Task Assessment Type Mark Learning Outcomes Assessed Assessment Criteria Due Tasks or Submissions Assignment 1 FE Fundamentals Engineering Report 10%1 Technical Results, Report writing and communication skills, creative problem solving Friday Week 4 Assignment 2 Good FE Practice Engineering Report 10%3 Thorough planning and execution, Report writing and communication skills. Friday Week 6 FE Fundamentals Exam Exam (2h)30%1,2 Correct answer, Correct working, Logical approach Thursday Week 8 Major Project Engineering Report 50%3,4 Monday Week 9 Friday Week 13 Total100%

Simulation Tool Response of a structure or system to the loads imposed Structural Analysis Tool Measuring critical loads or failure criteria Structural Design Tool Modifying the structure to improve performance Number of Edges Calculated circumference 2r Node Element Mesh Load Boundary Condition Material Property TrueGrid Compatibility Equilibrium Constitutive Law Things fit together with no gaps Each node and element boundary matches the one beside it. F1 F2 Fk Fn E o c =P P A P P A oA oA oPn 0limnAPAoPs oA 0limsAPAMegson Megson xy yxyz zyzx xz, ,x y zNormal components , , ,, ,xy yx yzzy zx xzShear components Perpendicular to this axis Parallel to this axis xyx xy xzyx y yzzx zy zxyzyzzxxyCareful with the shear terms, different sources use different order xyzyzzxxySigma Tau Epsilon Gamma E o c ={ } { }1 0 0 01 11 0 0 01 11 0 0 01 1(1 )(1 2 )(1 )(1 2 )0 0 0 0 02(1 )(1 2 )0 0 0 0 02(1 )(1 2 )0 0 0 0 02(1 )x xy yz zyz yzzx zxxy xEv vv vv vo cv vo cv vv v o cvvt v vvt vt vvv ( ( ( ( ( ( ( ( ( = ( `+ ( ( ( ( ) ( ( (( = Ey ` )Continuum(Displacement only) Structural(Displacement and Slope) 1D Bar (u) Truss (u,v,w) Beam (u,v,w, x,y,z) 2D Plane stress (u,v) Plane Strain (u,v) Plate (u,v,w, x,y) Shell (u,v,w, x,y,z) 3D Bricks (u,v,w) Real Structure Model of Structure Discretised Model Different Modelof Structure Discretised Model Magnitude of Errors Remember: Refinement does not make your closer to REALITY.

Refinement makes your results closer to your MODEL! (a) Von mises stress (b) Fore-aft load paths showing detail around mast step 11 12 1 121 22 2 2k k u Pk k u P ( = ` ` ( ) )P1P2 u1u2 E, A, L P1P2 u1=0u2=1 NN Node 1Node 2 What if we have multiple elements or loads? We can assemble multiple elements using equilibrium at the nodes (+ compatibility) IIIIII P1P2P3P4 1IP2IP2IIP3IIP3IIIP4IIIPIIIIII At Node 1: At Node 2: 1 11 2 12 2 21 2 3 200II I I II II III I I I II II II III I II IIP PA E A Eu u PL LP P PA E A E A E A Eu u u PL L L L + =| | = |\ . + =| || | + + = | | |\ .\ . In matrix form 1 12 23 34 40 0000 0I I I II II I I I II II II III I II IIII II II II III III III IIIII II III IIIIII III III IIIIII IIIA E A EL Lu PA E A E A E A EL L L L u Pu P A E A E A E A EL L L Lu PA E A EL L ( ( ( ( + ( (= ` ` ( + ( ) ) ( ( ( u1u3u2u4 IIIIII = Ku PK u P (( ( (( (= (( ( (( ( Steel Bar, A = 100mm2, L = 1mAl Bar, A = 50mm2, L = 600mm F1 = 3kNF2 = 1kN X=0X=1000mmX=1600mm ( )ij j iijEAF A u uLo = = ( )j iijiju uLc=( )ij j iijEE u uLo c = = 1 kN 2 kN 100010001000 A = 50 mm2 E = 70 GPa P1 P2 u1 u2 Q2 Q1 v1 v2 2 21 12 21 12 22 22 22 2coswheresinj ij iu P l lm l lmv Q lm m lm mAEu P L l lm l lmv Q lm m lm mx xlLy ymLoo ( ( (= ` ` ( ( ) ) = == =| |j iijj iiju uEAF l mv vL = ` ){ } { }{ } { }( ) ( ) ( )2 22 22 22 22 22 22 2 2[ ]Ti i i j j jTi i i j j jijj i j i j iij j i j i j iij iju v w u v wP Q R P Q Rl lm ln l lm lnlm m mn lm m mnln mn n ln mn nEAL l lm ln l lm lnlm m mn lm m mnln mn n ln mn nx x y y z zL x x y y z z l m nL L== ( ( ( ( = ( ( ( ( ( = + + = = =uPK| |ijj iij j iijj iLu uEAF l m n v vLw w = ` )http://bendingmomentdiagram.com/ v3v2 u1 u3 u2 IIIv1 Try with truss elements 11 122 21 100, 10 1 0 1 00 0 0 00 1 0 1 00 0 0 00 ???!?!?m luv QAEu Lv Qv Qo = = = ( ( (= ` ` ( ( ) ) =3 2 3 21 12 21 12 23 2 3 22 22 212 6 12 66 4 6 212 6 12 66 2 6 4EI EI EI EIL L L Lv PEI EI EI EIML L L Lv P EI EI EI EIL L L LMEI EI EI EIL L L Luu ( ( ( ( ( = ` ` ( ( ( ) ) (( P2P1 v2 M1M2 2 v1 1