ = [0.12π(100 π1 πππ ) ](πππππππ‘ππ Γ π)
(2000)(247)
1
= [0.12(1.9)(100 Γ 0.002 Γ 30)3 ](4303.89 Γ 247) = 440.43kN < ππ π·,πΆ πππ = 533.72kN
β΄ ππ π·,πΆ = 533.72kN > ππΈπ· = 360.21kN β΄ h is adequate. Step 7. Check the shear force at critical zone ( ππΈπ· ) ππΈπ· = Earth Pressure Γ βππππ‘πππ Γ [ = 141.26 Γ 2 Γ [
(2β 0.3) 2
(πππππ‘πππ β πππππ’ππ ) 2
β π]
β 0.247]
= 170.36kN 3
1
ππ π·,πΆ πππ = [0.035π 2 πππ 2 ](ππ€ Γ π) 3
1
= [0.035(1.9)2 (30)2 ](2000 Γ 247) = 248.02kN 1
ππ π·,πΆ = [0.12π(100 π1 πππ )3 ](ππ€ Γ π) 1
= [0.12(1.9)(100 Γ 0.002 Γ 30)3 ](2000 Γ 247) = 204.67kN < ππ π·,πΆ πππ = 248.02kN
β΄ ππ π·,πΆ = 248.02kN > ππΈπ· = 170.36kN β΄ No shear reinforcement is required.
4.0 CONCLUSION In this report, 4 footings were designed: 1. Footing A1, A3, D1, D3
2. Footing n B1, B3, C1, C3
3. Footing A2 and D2
4. Footing B2 and C2
The design of these footing is done using Eurocode 2. All the footing is singly reinforced.
21
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