LECTURE+9+COE-3001-A

# rotated stress element aligned with

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: AL AND SHEAR STRESSES shear stress maxima at Θ=- 45° max ⌧✓ = normal stress maxima at Θ=0° max ✓ = x x NORMAL STRESS 2 ✓ = x 2 (1 + cos 2✓) (2-29a) SHEAR STRESS ⌧✓ = x 2 (sin 2✓) (2-29b) shear stress “maxima” at Θ=45° 1/17/13 M. Mello/Georgia Tech Aerospace 7 MAXIMUM STRESSES FOR A BAR IN TENSION ²་  Luders’ Bands begin to appear as yield stress is reached ²་  Bands reveal that the material is failing in shear along planes where shear stress is maximum. 45° ELEMENT UNDER MAXIMUM SHEAR STRESS SLIP BANDS (LUDERS’ BANDS) FOR A POLISHED STEEL SAMPLE LOADED IN TENSION (W. LUDERS, 1860) 1/17/13 M. Mello/Georgia Tech Aerospace 8 MAXIMUM STRESSES FOR A BAR IN COMPRESSION 45° ELEMENT UNDER MAXIMUM SHEAR STRESS SHEAR FAILURE ALONG A 45° PLANE IN A BLOCK OF WOOD LOADED IN COMPRESSION 1/17/13 M. Mello/Georgia Tech Aerospace 9 EXAMPLE 2- 9: FAILURE LOAD ANALYSIS OF A COMPRESSION BAR GIVEN: ²་  Compression bar made out of structural plas\c must support P=35kN load ²་  Bar constructed of two pieces glued by a joint (scarf joint) inclined by 40°with respect to the ver\cal axis ²་  Allowable stresses in plas\c bar : •  max compression: 7.5MPa •  shear: 4.0MPa ²་  Allowable stresses in scarf joint : •  max compression: 5.2MPa •  shear: 3.4MPa GLUED JOINT (SCARF JOINT) DETERMINE: Minimum width (b) of the bar SOLUTION: ²་  Each allowable stress criteria has a corresponding σx ²་  Find σx for each allowable stress criteria and iden\fy the minimum x = P/A bmin = 1/17/13 p Amin Aminimum = P/ b allowable x bmin M. Mello/Georgia Tech Aerospace Note: Θ < 0 by sign conven\on 10 EXAMPLE 2- 9 (cont.) RECALL GENERAL STRESS FORMULAS: NORMAL STRESS SOLUTION: ²་  COMPUTE σx for each allowable stress criteria for GLUED JOINT x = = 5.2M P a = x = ✓ = ⌧✓ = ⌧✓ = SHEAR STRESS ²་  Max allowable compressive stress: max glue ✓ x x 2 cos2 ✓ (1 + cos 2✓) (2-29a) sin ✓ cos ✓ x (2-29b) 2 (sin 2✓ ) x cos2 ( 50 ) 12.6M P a ²་  Max allowable shear stress or GLUED JOINT: max ⌧glue = 3.4M P a = x sin( 50 ) cos( 50 ) NOTE: τ < 0 x = 6.9M P a NOTE: τ < 0 since shear will tend to rotate either element clockwise 1/17/13 M. Mello/Georgia Tech Aerospace 11 EXAMPLE 2- 9 (cont.) SOLUTION: ²་  Max allowable compressive stress: ma...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern