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Page 1: 10. Design of Footing

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Design of Footing :

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Foundation:The purpose of a foundation is to transfer the imposed load of a building or structure onto a suitable substratum.

Loadings:There are three main types of load:

1. Dead loads are the static or constant weight of the structure made up from the walls, ���������������

2. Live loads (also known as imposed loads) such as furniture, goods and people, are movable. In addition there are variable loads caused by the weather, such as snow and rain.

3. Wind loads can be positive or negative. They are not dead or imposed and therefore require speci ��������������

Loads can be calculated and expressed as a force (weight) that is to be transferred onto the substrata.

Substrata bearing capacity:The bearing capacity of the substrata is the amount of weight the ground (substrata) can support. For a structure to be stable, the force (load) bearing down must be opposed by an equal force pushing upwards. When both upward and downward forces are of the same magnitude (size) they are said to be in equilibrium. When the downward force is greater than the upward force, the structure will sink or subside. Conversely, a larger upward force than downward force can push the structure upward, causing it to heave as in Figure.

What causes heave?The main cause of movement of buildings is the weather; those on clay soils are most affected.If clay dries out during a period of drought ���� it will shrink, causing uneven support beneath a structure. Figure below shows how ground around a building has dried out from the effects of the Sun and access to air, whereas the ground beneath the building remains moist; the resulting uneven support for the building may cause the foundations to fail.

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In the past three decades there have been several severe droughts causing hundreds of millions of pounds worth of damage each year. With climate change and a predicted increase in frequency of periods of low rainfall the problem will continue. It’s requires foundations in clay to be at a minimum depth of 0.75m below ground level because it is considered that the effects of the Sun cannot penetrate down to that level. However, it is the building control inspectors who will decide the required depth on site; this is commonly at least 1.00m. If tree roots or soft spots are found, the required depth of foundations can increase signi ������� ��� ���� ��� -cohesive soils, such as sands and gravels, the minimum depth of foundations can be 0.45m.What else can affect the bearing capacity of the ground? Water can have signi �������������the ground; some clay soils can increase in height (swell) by up to 75mm between dry and saturated. As the moisture levels rise, the plasticity of clay increases allowing landslip.Another cause of water loss is the presence of trees, in particular broad leaved trees whichremove vast quantities of water from the ground in the spring and summer months during their growing period. Water and nutrients are extracted from the ground by the roots and transported through the plant to the leaves from which large volumes of water evaporate; this is known as the transpiration stream. Trees absorb carbon dioxide which combines with water in the leaves to produce sugars for plant growth, releasing oxygen as a waste product; this process is called photosynthesis. In drought conditions trees continue to evaporate water brought in via their root system. During continual drought the trees will drop many of their leaves to conservewater in the trunk; however, the ground around the tree will have been seriously depleted of water (becoming desiccated).When wetted the clays wells causing heave and softens (becomes plastic). When clay is saturated, its compressive strength is reduced considerably (by up to 50%) because the platelets become lubricated and easily move over each other. For example, after heavy rain site vehicles may not be able to operate because they would sink into the softened clay. Clay substratum can lose up to 50% of its load bearing capacity and buildings may subside (subsidence). In Figure4.3, structure A may be pushed evenly upward and therefore is unlikely to be damaged.However, if the ground strata change as beneath structure B, the upward force will be uneven and may result in reduced load bearing capacity causing slip and heave, so structural damage is

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likely. Frost can also cause heave in saturated clay, although generally only about the top 0.6m will be subject to movement. Horizontal pressure can cause sub-ground level walling to be moved if inadequate back ������������ ���������������Other causes of ground movement can include broken drains and changes to water courses Water abstraction for irrigation, manufacture or human use can have considerable effects on ground stability. Where the land is adjacent to tidal rivers or sea, the ground may swell and shrink with the tidal cycles. This is particularly common in marsh of ������������������Old mines which have not been charted, and therefore remain undiscovered in modern times, can be a major cause of subsidence; for example, mines for coal, iron ore, copper, tin, zinc and galena, stone extraction for building in and around Bath and parts of Derbyshire, or �ints in parts of Norfolk.

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Design for wall footing:1- Calculate the width of footing (b):The wall footing will be designed as one way cantilever slab for a strip of (1m) length

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���� = �������� � ( �� .) = �������� � � ���� . + �����2 � = � + ����� ������������� �� 0.25�

Where:�������� = �������� ���� �������� ��/�2H = depth of footing- mb = width of footing –mD = service dead loads -kN/mL = service live loads -kN/m���� . = density of concrete = 24 kN/m3����� = density of soil kN/m3

2- Thickness of footing (h):The thickness is calculated depending on shear capacity, the ACI-Code 318 limits the minimum thickness of 150mm and 300mm for pile cups.

���� = 1.2 � � + 1.6 � � � 1��� �� ������ (�) ���� > 150�� ���� ! = � 70 �� 75 (�� ��) !�� /2"#� = 0.75 � $%&�6 � !

For shear the critical section is at distance (d ) from face of wall:#�! = ���� (�2 !) ' "#�If the difference between the two shear is too much reduce the assumed (h).

3- Reinforcement required:Two type of reinforcement required the main and the shrinkage and temperature reinforcement.

For main reinforcement: * = ���� ��22 !����!��, �� ��� ���� ���� ��� �������� ������� %�� ��!��, �� ����� �����

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For concrete wall * = ���� �(�)28

For masonry wall * = ���� �(2�)232

-� = *�"!2 4 = %�0.85%&� 9 = 14 [1 :1 2�-��4%� ] ;� = 9 � � ! ��! = 1000��For secondary reinforcement the ACI-Code requirements:

;���� = 9��� � � � < = �;��;� ' 3h for main reinf. and ' 5h for minimum reinf. ' 450mm.

4- Development length: �! ��� �!�! = �2 70 �� 75���! �������! !����!��, �� ��� !������� �% ����! = 9 � ! � %� � @A�10 � $%&� B 300�� %�� ��� ����" > 19���! = 18 � ! � %� � @A�25 � $%&� B 300�� %�� ��� ����" < 19��If the required Ld is larger than provided use hooks:

�!� = 100 � !��$%&� B 8 � !�� B 150��Ex. 1: Design a concrete wall footing with 400mm thickness supports dead loads of 120kN/m and love loads of 100kN/m allowable soil bearing capacity at depth 1m equal 200kN/m2 �������= 16 kN/m3 ( use f’c = 20 MPa and fy = 400MPa with bars of 12mm)

���� = �������� � ( �� .) = 200 1 � 24 + 162 � = 180��/�2 = 120 + 100180 = 1.22 E 1.25� ���� = 1.2 � � + 1.6 � � � 1 = 1.2 � 120 + 1.6 � 1001.25 = 243.2��/�2Assume h= 300mm d = 300-70-12/2= 224mm

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"#� = 0.75 � $%&�6 � ! = 0.75 F206 � 1000 � 2241000 = 125.75��#�! = ���� � �2 !� = 243.2 �1.25 0.42 0.224� = 48.64�� ' "#� �. �.* = ���� � ( �)2

8 = 243.2(1.25 0.4)28 = 21.95��. �/�

-� = *�"!2 = 21.95�1060.9�1000�2242 = 0.486 4 = %�0.85%&� = 4000.85�20 = 23.35

9 = 14 G1 :1 2�-��4%� H = 0.00123 < 0.0018Then in both direction use ;���� = 9��� � � � = 0.0018 � 1000 � 300 = 540��2< = � ;��;� = 1000 � 113540 209�� ��� "IJKK@JMMKK ' 3 � 300 = 900�� �! ��� �!�! = �2 70 = 1250 4002 70 = 355��

�! = 18 � ! � %� � @A�25 � $%&� = 18 � 12 � 400 � 125F20 = 772.78�� > � ��� �!�! = 355��Use hooks �!� = 100 � !��$%&� = 100 � 12F20 = 268.3�� ��� 270�� B 8 � 12 = 96 B 150��Ex.2: Check the adequacy for the footing shown below, the bearing capacity of the soil is 100kN/m2 at depth of 0.75 for ����� . = 16 ��/�3 use f’c= 25 MPa and fy = 400Mpa. The service dead loads are 45 kN/m and the service live loads are 35kN/m?

���� = 100 0.25 � 24 16 � 0.5 =86 ��/�2 = 45 + 3586 = 0.93 ��� 1.0 � ���� = 1.2 � � + 1.6 � � � 1

= 1.2 � 45 + 1.6 � 351.0= 110 ��/�2Assume h= 250mm d = 250-70-14 / 2 = 173 mm

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"#� = 0.75 � $%&�6 � ! = 0.75 F256 � 1000 � 1731000 = 108.125 ��#�! = ���� � �2 !� = 110 �1.0 0.252 0.173� = 22.22 �� ' "#� �. �.* = ���� � (2 �)2

32 = 110(2 � 1.0 0.25)232 = 10.53��. �/�

-� = *�"!2 = 10.53�1060.9�1000�1732 = 0.0.391 4 = %�0.85%&� = 4000.85�20 = 18.83

9 = 14 G1 :1 2�-��4%� H = 0.00098 < 0.0018Then in both direction use ;���� = 9��� � � � = 0.0018 � 1000 � 250 = 450��2< = � ;��;� = 1000 � 154450 = 342.2 �� ' 3 � 250 = 750�� < = � ;��;� = 1000 � 79450 = 175.55 �� ' 5 � 250 = 1250�� �� NOM���! ��� �!�! = �2 70 = 1000 2502 70 = 305��

�! = 18 � ! � %� � @A�25 � $%&� = 18 � 14 � 400 � 125F25 = 806.4�� > � ��� �!�! = 305��Use hooks �!� = 100 � !��$%&� = 100 � 14F25 = 280�� B 8 � 14 = 112�� B 150��Single column footing:

A) Concentrically loaded single column footing area:

���� = �������� � ( �� .) = �������� � � ���� . + �����2 �; = P � � = �+����� ������������� �� 0.25� For square footing F; = P = �Thickness of footing assumed more than 150mm. Then d= h -70 – dav.

1- Beam shear(critical section at distance d from face of support)

/2

���� = 1.2 � � + 1.6 � �;"#� = 0.75 � $%&�6 � !

For shear the critical section is at distance (d ) from face of wall:

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#�! = ���� (P2 !) ' "#�2- Punching shear A1= (b+d)*(c+d) A2= B*H

#� = ����. (;2 ;1) �� VcAnd Vc shall be calculated according to ACI-Code 318-08:

Where:A = ���,�� � ������� B 2 is the ratio of long to short side of column for square column = 1� = ��������� �% ���� �������! � ������ ��! �%%���� � !���� = 2( + ! + � + !)Q = 1 for normal concrete weight.

Reinforcement required:

Long direction: * = ���� �(�)2�8-� = *�"�!2 4 = %�0.85%&� 9 = 14 G1 :1 2�-��4%� H ' 9��� = 0.75 9 ��! > 9��� = 1.4%�;� = 9 � � � ! ��. �% ��� = ;� �����;� �� Short direction: * = ���� �(P�)2�P8-� = *�"P!2 4 = %�0.85%&� 9 = 14 G1 :1 2�-��4%� H ' 9��� = 0.75 9 ��! > 9��� = 1.4%�;� = 9 � P � !��. �% ��� = ;� �����;� ��

Distribution of reinforcement according to ACI-Code 318-08:

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;�2 = �� � ;� ����� ;�1 = (1 ��);�

Bearing transfer of stresses at base of column :

RSTUS. = " VM. WO � X& S:YJYI � YIZ < " � 0.85 � X& S � J � YI and " = M. \ Pult.=1.2*D+1.6*L �� ������

A1= area of column (loaded area (b*c))A2 = area rounded by 2d around column(b+2d)(c+2d) according to figure below.

Development length: check in both direction as in wall footing.

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Development length: �! ��� �!�! = �2 70 �� 75���! �������! !����!��, �� ��� !������� �% ����! = 9 � ! � %� � @A�10 � $%&� B 300�� %�� ��� ����" > 19���! = 18 � ! � %� � @A�25 � $%&� B 300�� %�� ��� ����" < 19��If the required Ld is larger than provided use hooks:

�!� = 100 � !��$%&� B 8 � !�� B 150��Dowels:

;!���� = 0.005 ;(������ )�! ����. = ! � %�4$%&� B 0.04 ! � %� B 200��

Ex. 3: A circle tied column 500*500mm reinforced with 4"25mm support a service dead load of 1100kN and service live load of 800 kN. The allowable soil pressure at depth 1500mm is

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250kN/m2 and ����� = 16��/�3. Design a separate footing for this column, use f’c =27.5MPa and fy = 410MPa?

���� = �������� � ( �� .) = 250 1.5 � 24 + 162 � = 220 ��/�2; = �+����� = 1100+800220 = 8.64�2 , For square footing P = � = F; = F8.64 = 2.94 _ 3�Thickness of footing assumed more than 150mm.try h=600mm

Then d = 600 -75 – 25 = 500mmBeam shear (critical section at distance d from face of support)

���� = 1.2 � 1100 + 1.6 � 8003 � 3 = 288.89 _ 289 ��/�2"#� = 0.75 � $%&�6 � ! = 0.75 � F27.56 3000 � 5001000 = 983.26 ��

For shear the critical section is at distance (d ) from face of wall:

#�! = ���� �P 2 !� = 289 �3 0.52 0.5� � 3 = 650.25�� ' "#�Punching shear A1= (b+d)*(c+d) = (0.5+ 0.5)2 = 1 m2 A2A = � = 1 ��� 2 and � = 4 � (500 + 500) = 4000��= B*H =3*3=9m2

"#� = 0.75 � 0.17 `1 + 2Ab $%& � � � � ! = 0.75 � 0.17 �1 + 22� � F27.5 � 4000 � 5001000 = 2674.5 ��#� = ����. (;2 ;1) = 289 � (9 1) = 2312 �� �� VcReinforcement required:Long direction & Short direction (square column) :

* = ���� �(�)2�8 = 289�(3 0.5)28 = 225.78 ��. �

-� = 225.78�10609�3000�5002 = 0.334 4 = %�0.85%&� =17.54

9 = 14 G1 :1 2�-��4%� H = 0.00042 ' 9��� = 0.75 9 ��! > 9��� = 1.4%� = 1.4 410 =0.00341

;� = 0.00341 � 3000 � 500 = 5115��2��. �% ��� = ;� �����;� �� "20 = 5115314 = 16.2 ��� 17 ��� "20��Bearing transfer of stresses at base of column :

Pult.=1.2*D+1.6*L =1.2*100 +1.6*800 = 2600 kN �� ������

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A1= 500*500= 250000mm2 A2 = (500 + 4*500)(500 + 4*500) = 6250000mm2

;1;2 = j6250000250000 = 5 > 2 ��� 2P����. = " V0.85 � %& �:;2;1 � ;1Z < " � 0.85 � %& � � 2 � ;1 and " = M. \ P����. = 0.7 `0.85 � 27.5 � 2 � 2500001000 b = 8181�� < k��� = 2600�� �. �. pqvqwxyzq{| wq{}|~ �! ��� �!�! = � 2 70 = 3000 5002 70 = 1180��

�! = 9 � ! � %� � @A�10 � $%&� = 9 � 20 � 410 � 110F27.5 = 1407�� > � ��� �!�! = 1180��Use hooks �!� = 410�204�F27.5 = ��I�� B 8 � 20 = 160�� B 150��

Dowels:;!���� = 0.005 ;(������ )=0.005*500*500= 1250mm2Minimum 4 bars one at each corner 4" 25 mm same size as column reinforcement.

�! ����. = ! � %�4$%& � B 0.04 ! � %� B 200�� = 25�4104F27.5 = NWW. �KK B 0.04 � 25 � 410 = 410�� B 200��

Ex.4: Check the adequacy of the rectangular footing shown; support a service dead load of 900kN and service live load of 1400kN. the allowable soil pressure 150kN/m2 ,use f’c = 25MPa and fy = 400MPa?

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Beam shear (critical section at distance d from face of support)

���� = 1.2 � 900 + 1.6 � 14003.3 � 5.4 = 332017.82 = 186.31 ��/�2����� ���� ��� = 900 + 14003.3 � 5.4 = 129���2Wt. of footing = 24*0.8= 19kN/m2

����� = 129 + 19 = 148 < 150���2 �. �. !���, = 800 75 12.5 = 712.5�� ��! !����� = 800 75 25 12.5 = 687��A1= (b+d)*(c+d) = (0.6+ 0..7125)2 = 1.722 m2

A2A = � = 1 ��� 2= B*H =3.3 * 5.4 = 17.82 m2

and � = 4 � (0.6 + 0.35 + 0.35) = 5.2 �#� = ����. (;2 ;1) = 186.31 � (17.82 1.722)= 2995.84 �� "#� = 0.75 � 0.17 `1 + 2Ab $%& � � � � !

2995.84 = 0.75 � 0.17 �1 + 22� � F20 � 5.2 � !���!��� = 505.2�� < 712.5�� �. �.#�! = ���� �P 2 !� = 186.31 � 3.3 � 1.688 = 1037.82 �� "#� = 1037.82 = 0.75 � F206 � 3.3 � !��� then !��� =562.5�� < 712�� �. �.* = 186.31 � 3.3 � 2.42

2 = 1770.7 ��. �

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-� = 1770.7�1060.9�3300�7122 = 1.177 4 = %�0.85%&� =23.53

9 = 14 G1 :1 2�-��4%� H = 0.00162 ' 9��� = 0.75 9 ��! > 9��� = 1.4%� = 1.4 400 =0.0035

;� = 0.0035 � 3300 � 712 = 8223.6 ��2;� ��� �!�! = 1000�491�3.3175 = 9258.86��2 > 8223.6��2 �. �.!����� = 687��

* = 186.31 � 5.4 � 1.3522 = 916.78 ��. �

-� = 916.78�1060.9�5400�6872 = 0.399 4 = %�0.85%&� =23.53

9 = 14 G1 :1 2�-��4%� H = 0.001 ' 9��� = 0.75 9 ��! > 9��� = 1.4%� = 1.4 400 =0.0035

;� = 0.0035 � 5400 � 687 = 12984.3 ��2;� ��� �!�! = 1000 � 491 � 3.3150 + 1000 � 491 � 0.9 � 2300 = 13748��2 > 12984.3��2 �. �.Bearing transfer of stresses at base of column :

Pult.=1.2*D+1.6*L =1.2 * 900 + 1.6 * 1400 = 3320 kN �� ������A1= 600 * 600 = 360000mm2 A2 = (600 + 4 * 712)2 = 11888704 mm2

;2;1 = j11888704360000 = 5.74 > 2 ��� 2P����. = " V0.85 � %& �:;2;1 � ;1Z < " � 0.85 � %& � � 2 � ;1 and " = M. \ P����. = 0.7 `0.85 � 20 � 2 � 3600001000 b = 8568 �� < k��� = 3320 �� �. �. pqvqwxyzq{| wq{}|~ �! ��� �!�! = � 2 70 = 3300 6002 70 = 1280�� 5400 6002 70 = 2330��

�! = 9 � ! � %� � @A�10 � $%&� = 9 � 25 � 400 � 110F20 = 2012.46�� > � ��� �!�!Use hooks �!� = 400�254�F20 = 559�� B 8 � 25 = 200 �� B 150��

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B) Eccentrically loaded columns:

� = ;

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� = )�!"#$�%&'()*�+�,-h )�!./01 (b2"3�45"6�7&8'(91�:;06�%<="6�>?�@&8*��5A1 (

� = �3/12���/2 = �/6:;06�%<="6�>?��5A1�+�,-6�BC0�D�@*�EF��GH0�IHJ*�%<=9�K#<&,�L63�M�-6�NOP#Q�K�?����@*�/8A6�2R6P-6�ST�A�6U6

���� � 3�2 = k ���� = 2k3� � = �2 2Ex. 5: The diagram below shows a footing ( 1.2*1.8m) loaded with P= 890 kN and at distance of e= 15cm on the right side of the footing :1- Find soil pressure under the footing2- Find soil pressure if e= 38cm.

� = 15 < �6 = 18006 = 30 �� the footing will be under stress less than soil pressure.

���� = k; k. � � �� = 8901.2 � 1.8 + 890 � 0.15 � 1.821.2 � 1.8312= 618.05 ��/�2

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���� = k; k. � � �� = 8901.2 � 1.8 890 � 0.15 � 1.821.2 � 1.8312= 206.01 ��/�2

� = 38 < �6 = 18006 = 30��)�2?�)*�GH0�VP#(&,�3�M�-6�K�?3a2"3�45"6�W6CX(,�91 (

� = �2 � = 1.82 0.38 = 0.52���� = 2k3� = 2 � 8903 � 0.52 � 1.2 = 950.8��/�2