TRUSSES CHAPTER 4 § Trusses are multi-dimension structures composed of one-dimensional elements. We derived a 1-D element in Chapter 3, let’s see if it can be used to solve truss problems. 2-D Stiffness Matrix for a Rod ym 1 2 i j K21 Δ1=1 unit K11 3 4 xm K41 K31 § The (xm,ym) coordinates are local coordinates for each truss element. xm originates at node i and proceeds through node j. Local zm is always normal to the plane of the truss. Local ym is established using right hand rule. Direct Stiffness Method § The stiffness term KMN has the meaning: force at the M degree-offreedom due to unit displacement at the N degree-of-freedom. So first set D.O.F. #1 = 1 unit, and restrain all others. From strength of materials we know: ΔL F AE = and F = ΔL L AE L Thus: ym 0 Δ1=1 unit EA/L 0 EA/L EA K11 = L EA K13 = − L K12 = 0 K14 = 0 xm § Now set D.O.F. #2 = 1 unit, restrain all others (the rod has no bending stiffness). K 21 = 0 K 22 = 0 K 23 = 0 K 24 = 0 § Now set D.O.F. #3 = 1 unit, restrain all others. K 31 = − EA K 32 = 0 L EA K 33 = K 34 = 0 L § Finally, set D.O.F. #4 = 1 unit, restrain all others. K 41 = 0 K 42 = 0 K 43 = 0 K 44 = 0 Here u and v are local deformations defined with respect to the local xm, ym coordinates. § So ui vi EA [K ] = ⎡ 1 L ⎢ 0 ⎢− 1 ⎢⎣ 0 uj vj 0 − 1 0⎤ 0 0 0⎥ 0 1 0⎥ 0 0 0⎥⎦ § We have developed [K] in a local x-y coordinate system. We must transform [K] to global coordinates such that the direct stiffness assembly method can be used to form [K] for the entire structure in one coordinate system. x 3 4 y 4 Y 3 ⇒ 2 1 1 2 Local x-y Global X-Y Project vector X1 onto the x & y axes: x1 = X1 cos θ y1 = -X1 sin θ y Y x1 y1 X Project vectorY1 onto x & y axes: y1 = Y1 cos θ x1 = Y1 sin θ y x X1 θ X Y Y1 y1 x1 x θ X § In matrix form, we have: ⎧x1 ⎫ = ⎡ cos θ sin θ ⎤ ⎧X1 ⎫ ⎨ y ⎬ ⎢− sin θ cos θ⎥ ⎨ Y ⎬ ⎩ 1 ⎭ ⎣ ⎦ ⎩ 1 ⎭ [t ] where the transformation matrix [t] can transform a vector VXY with components X1 and Y1 in global coordinates (XY) to a vector v Vxy in a local coordinate system (xy) with components x1 and y1. § Define Displacement of a node ⎧δ x ⎫ Δ ⎨δ ⎬ = {δ}= in local coordinates ⎩ y ⎭ Displacement of a node ⎧Δ X ⎫ = {Δ}Δ = ⎨Δ ⎬ in global coordinates ⎩ Y ⎭ ⎧f x ⎫ Δ Forces at a node in = ⎨f ⎬ = {f } local coordinates ⎩ y ⎭ ⎧Fx ⎫ Forces at a node in Δ ⎨F ⎬ = {F}= global coordinates ⎩ y ⎭ Y y x δy δx ΔY ΔX X § The work performed by given forces must be the same regardless of the coordinate system chosen: 1 T 1 T {f } {δ} = {F} {Δ} 2 2 § Note: {δ} = [t ]{Δ} is the Displacement Transform and the Force Transform is: {f } = [t ]{F} § Also we developed K in local coordinates: {f } = [K ]{δ} § We seek: {F} = [K]{Δ} in global coordinates. § Substituting into our work equation yields: 1 1 T T T {δ} [K] {δ} = {Δ} [K]{Δ} 2 2 § Originally we defined our [t] to transform the displacement or forces at a node, since our force and displacement vectors for our element involve 2 nodes we must expand our transformation matrix: 0 0 ⎤ ⎡ cos θ sin θ ⎢ − sin θ cos θ 0 0 ⎥ ⎥ [T ] = ⎢ 0 cos θ sin θ ⎥ ⎢ 0 ⎢⎣ 0 0 − sinθ cos θ ⎥⎦ § So: T T T {Δ} [T] [K][T]{Δ} = {Δ} [K]{Δ} § And thus: [K] = [T]T [K][T] will transform our stiffness matrices in local coordinates to the global coordinate system. § This product can be determined symbolically as: ui vi uj vj ⎡ C2θ CθSθ −C2θ −CθSθ ⎤ ⎢ ⎥ S2θ −CθSθ −S2θ ⎥ EA ⎢ CθSθ ⎡ K ⎤ = ⎥ ⎣ ⎦ L ⎢ 2 2 ⎢ −C θ −CθSθ C θ CθSθ ⎥ ⎢ ⎥ 2 2 CθSθ S θ ⎥⎦ ⎢⎣ −CθSθ −S θ § The stiffnesses are now defined in terms of global DOF ui , vi , uj and vj and can be included into the global [K] for the structure. Example EA = 100 lb 3 1 lb For all elements φ = 60o 3 1 φ φ 2 φ 2 1 100" § For all 3 members: ⎡ 1 [K]xy = ⎢ 0 local ⎢ ⎢−1 ⎢⎣ 0 0 −1 0 ⎤ 0 0 0 ⎥ ⎥ 0 1 0 ⎥ 0 0 0 ⎥⎦ § For rod 1 Y y 1 1 2 θ = 0° [K]1 = x u1 v1 u2 ⎡ 1 0 −1 ⎢ 0 0 0 ⎢ 1 ⎢−1 0 ⎢⎣ 0 0 0 X v2 0 ⎤ 0 ⎥ ⎥ 0 ⎥ 0 ⎥⎦ § For rod 2 x 3 Y 2 θ =120° 2 y u2 [K] 2 = X v2 u3 v3 .433⎤ ⎡ .25 −.433 −.25 ⎢−.433 .75 .433 −.75 ⎥ ⎢ ⎥ .433 .25 −.433⎥ ⎢−.25 ⎢⎣ .433 −.75 −.433 .75 ⎥⎦ § For rod 3 x 3 Y 3 y 1 [K] 3 = θ = 60° X u1 v1 u3 v3 .433 −.25 −.433⎤ ⎡ .25 ⎢ .433 .75 −.433 −.75 ⎥ ⎢ ⎥ .433⎥ ⎢−.25 −.433 .25 ⎢⎣−.433 −.75 .433 .75 ⎥⎦ § What if we chose node 3 as the origin rather than node 1 on rod 3? 3 Y y 3 θ = 240° X 1 x u3 v3 u1 v1 .433 −.25 −.433 ⎤ ⎡ .25 ⎢ .433 ⎥ .75 − .433 − .75 ⎥ ⎡⎣K ⎤⎦ = ⎢ 3 ⎢ −.25 −.433 .25 .433 ⎥ ⎢ ⎥ .75 ⎦ ⎣ −.433 −.75 .433 § Note the indices on the degrees of freedom have changed. Also note Ku3,v1 from the previous slide is equal to Ku3,v1 on this slide, thus after assembly the structure [K] will be the same. § Now use direct stiffness assembly to form [K] for the entire truss in global coordinates! u1 v1 u2 v2 u3 v3 −1 0 −.25 −.433⎤ ⎡ 1.25 .433 ⎢ .433 .75 0 0 −.433 −.75 ⎥ ⎢ ⎥ 0 1.25 −.433 −.25 .433 ⎥ ⎢ −1 [K] = ⎢ 0 0 −.433 .75 .433 −.75 ⎥ ⎢ ⎥ .5 0 ⎥ ⎢ −.25 −.433 −.25 .433 ⎢⎣−.433 −.75 .433 −.75 0 1.5 ⎥⎦ {Δ} = ⎧ u1 ⎫ ⎪ v1 ⎪ ⎪⎪u ⎪⎪ 2 ⎨ v ⎬ ⎪ 2 ⎪ u ⎪ 3 ⎪ ⎪⎩ v3 ⎪⎭ ⎧0 ⎫ ⎪0 ⎪ ⎪ ⎪ ⎪0 ⎪ {F} = ⎨ ⎬ ⎪0 ⎪ ⎪1 ⎪ ⎪ ⎪ ⎩0 ⎭ § After constraining u1, v1 and v2 we can solve: ⎡ 1.25 − .25 .433⎤ ⎧⎪u 2 ⎫⎪ ⎧⎪0⎫⎪ 0 ⎥ ⎨ u 3 ⎬ = ⎨1⎬ ⎢− .25 .5 1.5 ⎥⎦ ⎪⎩ v3 ⎪⎭ ⎪⎩0⎪⎭ ⎢⎣ .433 0 ⎧⎪u 2 ⎫⎪ ⎧⎪ 0.5 ⎫⎪ ⎨ u 3 ⎬ = ⎨ 2.25 ⎬ in ⎪⎩ v3 ⎪⎭ ⎪⎩− .1443⎪⎭ Member Loads and Stresses § We now know {Δ} for our truss. How do we compute loads and stresses? Procedure § Extract the deformations for the nodes i and j from {Δ} for a given element. Now compute local deformations using [T ] : {δ} = [T][Δ] ⎧u xyi ⎫ ⎡ ⎤ ⎧u XYi ⎫ ⎪⎪ v ⎪⎪ ⎢ ⎥ ⎪⎪ v ⎪ xyi XYi ⎪ ⎨ u ⎬ = ⎢T ⎥ ⎨ u XYj ⎬ ⎪ xyj ⎪ ⎢ ⎥ ⎪ ⎪ ⎪⎩ v xyj ⎪⎭ ⎣ ⎦ ⎪⎩ v XYj ⎪⎭ § For member 1 ⎧ u1 ⎫ = ⎡1 0 0 ⎪⎪ v ⎪⎪ ⎢0 1 0 1 ⎢ ⎨ ⎬ u ⎢0 0 1 ⎪ 2 ⎪ ⎪⎩ v2 ⎪⎭ xy ⎢⎣0 0 0 0⎤ ⎧ 0 ⎫ ⎧ 0 ⎫ ⎪⎪ 0 ⎪⎪ 0⎥ ⎪⎪ 0 ⎪⎪ ⎥ ⎨ ⎬ = ⎨ ⎬ 0⎥ ⎪.5⎪ ⎪.5⎪ 1⎥⎦ ⎪⎩ 0 ⎪⎭ XY ⎪⎩ 0 ⎪⎭xy § We know: [K]xy{δ}={f}: ⎡ 1 ⎢ 0 ⎢− 1 ⎢⎣ 0 0 − 1 0⎤ ⎧ 0 ⎫ ⎧− .5⎫ 0 0 0⎥ ⎪ 0 ⎪ = ⎪ 0⎪ = {f } local 0 1 0⎥ ⎨.5⎬ ⎨ .5⎬ 0 0 0⎥⎦ ⎪⎩ 0 ⎪⎭ ⎪⎩ 0⎪⎭ 1 .5 1 2 .5 § So element 1 is subjected to .5 lbs of tension and a tensile stress . 5 of σ = PSI. A § Follow identical procedure for elements 2 and 3 . From a design perspective, we now have: 1. Deformations 2. Member loads (we should check buckling) 3. Stresses! Temperature Effects and Mismatch § In local coordinates, a truss element is a 1-dimensional element. As discussed earlier: ⎧−1⎫ ⎪⎪ 0 ⎪⎪ {θ}xy = E e Ae α e ΔT ⎨ ⎬ ⎪ 1⎪ ⎪⎩ 0 ⎪⎭ are the loads imposed by temperature at nodes in the local element coordinate system. Using our 2-D force transformation: ⎧−Cθ⎫ ⎪ −Sθ⎪ ⎪ ⎪ T {θ}XY = [T ] {θ}xy = Ee Ae αe ΔT ⎨ ⎬ ⎪ Cθ⎪ ⎪⎩ Sθ⎪⎭ We now have forces in global coordinates that can be assembled with the systeme force vector: F = ∑ f + T e + θe + P (e ) § We realize imperfection always exists to some degree in the manufacturing process. What if some of the truss members were not of the design lengths? This can be modeled by letting: Δ L ε0 = Le where ΔL is the error in length. This can now replace the thermal strain αΔT as: ⎧ −1⎫ ⎪ ⎪ 0 ΔL ⎪ ⎪ {θ}xy = E e Ae ⎨ ⎬ L e ⎪ 1⎪ ⎪⎩ 0 ⎪⎭ § This is the simple truss problem we worked earlier. EA = 100 lb 3 1 lb For all elements φ = 60o 3 1 φ φ 2 φ 2 1 100" § What would be the effect of heating one element or by having a manufacturing error of 0.01” in one of the 100” members? § The answer is very little. This truss is internally determinate. Heating one member or introducing some manufacturing will change the total deformations but the internal loads will be due only to the 1 lb force. § Thus the truss must be internally indeterminate before an expansion or contraction of a element will affect the internal loads in the truss. Symmetry § Many components we deal with have some degree of symmetry. § If the components are symmetric in terms of structure, loads, and constraints it may be possible to reduce the size of the problem we must solve. § If there is one axis of symmetry you may be able to model only half of a problem, if there are two axes of symmetry you may half to model only a quarter of the problem,etc. § You must watch to ensure that constraints are applied properly and that any members that are shared between the two symmetric pieces of structure are treated properly. § Problem 4.10 is good example: 30,000 lb 20 ft 20 ft (10) (10) (10) 15 ft (12.5) (12.5) (10) (10) (12.5) (12.5) 15 ft Axis of Symmetry § Symmetric Structures – share the symmetric load – share the stiffness of shared members – nodes on the axis of symmetry are allowed to displace only in the direction of the axis of symmetry. 15,000 lb 20 ft 4 5 (10) 3 3 15 ft (12.5) (5) 4 (10) 1 15 ft (12.5) 2 2 1 § Dealing with Constraints which are not aligned with the global coordinate system. 30,000 lb This is not a symmetric structure due to the constraints! 1 d 30o u1 = d cos30° v1 = d sin30° u1 v1 u1 v1 =d = or − =0 cos30° sin30° cos30° sin30° This is a multi-point constraint equation. Enforce it using Penalty or the Lagrange method. Space Trusses § 3D trusses are a simple extension of 2D planar trusses. § Each element has stiffness only on the local x axis defined by the 2 nodes which are coupled by the element. § In 3D global coordinates, the element will contribute to the structures stiffness in the X, Y and Z dimensions. ⎡ l2 lm ln −l2 −lm − ln ⎤ ⎢ ⎥ 2 2 ⎢ lm m mn −lm −m −mn ⎥ ⎢ ⎥ 2 2 mn n − ln −mn −n ⎥ Ee A e ⎢ ln K global = ⎢ ⎥ 2 Le ⎢ −l2 −lm − ln l lm ln ⎥ ⎢ ⎥ 2 2 −mn lm m mn ⎥ ⎢ −lm −m ⎢ 2 2 ⎥ − ln − mn − n ln mn n ⎣ ⎦ § You are given code with your text to solve 3D truss problems. § Cessna 421C twin Continental Lycoming 375 hp piston engines § Cessna 425 twin Pratt and Whitney 450 hp turbines § Example: A space truss for a turbine for a turboprop aircraft. § Analysis – Flight loads • Up gust, down gust, etc. • Thermal loads? • Seizure? – Landing loads – Manufacturing effects