Question

6) 2.5-16 2.5-16 A rigid bar ABCD is pinned at end A and supported by two cables at points Band C (see figure). The cable at B has a nominal diameter dg - 12 mm and the cable at Chas a nominal diameter dc = 20 mm. A load Pacts at end D of the bar. What is the allowable load P if the temperature rises by 60°C and each cable is required to have a factor of safety of at least 5 against its ultimate load? Note: The cables have an effective modulus of elasticity E= 140 GPa and coefficient of thermal expansion a = 12 x 10°/°C. Other properties of the cables can be found in Table 2-1, Section 2.2. E=140GPa a=12E-6 or AT = 60°C I de -dc- 20mm - 12mm c -2626 PROBLEM 2.5-16

Answer 1

obtain the area of the Cables and the ultimate. load from table 2-1 Properties of sted tables for. d=12mm & comm. А = 16.7 uu? 60 = 102 KN. A c = 173 mu? Ucz 231 KNO free body diagram of rigid bor. given L = Rx10b ul. E= 140 CPA DT = 60c EMATO To (2b) + Te Cub) - P(Sb)=0. 2Ta+Qта = se — 0 Draw the displacement diagram. D k 26 x 2b k b. t 8c2280 deflection. Tel +20TL = 2502 +220TL GAD GAC 2 TO AC- Te Ao + E2 (AT) AB ACTO. & 2T8 (173)-72667)+140x10x12x1õʻ(60) 6767)(173)

346 TB-7617 Tet 1337525 2820 Toro.222 Tc +3865.68=0 @. Solve exations ☺ a weget. FS. Es=5. Te = 1789.73 +1.12512 -0. Tr] =-3479.46+0.24977p © Determine the allowable loved out 'B' [Toallows to (Te)allow = 102x10 = 20400N. now, allowable Load at CC (Te) astow - lepeni = 2317109 - 4 6200N. Substitute chollow in equation ③ 46200= 1734.73 +1+1251p. Pc = 39520.2uN.

Similarly substitute (Toale in. equo. 204oo a 3479.46 to 24977P Po =95605.79 No. Since the load is less in Cable & so the Cable C goverus. Pallow = 39.520.24 N ox. 39.52 KN :. the allowable lood is 39.52 ko