chapter 13 answers

chapter 13 answers - Problem 13.1 The components of plane...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 13.1 The components of plane strain at point p are ε x = 0 . 003 , ε y = 0, and γ xy = 0 . If θ = 45 , what are the strains ε x , ε y , and γ xy at point p ? Solution: From Equation (13-7): ε x = ε x + ε y 2 + ε x ε y 2 (cos 2 θ )+ γ xy 2 (sin 2 θ ) = 0 . 003 + 0 2 + 0 . 003 0 2 (cos 90 )+ 0 2 (sin 90 ) ANS: ε x = 0 . 0015 From Equation (13-8): ε y = ε x + ε y 2 ε x ε y 2 (cos 2 θ ) γ xy 2 (sin 2 θ ) = 0 . 003 + 0 2 0 . 003 0 2 (cos 90 ) 0 2 (sin 90 ) ANS: ε y = 0 . 0015 From Equation (13-9): γ xy 2 = ε x ε y 2 (sin 2 θ )+ γ xy 2 (cos 2 θ ) = 0 . 003 0 2 (sin 90 )+ 0 2 (cos 90 ) ANS: γ xy = 0 . 003 Problem 13.2 The components of plane strain at point p are ε x = 0 , ε y = 0 , and γ xy = 0 . 004 . If θ = 45 , what are the strains ε x , ε y , and γ xy at point p ? Solution: From Equation (13-7): ε x = ε x + ε y 2 + ε x ε y 2 (cos 2 θ )+ γ xy 2 (sin 2 θ ) = 0 + 0 2 + 0 0 2 (cos 90 )+ 0 . 004 2 (sin 90 ) ANS: ε x = 0 . 002 From Equation (13-8): ε y = ε x + ε y 2 ε x ε y 2 (cos 2 θ ) γ xy 2 (sin 2 θ ) = 0 + 0 2 0 0 2 (cos 90 ) 0 . 004 2 (sin 90 ) ANS: ε y = 0 . 002 From Equation (13-9): γ xy 2 = ε x ε y 2 (sin 2 θ )+ γ xy 2 (cos 2 θ ) = 0 0 2 (sin 90 )+ 0 . 004 2 (cos 90 ) ANS: γ xy = 0 Problem 13.3 The components of plane strain at point p are ε x = 0 . 0024 , ε y = 0 . 0012 , and γ xy = 0 . 0012 , If θ = 25 , what are the strains ε x , ε y , and γ xy at point p ? Solution: From Equation (13-7): ε x = ε x + ε y 2 + ε x ε y 2 (cos 2 θ )+ γ xy 2 (sin 2 θ ) = 0 . 0024 + 0 . 0012 2 + 0 . 0024 0 . 0012 2 (cos 50 )+ 0 . 0012 2 (sin 50 ) ANS: ε x = 0 . 00222 From Equation (13-8): ε y = ε x + ε y 2 ε x ε y 2 (cos 2 θ ) γ xy 2 (sin 2 θ ) = 0 . 0024 + 0 . 0012 2 0 . 0024 0 . 0012 2 (cos 50 ) 0 . 0012 2 (sin 50 ) ANS: ε y = 0 . 00102 From Equation (13-9): γ xy 2 = ε x ε y 2 (sin 2 θ )+ γ xy 2 (cos 2 θ ) = 0 . 0024 0 . 0012 2 (sin 50 )+ 0 . 0012 2 (cos 50 ) ANS: γ xy = 0 . 00099
Image of page 1

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

View Full Document Right Arrow Icon
Problem 13.4 The components of plane strain at point p of a bit during a drilling operation are ε x = 0 . 00400 , ε y = 0 . 00300 , and γ xy = 0 . 00600 , and the com- ponents referred to the x y z coordinate system are ε x = 0 . 00125 , ε y = 0 . 00025 , and γ xy = 0 . 00910 . What is the angle θ ? Solution: From Equation (13-8): 0 . 00125 = 0 . 004 + ( 0 . 003) 2 + 0 . 004 ( 0 . 003) 2 (cos 2 θ )+ 0 . 006 2 (sin 2 θ ) Using a graphing calculator to determine the value of θ : 2 θ = 40 ANS: θ = 20 Problem 13.5 The components of plane strain at point p are ε x = 0 . 0024 , ε y = 0 . 0012 , and γ xy = 0 . 0048 . The extensional strains ε x = 0 . 00347 , and ε y = 0 . 00227 . Determine γ xy and the angle θ . Solution: From Equation (13-7): ε x = ε x + ε y 2 + ε x ε y 2 (cos 2 θ ) + γ xy 2 (sin 2 θ ) 0 . 00347 = 0 . 0024 + ( 0 . 0012) 2 + 0 . 0024 ( 0 . 0012) 2 (cos 2 θ )+ 0 . 0048 2 (sin 2 θ ) Using a graphing calculator: θ = 35 Equation (13-9): γ xy 2 = ε x ε y 2 (sin 2 θ )+ γ xy 2 (cos 2 θ ) = 0 . 0024 ( 0 . 0012) 2 (sin 70 )+ 0 . 0048 2 (cos 70 ) ANS: γ xy = 0 . 00174
Image of page 2
Problem 13.6 During liftoff, strain gauges attached to one of the Space Shuttle main engine nozzles determine that the components of plane strain at point p are strains ε x = 0 . 00665 , and ε y = 0 . 00825 , and γ xy = 0 . 00135 for a coordinate system oriented at θ = 20 . What are the strains ε x , ε y , and γ xy at that point?
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern