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Mathematics

1. If 𝑟 ≠ 𝑟′, no solution, if 𝑟 = 𝑟′ = 𝑛, unique solution if 𝑟 = 𝑟′ <
𝑛, many solutions. (non-homogeneous)
2. If 𝑟 = 𝑛, trivial solution, if 𝑟 < 𝑛 ,then (𝑛 − 𝑟) linearly independent solutions. (Many solutions) and if 𝑚 < 𝑛, then many solutions. 25. 𝐿(𝑒𝑎𝑡) = 1 𝑠−𝑎 26. 𝐿(sin 𝑎𝑡) = 𝑎 𝑠2+𝑎2 27. 𝐿(cos 𝑎𝑡) = 𝑠 𝑠2+𝑎2 28. 𝐿(sinh 𝑎𝑡) = 𝑎 𝑠2−𝑎2 29. 𝐿(cosh 𝑎𝑡) = 𝑠 𝑠2−𝑎2 3. 𝑓(𝑥 + ℎ) = 𝑓(𝑥) + ℎ𝑓′(𝑥) + ℎ 𝑓′′(𝑥) + 2! ℎ 𝑓 3! ′′′ (𝑥) + … … … ∞ 30. 𝐿{𝑒𝑎𝑡𝑓(𝑡)} = 𝑓̅(𝑠 − 𝑎) ∫𝑇 𝑒−𝑠𝑡 𝑓(𝑡)𝑑𝑡 4. If 𝑟𝑡 − 𝑠2 > 0 and 𝑟 < 0 𝑓(𝑥, 𝑦) have maximum, if 𝑟𝑡 − 𝑠2 >

31. 𝑓(𝑡 + 𝑇) = 𝑓(𝑡) then 𝐿{𝑓(𝑡)} = 0
1−𝑒−𝑠𝑇

0 and 𝑟 > 0 𝑓(𝑥, 𝑦) have minimum at(𝑎, 𝑏) and if 𝑟𝑡 − 𝑠2 < 0, then saddle point. If 𝑟𝑡 − 𝑠2 = 0, 𝑓𝑢𝑟𝑡ℎ𝑒𝑟 investigation is required to decide. 5. ∫ (∅𝑑𝑥 + 𝜓𝑑𝑦) = ∫ ∫ (𝜕𝜓 − 𝜕∅) 𝑑𝑥𝑑𝑦 (Green’s) 32. 𝐿{𝑓′(𝑡)} = 𝑠𝑓(̅ 𝑠) − 𝑓(0) 33. 𝐿{𝑓𝑛(𝑡)} = 𝑠𝑛𝑓(̅ 𝑠) − 𝑠𝑛−1𝑓(0) − 𝑠𝑛−2𝑓′(0) − ⋯ … … . . 𝑓𝑛−1(0) 34. 𝐿{ 𝑡 𝑓(𝑥)𝑑𝑥} = 1 𝑓̅(𝑠) 0 𝑆 𝑑𝑛 𝑐 𝐸 𝜕𝑥 𝜕𝑦 35. 𝐿{𝑡𝑛𝑓(𝑡)} = (−1)𝑛 . [𝑓̅(s)] 𝑑𝑠𝑛 6. ∫𝑐𝑭. 𝑑ℝ=∫𝑆𝑐𝑢𝑟𝑙𝑭. 𝑁𝑑𝑠 (Stokes) 7. ∫ 𝑭. 𝑁𝑑𝑠 = ∫ 𝑑𝑖𝑣𝑭𝑑𝑣 (Gauss) 36. 𝐿 {1 𝑓(𝑡)} = 𝑡 ∞ 𝑓̅(s) 𝑑𝑠 𝑆 𝑆 𝐸 37. 𝑓(𝑥) = 𝑎0 + ∑∞ 𝑎 cos 𝑛𝑥 + ∑∞ 𝑏 sin 𝑛𝑥 8. 𝑦𝑒∫ 𝑃𝑑𝑥 = ∫ 𝑄 𝑒∫ 𝑃𝑑𝑥𝑑𝑥 + 𝑐 2 𝑛=1 𝑛 𝑛=1 𝑛 38. 𝑎 = 1 𝛼+2𝜋 𝑓(𝑥)𝑑𝑥, 𝑎 = 1 𝛼+2𝜋 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥, 𝑏 = 9. If 𝑀 𝑑𝑥 + 𝑁 𝑑𝑦 = 0 be a homogeneous equation in 𝑥 and 𝑦, then 0 𝜋 ∫𝛼 𝑛 𝜋 ∫𝛼 𝑛 1 𝑖𝑠 an integrating factor 1 𝛼+2𝜋 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 𝑀 𝑥+𝑁 𝑦 10. If the equation of the type 𝑓 (𝑥𝑦)𝑦 𝑑𝑥 + 𝑓 (𝑥𝑦)𝑥 𝑑𝑦 = 0. If the 𝜋 ∫𝛼 39. 𝑓(𝑥) = 𝑎0 + ∑∞ 𝑎 cos 𝑛𝜋𝑥 + ∑∞ 𝑏 sin 𝑛𝜋𝑥 1 2 2 𝑛=1 𝑛 𝑐 𝑛=1 𝑛 𝑐 equation 𝑀 𝑑𝑥 + 𝑁 𝑑𝑦 = 0 be of this type then 1 is an 40. 𝑎 = 1 𝛼+2𝑐 𝑓(𝑥)𝑑𝑥, 𝑎 = 1 𝛼+2𝑐 𝑓(𝑥) cos 𝑛𝜋𝑥 𝑑𝑥, 𝑏 = integrating factor 𝑀 𝑥−𝑁 𝑦 0 𝑐 ∫𝛼 𝑛 1 𝛼+2𝑐 𝑓(𝑥) sin 𝑛𝜋𝑥 𝑑𝑥 𝑐 ∫𝛼 𝑐 𝑛 11. If 𝜕𝑀 𝜕𝑁 𝜕𝑦 −𝜕𝑥 be a function of x only = 𝑓(𝑥) say then 𝑒∫ 𝑓(𝑥)𝑑𝑥 is an 𝑐 ∫𝛼 41. 𝑓(𝑥) = ∑∞ 𝑐 𝑏 sin 𝑛𝜋𝑥 , where 𝑏 = 2 𝑐 𝑓(𝑥) sin 𝑛𝜋𝑥 𝑑𝑥 𝑁 𝑛=1 𝑛 𝑐 𝑛 𝑐 ∫0 𝑐 integrating factor 42. 𝑓(𝑥) = 𝑎0 + ∑∞ 𝑎 cos 𝑛𝜋𝑥 where, 𝑎 = 2 𝑐 𝑓(𝑥)𝑑𝑥, 2 𝑛=1 𝑛 𝑐 0 𝑐 ∫0 𝑎 = 2 𝑐 𝑓(𝑥) cos 𝑛𝜋𝑥 𝑑𝑥 𝜕𝑁 𝜕𝑀 12. If 𝜕𝑥− 𝜕𝑦 be a function of y only = 𝑓(𝑦) say then 𝑒∫ 𝑓(𝑦)𝑑𝑦 is an 𝑛 𝑐 ∫0 𝑐 ∞ ( ) 𝑀 43. 𝜇 = ∑𝑗 𝑥𝑗𝑓(𝑥𝑗) 𝑎𝑛𝑑 𝜇 = ∫−∞ 𝑥 𝑓 𝑥 𝑑𝑥 integrating factor. 44. 𝜎2 = ∑ ( 2 𝑎𝑛𝑑 𝜎2 = ∞ (𝑥 − 𝜇)2 𝑓(𝑥)𝑑𝑥 𝑗 𝑥𝑗 − 𝜇) 𝑓(𝑥𝑗) ∫−∞ 13. ∫𝑦=𝑐𝑜𝑛𝑡 𝑀𝑑𝑥 + ∫ terms of N not containing x dy = c 45. 𝑀𝑒𝑎𝑛: 𝑛𝑝 = 𝜇 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒: 𝜎2 = 𝜇 (Poisson’s 14. 𝑃. 𝐼. = 1 𝑓(𝐷) 𝑒𝑎𝑥 = 1 𝑓(𝑎) 𝑒𝑎𝑥 , 𝑓(𝑎) ≠ 0, if 𝑓(𝑎) = 0, 𝑡ℎ𝑒𝑛 𝑃. 𝐼. = distribution) 1 1 𝑥−𝜇 2 𝑥 1 𝑓𝘍(𝑎) 𝑒𝑎𝑥, 𝑓′(𝑎) ≠ 0 46. 𝑓(𝑥) = 𝜎√2𝜋 𝑒− 2( 𝜎 ) (Normal distribution) 1 ( ) 1 ( 2) 47. 𝑦 = 𝑎 + 𝑏𝑥, ∑ 𝑦 = 𝑛𝑎 + 𝑏 ∑ 𝑥, ∑ 𝑥𝑦 = 𝑎 ∑ 𝑥 + 𝑏 ∑ 𝑥2 15. 𝑃. 𝐼 = 𝑓(𝐷2) sin 𝑎𝑥 + 𝑏 𝑓(−𝑎2) = 0, = 𝑓(−𝑎2) , 𝑓 −𝑎 ≠ 0, if 48. 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2, ∑ 𝑦 = 𝑛𝑎 + 𝑏. ∑ 𝑥 + 𝑐. ∑ 𝑥2, ∑ 𝑥𝑦 = 𝑎 ∑ 𝑥 + 𝑏. ∑ 𝑥2 + 𝑐 ∑ 𝑥3, ∑ 𝑥2𝑦 = 𝑎 ∑ 𝑥2 + 𝑏. ∑ 𝑥3 + 𝑐 ∑ 𝑥4 then 𝑃. 𝐼. = 𝑥 1 sin(𝑎𝑥 + 𝑏), 𝑓′(−𝑎2) ≠ 0 (𝑥−𝑥 )(𝑥−𝑥 )…(𝑥−𝑥 ) (𝑥−𝑥 )(𝑥−𝑥 )…(𝑥−𝑥 ) 𝑓𝘍(−𝑎2) 49. 𝑦 = 𝑓(𝑥) = 1 2 𝑛 𝑦0 + 0 2 𝑛 𝑦1 + 16. 𝑃. 𝐼. = 1 𝑒 𝑎𝑥 𝑉 = 𝑒 𝑎𝑥 1 𝑉 (𝑥0−𝑥1)(𝑥0−𝑥2)…(𝑥0−𝑥𝑛) (𝑥−𝑥1)(𝑥−𝑥2)…(𝑥−𝑥𝑛−1) (𝑥1−𝑥0)(𝑥1−𝑥2)…(𝑥1−𝑥𝑛) 𝑓(𝐷) 𝑓(𝐷+𝑎) ⋯ + (𝑥 −𝑥 )(𝑥 −𝑥 )…(𝑥 −𝑥 ) 𝑦𝑛 17. 𝑃. 𝐼 = 1 𝑥𝑚 = [𝑓(𝐷)]−1𝑥𝑚, 𝑛 0 𝑛 1 𝑛 𝑛−1 𝑓(𝐷) 50. (𝑑𝑦) = 1 [∆𝑦 − 1 ∆2𝑦 + 1 ∆3𝑦 − 1 ∆4𝑦 + ⋯ ] 18. (1 + 𝑥)−1 = 1 − 𝑥 + 𝑥2 − ⋯ 𝑑𝑥 𝑥0 ℎ 2 3 0 4 0 19. (1 − 𝑥)−1 = 1 + 𝑥 + 𝑥2 + ⋯ 51. (𝑑𝑦) = 1 [∇𝑦 + 1 ∇2𝑦 + 1 ∇3𝑦 + 1 ∇4𝑦 + ⋯ ] 𝑑𝑛𝑦 𝑑𝑛−1𝑦 𝑑𝑦 𝑑𝑥 𝑥𝑛 ℎ 𝑛 2 𝑛 3 𝑛 4 𝑛 20. 𝑥𝑛 + 𝑘1𝑥𝑛−1 + ⋯ 𝑘𝑛−1𝑥 + 𝑘𝑛𝑦 = 𝑋, 𝑥 = 𝑒𝑡 , 𝑓(𝑥𝑛) 𝑑𝑥𝑛 𝑑𝑦 𝑑𝑥𝑛−1 2 𝑑2𝑦 𝑑𝑥 3 𝑑3𝑦 52. 𝑥𝑛+1 = 𝑥𝑛 − 𝑓𝘍(𝑥 ) (Newton-Raphson) 𝑥 = 𝐷𝑦, 𝑥 2 = 𝐷(𝐷 − 1)𝑦, 𝑥 3 = D(D − 1)(D − 2) 53. 𝑥0+𝑛ℎ 𝑓(𝑥)𝑑𝑥 = ℎ [𝑦 + 𝑦 + 2(𝑦 + 𝑦 + ⋯ . . +𝑦 )] 𝑑𝑥 𝑑𝑥 𝑑𝑥 ∫𝑥0 2 0 𝑛 1 2 𝑛−1 21. 𝜕 ( 𝑔(𝑥) ℎ(𝑡, 𝑥)𝑑𝑡) = 𝑔(𝑥) 𝜕 ℎ(𝑡, 𝑥)𝑑𝑡 + 𝑑𝑔 ℎ[𝑔(𝑥), 𝑥] − 𝜕𝑥 𝑑𝑓 ∫𝑓(𝑥) ∫𝑓(𝑥) 𝜕𝑥 𝑑𝑥 54. (Trapezoidal) 𝑥0+𝑛ℎ 𝑓(𝑥)𝑑𝑥 = ℎ [(𝑦 + 𝑦 ) + 4(𝑦 + 𝑦 + ⋯ 𝑦 ) + 2(𝑦 + ℎ[𝑓(𝑥), 𝑥] 𝑑𝑥 ∫𝑥0 3 0 𝑛 1 3 𝑛−1 2 22. 𝐿{𝑓(𝑡)} = 23. 𝐿(1) = 1 ∞ 𝑒−𝑠𝑡 𝑓(𝑡)𝑑𝑡 0 𝑦4 + ⋯ 𝑦𝑛−2)] (Simpson’s) 55. 𝐸𝑟𝑟𝑜𝑟 = − 𝑏−𝑎 ℎ2 𝑓′′(𝜉) = 𝑂(ℎ2) (Trapezoidal) 12 𝑠 24. 𝐿(𝑡𝑛) = 𝑛! 𝑠𝑛+1 56. 𝐸𝑟𝑟𝑜𝑟 = − 𝑏−𝑎 180 ℎ4𝑓 𝑖𝑣 (𝜉) = 𝑂(ℎ4) (Simpson’s) 57. 𝑦 = 𝑦 + ℎ. 𝑓(𝑡 , 𝑦 ) where 𝑑𝑦 = 𝑓(𝑡, 𝑦) (Euler’s) 𝑘+1 𝑘 𝑘 𝑘 𝑑𝑥 Strength of material 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 1. = 12. 𝜎 = 3𝑤𝑙 and 𝛿 = 𝜎𝑙2 (Leaf spring) 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 2. 𝑒 = 𝑓𝑥 −𝜇 𝑓𝑦 − 𝜇 𝑓𝑧 2𝑛𝑏𝑡2 13. 𝛿 = 64𝑊𝑅3𝑛 4𝐸𝑡 (Helical spring) 𝑥 𝐸 𝐸 𝐸 𝐺𝑑4 𝑓𝑦 𝑓𝑥 𝑓𝑧 3. 𝑒 = − 𝜇 − 𝜇 𝜎𝑥+𝜎𝑦 𝜎𝑥−𝜎𝑦 14. 𝜎 = + cos 2𝜃 + 𝑟 sin 2𝜃 𝑦 𝐸 𝐸 𝐸 𝑛 2 2 4. 𝑒 = 𝑓𝑧 − 𝜇 𝑓𝑥 − 𝜇 𝑓𝑦 15. 𝑟 = − (𝜎𝑥−𝜎𝑦) sin 2𝜃 + 𝑟 cos 2𝜃 𝑧 𝐸 𝐸 𝐸 𝑡 2 5. 𝐺 = 𝜎 +𝜎 𝜎 −𝜎 2 2(1+𝜇) 16. 𝜎 = 𝑥 𝑦 + √( 𝑥 𝑦) + 𝑟2 6. 𝐾 = 3(1−2𝜇) 1 2 𝜎 +𝜎 2 𝜎 −𝜎 2 𝑀 𝑓 𝐸 17. 𝜎 = 𝑥 𝑦 − √( 𝑥 𝑦) + 𝑟2 7. = = 𝐼 𝑦 𝑅 3 2 2 𝜋2𝐸𝐼 𝜋2𝐸𝐼 8. 𝑟 = 18. 𝑃 = (hinged and hinged)) ,𝑃 = 𝑙2 4𝑙2 (fixed and free) Ib 𝑇 𝑐 𝐺𝜃 9. 19. 𝑃 = 4𝜋2𝐸𝐼 (Fixed and fixed) ,𝑃 = 𝑙2 2𝜋2𝐸𝐼 𝑙2 𝐽 𝑟 𝐿 𝜎𝑐.𝐴 𝜎𝑐 10. 𝑀 = 1 [𝑀 + √𝑀2 + 𝑇2] 20. 𝑃 = 𝑙𝑒 2 where 𝛼 = 𝜋2𝐸 is Rankine’s constant 𝑒 2 1+𝛼(𝐾) 11. 𝑇𝑒 = √𝑀2 + 𝑇2 Structural analysis 1. 𝑀̅ = − 𝑃𝑎𝑏2 and 𝑀̅ = 𝑃𝑎2𝑏 (Point load) 9. 𝑀 = 𝑀̅ + 2𝐸𝐼 (2𝜃 + 𝜃 − 3∆) , right support sinks by ∆ 𝐴𝐵 𝑙2 𝐴𝐵 𝑙2 𝐴𝐵 𝐴𝐵 𝐿 𝐴 𝐵 𝐿 2. 𝑀̅ = − 𝑤𝑙2 and 𝑀̅ = 𝑤𝑙2 (udl) 10. 𝑀 = 𝑀̅ + 2𝐸𝐼 (2𝜃 + 𝜃 − 3∆), right support sinks by ∆ 𝐴𝐵 12 𝐴𝐵 12 𝐵𝐴 𝐵𝐴 𝐿 𝐵 𝐴 𝐿 ̅ 𝑤𝑙2 ̅ 𝑤𝑙2 11. 𝑀 𝑙 + 2𝑀 (𝑙 + 𝑙 ) + 𝑀 𝑙 = − 6𝑎1𝑥1 − 6𝑎2𝑥2 3. 𝑀𝐴𝐵 = − 30 and 𝑀𝐴𝐵 = 20 (uvl from left to right increase) 𝐴 1 𝐵 1 2 𝐶 2 𝑙1 𝑙2 4. 𝑀̅ = 𝑀𝑏 (3𝑎 − 𝑙) and 𝑀̅ = 𝑀𝑎 (3𝑏 − 𝑙) (clockwise moment) 12. 𝛿 = ∑ 𝑃𝑘𝐿 𝐴𝐵 𝑙2 𝑃2𝐿 𝐿 𝑀2 𝐴𝐵 𝑙2 𝑇2𝐿 𝐴𝐸 ∑𝑃𝑘𝐿 ∑(𝑃𝐿+𝐿𝛼𝑡)𝑘 5. 𝑈 = 2𝐴𝐸 , 𝑈 = ∫0 2𝐸𝐼 𝐿 𝑉2 𝑑𝑥 and 𝑈 = 2𝐺𝐽 13. 𝑄 = − 𝐴𝐸 ∑𝑘2𝐿 𝐿0 𝐴𝐸 𝐴0𝐸0 and 𝑄 = − 𝐴𝐸 ∑𝑘2𝐿 𝐿0 𝐴𝐸 𝐴0𝐸0 6. 𝑈 = 1.2 ∫0 2𝐺𝐴 𝑑𝑥 (rectangular beam) ∫𝑙𝑀𝑠𝑦 𝑑𝑥+𝖺𝑡𝑙−𝛿 14. 𝑘𝑖𝑗 = 𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑖 𝑑𝑢𝑒 𝑡𝑜 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑗, 0 𝐸𝐼 𝑙𝑦2𝑑𝑥 ∫0 𝐸𝐼 𝑤𝑙2 (two hinged arch) 8 𝑑2 𝛿𝑖𝑗 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑖 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑗 15. Far end fixed, Transverse displacement, 𝛿 = 𝐿 12𝐸𝐼 and 𝑘 = 12𝐸𝐼 𝐿3 8. H= (cable with udl) and Length of cable 𝐿 = 𝑙 + 8𝑑 3 𝑙 16. Far end hinged, Transverse displacement, 𝛿 = 𝐿3 and 𝑘 = 3𝐸𝐼 RCC 3𝐸𝐼 𝐿3 1. 𝑓𝑚 = 𝑓𝑐𝑘 + 1.65 𝜎 2. 𝐸𝑐 = 5000 √𝑓𝑐𝑘 and 𝑓𝑐𝑟 = 0.7 √𝑓𝑐𝑘 21. (𝑙 ) =7 for cantilever, 20 for SS and 26 for continuous beams. 𝑑 𝑏𝑎𝑠𝑖𝑐 If span is more than 10 m, multiply above values with 10/span for 3. 𝜀𝑠𝑡 = 0.002 + 0.87 𝑓𝑦 𝐸𝑠 SS and continuous beams. 4. 𝐸 = 2 × 105 𝑀𝑃𝑎 , 𝑚 = 280 22. 𝐿 = 𝑓𝑠∅ , multiply 𝑟 value with 1.6 for deformed bars and 1.25 𝑠 3𝜎𝑐𝑏𝑐 𝑑 4𝑐𝑏𝑑 𝑏𝑑 5. 𝑏 = 𝑙0 + 𝑏 + 6𝐷 for T-beams. for bars in compression. 𝑓 6 𝑤 𝑓 𝑀𝑢 𝑀𝑢 6. 𝑏 = 𝑙0 + 𝑏 + 3𝐷 for L-beams. 23. + 𝐿0 ≥ 𝐿𝑑 and 1.3 𝑉 𝑉 + 𝐿0 ≥ 𝐿𝑑 (If confinement exists) 𝑓 12 𝑤 𝑓 𝑙0 𝑢 24. 𝑟 = 𝑉𝑢 , 𝑉 𝑢 = 𝑉 + 𝑀 tan 𝛽 7. 𝑙0 + 𝑏𝑤 ≤ 𝑏 for isolated T-beams. 𝑣 𝑏𝑑 𝑢,𝑛𝑒𝑡 𝑢 𝑑 𝑏 +4 8. 0.5𝑙0 + 𝑏 ≤ 𝑏 for isolated L-beams. 25. 𝑉𝑢𝑠 = 0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑 𝑠𝑣 𝑙0+4 𝑤 𝑏 9. 𝑥𝑢 = 0.87 𝑓𝑦. 𝐴𝑠𝑡 and 𝑥𝑢,𝑚𝑎𝑥 = 0.0035 26. 𝐴𝑠𝑣 ≥ 𝑏 𝑠𝑣 0.4 0.87 𝑓𝑦 (minimum shear reinforcement) 𝑑 0.36 𝑓𝑐𝑘 𝑏𝑑 𝑑 0.87 𝑓𝑦 27. 𝑠 ≤ 0.75 𝑑 𝑎𝑛𝑑 300 𝑚𝑚 for vertical stirrups. 0.0055+ 𝐸𝑠 𝑣 10. 𝑀 = 0.87 𝑓 . 𝐴 (𝑑 − 0.42 𝑥 ) for 𝑥 ≤ 𝑥 28. 𝑉 = 𝑉 + 1.6 𝑇𝑢 and 𝑀 = 𝑀 + 𝑇𝑢 (1 + 𝐷) 𝑢 𝑦 𝑠𝑡 𝑢 𝑢 𝑢,max 𝑒 𝑢 𝑏 𝑒 𝑢 1.7 𝑏 11. 𝑀𝑢 = 0.36 𝑓𝑐𝑘. 𝑏. 𝑥𝑢,𝑚𝑎𝑥(𝑑 − 0.42 𝑥𝑢,𝑚𝑎𝑥) 29. ≤ 35 for ss and 40 for continuous slabs for Fe-250. For Fe-415 12. 𝑥𝑢,𝑚𝑎𝑥 = 0.87 𝑓𝑦 (𝑃𝑡,𝑙𝑖𝑚 ) 𝑑 𝑑 0.36 𝑓𝑐𝑘 100 multiply above values with 0.8 13. 𝑀 = 0.87 𝑓 𝐴 (𝑑 − 𝑓𝑦 𝐴𝑠𝑡) 30. 𝑟 = 𝑘 0.25√𝑓 , 𝑘 = 0.5 + 𝛽 ≤ 1.0 and 𝛽 = 𝑏 𝑢 𝑦 𝑠𝑡 𝑓𝑐𝑘 𝑏 𝑐2 𝑠 𝑐𝑘 𝑠 𝑐 𝑐 𝐷 14. 𝑦𝑓 = 0.15 𝑥𝑢 + 0.65 𝐷𝑓 when 𝐷𝑓 > 0.2
𝑑

31. 𝑒𝑥,𝑚𝑖𝑛

= 𝑙 500

+ 𝑑 , 20 𝑚𝑚 whichever is greater.
30

15. 0.87 𝑓 (𝐴

− 𝐴

) = (𝑓

− 0.447 𝑓

). 𝐴

32. 𝑒

= 𝑙 + 𝑏 , 20 𝑚𝑚 whichever is greater.

𝑦 𝑠𝑡

𝑠𝑡,𝑙

𝑠𝑐

𝑐𝑘

𝑠𝑐

𝑦,𝑚𝑖𝑛

500 30

16. 𝑀𝑢 = 𝑀𝑢,𝑙 + 0.87 𝑓𝑦. (𝐴𝑠𝑡 − 𝐴𝑠𝑡,𝑙) (𝑑 − 𝑑′)

33. 𝑃𝑢 = 0.4 𝑓𝑐𝑘 𝐴𝑐 + 0.67 𝑓𝑦𝐴𝑠𝑐

17. 𝐴𝑠𝑡 = 0.85 (minimum tension reinforcement)

34. 𝑉ℎ ≥ 0.36 (𝐴𝑔 − 1) 𝑓𝑐𝑘 , 𝑑

= 𝑑 − 2𝑐 and 𝑑

= 𝑑 − ∅

𝑏𝑑

𝑓𝑦

𝑉𝑐

𝐴𝑐

𝑓𝑦 𝑐

𝑚 𝑐

18. 𝐴𝑠𝑡,𝑚𝑖𝑛 = 0.12 % of 𝐴𝑔 for 𝐹𝑒-415 and 0.15 % of 𝐴𝑔 for 𝐹𝑒-250. (slabs)

35. 𝐶𝑟
36. 𝑓

= 1.25 − 𝑙𝑒
48𝑏

= 0.45 𝑓

𝐴1 , 1 < 𝐴1 ≤ 2 , 𝐴 =Largest frustum of a 19. 𝐴𝑠𝑡,𝑚𝑎𝑥 = 4% 𝑜𝑓 𝐴𝑔 (Beams) ,𝐴𝑠𝑐,𝑚𝑎𝑥 = 4% 𝑜𝑓 𝐴𝑔 (Beams) 𝑏𝑟,𝑚𝑎𝑥 𝑐𝑘√ 𝐴2 √ 1 𝐴2 20. (𝑙 ) 𝑑 𝑚𝑎𝑥 = (𝑙 ) 𝑑 𝑏𝑎𝑠𝑖𝑐 𝑘𝑡 𝑘𝑐 pyramid with side slopes 1 in 2, 𝐴2 =loaded area of column base Geotechnical Engineering 1. 𝐼𝑓 = 𝑊1−𝑤2 log10 𝑁2/𝑁1 𝐼𝑝 37. ∆𝑢 = 𝐵(∆𝜎𝑐) + 𝐴𝐵 (∆𝜎𝑑) (Skempton’s pore pressure parameters) 2. 𝐼𝑡 = 𝑓 38. 𝑞 = 𝑚𝑃 + 𝑘 is stress path equation where ∅ = tan−1 𝑚 and 𝑐 = 3. 𝐶𝑢 = 𝐷60 𝐷10 𝐷2 𝑘/ cos ∅ 39. 𝐶𝑐 = ∆𝑒 𝜎 +∆𝜎 4. 𝐶𝑐 = 30 𝐷60𝐷10 5. 𝐼𝑃 = 0.73 (𝑤𝐿 − 20) 6. 𝐼𝑃 = 𝑤𝐿 − 𝑤𝑃 log 0 𝜎0 40. 𝐶𝑐 = 0.009 (𝑤𝐿 − 10) (for normally consolidated soil) 41. 𝐶𝑐 = 0.007 (𝑤𝐿 − 10) (for over consolidated soil) 7. 𝐼 = 𝑤 − 𝑤 ∆𝑒 42. = ∆𝐻 𝑠 8. 𝐼𝐿 𝑃 𝑆 = 𝑤−𝑤𝑃 𝐼𝑃 1+𝑒0 43. 𝑚𝑣 = 𝐻 ∆𝐻/𝐻 9. 𝐼 = 𝑤𝐿−𝑤 ∆𝜎0 𝑐 𝐼𝑃 𝐼 44. 𝑐𝑣 = 𝑘 𝛾𝑤𝑚𝑣 10. 𝐴 = 𝑝 𝐹 where 𝐹 is clay fraction (Activity) 45. 𝑇 = 𝑐 𝑣𝑡 𝑒𝑚𝑎𝑥−𝑒 1/𝛾𝑑,𝑚𝑖𝑛−1/𝛾𝑑 𝑑2 11. 𝑅𝐷 = 𝑒 −𝑒 = 1/𝛾 −1/𝛾 46. 𝑇 = 𝜋 𝑈2 when 𝑈 ≤ 0.6 𝑚𝑎𝑥 𝑚𝑖𝑛 𝑑,𝑚𝑖𝑛 𝑑,𝑚𝑎𝑥 𝑣 4 12. 𝑘1 = tan 𝛼1 (non homogeneous) 𝑘2 tan 𝛼2 47. 𝑇𝑣 = −0.933 log10 (1 − 𝑈) − 0.085 when 𝑈 > 0.6

13. 𝑘 = 𝐶 𝛾 𝑒3 𝑑2
𝜇 1+𝑒
𝑘𝜇

48. 𝑆𝑓
49. 𝑆

= 𝐶𝑐𝐻 log
1+𝑒0
= 𝐶𝑟𝐻 log

𝜎0+∆𝜎
𝜎0
𝜎𝑐 + 𝐶𝑐𝐻 log

𝜎0+∆𝜎

14. 𝐾 =

𝛾𝑤

(absolute permeability)

𝑓 1+𝑒0
𝐷2−𝐷2

10 𝜎0

1+𝑒0

10 𝜎𝑐

𝑞 ln(𝑟2/𝑟1)

50. 𝐴𝑟 = 0 𝑖

15. 𝑘 =

2 2 (Permeability in unconfined aquifer)

𝐷2

16. 𝑘 =

𝜋
𝑞 2𝜋𝑏

𝑧2 −𝑧1
ln(𝑟2/𝑟1)
𝑧2−𝑧1

(Permeability in confined aquifer)

𝑖
51. 𝑆 = 𝑐𝑢
𝐹𝛾𝐻

17. 𝑘

= 𝑘1𝐻1+𝑘2𝐻2

(effective horizontal permeability in stratified

52. 𝑞𝑢 = 𝑐𝑁𝑐 + 𝑞 𝑁𝑞 + 0.5 𝛾 𝐵 𝑁𝛾 (Terzaghi’s strip)

ℎ 𝐻1+𝐻2

53. 𝑞 = 1.3 𝑐𝑁 + 𝑞 𝑁 + 0.4 𝛾 𝐵 𝑁

(Terzaghi’s square)

𝑢 𝑐 𝑞 𝛾

soils)
𝐻1+𝐻218. 𝑘 = (effective vertical permeability in stratified soils)

54. 𝑞𝑢 = 1.3 𝑐𝑁𝑐 + 𝑞 𝑁𝑞 + 0.3 𝛾 𝐵 𝑁𝛾 (Terzaghi’s circle)
𝐵 𝐵

𝑘1 +𝑘2
19. 𝑘𝑒 = √𝑘ℎ𝑘𝑣 (effective permeability)

55. 𝑞𝑢 = (1 + 0.3 𝐿) 𝑐𝑁𝑐 + 𝑞 𝑁𝑞 + (1 − 0.2 𝐿) 0.5 𝛾 𝐵 𝑁𝛾 (Terzaghi’s
rectangle)

20. 𝑘 = 𝑎𝐿 ln ℎ1

(falling head permeability test)

56. 𝑞𝑢 = 𝑐𝑁𝑐𝑆𝑐𝑑𝑐𝑖𝑐 + 𝑞 𝑁𝑞𝑆𝑞𝑑𝑞𝑖𝑞 + 0.5 𝛾 𝐵′ 𝑁𝛾 𝑆𝛾𝑑𝛾𝑖𝛾

𝐴𝑡
21. 𝑘 = 𝑞𝐿
𝐴ℎ

ℎ2
(constant head permeability test)

(Meyerhof)
𝐵′ = 𝐵 − 2𝑒𝑥 and 𝐿′ = 𝐿 − 2 𝑒𝑦

22. 𝑞 = 𝑘𝑒

ℎ 𝑁𝑓
𝑁𝑑

(seepage discharge)
5

57. 𝑞𝑛𝑢

= 𝑐𝑁𝑐 (Skempton)
𝐷𝑓 𝐵

3𝑄 1 1 2

𝑁𝑐 = 5 (1 + 0.2 𝐵 ) (1 + 0.2 𝐿)

23. 𝜎𝑧 = 2𝜋 𝑧2 ( 𝑟 2 ) (Boussinesq’s formula)

1+(𝑧)
3

Limiting value of 𝐷𝑓/𝐵 𝑖𝑠 2.5
𝑊ℎ 𝑦ℎ

24. 𝜎 = 𝑐𝑄 1 (
2𝜋 𝑧 2

1 2
𝑟 2 )

where 𝑐 = 1−2𝜇 (Wesrwegaard’s
2−2𝜇

58. 𝑄𝑢 =
𝐶 =

(ENR) where
𝑆+𝐶

formula)

𝑐 +(𝑧)

2.54 𝑐𝑚 𝑓𝑜𝑟 𝑑𝑟𝑜𝑝 ℎ𝑎𝑚𝑚𝑒𝑟 𝑎𝑛𝑑 0.254 𝑐𝑚 𝑓𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 ℎ𝑎𝑚𝑚𝑒𝑟

25. 𝜎𝑧 =

2𝑞 (
𝜋𝑧

2
1
𝑥 2)

(line load)

59. 𝑄𝑢

𝑊ℎ𝑦ℎ𝑦𝑏
𝐶
𝑆+2

(Hiley)

1+(𝑧)

where 𝐶 = 𝐶1 + 𝐶2 + 𝐶3

26. 𝜎

= 𝑞 (2𝜃 + sin 2𝜃) where 𝜃 = tan−1 𝑏

(stress under centre

𝐶 = 9.05 𝑅 with dolley and 𝐶

= 1.77 𝑅 without dolley and

𝑧 𝜋 𝑧

1 𝐴

1 𝐴

of strip load of width 2𝑏 ) 𝑅𝐿

27. 𝜎 = 𝑞 (2𝜃 + sin 2𝜃 sin 2∅) where 2𝜃 = 𝛽 − 𝛽 𝑎𝑛𝑑 2∅ = 𝛽 +

𝐶2 = 0.657 𝐴

𝑧 𝜋

1 2 1
𝑅

𝛽2 ( strip eccentric point)

𝐶3 = 3.55 𝐴

28. 𝜎𝑧

= 𝑞(1 − cos3 𝜃) where 𝜃 = tan−1 𝑅
𝑧

(stress

𝐿 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑖𝑙𝑒 𝑖𝑛 𝑚

under centre of circular load) 𝑅 = 𝑃𝑖𝑙𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑖𝑛 𝑡𝑜𝑛𝑛𝑒𝑠 = 0.1𝑄

29. sin ∅ = 𝜎1−𝜎3
𝜎1+𝜎3

(for cohesion less soils)

𝐴 = 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑖𝑙𝑒 𝑖𝑛 𝑐𝑚2
𝑊+𝑒2𝑃

30. sin ∅ = (𝜎1−𝜎3)/2 (for cohesive soils)

𝜂𝑏 =

when 𝑊 > 𝑃
𝑊+𝑃

𝑐 cot ∅+(𝜎1+𝜎3)/2

𝑊+𝑒2𝑃 𝑊−𝑒𝑃 2

31. 𝜎 = 2𝑐 tan 𝛼 + 𝜎 tan2 𝛼 where 𝛼 = 45 + ∅

𝜂𝑏 =

− ( )
𝑊+𝑃 𝐸+𝑃

when 𝑊 < 𝑃𝑒 1 3 2 32. tan ∅ = 𝑐 𝜎 (shear box test for cohesion less soils) 60. 𝑄𝑢 𝑊ℎ𝑦ℎ 𝑆0 (Danish) 𝑆0 = √2𝑦ℎ𝑊ℎ𝐿 𝐴𝐸 33. 𝑇 = 𝑐 𝜋𝐷2 (𝐻 + 𝐷 ) (if both top and bottom surfaces 61. 𝑄 𝑆+ 2 = 𝐴 𝑐𝑁 + 𝐴 𝛼 𝑐 (clays) 2 6 𝑢 𝑝 𝑐 𝑠 contributes) 34. 𝑇 = 𝑐 𝜋𝐷2 (𝐻 + 𝐷 ) (if only bottom surface contribute) 2 12 1−𝜇2 35. 𝑆𝑖 = 𝑞 𝐵 𝐸 𝐼𝑓 (immediate settlement ) 62. 𝑄𝑢 = 𝐴𝑝𝑐𝑁𝑐 + 𝐴𝑠 𝜆(𝜎̅ + 2𝑐) (clays) 63. 𝑄𝑢 = 𝐴𝑝𝜎̅ 𝑁𝑞 + 𝐴𝑠𝜎̅ 𝑘 tan 𝛿 (sands) 𝜎̅ 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑢𝑝𝑡𝑜 15 𝑑 𝑑𝑒𝑝𝑡ℎ 64. 𝑄 = 𝑁(𝐴 𝑐𝑁 + 𝐴 𝛼 𝑐) or 𝑄 = (𝐴 𝑐𝑁 + 𝐴 𝑐) (Group) 2 𝑢 𝑝 𝑐 𝑠 𝑢 𝑔𝑝 𝑐 𝑔𝑠 36. 𝑆 𝐵 𝐵 +0.3 = 𝑆 ( ) (settlement of footing based on plate 65. 𝑝 = 𝑘 𝜎̅ − 2𝑐√𝑘 + 𝑢 𝑓 𝑝 𝐵𝑝 𝐵𝑓+0.3 𝑎 𝑎 𝑎 settlement) 66. 𝑝𝑝 = 𝑘𝑝𝜎̅ + 2𝑐√𝑘𝑝 + 𝑢 67. 𝑘𝑎 = 1−sin ∅ and 𝑘 1+sin ∅ = 1+sin ∅ 1−sin ∅ 68. 𝐻𝑐 = 2𝑐 𝛾√𝐾𝑎 and unsupported vertical cut = 2𝐻𝑐 71. 𝑘 = cos 𝛽−√cos2 𝛽−cos2 ∅ and 𝑃 = cos 𝛽+√cos 𝛽−cos ∅ 𝑘𝑎𝛾ℎ2 cos 𝛽 2 69. 𝑘 sin2(𝛽+∅) 𝑎 2 sin2 𝛽 sin(𝛽−𝛿) (1+√ sin(∅+𝛿) sin(∅−𝑖)) (Inclined backfill) cos 𝛽+√cos2 𝛽−cos2 ∅ 𝑘 𝛾ℎ2 sin(𝛽−𝛿) sin(𝛽+𝑖) 72. 𝑘 = and 𝑃 = 𝑝 cos 𝛽 (Coulomb’s active ) sin2(𝛽−∅) 70. 𝑘𝑝 = 2 𝑝 cos 𝛽−√cos2 𝛽−cos2 ∅ 𝑝 2 (Inclined backfill) 73. 𝑁 = 15 + 1 (𝑁 − 15) 𝑤ℎ𝑒𝑛 𝑁 > 15 𝑎𝑛𝑑 𝑁 = 𝑁 𝑤ℎ𝑒𝑛 𝑁 ≤

sin2 𝛽 sin(𝛽+𝛿) (1−√ sin(∅+𝛿) sin(∅+𝑖))
sin(𝛽+𝛿) sin(𝛽+𝑖)

𝑐 2 𝑐

(Coulomb’s passive )

15 (dilatancy)

74. 𝑖𝑐

= 𝐺−1 1+𝑒

(Quick sand condition)

Hydrology

1. A tropical cyclone is a zone of low pressure with anticlockwise winds in the northern hemisphere.
2. Anticyclones cause clockwise wind circulations in the northern hemisphere.

11. 𝑓𝑝 = 𝑓𝑐 + (𝑓0 − 𝑓𝑐)𝑒−𝐾ℎ𝑡 (Horton’s equation)
12. ∅ 𝑖𝑛𝑑𝑒𝑥: It is the average rainfall above which the rainfall volume is equal to the runoff volume.
13. 𝑊 𝑖𝑛𝑑𝑒𝑥: 𝑊 = 𝑃−𝑅−𝐼𝑎

2 𝑚 ̅ 2

𝑡𝑒

3. 𝑁 = (𝐶𝑣)

, 𝐶

= 100×𝜎𝑚−1 and 𝜎

= √∑1 (𝑃𝑖−𝑃)

14. 𝑣̅ = 𝑣

and 𝑣̅ = 𝑣0.2+𝑣0.8

𝗌 𝑣

𝑃̅

𝑚−1

𝑚−1

0.6 2

4. 𝑃𝑥 = 1 [𝑃1 + 𝑃2 + ⋯ + 𝑃𝑚 ] (Normal ratio method)

15. 𝑄𝑡𝐶1 + 𝑄𝐶0 = (𝑄 + 𝑄𝑡)𝐶2 (Dilution technique)

𝑁𝑥 𝑀 𝑁1 𝑁2

𝑁𝑚

16. 𝑄 = 2.778 𝐴 , (Equilibrium discharge) A in 𝑘𝑚2 and D in h. 𝑄 in

5. 𝑃̅ = 𝑃1𝐴1+𝑃2𝐴2+⋯+𝑃𝑚𝐴𝑚
𝑃1+𝑃2+⋯+𝑃𝑚

(Thiessen-mean method)

𝑠
𝑚3/𝑠
17. 𝜎𝑛−1 (𝑦

𝐷

− 𝑦

) = 𝑥

− 𝑥

and 𝑦

𝑠

= − (𝑙𝑛. ln 𝑇 ) (Gumbels)

𝑃1+𝑃2

𝑃2+𝑃3

𝑃𝑛−1+𝑃𝑛

𝑆𝑛

𝑇1

𝑇2

𝑇1

𝑇2 𝑇

𝑇−1

6. 𝑃̅ = 𝑎1(

2 )+𝑎2(

2 )+⋯+𝑎𝑛−1(
𝐴

)
2 (Isohyetal method)

𝑛
18. 𝑅𝑖𝑠𝑘, 𝑅 = 1 − 𝑞𝑛 = 1 − (1 − 𝑃)𝑛 = 1 − (1 − )

7. 𝑃𝑟,𝑛 = 𝑛𝐶𝑟𝑝 𝑞 and 𝑃1 = 1 − 𝑞
8. 75% dependable annual rainfall is annual rainfall with probability

19. 𝑄2 = 𝐶0𝐼2 + 𝐶1𝐼1 + 𝐶2𝑄1, 𝐶0 =
𝑘−𝑘𝑥−0.5 ∆𝑡

−𝑘𝑥+0.5 ∆𝑡

𝑘−𝑘𝑥+0.5 ∆𝑡

𝑇
, 𝐶1 =

𝑘𝑥+0.5 ∆𝑡

𝑘−𝑘𝑥+0.5 ∆𝑡

and

𝑃 = 0.75, i.e. 𝑇 = 𝑁+1 = 1
𝑚 0.75

𝐶2 = 𝑘−𝑘𝑥+0.5 ∆𝑡 , 𝐶1 + 𝐶2 + 𝐶3 = 1
20. 𝑆 + 𝑆 = 𝜂

9. The chemical used as evaporation inhibitor is cetyl alcohol.
10. Evapotranspiration can be measured by Lysimeters.
21. 𝐾 = 𝐾0 = 1 ln 𝐻1 (recuperation test)

𝑦 𝑟

𝑠 𝐴

𝑇𝑟

𝐻2

Fluid mechanics

1. 𝑟 = 𝜇.
𝑑𝑦
2. 𝜕𝜌 + 𝜕 (𝜌𝑣 ) + 𝜕 (𝜌𝑣 ) + 𝜕 (𝜌𝑣 ) = 0 (continuity equation)

26. 𝐺𝑀 = 𝐼 − 𝐵𝐺
𝑉
27. Q= 8 𝐶 . √2𝑔. tan 𝜃/2 𝐻5⁄2

𝜕𝑡 𝜕𝑥 𝑥 𝜕𝑦 𝑦 𝜕𝑧 𝑧

15 𝑑

3. + 𝜌 ∇. 𝑉 =0 (continuity equation)
𝐷𝑡
4. 𝐹 − 𝐹 = 𝑀2−𝑀1 = 𝑚𝑣2−𝑚𝑣1 = 𝜌𝑄(𝑣

− 𝑣 )

28. 𝑣 = √𝑔𝑥²
2𝑦

𝑎 𝑎 𝑠

1 2 𝑡 𝑡 2 1

29. 𝑄 = Cd 1 2 . √2gh and ℎ = 𝑥 ( 𝑚 − 1)

5. 𝐹 = 𝜌𝑎(𝑣 + 𝑢)[(𝑣 + 𝑢) − 𝑢] = 𝜌𝑎(𝑣 + 𝑢)𝑣 (Jet propulsion
moving with 𝑢 velocity)

√𝑎12−𝑎22
30. 𝑄 = 𝜓1 − 𝜓2

𝑠𝑓

𝑝 𝑣2

𝑝 𝑣2

31. = −𝑣 and = −𝑣

6. 1 + 1 + 𝑧1 = 2 + 2 + 𝑧2

𝜕𝑥

𝑥 𝜕𝑦 𝑦

𝜌𝑔

2𝑔

𝜌𝑔

2𝑔

𝜕𝜓

𝜕𝜓

𝜕𝑐 𝜕𝑝 𝜕2𝑣 𝜕𝑝
7. = and 𝜇 =

32.

𝜕𝑥 = 𝑣𝑦 and 𝜕𝑦 = −𝑣𝑥

𝜕𝑦

𝜕𝑥

𝜕𝑦2

𝜕𝑥

𝜌𝑟𝑉𝑟𝐿𝑟

8. 𝑟 = − 𝜕𝑃 . 𝑟 and 𝑟 = 𝛾𝑤.ℎ𝐿 . 𝑟

33. (𝑅𝑒)𝑟 =

= 1
𝜇𝑟

𝜕𝑥 2

𝐿 2

34. (𝐹 )

= 𝑣𝑟

9. 𝑣 = 1 . (− 𝜕𝑝) . (𝑅2 − 𝑟2) = 𝑣

𝑟 2

𝑟 𝑟

√𝑔𝑟𝐿𝑟

4𝜇

𝜕𝑥

𝑚𝑎𝑥. [1 − (𝑅) ]

𝜕𝑣

𝜕𝑣

𝜕𝑣

𝜕𝑣

35. 𝑎𝑥 = 𝑣𝑥 . 𝑥 + 𝑣𝑦 . 𝑥 + 𝑣𝑧 . 𝑥 + 𝑥

10. 𝑝 − 𝑝

= 32𝜇𝑉𝐿 = 128𝜇𝑄𝐿

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝜕𝑡

1 2
64𝜇

𝐷2 64

𝜋𝐷4

36. 𝑎

= 𝑣 . 𝜕𝑣𝑦 + 𝑣

. 𝜕𝑣𝑦 + 𝑣 . 𝜕𝑣𝑦 + 𝜕𝑣𝑦

11. 𝑓 = =
𝜌𝑉𝐷 𝑅𝑒

𝑦 𝑥

𝜕𝑥

𝑦 𝜕𝑦

𝑧 𝜕𝑧

𝜕𝑡

𝑟 1⁄7

37. 𝑎

= 𝑣 . 𝜕𝑣𝑧 + 𝑣

. 𝜕𝑣𝑧 + 𝑣 . 𝜕𝑣𝑧 + 𝜕𝑣𝑧

12. 𝑣 = 𝑣𝑚𝑎𝑥. (1 − 𝑅)

(Turbulent)

𝑧 𝑥

𝜕𝑥

𝑦 𝜕𝑦

𝑧 𝜕𝑧

𝜕𝑡

13. ℎ

= 𝑓𝐿𝑣2

38. 𝜔 = 1 𝑐𝑢𝑟𝑙 𝑣 = 1 (∇ × 𝑣)

𝐿 2𝑔𝐷 2 2

14. 𝐿 = 𝐿1 + 𝐿2 + 𝐿3 (Compound pipe)
𝐷 𝐷 𝐷 𝐷
𝑝𝑎 𝑝𝑠 𝑓𝑙𝑣2 0.5𝑣2 𝑣2

𝑓𝐿𝑣2

1.5𝑣2

39. Vorticity = 2𝜔
40. Circulation = 𝑣𝑜𝑟𝑡𝑖𝑐𝑖𝑡𝑦 × 𝑎𝑟𝑒𝑎

15.

− ℎ = + +
𝛾 𝛾 2𝑔𝐷

+ and 𝐻 = +
2𝑔 2𝑔 2𝑔𝐷

(Siphon)
2𝑔

41. 𝐹 = 𝜌𝑎𝑣2

16. 𝛿

∗ = ∞ (1 − 𝑣) 𝑑𝑦
0 𝑉

42. Q

= Q [1 + cos θ] 𝑎𝑛𝑑 Q = Q [1 − cos θ] (θ is with plate)
2 2

17. 𝜃 =

∞ 𝑣 (1 − 𝑣) 𝑑𝑦

43. N

= N√Q , N

= N√P

∫0 𝑉 𝑉

s H3/4

s H5/4

∞ 𝑣

𝑣2

𝑄2 𝐴3
44. =

18. 𝛿𝐸 = ∫0 𝑉 (1 − 𝑉2) 𝑑𝑦

𝑔 𝑇

𝛿 5
19. Blassius boundary layer thickness

45. 𝐹 = 𝑣

𝛿∗

𝑥
1.729

√𝑅𝑒𝑥

𝐴
𝑔.𝑇
2 2 2 2

20. Displacement thickness =

3 2𝑦1 𝑦2

𝑦1 +𝑦1𝑦2+𝑦2

𝑥 √𝑅𝑒𝑥

46. 𝑦𝑐 = 𝑦 +𝑦

and 𝐸 =

𝑦 +𝑦

0.664

1 2 1 2

21. Momentum thickness =

47. 𝑑𝑦 = 𝑠𝑜−𝑠𝑓

𝑥 √𝑅𝑒𝑥

𝑑𝑥

𝑄2𝑇

2 1− 3

22. 𝑐𝑑,𝑥

= 0.664 and 𝑟
√𝑅𝑒𝑥

= 𝑐𝑑,𝑥

𝜌𝑉 2

𝑔𝐴
48. 𝑃1+𝑀1 = 𝑃2 + 𝑀2 (specific force=pressure force+momentum per

23. Drag force= 𝐶 𝜌𝐴𝑉2

𝑎𝑛𝑑 𝐹 = 𝑟

= 𝐶 𝜌𝑉2, 𝐶

= 1.328

sec)

𝑑 2

𝐴 0

𝑑 2

𝑑 √𝑅𝑒 𝐿

𝑦 𝑦

2𝑞2

24. 𝛿′ = =
√𝑐𝑜/𝜌

11.6𝖯

𝑣∗

(Laminar sub layer)

49.

2 [1 + 2] =
𝑦1 𝑦1
(𝑦2−𝑦1)3

𝑔𝑦3

(sequent depths for rectangular channel)

25. ℎ̅ = x̅ + IG
AX̅

and ℎ̅ = 𝑥̅ + 𝐼𝐺 . 𝑠𝑖𝑛2𝜃
𝐴𝑥̅

50. 𝐸 =

4𝑦1𝑦2

(Energy loss in jump)

Irrigation Engineering

𝑦 𝑄
1.

𝐷 𝑑+𝐷

𝑡 = 𝑓 log𝑒 (𝑄−𝑓𝐴)
𝑄

30. 𝐶 = 19 √
𝑏

[ ], 𝑑 = 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑝𝑖𝑙𝑒 𝑎𝑛𝑑 𝐷 =
𝑏

2. 𝐴𝑚𝑎𝑥 = 𝑓

𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑛𝑒𝑖𝑔ℎ𝑏𝑜𝑢𝑟𝑖𝑛𝑔 𝑝𝑖𝑙𝑒 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡.

𝑁𝑎+

𝐻 1

1+√1+𝛼2 𝑏

3. 𝑆𝐴𝑅 =

√𝐶𝑎+++𝑀𝑔++

31. 𝐺𝐸 = 𝑑 𝜋√𝜆 , 𝜆 =

and 𝛼 =
2 𝑑

2 1 −1

𝜆−2

1 −1

𝜆−1

𝐵
4. ∆= 8.64

32. ∅𝐷 = 𝜋 cos

( 𝜆 ) and ∅𝐸 = 𝜋 cos ( 𝜆 )

𝐷 1

−1 𝜆2+1

1 −1

𝜆2

5. 𝐴 = 𝑄𝐷

33. ∅𝐶 = 𝜋 cos

( 𝜆 ) , ∅𝐷 = 𝜋 cos

(𝜆 ) and ∅𝐸 =

𝑊𝑎𝑡𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑

1 𝜆2−1

√1+𝛼2+√1+𝛼2

√1+𝛼2−√1+𝛼2

6. 𝜂𝑐 =

cos−1 (

) , 𝜆1 = 1 2 and 𝜆2 = 1 2

7. 𝜂𝑎

𝑊𝑎𝑡𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
= 𝑊𝑎𝑡𝑒𝑟 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑟𝑜𝑜𝑡 𝑧𝑜𝑛𝑒
𝑊𝑎𝑡𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑

𝜋
, 𝛼1

𝜆1
= 𝑏1 𝑎𝑛𝑑 𝛼
𝑑

2 2
= 𝑏2
𝑑

8. 𝜂𝑠
9.

= 𝑊𝑎𝑡𝑒𝑟 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑟𝑜𝑜𝑡 𝑧𝑜𝑛𝑒
𝑊𝑎𝑡𝑒𝑟 𝑛𝑒𝑒𝑑𝑒𝑑 𝑖𝑛 𝑟𝑜𝑜𝑡 𝑧𝑜𝑛𝑒
𝑑

34. Non-modular modules: Drowned pipe outlet, masonry
sluice and wooden shoots.

𝜂𝑑 = 1 − 𝐷, 𝐷 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑑𝑒𝑝𝑡ℎ𝑠 and
𝑑 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠
10. 𝐶𝐼𝑅 = 𝐶𝑢 − 𝑅𝑒
11. 𝑁𝐼𝑅 = 𝐶𝐼𝑅 + 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑎𝑠 𝑝𝑒𝑟𝑐𝑜𝑙𝑎𝑡𝑖𝑜𝑛
𝑘𝑝

35. Semi modules or flexible modules: Pipe outlet, venturi flume or Kennedy’, open flume and orifice semi module.
36. Rigid modules: Gibb’s module, Kanna’s rigid module and
foote module.

12. 𝐶𝑢 = 40 (1.8𝑡 + 32)
13. 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 𝑏𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 =

𝛾𝑑 𝑑(𝐹 − 𝑂𝑀𝐶)

37. Flexibility 𝐹 = 𝑑𝑞/𝑞 = 𝑚
𝑑𝑄/𝑄 𝑛

. 𝑦
𝐻

, 𝑞 =discharge in outlet and

14. 𝑟𝑜 = 𝛾𝑤𝑅𝑆

𝛾𝑤

𝑄 = discharge in channel, 𝑚 =outlet index and
𝑛 =channel index. 𝐻 =working head of outlet and 𝑦 =

15. 𝑛 =

1 𝑑6, 𝑑 is in meters
24

depth of water in channel.

16. 𝑣 = 0.55 𝑚 𝑦0.64

38. Proportionality: 𝐹 = 1 𝑎𝑛𝑑 𝑚 = 𝐻

0
1+(23+0.00155)

𝑛 𝑦
𝐻

17. 𝑣 = [ 𝑛 𝑆 ] 𝑅𝑆
1+(23+0.00155) 𝑛

39. Setting =
𝑦

1
18. 𝑣 = (𝑄𝑓2 6

𝑆 √𝑅

40.

𝑚 >
𝑛

𝐻
, hyper proportional outlet.
𝑦

140

𝑚 𝐻
41. < , sub proportional outlet.

5 𝑣2

𝑛 𝑦

19. 𝑅 = ( )
2 𝑓

42. Sensitivity, 𝑆 = 𝑑𝑞/𝑞 = 𝑛 𝑑𝑞/𝑞 = 𝑛 𝐹

20. 𝑃 = 4.75√𝑄

𝑑𝑦/𝑦

𝑑𝑄/𝑄

5

21. 𝑆 = 𝑓3 3340𝑄6

1
𝑞2 3

43. Aqueduct: Canal over drainage with clear gap.
44. Syphon Aqueduct: Canal over drainage with syphonic action.
45. Super passage: Drain over canal with clear gap.

22. 𝐷𝑠 = 1.35 ( 𝑓 )
2 1 46. Canal Syphon: Drain over canal with syphonic action.

23. 𝑣 = 10.8𝑅3𝑆3

47. Principal stress in dam

24. 𝐴 = 𝑦2(𝜃 + cot 𝜃), 𝑃 = 2𝑦(𝜃 + cot 𝜃) for Lined

𝜎 = 𝑝𝑣

sec2 𝛼 − 𝑝′ tan2 𝛼

Triangular section.
25. 𝐴 = 𝐵𝑦 + 𝑦2(𝜃 + cot 𝜃), 𝑃 = 𝐵 + 2𝑦(𝜃 + cot 𝜃) for Lined Trapezoidal section.
26. Launching apron scour depth,
1
𝑄
𝐷 = 𝑥𝑅 − 𝑤𝑎𝑡𝑒𝑟 𝑑𝑒𝑝𝑡ℎ 𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑑, 𝑅 = 0.47 ( )
𝑓

where 𝑝𝑣 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 and 𝑝′ = water pressure of tail water and
𝛼 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑎𝑚 𝑓𝑎𝑐𝑒 𝑤𝑖𝑡ℎ 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙
48. 𝐵 = 𝐻
√𝑆𝑐−𝐶

27. Length of launching apron = √5 𝐷

49. 𝐻 = 𝑓
𝛾𝑤(𝑆𝑐−𝐶+1)

and 𝐻𝑚𝑎𝑥

= 𝑓
𝛾𝑤(𝑆𝑐+1)

28. 𝑡 = ℎ𝘍
𝐺

, where

1.85

0.85

ℎ′ = 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝐻𝐺𝐿 𝑎𝑏𝑜𝑣𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟
ℎ

50. 𝑥

= 2 𝐻𝑑 𝑦

29. 𝑡 =

𝐺−1

, where

ℎ = 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝐻𝐺𝐿 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟

Environmental Engineering

1. 𝑃𝑛 = 𝑃0 + 𝑛𝑥̅ , where 𝑥̅ =average of population increase (Arithmetic increase method)

23. PH range for alum: 6.5 to 8.3.
24. 𝐻𝑂𝐶𝑙 is most destructive disinfectant.

2. 𝑃𝑛 = 𝑃0 (1 +

𝑟 𝑛
)
100

1

, where 𝑟 = (𝑟1𝑟2 … 𝑟𝑡 )𝑡 (Geometric increase

25. Quick lime required in softening =Carbonate hardness in 𝑚𝑔/𝑙 as
𝐶𝑎𝐶𝑂 × 56/100 + 𝑀𝑔++ × 56 +𝐶𝑂 × 56

method)

3 24

2 44

3. 𝑃𝑛 = 𝑃0 + 𝑛𝑥̅ + 2 𝑦̅ , where 𝑦̅ =average of incremental

26. Soda required in softening =Non-carbonate hardness in 𝑚𝑔/𝑙 as
𝐶𝑎𝐶𝑂 × 106/100

increase

3
27. 𝑌𝑡 = 𝐿(1 − 10−𝑘𝐷𝑡)

4. Carbonate hardness=Total hardness or Alkalinity whichever is

28. 𝑘

𝐷,𝑇

= 𝑘

𝐷,20

× (1.047)𝑇−20

lesser.
5. Non-carbonate hardness = Total hardness – carbonate hardness
6. Total hardness = 𝐶𝑎++ in mg/l × 50 +𝑀𝑔++ in mg/l × 50 +𝐴𝑙3+ in

29. 𝑆 = 1 − (0.794)𝑡20 and 𝑆 = 1 − (0.630)𝑡37
30. 𝜂% = 100
1+0.0044√u

20 12
50

where 𝑢 =organic loading kg/ha-m/day

mg/l ×

100

1+R⁄I

9
7. Akalinity = 𝐶𝑂−− in mg/l × 50 + 𝐻𝐶𝑂− in mg/l × 50 + 𝑂𝐻− in

31. 𝜂% = 1+0.0044√u/F where 𝐹 = (1+0.1R⁄I)2

3 30 3 61

32. HRT = V = volume of Aeration tank

mg/l × 50

Q Rate of sewage flow.

17
8. Ammonia nitrogen and organic nitrogen is called Kjedahl nitrogen.
9. Colour: 5 Hazen units (max), PH: 6.5-8.5, Turbidity: 1 NTU (max), TDS: 500 mg/l (max), Chloride: 250 mg/l (max), Sulphate: 200 mg/l

33. Volumetric BOD loading = organic loading =
Mass of BOD applied per day to Aeration tank = QY0 volume of aeration tank. V
34. F = Q Yo

M V Xt

(max), Nitrate: 45 mg/l (max), fluoride : 1mg/l (max), Total alkalinity: 200 mg/l (max), Total hardness: 200 mg/l (max), Magnesium: 30 mg/l (max), Calcium: 75 mg/l (max), Zinc: 5 mg/l

35. 𝑀𝐶𝑅𝑇 = V.Xt = 𝐵𝑖𝑜𝑚𝑎𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 Qw.XR+(Q−Qw)XE. 𝐵𝑖𝑜 𝑚𝑎𝑠𝑠 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑝𝑒𝑟 𝑑𝑎𝑦
36. Sludge volume index (SVI) = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑙𝑢𝑑𝑔𝑒 𝑠𝑒𝑡𝑡𝑙𝑒𝑑 𝑖𝑛 30 𝑚𝑖𝑛 𝑖𝑛 𝑚𝑙
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑙𝑢𝑑𝑔𝑒 𝑠𝑜𝑙𝑖𝑑𝑠 𝑖𝑛 𝑔𝑚

(max), Iron: 0.3 mg/l (max), Free residual chlorine: 0.2 mg/l (min).

37. QR Q

= Xt
XR−X

Xt 106
−X

10. Toxic substances: Cadmium, Cyanide, lead, Mercury, Nickel,
Arsenic, chromium.

t
38. C = VsQs + VRQR
Qs + QR

SvI t

11. E-coli shall not be detectable in any 100 ml sample of drinking
water.
12. Standard sample of MPN: 10ml, 1 ml and 0.1 ml

39. ( 𝐿0 )𝑓−1 = {1 − (𝑓 − 1) 𝐷0} 𝑓
𝐷𝑐𝑓 𝐿0
40. 𝑡 = 1 . 𝑙𝑜𝑔 [{1 − (𝑓 − 1) 𝐷0} 𝑓]

𝑐 𝑘𝑟−𝑘𝑑

𝐿0

13. MPN/100 ml, by Thomas
= 100× 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑡𝑢𝑏𝑒𝑠

41. 𝐷 = 𝐿0 (10−𝑘𝐷𝑡 − 10−𝑘𝑅𝑡) + 𝐷 10−𝑘𝑅𝑡
𝑓−1

√𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑖𝑛 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑡𝑢𝑏𝑒𝑠×𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒
14. 𝑣 = 𝑔 (𝐺 − 1) 𝑑2

42. Primary pollutants: CO, 𝑆𝑂2, 𝑁𝑜𝑥, hydrocarbons and particulate
matter are primary air pollutants.

𝑠 18 𝖯

15. Percentage particle removal: 𝑣𝑠 × 100 (If 𝑣 < 𝑣 )

43. Secondary pollutants: Ozone, PAN (Peroxy acetyl nitrate),

𝑣0

𝑠 0

photochemical 𝑆mog, Aerosols and mists.

16. Chemical used in coagulation: Alum (Aluminium sulphate),
Copperas (Ferrous sulphate), Chlorinated copperas, Sodium

44. 𝐸𝐿𝑅 < 𝐴𝐿𝑅 : Sub adiabatic and stable. 45. 𝐸𝐿𝑅 > 𝐴𝐿𝑅 : Super adiabatic and unstable

aluminates.
17. Al2(SO4)3 18H2O + 3 Ca(HCO3)2 →
3CaSO +2Al(OH) +6CO +18H O

46. 𝐿𝑝

=20 log

𝑃𝑟𝑚𝑠
10 𝑃𝑜
𝑛

𝐿𝑖

where 𝑃𝑜 = 20 𝜇𝑃𝑎

4 3 2 2
18. Alkalinity requirement = 300 𝑚𝑔/𝑙 as 𝐶𝑎𝐶𝑂

per 1𝑚𝑔/𝑙 of Alum

47. 𝐿𝑒𝑞=10log ∑𝑖=1 1010 × 𝑡𝑖
48. Addition of sound levels:

666 3

19. Permanent hardness due to alum = 408 𝑚𝑔/𝑙 as 𝐶𝑎𝐶𝑂

per

𝐿 = 10 log

𝐿1 𝐿2
[10 + 10

𝐿𝑛

666 3

𝑝 10 10

10 + ⋯ + 1010]

1𝑚𝑔/𝑙 of Alum

49. Averaging of Sound Pressure Levels:

20. Sludge production = 156 𝑚𝑔/𝑙 as 𝐶𝑎𝐶𝑂 per 1𝑚𝑔/𝑙 of Alum

1 𝐿1

𝐿2

𝐿𝑛

666
21. 𝐶𝑂 release = 264 𝑚𝑔 as 𝐶𝑎𝐶𝑂
666 𝑙

3
per 1𝑚𝑔/𝑙 of Alum

̅𝐿̅𝑝̅ = 20 log10 { (1020 + 1020 + ⋯ + 1020 )}
50. 𝐿 = 𝐿 – 20 𝑙𝑜𝑔 (𝑟2)

1

22. 𝐺 = ( ) where P is in Watts
𝜇𝑉

2 1 10

𝑟1

Transportation Engineering

2𝑥2
1. 𝑦 = (parabolic camber)
𝑛𝑊

37. 𝐸𝑆𝑊𝐿: interpolate load for depth from line joining (𝑙𝑜𝑔𝑃, log 𝑧)
2

𝑣2
2. 𝑆𝑆𝐷 = 𝑣𝑡 +
2𝑔(𝑓±𝑛)

, 𝐼𝑆𝐷 = 2 × 𝑆𝑆𝐷

and (𝑙𝑜𝑔2𝑃, log 2𝑠).
1

3. 𝑂𝑆𝐷 = 𝑑1

+ 𝑑2

+ 𝑑3

, 𝑑1

= 𝑣𝑏

𝑡, 𝑑2

= 2𝑠 + 𝑣𝑏𝑇 , 𝑑3

= 𝑣𝑇 ,

𝐸ℎ3 4
38. Radius of relative stiffness 𝑙 = [ ]
12𝑘 1−𝜇 )

𝑇 = √4𝑠/𝑎
𝑣2 𝑛𝑙2 𝑉
4. = 𝑒 + 𝑓 , 𝑊 = +

( 𝑘𝑔
39. 𝑘 = 𝑐𝑚 (modulus of subgrade reaction)
0.125 𝑐𝑚

𝑅𝑔

𝑒 2𝑅

9.5√𝑅

40. 𝑘 = 𝐸

(𝑎 is rigid plate radius)

5. 𝐿𝑠

= 𝑣3 (comfort condition)
𝐶𝑅

1.18 𝑎
41. 𝑘 = 𝐸

(𝑎 is flexible plate radius)

6. 𝐿𝑠 = 𝑒𝑁(𝑊 + 𝑊𝑒) (rotated about inner edge)

1.5 𝑎

𝑉2

𝑉2

42. 𝑘1𝑎1 = 𝑘2𝑎2

7. 𝐿𝑠 = 2.7 𝑅 for plain and rolling terrain, 𝐿𝑠 = 𝑅 for steep and
mountainous terrain.
8. IRC recommends Spiral as transition curve.

43. 𝑅𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (below 7 𝑘𝑔/𝑐𝑚2 contact
𝑡𝑦𝑟𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
pressure is more)
𝛿

2
9. 𝑆 = 𝑠

44. 𝐿𝛼𝑡 =

(expansion joint)
2

24𝑅

45. 𝐿𝑐 𝑏ℎ𝛾 𝑓 = 𝑆 𝑏ℎ

10. 𝑚 = 𝑅 (1 − cos ), 𝜃 = 𝑆/𝑅 (single lane, 𝐿 > 𝑆)

2 𝑐 𝑐

2 46. 𝑏ℎ𝛾𝑐𝑓 = 𝐴𝑠𝑡 𝑆𝑠 (tie bar area of steel per meter)

11. 𝑚 = 𝑅 (1 − cos ) + 𝑆−𝐿 sin , 𝜃 = 𝐿 (Single lane, 𝐿 < 𝑆) 𝐿 𝜋∅2 2 2 2 𝑅 47. 2 𝜋∅𝑟𝑏𝑑 = 4 𝑆𝑠 (length of tie bar) 12. Curve resistance = 𝑇(1 − cos 𝛼) (𝜆𝑡)𝑥𝑒−𝜆𝑡 13. Grade compensation = 30+𝑅 or 75 whichever is less 48. 𝑃(𝑥) = 𝑥! (Poisson distribution) 14. 𝐿 = 𝑁𝑠2 (√2ℎ+√2𝐻) 𝑅 𝑅 when 𝐿 > 𝑠 (Summit curve for SSD), ℎ =

49. 𝑃(ℎ ≥ 𝑡) = 𝑒−𝜆𝑡
50. 𝑘 = 𝑘 (1 − 𝑣 ) and 𝑣 = 𝑣 (1 − 𝑘 )

0.15 𝑎𝑛𝑑 𝐻 = 1.2

𝑗 𝑣𝑓

𝑓
1000

𝑘𝑗

15. 𝐿 = 2𝑠 −

2
(√2ℎ+√2𝐻) when 𝐿 < 𝑠 (Summit curve for SSD), 𝑁 51. 𝑞 = 𝑘𝑣 and 𝑘 = 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 52. 𝑃𝐻𝐹 = 𝑝𝑒𝑎𝑘 ℎ𝑜𝑢𝑟 𝑓𝑙𝑜𝑤 or 𝑃𝐻𝐹 = 30𝑡ℎ ℎ𝑜𝑢𝑟𝑙𝑦 𝑣𝑜𝑙𝑢𝑚𝑒 ℎ = 0.15 𝑎𝑛𝑑 𝐻 = 1.2 𝑁𝑠2 4 ×𝑝𝑒𝑎𝑘 15 𝑚𝑖𝑛 𝑓𝑙𝑜𝑤 53. 𝑇𝑖𝑚𝑒 𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑 = 𝑣1+𝑣2+⋯+𝑣𝑛 𝐴𝐴𝐷𝑇 16. 𝐿 = (√2𝐻+√2𝐻) 2 when 𝐿 > 𝑠 (Summit curve for OSD), 𝐻 = 1.2

𝑛
54. 𝑆𝑝𝑎𝑐𝑒 𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑 = 𝑑

17. 𝐿 = 2𝑠 −

2
(√2𝐻+√2𝐻)
𝑁

, 𝐿 < 𝑠 (Summit curve for OSD), 𝐻 = 1.2 55. 𝑞 = 𝑛𝑎+𝑛𝑦, 𝑡̅ = 𝑡 − 𝑛𝑦 𝑡1+𝑡2+⋯+𝑡𝑛 𝑛 𝑁𝑠2 𝑡 +𝑡 𝑤 𝑞 18. 𝐿 = when 𝐿 > 𝑠 (valley curve), ℎ = 0.75 𝑎𝑛𝑑 𝛼 = 1°

𝑎 𝑤

2ℎ+2𝑠 tan 𝛼
2ℎ+2𝑠 tan 𝛼

56. Safe speed limit is 85th percentile speed

19. 𝐿 = 2𝑠 −
1
3

, 𝐿 < 𝑠 (valley curve), ℎ = 0.75 𝑎𝑛𝑑 𝛼 = 1°
𝑁

57. Geometric design is based on 98th percentile speed.
58. Road side facilities are based on 30th highest hourly volume.

20. 𝐿 = 2 (𝑁𝑣 )2 (valley curve comfort condition)
𝐶
21. 𝐶 = 80
75+𝑉

59. 𝐶0

= 1.5𝐿+5
1−𝑌0

(1+𝑒 )(

𝑝)

22. 𝐹𝑙𝑎𝑘𝑖𝑛𝑒𝑠𝑠 𝐼𝑛𝑑𝑒𝑥 =

60. 𝑄 = 280𝑤

𝑤 1−3
( 𝑤

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑎𝑘𝑦 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑡ℎ𝑖𝑛𝑛𝑒𝑟 𝑡ℎ𝑎𝑛 0.6 𝑑𝑚 × 100
𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡
23. 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑑𝑒𝑥 =
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑎𝑘𝑦 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑙𝑜𝑛𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 1.8 𝑑𝑚 × 100

1+ 𝑙 )
61. 𝑇 = 𝑇 + 𝑇 𝑚−𝑇𝑎
3
62. 𝐸𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = 7 % 𝑓𝑜𝑟 300 𝑚

6.5

𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡
100(𝑊
24. Angularity number= 67 − 𝐺
𝐶
cylinder, W is weight of aggregate packed in the cylinder.
25. Penetration test unit is 1/10th mm. Weight used 100 grams. Temperature 25℃.

63. 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑎𝑖𝑟𝑝𝑜𝑟𝑡, 𝑇𝑠 = 15 − 1000 ℎ
64. Temperature correction = 1% 𝑝𝑒𝑟 1℃ 𝑟𝑖𝑠𝑒 𝑜𝑓 𝑇𝑟 𝑎𝑏𝑜𝑣𝑒 𝑇𝑠
65. Gradient correction = 20% 𝑝𝑒𝑟 1% 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡.
66. 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑟𝑎𝑑𝑖𝑢𝑠 𝑅 = 𝑉2
125𝑓
67. 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑟𝑎𝑑𝑖𝑢𝑠 𝑅 = 0.388𝑊2 where 𝑠 = 6 + 𝑡𝑟𝑒𝑎𝑑

𝑊1+𝑊2+𝑊3
𝑎 𝑊1 𝑊2 𝑊3

𝑊𝑎+𝑊𝑏
𝑡 𝑊 𝑊

𝑇−𝑆 2
2

+ + 𝑎+ 𝑏

68. Turning radius for subsonic aircraft is 120 m and for supersonic it

𝐺1

𝐺2

𝐺3

𝐺𝑎

𝐺𝑏

27. 𝐺𝑚

= 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑜𝑢𝑙𝑑
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑜𝑢𝑙𝑑

is 180 m

28. 𝑣

= 𝐺𝑡−𝐺𝑚

, 𝑣

= 𝑊𝑏 × 𝐺𝑚

and 𝑉𝑀𝐴 = 𝑣

+ 𝑣

69. Grade compensation for BG is 0.04%, for MG is 0.03% and for NG

𝑎 𝐺𝑡

𝑏 𝑊𝑚

𝐺𝑏

𝑎 𝑏

is 0.02% per degree of curve.

29. 𝑉𝐹𝐵 = 𝑣𝑏
𝑉𝑀𝐴
30. Flow value units 1/4th mm

70. 𝐷 = 1720
𝑅

𝑐2

(1+𝑟)𝑛−1
31. 𝑁 = 365𝐴 [
𝑛

] 𝑉𝐷𝐹 × 𝐿𝐷𝐹 and 𝐴 = 𝑃(1 + 𝑟)𝑥

71. 𝑉𝑒𝑟𝑠𝑖𝑛𝑒 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒 =
8𝑅
72. 𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑎𝑛𝑡 = 𝑉2

× 𝐺

32. 𝐿𝐷𝐹 = 0.75 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑙𝑎𝑛𝑒𝑠 𝑎𝑛𝑑 0.4 𝑓𝑜𝑟 𝑓𝑜𝑢𝑟 𝑙𝑎𝑛𝑒 (single
carriageway)
33. LDF=0.75 for two lanes and 0.60 for three lane and 0.45 for four lane (dual carriageway)

127𝑅
73. Theoretical cant = 𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑎𝑛𝑡 + 𝑐𝑎𝑛𝑡 𝑑𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
74. Widening of gauge in cm, 𝑑 = 13(𝐵+𝐿)2 where B is wheel base in
𝑅

m, lap of flange in m, 𝐿 = 0.02√ℎ2 + 𝐷ℎ , h is depth of wheel

34. 𝑉𝐷𝐹 = (

𝑃 4
) , where P is in kN
80

flange below rail top level, D dia of wheel in cm.
𝑣3

35. 𝐶𝐵𝑅 𝑎𝑡 2.5 𝑚𝑚 =

𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑘𝑔/𝑐𝑚2 70

= 𝑙𝑜𝑎𝑑 𝑖𝑛 𝑘𝑔 1370

75. 𝐿𝑠 = 3.28 𝑅
𝑥3

(Transition curve)

36. 𝐶𝐵𝑅 𝑎𝑡 5 𝑚𝑚 =

𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑘𝑔/𝑐𝑚2

= 𝑙𝑜𝑎𝑑 𝑖𝑛 𝑘𝑔

76. 𝑦 =

6𝐿𝑅

(Transition curve)

105

2055

77. Usually adopted transition curve for railways is cubic parabola.