Fluid Mechanics-Frank M White Solution Ch2

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Fluid Mechanics-Frank M White Solution Ch2에 대한 보고서 자료입니다.

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2.1 For the two-dimensional stress field in Fig. P2.1, let σ xx =150 kPa σ yy =100 kPa σ xy =25 kPa Find the shear and normal stresses on plane AA cutting through at 30°. Solution: Make cut “AA” so that it just hits the bottom right corner of the element. This gives the freebody shown at right. Now sum forces normal and tangential to side AA. Denote side length AA as “L.” ∑Fn,AA =0= σ AAL −(150 sin 30 + 25 cos 30)L sin 30 −(100 cos 30 + 25 sin 30)L cos 30


Fig. P2.1

Solve for
σ AA ≈134 kPa Ans. (a) ∑Ft,AA =0= τ AAL−(150 cos 30−25 sin 30)L sin 30−(25 cos 30−100 sin 30)L cos 30 Solve for τ AA ≈34 kPa Ans. (b)

2.2 For the stress field of Fig. P2.1, change the known data to σ xx = 100 kPa, σ yy = 150 kPa, and σ n(AA) = 125 kPa. Compute σ xy and the shear stress on plane AA. Solution: Sum forces normal to and tangential to AA in the element freebody above, with σ n(AA) known and σ xy unknown: ∑Fn,AA =125L−( σ xycos30°+100sin30°)Lsin30° −( σ xysin30°+150cos30°)Lcos30°=0 Solve for σ xy =(125−25−112.5)/0.866≈ −14.4 kPa Ans. (a) In like manner, solve for the shear stress on plane AA, using our result for σ xy: ∑Ft,AA = τ AAL−(100cos30°+14.4sin30°)Lsin30° +(14.4cos30°+150sin30°)Lsin30°=0
Solve for τ AA =46.9−75.8≈ −28.9 kPa Ans. (b) This problem and Prob. 2.1 can also be solved using Mohr’s circle.

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