Explanation Given information: The diameter of the shaft ( d ) is 20 mm . The shaft is made of A-36 steel. The modulus of elasticity of the shaft ( E ) is 200 GPa The moment of inertia of the shaft is I The load act at first pulley of the shaft ( P 1 ) is 800 N . The load act at second pulley of the shaft ( P 2 ) is 350 N . The length of the shaft ( L ) is 1 , 000 mm . Calculation: Show the free body diagram of the shaft as in Figure 1. Refer to Figure 1. Calculate the vertical forces by applying the Equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 R A + R B − 800 − 350 = 0 R A + R B − 1 , 150 = 0 R A + R B = 1 , 150 (1) Calculate the moment about point A by applying the Equation of equilibrium: Sum of moments about point A is equal to 0. ∑ M A = 0 ( R B × 800 ) − ( 350 × 300 ) + ( 800 × 200 ) = 0 800 R A + 55 × 10 3 = 0 800 R A = − 55 × 10 3 R B = − 55 × 10 3 800 R B = − 68.75 N R B = 68.75 N ↓ Substitute 68.75 N for R B in Equation (1). R A − 68.75 = 1 , 150 R A = 1 , 150 + 68.75 R A = 1 , 218.75 N Show the moment area diagram of the shaft as in Figure 2. Refer to Figure 2. Show the bending monument calculation follows: Calculate the bending moment at D : M D = − ( 68.75 × 1 ) − ( 350 × 0.5 ) + ( 1218.75 × 0.2 ) = 0 Calculate the bending moment at A : M A = − ( 68.75 × 0.8 ) − ( 350 × 0.3 ) + ( 1218.75 × 0 ) = − 160 N ⋅ m Calculate the bending moment at C : M C = − ( 68.75 × 0.5 ) − ( 350 × 0 ) = 34.375 N ⋅ m Show the deflected shape of the shaft as in Figure 3. Refer to Figure 2. Calculate the vertical distance between the tangent at a point B and A ( t B / A ) : t B / A = { [ 1 2 × ( − 34.375 E I ) × 0.5 × 0.3333 ] + [ 1 2 × ( − 125.625 E I ) × 0.3 × 0.7 ] + [ ( − 34.375 E I ) × 0.3 × 0.65 ] } = − 22.75833 E I N ⋅ m 3 Calculate the vertical distance between the tangent at a point A and B ( t A / B ) : t A / B = { [ 1 2 × ( − 34

10th Edition

ISBN: 9780134583228

Author: HIBBELER

Publisher: PEARSON

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Design of Shafts and Shaft Components

The shaft is basically defined as the machine element that rotates, is in circular cross-section, it is majorly used to transfer the power from one part of the system to another, or from a power generating machine to a power absorbing machine. The shaft …

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Chapter 12.4, Problem 12.62P

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The slope of the shaft at the journal bearings at A and B (θA and θB).

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Chapter 12.4, Problem 12.61P

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The slope of the shaft at the journal bearings at A and B ( θ A and θ B ) .

Solution Summary: The author shows the free body diagram of the shaft as in Figure 1. Calculate the vertical forces by applying the Equation of equilibrium.

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The slope of the shaft at the journal bearings at A and B (θA and θB).

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Chapter 12 Solutions