Week Eight: Advanced structural analyses Tutorial Eight Part A: Objective questions (5 marks) 1. theorem is used to derive deflection of curved beams with small initial curvature (Castigliano's theorem) 2. variations are nonlinear in curved beams with large initial curvature (stress) 3. Stresses in are larger than that of extrados (intrados) 4. In curved beams with large initial curvature, stress distribution is nonlinear due to the fact that shifts towards the centre of curvature (Neutral axis) 5. Winkler-Bach equation is useful to estimate stresses in curved beams with large initial curvature only at (at the extreme skin layers) Part B: Descriptive questions (15 marks) 1. What is the sign convention used in estimating stresses in curved beams with large initial curvature? copyright 2. State simplified equations to estimate stresses in extreme fibres of curved beams with large initial curvature NPTEL 1
copyright 3. Determine the location of the shear centre for an I section: Flange: 150 x 30 mm Web: 20 x 200 mm NPTEL 2
Consider a section of thickness dx on the flange at a distance x from the end. V 1 = τ da 3
V 1 = VAy It da where, A = t x da = dx t y = h/2 V 1 = V b 1 (t x) (dx t) h It 0 2 = V It 2 = Vtb 1 2 h 4I t 2 h x 2 b 1 By symmetry, V 2 = Vtb 2 2 h 4I 2 0 = 0.0546V = 0.123V I = 9.886 10 7 mm 4 Taking moments about the centre of the web, 4. Estimate the stresses at points A, B, C and D. Ve + V 1 (h) V 2 (h) = 0 e = 13.66 mm 4
Given: Type of the section: Rectangular section Length of the section, l = 50 mm Breadth of the section, b = 25 mm 5
Inner radius of the beam, r 1 = 35 mm Load acting on the beam, P = 25 kn Angle of CD from centroidal axis, Ɵ = 30 o Solution: A. WINKLER BACH EQUATION (i) Calculation of Geometric properties: h= l/2 = 25 mm h i = h o = h = 25 mm 1. Radius of the curved beam, R = r₁+h = 60 mm 2. Outer radius of the curved beam, r₂ = r₁+l = 85mm 3. CS area of the section, A= l x b = 1250 mm² 4. Sectional property, m = 1 R. b. ln r2 A 1 = 0.0647 (no unit) r 5. Eccentricity, e = m R = 3.649 mm (The positive value of e indicates that the Neutral m 1 axis will shift towards the centre of curvature) 6. Moment of Inertia, I = 2.604 x 10 5 mm 4 (ii) Section AB: (a) Direct Stress, σ d = P = -20 N/mm2 A (b) Moment at CG, M =P x R = 1.5 knm Moment about CG will cause tensile stresses in extrados and compressive stresses in intrados. (c) Stress at intrados, σ i = M Ae (d) Stress at Extrados, σ i = M Ae h i e = -200.582 N/mm 2 (compressive) r i h o e = 110.828 N/mm 2 (tensile) r o (e) Total stress at intrados,σ A = σ d + σ i = -220.582 N/mm 2 (compressive) (f) Total stress at extrados, σ B = σ d + σ o = 90.828 N/mm 2 (tensile) 6
(iii) Section CD: Since the section CD is at an angle from the centroidal axis, the normal load at the section is given by, P n = Pcos θ = 21.651 kn. (a) Direct Stress, σ d = P n A = -17.321 N/mm2 (b) Moment at CG, M = M x cosɵ = 1.299 knm (c) Stress at intrados, σ i = M Ae (d) Stress at Extrados, σ i = M Ae h i e r i h o e r o = -173.709 N/mm 2 (compressive) = 95.98 N/mm 2 (tensile) (e) Total stress at intrados,σ A = σ d + σ i = -191.029 N/mm 2 (compressive) (f) Total stress at extrados, σ B = σ d + σ o = 78.659 N/mm 2 (tensile) B. WILSON AND QUEREAN EQUATION: R/h = 2.4 From the table, k i = 1.432. k o = 0.762 M (a) Stress at intrados, σ i = k i h= -206.208 Ae N/mm2 (compressive) 7
(b) Stress at Extrados, σ o = k o M Ae h = 109.728 N/mm2 (tensile) (c) Total stress at intrados,σ A = σ d + σ i = -226.208 N/mm 2 (compressive) (d) Total stress at extrados, σ B = σ d + σ o = 89.728 N/mm 2 (tensile) 5. A 10 ton crane hook is used to lift an object during commissioning of an offshore deck. Find the stresses at the intrados and extrados. Given: Type of the section: Trapezoidal section 8
Height of the section, l = 100 mm Breadth of the section, b 1 = 80 mm b 2 = 40 mm Inner radius of the beam, r 1 = 60 mm Load acting on the beam, P = 50 kn Solution: A. WINKLER BACH EQUATION (i) Calculation of Geometric properties: b 3 = 20 mm Location of the neutral axis, x = Ax A h i = x = 44.44 mm h o = h-x = 55.56 mm = 44.44 mm 1. Radius of the curved beam, R = r₁+x = 104.44 mm 2. Outer radius of the curved beam, r₂ = r₁+h= 160 mm 3. CS area of the section, A = 6000 mm² 4. Sectional property, m= 1 R b A 2 + r 2 (b 1 b 2 ). ln r2 (r 2 r 1 ) r 1 (b 1 b 2 ) = -0.0794 (no unit) 5. Eccentricity, e = m R = 7.679 mm (The positive value of e indicates that the Neutral m 1 axis will shift towards the centre of curvature) 6. Moment of Inertia, I = 4.814 x 10 6 mm 4 (ii) Section AB: r i = r 1 = 60 mm r o = r 2 = 100 mm (a) Direct Stress, σ d = P = 8.333 N/mm2 A 9
(b) Moment at CG, M = -P x R = -5.222 knm (c) Stress at intrados, σ i = M Ae (d) Stress at Extrados, σ i = M Ae h i e = 69.442 N/mm 2 (tensile) r i h o e = -44.79 N/mm 2 (compressive) r o (e) Total stress at intrados,σ A = σ d + σ i = 77.775 N/mm 2 (tensile) (f) Total stress at extrados, σ B = σ d + σ o = -36.457 N/mm 2 (compressive) B. STRAIGHT BEAM FORMULA: σ = M y I y i = 44.44 mm y o = 55.56 mm (a) Stress at intrados, σ i = M y I i = 48.205 N/mm 2 (tensile) (b) Stress at Extrados, σ o = M y I o = -60.256 N/mm 2 (compressive) (c) Total stress at intrados,σ A = σ d + σ i = 56.539 N/mm 2 (tensile) (d) Total stress at extrados, σ B = σ d + σ o = -51.923 N/mm 2 (compressive) 10