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chapter 9 answers

chapter 9 answers - Problem 9.1 The prismatic bar has a...

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Problem 9.1 The prismatic bar has a circular cross section with 50-mm radius and is subjected to 4-kN axial loads. Determine the average normal stress at the plane P . Free Body Diagrams: Solution: The cross-sectional area of the bar is: A = πr 2 = π (0 . 05 m) 2 = 0 . 007854 m 2 The average normal stress is the load distributed across the face of the cross section. σ AV = P/A = (4000 N) / (0 . 007854 m 2 ) ANS: σ AV = 509 kPa Problem 9.2 In Problem 9.1, what is the average shear stress at the plane P ? Solution: Free Body Diagram: We see from the FBD that the shearing force is the only force with any vertical component. Summing vertical forces: Σ F y = 0 = τA = τ [ π (0 . 025 m) 2 ] ANS: τ = 0 Problem 9.3 The prismatic bar has a cross-sectional area A = 30 in 2 and is subjected to axial loads. De- termine the average normal stress (a) at plane P 1 ; (b) at plane P 2 . Free Body Diagrams: Solution: At Plane P 1 , the average normal stress is: ( σ AV ) 1 = P/A = (2000 lb) / (30 in 2 ) ANS: ( σ AV ) 1 = 66 . 7 psi At Plane P 2 , the average normal stress is: ( σ AV ) 2 = P/A = (2000 lb) / (30 in 2 ) ANS: ( σ AV ) 2 = 66 . 7 psi
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Problem 9.4 The prismatic bar has a solid circular cross section with 2-in. radius. Determine the average normal stress (a) at plane P 1 ; (b) at plane P 2 . Free Body Diagram: Solution: The cross-sectional area of the prismatic bar is: A = πr 2 = π (2 in) 2 = 12 . 566 in 2 Note that the loads are NOT the same at P 1 and P 2 . The bar is in tension at P 1 and in compression at P 2 . (a) The average normal stress at P 1 is: ( σ AV ) 1 = P/A = (4000 lb) / (12 . 566 in 2 ) ANS: ( σ AV ) 1 = 318 . 3 psi (b) The average normal stress at P 2 is: ( σ AV ) 2 = P/A = ( 8000 lb) / (12 . 566 in 2 ) ANS: ( σ AV ) 2 = 636 . 6 psi Problem 9.5 A prismatic bar with cross-sectional area A is subjected to axial loads. Determine the average normal and shear stresses at the plane P if A = 0.02 m 2 , P = 4 kN , and θ = 25 . Free Body Diagram: Solution: The cross-sectional area of this section is larger than 0 . 02 m 2 because the bar is cut at an angle. The actual area of the angled cut is: A = (0 . 02 m 2 ) / (cos 25 ) = 0 . 02206 m 2 Summing horizontal forces on the bar: Σ F x = 0 = 4000 N+[ σ AV (0 . 02206 m 2 )](cos 25 ) τ AV (0 . 02206 m 2 )(sin 25 ) [1] τ AV = 429 , 048 N + 2 . 145 σ AV Summing vertical forces on the bar: Σ y = 0 = σ AV (0 . 02206 m 2 )(sin 25 )+ τ AV (0 . 02206 m 2 )(cos 25 ) [2] τ AV = 0 . 4663 σ AV Solving Equations [1] and [2] together: ANS: σ AV = 164 . 3 kPa ANS: τ AV = 76 . 6 kPa Note: The ( ) sign indicates the wrong direction assumed for the FBD.
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Problem 9.6 Suppose that the prismatic bar shown in Problem 9.5 has cross-sectional area A = 0 . 024 m 2 . If the angle θ = 35 and the average normal stress on the plane P is σ AV = 200 kPa , what are τ AV and the axial force P ? Free Body Diagram: Solution: The cross-section area at plane P is: A = (0 . 024 m 2 ) / (cos 35 ) = 0 . 0293 m 2 Summing vertical forces on the FBD: Σ F y = 0 = τ AVG A (sin 55 ) + σ AVG A (sin 35 ) τ AVG = 0 . 7 σ AVG = (0 . 7)(200 kPa) ANS: τ AVG = 140 kPa ( ) indicates opposite direction.
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