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3. (30 marks) The bar shown in the figure below is cylindrical in shape. When laying flat on the ground, it has a length of 2m. When supported from a ceiling mount, the bar wil deflect under its own weight. The bar is made from HDPE with E=10 MPa and a density of 0.95 g/cm?. By breaking the bar into 8 finite elements, each with a free length of 250mm, determine the displacement of the bar tip, and the reaction force at the ceiling support: a. Due to the self weight of the bar only, and b. Ifa 1 kN load is applied centrally at the tip. (Note: Matlab may be used to solve equations, please submit a basic code with your submission. All “process calculations" should be done by hand) © 100mm «> "Flat length" = 2m PowerPoint Presentation MECH4410 – Mechanics of Solids 2 and FEA Bar Elements and 2D Transformations Peter Robinson Lecturer, School of Engineering The University of Newcastle 2 Lecture Objectives • Present the equations used to determine the behaviour of bar elements • Derive the stiffness matrix of the bar element using the direct method • Interpretation of a stiffness matrix • Define “shape function” • Express the shape function matrix, N, of a 2-noded bar element • Express the strain-displacement matrix, B, of a 2-noded bar element • Derive the stiffness matrix of the bar element using the formal procedure • Able to manually solve for axial displacements and stresses for a simple assembly of bar Elements • Determine the work equivalent loading for a bar subject to a distributed load • Transforming the stiffness of a bar element in 2D space. 3 Bar or Truss Elements • A 1D element that can only transmit axial force (no bending). • Consider the bar with a cross-sectional area A, modulus E and initial length L • Elastic behaviour – consider analogies with springs. ???? = ?????? = ?? ???? ???? = ?? ?? • Force in the bar: ?? = ???? ???? ???? = ???? ??2 − ??1 ?? 4 Bar or Truss Elements • The bar cannot sustain shear force or bending moment. • Any effect of transverse displacement is ignored. • No intermediate applied loads (only applied at the ends). 5 Global Stiffness Matrix – Direct Method • Consider a bar with uniform cross-section of length L as shown below: • At node 1: ??1 + ???? = 0 • At Node 2: ??2 − ???? = 0 • So ???? ?? ??1 − ??2 = ??1 ???? ?? ??2 − ??1 = ??2 • In matrix form: ???? ?? 1 −1 −1 1 ??1 ??2 = ??1 ??2 ?? ≡ ???? ?? , ?? −??−?? ?? 6 Interpretation of a Stiffness Matrix • What do the numbers in a stiffness matrix mean? A column of k is a vector of nodal loads that must be applied to the element at its nodes to sustain a deformation state in which the corresponding degree of freedom (d.o.f.) has unit value and all other d.o.f. are zero. 7 Interpretation of a Stiffness Matrix • What do the numbers in a stiffness matrix mean? A column of k is a vector of nodal loads that must be applied to the element at its nodes to sustain a deformation state in which the corresponding degree of freedom (d.o.f.) has unit value and all other d.o.f. are zero. • An example: If u1 = 1 and u2 = 0 then ???? ?? 1 −1 −1 1 1 0 = ??1 ??2 ,ℎ???????? ??1??2 = ???? ?? 1 −1 = ?? −?? 8 Interpretation of a Stiffness Matrix • Example from exercises: • Therefore rows and columns sum to zero: static equilibrium. 9 Global Stiffness Matrix – Formal Method • Direct method works for simple systems only. • For complex systems, the stiffness matrix is derived by stating the work done by external loads is stored in the element as elastic strain energy. • General formula is given by K = ∫???????????? B is the strain-displacement matrix E is the material matrix dV is the increment of the element volume V • To obtain B we first need to know the relationship between the displacement of an arbitrary point on the bar and the displacement of the nodes - this relationship defines what is happening within a body, based on surface movement. Best to show with an example. 10 Global Stiffness Matrix – Formal Method • Case 1: bar element subject to axial loading • Move node 1 by +1 unit to the right, and hold node 2 fixed. • How far does a point x = L/2 move? • How far does a point x = 3L/4 move? 11 Global Stiffness Matrix – Formal Method • Case 1: bar element subject to axial loading • Move node 1 by +1 unit to the right, and hold node 2 fixed. • How far does a point x = L/2 move? Ans: ½ unit • How far does a point x = 3L/4 move? Ans: ¼ unit 12 • This allows a relation to be developed, showing the displacement along the bar, with respect to movement of node 1: ?? = ?? − ?? ?? ??1 This is called a ‘shape function’ Global Stiffness Matrix – Formal Method • Case 1: bar element subject to axial loading • Move node 1 by +1 unit to the right, and hold node 2 fixed. 13 Global Stiffness Matrix – Formal Method • Case 2: bar element subject to axial loading • Move node 2 by +1 unit to the right, and hold node 1 fixed. • How far does a point x = L/2 move? • How far does a point x = 3L/4 move? 14 Global Stiffness Matrix – Formal Method • Case 2: bar element subject to axial loading • Move node 2 by +1 unit to the right, and hold node 1 fixed. • How far does a point x = L/2 move? Ans: ½ unit • How far does a point x = 3L/4 move? Ans: ¾ unit 15 • This allows a relation to be developed, showing the displacement along the bar, with respect to movement of node 2: ?? = ?? ?? ??2 This is called a ‘shape function’ Global Stiffness Matrix – Formal Method • Case 2: bar element subject to axial loading • Move node 2 by +1 unit to the right, and hold node 1 fixed. 16 Global Stiffness Matrix – Formal Method • Combining gives: ?? = ?? − ?? ?? ??1 + ?? ?? ??2 ?? = ?? − ?? ?? ?? ?? ??1 ??2 ?? = ?? � ?? • ?? is a shape function matrix • ?? is a vector of element d.o.f 17 Shape Functions ?? = ?? − ?? ?? ??1 + ?? ?? ??2 • This is the sum of two shape functions: ??1 = ?? − ?? ?? , ??2= ?? ?? • The shape function defines how displacement (u) varies with position (x), when a unity displacement is applied at one end, and all other degrees of freedom are fixed. • Now – how do calculate strains, and stresses? We create B, the strain-displacement matrix. 18 Strain Displacement Matrix Axial strain, ????, is the gradient of axial displacement: ?? = ?? � ?? ???? = ???? ???? = ???? ???? ?? = ???? And for stress: ?? = ?????? = ?????? If we know the shape functions: ?? = ?? − ?? ?? ?? ?? ?? = − 1 ?? 1 ?? This is the B matrix for a bar-type element. B matrix is the derivative of the shape function matrix. 19 Strain Displacement Matrix • For a constant cross-sectional area: ???? = ?????? • Therefore: ?? = �???????????? ?? = � 0 ?? − �1 ?? �1 ?? ?? − �1 ?? � 1 ?? ?????? ?? = ???? ?? 1 −1 −1 1 • Same stiffness matrix as the direct method. • Proof that it works for a bar element. 20 Example 5.1 Determine the stresses in the two-bar assembly shown in the figure below. 21 Example 5.1 Stiffness matrices for each element: ??1 = 2???? ?? 1 −1 −1 1 and ??2 = ???? ?? 1 −1 −1 1 Global matrix: ?? = ???? ?? 2 −2 0 −2 3 −1 0 −1 1 22 Example 5.1 Knowing ???? = ?? ???? ?? 2 −2 0 −2 3 −1 0 −1 1 ??1 ??2 ??3 = ??1 ??2 ??3 Boundary conditions: ??1 = ??3 = 0 ??2 = ?? ???? ?? 2 −2 0 −2 3 −1 0 −1 1 0 ??2 0 = ??1 ?? ??3 23 Example 5.1 ???? ?? 2 −2 0 −2 3 −1 0 −1 1 0 ??2 0 = ??1 ?? ??3 Solving the 2nd line yields: 3??????2 ?? = ?? ⇒ ??2 = ???? 3???? 24 Example 5.1 What about the stresses? Solve for ??1 and ??3 and divide by the respective, ‘A’, or: ??1 = ????1 = ?????? = ?? −1 ?? 1 ?? ??1 ??2 = ?? ??2 − ??1 ?? = ?? ?? ???? 3???? − 0 = ?? 3?? ??2 = ????2 = ?????? = ?? −1 ?? 1 ?? ??2 ??3 = ?? ??3 − ??2 ?? = ?? ?? 0 − ???? 3???? = − ?? 3?? 25 Uniformly Distributed Loads on a Bar Element • Distributed, or varying loads along the bar length – convert UDL into equivalent nodal loads. • If uniform – half the total load at each node • If linear, quadratic (or complex) – not so easy. • Calculate the ‘work equivalent’ nodal loads – look at the work done by the loads: • For a distributed load of ?? (N/m): ?? = � 0 ?? ????????, ??ℎ?????? ?? = ???? 26 Uniformly Distributed Loads on a Bar Element This work, is now equivalent to loads at the ends, multiplied by displacement: ?? = ???? = � 0 ?? ??????????, where ?? are nodal forces If ?? is constant, we know N = ??−???? ?? ?? therefore: ???? ???? ???? ???? = ??� 0 ?? ?? − ?? ?? ?? ?? ???? ???? ???? ???? ???? ???? ???? = ?? ?? − ??2 2?? ??2 2?? 0 ?? ???? ???? = ???? 2 ???? 2 ???? ???? Force is equal at each end…! 27 Uniformly Distributed Loads on a Bar Element What if there are multiple nodes in a bar? 28 Uniformly Distributed Loads on a Bar Element What if the distributed force is not constant – ie ?? = ?? ?? ? • Use the same technique: ?? = ∫0 ?? ?? ?? ?????? where ?? = ???? we know N = ??−???? ?? ?? therefore: ???? ???? ???? ???? = � 0 ?? ??(??) ?? − ?? ?? ?? ?? ???? ???? ???? Or: ???? = � ?? ?? ?? ?? ?? − ?? ?? ????, ????= � ?? ?? ?? ?? ?? ?? ????, If the force is linear – ie ?? = ???? ???? = ??∫0 ?? ??−?? ?? ?????? =?? ?? 2 6 and ???? = ??∫0 ?? ?? ?? ?????? =?? ?? 2 3 Pi/Pj split = 1:2 29 How is this used in FEA? • FEA packages assign node numbers in a mesh to minimise sparcity (and bandwidth) in a stiffness matrix. • Aims to solve: Kd = F • Solve for unknown displacements by: • Direct method – some form of gaussian elimination. Easiest to solve for multiple load cases. Computational time proportional to nb2, n≡order of K, b ≡bandwidth. • Iterative method • What about 2D space? 30 Bar Elements in 2D space • Examples up until now have focused on 1-D bar elements (or series of). • What if bars are in 2-D space, and not in line? What is K? • Transform from 2-D space to 1-D space by transformation matrix • Never fear – this is done in the software. 31 Bar Elements in 2D space • Skipping some maths (you can thank me later), the element stiffness matrix, K, is given by: ?? = ????????? = ???? ?? ??2 ???? −??2 −???? ???? ??2 −???? −??2 −??2 −???? ??2 ???? −???? −??2 ???? ??2 for DOF ?? = ??1 ??1 ??2 ??2 • Where k’ is the 1-D stiffness matrix: ??′ = ???? ?? 1 −1 −1 1 If ∅ = 90° ???? 0°, this matrix reduces to the 1-D version 32 Example 5.2 • The figure shows a simple planar pin-jointed truss. Determine the displacement at node 2 and the stress in each bar. • No bending stresses (pin joints) • Global matrix will be 6 x 6 (3 u’s, 3 v’s) 33 Example 5.2 • The figure shows a simple planar pin-jointed truss. Determine the displacement at node 2 and the stress in each bar. • In local coordinates, we know: ??′ = ???? ?? 1 −1 −1 1 • As each bar is misaligned, we need global coordinates. • For bar 1 - ∅ = 45°: sin 45° = cos 45° = �1 2 • Therefore ??2 = ??2 = ???? = ⁄1 2 • For bar 1: −− −− −− −− = 22 22 22 22 scsscs csccsc scsscs csccsc L AEk −− −− −− −− = 1111 1111 1111 1111 2 2211 L AEk vuvu 34 Example 5.2 • The figure shows a simple planar pin-jointed truss. Determine the displacement at node 2 and the stress in each bar. • For bar 2 - ∅ = 135°: sin 135° = −cos 135° = �1 2 • Therefore ??2 = ??2 = ⁄1 2 ,