Question

Arigid bar of negligible weight is supported as shown in the Figure. find the stress in each rod if the temperature rises 30°C after a load W = 120 kN is applied. (hint: for compatibility equation use the ratio and proportional relationship between material and 8t). Bronze L=2.5 m A=1300 mm E=83 GP Steel a=18.9 x10 m/m.°C L=1.25 m A=320 mm2 E=200 GPa a=11.7 x10-6m/m.°C -1.2 m !! -3 m 2 m w

please show all work

Answer 1

The total deflection in the rod is equal to the deflection due to axial load plus thermal deflection. The compatibility equations are find out by taking the deflections in the bar and use the similar triangles to formulate the relationship between deflection in both of the rods. The unit of load in the rod is taken as kilo Newton, do not forget to change it for calculating deflections.
Solution P is the load in in the rod. and subscript and B use for steel and bronze are wo O orod suspectively. B 1 케 1.2 m 1 * Moment at 120 KN point o 9 thart is hinge K 3m joint is zero. 14 sm Mo - o Ps(1.2) - 120(3) + 18 (5) 30 Ps (1.2) + Pp (5) » 36050 1.2 from figure (2) & SB Ss coo → 112 5 5 588 =) 1.2 SB figwe (2) So total deflection in the rod. Ss Esp. + EST un deflection deflection due to load thermal expansion due to com 4s (BT) Ls + As Es Similarly 88 » PBLB AB EB + % (OT) LB

From eqh Ps x (1.25x23x103 11.7 X106(30) (1.2581033] 112 320X 200x lo 3 X 2.5XL03x10 1300x 83 x 3 18.9X166(30) (2.58/03) it (1.25) 0,4387 + => 0.24) (7.5x/ 3 64000 0.3402 + to 7900 Ps (8.25 7x103 PR (0.61X/63 키 0,3402 20.4387 > 3 U 64000 107900 From ean Ps 360 2015) 300 (4.1667) PB UL 1.2 Put in eah Boo 4.1667 P8) 1.2.58163 PB (0.63x10² -0.0985 64000 107900 0.6 5.268 640oo tora 4436 -5.9578 PB 107900 PB » 68.53 KN Ps » 360 68.53(5) 14.458 KN 1-2 TS » Ps 3 14.458X0 320 45.18 Meal Ary AS B 68.538103 52.75 MP E PB AB Ay 1300