The document contains solutions to 3 problems determining the size of square footings. For each problem, the bearing capacity equation is set up and solved for B, the length of one side of the square footing. For problem 1, B is found to be 2.45m. For problem 2, B is found to be 2.078m. Problem 3 follows the same process but the given information is different.

1. `Problem 1 Determine the size of the footing.
Solution:
𝑞𝑞 = 𝐷𝐷𝑓𝑓 × 𝛾𝛾
= 1.2 × 16
= 19.2 𝑘𝑘𝑘𝑘/𝑚𝑚2
𝑁𝑁𝑞𝑞 = tan2
�45° +
∅
2
� 𝑒𝑒 𝜋𝜋 tan ∅
= tan2
�45° +
30°
2
� 𝑒𝑒 𝜋𝜋 tan 30°
= 18.4
𝑁𝑁𝛾𝛾 = 2�𝑁𝑁𝑞𝑞 + 1� tan ∅
= 2(18.4 + 1) tan 30°
= 22.4
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 +
𝐵𝐵
𝐿𝐿
tan ∅
= 1 +
𝐵𝐵
𝐵𝐵
tan 30°
= 1.577
𝐹𝐹𝑞𝑞 𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
1.2
𝐵𝐵
= 1 +
0.346
𝐵𝐵
𝐹𝐹𝛾𝛾𝛾𝛾 = 1 − 0.4
𝐵𝐵
𝐿𝐿
= 1 − 0.4
𝐵𝐵
𝐵𝐵
= 0.6
𝐹𝐹𝛾𝛾𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = �1 −
𝛽𝛽
90°
�
2
= �1 −
20°
90°
�
2
= 0.605
𝐹𝐹𝛾𝛾𝛾𝛾 = �1 −
𝛽𝛽
∅
�
2
= �1 −
20°
30°
�
2
= 0.11
Since 𝑐𝑐′
= 0 general bearing capacity equation becomes ,
20°
400 𝑘𝑘𝑘𝑘
𝐵𝐵 × 𝐵𝐵
1.2 𝑚𝑚
𝛾𝛾 = 16 𝑘𝑘𝑘𝑘 𝑚𝑚3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
Water table
𝛾𝛾𝑠𝑠𝑠𝑠𝑠𝑠 = 19 𝑘𝑘𝑘𝑘 𝑚𝑚3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
Created by
Ragib Nur Alam Shuvo,
CE13
ragibnur.ce@gmail.com

2. 𝑞𝑞𝑎𝑎𝑎𝑎𝑎𝑎 =
1
𝐹𝐹𝐹𝐹
∗ �𝑞𝑞𝑁𝑁𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 𝐹𝐹𝑞𝑞 𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 +
1
2
𝛾𝛾𝛾𝛾𝑁𝑁𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾 𝐹𝐹𝛾𝛾 𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾�
⟹
400
𝐵𝐵 × 𝐵𝐵
=
1
6
�19.2 × 18.4 × 1.577 × �1 +
0.346
𝐵𝐵
� 0.605 +
1
2
× (19 − 9.81) × 𝐵𝐵 × 22.4 × 0.6 × 1 × 0.11�
⟹
2400
𝐵𝐵2
= 337.059 �
𝐵𝐵 + 0.346
𝐵𝐵
� + 6.79𝐵𝐵
⟹ 6.79𝐵𝐵3
+ 337.059𝐵𝐵2
+ 116.62𝐵𝐵 − 2400 = 0
⟹ 𝐵𝐵 = 2.45 𝑚𝑚
Check:
𝐷𝐷𝑓𝑓
𝐵𝐵
=
1.2
2.45
= 0.489 < 1
ok
Size of Footing = 2.45 𝑚𝑚 × 2.45 𝑚𝑚
Problem 2:
A square footing is shown in the above figure use F.S=6. Determine the size of the footing
Solution:
𝑞𝑞 = 𝐷𝐷𝑓𝑓 × 𝛾𝛾
= 1.2 × 16
= 19.2 𝑘𝑘𝑘𝑘/𝑚𝑚2
𝐵𝐵′
= 𝐵𝐵 − 2𝑒𝑒
= 𝐵𝐵 − 2 × 0.15
= 𝐵𝐵 − 0.30
𝑒𝑒 =
𝑀𝑀
𝑄𝑄
=
58 × 1.2
450
= 0.15 𝑚𝑚
L’=B;
450 𝑘𝑘𝑘𝑘
𝐵𝐵 × 𝐵𝐵
1.2 𝑚𝑚
𝛾𝛾 = 16 𝑘𝑘𝑘𝑘 𝑚𝑚3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
Water table
𝛾𝛾𝑠𝑠𝑠𝑠𝑠𝑠 = 19 𝑘𝑘𝑘𝑘 𝑚𝑚3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
58 𝑘𝑘𝑘𝑘
Ragib CE’13
130095

3. 𝑁𝑁𝑞𝑞 = tan2
�45° +
∅
2
� 𝑒𝑒 𝜋𝜋 tan ∅
= tan2
�45° +
20°
2
� 𝑒𝑒 𝜋𝜋 tan 30°
= 18.4
𝑁𝑁𝛾𝛾 = 2�𝑁𝑁𝑞𝑞 + 1� tan ∅
= 2(18.4 + 1) tan 30°
= 22.4
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 +
𝐵𝐵′
𝐿𝐿′
tan ∅
= 1 +
𝐵𝐵 − 0.30
𝐵𝐵
tan 30°
= 1 +
𝐵𝐵 − 0.30
𝐵𝐵
× 0.577
=
𝐵𝐵 + 0.577𝐵𝐵 − 0.173
𝐵𝐵
=
1.577𝐵𝐵 − 0.173
𝐵𝐵
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
1.2
𝐵𝐵
= 1 +
0.346
𝐵𝐵
=
𝐵𝐵 + 0.346
𝐵𝐵
𝐹𝐹𝛾𝛾𝛾𝛾 = 1 − 0.4
𝐵𝐵′
𝐿𝐿′
= 1 − 0.4
𝐵𝐵 − 0.30
𝐵𝐵
=
𝐵𝐵 − 0.4𝐵𝐵 − 0.12
𝐵𝐵
=
0.6𝐵𝐵 − 0.12
𝐵𝐵
𝐹𝐹𝛾𝛾 𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = �1 −
𝛽𝛽
90°
�
2
= �1 −
0°
90°
�
2
= 1
𝐹𝐹𝛾𝛾𝛾𝛾 = �1 −
𝛽𝛽
∅
�
2
= �1 −
0°
30°
�
2
= 1
Since 𝑐𝑐′
= 0 general bearing capacity equation becomes
𝑞𝑞𝑎𝑎𝑎𝑎𝑎𝑎 =
1
𝐹𝐹𝐹𝐹
∗ �𝑞𝑞𝑁𝑁𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 𝐹𝐹𝑞𝑞 𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 +
1
2
𝛾𝛾′𝐵𝐵′𝑁𝑁𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾 𝐹𝐹𝛾𝛾 𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾�
⟹
450
𝐵𝐵 × (𝐵𝐵 − 0.30)
=
1
6
�19.2 × 18.4 × �
1.577𝐵𝐵 − 0.173
𝐵𝐵
� × �
𝐵𝐵 + 0.346
𝐵𝐵
� × 1 +
1
2
× (19 − 9.81) × (𝐵𝐵 − 0.30) × 22.4 × �
0.6𝐵𝐵 − 0.12
𝐵𝐵
� × 1 × 1�
⟹
2700
𝐵𝐵 × (𝐵𝐵 − 0.30)
= 353.28 ×
1
𝐵𝐵2
(1.577𝐵𝐵2
− 0.173𝐵𝐵 + 0.546𝐵𝐵 − 0.06) + 102.928 ×
1
𝐵𝐵
× (0.6𝐵𝐵2
− 0.18𝐵𝐵 − .12𝐵𝐵 + 0.036)
⟹ 2700 = 353.28 × (1.577𝐵𝐵2
+ 0.373𝐵𝐵 − 0.06) + 102.928 × 𝐵𝐵 × (0.6𝐵𝐵2
− 0.3𝐵𝐵 + 0.044) − 0.3 �353.28 ×
1
𝐵𝐵
(1.577𝐵𝐵2
− .373𝐵𝐵 − 0.06) + 102.928 × (0.6𝐵𝐵2
− 0.3𝐵𝐵 + 0.036) �
⟹ 2700 = 577.123𝐵𝐵2
+ 131.773𝐵𝐵 − 21.2 + 61.76𝐵𝐵3
− 30.878𝐵𝐵2
+ 3.705𝐵𝐵 − ��167.136𝐵𝐵 − 39.53 − 6.359
1
𝐵𝐵
� + (18.527𝐵𝐵2
− 9.263𝐵𝐵 + 1.112)�
⟹ 2700 = 61.76𝐵𝐵3
+ 546.245𝐵𝐵2
− 22.395𝐵𝐵 + 17.218 + 6.359
1
𝐵𝐵
⟹ 61.76𝐵𝐵4
+ 546.245𝐵𝐵3
− 22.395𝐵𝐵2
− 2682.782𝐵𝐵 + 6.359 = 0
Using trial error method
B = 2.078
Check:
Ragib CE’13
130095

4. 𝐷𝐷𝑓𝑓
𝐵𝐵
=
1.2
2.078
= 0.577 < 1 (𝑂𝑂𝑂𝑂)
Size of Footing = 2.078 𝑚𝑚 × 2.078 𝑚𝑚
Problem 3:
A square footing is shown in the figure below. Use F.S=6. Determine the size of the footing.
Solution:
𝑞𝑞 = 𝐷𝐷𝑓𝑓 × 𝛾𝛾
= 1.2 × 16
= 19.2 𝑘𝑘𝑘𝑘/𝑚𝑚2
𝑒𝑒 =
𝑀𝑀
𝑄𝑄
=
75
450
= 0.167 𝑚𝑚
𝐵𝐵′
= 𝐵𝐵 − 2𝑒𝑒
= 𝐵𝐵 − 2 × 0.167
= 𝐵𝐵 − 0.333
L’=B;
𝑁𝑁𝑞𝑞 = tan2
�45° +
∅
2
� 𝑒𝑒 𝜋𝜋 tan ∅
= tan2
�45° +
20°
2
� 𝑒𝑒 𝜋𝜋 tan 30°
= 18.4
𝑁𝑁𝛾𝛾 = 2�𝑁𝑁𝑞𝑞 + 1� tan ∅
= 2(18.4 + 1) tan 30°
= 22.4
𝐹𝐹𝑞𝑞 𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
1.2
𝐵𝐵
= 1 +
0.346
𝐵𝐵
=
𝐵𝐵 + 0.346
𝐵𝐵
𝐹𝐹𝛾𝛾𝛾𝛾 = 1 − 0.4
𝐵𝐵′
𝐿𝐿′
= 1 − 0.4
𝐵𝐵 − 0.333
𝐵𝐵
=
𝐵𝐵 − 0.4𝐵𝐵 − 0.133
𝐵𝐵
=
0.6𝐵𝐵 − 0.133
𝐵𝐵
𝐹𝐹𝛾𝛾 𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = �1 −
𝛽𝛽
90°
�
2
= �1 −
0°
90°
�
2
= 1
450 𝑘𝑘𝑘𝑘
𝐵𝐵 × 𝐵𝐵
1.2 𝑚𝑚
𝛾𝛾 = 16 𝑘𝑘𝑘𝑘 𝑚𝑚3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
Water table
𝛾𝛾𝑠𝑠𝑠𝑠𝑠𝑠 = 19 𝑘𝑘𝑘𝑘 𝑚𝑚3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
75 𝑘𝑘𝑘𝑘
Ragib CE’13
130095
Created by
Ragib Nur Alam Shuvo,
CE13
ragibnur.ce@gmail.com

5. 𝐹𝐹𝑞𝑞𝑞𝑞 = 1 +
𝐵𝐵′
𝐿𝐿′
tan ∅
= 1 +
𝐵𝐵 − 0.333
𝐵𝐵
tan 30°
= 1 +
𝐵𝐵 − 0.333
𝐵𝐵
× 0.577
=
𝐵𝐵 + 0.577𝐵𝐵 − 0.192
𝐵𝐵
=
1.577𝐵𝐵 − 0.192
𝐵𝐵
𝐹𝐹𝛾𝛾𝑖𝑖 = �1 −
𝛽𝛽
∅
�
2
= �1 −
0°
30°
�
2
= 1
Since 𝑐𝑐′
= 0 general bearing capacity equation becomes
𝑞𝑞𝑎𝑎𝑎𝑎𝑎𝑎 =
1
𝐹𝐹𝐹𝐹
∗ �𝑞𝑞𝑁𝑁𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 𝐹𝐹𝑞𝑞 𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 +
1
2
𝛾𝛾𝛾𝛾′𝑁𝑁𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾 𝐹𝐹𝛾𝛾 𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾�
⟹
450
𝐵𝐵 × (𝐵𝐵 − 0.333)
=
1
6
�19.2 × 18.4 × �
1.577𝐵𝐵 − 0.192
𝐵𝐵
� × �
𝐵𝐵 + 0.346
𝐵𝐵
� × 1 +
1
2
× (19 − 9.81) × (𝐵𝐵 − 0.333) × 22.4 × �
0.6𝐵𝐵 − 0.133
𝐵𝐵
� × 1 × 1�
⟹
2700
𝐵𝐵 × (𝐵𝐵 − 0.333)
= 353.28 ×
1
𝐵𝐵2
(1.577𝐵𝐵2
− 0.192𝐵𝐵 + 0.546𝐵𝐵 − 0.066) + 102.928 ×
1
𝐵𝐵
× (0.6𝐵𝐵2
− 0.2𝐵𝐵 − .133𝐵𝐵 + 0.044)
⟹ 2700 = 𝐵𝐵 × (𝐵𝐵 − 0.333) �353.28 ×
1
𝐵𝐵2
(1.577𝐵𝐵2
+ 0.373𝐵𝐵 − 0.066) + 102.928 ×
1
𝐵𝐵
× (0.6𝐵𝐵2
− .333𝐵𝐵 + 0.044)�
⟹ 𝐵𝐵 = 2.12 𝑚𝑚
Check:
𝐷𝐷𝑓𝑓
𝐵𝐵
=
1.2
2.12
= 0.566 < 1
ok
Size of Footing = 2.12 𝑚𝑚 × 2.12𝑚𝑚
Problem 4:
Determine the size of footing if (i) h = 0’ (ii) h = 2’ (iii) h=5’
Solution:
When h = 0, let the footing be square.
∴ Footing dimention = 𝐵𝐵 × 𝐵𝐵
40000 𝑙𝑙𝑙𝑙
𝐵𝐵 × 𝐵𝐵
5′
𝛾𝛾 = 100 𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓3⁄
𝑐𝑐′
= 0 ; 𝜑𝜑′
= 30°Water table
𝛾𝛾𝑠𝑠𝑠𝑠𝑠𝑠 = 120 𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓3⁄
𝑐𝑐′
= 0
𝜑𝜑′
= 30°
ℎ
Ragib CE’13
130095

6. 𝑞𝑞 = 𝐷𝐷𝑓𝑓 × (𝛾𝛾𝑠𝑠𝑠𝑠𝑠𝑠 − 𝛾𝛾𝑤𝑤)
= 5 × (120 − 62.4)
= 288 𝑙𝑙𝑙𝑙/𝑓𝑓𝑓𝑓2
𝑁𝑁𝑞𝑞 = tan2
�45° +
∅
2
� 𝑒𝑒 𝜋𝜋 tan ∅
= tan2
�45° +
20°
2
� 𝑒𝑒 𝜋𝜋 tan 30°
= 18.4
𝑁𝑁𝛾𝛾 = 2�𝑁𝑁𝑞𝑞 + 1� tan ∅
= 2(18.4 + 1) tan 30°
= 22.4
𝐹𝐹𝛾𝛾𝑠𝑠 = 1 − 0.4
𝐵𝐵
𝐿𝐿
= 1 − 0.4
𝐵𝐵
𝐵𝐵
= 0.6
𝐹𝐹𝛾𝛾𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = �1 −
𝛽𝛽
90°
�
2
= �1 −
0°
90°
�
2
= 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 +
𝐵𝐵
𝐿𝐿
tan ∅
= 1 +
𝐵𝐵
𝐵𝐵
tan 30°
= 1.577
Assuming,
𝐷𝐷𝑓𝑓
𝐵𝐵
≤ 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
5
𝐵𝐵
= 1 +
1.443
𝐵𝐵
𝐹𝐹𝛾𝛾𝛾𝛾 = �1 −
𝛽𝛽
∅
�
2
= �1 −
0°
30°
�
2
= 1
Since 𝑐𝑐′
= 0 general bearing capacity equation becomes ,
𝑞𝑞𝑎𝑎𝑎𝑎𝑎𝑎 =
1
𝐹𝐹𝐹𝐹
∗ �𝑞𝑞𝑁𝑁𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 𝐹𝐹𝑞𝑞 𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 +
1
2
𝛾𝛾𝛾𝛾𝑁𝑁𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾 𝐹𝐹𝛾𝛾 𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾�
⟹
40000
𝐵𝐵 × 𝐵𝐵
=
1
3
�288 × 18.4 × 1.577 × �1 +
1.443
𝐵𝐵
� × 1 +
1
2
× (120 − 62.4) × 𝐵𝐵 × 22.4 × 0.6 × 1 × 1�
⟹
120000
𝐵𝐵2
= 8356.838 �
𝐵𝐵 + 1.443
𝐵𝐵
� + 387.072𝐵𝐵
⟹ 387.072𝐵𝐵3
+ 8356.838𝐵𝐵2
+ 12058.917𝐵𝐵 − 120000 = 0
⟹ 𝐵𝐵 = 2.975 𝑓𝑓𝑓𝑓
Check:
𝐷𝐷𝑓𝑓
𝐵𝐵
=
5
2.975
= 1.681 > 1
Not ok.
𝐿𝐿𝐿𝐿𝑡𝑡,
𝐷𝐷𝑓𝑓
𝐵𝐵
> 1
𝐹𝐹𝑞𝑞 𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
tan−1
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
tan−1
5
𝐵𝐵
= 1 + 0.289 tan−1
5
𝐵𝐵
Ragib CE’13
130095

7. From bearing capacity equation,
40000
𝐵𝐵 × 𝐵𝐵
=
1
3
�288 × 18.4 × 1.577 × �1 + 0.289 tan−1
5
𝐵𝐵
� × 1 +
1
2
× (120 − 62.4) × 𝐵𝐵 × 22.4 × 0.6 × 1 × 1�
⇒
120000
𝐵𝐵2
= 8356.838 �1 + 0.289 tan−1
5
𝐵𝐵
� + 387.072𝐵𝐵
⇒ 120000 = 8356.838𝐵𝐵2
�1 + 0.289 tan−1
5
𝐵𝐵
� + 387.072𝐵𝐵3
⇒ 𝐵𝐵 = 3.16 𝑓𝑓𝑓𝑓
Now,
𝐷𝐷𝑓𝑓
𝐵𝐵
=
5
3.16
= 1.58 > 1
ok.
Size of footing 3.16 ft x 3.16 ft
When h = 2’,
let the footing be square.
∴ Footing dimention = 𝐵𝐵 × 𝐵𝐵
𝑞𝑞 = ℎ𝛾𝛾 + (𝐷𝐷𝑓𝑓 − ℎ) × (𝛾𝛾𝑠𝑠𝑠𝑠𝑠𝑠 − 𝛾𝛾𝑤𝑤)
= 2 × 100 + (5 − 2) × (120 − 62.4)
= 372.2 𝑙𝑙𝑙𝑙/𝑓𝑓𝑓𝑓2
𝑁𝑁𝑞𝑞 = 18.4
𝑁𝑁𝛾𝛾 = 22.4
𝐹𝐹𝛾𝛾𝛾𝛾 = 0.6
𝐹𝐹𝛾𝛾 𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1
𝐹𝐹𝛾𝛾𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1.577
Assuming,
𝐷𝐷𝑓𝑓
𝐵𝐵
> 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
tan−1
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
tan−1
5
𝐵𝐵
= 1 + 0.289 tan−1
5
𝐵𝐵
Since 𝑐𝑐′
= 0 general bearing capacity equation becomes ,
𝑞𝑞𝑎𝑎𝑎𝑎𝑎𝑎 =
1
𝐹𝐹𝐹𝐹
∗ �𝑞𝑞𝑁𝑁𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 𝐹𝐹𝑞𝑞 𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 +
1
2
𝛾𝛾𝛾𝛾𝑁𝑁𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾 𝐹𝐹𝛾𝛾 𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾�
⟹
40000
𝐵𝐵 × 𝐵𝐵
=
1
3
�372.2 × 18.4 × 1.577 × �1 + 0.289 tan−1
5
𝐵𝐵
� × 1 +
1
2
× (120 − 62.4) × 𝐵𝐵 × 22.4 × 0.6 × 1 × 1�
⟹ 120000 = 10820.59𝐵𝐵2
�1 + 0.289 tan−1
5
𝐵𝐵
� + 387.072𝐵𝐵3
⟹ 𝐵𝐵 = 2.81 𝑓𝑓𝑓𝑓
Now,
𝐷𝐷𝑓𝑓
𝐵𝐵
=
5
2.81
= 1.78 > 1
Created by
Ragib Nur Alam Shuvo,
CE13
ragibnur.ce@gmail.com

8. ok.
Size of footing 2.81 ft x 2.81ft
When h = 5’, let the footing be square.
∴ Footing dimention = 𝐵𝐵 × 𝐵𝐵
𝑞𝑞 = ℎ𝛾𝛾
= 5 × 100
= 500 𝑙𝑙𝑙𝑙/𝑓𝑓𝑓𝑓2
𝑁𝑁𝑞𝑞 = 18.4
𝑁𝑁𝛾𝛾 = 22.4
𝐹𝐹𝛾𝛾𝛾𝛾 = 0.6
𝐹𝐹𝛾𝛾 𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1
𝐹𝐹𝛾𝛾𝛾𝛾 = 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1.577
Assuming,
𝐷𝐷𝑓𝑓
𝐵𝐵
> 1
𝐹𝐹𝑞𝑞𝑞𝑞 = 1 + 2 tan ∅ (1 − sin ∅)2
tan−1
𝐷𝐷𝑓𝑓
𝐵𝐵
= 1 + 2 tan 30° (1 − sin 30°)2
tan−1
5
𝐵𝐵
= 1 + 0.289 tan−1
5
𝐵𝐵
Since 𝑐𝑐′
= 0 general bearing capacity equation becomes ,
𝑞𝑞𝑎𝑎𝑎𝑎𝑎𝑎 =
1
𝐹𝐹𝐹𝐹
∗ �𝑞𝑞𝑁𝑁𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 𝐹𝐹𝑞𝑞 𝑞𝑞 𝐹𝐹𝑞𝑞𝑞𝑞 +
1
2
𝛾𝛾𝛾𝛾𝑁𝑁𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾 𝐹𝐹𝛾𝛾 𝛾𝛾 𝐹𝐹𝛾𝛾𝛾𝛾�
⟹
40000
𝐵𝐵 × 𝐵𝐵
=
1
3
�500 × 18.4 × 1.577 × �1 + 0.289 tan−1
5
𝐵𝐵
� × 1 +
1
2
× (120 − 62.4) × 𝐵𝐵 × 22.4 × 0.6 × 1 × 1�
⟹ 120000 = 14508.4𝐵𝐵2
�1 + 0.289 tan−1
5
𝐵𝐵
� + 387.072𝐵𝐵3
⟹ 𝐵𝐵 = 2.44 𝑓𝑓𝑓𝑓
Now,
𝐷𝐷𝑓𝑓
𝐵𝐵
=
5
2.44
= 2.05 > 1
ok.
Size of footing 2.44 ft x 2.44 ft
Prepared By-
Ragib CE’13
130095