LECTURE+7+COE-3001-A

# Axial force into a normal force n perpendicular

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Unformatted text preview: onstructed of two pieces glued by a joint (scarf joint) inclined by 40°with respect to the verOcal axis ²་  Allowable stresses in plasOc 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 idenOfy 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 convenOon 31 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 32 EXAMPLE 2- 9 (cont.) SOLUTION: ²་  Max allowable compressive stress: max plastic = 7.5M P a = x ²་  Max allowable shear stress: max ⌧plastic = 4.0M P a = x = x 2 8.0M P a ²་  Finally, select minimum σx from the among the four values just calculated and use this value to determine Aminimum and then bminimum Aminimum = P/ bmin = 1/17/13 p min.allowable x Aminimum Aminimum = 35000N/6.9M P a = 5072mm2 bminimum = p M. Mello/Georgia Tech Aerospace 5072mm2 = 71.2mm 33...
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