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Problem_09_25

# Problem_09_25 - Problem 9.25 Given the tirangular cross...

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Problem 9.25 Given the tirangular cross section curved beam in Problem 9.12 is made of steel (E=200 GPa and G=77.5 GPa) Find: the separation of the points of application of the load. Let k=1.5. Solution: E 200 GPa := G 77.5 GPa := k 1.5 := P4 0 k N := A 60 mm 90 mm 2 2700 mm 2 = := R 245 mm 135 mm + 3 75 mm = := P R N V Mx 120 mm R sin θ A m 60 mm 135 mm 135 mm 45 mm ln 135 mm 45 mm 60 mm 38.875 mm = := I 60 mm 90 mm () 3 36 1.215 10 6 × m 4 = := For 0 < z < 120 mm MP z = N0 = VP = For 0 < θ < π M P 0.12 m R sin θ + = N P sin θ = V P cos θ = δ P 2 0 .12 m z M EI P M d d d 2 0 .12 m z kV GA P V d d d + 0 π θ N AE P N d d R d + 0 π θ P V d d R d 0 π θ A m M RA m A P M d d d + 0 π θ 1 EA P MN d d

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Problem_09_25 - Problem 9.25 Given the tirangular cross...

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