Unformatted text preview: 2.15 The specimen show is made from a lin.diameter cylindrical steel rod with two
PROBLEM 2°15 1.5in.—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‘XO002 Kean)" = msea'nozJib. 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 ‘
20mmdinmeter 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, 60mm diameter \ Force m member BC is Q' P ConprtsSt‘on . _  coP>L.5_ CQP){o.5)
“WWW 33‘ ‘ E nu ‘(7owowx2327m5n .. 2,5243 Ho" (QP)
FOJ‘ zem derjea‘l'iov‘ av'l‘ A $3; = 3m 2.52g3xio""(QP) = 72.75am)“ .1 62 P: 288)!103 N G. = 28.3 Ho‘ + 4x103 = 329 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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