Explanation Write the expression for the moment of inertia of the shaft. I = π d 4 64 (I) Here, the diameter of the shaft is d and the moment of inertia is I . Draw the free body diagram of the beam. Figure-(1) The free body diagram of the beam in the direction of y- axis is shown in Figure-(1). Write the expression for the deflection along y -axis for beam 6 using Table A-9. y A = F 1 b 1 x 6 E I l ( x 2 + b 1 2 − l 2 ) (II) Here, the deflection of the beam is y A , the net force at point A is F 1 , the location of point A from right end is b 1 , the elongation distance of any given point from left end is x , the total length of the beam is l , the Young modulus of the material is E and moment of inertia of the beam is I . The free body diagram of the beam in the direction of z- axis is shown below. Figure (2) Write the expression for the deflection along z -axis for beam 6 using Table A-9. z A = F 2 b 2 x 6 E I l ( x 2 + b 2 2 − l 2 ) (III) Here, the deflection of the beam is z A , the net force at point B is F 2 and the location of point B from right end is b 2 . Write the expression for net deflection at point A . δ = y A 2 + z A 2 (IV) Here, the net deflection at point A is δ . Write the expression for slope along z- axis at point A . ( θ A ) z = ( d y d x ) = { d d x [ F 1 b 1 x 6 E I l ( x 2 + b 1 2 − l 2 ) ] } = F 1 b 1 6 E I l ( 3 x 2 + b 1 2 − l 2 ) (V) Here, the slope along z- axis at point is ( θ A ) z . Write expression for the slope along y- axis at point A . ( θ A ) y = ( d z d x ) = { d d x [ F 2 b 2 x 6 E I l ( x 2 + b 2 2 − l 2 ) ] } = F 2 b 2 6 E I l ( 3 x 2 + b 2 2 − l 2 ) (VI) Here, the slope along z- axis at point is ( θ A ) y . Write the expression for the net slope at point A . Θ A = ( θ A ) z 2 + ( θ A ) y 2 (VII) Here, the net slope is Θ A . Write the magnitude of net force at point A along y -axis. F 1 = T 1 + T 2 (VIII) Here, the net force at point A along y -axis is F 1 , the tension at point A is T 1 and the tension at point A is T 2 Write the magnitude of net force at point B along z -axis. F 2 = T 1 B + T 2 B . (IX) Here, the net force at point B along y -axis is F 2 , the tension at point B is T 1 B and the tension at point B is T 2 B . Conclusion: Substitute 1 in for d in the equation (I). I = π ( 1 mm ) 4 64 = 3.1415 64 mm 4 = 0.04909 mm 4 Substitute − 300 lbf for T 1 and − 50 lbf for T 2 in Equation (VIII). F 1 = ( − 300 lbf ) + ( − 50 lbf ) = − 350 lbf Substitute − 392.16 lbf for T 1 B and − 58.82 lbf for T 2 B in Equation (IX). F 2 = ( − 392.16 lbf ) + ( − 58.82 lbf ) = − 450.98 lbf Substitute − 350 lbf for F 1 , 8 in for x , 14 in for b 1 , 22 in for l , 30 × 10 6 psi for E and 0.04909 in 4 for I in the Equation (II). y A = ( − 350 lbf ) ( 14 in ) ( 8 in ) 6 ( 30 × 10 6 psi ) ( 0.04909 in 4 ) ( 22 in ) ( ( 8 in ) 2 + ( 14 in ) 2 − ( 22 in ) 2 ) = { ( − 2.016 × 10 − 4 ) ( − 224 ) lbf / psi ⋅ in } ( 1 psi 1 lbf / in 2 ) = 0

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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Chapter 4, Problem 25P

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The deflection of the shaft at point A.

The slope of the shaft at point A.

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Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 4 - The figure shows a torsion bar OA fixed at O,...Ch. 4 - For Prob. 41, if the simple support at point A...Ch. 4 - A torsion-bar spring consists of a prismatic bar,...Ch. 4 - An engineer is forced by geometric considerations...Ch. 4 - A bar in tension has a circular cross section and...Ch. 4 - Prob. 6P Ch. 4 - Prob. 7P Ch. 4 - Derive the equations given for beam 2 in Table A9...Ch. 4 - Derive the equations given for beam 5 in Table A9...Ch. 4 - The figure shows a cantilever consisting of steel...

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The deflection of the shaft at point A . The slope of the shaft at point A .

Solution Summary: The author explains the deflection of the shaft at point A and the moment of inertia for the beam.

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The deflection of the shaft at point A.

The slope of the shaft at point A.

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The deflection of the shaft at point A . The slope of the shaft at point A .

Solution Summary: The author explains the deflection of the shaft at point A and the moment of inertia for the beam.

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The deflection of the shaft at point A. The slope of the shaft at point A.

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Chapter 4 Solutions

The deflection of the shaft at point A.

The slope of the shaft at point A.