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Unformatted text preview: Read Chapter 1 Basic Elasticity Equilibrium Equations Plane Stresses Principal Stresses Mohr’s Circle Stress Strain Relationships Stress Stress Assumptions: • Equilibrium under external forces • Continuous and deformable material • Forces are transmitted throughout its volume Stress Stress Resultant force at point O δP Force must be in equilibrium: Equal and opposite force δP on the particle Divide the particle along plane nn containing O δP can be considered as uniformly distributed over a small area δA A P Stress A δ δ δ lim → = Stress Stress A P n A δ δ σ δ lim → = Normal or direct stress A P s A δ δ τ δ lim → = Shear stress 2 2 stress Resultant τ σ + = Equilibrium Equations Equilibrium Equations = + ∂ ∂ + ∂ ∂ + ∂ ∂ = + ∂ ∂ + ∂ ∂ + ∂ ∂ = + ∂ ∂ + ∂ ∂ + ∂ ∂ Z y x z Y z x y X z y x zy zx z yz yx y xz xy x τ τ σ τ τ σ τ τ σ Plane Stresses • First two terms Taylor Series Expansion • Derivation on Page 9 (xdirection equilibrium) Such as thin plates loaded in plane (Thickness direction << plane directions) Stresses are negligible (thickness direction) as they can not develop within the material Equilibrium Equations Equilibrium Equations • Taking sum of the moments about an axis through the centre line of the element parallel to the z axis it would be found that: zy yz zx xz yx xy τ τ τ τ τ τ = = = “Equal and Complementary” Determination of stresses Determination of stresses on an incline plane on an incline plane Body forces ignored (second order term) δx and δy are small stress distribution is assumed uniform This is done to study stress severity in different planes Determination of stresses Determination of stresses on an incline plane on an incline plane Sum of the Forces perpendicular to ED: Sum of the Forces parallel to ED: θ τ θ τ θ σ θ σ σ cos sin sin cos CD EC CD EC ED xy xy y x n + + + = θ τ θ σ θ σ σ 2 sin sin cos 2 2 xy y x n + + = θ τ θ τ θ σ θ σ τ sin cos cos sin CD EC CD EC ED xy xy y x + = ( 29 θ τ θ σ σ τ 2 cos 2 sin 2 xy y x = Example Example • Cantilever beam of solid crosssection • Compressive load of 50 KN, 1.5 mm below horizontal diameter plane • Torque of 1200 Nm • Calculate direct and shear stresses on a plane 60 to the axis of the cantilever beam on a point located the lower edge of the vertical plane of symmetry Example cont’d...
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This note was uploaded on 10/28/2009 for the course AERSP 301 at Penn State.
 '09
 GANDHI,FARHANLESIEUTRE,GEORGE

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