cee130-sp06-mt1-Lubliner-soln

cee130-sp06-mt1-Lubliner-soln - 0,0625.(29’ 1 ”...

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Unformatted text preview: 0,0625.» (29’ 1 ” Aluminum tube, % in. o.d., % in. thick (350.25?» a :(n u" may» v «42m.- # 6 steel wire (diam. = 0.192 in.) Esteel = 30 X 106 pSi, EN = 10 X 106 psi (a) Find the magnitude of the force P that will produce a total elongation of 0.020 in.. and find the resulting stresses in the steel and aluminum. (b) If the tensile yield stresses are 15 ksi in aluminum and 36 ksi in steel, find the ultimate tensile load Pu for the system, and compute the safety factor under which it is working as i a. n() LL , -13— -r 3 (t) A’ P 9W in 8 - v . 2-, f ’- o.0 W 0 a, rt. ) J I 'K '5 +____———~tw- ) 93mm? , ° 0.01 La 7 P 59% mow/3 Q lono‘ 50,0953 ‘5 6,, WV ¢ 7/1,; pg; 63,—, 7.4% Es“: ' 0.02/30“ ’ a P“; RNA 6Y|A1)25:(VAL)//'M3w(lohlq} i 9 ’Iou‘lm ,L Seesaw/Takm/ 63 2. The steel plate shown is restrained from expanding or contracting in the y-direction. and. in addition to the stress a”, it experiences a temperature drop of 20°C . Determine the stress am, and the strain 532 if E = 208 GPa. G = 80 GPa and a = 11.7 x 10‘“/°C. y 0m a" : —75 \IPa / 991%?)- _—>, 95%,13 32% —! :0 3; ATE-5117C Flaw; (Pixie) o E)3 ., 0:075 M M ’ loam)" Mm mno‘rfil”) 4,145 4/4817) want 5 26.; MP4 7 £22. 79G @i‘) "7‘ " ya)? + MT ,, 4941.,— (+75 m ’ 7020* ’1540 H0 ' t ...
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cee130-sp06-mt1-Lubliner-soln - 0,0625.(29’ 1 ”...

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