Question

Q2. A simply supported beam AB (Figure 2) supports a uniformly distributed load of q = 18kN/m and a concentrated load of P = 23kN at the centre. Consider length of the beam, L = 3m, Young's modulus, E = 200GPa and moment of inertial, I = 30 x 10 mm-. Assume the deflection of the beam can be expressed by elastic curve equations of the form: y(x) = Ax4 + Bx3 + Cx2 + Dx + E. 1) Sketch the bending moment, Mas a function of length of the beam and comment on the distribution of M over the beam. [2.5 marks 2) Write the elastic curve equations describing the slope, v'and deflection, v of the beam and list the boundary conditions, which can be used to calculate the constants in the equation. [2.5 marks) 3) Determine the total elastic strain energy, U stored in the beam. [2.5 marks] 4) Calculate the deflection at point C using Castegliano's theorem/method. [2.5 marks] Figure 2

Answer 1

Agter finding unknown reactions we can write moment equation in integral form

then we can draw moment curve by assuming symmetry from both sides

apply boundary condition and then for strain energy integrate to find total strain energy (upto 0 to 0.5L then multiply by 2

i have shown calculations in my solution

Thanks

page-ol RA + RB = P +2.L RAT. RR = 23 + 18x3 RA+RB = 77 emp =0 PAXL = P x L - 2.2 XL ARA = 2 + 2 = 23 + 1883 ARA = 155 in 38.5kN RB = 38.5KN symmetrical with y-Component forces. from A, at any distance x, O. Mx - RAXX-9.X.X - 38.5x9x² when I CXCL me = RAX - 2012 - P(x - 2) = 38.50- 900²_ 23 (X -1.5) Ele - oder G (-) for OLX < Ce să (38.530-909 so in general at distance x where Þj also considered 38.56-982-23x+34.57

Page 02 Indlgorting, 2 on [ 38.692-393 – 23x2 +39.51) & I [7.75 x²-3x² + 34.50]+6, again trezorting y = 4 [7.75x3 2x + 34.500 J + G x + (₂ Bending moment dingrum 37.5 km Boundary conditions are at x = 0, y = o atx=L - Bm y=0 ad x = 4, y = 0 Total strain everso U=1 mdx efl (38 ZEI 201=2138.59-939dbd due to Summetsz = 21 (1482.25.0c²+8100 - 69303) doc. ZEL

Page 03 [995, (900X107 y 2oxišx 1080/1982232.597 20:55 * LENGU) - 1993 (1:5) -0.01522 Khesario ir TOTTEN 15.82 k m = 15.22 krom since at C M = RAX - 987 _ P (x - 2) no U=(PxC - 2012 -p (x-4) 272 and D EI For finding deflection at Ĉ using Categliano's theorem we remove another load and moment except the term with P. IU = Ep (x-4) ²x o EI - 1 p? (x² + 4 / _ x dx +- de election - 30 025 P2) - (215] - 0.8625 m