Explanation The figure below shows the forces and torque acting on the centre of the shaft 3 . Figure-(1) In Figure-(1), the force on bearing D in z direction is F D z , the force on bearing D in y direction is F D y , the force on bearing C in x direction is F C x , the force on bearing C in y direction is F C y , the tangential load on G is W t , the radial load on G is W r and the axial load on G is W a . Write the transverse pitch. P t = P n cos ψ (I) Here, the normal pitch is P n and the helix angle is ψ . Write the transverse pressure angle. ϕ t = tan − 1 ( tan ϕ n cos ψ ) . (II) Here, the normal pressure angle is ϕ n . Write the diameter of the gear 3. d 3 = N 3 P t (III) Here, the number of teeth on gear 3 is N 3 . Write the diameter of the gear 2. d 2 = N 2 P t (IV) Here, the number of teeth on gear 2 is N 2 . Write the tangential load on the gear 2. W t = 33000 H V 2 (V) Here, the power is H and the pitch line velocity of gear 2 is V 2 . Write the pitch line velocity of gear 2. V 2 = π d 2 n 2 Here, the rotational speed of gear 2 is n 2 . Substitute π d 2 n 2 for V 2 in Equation (V). W t = 33000 × 12 H π d 2 n 2 . (VI) Write the input torque on gear 2. T 2 = W t ( d 2 2 ) (VII) Substitute 33000 × 12 H π d 2 n 2 for W t in Equation (VII). r 3 T 2 = 33000 × 12 H π d 2 n 2 ( d 2 2 ) = 63025 H n 2 (VIII) Write the expression for the radial load. W r = W t tan ϕ t (IX) Write the expression for the axial load. W a = W t tan ψ (X) Write the net load at G . W = − W r i ^ − W a j ^ + W t k ^ (XI) Write the position vector of D G . r → D G = l B G i ^ − l B D j ^ (XII) Here, the length between the points B and G is l B G and the length between the points B and D is l B D . Write the position vector of D C . r → D C = ( l B G + l G A ) i ^ (XIII) Here, the length between the points A and G is l A G . Write the moment about D . r → D C × F C + r → D G × W + T = 0 (XIV) Write the force balance equation for the shaft C and D in vector form. F D y j ^ + F D z k ^ + F C x i ^ + F C y j ^ + F C z k ^ + W → = 0 (XV) Conclusion: Substitute 6 in − 1 for P n and 30 ° for ψ in Equation (I). P t = ( 6 in − 1 ) cos ( 30 ° ) = ( 6 in − 1 ) ( 0.866 ) = 5.196 in − 1 Substitute 20 ° for ϕ n and 30 ° for ψ in Equation (II). ϕ t = tan − 1 ( tan 20 ° cos 30 ° ) = tan − 1 ( 0.3639 0.866 ) = tan − 1 ( 0.42 ) = 22.8 ° Substitute 42 for N 3 and 5.196 in − 1 for P t in Equation (III). d 3 = 42 5.196 in − 1 = 8.083 in Substitute 16 for N 2 and 5.196 in − 1 for P t in Equation (IV). d 2 = 16 5.196 in − 1 = 3.079 in Substitute 25 hp for H , 3.079 in for d 2 and 1720 rev / min for n 2 in Equation (VI). W t = ( 33000 ) ( 12 ) ( 25 hp π ( 3.079 in ) ( 1720 rev / min ) ) = 25 hp π ( 3.079 in ) ( 0.00434 rev / min ) = 595.04 lbf ≈ 595 lbf Substitute 595 lbf for W t and 3.079 in for d 2 in Equation (VII). T 2 = 595 lbf × ( 3.079 in 2 ) = 595 lbf × ( 1.5395 in ) = 916.0025 lbf ⋅ in ≈ 916 lbf ⋅ in Substitute 595 lbf for W t and 22.8 ° for ϕ t in Equation (IX). W r = 595 lbf × tan 22.8 ° = 595 lbf × 0.42 = 249.9 lbf ≈ 250 lbf Substitute 595 lbf for W t and 30 ° for ψ in Equation (X). W a = 595 lbf × tan 30 ° = 595 lbf × 0.577 = 343.5234 lbf ≈ 344 lbf Substitute 344 lbf for W a , 250 lbf for W r and 595 lbf for W t in Equation (XI). W → = ( 344 lbf ) i ^ + ( 250 lbf ) j ^ + ( 595 lbf ) k ^ Substitute 3 in for l B G and 4.04 in for l B D in Equation (XII). r → D G = ( 3 in ) i ^ − ( 4.04 in ) j ^ Substitute 3 in for l B G and 3 in for l G A in Equation (XIII)

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ISBN: 9781260058499

Author: BUDYNAS

Publisher: INTER MCG

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Chapter 13, Problem 48P

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The reaction exerted by bearing C on shaft b.

The reaction exerted by bearing D on shaft b.

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Chapter 13, Problem 47P

Chapter 13, Problem 49P

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Chapter 13 Solutions

CONNECT F/SHIGLEY'S MECH.ENGR.DESIGN>I<

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The reaction exerted by bearing C on shaft b . The reaction exerted by bearing D on shaft b .

Solution Summary: The author explains the reaction exerted by bearing C on shaft b and the torque acting on the centre of the shaft.

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The reaction exerted by bearing D on shaft b.

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The reaction exerted by bearing C on shaft b . The reaction exerted by bearing D on shaft b .

Solution Summary: The author explains the reaction exerted by bearing C on shaft b and the torque acting on the centre of the shaft.

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The reaction exerted by bearing C on shaft b. The reaction exerted by bearing D on shaft b.

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Chapter 13 Solutions

The reaction exerted by bearing C on shaft b.

The reaction exerted by bearing D on shaft b.