Solutions to Exercises omework Exercises for Chapter TL Bar Elements: Truss Analysis Note: not all solutions fully worked out. EXECISE. The strain measures (engineering strain versus GL strain) are different. Since a snapping truss actually goes through large strains (of about 5% if α = ) that difference is reflected in limit-point s that differ by noticeable amounts. EXECISE. Three choices for the optimality criterion are studied below. For all cases volume of truss material, cross section and geometric dimensions are linked by the relations V = A L = A (S/) +, = V /(A ) (S/), A = V / (S/) + (E.) We recall the results (.9) for the limit and bifurcation points, in which for convenience we will take the + sign for the primary path critical points: 6EA L =+, B =+ EA S S. (E.) ( + S ) ( + S ) In all cases we eventually set S = and E = to simplify expressions. Since is dimensionless, this makes no difference in the results. ClearAll[Em,A,V,S,,uX,uY,pX,pY,B,L]; S=; Em=; A=; =Sqrt[-(A^*S^)+V^]/(*A); L=-(6/(*Sqrt[]))*A*Em*^/(^+S^)^(/); B=-*Sqrt[]*A*Em*S^*Sqrt[*^-S^]/(^+S^)^(/); Print["L/V=",L/V," B/V=",B/V]; style={{absolutethickness[],gbcolor[,,]}, {AbsoluteThickness[],GBColor[,,]}}; Plot[{-L,-B},{V,,6},Plotange->All,AxesOrigin->{,}, Frame->True,ImageSize->,PlotStyle->style]; Plot[{-L/V,-B/V},{V,,6},Plotange->All,AxesOrigin->{,}, Frame->True,ImageSize->,PlotStyle->style]; sol=simplify[solve[d[l/v,v]==,v]]; Print[sol,N[sol]]; sol=simplify[solve[d[b/v,v]==,v]]; Print[sol,N[sol]]; =(Sqrt[-(A^*S^)+V^]/(*A))/.V->; Print["=",," = ",N[]]; Figure E.. Mathematica script to solve Exercise 9., criterion (I)...5..5..5 B L 5 6.6.5.... L V B V 5 6 Figure E.. esults from running the script of Figure E.. 9
Chapter : TL BA ELEMENTS: TUSS ANALYSIS (I) Maximize load capacity max = min( L, B ) with respect to unit volume. The area A is varied as per the last of (E.). Then L = 6E V /( ( +S ) ) and B = ES V S /( +S ). Both attain their maxima with respect to at exactly the same tan α = /( S) =, hence α opt = 6. This happens to be the angle at which L and B coincide. This is called a simultaneous failure mode design and is typical of optimization criteria that involve critical loads. The Mathematica script shown in Figure E. solves this problem. unning the script gives the results shown in Figure E.. The results include plots of ( L, B ) as well as ( L /V, B /V ) with respect to. The critical load coalescence is clear in both plots. That of ( L /V, B /V ) versus shows that the load capacity is maximized at the coalescence (horizontal tangent there). ClearAll[Em,A,S,,uX,uY,pX,pY,B,L]; S=; Em=; A=; V=A**Sqrt[(S/)^+^]; L=-(6/(*Sqrt[]))*A*Em*^/(^+S^)^(/); B=-*Sqrt[]*A*Em*S^*Sqrt[*^-S^]/(^+S^)^(/); Print["L=",L," B=",B]; style={{absolutethickness[],gbcolor[,,]}, {AbsoluteThickness[],GBColor[,,]}}; Plot[{-L,-B},{,,},Plotange->All,AxesOrigin->{,}, Frame->True,ImageSize->,PlotStyle->style]; sol=simplify[solve[d[l,]==,]]; Print[sol,N[sol]]; sol=simplify[solve[d[b,]==,]]; Print[sol,N[sol]]; Figure E.. Mathematica script to solve Exercise 9., criterion (II)..5..5..5 L B.5.5.5 Figure E.5. esults from running the script of Figure E.. (II) Maximize load capacity with respect to V (or equivalently ) while keeping S and A fixed The cross section area is given in terms of V and by A = V /( (S/) + ). Inserting S = and E = one gets 6 L = ( + ), / B = 8, (E.5) ( + ) / Maximizing max = min( L, B ) with respect to yields = 7/, or tan α = 7/ =.878 and α opt 6.9. Failure occurs by bifurcation. The Mathematica script shown in Figure E. solves this problem. unning the script gives the results shown in Figure E.5. The results include a plot of ( L, B ) versus. ere the maximum load capacity occurs slightly after coalescence. (III) Maximize the load capacity per unit volume of material, with respect to the total volume. Mathematically we maximize max /V with respect to V (or ), while keeping S and A fixed. Then
Solutions to Exercises ClearAll[Em,A,V,S,,uX,uY,pX,pY,L,B]; S=; Em=; V=; A=V/Sqrt[(S/)^+^]; L=-(6/(*Sqrt[]))*A*Em*^/(^+S^)^(/); B=-*Sqrt[]*A*Em*S^*Sqrt[*^-S^]/(^+S^)^(/); Print["L=",L," B=",B]; style={{absolutethickness[],gbcolor[,,]}, {AbsoluteThickness[],GBColor[,,]}}; Plot[{-L,-B},{,,},Plotange->All,AxesOrigin->{,}, Frame->True,ImageSize->,PlotStyle->style]; sol=simplify[solve[d[l,]==,]]; Print[sol,N[sol]]; sol=simplify[solve[d[b,]==,]]; Print[sol,N[sol]]; Figure E.6. Mathematica script to solve Exercise 9., criterion (III).5..5. L B.5.5.5.5 Figure E.7. esults from running the script of Figure E.6. = V A S /(A ). Inserting S = and E = one gets L V = (V )/ V, B = 8 (V ) V, (E.6) V Note that V to get real values. Maximizing max /V = min( L /V, B /V ) with respect to V yields V =, whence = sqrt and α o pt = 6. This is the same result of criterion (a). The Mathematica script shown in Figure E.6 solves this problem. unning the script gives the results shown in Figure E.7. The results include a plot of ( L, B ) versus. ere the maximum load capacity occurs again at coalescence, as in citerion (I). Credit is given if the optimality criterion (whatever is chosen) is clearly stated and answers for that criterion are correct. Some points deducted if only L is checked against and B ignored,
Chapter : TL BA ELEMENTS: TUSS ANALYSIS EXECISE. The internal force expressions are given in the Exercise statement. On replacing the numerical data of Figure E.7 into the internal force expressions yields px = u X ( + u X + + u Y ) 5 5, py = ( + )(u X + ( + )) 5. 5 The equilibrium paths are obtained by solving the residual equations p X = and p Y = f ref = / for u X and while keeping as independent variable. Solving this system gives five segments. These are identified as and PP for the primary (fundamental) path, and, SP and SP for the secondary path. Their expressions are (purple) : = + c /, u X = 5 /, PP (green) : = c /, u X = 5 /, (black) : =, u X = 6/ + (9 5 + c ) / 6 / (9, 5 + c ) / SP (red) : =, u X = 6/) ( + j ) ( j )(9 5 + c ) / 6 / (9 5 + c ) /, SP (blue) : =, u X = 6/ ( j ) ( + j )(9 5 + c ) / 6 / (9 5 + c ) /. (E.7) in which j =, c = 8 + 5 and c = 6 5. The color indicated after the name in (E.7) is that used in Figures E.8 and E.9 for equilibrium paths, which are drawn as solid lines. The paths of (E.7) are plotted in Figure E.8 in various configurations as identified in plot titles. The D plot of Figure E.8(a) on state space (u X, ) plainly shows that the secondary path, composed by segments SS, SS, and SS, lies on the = plane, which corresponds to a fully flatenned truss. Figures E.8(b,c) show paths projected on (u X,)and (,), respectively. Figure E.8(b) suggests two limit points at the junctions of with PP, and of PP with PP, respectively; this is corroborated by Figure E.8(d). Figure E.8(c) suggest two limit points on the primary path at the junctions of and PP. The three-dimensional control-state plots in Figure E.8(e,f) clarify the foregoing suggestions further and in addition show two bifurcation points at the junctions of the primary and secondary paths, which coalesce with limit points. Next we proceed to verify the foregoing visual guesses analytically. Differentiating p X and p Y gives the tangent stiffness matrix K = [ + u ] X + + u Y u X ( + ) 5 u X ( + ) 8 + u X + + u. Y The determinant is det(k) = (6 + u X + 56 + + + u X (5 + + u Y ))/5. Solving the quartic equation det(k) = for in terms of u X yields four expressions: u c Y = c c, u c Y = + c c, u c Y = c + c, u c Y = + c + c, in which c = 5u X and c = +u X. These are potential critical point locations, and consequently called critical point loci. These are plotted in Figure E.9(a) over the (u X, ) state space. As can be observed, segment coalesce to form two closed curves. The intersection of these curves with equilibrium paths will define the actual critical points. This is done graphically in Figure E.9(b), in which Figures E.9(a) and
(a) Equilibrium paths projected on the (u X,) state space.5.5.5.5 PP PP u X (c) Equilibrium paths projected on the (u,) plane Y PP.5.5.5.5 (e) Three-D plot of equilibrium paths in the (u X,,) control-state space. Viewpoint: (5,, 5) PP SP PP PP SP SP,SP,SP (overlaid) SP u X u X Solutions to Exercises (b) Equilibrium paths projected on the (u, ) plane X (d) Secondary paths projected on the = plane u X SP (f) Three-D plot of equilibrium paths in the (u X,,) control-state space. Viewpoint: (5, 5, 5) SP PP,PP (overlaid) SP SP SP SP SP SP Figure E.8. Equilibrium paths for Mises truss under horizontal load. u X uy.5.5.5.5 (a) Critical point loci (a.k.a. critical point surfaces) Bifurcation pt loci Limit pt loci u X u.5 Y.5.5.5 (b) Superimposing equilibrium paths and critical point loci in state space L+B L L+B u X L Figure E.9. Geometric determination of critical points.
Chapter : TL BA ELEMENTS: TUSS ANALYSIS E.8(a) have been combined. Obviously intersections can occur only for =. Setting this value in the tangent stiffness matrix simplifies it to a diagonal form: ˆK = 5 [ + u X + u X ] def = [ ] ˆK XX. ˆK YY Setting u X = / ±.5 so that ˆK XX vanishes makes ˆK singular, with null eigenvector z = [, ] T. The corresponding load factors are cr = /5/5 = ±.977. Setting u X = = ± so that ˆK YY vanishes makes ˆK singular, with null eigenvector z = [, ] T. The corresponding load factors are cr = /(5 5) =±.5777. Since the incremental load vector is q = r/ = [/, ] T, the critical point indicators are q T z = / (limit point) and q T z = (bifurcation). This explains the labels in Figure E.9(a). Four of the six critical point labels in Figure E.9(b) are thereby verified. The fact that both bifurcation points are also limit points cannot be discerned by the q T z indicator, since that is valid only for isolated critical points. The practical value of this Exercise is limited since the occurrence of any critical point requires the truss to be fully flattened. The vertical load case is far more instructive.