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A 60mm diameter shaft is subjected to an axial force of 10kN. The collar of the shaft (A) has a diameter of 100mm, and is supported by a rigid steel support. a. Determine the axial stress in the shaft. b. Determine the bearing stress between the collar (A) and the support. c. Determine the maximum shear stress between the collar and shaft.

A 60mm diameter shaft is subjected to an axial force of 10kN. The collar of the shaft (A) has a diameter of 100mm, and is supported by a rigid steel support. a. Determine the axial stress in the shaft. b. Determine the bearing stress between the collar (A) and the support. c. Determine the maximum shear stress between the collar and shaft.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A 60mm diameter shaft is subjected to an axial force of 10KN. The collar of the shaft (A) has a diameter of 100mm, and is supported by a rigid steel support. a. Determine the axial stress in the shaft. b. Determine the bearing stress between the collar (A) and the support. c. Determine the maximum shear stress between the collar and shaft. 2.5 mm 60 mm 100 mm 10 kN 2.5 mm 15 mm
Transcribed Image Text:A 60mm diameter shaft is subjected to an axial force of 10KN. The collar of the shaft (A) has a diameter of 100mm, and is supported by a rigid steel support. a. Determine the axial stress in the shaft. b. Determine the bearing stress between the collar (A) and the support. c. Determine the maximum shear stress between the collar and shaft. 2.5 mm 60 mm 100 mm 10 kN 2.5 mm 15 mm
Formula Sheet Centre of Gravity and Moment of Inertia ΣΑ ΣΑ EA, ΣΑ I, = I, + Ad² I¸ =E(!.)+ E(Ad²) 1, = E(!.)+ y Axial Stress and Internal Forces ΣΕ, -0 Σ , =0 ΣΜ=0 o = P/ A T =V | A S.F. = o alow /o. allow actual Strain PL 8 = AE · E \ateral V = E = E = L E axial
Transcribed Image Text:Formula Sheet Centre of Gravity and Moment of Inertia ΣΑ ΣΑ EA, ΣΑ I, = I, + Ad² I¸ =E(!.)+ E(Ad²) 1, = E(!.)+ y Axial Stress and Internal Forces ΣΕ, -0 Σ , =0 ΣΜ=0 o = P/ A T =V | A S.F. = o alow /o. allow actual Strain PL 8 = AE · E \ateral V = E = E = L E axial
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