File Lecture9_Slides.pdf - Mechanics of Solids Lecture 9 by...

Info iconThis preview shows pages 1–12. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Mechanics of Solids Lecture 9 by Dr Emre Erkmen Lecturer, School of Civil and Environmental Engineering, The University of Technology Sydney Office: 2.520 Phone:9514 9769 Email: emre.erkmen@uts.edu.au Lecture hours: Tuesdays 11:00-14:00, Wednesdays 15:00-18:00 Office hours: Tuesdays 15:00-17:00 1 Mechanics of Solids - Lecture 9 Outline Outline- Stress transformations- Maximum and minimum stresses- Mohrs circle 2 Mechanics of Solids - Lecture 9 Body in Static Equilibrium Y X F 1 F 2 3 F 3 Cube in static equilibrium Triangular prism in static equilibrium Mechanics of Solids - Lecture 9 Stresses due to internal forces 4 N My A I = VQ T Ib J = + V M N static equilibrium of a free body cut from the beam Mechanics of Solids - Lecture 9 Transformation of plane stresses Uniaxial stress 5 x P A = F = x Mechanics of Solids - Lecture 9 Area of the surfaces b cos tan 6 Mechanics of Solids - Lecture 9 Transformation of plane stresses Uniaxial stress 7 x P A = F = F = x cos cos x A A = sin cos x A A + = 2 cos x = sin cos x = Eq. ( 1 ) Eq. ( 2 ) Mechanics of Solids - Lecture 9 Transformation of stresses Only y acting y 8 F = F = tan sin cos y A A = tan cos cos y A A = 2 sin y = sin cos y = Eq. ( 3 ) Eq. ( 4 ) Mechanics of Solids - Lecture 9 Transformation of stresses Both x acting 2 2 cos sin x y = + sin cos sin cos x y = + y and From the superposition of the previous equations Eq. ( 1 ) + Eq. ( 3 ) Eq. ( 2 ) + Eq. ( 4 ) 9 From trigonometry ( ) 2 1 cos 1 cos2 2 = + ( ) 2 1 sin 1 cos2 2 = 1 cos sin sin 2 2 = ( ) ( ) 1 1 cos2 2 2 x y x y = + + ( ) 1 sin 2 2 x y = Eq. ( 5 ) Eq. ( 6 ) Mechanics of Solids - Lecture 9 Transformation of stresses Pure shear xy A xy A cos xy A sin 10 F = F = sin tan cos cos xy xy A A A = + cos tan sin cos xy xy A A A + = 2 sin cos xy = ( ) 2 2 cos sin xy = Using trigonometric relations we obtain sin 2 xy = cos2 xy = Eq. ( 7 ) Eq. ( 8 ) Mechanics of Solids - Lecture 9 Transformation of stresses All x acting y and From the superposition of the previous equations ( ) ( ) 1 1 cos2 sin 2 2 2 x y x y xy = + + + xy , ( No stresses on the z surface ) Eq. ( 9 ) 11 ( ) 1 sin 2 cos2 2 x y xy = + Given stresses on 2 perpendicular planes at a point x , y , xy , we can calculate stresses , on another plane (at the point) making an angle, , to one of the perpendicular planes....
View Full Document

This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.

Page1 / 48

File Lecture9_Slides.pdf - Mechanics of Solids Lecture 9 by...

This preview shows document pages 1 - 12. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online