Tutorial Three solution 2011

# Tutorial Three solution 2011 - Solution for Tutorial Three...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution for Tutorial Three MECH2100, 2010 1 The University of Queensland School of Mechanical & Mining Engineering MECH2100 - Machine Element Design Solution for Tutorial Three 2011 Q1: Answer: C & D Q2 & Q3: 4 2 2 7 4 2 D Cross sectional area of the rod A= 30 mm , I = 6.36172(10 ) m , J=2I. 64 F 12000 Axial stress: = 4.244132699 MPa (for all points) A (30) Bending stress: axial bending and Mc I π π σ π σ- = = = = 7 12000(0.025)(0.03) 14.1471175 MPa 6.36172(10 ) (tension for points C & D, compression for points A & F) 120(0.03) Torsion stress: 2.8294 MP 2( ) torsion Tc J I τ- = = = = = a (for all points) Please note all points experience biaxial stresses. For points C & D: y 4.244 14.1471175 18.39125 MPa 2.829 MPa x axial bending xy σ σ σ σ τ = + = + = = = 2 1 2 2 1 2 1 2 max & [ ] 14.85447 15.12146 MPa 2 2 18.81658 MPa [answer] 0.425329 MPa 9.620954 MPa [answer] x y x y xy σ σ σ σ σ σ τ σ σ τ +- = ± + = ± ∴ = = = Q4: In the maximum normal stress theory, failure is expected whenever the maximum tensile...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

Tutorial Three solution 2011 - Solution for Tutorial Three...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online