Lecture 4 3 d stress tensor and equilibrium equations
Upcoming SlideShare
Loading in...5
×
 

Like this? Share it with your network

Share

Lecture 4 3 d stress tensor and equilibrium equations

on

  • 8,028 views

 

Statistics

Views

Total Views
8,028
Views on SlideShare
8,028
Embed Views
0

Actions

Likes
1
Downloads
604
Comments
1

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
  • This may be a really simple thing but I'm lost as to how you get the terms for the green arrows. I don't see why the partial of derivatives of sigma are involved.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Lecture 4 3 d stress tensor and equilibrium equations Presentation Transcript

  • 1. Unit 1- Stress and StrainTopics Covered  Lecture -1 - Introduction, state of plane stress  Lecture -2 - Principle Stresses and Strains  Lecture -3 - Mohrs Stress Circle and Theory of Failure  Lecture -4- 3-D stress and strain, Equilibrium equations and impact loading  Lecture -5 - Generalized Hooks law and Castiglionos
  • 2. 3-D Stress and Strain stress vector σ that represents the force per unit area acting at a given location on the bodys surface. In other words, a stress vector cannot be fully € described unless both the force and the surface where the force acts on has been specified. ΔF dF σ = lim = Δs−>0 Δs ds €
  • 3. 3-D Stress and Strain Suppose an arbitrary slice is made across the solid shown in the above figure, leading to the free body diagram shown at left. Stress would appear on the exposed surface, similar in form to the external stress applied to the bodys exterior surface. The stress at point P can be defined using the same above equation
  • 4. 3-D Stress and Strain Stresses acting on an plane, are typically decomposed into three mutually orthogonal components. One component is normal to the surface and represents direct stress. The other two components are tangential to the surface and represent shear stresses. Normal component = σxx ,σyy ,σzz Tangential component = xy ,σyx ,σxz ,σzx ,σyz ,σzy σ € €
  • 5. 3-D Stress and Strain Since each point on the cube is under static equilibrium (no net force in the absense of any body forces), only nine stress components from three planes are needed to describe the stress state at a point P. These nine components can be organized into the matrix: ⎡σ σxy σxz ⎤ xx ⎢ ⎥ ⎢σyx σyy σyz ⎥ ⎢σ ⎣ zx σzy σzz ⎥ ⎦In this course we are also where shear stresses across the diagonal are identicaldenoting shear stresses as τ as a result of static equilibrium (no net moment). This grouping of the nine stress components is known as the stress tensor (or stress matrix). € €
  • 6. 3-D Stress and Strain Shear stresses across the diagonal are identical as a result of static equilibrium (no net moment). The six shear stresses reduces to 3 shear stresses. This grouping of the six stress components is known as the stress tensor (or stress matrix). The off diagonal elements are equal i.e σxy = σyx ⎡σ σxy σxz ⎤ xx ⎢ ⎥ ⎢σxy σyy σyz ⎥ € ⎢σ ⎣ xz σyz σzz ⎥ ⎦ €
  • 7. Equilibrium equations ∂σyy σyy + dy ∂y X, Y – body force such as weight of the body Y ∂σyx σyx + dy ∂y σxy Y € ∂σxx dy σ σxx + dx xx € X ∂x€ σxy + ∂σxy ∂x dx ∑F x =0 σyx σyy€€ X ⎛ ∂σxx ⎞ dx ⎜σxx + dx ⎟(dy × 1) − σxx ( dy × 1) + € ⎝ ∂x ⎠ € € ⎛ ∂σyx ⎞ € ⎜σyx + dy ⎟( dx × 1) − σyx ( dx × 1) + X ( dxdy × 1) = 0 ⎝ ∂y ⎠ €
  • 8. Equilibrium equations ∂σyy σyy + dy ∂y ∂σyx For 2 dimension y σyx + dy ∂y σxy € Y ∂σ xx ∂σ xy dy σ σxx + ∂σxx dx + +X =0 xx € X ∂x ∂x ∂y€ ∂σxy σxy + ∂x dx ∂σ yx ∂σ yy σyx σyy€ + +Y = 0€ x ∂x ∂y dx € € € X, Y – body force such as weight of the body €
  • 9. Equilibrium equations For 3 dimension ∂σ xx ∂σ xy ∂σ xz + + +X =0 ∂x ∂y ∂z ∂σ yx ∂σ yy ∂σ yz + + +Y = 0 ∂x ∂y ∂z ∂σ zx ∂σ zy ∂σ zz + + +Z =0 ∂x ∂y ∂z
  • 10. Impact Load  Definitions   Resilience – Total strain energy stored in the system.   Proof resilience – Maximum strain energy stored in a body is known as proof resilience. Strain energy in the body will be maximum when the body is stressed upto elastic limit   Modulus of resilience- Proof resilience of a material per unit volume. Pr oof _ resilience Modulus of resilience = Volume _ of _ the _ body €
  • 11. Impact Load  Strain energy when load is applied gradually. M 2 Energy stored in a body= σV Load 2E P σ 2 AL N = 2E O Extension € x €
  • 12. Impact Load   Strain energy when load is applied suddenly. M 2 Energy stored in a body= σ AL Load 2E P σ 2 AL σ =P×x=P× ×L N 2E E O Extension € P x σ =2× A derivation in book - R.K Bansal€
  • 13. Impact Load  PROBLEM- A steel rod is 2m long and 50mm in diameter. An axial pull of 100 kN is suddenly applied to the rod. Calculate the instantaneous stress induced and also the instantaneous elongation produced in the rod. Take E=200GN/mm2
  • 14. Impact Load   Strain energy when load is applied with impact. Energy of impact = Potential energy of the falling load σ 2 AL Energy of impact = 2E Potential energy of the falling load = P ( h + δL ) € P ⎛ €2AEh ⎞ σ = ⎜1+ 1+ ⎟ A ⎝ PL ⎠€
  • 15. Impact Load   PROBLEM- A vertical compound tie member fixed rigidly at its upper end 20 mm consists of a steel rod 2.5 m long and2.5 m P=10kN 30mm external diameter. The rod and 1 2 the tube are fixed together at the ends. 3 mm The compound member is then suddenly loaded in tension by a weight 21 mm of 10 kN falling through a height of 3 mm on to a flange fixed to its lower 30 mm end. Calculate the maximum stresses in steel and brass. Assume Es=2x105 N/mm2 and Eb=1.0x105 N/mm2
  • 16. Impact Load  Strain energy in shear loading. τ 2 AL Strain energy stored = 2C P D D1 C C1 € h φ φ A l B € €
  • 17. Impact Load  PROBLEM- The shear stress in a material at a price is given as 50N/mm2. Determine the local strain energy per unit volume stored in the material due to shear stress. Take C=8x104 N/mm2