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# chapter 11 answers - Problem 11.1 A cube of material is...

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Problem 11.1 A cube of material is subjected to a pure shear stress τ = 9 MPa . The angle β is measured ad determined to be 89 . 98 . What is the shear modulus G of the material? Diagram: Solution: Converting the shear strain angle into radians: γ = (90 89 . 98 ) 180 π = 3 . 49 × 10 4 radians Using the definition of the shear modulus: G = τ γ = 9 × 10 6 N / m 2 3 . 49 × 10 4 ANS: G = 25 . 8 GPa Problem 11.2 If the cube in Problem 11.1 consists of material with shear modulus G = 4 . 6 × 10 6 psi and the shear stress τ = 8000 psi , what is the angle β in degrees? Free Body Diagram: Solution: The shear strain will be: γ = τ G = 8 , 000 lb / in 2 4 . 6 × 10 6 lb / in 2 = 1 . 739 × 10 3 radians = 0 . 0996 The angle β is: β = 90 γ = 90 0 . 0996 ANS: β = 89 . 9 Problem 11.3 If the cube in Problem 11.1 consists of aluminum alloy that will safely support a pure stress of 270 MPa and G = 26 . 3 GPa , what is the largest shear strain to which the cube can safely be subjected? Solution: The shear strain will be: γ = τ G = 270 × 10 6 N / m 2 26 . 3 × 10 9 N / m 2 = 0 . 010266 ANS: γ = 0 . 0103

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Problem 11.4 The cube of material is subjected to a pure shear stress τ = 12 MPa . What are the normal stress and the magnitude of the shear stress on the plane P ? Free Body Diagram: Solution: Summing vertical forces on the free body diagram: Σ F Y = 0 = (12 × 10 6 N / m 2 ) A (cos 30 )+ τ P A (cos 30 )+ σA (sin 30 ) [1] 0 . 866 τ P + 0 . 5 σ P = 10 . 39 × 10 6 N / m 2 Summing horizontal forces on the free body diagram: Σ F X = 0 = (12 × 10 6 N / m 2 ) A (sin 30 ) τ P A (sin 30 )+ σ P A (cos 30 ) [2] 0 . 5 τ P + 0 . 866 σ P = 6 × 10 6 N / m 2 Solving Equations [1] and [2] together: ANS: τ P = 6 MPa ANS: σ P = 10 . 4 MPa Problem 11.5 In Problem 11.4, what are the magni- tudes of the maximum tensile, compressive, and shear stresses to which the material is subjected? Free Body Diagram: Solution: Summing forces in the x -direction on the element: Σ F x = 0 = σ A A (cos θ ) (12 × 10 6 N / m 2 ) A (sin θ ) τA (sin θ )[1] Solving Equation [1] for σ A : τ = σ A (cos θ ) (12 × 10 6 N / m 2 )(sin θ ) sin θ = σ A (cot θ ) 12 × 10 6 N / m 2 We see that σ A is maximum when cot θ is minimum ( θ = 0 ), or: ANS: σ MAX = τ = 12 MPa Summing forces in the y -direction on the element: Σ F y = 0 = (12 × 10 6 N / m 2 )( A )(cos θ )+ τA (cos θ )+ σA (sin θ ) [2] Solving Equation [2] for τ τ = 12 MPa σ (tan θ ) We see that τ is maximum when tan θ is minimum ( θ = 0) , or: ANS: τ MAX = τ = 12 MPa
Problem 11.6 The cube of material shown in Prob- lem 11.4 is subjected to a pure shear stress τ . If the normal stress on the plane P is 14 MPa, what is τ ? Free Body Diagram: Solution: Summing forces in the x -direction on the element: Σ F x = 0 = (14 × 10 6 N / m 2 ) A (cos 30 ) τ A (sin 30 ) τA (sin 30 ) τ = 24 . 25 × 10 6 N / m 2 τ [1] Summing forces in the y -direction on the element: Σ F y = 0 = (14 × 10 6 N / m 2 ) A (sin 30 )+ τ A (cos 30 ) τA (cos 30 ) τ = 8 . 083 × 10 6 N / m 2 + τ [2] Solving Equations [1] and [2] together: 24 . 25 × 10 6 N / m 2 τ = 8 . 083 × 10 6 N / m 2 + τ ANS: τ = 16 . 17 MPa Problem 11.7 The cube of material shown in Prob- lem 11.4 is subjected to a pure shear stress τ . The shear modulus of the material is G = 28 GPa . If the normal

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chapter 11 answers - Problem 11.1 A cube of material is...

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