Question

TIN Kasus 4 A bending rod is tested with two centralized loads as in Case 4 in the analysis data given is: P4 = [60] KN L41 = [2 x 1] m L42 = [10-8/21m E = [81] GPS thickness of the trunk cross section box, t = [6-11 mm length and height of the outermost box section, lw = [60] mm and lh = [60 + 180 mm a. Draw a free body diagram of the bending rod, then draw the force distribution and deep moments along the stem. b. Calculate the reaction force and moment, the maximum deformation, and the magnitude and location of the maximum stress which occur analytically on the stem. C. Make a numerical simulation of the bending bar of the case. Determine deformation and stress values maximum that occurs, then compare it with the analytic results obtained.

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Answer 1

(4) y at no 5m. u 120x25 + 600x5960 EI - 100 + 3000 - 960 - EI 540 EI Iyo = 6.27 mm Bending storm o= M.y 120 xlood X3 Xlo I To= 4.4 GPa) Max. shear Force = 60 KN Max. Bending moment = 120 KN-m Answer Max. Bending stren = 4.4 Gla Max Deflection at A do = 11.15 mm Deflection at cente = 6.27 mm

78 mm Go KN 60KN Smin A RB ren 2m 6 m E = 81 GPa = 81x10² mpa I Area moment of inertia Gokn GORN (SFD) 60KN GOKN (120 KN ) (120 KN-m) (BMD) = 10.62x105 mm ca since, it us a case of overhanging beam. Mark, A, B, C & D on beam. looding in symmetric, : Reaction at B4C will be same in wertical direction. Taking moment about B. - RB X 6 + 60x2 = 60x8 → RB= 60 (8-2) (a) SFD of BMD RB= 60 KN | Also, RA= 60 KN SFP for AB for Fore=-60KN (Downward) BC Fore -60+60=o constant)

(2) for co Force = = -60+60+60 60 KN Fc at left = 0 KN E at right = 60KN Bending moment diagram For AB which gines, Moc - 60xx MA=0 (linear Profile) Mg=-60x2= -120 (KN-m) For BC Mac = -60xx + 60x(x-2) = -60x+ 600 - 120 =-120 KN-m ( constant Profile) 27 c 2 B L 2 60KN CORN For co 60KN Moc = -60xX+ 60 (0-2) +6012-8) --6066 + 6056-120+600-480 = 600-600 (linear Profile) Mc = 608-600 = -120 KN Mo = 60x10-600 = 0 KN Strim Bending stress (Ob man = max X Y man I 120x 108 x 39 10.62x 105 = 4406 MPa = 4.4 G Pa

(3) Deflection of Beam doc Measuring from A, upto o, middle of BC, I dy -Mn = - 60 xx + 60 x(x - 2) = -60x+600-120 = -120 ET -126 dy = - 120x+a, - EI y = - 120 e 3 + ( x + C2 At, x= 24 2+3 = 5m, dy =o. 0= 120x5tc → ci = 600 At x=2 y=0 - 0= -120x2 + 600x 2 + C2 C = 240-12.00 = -960 | EI Y = - 120 x x + 600xx - 960 YA at x=0 YA = - 960 Holy 8nOXTO EXTE EI - 960 81 X1 X 0.00 0010626 YA = 11.15 mm