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Geotech Notes -1 ( Important problem solve)

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Problems related to foundation.

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Geotech Notes -1 ( Important problem solve)

  1. 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. 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. 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. 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. 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. 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. 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. 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

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