design foundation that carry just axial load

1.  4.11 Design of Isolated footing.
This report includes the results of the laboratory tests results and recommendations to choose the
type and depth of foundations.
“For shallow foundation, the bearing capacity calculations from shear test results using
conservative values are 3.2 and 2.9 kg/cm2 for isolated and strip footings respectively at a
minimum depth of -2.0 meters from current ground level. “allowable bearing capacity of the soil
for isolated footing is 314 KN/𝑚2
for minimum depth of 2 meter below the ground level.
4.11.1 Determine the base area and overall thickness.
Determine the base area and overall thickness for a square spread footing with the following
design conditions:
𝑆𝑒𝑟𝑣𝑖𝑐𝑒 𝐷𝑒𝑎𝑑 𝑙𝑜𝑎𝑑 𝐷𝐿 = 1230.3 𝐾𝑁 𝑎𝑛𝑑 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝐿𝑖𝑣𝑒 𝑙𝑜𝑎𝑑 𝐿𝐿 = 210.6 𝐾𝑁.
Assume Service surcharge 5 KN/𝑚2
, , 𝑃𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 ( 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒) 𝑠𝑜𝑖𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑞 𝑎 =
314 KN/𝑚2
𝑆𝑜𝑖𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝛾𝑠𝑜𝑖𝑙 = 16 KN/𝑚3
Calculating the weight of footing, soil, and the surcharge floor load:
Assume ℎ 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 = 60 𝑐𝑚, 𝑊𝑓𝑜𝑜𝑡𝑖𝑛𝑔 = 0.6𝑥25 = 15 KN/𝑚2
,𝑊𝑆𝑜𝑖𝑙 = 1𝑥16 = 16 KN/𝑚2
𝑁𝑒𝑡 𝑠𝑜𝑖𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑞 𝑎.𝑛𝑒𝑡 = 314 − 15 − 16 − 5 = 278 KN/𝑚2
4.11.2 Requiredsizes of footing.
𝐴 =
𝑃 𝑛
𝑞 𝑎.𝑛𝑒𝑡
=
1230.3+210.6
278
= 5.18 𝑚2
, 𝐴 = 𝐿2
→ 𝐿 = √ 𝐴 = √5.15 = 2.28 𝑚 , take L = 2.3 m
4.11.3 Depthof footing and shear design.
𝑃𝑢 = 1.2𝑑𝑙 + 1.6𝑙𝑙 = 1.2𝑥1230.3 + 1.6𝑥210.6 = 1813.32 𝐾𝑁
𝑞 𝑢 =
1813.32
2.32
= 342.78 KN/𝑚2
4.11.3.1 One-way shear (Beam Shear).
𝑉𝑢 = 𝑞 𝑢 𝑏 (
𝑙
2
−
𝑎
2
− 𝑑) = 342.78𝑥2.3(
2.3
2
−
0.45
2
− 𝑑) , 𝐿𝑒𝑡 𝑉𝑢 = ∅𝑉𝑐 (∅ = 0.75)

2. 𝑉𝑐 =
1
6
√𝑓𝑐`𝑏 𝑤 𝑑 =
1
6
√24 𝑥 2300 𝑥 𝑑
342.78𝑥2.3
0.75
(
2.3
2
−
0.45
2
− 𝑑) =
1
6
√24 𝑥 2300 𝑥 𝑑 → 𝑑 = 0.332 𝑚
Assume cover 75 mm, and steel bars of ∅20
Generally, the thickness of spread footing is governed by two-way shear. The shear will be
checked on the critical perimeter at 𝑑/2 from the face of the column and, if necessary, the
thickness will be increased or decreased. Because there is reinforcement in both directions, the
average 𝑑 will be used:
ℎ = 332 + 75 + 20 = 427 𝑚𝑚
𝑡𝑎𝑘𝑒 ℎ = 500 𝑚𝑚 , 𝑡ℎ𝑒𝑛 𝑑 = 500 − 75 − 20 = 405 𝑚𝑚
4.11.3.2 Two-way shear (Punching Shear).
𝐿𝑒𝑡 𝑉𝑢 = ∅𝑉𝑐 (∅ = 0.75)
𝑉𝑢 = 𝑞𝑢(𝑏𝑙 − (ℎ. 𝑐𝑜𝑙𝑢𝑚𝑛 + 𝑑)( 𝑏. 𝑐𝑜𝑙𝑢𝑚𝑛 + 𝑑)) =
𝑉𝑢 = 342.78(2.3𝑥2.3 − (0.45 + 0.405)(0.5 + 0.405)) = 1548.07 𝐾𝑁
𝛽 =
500
450
= 1.11 , 𝛽 = 𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝑙𝑜𝑛𝑔 𝑠𝑖𝑑𝑒 𝑡𝑜 𝑠ℎ𝑜𝑟𝑡 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑐𝑜𝑙𝑢𝑚𝑛
𝑏0 𝑖𝑠 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑡𝑎𝑘𝑒𝑛 𝑎𝑡
𝑑
2
𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑𝑒𝑑 𝑎𝑟𝑒𝑎 .
𝑏0 = 2(ℎ. 𝑐𝑜𝑙𝑢𝑚𝑛 + 𝑑) + 2( 𝑏. 𝑐𝑜𝑙𝑢𝑚𝑛 + 𝑑) = 2(0.45 + 0.405) + 2(0.5 + 0.405) = 3.52 𝑚
𝛼 𝑠 𝑖𝑠 𝑎𝑠𝑠𝑢𝑚𝑒𝑑 𝑡𝑜 𝑏𝑒 ∶
 𝛼 𝑠 = 40 𝑓𝑜𝑟 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 − 𝑐𝑜𝑛𝑡𝑟𝑜𝑙
 𝛼 𝑠 = 30 𝑓𝑜𝑟 𝑒𝑑𝑔𝑒 𝑐𝑜𝑙𝑢𝑚𝑛𝑠
 𝛼 𝑠 = 20 𝑓𝑜𝑟 𝑐𝑜𝑟𝑛𝑒𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠
𝑡ℎ𝑒 𝐴𝐶𝐼 𝑐𝑜𝑑𝑒 , 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
− 𝑎𝑙𝑙𝑜𝑤𝑠 𝑎 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ , 𝑉𝑐 𝑖𝑛 𝑓𝑜𝑜𝑡𝑖𝑛𝑔𝑠 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠ℎ𝑒𝑎𝑟 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡
𝑓𝑜𝑟 𝑡𝑤𝑜 𝑤𝑎𝑦 𝑠ℎ𝑒𝑎𝑟 𝑎𝑐𝑡𝑖𝑜𝑛 , 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑜𝑓
 𝑉𝑐 =
1
6
(1 +
2
𝛽
) 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 → 𝑉𝑐 =
1
6
(1 +
2
1.11
) 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 =
0.467 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 𝐾𝑁
 𝑉𝑐 =
1
12
(
𝛼 𝑠 𝑑
𝑏0
+ 2) 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 → 𝑉𝑐 =
1
12
(
40𝑋0.405
3.52
+ 2) 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 =
0.5502 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 𝐾𝑁

3.  𝑉𝑐 =
1
3
𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 → 𝑉𝑐 =
1
3
𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 = 0.333 𝜕√ 𝑓𝑐
′ 𝑏0 𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙
𝑉𝑐 =
1
3
𝜕√𝑓𝑐
′ 𝑏0 𝑑 =
1
3
𝑥1.0𝑥√24 𝑥3520 𝑥405 = 2327.995 𝐾𝑁
∅𝑉𝑐 = 0.75𝑋2327.995 = 1745.996 > 𝑉𝑢 = 1548.07 𝐾𝑁 𝑖𝑡𝑠 𝑂𝐾
The thickness of 50 cm is adequate enough
Check again for One-way shear and the result is ∅𝑉𝑐 = 570.42 𝐾𝑁 > 𝑉𝑢 = 409.96 𝐾𝑁 𝑖𝑡 𝑂𝐾
4.11.4 Depthfor flexure in long direction.
Take steel bar of ∅20
b = 2.3 m ,h = 500 mm ,d = 500 − 75 −
20
2
= 415 𝑚𝑚
𝑓𝑐
′
= 24 𝑀𝑝𝑎, 𝑓𝑦 = 420 𝑀𝑝𝑎
𝑀 𝑢 =
𝑤𝑙2
2
= 342.78 𝑥 2.3 𝑥 0.925 𝑥
0.925
2
= 337.285 𝐾𝑁. 𝑚
𝑅 𝑛 =
Mu
∅𝑏d2 =
337.285 𝑥 106
0.9𝑥2300𝑥4152 = 0.946 Mpa
𝑚 =
𝐹𝑦
0.85 𝐹𝑐
=
420
0.85 x 24
= 20.59
ρ =
1
𝑚
(1 − √1 −
2𝑅 𝑛 𝑚
𝑓𝑚
) =
1
20.59
(1 − √1 −
2 𝑥 20.59 𝑥 0.946
420
) = 0.002307
𝐴 𝑆 = 𝜌𝑏𝑑 = 0.002307 𝑥 2300 𝑥 415 = 2202.205 𝑚𝑚2
𝐴 𝑆,𝑚𝑖𝑛 = 0.0018𝑏ℎ = 0.0018 𝑥 2300 𝑥 500 = 2070 𝑚𝑚2
𝐴 𝑠 = 2202.205 𝑚𝑚2
> 𝐴𝑠,𝑚𝑖𝑛 = 2070 𝑚𝑚2
− 𝑜𝑘
U𝑠𝑒 12∅16 𝑤𝑖𝑡ℎ 𝐴 𝑠 = 2412 𝑚𝑚2
> 𝐴𝑠,𝑟𝑒𝑞 = 2202.205 𝑚𝑚2
− 𝑜𝑘
Using bars of ∅16 instead of ∅20 as assumed before makes the effective depth 𝑑 larger. So, no
need to check for 𝑀 𝑛:
𝑆 =
2300 − 75𝑥2 − 12𝑥16
11
= 178 𝑚𝑚
Step (s) is the smallest of:
1. 3ℎ = 3𝑥 500 = 1500 𝑚𝑚

4. 2. 450 𝑚𝑚
3. 𝑠 = 380(
280
𝑓𝑠
) − 2.5𝐶𝑐 = 380 (
280
(
2
3
)420
) − 2.5 ∙ 75 = 192.5 𝑚𝑚 𝑐𝑜𝑛𝑡𝑟𝑜𝑙
𝑆 = 178 𝑚𝑚 < 𝑆 𝑚𝑎𝑥 = 192.5 𝑚𝑚 𝑜𝑘
4.11.5 Depthfor flexure in Short direction.
Take steel bar of ∅16
h = 500 mm ,d = 500 − 75 − 16 − 16/2 = 401 𝑚𝑚
𝑓𝑐
′
= 24 𝑀𝑝𝑎, 𝑓𝑦 = 420 𝑀𝑝𝑎
𝑀 𝑢 =
𝑤𝑙2
2
= 342.78 𝑥 2.3 𝑥 0.9 𝑥
0.90
2
= 319.3 𝐾𝑁. 𝑚
𝑅 𝑛 =
Mu
∅𝑏d2 =
319.3 𝑥 106
0.9𝑥2300𝑥4012 = 0.959 Mpa
𝑚 =
𝐹𝑦
0.85 𝐹𝑐
=
420
0.85 x 24
= 20.59
ρ =
1
𝑚
(1 − √1 −
2𝑅 𝑛 𝑚
𝑓𝑚
) =
1
20.59
(1 − √1 −
2 𝑥 20.59 𝑥 0.959
420
) = 0.0023397
𝐴 𝑆 = 𝜌𝑏𝑑 = 0.0023397 𝑥 2300 𝑥 401 = 2157.91 𝑚𝑚2
𝐴 𝑆,𝑚𝑖𝑛 = 0.0018𝑏ℎ = 0.0018 𝑥 2300 𝑥 500 = 2070 𝑚𝑚2
𝐴 𝑠 = 2157.91 𝑚𝑚2
> 𝐴𝑠,𝑚𝑖𝑛 = 2070 𝑚𝑚2
− 𝑜𝑘
U𝑠𝑒 12∅16 𝑤𝑖𝑡ℎ 𝐴 𝑠 = 2412 𝑚𝑚2
> 𝐴𝑠,𝑟𝑒𝑞 = 2157.91 𝑚𝑚2
− 𝑜𝑘
Using bars of ∅16 instead of ∅20 as assumed before makes the effective depth 𝑑 larger. So, no
need to check for 𝑀 𝑛:
𝑆 =
2300 − 75𝑥2 − 12𝑥16
11
= 178 𝑚𝑚
Step (s) is the smallest of:
1. 3ℎ = 3𝑥 500 = 1500 𝑚𝑚
2. 450 𝑚𝑚
3. 𝑠 = 380(
280
𝑓𝑠
) − 2.5𝐶𝑐 = 380 (
280
(
2
3
)420
) − 2.5 ∙ 75 = 192.5 𝑚𝑚 𝑐𝑜𝑛𝑡𝑟𝑜𝑙
𝑆 = 178 𝑚𝑚 < 𝑆 𝑚𝑎𝑥 = 192.5 𝑚𝑚 − 𝑜𝑘