5.2 Elastic Strain Energy

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Section 5.2 The shear stresses arising in a beam at location y from the neutral axis are given by Eqn. 4.6.27, τ ( y ) = Q( y) V / Ib( y) , where Q is the first moment of area of the section of beam from y to the outer surface, V is the shear force, I is the moment of inertia of the complete cross-section and b is the thickness of the beam at y. From Eqns. 5.2.19a and 5.2.14 then, the total strain energy in a beam of length L due to shear stress is 2 L 2 2 τ 1 V ⎡ Q ⎤ U = ∫ dV = ∫ ⎢ dA⎥dx 2 V I ∫ 2 2μ 2 0 μ ⎣ A b ⎦ Solid Mechanics Part I 187 Kelly (5.2.20) Here V , μ and I are taken to be constant for any given cross-section but may vary along the beam; Q varies and b may vary over any given cross-section. Expression 5.2.20 can be simplified by introducing the form factor for shear f s , defined as so that 2 A Q f s ( x) = dA 2 I ∫ (5.2.21) 2 b U The form factor depends only on the shape of the cross-section. For example, for a rectangular cross-section, using Eqn. 4.6.28, = 1 2 L ∫ 0 A f V s 2 μA dx 2 3 2 ( / 12) ∫ 2 ⎢ 2 ⎜ − y ⎥ = 4 ⎟ dy dz bh 5 −h / 2 b ⎣ ⎝ ⎠⎦ −b / 2 (5.2.22) + h / 2 2 + b / 2 bh 1 ⎡b ⎛ h ⎞⎤ 6 f s ( x) = ∫ (5.2.23) In a similar manner, the form factor for a circular cross-section is found to be 10 / 9 and that of a very thin tube is 2. 5.2.4 Castigliano’s Second Theorem The work-energy method is the simplest of energy methods. A more powerful method is that based on Castigliano’s second theorem 2 , which can be used to solve problems involving linear elastic materials. As an introduction to Castigliano’s second theorem, 2 consider the case of uniaxial tension, where U = P L / 2EA. The displacement through which the force moves can be obtained by a differentiation of this expression with respect to that force, dU dP PL = EA = Δ 2 Casigliano’s first theorem will be discussed in a later section 2 (5.2.24)
Section 5.2 2 Similarly, for torsion of a circular bar, U = T L / 2GJ , and a differentiation gives dU / dT = TL / GJ = φ . Further, for bending of a beam it is also seen that dU / dM = θ . These are examples of Castigliano’s theorem, which states that, provided the body is in equilibrium, the derivative of the strain energy with respect to the force gives the displacement corresponding to that force, in the direction of that force. When there is more than one force applied, then one takes the partial derivative. For example, if n independent forces P 1, P2 , K , Pn act on a body, the displacement corresponding to the ith force is Before proving this theorem, here follow some examples. Example ∂U Δ i = (5.2.25) ∂P The beam shown in Fig. 5.2.11 is pinned at A, simply supported half-way along the beam at B and loaded at the end C by a force P and a moment M 0 . Figure 5.2.11: a beam subjected to a force and moment at C The moment along the beam can be calculated from force and moment equilibrium, ⎧− Px − 2M 0 x / L, ⎪ M = ⎨ ⎪⎩ − M 0 − P( L − x), Solid Mechanics Part I 188 Kelly i 0 < x < L / 2 L / 2 < x < L The strain energy stored in the bar (due to the flexural stresses only) is 2 L / 2 L 1 ⎪⎧ ⎛ 2M 0 ⎞ 2 U = ⎨⎜ P + ⎟ x dx 2EI L ∫ + ∫ ⎪⎩ ⎝ ⎠ 0 L / 2 2 3 2 2 P L 5PM 0L M 0 L = + + 24EI 24EI 3EI L A C B L / 2 L / 2 ( M + P( L − x) ) 0 2 ⎪⎫ dx⎬ ⎪⎭ (5.2.26) (5.2.27) In order to apply Castigliano’s theorem, the strain energy is considered to be a function of the two external loads, U = U ( P, M 0 ) . The displacement associated with the force P is then P M 0
Page 1 and 2: 5.2 Elastic Strain Energy Section 5
Page 3 and 4: ΔW Figure 5.2.4: torque - angle of
Page 5 and 6: Figure 5.2.7: a beam subjected to a
Page 7: Section 5.2 5.2.10. The element def
Page 11 and 12: But θ A = 0 and so Eqn. 5.2.33 can
Page 13 and 14: 1 2 EA w ( h + Δ ) = kΔ max , k =

5.2 Elastic Strain Energy

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