Question

4 The figure below shows a 2D symmetric beam loaded as shown. Determine using the finite element method: a. The displacements at all nodes (both in translation and rotation), and b. If the distance from the centroid of the beam to its outside surface is 100 mm, determine the location of the point of highest and lowest stress in the beam. 8 KN -2 m- 5 E - 70 GPa 1 = 1 10m

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Answer 1

1 Page Solo to Problemſ Finite element method. Step : free body diagrom. to fotojah, 2 Step ③: Element stiffness metrix K-EI 3 12 6L -12 6L 6L 42² -6L 22² -12 -6L 12 -6L L6L 22 -6L 462 Given. E= 7oGPa = 70x10 N/m Iz 1x10 umu Element K, 2 70x109 XIX104 12 32 GL 1-12 6L -12 6L 42² -6L 22 GL 12 -6L لا اے 2 6L K = 106 19.34 13.86 13.86 27.72 1-9.34 -13.86 13.86 13.86 -9.34 13.86 4. -13.86 13.86 o 9.34 -13.86 13.86 27.72 1 soo

Poze Elenes | 12 62 -12 6L K: 70x10?x1x10 64 42² -bL 21² 1-12 GL 12 -6L L6L 21² L UL² U 0. Uz 03 21 21-21 21 lu K, 106 21 28 -2) 14 02 1-21 -21 21 - 21 L21 14 21 28 oz - Element Kz = 70x10 XIXIO4 12 6L -12 62] 6L 4 2² -6L 21² 42-6L 12 TL L6L 21² 6L 4L² Uz z uy Ou 121 21 21 21 uz k = 1062128 21 14 03 1-21 -21 21 -21 L21 14 21 28 - Element ka = 70xc0lxlX0Y 6L 46² 62 21² 1-12 -6L 126L. L6L 21² 6L 42 Uu Oy us or 9.34 13.86 -9.34 13.86 - 0 113.84 27.72 -13.86 13.86 10° 1-9.34 -13.86 9.34 13.86 4. 113.86 12.86 – 13.-26. 27.7200

Olxal sortita 9881988198941 O P 1816171 172 98.81-581- o no 986 98-81-ss hitti bn 98161 heb het te O o l ooo os lolo ooo marix 7988794 : Global stiffnen h o 0 ho 14 Pol O o o o 0 0 o o M L M hi te tugis hit 98.81 98.81 o 16- hit heog 198461hek 0 0 98021 98001 ts 9841 o o 18:47 14:4+ 981 heb Fell En Tol th! 'ol 'n sup Pase

Step : Applying boundary conditions At node ① & the beam is simply supported with pin support. O in Displauments=0 a Moments to So 4,0 450 At node ③ E (4) the beam is on tolles support so the transverse displacement is zero. i Displacements zo a Moments to to the reduces to global stiffness O O 127.72 13,86 | 13086 55.72 métrix Us 0 - 21 Og 04 05 o 14 o o 00908255 K=100 o 0 o 21 0 14 0 50-72 13.86 13.86 ou 27.920 0 8686

Page step O: To find displacements & moments Ef [k]{us Mizo M₂:0 F2=8000/= 10% M₂=0 My o LM 20 127.72 13.86 oooo 13.86 55.72 -2114 o o o -21 420210 lo 14 0 49 14 o 0 0 21 14 SCA213-86 00 13-86 27.72 10" (29.420, +13.860)-0 =0 10 (13.860,755-728, -214g +148)=0 ② 106 (-210, +4243 +21 D4 ) = 8000 609 (1404+4903 + 14 04) 20 - 100(2148 +1483 +55-7204 + 13-860, 720 106 ( 13.26 04 + 27.72 Os)=0 (6) From equation O = -20, from equation 6 Os 20.50 -8 Substitution into 13.860 ~ 111.440,- 2142+ 140 =0 -97,580, 2142 +140₂ = 0 Ug = -4.640, + 0.67 0 1

Page 6 quation ~210, +424₂ + 2104 = 0.008 - 10 Now Lubyceutic isto © +42(4.640, +420, +42(~4.640, + 0,670 ) + 210,= 0.008 -152.880, +28.1483 +2101=0.008 quation 6 - 140, + 490 +1Gdy = 0 quation & ) 21 4g +140, +557204 + 13.860 =0 Now substituting & & & into (12) 216-4640, +0.6203) +1403 +55.7284+13.26x0X0 97.440, +14.0743 +140, +55.9204-6.930,20 -97.440,+28.070, +48.790,200 Consider equations (@, & ( -152.880,+28.14 03 +2104 = 0.008 140 +4903 + 1404 - o -97.640, +28.070 +48 79 040 sowing these 3 equations wing Calubtory We get 0,= - 6.2796810 With O = 6.435 1 X 10MM 04 = -106243 810-4Nm

I Page 1 (0 = 12.5592 x 10 m, / 05-2 0.81215 Xco-Wm 7 u = 3.3448810-4 m / The displacements & moments (translation & rotation) at all nodes are U₂ = 3.3448810-4 m/ TQ = -6,2796410-Nm) 102 - 12.5592X10 Nm O₂ z 6,4551 X10- Nm / TO4 = -1,6243x10-4 Nm Oca 0.81215 X10 Nm