Question

P1 P1 P1 The cage system shown in the figure, EB “Force Method” by taking the bar unknown Solve with. Find the internal forces. (E constant, C moving support.) please make the solution readable B Ly1 LX1 LX2 5,3m 3,4m LY1 P1 3,1m 11,3kN E D Lx2 LX1 X *

Answer 1

solution :- The giren truss is 11.3kN 11:3 KN 11.5kn 1.3k A VB C 02 Ten 3:1), 0 D E * 3.4m 5.3m @z = fon't_3:) = 30.39 tan (3:46) = 42.35 Og => Ds=mtr-uj where M = 8 r= 3 &j=5 si Ds= 8+3 -10 =) .. The given truss is statically indeterminate to one degree. Slet us consider BE as redundant. Removing this member lerus determine forces in members of structure. . calculation of reactions 11:36N 11.3k 11.3KN 7.55TKN 7.55T IN A Eme=0 11.3x3.4 41.3C3,475.3)- Vexã.4+5-3) 11.3 kN n =0 101225KN 4.41 KN Vc=150716kN 82752KN " Vc = 15.716 KN ( ) HE t Efyzo 116= -4.416 EN CH) VE = 4.416 KN

consider Joint c 11.3 KM Foc EfY=0 .:-11:3+157716-fco Sin 30.320 FcD = 8.752 KN © LAR 151716 KN fcd EGZ- 0 FBC- fap CO 530.3°=0 fBc = 70557 IN - . consider 11:3 Joint B toa FAB 7.557 KN Efx=0 infAB =70557 KN Efy=0 FBP = 11.3 KN. } FBD consider Joint A 1103 Efx=0 -70557+FAD COS 42.35=0 7.557KN ,FAD 10.225KN. FAD FAE Efry=0 11:3-faD Sin42:35-FAE=0 FAE = 4.,411 en

=> Let us now apply a pair of unit forces in place of Member Be & consider the independent effect of pair of unit forces alone. 'Let us assume the member BE to be a compressive members consider Joint B A Ở : 39KNB © B Ik YEN. 0.673 KN 10.673KN IKN KBD Efze=0 .:locos 42-35-KAB=0 .: KAB=0.739kn. Efy=0 losino 2:35- KBD=0 KBD = 0.673 kN VIEN 0.739 IN. consider joint A A 0.739 KN K KAD KAE Efaro - Kad cos 62.35 +0.73920 ci KAD= I KN: Efy=0 -KAD Sin42:35+ KAE=0 KAE -0.673 KN. consider Joint o IKN 20.673 © KED Efxzo la licos 42-35-KAG= :: Kae: 0.739 KN

ashet us tabulate the above results as shown below 4) comp. =s tre e Tensions-ve. Member P k L PKL kal AB 70557 |-0.739 -18.987 108568 3.4 Bc 7.557 0 5.3 CD -8.752 G. 14 DE -0.739 304 01 1.8568 A E -4.416 -0.673 321 g. 213 1.404 AD -10.225 4.60) -47.04 4.80) EB - 4.601 4.601 BD 11.3 -0.673 3.1 -23.57 1.404. E -80,384 -150723. The force in any member is given by S= P + K: X where X=- E PRL (-88384) 150723 {k²e .! X= 5.1128

The internal forces in members ane SAB -7.517 - 0.739 25:112 - 3.78 KNCC). SBC - 7.557 +0 - 7.5T1 KJU. CC) ср -8.752+0 = -8.752 KN (T) SOE = 0 - 0.739 x 5:112 = -3.778 KN CT) SAE = -4,416-0167345:112 = -7.856 KNCT) SAD = -10.225+1x5112 = -5.113 KN CT) SAB o tix 50112 = 5:112 Kucc) SBD 11.3 -0.673 X5:1|2 - 7.859 KN CC) The forces in members can be shown as. 11:3kN 11:3 KN 11.3 KV A 3.78KN 70557 KY B -Sulli3 KN 7.8 59KN. -7.850 KN 15.716 KN. -8.752 kW. cel 1510112KN -3.77841 D 4.416 KN