Question

Both portions of the rod ABC are made of an aluminum for whichE = 70 GPa. 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 correspondingdeflection of B.0.4 m0.5 m

Answer 1

Explanation:

Δ$L_{BC}$ = Δ$L_{AB}$

$\frac{(Q - 4000)(0.5)}{3.14* 0.03 *0.03 *70*10^{9} }$     (1)

= $\frac{4000*0.4}{3.14*0.01*0.01*70*10^{9} }$

Q = 32,800 N

now put this value in equation 1.

Deflection of B = $\frac{(32800-4000)(0.5)}{3.14*0.03*0.03*70*10^{9} }$

                       = 0.0728 mm