2.080J Structural Mechanics: Solutions to Problem Set 10

Lecture 12 Fundamental Concepts in Structural Plasticity Problem 12-1: Stress yield condition Consider the plane stress yield condition in the principal coordinate system a) 1 , 2 between the Von-Mises and Tresca yield Calculate the maximum difference condition b) Show the locations on the plane stress graph where the maximum difference occurs Problem 12-1 Solution: Through observation, the maximum difference occurs at either A or B, as indicated in the below figure Mises yield condition can be expressed as 2 1 2 1 2 2 2 y 1 The difference at A At Tresca yield surface Tresca 2A y At Mises Yield surface 1 2 1 Substitute 1 1 2 y into Mises yield condition, we have 1 2 2 1 2 y 2 y Mises 2A The difference 2 2 2 y 1.15 y 1.15 y at A is Mises A Tresca 2A 2A 0.15 A The difference y y 0.15 y y at B Tresca yield surface in the second quadrant is 2 2 1 2 1 2 2 y The distance from Tresca yield surface B to the origin is d BTresca 2 1 2 2 2 2 y Mises yield surface in the second quadrant is 2 1 2 1 2 2 2 y 2 Also, at Mises yield surface 1 0 2 Combining the above two equations, we have the coordinate of Mises yield surface B 1 3 1 y , 1 3 2 y The distance from Tresca yield surface B to the origin is d B Mises The difference 2 1 d B Mises d BTresca B A y at B is B Compare 2 3 2 2 and B 2 2 0.11 2 3 y 0.15 0.11 y y , the maximum difference max y occurs at A, where y 3 Problem 12-2: In the early twenties, passenger cars did not have electric starters. crank to start the engine. is shown below. The crank is a solid rod of radius r and the geometry of the crank 3 Define the equivalent stress by stress produced by bending and a) The driver had to use a 3 2 where sigma is the is the shear stress due to torsion. Find the relationship between the maximum equivalent stress in the crank and the magnitude of the crank load P. (Use the principle of superposition) b) Derive a formula for the elastic deflection under the load P in the direction of the load P. Problem 12-2 Solution: a) Determine the magnitude of the load P causing first yield in the mostly stressed point on the crank We will use the Mises yield condition which simplified to 2 o from the moment, and We calculate 2 3 2 from the torsion Mr where I I Tr where J J r4 4 r4 2 Because we are asked to find P causing first yield at the maximum stressed point, we need to find max and max From physics we can see that the M max and Tmax will occur at the support 4 Pl M max Tmax M max r Plr I r4 4 Tmax r PRr J r4 2 max PR 4 Pl r3 2 PR r3 max max 2 2 max o 4 Pl r3 2 max 2 2 PR 3 r3 2 4 P 2 4l 2 3R 2 r 2 o P 3 2 6 o 2 r2 1 l 4 r 2 R 3 r 2 b) Calculate the elastic deflection in the direction of the load P Beam AB Beam CD MA M T P L R PR 0 PL 0 Mo M M M Px1 at 0 Px1 x1 0 R Mx Mx PL Px3 0 PL Px3 at 0 x3 L R 5 Beam CB M T P R x2 at 0 PR at 0 x2 x2 R R In summary, here are the moment distributions Torsion is constant along Beams AB and BC: T PR Beam CD: T 0 6 M2 dx 2EI U bending l 0 2 Px1 dx1 2 EI P 2 x13 2 EI 3 2 P R x2 R dx2 2 EI 0 R 2 Rx2 2 2 2 R x2 0 P L x3 L R 2EI 0 R x23 3 2 L x3 0 2Lx32 2 2 dx3 x33 3 L R 0 after lengthly algebra P2 R3 6EI L3 P2 R3 6EI U bending L3 T2 dx 2GJ U torsion R 0 2 PR dx2 2GJ PR 2GJ L R 0 2 PR dx2 2GJ 2 R l R 2 U torsion PR L 2GJ Use Castigliano’s Theorem to calculate the deflection where the load is applied, in the direction of the load U P U torsion U bending P 2 P R3 P 6 EI P R3 L3 3EI 2 3 L PR L 2GJ R2 L GJ 7 r4 and J 4 Recall that for solid circular cross-section I Also, G E 1 2 P R3 L3 R2 L GJ 3EI R2 L 1 R 3 L3 P 3E r 4 4 Note, if R r4 2 E r4 4 4 P R 3 L3 Er 2 3r 2 R2 L 1 r2 4 P R3 L3 Er 2 3r 2 R2 L 1 r2 0 , we have a cantilever beam whose deflection is PL3 3EI 8 Problem 12-3: Consider a thin-walled tube of radius r, thickness t and length L. The tube is fully clamped on one end and free on the other. It is twisted at the free end by an axial torque T. (a) Derive an expression between torsional moment and the relative end rotation. (b) Assuming L/R=10 and R/t=10, give the expression for the critical torque that will cause the tube stress to reach yield in shear. Problem 12-3 Solution: (a) Express T as a function of T x rdA d dA dx Gr 2 d r 2 dA dx d GJ dx G x 0 0 max x L T x GJ L (b) The distribution of torsional shear stress can be expressed as x Tr J Where 9 J ro 4 2 ri 4 ro 4 2 ro t 4 Given r t J 10 ro 2 4 t r 10 ro ro 10 0.071 ro 4 J Tr 0.071 ro 4 x 4 5.8T ro 3 Plane stress yield condition states: 2 xx 2 xx yy yy 3 xx yy 0 2 xy y In pure shear y xy 3 5.8T ro3 x y 3.2 y 3 T ro3 10 Problem 12-4: Consider the following key ring problem a) b) c) d) Derive the out of plane displacement where the force is applied. Determine the magnitude and distribution of the bending stress and the shear stress along the ring. Find the location of the maximum equivalent plastic strain. Determine the critical opening force for which first yield would occur. Consider the plane stress Problem 12-4 Solution: a) Derive the out of plane displacement where the force is applied. We can use Castigliano’s Theorem to calculate the displacement where the force is applied. In addition, to calculating the strain energy contribution from the moment, we must also account for the contribution from torsion U 1. 1 M2 dx 2 EI 1 T2 dx 2 GJ Calculate M, T From the geometry above, we can see that M Pa T Pb where a b R sin R R cos 11 So M PR sin T 2. PR 1 cos Calculate strain energy U 1 M2 dx 2 EI 0 1 T2 dx 2 GJ 2 1 PR sin 2 EI Rd 2 PR R 2 EI 0 PR R GJ 0 PR R GJ 0 3 GJ Apply Castigliano Theorem, the out of plane displacement where the force is applied is U P PR 3 1 EI 3 GJ Determine the magnitude + distribution of bending stress + shearing Mz I Tr J c) 2 sin 2 4 2 P 2 R3 1 2 EI U b) 2sin 2 PR R 2 EI 3. Rd 2 sin 2 4 2 2 1 PR 1 cos 2 GJ PR sin r r4 4 PR 1 cos r4 2 4 PR sin r3 r 2 PR 1 cos r3 Find the location of maximum plastic strain Recall that relation of plastic strain and plastic stress is n A The maximum plastic strain occurs at location of the maximum Mises equivalent stress, 1. Calculate the Mises equivalent stress The stress on the outer surface is planar, so we will use the plane stress condition 2 xx Where o 2 xx yy yy 3 2 xy 2 o is Mises equivalent stress 12 In our case, 0 yy So 2 2 3 o 2 2 4 PR sin r3 2. 3 Calculate the Maximum Mises equivalent stress omax will occur when we have setting it equal to 0, find d d 2 8 d Set d PR r3 omax . We can find the maximum by taking the derivative, 2 where 4 PR r3 o 2 omax 2 2 PR 1 cos r3 occurs and then go back to get omax 2 2sin cos 3 2 PR r3 omax 2 2 1 cos sin 2 sin cos 3 sin cos 2 o 0 , we have 3 0 0, 0 , where Minimum Mises equivalent stress occurs at Maximum Mises equivalent stress occurs at d) , where 0 o o The critical opening force for which first yield would occur when y o 4 3 PR r3 o y , that is 4 3 Pcr R r3 So we have Pcr r3 4 3R y 13 Problem 12-5: Plasticity Consider the four –point bending of a beam of length L. The beam is loaded by two rollers parted by a distance of L/3. The material of the beam is rigid, perfectly plastic. Determine the load capacity of the beam under two different end conditions. a) Write an expression for fully plastic bending moment of a beam of rectangular cross-section b h . b) Ends of the beam are simply supported c) Ends of the beam are clamped Problem 12-5 Solution: y h 2b a) Mo b) Moment distribution M M M 4 Px 2 Pl 6 P l x 2 0 x l x 3 2l x l 3 We can assume that within M Mo l 3 2l 3 y l 3 x 2l 3 h 2b 4 14 The rate of change of internal energy is 2l 3 U l 3 2l 3 MKdx Where geometrically 2 o l 3 wo l 3 Mo 6 d dx dx Mo x 2l 3 x l 3 Mo o wo l Rate of work balance Mo o Mo 6 wo l Pc c) 2 Pc wo 2 Pc wo 6M o l If the ends are clamped, we have two plastic hinges at the supports 1 wo l 3 2 3 wo l Rate of work balance i M oi Mo o Mo M o 12 Pc o 2 1 Mo wo l Pc wo 2 2 Pc wo Pc wo 12M o l 15