Explanation The free body diagram of Gear F is as follows. Figure-(1) The free body diagram of the shaft is as follows. Figure-(2) Write the expression for tangential force acting on the shaft. F t = T ( c 2 ) (I) Here, the tangential force on the shaft is F t , torque acting on the shaft is T and the diameter of the shaft is c . Write the expression for normal force acting on the shaft. F n = F t tan ϕ (II) Here, F n is the normal force acting on the shaft and ϕ is the pressure angle. Write the expression for torque at gear C . T c = F t ( b 2 ) (III) Here, torque at gear C is T c . Write the expression for load at sprocket B . P = T c ( a 2 ) (IV) Here, the load at sprocket B is P . Write the expression for moment about A in z - direction. ∑ ( M A ) z = 0 ( e + f + g ) R D y − F n ( f + g ) = 0 R D y = F n ( f + g ) ( e + g + f ) (V) Here, the moment about A in z - direction is ( M A ) z , the reaction at D in y - direction is R D y , the distance between sprocket B and pivot A is e , the distance between sprocket B and gear C is f and the distance between gear C and pivot D is g . Write the expression for moment about A in y - direction. ∑ ( M A ) y = 0 − ( e + f + g ) R D z + F t ( f + g ) − P ( e ) = 0 R D z = F t ( f + g ) − P ( e ) ( e + g + f ) (VI) Here, the moment about A in y - direction is ( M A ) y and the reaction at D in z direction is R D z . Write the expression for forces in y - direction. ∑ F y = 0 R A y + R D y − F n = 0 R A y = F n − R D y (VII) Here, the reaction at A in y - direction is R A y . Write the expression for forces in z - direction. ∑ F z = 0 R A z + R D z − F t + P = 0 R A z = F t − P − R D z (VIII) Here, the reaction at A in z - direction is R A z . From given figure, the maximum bending moment will be either at B or C . Write the expression for bending moment at B . M B = e R A y 2 + R A z 2 (IX) Here, M B is the bending moment at B . Write the expression for bending moment at C . M C = g R D y 2 + R D z 2 (X) Here, M C is the bending moment at C . Write the expression for bending stress at B . σ B = 32 M B π d 3 (XI) Here, the diameter of the shaft is d and the bending stress at B is σ B . Write the expression for shearing stress at B . τ B = 16 T C π d 3 (XII) Here, the shearing stress at B is τ B . Calculate the maximum principal normal stresses in the shaft. σ max = σ B 2 + ( σ B 2 ) 2 + τ B 2 (XIII) Here, the maximum principal normal stress in the shaft is σ max . Calculate the maximum principal shear stresses in the shaft. τ max = ( σ B 2 ) 2 + τ B 2 (XIV) Here, the maximum principal shear stress in the shaft is τ max . Conclusion: Substitute 100 N ⋅ m for T and 125 mm for c in Equation (I). F t = 100 N ⋅ m ( 125 mm 2 ) = 100 N ⋅ m ( 125 mm ( 1 m 1000 mm ) 2 ) = 1600 N Substitute 1600 N for F t and 20 ° for ϕ in Equation (II). F n = ( 1600 N ) ( tan 20 ° ) = 582.352 N ≈ 582.4 N Substitute 1600 N for F t and 250 mm for b in Equation (III). T c = ( 1600 N ) ( 250 mm ) 2 = ( 1600 N ) ( ( 250 mm ) ( 1 m 1000 mm ) ) 2 = 200 N ⋅ m Substitute 200 N ⋅ m for T c and 150 mm for a in Equation (IV). P = 200 N ⋅ m ( 150 mm 2 ) = 200 N ⋅ m ( 150 mm × 1 m 1000 mm 2 ) ≈ 2667 N Substitute 582

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 3, Problem 78P

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The maximum principal normal stresses in the shaft.

The maximum principal shear stresses in the shaft.

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Chapter 3, Problem 77P

Chapter 3, Problem 79P

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6/86 The connecting rod AB of a certain internal-combustion engine weighs 1.2 lb with mass center at G and has a radius of gyration about G of 1.12 in. The piston and piston pin A together weigh 1.80 lb. The engine is running at a constant speed of 3000 rev/min, so that the angular velocity of the crank is 3000(2)/60 = 100л rad/sec. Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle 0 = 90°. (Suggestion: Use the alternative moment relation, Eq. 6/3, with B as the moment center.) Answer A = 347 lb 3" 1.3" B 1.7" PROBLEM 6/86

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

Ch. 3 - 31 to 34 Sketch a free-body diagram of each...Ch. 3 - 31 to 34 Sketch a free-body diagram of each...Ch. 3 - Sketch a free-body diagram of each element in the...Ch. 3 - 3-1 to 3-4 Sketch a free-body diagram of each...Ch. 3 - 35 to 38 For the beam shown, find the reactions at...Ch. 3 - 35 to 38 For the beam shown, find the reactions at...Ch. 3 - 35 to 38 For the beam shown, find the reactions at...Ch. 3 - For the beam shown, find the reactions at the...Ch. 3 - For the beam shown, find the reactions at the...Ch. 3 - Repeat Prob. 36 using singularity functions...

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The maximum principal normal stresses in the shaft. The maximum principal shear stresses in the shaft.

Solution Summary: The author explains the principal normal stresses and shear stresses in the shaft. The free body diagram of Gear F is as follows.

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The maximum principal normal stresses in the shaft. The maximum principal shear stresses in the shaft.

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Chapter 3 Solutions

The maximum principal normal stresses in the shaft.

The maximum principal shear stresses in the shaft.