Explanation Given information: The torque is applied alone to a solid bar with a circular cross section. It cause the distortion energy as the combination of an applied bending moment M and torque T . Calculation: Consider that the distortion energy is the torque applied to the solid bar. The shear stress applied to the solid bar. τ = T e c J Here, T e is the equivalent torque, c is the radius of the bar, and J is the polar moment of inertia. Find the principal stresses in complex system as shown below: σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 (1) Here, σ x is the normal stress in x direction, σ y is the normal stress in y direction, and τ x y is the shear stress. Substitute 0 for σ x , 0 for σ y , and τ for τ x y in Equation (1). σ 1 , 2 = 0 ± 0 + τ 2 = ± τ σ 1 = τ σ 2 = − τ Apply the maximum distortion theory for plane stress as shown below. σ 1 2 − σ 1 σ 2 + σ 2 2 = σ Y 2 Here, σ 1 and σ 2 are the principal stresses and σ Y is the principal yield stress. Apply the strain-energy density as shown below: u d = 1 + ν 3 E ( σ 1 2 − σ 1 σ 2 + σ 2 2 ) (2) Here, E is the modulus of elasticity and ν is the Poisson’s ratio. Substitute τ for σ 1 and − τ for σ 2 in Equation (2). ( u d ) 1 = 1 + ν 3 E ( τ 2 − τ ( − τ ) + ( − τ ) 2 ) = 1 + ν 3 E ( 3 τ 2 ) = 1 + ν E τ 2 Substitute T e c J for τ . ( u d ) 1 = 1 + ν E ( T e c J ) 2 = 1 + ν E ( T e 2 c 2 J 2 ) (3) Consider that the distortion energy is the combination of the bending moment and torque applied to the solid bar. Consider the normal stress and shear stress applied to the solid bar. σ = M c I Here, M is the bending moment and I is the moment of inertia. τ = T c J Here, T is the torque. Substitute σ for σ x , 0 for σ y , and τ for τ x y in Equation (1). σ 1 , 2 = σ + 0 2 ± ( σ − 0 2 ) 2 + τ 2 = σ 2 ± σ 2 4 + τ 2 σ 1 = σ 2 + σ 2 4 + τ 2 σ 2 = σ 2 − σ 2 4 + τ 2 Consider a = σ 2 and b = σ 2 4 + τ 2 . Hence, σ 1 = a + b and σ 2 = a − b . Substitute ( a + b ) for σ 1 and ( a − b ) for σ 2 in Equation (2)

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

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Chapter 10.7, Problem 10.65P

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To derive: An expression for an equivalent torque Te.

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The tubular rotor shaft of a wind turbine shown in Figure 3(a) is of 50 mm and 40 mm outer and inner diameters respectively, and it is expected to be subjected to the following extreme loading conditions which act simultaneously: A torque of 2.5kNm about the axis. A bending moment of 3.6kNm acting on the vertical plane containing the shaft axis. 3.6 kNm 2-5 kNm Figure 3(a) Calculate the following: i, The polar second moment of area, J The shear stress at point A on the outer surface of the shaft (where the vertical plane containing the shaft axis intersects the outer periphery of the shaft). ii. The second moment of area I about a diametric axis (i.e., an axis across a diameter of the cross-section). ii.

3. (a) The tubular rotor shaft of a wind turbine shown in Figure 3(a) is of 50 mm and 40 mm outer and inner diameters respectively, and it is expected to be subjected to the following extreme loading conditions which act simultaneously: A torque of 2.5kNm about the axis. • A bending moment of 3.6kNm acting on the vertical plane containing the shaft axis. 3.6 ENm 25m Figure 3(a) Calculate the following: i. The polar second moment of area, J ii. The shear stress at point A on the outer surface of the shaft (where the vertical plane containing the shaft axis intersects the outer periphery of the shaft). i. The second moment of area I about a diametric axis (i.e., an axis across a diameter of the cross-section). iv. The bending stress at point A. (b) The spherical LPG storage tank shown in Figure 3(b) has the following design specifications: Tank diameter of 22 m, shell plate thickness of 24 mm, yield stress of steel plates 550 MPa, a factor of safety of 2. Calculate the design pressure,…

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Mechanics of Materials (10th Edition)

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To derive: An expression for an equivalent torque T e .

Solution Summary: The author explains that the distortion energy is the torque applied to the solid bar with a circular cross section.

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