LECTURE+21+COE-3001-A

550 he 600 he 450 he 550 he 360 he 450 weak axis 2 2

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

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

Unformatted text preview: 180 160 140 120 140 8.5 8 7 6.5 5.5 14 13 12 11 8.5 3831 2492 1509 864.4 1033 425.7 311.5 215.6 144.1 155.4 7.66 6.78 5.93 5.04 5.73 1363 889.2 549.7 317.5 389.3 151.4 111.2 78.52 52.92 55.62 4.57 4.05 3.58 3.06 3.52 HE 100 B HE 100 A 20.4 16.7 26.04 21.24 100 96 100 100 6 5 10 8 449.5 349.2 4.16 4.06 167.3 133.8 33.45 26.76 2.53 2.51 89.91 72.76 Note: Axes 1-1 and 2-2 are principal centroidal axes M. Mello/Georgia Tech Aerospace 10 2/22/13 M. Mello/Georgia Tech Aerospace 11 EXAMPLE 5- 1 Given: •  Simply supported beam AB •  Pure bending (note the couples M0) •  L=4.9m •  h=300mm •  εx = 1250x10- 6 (strain gage reading) Determine: •  Radius of curvature (ρ) •  h=curvature (κ) •  deflecTon (δ) x 150mm strain gage: ✏x = 1250µe 2/22/13 M. Mello/Georgia Tech Aerospace 12 EXAMPLE 5- 1 (cont.) ⇢ cos ✓ ⇢ x 150mm strain gage: ✏x = 1250µe SOLUTION: ⇢= ✏= y /⇢ ⇢= RADIUS OF CURVATURE: y /✏ ( 150mm)/1250 ⇥ 10 6 beam deflection: = 120000mm ⇢ = 120m curvature: = 1/⇢ = 1/120m = 8.33 ⇥ 10 2/22/13 m 1 ⇢ cos ✓ (need to estimate ✓) sin ✓ = 3 =⇢ (L/2) ⇢ sin ✓ = (4.9m/2) 120m ✓ = 0.0200 rad = 1.146 M. Mello/Georgia Tech Aerospace = 120m(1 cos 1.146 ) = 24mm 13 EXAMPLE 5- 2 Given: •  High- strength steel wire wrapped around cylindrical drum •  wire diameter: d = 4mm •  drum radius R0= 0.5m •  Ewire = Ew = 200GPa •  Steel wire proporTonal limit: σpl = 1200MPa Steel wire wrapped around drum Determine: •  Bending moment (M) •  Maximum bending stress: (σmax) Maximum bending stress: section modulus: S = max max Radius of curvature: ⇢ = Ro + d/2 Bending moment: M = EI ⇢ (200GP a)(4mm) 2(0.5m)+4mm ⇡ Ed4 32(2R0 +d) M= ⇡ d3 32 = 797M P a d moment of inertia of a circular cross section: I = 2/22/13 = = M S Ed 2R0 +d wire SOLUTION: M= = I d/2 = ⇡r4 4 ⇡ (200GP a)(4mm)4 32[2(0.5m+4mm] = ⇡ d4 64 = 5.01N · m M. Mello/Georgia Tech Aerospace 14 EXAMPLE 5- 4 Given: •  span length: L = 3.0m •  overhang length: 1.5m •  uniform load of intensity: q=3.2kN/m Determine: •  Maximum tensile and compressive stresses in the beam under load Beam with simple supports at AB and overhang from B to C web thickness avg.thickness of sloping flanges Channel- shaped Cross- secTon of beam ABC 2/22/13 M. Mello/Georgia Tech Aerospace 15 SOLUTION: EXAMPLE 5- 4 (cont.) I.  REACTIONS, SHEAR FORCES, AND BENDING MOMENTS •  STATIC EQBM. OF ENTIRE STRUCTURE LEADS TO: RA = 3.6kN and RB = 10.8kN •  From these two reacTons and knowledge of q(x) we easily construct the shear...
View Full Document

Ask a homework question - tutors are online