A0 sec free body diagram element in

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Unformatted text preview: A0 ⌧✓ A0 tan ✓ ⌧✓ A0 sec ✓ Y ⌧✓ A0 sec ✓ sec ✓ ✓ ⌧ A0 TRIANGULAR ELEMENT IN FORCE EQUILIBRIUM X X FX = 0 ⌧✓ A0 sec ✓ = ⌧ A0 cos ✓ ⌧✓ A0 sec ✓ = ⌧ A0 cos ✓ ⌧✓ = ⌧ (cos2 ✓ FY = 0 ✓ A0 sec ✓ = ⌧ A0 sin ✓ + ⌧ Ao tan ✓ cos ✓ sin2 ✓) ✓ = 2⌧ sin ✓ cos ✓ ✓ ⌧ A0 tan ✓ sin ✓ = ⌧ sin 2✓ (3- 30b) ⌧✓ = ⌧ cos 2✓ (3- 30a) ⌧✓ = ⌧ cos 2✓ 2/1/13 M. Mello/Georgia Tech Aerospace ✓ = ⌧ sin 2✓ 8 MAX AND MIN PRINCIPAL STRESSES FOR AN ELEMENT IN PURE SHEAR TENSILE AND COMPRESSIVE STRESSES ACTING ON A STRESS ELEMENT ORIENTED AT 45° TO THE LONGITUDINAL AXIS maximum principal stress ( ⌧ ✓ 0) = minimum principal stress ( = ) ⌧✓ 0 ⌧✓ = ⌧ cos 2✓ ✓ = ⌧ sin 2✓ Crack propagates along A 45° helical surface 2/1/13 M. Mello/Georgia Tech Aerospace 9 ANALYSIS OF STRAINS IN PURE SHEAR STRESS ELEMENT ORIENTED AT Θ=0° STRAINS IN PURE SHEAR: ELEMENT DISTORTION AT Θ=0° shear strain for element oriented at Θ=0° ✏max = = ⌧ G (3- 31) STRESS ELEMENT STRAINS IN PURE SHEAR: ORIENTED AT Θ=45° ELEMENT DISTORTION AT Θ=45° stress element oriented at Θ=45° G(1 + ⌫ ) E so, ✏max M. Mello/Georgia Tech Aerospace ✏max max E ⌫ ✏min ⌧ = E ✓E◆ ⌧ ⌧ = ⌫ E E ⌧ = (1 + ⌫ ) E but, ✏min = Useful relaVon! But we can do beker than this.. This is a strong hint of a fundamental connecVon between G, E, and ν. 2/1/13 ✏max = min 10 RELATIONSHIP BETWEEN ELASTIC MODULUS (E) AND SHEAR MODULUS (G) GEOMETRY OF DEFORMED ELEMENT IN PURE SHEAR = undeformed Ldb = p 2h 1 \adc 2 deformed 0 Ldb = 0 Ldb p p 2h + ✏max ( 2h) p = 2h(1 + ✏max ) (a) INVOKE LAW OF COSINES ✓ ◆ 0 ⇡ (Ldb )2 = h2 + h2 2h2 cos + 2 0 SUBSTITUTE FOR Ldb FROM (a) 2/1/13 M. Mello/Georgia Tech Aerospace ✓ ⇡ (1 + ✏max )2 = 1 cos + 2 1 + 2✏max + ✏2 max = 1 + sin ⇡0 ✏max = 2 G(1 + ⌫ ) recall, ✏max = E Hence, G = E 2(1 + ⌫ ) 11 ◆...
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This note was uploaded on 09/19/2013 for the course CEE 3001 taught by Professor Zhu during the Spring '09 term at Georgia Institute of Technology.

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