Week 8, Question 2: Member AC of the following truss is subjected to a temperature change…
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Question “Week 8, Question 2: Member AC of the following truss is subjected to a temperature change…”
Week 8, Question 2: Member AC of the following truss is subjected to a temperature change of +50 °C. Calculate the displacements of node A and the forces in each member using the stiffness method. Take: a= 2x 10-5; EA= 2x 104 kN; the cross section area of AC as A; the cross section area of AD as AV2; the cross section area of AB as 1A. B 1.732 m A 1 m D 1 m
Part 1. The displacements at joint A: a) Az = mm b) Ay mm Part 2. The force in each member: a) NAB kN b) NAC – kN c) NAD = KN
Answer
Sos. Given UzP3 A Degrees in freedom = 2 1.732m V6 Ch.R. AEAL 2x104x0.001 20 KN AEAL – in global coordinate Force in member a because of rising temp. AEAL 2x104x0.001 20KN AEAL – in global coordinate Force within member 2 due to rising temp.
Stiffness matrix for each memben within global coordinate systery 2 Cs 2 CS CS c2 S2 CS as? Cs FA L cs 2 -c2-cs-CS s2 -s2 cs =2m; A – BA 12+1.732 L,= Element 1:- 0,= -60 ; -0.433 0.25 1x &x 104 -0.433 -0.25′ 0.433 0.433 -0.75 0.25 -0.433 -0.433 0.75 i’ -0.25 0.433 0.75 0.433 -0.75 ca cony im ; A LA o ; Lai 10 02 Element 2: – 2 2 O K? 104 oo :12m A-12 A =45 ; 5 3:- Element c k; lou matnie Globool stiffness – 2 0.433 -0.25 -1 -1 0.567 [3.25 -0.75 0.433 0 1.75 0.567 -0.433 0.25 0.433 -0.25 0.75 K: 104 -0.433 0.433 -0.75 2 2 – 0 –
equ. Pk ny p] Kal Kh22 -l = 0.25 0.433 3.25 Po 0.75 0.433 ol143 1.75 Po 0.75 104 0.75 0.433 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.75 104 0.75 104 0.75 0.433 ol143 1.75 Po 0.75 104 0.75 0.433 of us 2 O 0.433 Pu 46 Pp Po ; P2, = P2:0 Boundary =uu = Ug = 46 = 4y = 48 = 0 Uz 8] = 101[***] :]+ 2013] (*:).-10[0535 153] 2017 ] 0.000 6522 -0.000 2113 () =(26 [*][.. o. 6522 mm -0.2113
Co2 . membon Fonces [Xin] – [4]] PIP2 [2] Pon : “J[18] AEL L S ULAF [-c 60 ; = 2mA-A. Memben – | – bbeun – 866 | x 20109 os 5 0.866 0.5 0.000 65 21 -0.000 2 3 2 (r) S. 09 kW 2 stement 2: = 0; l = Im ; A – A. 00 eum20 9 2x to 1. 000 6S 22 -0.000 2113 N, ? – 6.456 kW (c). 000 65 22 Elemeot 3: 6 = us ; L = UI m ; A = 3 4 . Fexrx104 [o – 07070707 0.707 (t), 6.923 kW each force – The Displacement out joint A – NAB A: 0. 6522 pm NAC 5.09 k (T) – 6.25 Ik~ (c) 6.42 3 ko (T) Ay – -0.2113 mm) NAD
Conclusion
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