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Question Number 68418 by necxxx last updated on 10/Sep/19

Commented by necxxx last updated on 10/Sep/19

$T h e U D L i s 2.5 K N / m . T h e f i r s t p o i n t$ $l o a d i s 35 K N . I^{'} v e t r i e d s o l v i n g b u t I$ $e n c o u n t e r e d s o m e i s s u e s w i t h t h e d i a g r a m .$ $I^{'} l l b e r e a l l y g r a t e f u l t o g e t t h i s s o l v e d$ $w i t h t h e p o s s i b l e d i a g r a m s .$
Commented by mr W last updated on 10/Sep/19

$p o i n t l o a d s a r e 35 (n o t 85), 25, 5 K N ?$
Commented by necxxx last updated on 10/Sep/19

$e x a c t l y s i r . I a l s o s t a t e d t h a t t h e U D L$ $i s 2.5 K N / m$
	Answered by mr W last updated on 10/Sep/19

	Commented by mr W last updated on 10/Sep/19

	$12 R_{2} = 5 \times 14 + 25 \times 9 + 2.5 \times 6 \times 6 + 35 \times 3 = 490$ $\Rightarrow R_{2} = \frac{245}{6}$ $12 R_{1} + 5 \times 2 = 35 \times 9 + 2.5 \times 6 \times 6 + 25 \times 3 = 480$ $\Rightarrow R_{1} = \frac{480 - 10}{12} = \frac{235}{6}$ $c h e c k : R_{1} + R_{2} = \frac{245 + 235}{6} = 80$ $35 + 2.5 \times 6 + 25 + 5 = 80 \Rightarrow o k$
	Commented by mr W last updated on 10/Sep/19

	Commented by mr W last updated on 10/Sep/19

	$t o d e t e r m i n e M_{m a x}$ $p o s i t i o n w i t h s h e a r f o r c e = 0 :$ $d i s t a n c e t o p o i n t l o a d 35 K N :$ $\frac{25}{25 + 65} \times 3 = \frac{5}{6} m$ $M_{m a x} = \frac{235}{6} \times (3 + \frac{5}{6}) - 35 \times \frac{5}{6} - \frac{2.5}{2} \times {(\frac{5}{6})}^{2}$ $= \frac{5765}{48} = 120.1 K N m$
	Commented by necxxx last updated on 10/Sep/19

	$T h a n k y o u s o m u c h m r W . I^{'} m v e r y g r a t e f u l$
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