Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A tractor mounted off-set type reciprocating mower is driven by the PTO shaft. The maximum inertia force of 3.2 kN occurs along the pitman at 32° crank angle and 27° pitman angle with the horizontal plane. The knives of the cutterbar are riveted to the slider. If each of the allowable tensile and compressive stresses of the slider material is 50 MPa, the minimum cross-sectional area of the slider in mm² is
Options:
(A) 29.05
(B) 33.91
(C) 54.27
(D) 57.02
Official Answer Key:
D
Step-by-Step Solution & Explanation
The slider (attached to the knife, moving horizontally along the cutterbar) experiences the horizontal component of the pitman's inertia force. Resolving the 3.2 kN force (acting along the pitman, at 27° to the horizontal plane) onto the horizontal (slider) direction:
F_axial = 3200×cos(27°) = 3200×0.8910 = 2851.2 N.
Minimum cross-sectional area = F_axial/allowable stress = 2851.2/50×10⁶ = 5.702×10⁻⁵ m² = 57.02 mm². Answer: (D).
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