Stresses in Curved Beam

Transcription

1 Stresses in Curved Beam Consider a curved beam subjected to bending moment M b as shown in the figure. The distribution of stress in curved flexural member is determined by using the following assumptions: i) The material of the beam is perfectly homogeneous [i.e., same material throughout] and isotropic [i.e., equal elastic properties in all directions] ii) The cross section has an axis of symmetry in a plane along the length of the beam. iii) The material of the beam obeys Hooke's law. iv) The transverse sections which are plane before bending remain plane after bending also. v) Each layer of the beam is free to expand or contract, independent of the layer above or below it. vi) The Young's modulus is same both in tension and compression. Derivation for stresses in curved beam Nomenclature used in curved beam C i =Distance from neutral axis to inner radius of curved beam C o =Distance from neutral axis to outer radius of curved beam C 1 =Distance from centroidal axis to inner radius of curved beam C 2 = Distance from centroidal axis to outer radius of curved beam F = Applied load or Force A = Area of cross section L = Distance from force to centroidal axis at critical section σ d = Direct stress σ bi = Bending stress at the inner fiber σ bo = Bending stress at the outer fiber σ ri = Combined stress at the inner fiber σ ro = Combined stress at the outer fiber

2 F CA NA c 1 c 2 c i co e F F M b F M b r i r n C L rc r o Stresses in curved beam M b = Applied Bending Moment r i = Inner radius of curved beam r o = Outer radius of curved beam r c = Radius of centroidal axis r n = Radius of neutral axis C L = Center of curvature In the above figure the lines 'ab' and 'cd' represent two such planes before bending. i.e., when there are no stresses induced. When a bending moment 'Mb' is applied to the beam the plane cd rotates with respect to 'ab' through an angle 'dθ' to the position 'fg' and the outer fibers are shortened while the inner fibers are elongated. The original length of a strip at a distance 'y' from the neutral axis is (y + rn)θ. It is shortened by the amount ydθ and the stress in this fiber is, σ = E.e Where σ = stress, e = strain and E = Young's Modulus

3 We know, stress σ = E.e We know, stress e Ѳ Ѳ i.e., σ = E θ θ... (i) Since the fiber is shortened, the stress induced in this fiber is compressive stress and hence negative sign. The load on the strip having thickness dy and cross sectional area da is 'df' i.e., df = σda = θ θ da From the condition of equilibrium, the summation of forces over the whole cross-section is zero and the summation of the moments due to these forces is equal to the applied bending moment. Let M b = Applied Bending Moment r i = Inner radius of curved beam r o = Outer radius of curved beam

4 r c = Radius of centroidal axis r n = Radius of neutral axis C L = Centre line of curvature Summation of forces over the whole cross section i.e. df 0 θ θ =0 As θ θ 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.

5 M b = θ θ A.e... (iii) From equation (i) θ = σ θ 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 = c o Bending stress at the outer fiber σ bo 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.

6 Difference between a straight beam and a curved beam Sl.no straight beam curved beam 1 In Straight beams the neutral In case of curved beams the axis of the section coincides neutral axis of the section is with its centroidal axis and the shifted towards the center of stress distribution in the beam curvature of the beam causing is linear. a non-linear stress distribution.

7 2 3 Neutral axis and centroidal Neutral axis is shifted axis coincides towards the least centre of curvature Location of the neutral axis By considering a rectangular cross section

8 Centroidal and Neutral Axis of Typical Section of Curved Beams

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10 Why stress concentration occur at inner side or concave side of curved beam Consider the elements of the curved beam lying between two axial planes ab and cd separated by angle θ. Let fg is the final position of the plane cd having rotated through an angle dθ about neutral axis. Consider two fibers symmetrically located on either side of the neutral axis. Deformation in both the fibers is same and equal to ydθ.

11 Since length of inner element is smaller than outer element, the strain induced and stress developed are higher for inner element than outer element as shown. Thus stress concentration occur at inner side or concave side of curved beam The actual magnitude of stress in the curved beam would be influenced by magnitude of curvature However, for a general comparison the stress distribution for the same section and same bending moment for the straight beam and the curved beam are shown in figure. It is observed that the neutral axis shifts inwards for the curved beam. This results in stress to be zero at this position, rather than at the centre of gravity. In cases where holes and discontinuities are provided in the beam, they should be preferably placed at the neutral axis, rather than that at the centroidal axis. This results in a better stress distribution. Example: For numerical analysis, consider the depth of the section ass twice the inner radius.

12 For a straight beam: Inner most fiber: Outer most fiber: For curved beam: h=2r i e = r c - r n = h 0.910h = h c o = r o - r n = h 0.910h = 0590h c i = r n - r i = 0.910h - = 0.410h

13 Comparing the stresses at the inner most fiber based on (1) and (3), we observe that the stress at the inner most fiber in this case is: σ bci = 1.522σ BSi Thus the stress at the inner most fiber for this case is times greater than that for a straight beam. From the stress distribution it is observed that the maximum stress in a curved beam is higher than the straight beam. Comparing the stresses at the outer most fiber based on (2) and (4), we observe that the stress at the outer most fiber in this case is: σ bco = 1.522σ BSi Thus the stress at the inner most fiber for this case is times that for a straight beam. The curvatures thus introduce a non linear stress distribution. This is due to the change in force flow lines, resulting in stress concentration on the inner side. To achieve a better stress distribution, section where the centroidal axis is the shifted towards the insides must be chosen, this tends to equalize the stress variation on the inside and outside fibers for a curved beam. Such sections are trapeziums, non symmetrical I section, and T sections. It should be noted that these sections should always be placed in a manner such that the centroidal axis is inwards. Problem no.1 Plot the stress distribution about section A-B of the hook as shown in figure. Given data: r i = 50mm r o = 150mm F = 22X10 3 N b = 20mm h = = 100mm A = bh = 20X100 = 2000mm 2