Question

3. (a) Draw qualitatively the deflection curve based on the bending moment diagram. (6) Determine the deflections at B and D, and the slope at Cby the moment area method. (c) Locate the point of maximum deflection in AB and compute its magnitude. (E = 200GPa, I= 10 mm) 25N 3kN/m TI) А в |u2m ofe4m of 3m4

Answer 1

25KN 4m . 3 kN/m A amt A B bo zm 1 Rc RA+ Rc = 25+ 3x3 RAT Re = 34 KN - equ ③ E Mc=0 i.e. GRA- (25x4) +(3x8x1:5)=0 RA = 14.417 KN Rc = 19.583 KN Bending moment diagrammes - 28 Q3 4 KNM sm 2. ma "2 T24m 13.5 Kalm deflection Cune: -

- 6 M diagramme is given by: - 28.834 EI В м — 37) K2m . 2.724 ---- 13.5 EC • Calculation of Oc (OCX 6) = de flection of A e a moment of area of in fig about A. 28.834 2.7241 28.834 Ет EI 3 Dx6 = 4x 28:824x 283x3) + x 28:824, 2724 x2 +2434] - 1x L 12 +6X [1276x2 4-134] - x 485 x ®x (4 124 +1.276+] Oc= 13.508 mit EC

2999999 Calculation of OA s. (OAX6) = deftution of c = moment of area of n fig about Dax 6 = 5 x 28:8242 2x(4+ 3) + 3x 28194x 2.424x{1-276+3x27 3x18 5x1-246x 1276 - 5 x 18 x3 x3 13.5 ix 1.276 EE VAZ 242.20 ED Calculation of defection at Bir SB - (OBX) . moment of in fig. blul A and B about B. Et 232.59 mm SR EL 88 - 242.2022 - IX 28:89 * xox . Sa - 465.18 465.18% x100 bonne (9.00*10*x157) Calculation of deflection at Do- So= (- 0 cх3) + moment of the fig blue cand o about a So = -18-38x3+ *** x3] SD = 5.0745mm So = -2 -10.149 EI - - 200x/0x107

② point of maximum deflection blul A-Bi- of Maximum deflection is at distance. Let x The point from A 28-834 VA 1.276m JB 2me 14 - 2.124 Y V185 ! O 242-20 ΕΓ o at distance & from A, de Frelion is Marimum Therefore angle at under Maximum de flection will be zero .so, OA = 242-20 = area under y curve bla A and EI 242.20 1.128.834.). I EI a = 5.796 m 2 meaus max depletion post is not blut A and B. lot max defuntion point is at distance from M. . O 19.508 = fx 28.834 xax a]-[2x EL 2.724 bly X = 2.045m Max. de fiertion point is 2.045+1-276 m = 3.82m from - 2.68 m fram A