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68_Mechanics Homework Mechanics of Materials Solution

# 68_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 2.15 The specimen show is made from a l-in.-diameter cylindrical steel rod with two PROBLEM 2°15 1.5-in.—outer—diameter sleeves bonded to the rod as shown. Knowing thatE= 29 X 106 psi, determine (a) the load P so that the total deformation is 0.002 in., (b) the corresponding deformation of the central portion BC. SOLUTION L- (a) g = g 31L .. g: E If P=U~9>¢IO‘XO-002 Kean)" = msea'noz-Jib. 53L: 3 2:45:53? (3.3m) -_- 2.16 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. PROBLEM 2'16 Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deﬂection at A is zero, (b) the corresponding deﬂection ofB. SOLUTION (0.) Am = \$6}; = %E(0.02o)1 7: 3H,!é x10" m1 I: ‘t l .3 ‘ 20-mmdinmeter A3: .. a“: a 1&(0.0¢o)z= 2.3274 N10 M -. Force in member AB .‘s ? +eusion EJ’ al‘ 5 _ _l,—_n_e __ (ergo‘xoﬁil o a 0.. A“ E Am (3)01107X3HJ4xm“) = 72.756 VIC)“ m ‘ j, 60-mm diameter \ Force m member BC is Q'- P ConprtsSt‘on . _ - co-P>L.5_ CQ--P){o.5) “WWW 33‘ ‘ E nu ‘(7owowx2327m5n .-.- 2,5243 Ho" (Q-P) FOJ‘ zem derjea‘l'iov‘ av'l‘ A \$3; = 3m 2.52g3xio""(Q-P) = 72.75am)“ .1 62- P: 28-8)!103 N G. = 28.3 Ho‘ + 4x103 = 32-9 x10” N u (A) I“ 00 x Z I (b) S“; r 5%,, .- \$3 ? 72.7\$L¥10"‘m = 0.0723 mm 4 ...
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