Question

Consider the truss subjected to a single point load P = 12 kN downward at point C (Figure 1). The members of the truss all have E = 200 GPa and cross-sectional area 25 cm2.

<ME316 HW9 - Energy Methods Conservation of Energy View Available Hint(s) Learning Goal: To use conservation of energy to calculate the displacement of a point in a truss When a structure is subjected to a single point load and deforms in a linear- elastic fashion, if the deflection of interest is at the location and in the direction of the applied load, then the conservation of energy can be used to calculate the deflection. As the load is slowly increased from zero to some value P, and the final deflection in the direction of the load is A tho load does work U = -PA on the structure. By conservation of energy, this work is stored in the elastic deformation energy of the structure, U: 2 ÅR O ? Value Units U;,AC Submit For a truss, the internal strain energy is the sum of the axial strain energies in all the elements. This gives the equation Part B - Strain energy in BC ND PA= E ME 2AE What is the strain energy in truss member BC due to the applied load? Express your answer with appropriate units to three significant figures. where N is the internal axial force in each member View Available Hints) C! Å R 0 m ? Value Units U1,80 = Submit Part C - Deflection at C Figure < 1 of 1 > Use the conservation of energy to determine the vertical deflection downward at point C. Express your answer with appropriate units to three significant figures. View Available Hint(s) Ac = Value Units 2.4 m Submit A 1.8m+ 3.2 m

Answer 1

Consider joint c :- te 0 Tano = 18 - 36.870 - FBC. - FAC Tane = 3:2 = 8 = 53.130 2.9 (+) EF-o → FBc sinB - Fac sin@ -0. FBC sin 53.130 - FAC Sin 36.87"=0 08 FBC -0.6 FAC =0 FBC = 0.75 FAC (9+) Fy-0 * -p- Fac cose – Foc cos =0 -p - Fac Cos 36.870 - (0.75 Fac) C0553-13° = 0 -P-0.8 FAC 0.45 Fac -O - 1.25 Fac=-P

Given, P= 12 KN 1.25 Fac=-P 1.25 FAC = -12 Fac = -9.6 KN FBC = 0.75 X-9.6 = -7.2 kN. Consider joint B 3.130 KIRS $ - 90=8 = 980-63-130 Fab $- 36.870 (5) Z Fy=0 → - Fac cosb- FAB = 0 FAB=- FBC cost = -67.2) COS 36.870 FAB = 5.76 KN energy in member AC FACELAC Part A :- strain or = 2 AE

LAC = √2.4r+ 18r = 3m Ui, ac = [9.6X10% gr x3 2x (25 x104) x 200 x109 1 Ui, AC = 0.21648 J | Part Bi B: Ujbe = FBCT LBC 2AE LB c = √2.47€ 3.25 = 4m U BC = (-7.2x108x4 2x (25x174) x200x109 UiB = 0.207365) Part C - strain energy in member AB YAB = FAR LAB 2AE LAB = 18+ 3.2 = 15m YAB = (5-76 xio jYx5 2x 25X164 X200X109 | YAB = 0.16588857

Total strain energy (Ui) = U,, ABT U;, 8c t ,,AC ų; : 0.27468 + 0120736+ 0.165888 ų = 0.648 J work done due to external load (ue) = {pa Conservation of energy is U = Ue 0.648= £ PO 0.648 = $x 12000 XA A = 1.01988 x1o9 m A = 1.01988 x164 mx 1000mm A = 0.108 mm