Explanation Given information: The force F = 12 lb The angle of bend shaft is θ = 45 ° Calculation: Show the free body diagram of bent shaft as shown in Figure 1. V x is shear in x direction, M z is moment in z direction, N y is normal force in y direction, and T is the torque. Refer Figure 1, Determine the internal loadings as follows: Refer Figure 1, Apply equilibrium conditions the force in x direction. ∑ F x = 0 V x − 12 cos 45 ° = 0 V x − 8.4853 = 0 V x = 8.485 lb Apply equilibrium conditions the force in x direction. ∑ F y = 0 N y − 12 sin 45 ° = 0 N y − 8.485 = 0 N y = 8.485 lb Take a moment about x axis. ∑ M x = 0 M x − 12 sin 45 ° ( 3 ) = 0 M x − 25.4558 = 0 M x = 25.456 lb ⋅ in Take a moment about y axis. ∑ M y = 0 − T y + 12 cos 45 ° ( 3 ) = 0 − T y + 25.456 = 0 T y = 25.46 lb ⋅ in Take a moment about z axis. ∑ M z = 0 M z − 12 cos 45 ° ( 8 ) = 0 M z − 67.882 = 0 M z = 67.882 lb ⋅ in Calculate the moment of inertia ( I ) of the bend shaft as follows: I = 1 4 π r 4 (1) Here, r is the radius of the bent shaft. Substitute 0.625 in. for r in Equation (1). I = 1 4 π ( 0.625 ) 4 = 0.1198 in 4 Calculate the Area ( A ) of the bend shaft as follows: A = π r 2 (2) Substitute 0.625 in. for r in Equation (2). A = π ( 0.625 ) 2 = 1.2272 in 4 Calculate the polar moment of inertia ( J ) of the bend shaft as follows: J = 1 2 π ( r 4 ) (3) Substitute 0.625 in. for r in Equation (3). J = 1 2 π ( 0.625 4 ) = 1 2 π ( 0.1525 ) = 0.2397 in 4 Calculate the first moment area at point D along y axis. ( Q D ) y = ( 4 r 3 π ) 1 2 π ( r 2 ) (4) Substitute 0.625 in. for r in Equation (4). ( Q D ) y = ( 4 ( 0.625 ) 3 π ) 1 2 π ( 0.625 2 ) = 0.26525 × 0.61359 = 0.1628 in 3 Calculate the normal stress ( σ D ) at point D along y axis as follows: ( σ D ) y = N z A − M x z I (5) Here, N z is normal force in z direction, A is the cross sectional area, I is the moment of inertia, M x moment about x direction, and z is the centroid distance. Substitute 8.485 lb for N z , 1.2272 in 4 for A , 0.1198 in 4 for I , 25.456 lb ⋅ in for M x , and 0.625 in. for x in Equation (5). ( σ D ) y = 8.485 1.2272 − 25.456 × 0.625 0.1198 = 6.8921 − 132.80 = − 125.9 = − 126 psi Express to find the shear stress at point ( τ D ) y x . ( τ D ) y x = ( τ D ) V − ( τ D ) twist (6) Find the transverse shear ( τ D ) V as follows: ( τ D ) V = V x ( Q D ) z I t (7) Here, V x is shear force in x direction, I is the moment of inertia, t is the thickness, and ( Q D ) z is the first moment area at point D along z axis. Substitute 8.485 lb for V x , 0.1628 in 3 for ( Q D ) z , 1

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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Chapter 8.2, Problem 8.49P

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The stress components (σD) at point D of bent shaft.

The stress components (σE) at point E of bent shaft.

The shear stress (τD) at point D of bent shaft.

The shear stress (τE) at point E of bent shaft.

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Chapter 8.2, Problem 8.48P

Chapter 8.2, Problem 8.50P

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The stress components ( σ D ) at point D of bent shaft. The stress components ( σ E ) at point E of bent shaft. The shear stress ( τ D ) at point D of bent shaft. The shear stress ( τ E ) at point E of bent shaft.

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The stress components (σD) at point D of bent shaft. The stress components (σE) at point E of bent shaft. The shear stress (τD) at point D of bent shaft. The shear stress (τE) at point E of bent shaft.

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Chapter 8 Solutions

The stress components (σD) at point D of bent shaft.

The stress components (σE) at point E of bent shaft.

The shear stress (τD) at point D of bent shaft.

The shear stress (τE) at point E of bent shaft.