Høgskolen i Narvik Sivilingeniørutdanningen

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1 Høgskolen i Narvik Sivilingeniørutdanningen Eksamen i Faget STE66 ELASTISITETSTEORI Klasse: 4.ID Dato: Tid: Kl Tillatte hjelpemidler under eksamen: Kalkulator Kopi av Boken Mechanics of Aircraft Structures Kopi av transparenter Elastisitetsteori Forelesningsnotater Engelsk/Norsk, Norsk/Engelsk ordbok Faglig kontaktperson under eksamen: Professor II Gregory A. Chechkin tel Narvik 009

2 Task 1. The state of stress in a unit cube (see Figure 1) is uniform and given by σ xx = 1 MPa τ xy = 1 MPa τ xz = 0 MPa σ yy = 1 MPa τ yz = 0 MPa σ zz = 0 MPa. Figure 1: The crosssection of the unit cube. a) Find three components of the stress vector t on the surface ABCD. Solution. The unit normal to the surface ABCD is ) n = ( 1 6, 1 6, and respectively t = = b) Find the normal component σ n of the stress vector on the surface ABCD. Solution. By the definition ( t ) σ n =, n = 0.

3 c) Find the principal stresses and the respective principal directions for the stresses. Solution. To find principal stresses we consider the equation (1 σ) (1 σ) σ = 0 or equivalently Hence, For σ 1 = we have ( σ) σ = 0. σ 1 =, σ = σ = 0. n (1) x + n (1) n (1) x n (1) y = 0 y = 0 n (1) z = 0 and n (1) = ( 1 1 ; ; 0). For σ = σ = 0 we have n () x + n () n () x + n () y = 0 y = 0 0n () z = 0 and n () = (0; 0; 1) ( 1 n () = ; 1 ) ; 0. Task. A cantilever beam of a rectangular cross section is subjected to a shear force V as shown in Figure. The bending stress is given by σ xx = Mz I,

4 Figure : Cantilever beam subjected to a shear force. where M = V Lx. Assume a state of plane stress parallel to the x z plane, i.e. σ yy = τ xy = τ yz = 0. Find the normal stress σ zz by integrating the equilibrium equations over the beam thickness and applying the boundary conditions τ xz = 0 at z = ± h and σ zz = 0 at z = 0. Solution. Keeping in mind that I = bh 1 and M = V Lx, we write The first equilibrium equation reads as follows: σ xx = 1V L x z bh. (1) σ xx x + τ xz z = 0. Substituting (1) into the equilibrium equation, we get Hence, integrating, we derive 4V L xz + τ xz bh z = 0. τ xz = 1V L xz bh + C(x). Using boundary condition, namely, τ xz = 0 at z = ± h, we obtain τ xz = 1V L xz bh 4 V L x bh.

5 or It is time to use the second equilibrium equation 1V L z bh Hence, integrating, we derive σ zz = V L z bh τ xz x + σ zz z = 0 V L bh + σ zz z = 0. 4V L z bh + C(x) and, using boundary condition σ zz = 0 at z = 0, finally we deduce σ zz = V L z bh 4V L z bh. Task. Consider a thin panel (a square) loaded as shown in Figure. Show that the Airy stress function φ(x, y) = c 1 x + c xy + c y solves the problem. Find the constants c 1, c, c. Figure : Thin panel subjected to uniform loads. Solution. The Airy stress function given by the formula, satisfies the equation 4 φ φ = 0 or x + 4 φ 4 x y + 4 φ y =

6 By the definition consequently, σ xx = φ y, σ yy = φ x, τ xy = φ x y, σ xx = c, σ yy = c 1, τ xy = c. It remains to consider the boundary conditions. On I we have 1 1 n = (, ), t = ( σ 0, σ 0 ). On II we have On III we have On IV we have 1 1 n = (, ), n = ( 1, 1 ), n = ( 1, 1 ), t = (0, 0). t = ( σ 0, σ 0 ). t = (0, 0). Hence, by the formula ( tx t y ) ( σxx τ = xy we deduce the boundary conditions τ xy σ yy ) ( nx n y ) σ xx = σ 0, σ yy = σ 0, τ xy = σ 0. and the respective Airy stress function is φ(x, y) = σ 0 ( 1 4 x + 1 xy y ). Task 4. Denote ( by u ) = (u 1, u, u ) the displacement vector, by e ij (u) = 1 ui x j + u j x i the strain components, by σ ij (u) = µe ij (u) + δ ij λ(e 11 (u) + e (u) + e (u)) 6

7 Figure 4: The disc. the respective stress components, where x = (x 1, x, x ) R. Consider the boundary value problem for the elasticity system in the cylinder (shown in Figure 4) with lateral boundary S and upper and lower parts of the boundary Γ ±, in the following form: µ u (λ + µ) u = f in, u = 0 on S, σ 1 (u) = 0; σ (u) = 0; σ (u) = 0 on Γ ±, where f = (f 1, f, f ) is a given body force. Find the weak formulation for classical problem (). Recall that u = u and u = grad (div u). Solution. Note that (µ u + µ u) v ( ( u1 µ ( u 1 v 1 + u v + u v ) + µ + u + u ) ) v j = x j=1 j x 1 x x ( ( ) ) u i = µ v i + u j v i = µ (e ij (u))v i x j x i x j x j () () 7

8 and (λ u) v λ = λ ( j=1 x j ( u1 + u + u ) ) v j = x 1 x x δ ij x i (e 11 (u) + e (u) + e (u))v j. Let us define the space H 1 (, S) as the set of vector-functions from the Sobolev space H 1 () vanishing on S. Thus, multiplying the equation by the vector test-function v H 1 (, S), and integrating over the domain, we have (µ u + (λ + µ) u) v dx = f v dx. (4) Rewriting this identity by means of (), (4) and we get σ ij (u) = µe ij (u) + δ ij λ(e 11 (u) + e (u) + e (u)), (σ ij (u)) v i dx = x j f v dx. Let us apply the integration by parts (Green s formula) for the left-hand side of the equation in the form (σ ij (u)) v i dx = x j σ ij (u) e ij (v) dx σ ij (u) v i ν j ds. Bearing in mind the boundary conditions of problem (), we have σ ij (u) v i ν j ds = Γ ± (5) σ ij (u) v i ν j ds = 0. (6) Finally, using (6), we get the identity of the problem σ ij (u) e ij (v) dx = f v dx v H 1 (, S). 8