Design procedure for stepped footing. Important topic for civil engineering student.

1. Design of
STEPPED Footing
PRITESH PARMAR
Department of Civil Engineering
Faculty of Technology, DDU.
priteshparmar5.6.1998@gmail.com

2. INTRODUCTION
• The loads from the supporting members of the structure like columns, wall, etc.,
known as superstructure should be safely transmitted to the soils. The
permissible pressure on soil is much less than that on the concrete column or a
wall.
• Therefore it becomes necessary to spread the load over a sufficiently larger area.
• The substructure which is provided to transmit the loads from superstructure to
the soil is known as the foundation.
• In general a spread constructed in brick work, stone masonry or concrete under
the base of a wall or column for the purpose of distributing the load over a large
area is known as footing.

3. TYPE OF FOUNDATION
1) SHALLOW FOUNDATION :
• Depth of foundation is less than or equal to two times width of footing.
• Types of Shallow foundation :
i. Isolated or Spread Footing
ii. Strip or continuous wall
iii. Combined Footing
iv. Raft or Mat Foundation
v. Strap Footing
vi. Floating Rafts

4. TYPE OF FOUNDATION
2) DEEP FOUNDATION :
• Depth of foundation is greater than two times width of footing.
• Types of Deep foundation :
i. Piles
ii. Piers
iii. Well
iv. Caission
In this presentation isolated footing is focused, specially stepped footing.

5. ISOLATED FOOTING
Fig. : Type of Isolated footing (a) Pad footing, (b) Stepped footing, (c) Sloped footing
(a) (b) (c)

6. ADVANTAGE & DISADVANTAGE OF TYPES OF ISOLATED FOOTING
1) Isolated Pad Footing :
• Essential when wall is required to construct over the footing.
• When depth of footing is restricted i.e. when both top and bottom reinforcements
are req. pad footing is essential.
• Ease in construction and compaction of concrete.
• Waste of concrete is more when only bottom reinforcement is provided, because
of cantilever nature of element.
• Lesser steel compared to sloped and stepped footing, but concrete waste point of view
pad footing is less preferably.

7. ADVANTAGE & DISADVANTAGE OF TYPES OF ISOLATED FOOTING
2) Isolated Sloped Footing :
• Advantageous when wall is not required to construct over the footing.
• When depth of footing is restricted i.e. when both top and bottom reinforcements
are req. sloped footing some times causes extra efforts for bar bending and some
times avoided.
• Difficulty in construction and compaction of concrete when slope is more than 1:3
and difficult to finish the top without having concrete slump too much.
• Waste of concrete is less as compared to pad or stepped footing.
• Easier in execution as compare to stepped footing.

8. STEPPED FOOTING
3) Isolated Stepped Footing :
• Can be use when wall is required to construct over the footing.
• When depth of footing is restricted i.e. when both top and bottom reinforcements
are req. stepped footing some times causes large efforts for bar bending.
• To avoid difficulty in construction and compaction of concrete when slope is more
in sloped footing, in such cases stepped footing is prefered.
• Waste of concrete is less as compared to pad footing, more as compared to
slopped footing.
• Difficult in execution as compare to slopped footing.
• Less steel as compared to sloped footing.
• Difficult to bond if it is cast in two different operations.
• Extra shuttering is needed as compared to sloped and pad footing.

9. STEPPED FOOTING
Design procedure:
1) Proportining of footing :
• Proportining is done for unfactored load only.
• 10% of axial load on column is considered as self weight of footing.
• Proportining is done in such a way that maximum base pressure doesn’t exceeds
safe soil bearing pressure and minimum pressure should be such that tension
should not occur i.e no contact between footing and subsoil.
• 𝑃𝑚𝑎𝑥 =
𝑃
𝐴 𝑓
+
𝑀 𝑥
𝑍 𝑥
+
𝑀 𝑦
𝑍 𝑦
≤ 𝑆𝐵𝐶 𝑎𝑛𝑑 𝑃 𝑚𝑖𝑛=
𝑃
𝐴 𝑓
−
𝑀 𝑥
𝑍 𝑥
−
𝑀 𝑦
𝑍 𝑦
≥ 0
Where, P = Axial load + self weight, Af = area of footing = Bf x Lf
Mx = Major axis moment, Zx = (Bf x Lf
2) / 6
My = Minor axis moment, Zy = (Lf x Bf
2) / 6

10. STEPPED FOOTING
Design procedure:
1) Proportining of footing :
• Minimum depth at edge(step) should be 150 mm (Cl. 34.1.2 of IS:456).
• Depth at edge(thickness of each step) is kept 30 to 50% of overall depth.
• Depth should be such that shear reinforcement not required.

11. STEPPED FOOTING
Design procedure:
2) Design Bending Moment:(Cl. 34.2.3 of IS:456)
• The greatest bending moment to be used in the design of an isolated concrete
footing which supports a column, pedestal or wall shall be the moment computed
by passing through a vertical section which extends completely across the footing
at sections at face of column for footing supporting columns.
• Based on depth decide whether section is singly or doubly reinforced and
determine reinforcement as per limit state design.
• Nominal reinforcement equal to 0.15 % of gross c/s area for mild steel and 0.12
% of gross c/s area for HYSD bar. Spacing of these reinforcement shall not
exceed 3d or 300 mm whichever is smaller for main bars and 5d or 300 mm
whichever is smaller for secondary bar.

12. STEPPED FOOTING
Design procedure:
3) Check for One way Shear:(Cl. 22.6 of IS:456)
• The sum of the vertical forces due to soil pressure on footing outside the critical
section is called one way shear.
• The critical section for one-way shear shall be assumed a vertical section located
from the face of the column, pedestal or wall at a distance equal to effective
depth of footing in case of footing on soils, and at distance equal to half the
effective depth of footing on piles.
• Punching shear stress, 𝜏 𝑣𝑝 =
𝑃𝑢𝑛𝑐ℎ𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
𝐶/𝑆 𝑎𝑟𝑒𝑎
=
𝑉𝑢
𝑏𝑑
.
• Based on concrete grade and Pt(%) at critical section allowable shear stress 𝜏 𝑐 is
found from Table 23 of IS : 456 and 𝜏 𝑐,𝑚𝑎𝑥 and 𝜏 𝑐 should be greater than 𝜏 𝑣 .
• Punching shear stress should be less than Design shear strength.

13. STEPPED FOOTING
Design procedure:
4) Check for Two way Shear:(Cl. 31.6 of IS: 456)
• The sum of the vertical forces outside the appropriate perimeter as defined by IS
: 456 is called two way shear. The critical section for shear in this case is at a
distance d/2 from the periphery of the column where d is effective depth of
footing.
• Punching shear stress, 𝜏 𝑣𝑝 =
𝑃𝑢𝑛𝑐ℎ𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑎𝑟𝑒𝑎
.
• The design shear strength in this case shall be taken equal to 𝑘 𝑠 𝜏 𝑐 should be
greater than 𝜏 𝑣𝑝 , where 𝜏 𝑐 = 0.25 𝑓𝑐𝑘.
• 𝑘 𝑠 = (0.5 + 𝛽𝑐) but not greater than 1, 𝛽𝑐 being the ratio of short to long side of
the column or pedestal.
• Punching shear stress should be less than Design shear strength.

14. STEPPED FOOTING
Design procedure:
5) Check for Development Length:
• Critical section for Length Development Check is same as Flexure i.e. at the face
of column.
• Check is done as per Cl.26.2.1 of IS:456.
• Length development required should be greater than Length provided, if not
satisfy than 90o bent required at edge.
• Development length check req. for each steps of footing.
For each steps of footing check for moment, one & two way shear and length
development check is required as illustrated example.

15. EXAMPLE
Design a stepped footing for following data :
• Unfactored load on column : 800 kN
• Column size : 350 mm x 350 mm
• Safe Bearing Capacity of soil : 200 kN/m2
• Concrete Grade : M20
• Steel Grade : Fe415
Solution :
Size of footing = Load / SBC
Areq. = 1.1*800 / 200 = 4.4 m2

16. EXAMPLE
Considering square footing,
Bf = Lf = (4.4)0.5 ≈ 2.1 m
Apro. = 2.1*2.1= 4.41 m2 > Areq. = 4.4 m2 O.K.
Net upward pressure = Factored load / Area of footing
qu =
1.5∗800
2.1 𝑥 2.1
= 272.10 kN/m2
Design for Flexure :
Moment span, lx = (2100 – 350) / 2 = 875 mm
Design Bending Moment, Mu = qu * Bf * lx2/2
= 272.10 *2.1*0.8752/2 = 218.75 kNm.
Fig. : Stepped Footing

17. EXAMPLE
Fig. : Critical Section For Flexure Fig. : Base Pressure Distribution

18. EXAMPLE
Considering two step footing,(no of step are decided based on trial and error)
Width of footing at critical section for bending moment = top width of footing = 1050 mm
For Fe 415 Balanced Limiting Moment,
Mulim. = 0.138 fck b d2
dreq. = (218.75 * 106 /0.138/20/1050)0.5 = 274.14 mm
Providing overall depth, D = 600 mm (reason behind considering higher depth is lesser
depth will fail the section in One or Two way shear, based on trial and error depth can be
reduced).
Two step of equal thickness 300 mm.
Effective depth, dx = 600-50(clear cover)-12/2 = 544 mm.
𝑃𝑡 =
50𝑓𝑐𝑘
𝑓𝑦
(1 − 1 −
4.6 𝑀𝑢
𝑓𝑐𝑘 𝑏 𝑑2) =
50∗20
415
(1 − 1 −
4.6 ∗218.75∗106
20∗1050∗5442 ) = 0.2036 %.
Ast req. = 0.2036*1050*544/100 = 1162.96 mm2.
Ast req. per meter width = 1162.96 / 2.1 = 553.80 mm2.

19. EXAMPLE
Ast min. = 0.12*1000*600/100 = 720 mm2.
𝑆𝑝𝑎𝑐𝑖𝑛𝑔 =
𝜋
4
122
720
*1000 = 157 mm
Providing #12 @ 150 c/c on both directions.
Ast pro. per meter width = 754 mm2
Fig. : Critical Section for One way shear

20. EXAMPLE
One way shear check :
Critical section for two way shear is at 544 mm from face of column.
Shear span = 875-544 = 331 mm.
b = 2100 mm.
d = 244 mm.
Shear force = 272.10 * 0.331 = 90 kN
𝜏 𝑣 =
90 ∗103
2100∗244
= 0.1756 N/mm2
Pt at critical section = 754*100/(1000*244) = 0.31%.
For Pt = 0.31 % and M20 grade 𝜏 𝑐 = 0.39 N/mm2 >> 𝜏 𝑣 O.K

21. EXAMPLE
Two way shear check :
Critical section for two way shear is at 544/2 = 272 mm from face of column.
Bo = bc + d/2 + d/2 = 350 + 544 = 894 mm.
Lo = lc + d/2 + d/2 = 350 + 544 = 894 mm.
Punching shear force = 272.10 * (2.12 – 0.8942) = 982.48 kN
Resisting area = 2(Bo +Lo)*d = 2*(0.894+0.894)*0.544 = 1.95 m2
Fig. : Critical Section for Two way shear

22. EXAMPLE
𝜏 𝑣 =
982.48 ∗103
1.95 ∗ 106 = 0.5 N/mm2
𝜏 𝑐
′ = 𝑘 𝑠 𝜏 𝑐
𝑘 𝑠 = 0.5 + 𝛽 𝛽 =
𝑆ℎ𝑜𝑟𝑡𝑒𝑟 𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
𝐿𝑜𝑛𝑔𝑒𝑟 𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
= 1
𝑘 𝑠 = 0.5 + 1 = 1.5 which is greater than 1.
𝜏 𝑐 = 0.25 𝑓𝑐𝑘 = 1.12 N/mm2
𝜏 𝑐
′
= 1 ∗ 1.12 = 1.12 ≫ 𝜏 𝑣 O.K.
Length development check :
Ldreq. = 47∅ for M20 & Fe415 steel grade.
= 47*12 = 564mm.
Ldpro. = (875-50) = 825 >> Ldreq. O.K.
Step-1 is O.K now check for Step-2 is required.

23. EXAMPLE
Design for Flexure for Step-2:
Moment span, lx = (2100 – 1050) / 2 = 525 mm
Design Bending Moment, Mu = qu * Bf * lx2/2
= 272.10 *2.1*0.5252/2 = 78.75 kNm.
Fig. : Critical Section For Flexure for step-2

24. EXAMPLE
As steel remain same, we have to check flexure capacity of section only.
Finding Neutral Axis for step-2,(as section width changes hence N.A will be different from
balanced N.A)
Total compression = Total tension
0.36 fck b xu = 0.87 fy Ast
Ast =
𝜋
4
122
150
*2100 = 1583.36 mm2
0.36 * 20 * 2100 * xu = 0.87 * 415 * 1583.36
xu = 37.80 mm
Moment capacity = 0.36*fck*b*xu*(d-0.42*xu)
= 0.36*20*2100*37.80*(244-0.42*37.80)
= 130.38 kNm >> 78.75 kNm O.K.

25. EXAMPLE
Check for one way shear :
Critical section for one way shear is at 244 mm from face of step-1(i.e. from critical section
for flexure)
Shear span = 525-244 = 281 mm.
Fig. : Critical Section For One way shear for step-2

26. EXAMPLE
b = 2100 mm.
d = 244 mm.
Shear force = 272.10 * 0.281 = 76.46 kN
𝜏 𝑣 =
76.46 ∗103
2100∗244
= 0.1492 N/mm2
Pt at critical section = 754*100/(1000*244) = 0.31%.
For Pt = 0.31 % and M20 grade 𝜏 𝑐 = 0.39 N/mm2 >> 𝜏 𝑣 O.K
Two way shear check :
Critical section for two way shear check is 122 mm from face to step-1.
Bo = 1050+244 = 1294 mm Lo = 1050+244 = 1294 mm
Punching shear force = 272.10*(2.12-1.2942) = 774.35 kN
Resisting area = 2*(1.294+1.294)*0.244 = 1.26 m2

27. EXAMPLE
𝜏 𝑣 =
774.35 ∗103
1.26 ∗ 106 = 0.61 N/mm2
𝜏 𝑐
′ = 𝑘 𝑠 𝜏 𝑐
𝑘 𝑠 = 0.5 + 𝛽 𝛽 =
𝑆ℎ𝑜𝑟𝑡𝑒𝑟 𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
𝐿𝑜𝑛𝑔𝑒𝑟 𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
= 1
𝑘 𝑠 = 0.5 + 1 = 1.5 which is greater than 1.
𝜏 𝑐 = 0.25 𝑓𝑐𝑘 = 1.12 N/mm2
𝜏 𝑐
′
= 1 ∗ 1.12 = 1.12 ≫ 𝜏 𝑣 O.K.
Length development check :
Ldreq. = 47∅ for M20 & Fe415 steel grade.
= 47*12 = 564mm.
Ldpro. = (525-50) = 475 < Ldreq. Hence,90o bent is req. full edge thickness or less dia. Bar
is chosen.

28. EXAMPLE
Based on trial and error one more step in footing may be added and should satisfy all check, main
purpose of this presentation is to give idea about design procedure for stepped footing.
Fig. : Detail of Footing

29. EXAMPLE
Another drawing for stepped footing for reference is provided, due to restricted depth section is
design as doubly reinforced section.
Fig. : Section Fig. : Plan

30. THANK YOU