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Question Number 31114 by rahul@ last updated on 02/Mar/18

Commented by rahul@ last updated on 02/Mar/18

$p l z s o l v e i t . . i t s u r g e n t$
Commented by rahul@ last updated on 03/Mar/18

$t h a n k u s i r$
	Answered by mrW2 last updated on 02/Mar/18

	$f o r c e i n a l u a n d i n s t e e l = F$ $f i n a l l e n g t h = L$ $c r o s s s e c t i o n o f a l u r o d = A_{A}$ $c r o s s s e c t i o n o f s t e e l r o d s = A_{S} = 2 A_{A}$ $l_{0} (1 + α_{A} θ) - L = {F l}_{0} / A_{A} Y_{A} . . . (i)$ $L - l_{0} (1 + α_{S} θ) = {F l}_{0} / A_{S} Y_{S} . . . (i i)$ $(i) / (i i) :$ $\frac{l_{0} (1 + α_{A} θ) - L}{L - l_{0} (1 + α_{S} θ)} = \frac{A_{S} Y_{S}}{A_{A} Y_{A}} = \frac{2 Y_{S}}{Y_{A}}$ $2 {L Y}_{S} - 2 l_{0} (1 + α_{S} θ) Y_{S} = l_{0} (1 + α_{A} θ) Y_{A} - {L Y}_{A}$ $L (2 Y_{S} + Y_{A}) = [2 (1 + α_{S} θ) Y_{S} + (1 + α_{A} θ) Y_{A}] l_{0}$ $L = \frac{[2 (1 + α_{S} θ) Y_{S} + (1 + α_{A} θ) Y_{A}] l_{0}}{2 Y_{S} + Y_{A}}$ $L = \frac{[2 Y_{S} + 2 α_{S} θ Y_{S} + Y_{A} + α_{A} θ Y_{A}] l_{0}}{2 Y_{S} + Y_{A}}$ $\Rightarrow L = [1 + \frac{(2 α_{S} Y_{S} + α_{A} Y_{A}) θ}{2 Y_{S} + Y_{A}}] l_{0}$
	Commented by Tinkutara last updated on 05/Mar/19
	Sir, why in (i) L is subtracted but in (ii) it is subtracted from L?
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