Question

For the truss shown in the following figure, the temperature of member BC is raised by 10 °C, and member BD is raised by 150 °C. EA= 300000 N for all members and a= 1/75000 1/°C. Use the st do the following: А B 5 m A с D 5 m Part 1. Calculate the displacements at the joints: a) Ax = mm b) Ay = mm c) B. = mm d) B = mm e) Ct = mm Part 2 Calculate the force in each member: a) N = N b) NAC = N N c) NBD = d) NBC = e) NCD = N N

Answer 1

Solution Given. A TBC= 10°C ATBD = 150°C EA = 300000 N = 300 KN 1 loc 75ooo 2 3 → 1 B (5) 2 oo 3 C In this problem, u deflection U = Uz = Ug = 0 due to support & Uų, Us, Ua, us represent Unconstrained degree of freedom . temperature of the member (2) is raised by 150°C AL² = 2LAT So, 1 xsx 150 = 0.01 m 75000 The forces forces are developed in member '2' as it prevented from expansion. Pr ) Cosol (PP) من که co 300 x 0.01/ Sino (Pi) - Cos o - Sino CSScanned with CamScanner

The displacement of the member (5) was raised by gooc ... ALS & LAT LS10 = 9.42 *10-4 The forces developed in member (5) as it was not allowed to expand is 0.27 m 25000 0.707 = 300 x 9.48X10-40.407 1.). (P.)? (C.). 1- 0.707 -0.2 -0.707 global force vector due to the thermal Load is, [Pilt 0.3. o O. ? 0.. (Ps4 (Polt Pad Ped ooriting the Load - displacement relation for the entire is given as tross -0.071 -0.071 P 0.271 0.071 -0.20 -0.3 P2 -०२/ 0.071 0.271 -0.2 03 -0.071 -0.071 0.271 -0.071 -0.20 -0.071 0.071 pu o -0.071 0.129 -0.2 0.071 0.071 Ps. = AE -0.071 -0.071 0.271 6,071 -0.2 PC 1-0.071 -0.071 - 0.7 0.071 0.271 Us Pq -0.071 0 0.071 -0.2 o 0.271 -0.071 Ub Uy 3 6.071 -0.071 -0.071 0.271 Us CS scanned with CamScanner

In this problem P = P₂ = P₂ = Py = P = P = Py = Pg=0 and Uz = Ug = 0 UG Thus 0.271 0.071 -0.071 -0.2 -0.2 -0.071 0.071 U2 0.271 1 U3 AE10.20 0.271 -0.071 Uu -0.071 0.129 US -0.071 -0.071 0.271 on, solving we get (10.27140.2)+(0.071X0.25 + 10.07180. ?)) vi 1 AE -4 2.75 x 10 m U2 - 1 ((0.071X0.2) + (0-271 X0.2) + (0.071 X0,2)) AE 2.75 x 10 M 1 03 (1 0.20 x0.2)) НЕ 1.33 x 10 м UHF 1 AE (0) Us AE AE (l (1 -0.071X0.2)+ (-0.071X0.8)+10-271x60.2))) m :- 2.75 x 10 Hence, Ax 11 Ug = 0. 133 mm 11 Ay - Uu U = 0.275 mm Bac С 0.275 mm By 0.275 mm ca U. CS Scanned with CamScanner

Now, PE reactions are computed as 1.0.071 -0.071 -0.2 0.071 -0.2 02 درد اور Pz -0.071 0.071 و بي) P - 0.2 0.071 -0.071 Un US Member AB, L=5m, oro [-1 0 1 o AE Uz Vu so, Pa 8.5 20 N Member BD I-5m Pe' AE [o-1 -1 o 1 Og 5 300 X 9.42 x 10 03 -266.1 N Member CD 2=5m Po Us AE € (-1 . 1 o s Uq 16.5 N Sember Ac;L=5, o=90' US PA A 10 -1 0 1 S Us UH Dember c3 06 [=552 0 = 45° P, AE AE (-0.707 - 0.707 0.707 0.707 1-300x 0,01 CS canned with Cam 31.9965 N

Demben AD ; 2=552, 0 = 6.4.5 Uy 0.707 -0.707 -0.707 0,707 0.707] 5 Uu - - 5.6418 N Hence, NABE 8.52 N NACE o NBD = -Q66.1 N NBC = 32 N Nco 16.5 N CScanned with CamScanner