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

Strength of material

1. 𝜇 = 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

2. 𝑒𝑥 =𝑓𝑥𝐸 −𝜇𝑓𝑦𝐸 − 𝜇𝑓𝑧𝐸

3. 𝑒𝑦= 𝑓𝑦𝐸 − 𝜇𝑓𝑥𝐸 − 𝜇𝑓𝑧𝐸

4. 𝑒𝑧= 𝑓𝑧𝐸 − 𝜇𝑓𝑥𝐸 − 𝜇𝑓𝑦𝐸

5. 𝐺= 𝐸2(1+𝜇)

6. 𝐾=𝐸 3(1−2𝜇)

7. 𝑀𝐼=𝑓𝑦=𝐸𝑅

8. 𝜏 =VQIb

9. 𝑇𝐽 = 𝜏𝑟 = 𝐺𝜃𝐿

10. 𝑀𝑒 = 12[𝑀+√𝑀2+𝑇2]

11. 𝑇𝑒= √𝑀2+𝑇2

12. 𝜎= 3𝑤𝑙2𝑛𝑏𝑡2 and 𝛿 = 𝜎𝑙24𝐸𝑡 (Leaf spring)

13. 𝛿 = 64𝑊𝑅3𝑛𝐺𝑑4 (Helical spring)

14. 𝜎𝑛=𝜎𝑥+𝜎𝑦2+𝜎𝑥−𝜎𝑦2cos2𝜃+𝜏sin2𝜃

15. 𝜏𝑡=−(𝜎𝑥−𝜎𝑦2)sin2𝜃+𝜏cos2𝜃

16. 𝜎1=𝜎𝑥+𝜎𝑦2+√(𝜎𝑥−𝜎𝑦2)2+𝜏2

17. 𝜎3=𝜎𝑥+𝜎𝑦2−√(𝜎𝑥−𝜎𝑦2)2+𝜏2

18. 𝑃= 𝜋2𝐸𝐼𝑙2 (hinged and hinged)) ,𝑃 = 𝜋2𝐸𝐼4𝑙2 (fixed and free)

19. 𝑃 = 4𝜋2𝐸𝐼𝑙2 (Fixed and fixed) ,𝑃 = 2𝜋2𝐸𝐼𝑙2

20. 𝑃 = 𝜎𝑐.𝐴1+𝛼(𝑙𝑒𝐾)2 where 𝛼=𝜎𝑐𝜋2𝐸 is Rankine’s constant

Civil Engineering Basic formula for strength of material,structure analysis and RCC
Engineering Formula
Structural analysis 

1. 𝑀̅𝐴𝐵=−𝑃𝑎𝑏2𝑙2 and 𝑀̅𝐴𝐵=𝑃𝑎2𝑏𝑙2 (Point load)

2. 𝑀̅𝐴𝐵=−𝑤𝑙212 and 𝑀̅𝐴𝐵=𝑤𝑙212 (udl)

3. 𝑀̅𝐴𝐵=−𝑤𝑙230 and 𝑀̅𝐴𝐵=𝑤𝑙220 (uvl from left to right increase)

4. 𝑀̅𝐴𝐵=𝑀𝑏𝑙2(3𝑎−𝑙) and 𝑀̅𝐴𝐵=𝑀𝑎𝑙2(3𝑏−𝑙) (clockwise moment)

5. 𝑈=𝑃2𝐿2𝐴𝐸 , 𝑈=∫𝑀22𝐸𝐼𝐿0 𝑑𝑥 and 𝑈=𝑇2𝐿2𝐺𝐽

6. 𝑈=1.2∫𝑉22𝐺𝐴𝐿0𝑑𝑥 (rectangular beam)

7. 𝐻= ∫𝑀𝑠𝑦𝐸𝐼 𝑑𝑥+∝𝑡𝑙−𝛿𝑙0∫𝑦2𝑑𝑥𝐸𝐼𝑙0 (two hinged arch)

8. H=𝑤𝑙28𝑑 (cable with udl) and Length of cable 𝐿= 𝑙+83𝑑2𝑙

9. 𝑀𝐴𝐵=𝑀̅𝐴𝐵+2𝐸𝐼𝐿(2𝜃𝐴+𝜃𝐵−3Δ𝐿) , right support sinks by Δ

10. 𝑀𝐵𝐴=𝑀̅𝐵𝐴+2𝐸𝐼𝐿(2𝜃𝐵+𝜃𝐴−3Δ𝐿), right support sinks by Δ

11. 𝑀𝐴 𝑙1+2𝑀𝐵(𝑙1+𝑙2)+𝑀𝐶𝑙2=−6𝑎1𝑥1𝑙1−6𝑎2𝑥2𝑙2

12. 𝛿=Σ𝑃𝑘𝐿𝐴𝐸

13. 𝑄=−Σ𝑃𝑘𝐿𝐴𝐸Σ𝑘2𝐿𝐴𝐸+𝐿0𝐴0𝐸0 and 𝑄=−Σ(𝑃𝐿𝐴𝐸+𝐿𝛼𝑡)𝑘Σ𝑘2𝐿𝐴𝐸+𝐿0𝐴0𝐸0

14. 𝑘𝑖𝑗 =𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑖 𝑑𝑢𝑒 𝑡𝑜 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑗, 𝛿𝑖𝑗=𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑖 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑗

15. Far end fixed, Transverse displacement, 𝛿=𝐿312𝐸𝐼 and 𝑘=12𝐸𝐼𝐿3

16. Far end hinged, Transverse displacement, 𝛿=𝐿33𝐸𝐼 and 𝑘=3𝐸𝐼𝐿3

RCC 

1. 𝑓𝑚=𝑓𝑐𝑘 +1.65 𝜎

2. 𝐸𝑐=5000 √𝑓𝑐𝑘 and 𝑓𝑐𝑟=0.7 √𝑓𝑐𝑘

3. 𝜀𝑠𝑡=0.002+0.87 𝑓𝑦𝐸𝑠

4. 𝐸𝑠=2× 105 𝑀𝑃𝑎 , 𝑚=2803𝜎𝑐𝑏𝑐

5. 𝑏𝑓=𝑙06+𝑏𝑤+6𝐷𝑓 for T-beams.

6. 𝑏𝑓=𝑙012+𝑏𝑤+3𝐷𝑓 for L-beams.

7. 𝑙0𝑙0𝑏+4+𝑏𝑤 ≤𝑏 for isolated T-beams.

8. 0.5𝑙0𝑙0𝑏+4 +𝑏𝑤 ≤𝑏 for isolated L-beams.

9. 𝑥𝑢𝑑=0.87 𝑓𝑦. 𝐴𝑠𝑡0.36 𝑓𝑐𝑘 𝑏𝑑 and 𝑥𝑢,𝑚𝑎𝑥𝑑= 0.00350.0055+0.87 𝑓𝑦𝐸𝑠

10. 𝑀𝑢=0.87 𝑓𝑦.𝐴𝑠𝑡 (𝑑−0.42 𝑥𝑢) for 𝑥𝑢 ≤𝑥𝑢,max

11. 𝑀𝑢=0.36 𝑓𝑐𝑘.𝑏.𝑥𝑢,𝑚𝑎𝑥(𝑑−0.42 𝑥𝑢,𝑚𝑎𝑥)

12. 𝑥𝑢,𝑚𝑎𝑥𝑑= 0.87 𝑓𝑦0.36 𝑓𝑐𝑘 (𝑃𝑡,𝑙𝑖𝑚 100)

13. 𝑀𝑢=0.87 𝑓𝑦 𝐴𝑠𝑡 (𝑑−𝑓𝑦 𝐴𝑠𝑡𝑓𝑐𝑘 𝑏)

14. 𝑦𝑓=0.15 𝑥𝑢+0.65 𝐷𝑓 when 𝐷𝑓𝑑>0.2

15. 0.87 𝑓𝑦 (𝐴𝑠𝑡−𝐴𝑠𝑡,𝑙)=(𝑓𝑠𝑐−0.447 𝑓𝑐𝑘).𝐴𝑠𝑐

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

17. 𝐴𝑠𝑡 𝑏𝑑= 0.85𝑓𝑦 (minimum tension reinforcement)

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

19. 𝐴𝑠𝑡,𝑚𝑎𝑥=4% 𝑜𝑓 𝐴𝑔 (Beams) ,𝐴𝑠𝑐,𝑚𝑎𝑥=4% 𝑜𝑓 𝐴𝑔 (Beams)

20. (𝑙𝑑)𝑚𝑎𝑥=(𝑙𝑑)𝑏𝑎𝑠𝑖𝑐𝑘𝑡 𝑘𝑐

21. (𝑙𝑑)𝑏𝑎𝑠𝑖𝑐=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=𝑓𝑠∅4𝜏𝑏𝑑 , multiply 𝜏𝑏𝑑 value with 1.6 for deformed bars and 1.25 for bars in compression.

23. 𝑀𝑢𝑉𝑢+𝐿0≥𝐿𝑑 and 1.3𝑀𝑢𝑉𝑢+𝐿0≥𝐿𝑑 (If confinement exists)

24. 𝜏𝑣=𝑉𝑢𝑏𝑑 , 𝑉𝑢,𝑛𝑒𝑡=𝑉𝑢+𝑀𝑑tan𝛽

25. 𝑉𝑢𝑠=0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑𝑠𝑣

26. 𝐴𝑠𝑣𝑏 𝑠𝑣≥0.40.87 𝑓𝑦 (minimum shear reinforcement)

27. 𝑠𝑣≤0.75 𝑑 𝑎𝑛𝑑 300 𝑚𝑚 for vertical stirrups.

28. 𝑉𝑒=𝑉𝑢+1.6𝑇𝑢𝑏 and 𝑀𝑒=𝑀𝑢+𝑇𝑢1.7(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. 𝜏𝑐2=𝑘𝑠 0.25√𝑓𝑐𝑘 , 𝑘𝑠=0.5+𝛽𝑐≤1.0 and 𝛽𝑐=𝑏𝐷

31. 𝑒𝑥,𝑚𝑖𝑛=𝑙500+𝑑30 ,20 𝑚𝑚 whichever is greater.

32. 𝑒𝑦,𝑚𝑖𝑛=𝑙500+𝑏30 ,20 𝑚𝑚 whichever is greater.

33. 𝑃𝑢=0.4 𝑓𝑐𝑘 𝐴𝑐+0.67 𝑓𝑦𝐴𝑠𝑐

34. 𝑉ℎ𝑉𝑐≥0.36(𝐴𝑔𝐴𝑐−1)𝑓𝑐𝑘𝑓𝑦 , 𝑑𝑐=𝑑−2𝑐 and 𝑑𝑚=𝑑𝑐−∅

35. 𝐶𝑟=1.25−𝑙𝑒48𝑏

36. 𝑓𝑏𝑟,𝑚𝑎𝑥=0.45 𝑓𝑐𝑘√𝐴1𝐴2 , 1<√𝐴1𝐴2≤2 , 𝐴1=Largest frustum of a pyramid with side slopes 1 in 2, 𝐴2=loaded area of column base