Problem 2.40

# Problem 2.40 - = The upper plate is a minus y surface....

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Problem 2.40 [Difficulty: 2] Given: Velocity distribution between flat plates Find: Shear stress on upper plate; Sketch stress distribution Solution: Basic equation τ yx μ du dy = du dy d dy u max 1 2y h 2 = u max 4 h 2 2 y = 8u max y h 2 = τ yx 8 μ u max y h 2 = At the upper surface y h 2 = and h 0.1 mm = u max 0.1 m s = μ 1.14 10 3 × Ns m 2 = (Table A.8) Hence τ yx 8 1.14 × 10 3 × Ns m 2 0.1 × m s 0.1 2 × mm 1m 1000 mm × 1 0.1 mm 1000 mm 1m × 2 × = τ yx 4.56 N m 2
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Unformatted text preview: = The upper plate is a minus y surface. Since yx < 0, the shear stress on the upper plate must act in the plus x direction. The shear stress varies linearly with y yx y ( ) 8 u max h 2 y = 5 4 3 2 1 1 2 3 4 5 0.05 0.04 0.03 0.02 0.01 0.01 0.02 0.03 0.04 0.05 Shear Stress (Pa) y (mm)...
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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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