CH-1-Stress-strain-08-1La - Chapter 1 Final Exam: 80% CA:...

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Chapter 1 Final Exam: 80% CA: 20%
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ANALYSIS OF STRESS AND STRAIN n is normal to dA S 1 , S 2 are tangential (in plane) Apply general force dF on dA Defn: As dA 0, stress state is at the point P. ( 29 ( 29 dA dF Shear dA dF Normal dA dF S dA S S dA S n dA n 2 0 2 1 0 1 0 lim lim lim = = = τ σ Note: Stress values depend on magnitude of dF and also the direction of dF. STRESSES dA P s 1 n dF s 2
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y x z σ yy τ yx τ yz τ zx τ zy σ zz σ xx τ xy τ xz Stresses shown are all positive on a cube of 1 unit length For a small isolated element with planes perpendicular to coordinate axes and surrounding a point P, there exist 9 stress components. They are 3 normal stresses 6 shear stresses As size of parallelepiped reduces, in the limit, these 9 stress components will define completely, the state of stress at the point P. P
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y x z σ yy τ yx τ yz τ zx τ zy σ zz σ xx τ xy τ xz Take Moment about Z-axis τ xy · 1x1 - τ yx · 1x1 = 0 i.e. τ xy = τ yx
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From equilibrium (i.e. taking moment about any axis), we can show that: zy yz zx xz yx xy τ = = = NO. of “unknown” stresses reduced to 6. (u only need 6 unknown stresses for 3D body) i.e. σ xx , σ yy , σ zz, National University of Singapore National University of Singapore yz xz xy , ,
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Stress Component on Arbitrary plane (2 dimensional case) z σ y θ τ yx σ x τ x’y’ σ x σ y τ xy σ x x y Arbitrary plane whose normal makes an angle θ with horizontal 2 thing are weld together , we interested to know τ x’y’ and σ x
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Let T x and T y be the stress components of σ x , τ x’y’ in the x- and y- directions. Equilibrium of forces in x- direction: θ τ σ sin cos xy x x T + = y- direction: BC AC BC AB sin cos y xy y T + = sin cos ' y x x T T + = sin cos ' ' x y y x T T - = x θ C B y x ' y ' A σ x τ x’y’ τ xy τ yx σ y T y T x σ x σ x *AC*1 + τ xy *AB*1 = T x * BC *1 (resolve)- \ sigma_x’*BC*1+ T_x*BC*1*cos\theta = T_y*BC*1*sin\theta (re solve)- (cancel out BC*1) Tau_x’y’*BC*1= T_y*BC*1*cos\theta – T_x*BC*sin\theta (resolve) (cancel out BC*1) \theta
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Substitute for T x and T y θ τ σ cos sin 2 sin cos ' 2 2 xy y x x + + = cos sin ) ( ) sin (cos 2 2 ' ' x y xy y x - + - = σ y can be found by substituting θ + π /2 for θ in expression for σ x ’ , i.e. cos sin 2 cos sin ' 2 2 xy y x y - + = National University of Singapore National University of Singapore etc 2 2 2 2 sin 2 1 1 cos 2 sin cos 2 cos - = - = - = Using the relation
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Rewriting, (i) (ii) (iii) θ τ σ 2 sin 2 cos 2 ) ( 2 ) ( ' xy y x y x x + - + + = 2 sin 2 cos 2 ) ( 2 ) ( ' xy y x y x y - - - + = 2 cos 2 sin 2 ) ( , , xy y x y x + - - = σ y y τ xy σ x x σ y y' τ x’y’ σ x x' θ x σ y σ x τ xy τ xy y’ Stress Transforming Equation
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When , ) ( 2 tan 2 1 ) ( 2 2 tan . . 0 2 cos 2 sin 2 ) ( 1 y x xy y x xy xy y x or e i σ τ θ - = - = = + - - - 0 , , = y x σ x which is denoted as σ 1 is known as the maximum principal stress σ y’ which is denoted as σ 2 is known as the minimum principal stress and θ , which is denoted as α is known as the principal angle and Eq. a
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y x z σ yy τ yx τ yz τ zx τ zy σ zz σ xx τ xy τ xz Convention for denoting stress 1. Normal stress, σ ij
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This note was uploaded on 04/21/2010 for the course ME ME2004 taught by Professor Shah during the Spring '10 term at National University of Juridical Sciences.

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CH-1-Stress-strain-08-1La - Chapter 1 Final Exam: 80% CA:...

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