Question

Example T: Determine the support reaction at C (ball and socket type) if the applied force at B is F = 3.6 kN. 0.8 m 0.8 D 0.6 m 1.2 m 1.2 m 3.6 kN 1.2 m

Answer 1

Solution: Drawing the FBD 0. 8m (-0.80.60) 0.8m. 2 (-i) E (0.8, 102,0) (0,0,2.4) CZ 1.2m А (0,0,12 x (2) Z 1. 2m 3.6KN T 1.2m Support c is a ball and socket type. → In this type of support moment is always zero. this type of supports only holds reaction components along 3 anis (cie x band Z) thighs we have to do: We have to find the tension forcce TAD and TAE at first. - To do that we will use the criteria moment about support "c" is zero. Available Heading Moments Moment due to 3.6 kN about a → Moment due to TAD about c → Moment due to TAE about a Sum of all IL 19

Moment due to 3.6kN force : Here we can clearly see that the 3.6kN force is directing towards negative barus so, we am write the force vector as F = – 3.6 from c to 3.6kn force initiating point (B) position vectore is TEB Now rio - (on tos + 1.2M)a Cortos tou) MEB = 1.2 Also Moment Marxł so, in our case; M = TEB X F os, Ñ - (1.2 Û) *(-3.65) O , M 0912 om, mis în (o3•Como) + üv(oro) on M = 4.32 Moment due to TAD forece: We need to find the force vector TAD as we know Force Vector - Magnitude • Unit CE) СЕ) vector

Force TAD is directing towards D from A so; direction vectore AD = 5 - 0 AD= (-0.80 +0.6) - (2.40) AD = -0.8² +0.65 -2.40 so, Ao la (0.8²€ (0.69² + (-2.472 laol= 2.6 therefore a = - 08 + 0.6 m 2 4 m so TAD - TAD ( - 0B + Bob ] - 24 ) From c to TA force initiating point (A) position vector will be => MCA - A - r = Coi ton + 2.4ů) - (oiton tou) ra - 2.4 so, Moment M2 = Tax T on M2 = 1 l û 0 0 204 & S e 6 AD 206 ) 01, Me = a (o- 24x aib TAD) - 3x (0 - (- 0.872 I tuxCoro) on, m2 = - low STAD E - 1.92 TAD Å

Moment due to TAE force: Force Th is directing towards E from A so, direction rector & A & E A = (0.8E +1.25) - 24 û At = 0.8i + 1.2 - 2.4 û so, l= V (0.8) ²4 (1.2)2 + (-2432 IF I = 28 therefore not = DE AS- so, TAE = TAE ( 28 2 + 1 1 3 2 5 - 2 ) From c to TAE force initiating point (A) position vector will be -> re - 2.4 û (Already Calendated) so, Moment M3 - MAX TAE 091, Moi = 1 Ś ū To 0 2.4 0.8 ITA 17 TAE - 24 ME 1.2 2 0.8 2.8C 니 1.2x2.4. 21480.8 ( X 29 2.8 at hx (oro) 2.8 Mis - - 2.88 The ê + 10 1.92 me 2.8

so that; Mix M2 + M3 =0 (4.32 ") + (-hen To E - 1.92 TAD) oiton = to 2 .6 2 .8 Com 2.88 uo 1044 2.6 2.8 AE + (- 1.32 and 1682 TAE) By composing both side we geti 4.32 - 10l s TAD - 2.86 TA 0 -- -- (1) - 1 TAD + 1.92 THE =O – – –- (2) from Equn (2) we get - 1.02 D + 1:32 THEO on, TAD = 206 TAE - - - - (3) putting the value of Tas in equin (1) we get; 4.32 - lovely on 26 TAE - 2.88 TAE=0 on 4.32 = ( 1.94 + 288) THE 2.88 louy 2.6 AEO or, louy 2.8 23 2.8

09, THE = 2.8 KM putting the value of TAE in equ? (3) we get TAD = 206 2.8 2.8% TAD = 206 KN Ca Since the system is in equilibriuon therefore sum total of all acting forces will be Zero. Acting forces are FC; F ; TAB; TAE We assumed Calong x amis & Con at support ch along bonis fonce component along Zaris - Cz so that; F = Calt colt Ca û Therefore; + F + TAD + TAE 0 ah (ccê + cy} + (ah) + (-3-6) 1 + (-0.8 TADE + 0.6 TD 5 - 24 h = ot of 1 + (08 The E+ 12 The J - 24 ITEM 2.6 AD of ou 2.6 2 .8 + (cy - 3-6 + 0.6 2.6 at 12 8 2.8 ) + ( C₂ - 2.4 2.6 - 204 208) û - to 6 1 254 2.6 2 2 .4 .8

By composing both sides we get Cx - (0.8 x 2.6) + (0.8 x 2.8) =0 09) or Cx= 0.8 -0.8 x = 0 KN (Nox-component is present) on, (y= 3.6-0. 6-102 on Cy= 108kN 2-component positive along, Danis) 2 or or Cz= 2.4 +2.4. C2 = 4.8 kN (2-component along positive Z anus) Answer Ca= 0 KN Support Cy=108 KN Reaction (2=4.8KN Jo a