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128/3337–11FE Reference Handbook 10.4 · Mechanics of Materials · Energy Methods
1. Axial Stress & Deformation2. Elastic Flexure Formula3. Circular Shaft Torsion4. Axial Deformation5. Hooke’s Law (Uniaxial)6. Transverse Shear Stress in Beams7. Angle of Twist8. Thermal Expansion9. In-Plane Principal Stresses10. Euler Buckling Load11. Combined Axial and Bending Stress12. Shear Modulus from E and ν13. Mohr's Circle for Plane Stress14. Plane Stress Transformation15. Thin-Wall Pressure Vessel16. Thermal Stress, Fixed-End Bar17. von Mises (Distortional Energy) Stress18. Maximum Shear (Tresca) Stress19. Cantilever Tip Deflection and Slope20. Combined Axial and Torsion (Shaft)21. Single vs Double Shear on a Pin22. Axial Elastic Strain Energy

Handbook formula

Axial Elastic Strain Energy

Castigliano: δ = ∂U/∂P. For axial bars in series, sum U_i; for trusses, U=Σ N_i² L_i /(2 A_i E). Bending: ∫ M² ds /(2 EI).

(J)
Stored elastic energy
End displacement

Step-by-step solved example

Bar A=400 mm², L=2 m, E=200 GPa, P=80 kN. U and δ?

Lbar
U = P²L/2AE stored in the bar; δ = ∂U/∂P.

  1. 1. δ

    PL/AE=80000×2000/(400×200e3)=2.0 mm.

  2. 2. U

    ½ P δ=0.5×80000×0.002=80 J (or P²L/2AE).

Answer: δ = 2 mm, U = 80 J

10 practice questions

0/10 correct

1.Castigliano δ_j =

2.Two identical bars in series (same P) store

3.Bending energy for constant M on L is

4.Torsion energy is

5.Units of U are

6.If P doubles, U

7.Dummy-load method is Castigliano with

8.Resilience is U/volume at yield:

9.Statically indeterminate axial bars: energy or

10.A=500 mm², L=1 m, E=200 GPa, P=100 kN: δ =