LECTURE+9+COE-3001-A

rotated stress element aligned with

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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...
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This note was uploaded on 09/19/2013 for the course CEE 3001 taught by Professor Zhu during the Spring '09 term at Georgia Tech.

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