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

See similar textbooks

Concept explainers

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 …

Question

Chapter 7, Problem 20P

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7, Problem 19P

Chapter 7, Problem 21P

Students have asked these similar questions

The solid steel shaft is supported by frictionless bearings in D and E. This system is connected to a motor that provides 5 kW of power to the shaft while rotating at w = 75 rpm. If A and B gears get 2 kW and 3 kW respectively, determine the minimum shaft diameter. w rotation speed and safety shear stress τem = 35MPa are known and given in the table with your names below.

A propeller shaft is required to transmit 45 kW power at 500 rpm. It is a hollow shaft having inside diameter 0.6 times the outer diameter. It is made of plain carbon steel and the permissible shear stress is 84 N/mm^2. Determine the inner and outer diameters of the shaft.

The figure (attached) shows a belt pulley mechanism which is loaded statically. The shaft is made of AISI 1030 steel with the yield strength of 480 MPa. Using distortion energy theory (DET), determine the diameter of the shaft with a factor of safety of 2.

Chapter 7 Solutions

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

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.

Concept explainers

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

Want to see the full answer?

Chapter 7 Solutions