This preview shows page 1. Sign up to view the full content.
Problem 6.17
A torsion member has and elliptical cross section with major and minor dimensions of 50.0 mm and
30.0 mm, respectively.
The yield stress of the material in the torsion member is Y=400 MPa.
Determine the maximum torque that can be applied to the torsion member based on a factor of safety of
SF=1.85 using the maximum shear stress criterion of failure.
Solution:
h
30 mm
⋅
2
15 mm
⋅
=
:=
b
50 mm
⋅
2
25 mm
⋅
=
:=
Y
400 MPa
⋅
:=
SF
1.85
:=
τ
max
2T
⋅
π
b
⋅
h
2
⋅
=
Eq. 6.47, page 215
All other stresses are zero.
SF
Y
2
τ
max
=
Y
2
⋅
π
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/11/2011 for the course MAE 3040 taught by Professor Mechanicsofsolids during the Spring '10 term at Utah State University.
 Spring '10
 MechanicsofSolids
 Stress, Torsion

Click to edit the document details