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Question Number 159297 by rs4089 last updated on 15/Nov/21

Commented by rs4089 last updated on 15/Nov/21

$h o w c a n i f i n d s l o p e a n d d e f l e c t i o n$ $o f t h i s c a n t i l e v e r b e a m a t f r e e e n d$ $p o i n t . b y u s i n g d o u b l e i n t e g r a l m e t h o d$
	Answered by mr W last updated on 15/Nov/21

	Commented by mr W last updated on 15/Nov/21

	$M_{1} = P_{1} a$ $M_{2} = P_{1} (a + b) + P_{2} b$ ${E I f}_{A} = \frac{M_{1} a}{2} \times \frac{2 a}{3} + M_{1} b \times (a + \frac{b}{2}) + \frac{(M_{2} - M_{1}) b}{2} \times (a + \frac{2 b}{3})$ ${E I f}_{A} = \frac{M_{1} a^{2}}{3} + \frac{M_{1} b (2 a + b)}{2} + \frac{(M_{2} - M_{1}) b (3 a + 2 b)}{6}$ ${E I f}_{A} = \frac{(a^{3} + 3 a^{2} b + 3 {a b}^{2} + b^{3}) P_{1}}{3} + \frac{(3 a + 2 b) b^{2} P_{2}}{6}$ ${E I f}_{A} = \frac{{(a + b)}^{3} P_{1}}{3} + \frac{(3 a + 2 b) b^{2} P_{2}}{6}$ $\Rightarrow f_{A} = \frac{{(a + b)}^{3} P_{1}}{3 E I} + \frac{(3 a + 2 b) b^{2} P_{2}}{6 E I}$ $E I φ_{A} = \frac{M_{1} a}{2} + \frac{(M_{1} + M_{2}) b}{2}$ $E I φ_{A} = \frac{P_{1} a^{2}}{2} + \frac{P_{1} a b + P_{1} a b + P_{1} b^{2} + P_{2} b^{2}}{2}$ $E I φ_{A} = \frac{{(a + b)}^{2} P_{1} + P_{2} b^{2}}{2}$ $\Rightarrow φ_{A} = \frac{{(a + b)}^{2} P_{1} + b^{2} P_{2}}{2 E I}$
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