1.050 Solid Mechanics: Problem Set #5 Solutions

Problem Set #5 1.050 Solid Mechanics Fall 2004 (Due Friday, 15 October) Problem 5.1 A closed can of soda is under pressure equivalent to that in an automobile tire. We measured the wall thickness in lab on Thursday to be 0.0025 inches. Estimate the axial stress σa and the hoop stress σθ? acting at a point in this closed, thin cylindrical shell, away from the ends. C y x From class and textbook, p. 95. R R σ θ = p i --and σ a = p i ----­ t 2t The pressure (within an automobile tire) I estimate to be 30- 40 psi. The thickness t is .0025 inches and the diameter is around 2 inches. With these we and σ a = 6,000 to 8,000 psi obtain σ θ = 12,000 to 16,000 psi σθ ( = σy ) σa ( = σx ) σyx Construct a Mohr’s circle for this state of stress with components with respect to the xy axis shown. (σx = σa; σy = σθ). = σx y = 0 We show the Mohr’s circle at the right. Note how we identify the axial stress with σx and the hoop stress with σy. The shear stress components σyx = σxy are zero so we plot the points corresponding to this orientation of the element on the line σyx = σxy = 0. The diameter of the Mohr’s circle lies on this axis and its radius is just R = (σy. - σx )/2 = 4,000 psi. σyx Determine the rotation of axis at the point which would bring you to planes of maximum shear stress. What is the magnitude of the maximum shear stress component at the point. σx =8,000 σy =16,000 σxy i ps 00 i ,0 0 ps -4 At the left we show the element at the surface of the cylinder before and after rotation of ref­ erence axes. = 12 , 00 y= y = σ’ 12, 00 yx 0 ps i σ’ x σ’ o σx ,σy R The rotation of axis to reach planes of maximum shear components is 2φ = 90o on Mohr’s circle so φ = 45o in the “material” or “physical” plane. σθ ( = σy ) 2φ = 90 max shear magnitude σ’ x Note that the normal stress components w.r.t this rotated frame are equal, evident from the Mohr’s circle σa ( = σx ) σyx = σx y 45o NB: We shall see that if we consider a three dimen­ sional element at the point, the max shear acts on a x or y face, but in the z direction. =0 The same results could have been obtained without the use of Mohr’s circle from the relationships which define the transformation of components of stress (in 2D) which appear on p. 118 in the text. (σ x + σ y ) (σ x – σ y ) σ' x = ------------------------ + ------------------------ ⋅ cos2φ + σ xy sin2 φ 2 2 (σ x + σ y ) (σ x – σ y ) σ' y = ------------------------ – ------------------------ ⋅ cos 2φ–σ xy sin2φ 2 2 (σ x – σ y ) σ' xy = – ------------------------ ⋅ sin2φ + σ xy cos2φ 2 To find the orientation of axes which shows a maximum shear stress component we set the derivative of σ’ xy to zero. This gives (σ x – σ y ) tan 2φ = – -----------------------2σ xy Problem Set #5 1.050 Solid Mechanics 10/22/04 LL Bucciarelli For our original components, with σx - σy finite and σxy zero, or approaching zero, the tangent of 2φ approaches a negative, very large number. This will be the case when 2φ approaches - 90o. And this, as before is the angle we rotate in the Mohr’s circle plane. Setting 2φ = 90o in the other two relations, we obtain that σ’ x and σ’ y are both equal to (σ x + σ y ) ------------------------- = 12,000psi 2 Problem 5.2 A rigid beam is pinned supported at its left end and at midspan and the right end by two springs, each of stiffness k (force/displacement). The beam supports a weight P at mid span. i) Construct a compatibility condition, relating the displacements of the springs to the rotation of the rigid beam. P L/2 1 2 θ L L/2 k We draw the system in its deflected state. From the geometry, ie. similar triangles, we have , assuming small rotations and displacements L 0 P k ∆1 ∆1 ∆ --------- = -----2- = θ L ⁄2 L ∆2 ii) Draw an isolation of the rigid beam and write out the consequences of force and moment equilibrium. This requires drawing another, quite different diagram - a free body diagram or isolation of the beam showing all the forces acting. P L L/2 Force and moment equilibrium yield ∑F = 0 ∑ M about0 k Ro F1 F2 Ro + F 1 + F 2 – P = 0 F 1 ⋅ L ⁄2 + F 2 ⋅ L – ( P ⋅ L ⁄2 ) = 0 = 0 iii) Using the force/deformation relations for the linear springs, express equilibrium in terms of the angle of rotation of the beam. F 1 = k ⋅ ∆ 1 = ( kL ⁄2 )⋅ θ F 2 = ( k ⋅ L) ⋅ θ so moment equilibrium gives [( kL ⁄2 ) ⋅ L ⁄2 + ( k ⋅ L ) ⋅ L]⋅ θ = P ⋅ L ⁄2 or 1 5 --- ⋅kL ⋅ θ = P ⋅ --2 4 iv) Solve for the rotation, then for the forces of reaction at the three support points. From above, we solve for theta, then back substitute into the force/deformation relations for the springs 2 θ =  --- P ⁄ ( kL) 5  1 F 1 = kL ⁄2 ⋅ θ = --- P 5 2 F 2 = k ⋅ L ⋅ θ = --- P 5 and, with these, the reaction at the pin is (2/5)P (Due Friday, 15 October) 10/22/04 LL Bucciarelli v) Sketch the shear force and bending moment diagram. We make two isolations, one for the portion of the beam left of center, another for right of center: for x > L/2 x Mb P Mb L/2 V V (2/5)P (2/5)P (1/5)P x for x < L/2 Force and moment equilibrium (about x, ccw > 0) gives  --2- P + V +  1-­- P – P = 0  5  5  2--- P + V = 0  5 2 M b –  --- P ⋅ x = 0  5 2 1 L M b –  --- P ⋅ x +  P –  --- P ⋅  x – --- = 0   5  5   2 We plot the shear force and bending moment distribution: P L L/2 k Ro =+2P/5 F1=+1P/5 F2=+2P/5 V +2P/5 -2P/5 Mb (Due Friday, 15 October) 10/22/04 Mb|max = PL/5 LL Bucciarelli Problem 5.3 The figure shows one of the eight ribs of the umbrellas, the unfurling of which we have studied in Class Exercise #7, the results of which have been posted on our MIT server at two places under the heading of “Class Exercises”). We now want to finish our analysis and pick a size, a spring constant, for the spring con­ necting the “pins” C and D. Set up a spread sheet, the final column of which com­ putes the (nondimensional) value of the force P ( as P/kb), the spring change-in-length, call it D, and the position of pin D rela­ tive to the top pin as functions of the angle θ. b cosφ a sinθ A c a cosθ a b b sinφ B d θ φ C Ψ Plot P/kb versus θ P l In this use the following values for lengths: a = 5.50 in. b = 5.25 in. c = 2.50 in. D and x d = 3.50 in. Take the initial value for theta to be 20o , θ0 = 20. deg. and increment by 2 to 5 degrees A plot of P/kb versus theta is shown. The full spread sheet is posted on our MIT server. (Due Friday, 15 October) 10/22/04 LL Bucciarelli