MP2001 Chap 7 part 2

# X y 2 2 3 and using eqn 744 x y x y 2 cos

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Unformatted text preview: at origin becomes < 90o • θ +ve when rotated anti-clockwise 17 • Using trigonometric relations as in transform. of stress, εx + εy εx − εy γ xy + ε x′ = cos 2θ + sin 2θ ( 7.44 ) 2 2 2 εx + εy εx − εy γ xy εy ′ = − cos 2θ − sin 2θ (7.45 ) 2 2 2 γ x ′y ′ εx − εy γ xy (7.49 ) 18 =− sin 2θ + cos 2θ 2 2 2 9 Example: Using a 60o rosette, the strains determined are ε1 = 40 µ, ε2 = 40 µ, ε3 = 40 µ. Find εx, εy and γ xy. 60o θ1 = 0, θ2 = 60o, θ3 = 120o. εx + εy 2 + 2 3 and using eqn (7.44), ε x′ = y εx − εy 2 cos 2θ + 60o 1 γ xy 2 x sin 2θ ( 7 .44 ) ε1 = ε x ε 2 = 0 .25 ε x + 0 .75 ε y + 0 .433 γ xy ε 3 = − 0 .25 ε x + 0 .75 ε y − 0 .433 γ xy Subst values of of , ε1 and εsolving gives values ε1 ε2 and 3 and solving give ε x = 40 μ ε y = 860 μ γ xy = 750 μ 19 8.4 Stresses Under Combined Loadings • Ch 1, 2: Normal stress due to axial load (σ = P A) • Ch 3 : Shearing stress due to torque (T J = τ ρ ) • Ch 4,5 : Normal stress...
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## This note was uploaded on 06/28/2013 for the course MEC 2001 taught by Professor Tansoonhuat during the Spring '10 term at Nanyang Technological University.

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