Question

Solve Using Castigliano's Theorem Considering that Castigloiano's is applicable to angular deflection as well.

Determine the deflection at the load P for the three-sided bracket shown in the Figure below. The bracket has the same material and cross section throughout. With solid steel rods of 56 mm diameter, using E = 200 GPa, G= 77 GPa, a = 300 mm, b = 800 mm, and P=5.5 kN. Р 3

Answer 1

Page 1 lovestion 1 Solution » Given that Hinge eff 6/2- k diameter of steel sud =d = 56mm E=200 hpa Passkn=5500N 42 77 6pa a = 300 mm 6= 8 no mom

page-(2) we have to find deflection at the load ep' by Castigliano's Theaston: Istitp-2)> Brake the whole solid part am Three section, and force analysis T=MR 3 at section X-X IM₂ = 1/2x MR x P Y 112 ppp MR Time udha 1/2- MR at section (xx ) Max = M₂= -MR +P/22 TOMR

page-(3) since middle Middle na portion ha portion of thig assembly will under goes asending moment and other two side part will go under Twisting Moment due to (MR) Mo= Reaction Moment since at fixed eard hinge cord (ie, at.q' 08 3 erd) - stope zo Step-(2) Aupply castigliano's Theorem for angular deflection (ie, for slope ) ( at fixed eard 2 M2 JM2 dx + 21 mi ami dnt El ama IMR 1012 ET 2. Ta UJ

page-(A) " M2= EMR + P2 x ma FMR -1 and P/2x o ami DMR sol 29-1 2012 - 1 m2 de tot 2 Tao El HJ 2 -2 [- Me + P32x] dat 2Ta 012 MR -(2/2)x] dose + Ta ज U. 습 [ MR 1/2 x 1 (6123 it (MR) a us Maib. 261 P62 16 FI + Mra HT => e. MR ART ta at 2ET 16 BI

+ а 2х2 vxt xad+ + (Wooy-(5) M& Р. 1ЬЕТ ße Го ПЕТ -) MA ß e HI-ex, $ = , 1, ET 6 - С now, calculate ß ст 2ET я 1ЊЕТ 1612x3XLo4 64 4 а Жо 2BT 3. Зо 14 р, ? (20) 64 32. (3) BG (bx 2лгох 46 х х 1 dwo 3 x + ахлухуваад яах халык 1) (8) 16X 2 р goox2 30 9х2 сло ㅋ가

page-(6) pa 2o 2+ 7.792 2v 20:42) g. 792 ß (20.42) MR2 ß P 2 (20.424 5500) Na mm MR= (112310) roman Step-3 Apply Castigliano's Theorem for som de flection at point of application of Load. a J012 Sp= 2 2 M2 am da + 2 Mi ami de EI JP op EI 8p+px Hex, MR = ßpe M2= and Me=(Plan)

Рах-(+) ., Әм, ОО Еe в + ami п ЭР S») М. де [-Р + for] ( -x) 4х + (Cerv) (%) от . а Jor 2 е (А. 8 = 2 P 4 +g -nB] dn 4 (2- (m3 14 1210 (я + е* - ау" о БТ 2 Р 4,3 в. в“. + (2 ( 4 2 2 FT

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