Basic formula for structural analysis

Looking at your requirement, we are sharing with you some of Important Civil EngineeringΒ Formulas & Shortcuts for Competitive Exam as well as Engineering exam.

Strength of material

  1. πœ‡Β =Β πΏπ‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™Β π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›πΏπ‘œπ‘›π‘”π‘–π‘‘π‘’π‘‘π‘–π‘›π‘Žπ‘™Β π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›
  2. 𝑒π‘₯Β =𝑓π‘₯πΈΒ βˆ’πœ‡π‘“π‘¦πΈΒ βˆ’Β πœ‡π‘“π‘§πΈ
  3. 𝑒𝑦=Β π‘“π‘¦πΈΒ βˆ’Β πœ‡π‘“π‘₯πΈΒ βˆ’Β πœ‡π‘“π‘§πΈ
  4. 𝑒𝑧=Β π‘“π‘§πΈΒ βˆ’Β πœ‡π‘“π‘₯πΈΒ βˆ’Β πœ‡π‘“π‘¦πΈ
  5. 𝐺= 𝐸2(1+πœ‡)
  6. 𝐾=𝐸 3(1βˆ’2πœ‡)
  7. 𝑀𝐼=𝑓𝑦=𝐸𝑅
  8. 𝜏 =VQIb
  9. 𝑇𝐽 =Β πœπ‘ŸΒ =Β πΊπœƒπΏ
  10. 𝑀𝑒 = 12[𝑀+βˆšπ‘€2+𝑇2]
  11. 𝑇𝑒= βˆšπ‘€2+𝑇2
  12. 𝜎= 3𝑀𝑙2𝑛𝑏𝑑2 and 𝛿 =Β πœŽπ‘™24𝐸𝑑 (Leaf spring)
  13. 𝛿 = 64π‘Šπ‘…3𝑛𝐺𝑑4 (Helical spring)
  14. πœŽπ‘›=𝜎π‘₯+πœŽπ‘¦2+𝜎π‘₯βˆ’πœŽπ‘¦2cos2πœƒ+𝜏sin2πœƒ
  15. πœπ‘‘=βˆ’(𝜎π‘₯βˆ’πœŽπ‘¦2)sin2πœƒ+𝜏cos2πœƒ
  16. 𝜎1=𝜎π‘₯+πœŽπ‘¦2+√(𝜎π‘₯βˆ’πœŽπ‘¦2)2+𝜏2
  17. 𝜎3=𝜎π‘₯+πœŽπ‘¦2βˆ’βˆš(𝜎π‘₯βˆ’πœŽπ‘¦2)2+𝜏2
  18. 𝑃=Β πœ‹2𝐸𝐼𝑙2 (hinged and hinged)) ,𝑃 =Β πœ‹2𝐸𝐼4𝑙2 (fixed and free)
  19. 𝑃 = 4πœ‹2𝐸𝐼𝑙2 (Fixed and fixed) ,𝑃 = 2πœ‹2𝐸𝐼𝑙2
  20. 𝑃=Β πœŽπ‘.𝐴1+𝛼(𝑙𝑒𝐾)2 where 𝛼=πœŽπ‘πœ‹2𝐸 is Rankine’s constant

Structural analysisΒ 

  1. 𝑀̅𝐴𝐡=βˆ’π‘ƒπ‘Žπ‘2𝑙2 and 𝑀̅𝐴𝐡=π‘ƒπ‘Ž2𝑏𝑙2 (Point load)
  2. 𝑀̅𝐴𝐡=βˆ’π‘€π‘™212 and 𝑀̅𝐴𝐡=𝑀𝑙212 (udl)
  3. 𝑀̅𝐴𝐡=βˆ’π‘€π‘™230 and 𝑀̅𝐴𝐡=𝑀𝑙220 (uvl from left to right increase)
  4. 𝑀̅𝐴𝐡=𝑀𝑏𝑙2(3π‘Žβˆ’π‘™) and 𝑀̅𝐴𝐡=π‘€π‘Žπ‘™2(3π‘βˆ’π‘™) (clockwise moment)
  5. π‘ˆ=𝑃2𝐿2𝐴𝐸 ,Β π‘ˆ=βˆ«π‘€22𝐸𝐼𝐿0 𝑑π‘₯Β andΒ π‘ˆ=𝑇2𝐿2𝐺𝐽
  6. π‘ˆ=1.2βˆ«π‘‰22𝐺𝐴𝐿0𝑑π‘₯Β (rectangular beam)
  7. 𝐻= βˆ«π‘€π‘ π‘¦πΈπΌΒ π‘‘π‘₯+βˆπ‘‘π‘™βˆ’π›Ώπ‘™0βˆ«π‘¦2𝑑π‘₯𝐸𝐼𝑙0 (two hinged arch)
  8. H=𝑀𝑙28𝑑 (cable with udl) and Length of cable 𝐿= 𝑙+83𝑑2𝑙
  9. 𝑀𝐴𝐡=𝑀̅𝐴𝐡+2𝐸𝐼𝐿(2πœƒπ΄+πœƒπ΅βˆ’3Δ𝐿) , right support sinks by Ξ”
  10. 𝑀𝐡𝐴=𝑀̅𝐡𝐴+2𝐸𝐼𝐿(2πœƒπ΅+πœƒπ΄βˆ’3Δ𝐿), right support sinks by Ξ”
  11. 𝑀𝐴 𝑙1+2𝑀𝐡(𝑙1+𝑙2)+𝑀𝐢𝑙2=βˆ’6π‘Ž1π‘₯1𝑙1βˆ’6π‘Ž2π‘₯2𝑙2
  12. 𝛿=Ξ£π‘ƒπ‘˜πΏπ΄πΈ
  13. 𝑄=βˆ’Ξ£π‘ƒπ‘˜πΏπ΄πΈΞ£π‘˜2𝐿𝐴𝐸+𝐿0𝐴0𝐸0 and 𝑄=βˆ’Ξ£(𝑃𝐿𝐴𝐸+𝐿𝛼𝑑)π‘˜Ξ£π‘˜2𝐿𝐴𝐸+𝐿0𝐴0𝐸0
  14. π‘˜π‘–π‘—Β =π‘“π‘œπ‘Ÿπ‘π‘’Β π‘Žπ‘‘Β π‘–Β π‘‘π‘’π‘’Β π‘‘π‘œΒ π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘Β π‘Žπ‘‘Β π‘—, 𝛿𝑖𝑗=π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘Β π‘Žπ‘‘Β π‘–Β π‘‘π‘’π‘’Β π‘‘π‘œΒ π‘“π‘œπ‘Ÿπ‘π‘’Β π‘Žπ‘‘Β π‘—
  15. Far end fixed, Transverse displacement, 𝛿=𝐿312𝐸𝐼 andΒ π‘˜=12𝐸𝐼𝐿3
  16. Far end hinged, Transverse displacement, 𝛿=𝐿33𝐸𝐼 andΒ π‘˜=3𝐸𝐼𝐿3


  1. π‘“π‘š=π‘“π‘π‘˜+1.65 𝜎
  2. 𝐸𝑐=5000 βˆšπ‘“π‘π‘˜Β andΒ π‘“π‘π‘Ÿ=0.7 βˆšπ‘“π‘π‘˜
  3. πœ€π‘ π‘‘=0.002+0.87 𝑓𝑦𝐸𝑠
  4. 𝐸𝑠=2Γ— 105Β π‘€π‘ƒπ‘ŽΒ ,Β π‘š=2803πœŽπ‘π‘π‘
  5. 𝑏𝑓=𝑙06+𝑏𝑀+6𝐷𝑓 for T-beams.
  6. 𝑏𝑓=𝑙012+𝑏𝑀+3𝐷𝑓 for L-beams.
  7. 𝑙0𝑙0𝑏+4+𝑏𝑀 ≀𝑏 for isolated T-beams.
  8. 0.5𝑙0𝑙0𝑏+4 +𝑏𝑀 ≀𝑏 for isolated L-beams.
  9. π‘₯𝑒𝑑=0.87 𝑓𝑦. 𝐴𝑠𝑑0.36Β π‘“π‘π‘˜Β π‘π‘‘Β andΒ π‘₯𝑒,π‘šπ‘Žπ‘₯𝑑= 0.00350.0055+0.87 𝑓𝑦𝐸𝑠
  10. 𝑀𝑒=0.87 𝑓𝑦.𝐴𝑠𝑑 (π‘‘βˆ’0.42Β π‘₯𝑒) forΒ π‘₯𝑒 ≀π‘₯𝑒,max
  11. 𝑀𝑒=0.36Β π‘“π‘π‘˜.𝑏.π‘₯𝑒,π‘šπ‘Žπ‘₯(π‘‘βˆ’0.42Β π‘₯𝑒,π‘šπ‘Žπ‘₯)
  12. π‘₯𝑒,π‘šπ‘Žπ‘₯𝑑= 0.87 𝑓𝑦0.36Β π‘“π‘π‘˜Β (𝑃𝑑,π‘™π‘–π‘šΒ 100)
  13. 𝑀𝑒=0.87 𝑓𝑦 𝐴𝑠𝑑 (π‘‘βˆ’π‘“π‘¦Β π΄π‘ π‘‘π‘“π‘π‘˜Β π‘)
  14. 𝑦𝑓=0.15Β π‘₯𝑒+0.65 𝐷𝑓 when 𝐷𝑓𝑑>0.2
  15. 0.87 𝑓𝑦 (π΄π‘ π‘‘βˆ’π΄π‘ π‘‘,𝑙)=(π‘“π‘ π‘βˆ’0.447Β π‘“π‘π‘˜).𝐴𝑠𝑐
  16. 𝑀𝑒=𝑀𝑒,𝑙 + 0.87 𝑓𝑦.(π΄π‘ π‘‘βˆ’π΄π‘ π‘‘,𝑙) (π‘‘βˆ’π‘‘β€²)
  17. 𝐴𝑠𝑑 𝑏𝑑= 0.85𝑓𝑦 (minimum tension reinforcement)
  18. 𝐴𝑠𝑑,π‘šπ‘–π‘›=0.12 % of 𝐴𝑔 for 𝐹𝑒-415 and 0.15 % of 𝐴𝑔 for 𝐹𝑒-250. (slabs)
  19. 𝐴𝑠𝑑,π‘šπ‘Žπ‘₯=4%Β π‘œπ‘“Β π΄π‘”Β (Beams) ,𝐴𝑠𝑐,π‘šπ‘Žπ‘₯=4%Β π‘œπ‘“Β π΄π‘”Β (Beams)
  20. (𝑙𝑑)π‘šπ‘Žπ‘₯=(𝑙𝑑)π‘π‘Žπ‘ π‘–π‘π‘˜π‘‘Β π‘˜π‘
  21. (𝑙𝑑)π‘π‘Žπ‘ π‘–π‘=7 for cantilever, 20 for SS and 26 for continuous beams. If span is more than 10 m, multiply above values with 10/span for SS and continuous beams.
  22. 𝐿𝑑=π‘“π‘ βˆ…4πœπ‘π‘‘Β , multiplyΒ πœπ‘π‘‘Β value with 1.6 for deformed bars and 1.25 for bars in compression.
  23. 𝑀𝑒𝑉𝑒+𝐿0β‰₯𝐿𝑑 and 1.3𝑀𝑒𝑉𝑒+𝐿0β‰₯𝐿𝑑 (If confinement exists)
  24. πœπ‘£=𝑉𝑒𝑏𝑑 , 𝑉𝑒,𝑛𝑒𝑑=𝑉𝑒+𝑀𝑑tan𝛽
  25. 𝑉𝑒𝑠=0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑𝑠𝑣
  26. 𝐴𝑠𝑣𝑏 𝑠𝑣β‰₯0.40.87 𝑓𝑦 (minimum shear reinforcement)
  27. 𝑠𝑣≀0.75Β π‘‘Β π‘Žπ‘›π‘‘Β 300Β π‘šπ‘šΒ for vertical stirrups.
  28. 𝑉𝑒=𝑉𝑒+1.6𝑇𝑒𝑏 and 𝑀𝑒=𝑀𝑒+𝑇𝑒1.7(1+𝐷𝑏)
  29. 𝑙𝑑≀35 for ss and 40 for continuous slabs for Fe-250. For Fe-415 multiply above values with 0.8
  30. πœπ‘2=π‘˜π‘ Β 0.25βˆšπ‘“π‘π‘˜Β ,Β π‘˜π‘ =0.5+𝛽𝑐≀1.0 and 𝛽𝑐=𝑏𝐷
  31. 𝑒π‘₯,π‘šπ‘–π‘›=𝑙500+𝑑30 ,20Β π‘šπ‘šΒ whichever is greater.
  32. 𝑒𝑦,π‘šπ‘–π‘›=𝑙500+𝑏30 ,20Β π‘šπ‘šΒ whichever is greater.
  33. 𝑃𝑒=0.4Β π‘“π‘π‘˜Β π΄π‘+0.67 𝑓𝑦𝐴𝑠𝑐
  34. π‘‰β„Žπ‘‰π‘β‰₯0.36(π΄π‘”π΄π‘βˆ’1)π‘“π‘π‘˜π‘“π‘¦Β , 𝑑𝑐=π‘‘βˆ’2𝑐 andΒ π‘‘π‘š=π‘‘π‘βˆ’βˆ…
  35. πΆπ‘Ÿ=1.25βˆ’π‘™π‘’48𝑏

36.Β π‘“π‘π‘Ÿ,π‘šπ‘Žπ‘₯=0.45Β π‘“π‘π‘˜βˆšπ΄1𝐴2 , 1<√𝐴1𝐴2≀2 , 𝐴1=Largest frustum of a pyramid with side slopes 1 in 2, 𝐴2=loaded area of column base

Hello friends, my name is Bipin Kumar, I am the Writer and Founder of this blog and share all the information related to Civil Engineering, Civil practical Knowledge, Site Execution Knowledge, latest information about construction and more through this website.

Leave a Comment