Explanation The following figure shows the free body diagram of the gear assembly. Figure-(1) Write the expression for equilibrium condition for the moments about x- axis. ∑ M x = 0 F 1 cos θ × r 1 + F B cos ϕ × r 2 = 0 (I) Here, the force on the gear A is F 1 , the angle made by force F 1 is θ , the radius of gear A is r 1 , the force at gear B is F B , the angle made by force F B is ϕ , and the radius of gear B is r 2 . Write the expression for moment equilibrium about point o . ∑ M o = 0 F 1 cos θ × l o A − F B sin θ × l o B + R c y × l o C = 0 (II) Here, the length from point o to A is l o A , the length from point o to B is l o B , the length from point o to C is l o C , and the reaction is R c y . Write the expression for equilibrium of forces along y- axis. ∑ F y = 0 R o y + F 1 cos θ + R c y − F B sin ϕ = 0 (III) Write the expression for moment about point A . ( M A ) 1 = R o y × l o A (IV) Write the expression for moment about point C . ( M C ) 1 = F B sin ϕ × l C B (V) Write the expression for moment equilibrium about point O . ∑ M o = 0 F 1 sin θ × l o A − R c z × l o C − F B cos ϕ × l o B = 0 (VI) Write the expression for equilibrium of force in the z- direction. ∑ F z = 0 R o z − F 1 sin θ + R c z + F B cos ϕ = 0 (VII) Write the expression for moment at point A . ( M A ) 2 = R o z × l o A (VIII) Write the expression for moment at point C . ( M C ) 2 = F B × l C B (IX) Write the expression for net moment. M A = ( M A ) 1 2 + ( M A ) 2 2 (X) Write the expression for net moment at point C . M C = ( M C ) 1 2 + ( M C ) 2 2 (XI) Write the expression for torque transmitted through the shaft from A to B . T = F 1 cos θ × r 1 (XII) Write expression for moment. M x y = ( − 208.5 lbf ) x + ( 281.9 lbf ) ( x − 16 in ) + ( 183.1 lbf ) ( x − 30 in ) = ( 256.5 lbf ) x − ( 10003 lbf ⋅ in ) Rewrite equation moment Equation. M x y = ( 256.5 lbf ) x − ( 10003 lbf ⋅ in ) (XIII) Write expression for equation deflection. E I d 2 y d x 2 = M x y (XIV) Here, modulus of elasticity is E and moment of inertia is I . Write expression for equation deflection. y = 1 E I [ ( 256.5 lbf ) x 3 6 − ( 10003 lbf ⋅ in ) x 2 2 + x C 1 + C 2 ] (XV). Substitute 111570 lbf ⋅ in 2 for C 1 , 0 for C 2 in Equation (XV). E I y = [ ( 256.5 lbf ) x 3 6 − ( 10003 lbf ⋅ in ) x 2 2 + x ( 111570 lbf ⋅ in 2 ) ] (XVI) Write expression for equation slope. E I y ′ = [ ( 256.5 lbf ) x 2 2 − ( 10003 lbf ⋅ in ) x + ( 111570 lbf ⋅ in 2 ) ] (XVII) Write expression for moment of inertia. I 1 = π d 1 4 64 (XVIII) Write expression for moment of inertia. I 2 = π d 2 4 64 (XIX) Write expression for slope at A . y ′ B = − 73096 E I 1 (XX) Write expression for slope cantilever part BC . y ′ B / A = F x ( x − 2 l ) 2 E I 2 (XXI) Write expression for slope at C . y ′ B = y ′ B + y ′ B / A (XXII) Write expression for moment. M x y = ( − 259.3 lbf ) x + ( 102.6 lbf ) ( x − 16 in ) + ( 861.5 lbf ) ( x − 30 in ) Rewrite the moment Equation. M x y = ( 704.8 lbf ) x − ( 27486 lbf ) (XXIII) Write expression for deflection. E I d 2 y d x 2 = M x y (XXIV) Write the expression for deflection. y = 1 E I [ ( 704.8 lbf ) x 3 6 − ( 27486 lbf ) x 2 2 + x C 1 + C 2 ] (XXV) Substitute 306570 lbf ⋅ in 2 111570 lbf ⋅ in 3 for C 1 , 0 for C 2 in Equation (XXV). y = 1 E I [ ( 704.8 lbf ) x 3 6 − ( 27486 lbf ⋅ in ) x 2 2 + x ( 306570 lbf ⋅ in 2 ) ] (XXVI) Write expression for slope. E I y ′ = [ ( 704.8 lbf ) x 2 2 − ( 27486 lbf ⋅ in ) x + ( 306570 lbf ⋅ in 2 ) ] (XXVII) Write expression for slope at B . y ′ ′ B = − 200850 E I 1 (XXVIII) Write expression for slope cantilever part BC . y ′ ′ B / A = F x ( x − 2 l ) 2 E I 2 (XXIX) Write expression for slope at C . y ′ ′ B = y ′ ′ A + y ′ ′ B / A (XXX) Write expression for resultant deflection at B . θ B = ( y ′ B ) 2 + ( y ′ ′ B ) 2 (XXXI) Write expression for the new diameter for bearing O . d n e w d o l d = [ n d ( d y d x ) o l d θ a l l ] 1 4 (XXXII) Here, diameter of new shaft is d n e w , diameter of old shaft is d o l d , old slope is ( d y d x ) o l d and allowable slope is θ a l l . Conclusion: Substitute − 300 lbf for F , 20 ° for θ , 10 in for r 1 , 20 ° for ϕ and 4 in for r 2 in Equation (I). − 300 ( cos 20 ° ) ( 10 in ) + F B ( cos 20 ° ) ( 4 in ) = 0 F B ( 3.75 in ) = 2819.07 lbf ⋅ in F B = 2819.07 lbf ⋅ in 3.75 in F B = 751.752 lbf Substitute 300 lbf for F 1 , 20 ° for cos θ , 16 in for l o A , 751.752 lbf for F B , 20 ° for ϕ , 39 in for l o B and 30 in for l o C in Equation (II). ( 300 lbf ) ( cos 20 ° ) ( 16 in ) − 750 lbf ( sin 20 ° ) ( 39 in ) + R C y ( 30 in ) = 0 ( 300 lbf ) ( 0.94 ) ( 16 in ) − ( 750 lbf ) ( 0.342 ) ( 39 in ) + R C y ( 30 in ) = 0 R C y = 5493 lbf ⋅ in 30 in R C y = 183.1 lbf Substitute 300 lbf for F 1 , 20 ° for θ , 183.1 lbf for R C y , 750 lbf for F B and 20 ° for ϕ in Equation (III). R o y + ( 300 lbf ) ( cos 20 ° ) + ( 183.1 lbf ) − ( 750 lbf ) ( sin 20 ) = 0 R o y + ( 300 lbf ) ( 0.94 ) + ( 183.1 lbf ) − ( 750 lbf ) ( 0.342 ) = 0 R o y + ( 465 lb ) f − ( 256.51 lbf ) = 0 R o y = − 208.49 lbf Substitute − 208.49 lbf for R o y and 16 in for l o A in Equation (IV). ( M A ) 1 = ( − 208.49 lbf ) ( 16 in ) = 3335.84 lbf ⋅ in Substitute 750 lbf for F B , 20 ° for ϕ , 9 in for l c B in Equation (V). ( M c ) 1 = ( 750 lbf ) sin ( 20 ° ) × ( 9 in ) = ( 256.51 lbf ) × ( 9 in ) = 2308.59 lbf ⋅ in Substitute 300 lbf for F 1 , 20 ° for θ , 16 in for l o A , 30 in for l o C , 750 lbf for F B , 20 ° for ϕ and 39 in for l o B in Equation (VI). ( 300 lbf ) ( sin 20 ° ) ( 16 in ) − R C z ( 30 in ) − ( 750 lbf ) ( cos 20 ° ) ( 39 in ) = 0 ( 300 lbf ) × 5.4723 in − R C z ( 30 in ) − ( 750 lbf ) ( 36.64 in ) = 0 R C z = 25838.31 lbf ⋅ in 30 in R C z = 861.27 lbf Substitute 300 lbf for F 1 , 20 ° for θ , 861.27 lbf for R C z , 750 lbf for F B and 20 ° for ϕ in Equation (VII). R o z − ( 300 lbf ) ( sin 20 ) − ( 861.5 lbf ) + ( 750 lbf ) ( cos 20 ) = 0 R o z − 102.60 lbf − 861.5 lbf + 704.76 lbf = 0 R o z = 861.5 lbf + 102.60 lbf − 704.76 lbf R o z = 259.34 lbf Substitute 259.34 lbf for R o z and 16 in for l o A in Equation (VIII). ( M A ) 2 = ( 259.34 lbf ) ( 16 in ) = 4149.44 lbf ⋅ in Substitute 750 lbf for F B , 9 in for l c B in Equation (IX). ( M C ) 2 = ( 750 lbf ) ( 9 in ) = 6750 lbf ⋅ in Substitute 3335.84 lbf ⋅ in for ( M A ) 1 and 4149.44 lbf ⋅ in for ( M A ) 2 in Equation (X). M A = ( 3335.84 lbf ⋅ in ) 2 + ( 4149.44 lbf ⋅ in ) 2 = 11127828.51 ( lbf ⋅ in ) 2 + 17217852.31 ( lbf ⋅ in ) 2 = 28345680.82 ( lbf ⋅ in ) 2 = 5324.06 lbf ⋅ in Substitute 2308.59 lbf ⋅ in for ( M C ) 1 and 6750 lbf ⋅ in for ( M C ) 2 in Equation (XI)

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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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 7, Problem 20P

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Weather the defection and slop is within the limit or not.

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Chapter 7, Problem 19P

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

Ch. 7 - A shaft is loaded in bending and torsion such that...Ch. 7 - The section of shaft shown in the figure is to be...Ch. 7 - The rotating solid steel shaft is simply supported...Ch. 7 - A geared industrial roll shown in the figure is...Ch. 7 - Design a shaft for the situation of the industrial...Ch. 7 - The figure shows a proposed design for the...Ch. 7 - For the problem specified in the table, build upon...Ch. 7 - Prob. 8P Ch. 7 - For the problem specified in the table, build upon...Ch. 7 - For the problem specified in the table, build upon...

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Weather the defection and slop is within the limit or not.

Solution Summary: The author shows the free body diagram of the gear assembly. Write the expression for equilibrium condition for the moments about x-axis.

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Weather the defection and slop is within the limit or not.

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