Question

(0, 2) (4,2) (Coordinates in inch units) (4,0) (0, 0) 3. (5 point) For the element shown in Figure above (Problem 2), determine the stress matrix at s = 0.5773, t = 0.5773 using isoparametric formulation. The nodal displacements are given as Uj = 0) Vj = 0 U2 = 0.005 in. v2 = 0.0025 in. Uz = 0.0025 in. v3 = -0.0025 in. U4 = 0 04 = 0

Answer 1

Given, 7 (42) . (4,0) (4:) 2 - shape functions NI = (1-3)(1-1) Ng = (+) (1-4) Nze (+)(vtt) Ny = (1-3)(14) 4 Here x=N, Xi+ Not +N32/3 +Nq44 | Y=NY, TN2 Y +N34₃ 7N4d = N, -(O) +44 + 4N3 +O (N) -N (0) + (0) +N, (2) +M412) = 402 403 = 2N₃+ana 2 = (1+3)()-)+(1+3)(1+t) = (+3)(lt) + (5) ( a = (+)+(1+t) = 2. (It) + (+5) =0 B-matrix? on these +) 20 (19) Cantot

。 地。 無。 |-- )- (G) : : (; 。 。 。 。 ? 成 st =0.5133 。 o -- Basic25 1 ob3335 |-28.315 0 |-oics 。 0.425 eBBS 0 0.1978/25 -0. 15 o +0.058375 0.39425 -0.07] 。 - $25 15 -0.155 0. KosY -0.4938/25] 0.198625 0.10S15

Given, ع 0 d ( o ه ( = که 0 گم گه.0 0 . 0025 من كه عي كه عم 0.035 |-6.0025 كما 0 Stoess to-fbfit [0] = 0 [e] is the strain, E = (B) (d] is material property matrix: (plane stress theory, As thickness =18 e D م ا ا ه ه ه = 30x108 psi 10 =0.25 ه 30x) | 0.35 11 10.5 5 کند 36 ) 525 ه| 0 . 25 اه ا | ه . oo. gs

Strain. . ex) fo.o528315 Gyao 6- B] [4] 0 20528375 0.1056750 -0.1528375 -0.105675 0 0.19781625 0.105675 0.0528375 0.394305 0 0.394325 0.1978105 Byl -0.105675 lo -0.19781625 0 ? o.105875 0.105675 -0.19781625 0.005 10.0025 10.025 -0.0025 ſo. Ovo 7587281 -0.00125 00004 49906 stress 5 = [0] (e) in 0.25 32000000 0.25 i 0 0.0007587281 7 0 11-0.00125 0.375 0.0000949906 (14279.3 1 -33930.175 11139.8875

I hope this will help you.