Curved Beam - VTU e-Learning

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c = Radius of centroidal axis r n = Radius of neutral axis CL= Centre line of curvature Summation of forces over the whole cross section i.e. dF 0 As θ θ ∴ θ θ =0 is not equal to zero, ∴ = 0 ..... (ii) The neutral axis radius 'rn' can be determined from the above equation. If the moments are taken about the neutral axis, M b = – ydF Substituting the value of dF, we get M b = θ θ dA = θ θ y dA = θ θ ydA 0 Since ydA represents the statical moment of area, it may be replaced by A.e., the product of total area A and the distance 'e' from the centroidal axis to the neutral axis.
∴ M = b θ A.e ..... (iii) θ From equation (i) θ θ = – σ Substituting in equation (iii) M = – b σ . A. e. ∴ σ = ..... (iv) This is the general equation for the stress in a fiber at a distance 'y' from neutral axis. At the outer fiber, y = co ∴ Bending stress at the outer fiber σbo i.e., σ = bo ( rn + co = ro) ..... (v) Where co = Distance from neutral axis to outer fiber. It is compressive stress and hence negative sign. At the inner fiber, y = – ci ∴ Bending stress at the inner fiber σ bi = i.e., σ bi = ( rn – ci = ri) ..... (vi) Where ci = Distance from neutral axis to inner fiber. It is tensile stress and hence positive sign.
Page 1 and 2: CURVED BEAMS CONTENT: WHAT’S A C
Page 3 and 4: G1H1 = (R+Y) Ѳ Also Stress OR dF =
Page 5 and 6: Stresses in Curved Beam Consider a
Page 7: We know, stress σ = E.e We know, s
Page 11 and 12: 2 3 Neutral axis and centroidal axi
Page 14 and 15: Why stress concentration occur at i
Page 16 and 17: For a straight beam: Inner most fib
Page 18: e = rc - rn = 100 - 91.024 = 8.976m
Page 21 and 22: = i.e., 4674.069+83.61t +83.61t = 4
Page 23 and 24: σri = σd + σbi = 7.0736 105.00
Page 25 and 26: Bending stress at the outer fiber
Page 27 and 28: =150.34mm To find the centroidal ax
Page 29 and 30: Distance of neutral axis to inner r
Page 31 and 32: Problem no.7 Determine the maximum
Page 33 and 34: Maximum shear stress max = 0.5x σm
Page 35 and 36: ∴ Radius of centroidal axis rc =
Page 37 and 38: Combined stress =-198.189 N/mm σ r
Page 39 and 40: x = = 101.785mm Radius of
Page 41 and 42: Problem no.10 Compute the combined
Page 43 and 44: Bending stress at the inner fiber
Page 45 and 46: C2 = = 50mm c1 = 90-50= 40mm Mome
Page 47 and 48: Problem no.11 The section of a cran
Page 49 and 50: Problem no.12 Design of steel crane
Page 51 and 52: Applied force F = 10 5 N Area of cr
Page 53 and 54: Problem no.13 The figure shows a lo
Page 55 and 56: Draw the critical section at PP-Q a
Page 57 and 58: 25 ro = 75 + 2 = 87.5 mm A = π 4 d
Page 59 and 60:
The ring is symmetrical and is load
Page 61 and 62:
As this quantity is positive the di
Page 63 and 64:
The stress at inner (σ1Ai) ) and o
Page 65 and 66:
As per Castigliano’s theorem \
Page 67 and 68:
The stress at any angle Ѳ can be f
Page 69 and 70:
Distance of neutral axis to outer r
Page 71 and 72:
Problem 16 Determine the stress ind
Page 73 and 74:
Consider the cross section B - B At
Page 75 and 76:
Distance of neutral axis to outer r
Page 77 and 78:
Using usual notations prove that th
Page 79 and 80:
Substituting the value of the σx f
Page 81 and 82:
1 ∆ = 1 {From equation (x
Page 83 and 84:
VTU,Jan/Feb.2005 Fig.1.35 5. Determ
Page 85:
REFERENCE BOOKS • “MECHANICAL E
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Curved Beam - VTU e-Learning

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