2.1 can also be solved using Mohr’s circle. Fluid Mechanics, Fifth Edition xySolve for (25)/0.866 (a)Ans.σ = − − ≈ − 2289 lbf/ft In like manner, solve for the shear stress on plane AA, using our result for σxy: t,AA AAF L (2000 cos30 289sin 30 )L sin 30 (289cos30 3000sin 30 )L cos30 0 τ∑ = − ° + ° ° + ° + ° ° = AASolve for 938 1515 (b)Ans.τ = − ≈ − 2577 lbf/ft This problem and Prob.Solution: Sum forces normal to and tangential to AA in the element freebody above, with σn(AA) known and σxy unknown: n,AA xy xy F 2500L ( cos30 2000sin30 )Lsin30 ( sin30 3000cos30 )L cos30 0 σ σ ∑ = − ° + ° ° − ° + ° ° = 72 Solutions Manual Compute σxy and the shear stress on plane AA. (b)Ansτ ≈ 2683 lbf/ft 2.2 For the stress field of Fig. Now sum forces normal and tangential to side AA. Solution: Make cut “AA” so that it just hits the bottom right corner of the element. P2.1, let xx y圓0 psfσ σ= xy 500 psfσ = Find the shear and normal stresses on plane AA cutting through at 30°. Pressure Distribution in a Fluid 2.1 For the two-dimensional stress field in Fig.To find the area, we will multiply all of that by the change in height, which is Δx.Download Fluid Mechanics Chapter 2 and more Numerical Methods in Engineering Cheat Sheet in PDF only on Docsity! Chapter 2 Now we can find the area of the trapezoid, which is the base added to double the width of the triangle (a), since the triangle is on both sides of the trapezoid. This is equal to the line from the base to the surface of the water (2 - x) divided by a, which is the width at 2-x. Set up the equation so that you divide the total height (4m) by the maximum width of this section (2m since (8m - 4m)/2, there's 2m on each side). First you need to find an area equation for the triangle section of the trapezoid where the width is increasing from 4m to 8m, keeping in mind that we're only interested in the area that is submerged in water (2m).
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