Question

1. A metal bar of length, L = 100 mm, and a constant cross-sectional area of A = 10 mm? is shown in figure Q1. The bar material has an elastic modulus, E = 200,000 N/mm2 with an applied load P at one end. The governing equation for elastostatic problems with no distributed loads is given as: d AE du dx = 0 dx where u is displacement. P 100 mm • Finite element nodes Figure Q1: Geometry and discretisation of a one-dimensional metal bar. (a) Using the weighted residual form and the Galerkin finite element method prove that the element stiffness matrix is APP [L1 EA) where is the element length and A) is the average area of a finite element. Assume a linear finite element. [18 marks) (b) The additional element load vector due to the thermal expansion effect is given as: {f} = AⓇE{{"}aat where thermal expansion coefficient of a = 10.0 x 1061/C. By dividing the domain into 2 equal elements, determine the displacement in the bar with a temperature increase AT = 50 °C and an applied load P = 500N. [7 marks) (TOTAL 25 MARKS)

need to solve the mathematical model to prove
that we can get the equations i Q1 a methematically

please use only the weighted resedual and gerkins methods to prove it

Answer 1

! soln: 100 MM pt load applied → finite element Area A = 10mm nodes Elastie modulus, E: 200,000 N/mm" displacement a) We have load Strees, To ak Close Area strain or M PL Elastic modulus E- bi E Alz displacement, Mr PL - (A) My EA P where EA -k stiffness P-1CH Now it we talk about matrix form, EA ko [.

Vector - 2) Additional {tr} (e) E {:} AT a - 10.0 X10 expansion coefficiene AT : 50°C Po 500N applied load alow, we have {ting + poklicy M). = -6 M2 — EA M₂ L A EX AT +P = 쁜 By putting the values we jox 200000 x10x10 have -€ 20000 -X 10 -M2 x 50 + 500 - 2x106 X 500 XLD 500 x 100+ t 500 - 20000 Hz a-20.000 ML 1000 + 500 I Soo 2 - 20000 M2 Isbo 2 M2 20000 M₂ = 0.075 mm m)