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Question Number 210180 by BHOOPENDRA last updated on 02/Aug/24

Commented by mr W last updated on 02/Aug/24

$i s t h e l o a d a c t i n g i n t h e s h e a r c e n t e r$ $i n b o t h c a s e s ?$
Commented by BHOOPENDRA last updated on 02/Aug/24

$Y e s w e c a n a s s u m e t h a t$
	Answered by mr W last updated on 02/Aug/24

	$Q u e r s c h n i t t 1 :$ $I = \frac{t {(4 a)}^{3}}{12} + 2 (2 a) t {(2 a)}^{2} = \frac{40 a^{3} t}{3}$ $τ_{A_{1}} = \frac{V (4 a) (a / 2)}{\frac{40 a^{3} t}{3}} = \frac{3 V}{20 a t}$ $τ_{B_{1}} = \frac{V (2 a) (2 a)}{\frac{40 a^{3} t}{3}} (1 + \frac{2 a}{2 a}) = \frac{3 V}{5 a t}$ $Q u e r s c h n i t t 2 :$ $I = \frac{t {(4 a)}^{3}}{12} + 2 (a) t {(2 a)}^{2} = \frac{28 a^{3} t}{3}$ $τ_{A_{2}} = \frac{V (4 a) (a / 2)}{\frac{28 a^{3} t}{3}} = \frac{3 V}{14 a t} > τ_{A_{1}}$ $τ_{B_{2}} = \frac{V (2 a) (a)}{\frac{28 a^{3} t}{3}} (1 + \frac{2 a}{a}) = \frac{9 V}{14 a t} > τ_{B_{1}}$ $\Rightarrow o n l y b i s t r u e .$
	Commented by BHOOPENDRA last updated on 02/Aug/24

	$T h a n k s s i r$
	Commented by mr W last updated on 02/Aug/24

	Commented by BHOOPENDRA last updated on 02/Aug/24

	$C a n y o u p l e a s e t e l l M r . W w h i c h f o r m u l a$ $y o u u s e d t o c a l c u l a t i n g m o m e n t o f$ $i n t e r i a$
	Commented by mr W last updated on 02/Aug/24

	Commented by mr W last updated on 02/Aug/24

	$f o r t h i n - w a l l e d c r o s s s e c t i o n :$ $t ≪ b, h$ $p a r t 1 : I = \frac{{t h}^{3}}{12}$ $p a r t 2 & 3 : I = 2 \times b t \times {(\frac{h}{2})}^{2}$ $Σ : I = \frac{h^{3} t}{12} + \frac{{b t h}^{2}}{2}$
	Commented by BHOOPENDRA last updated on 02/Aug/24

	$T h a n k y o u s i r$ $m e a n s I_{t o t a l} = Σ ({\bar{I}}_{i} + A_{i} d_{i}^{2})$ $W h e r e I_{t o t a l} m o m e n t o f i n e r t i a o f c o m p o s i t e$ $s e c t i o n$ $a n d I_{i}^{} i s t h e m o m e n t o f i n e r t i a o f$ $i n d i v i d u a l s e g m e n t a b o u t i t s o w n$ $c e n t r o i d a l a x i s$ $A_{i} i s a r e o f t h e i n d i v i d u l s e g m e n t$ $a n d d_{i} t h e v e r t i c l e d i s t a n c e f r o m$ $c e n t r o i d o f t h e s e g m e n t t o t h e$ $n e u t r a l a x i s (N A)$ $R i g h t M r . W s i r ?$
	Commented by mr W last updated on 02/Aug/24

	$y e s . f o r p a r t 2 & 3 : I_{i, 0} = \frac{{b t}^{3}}{12} \approx 0 d u e t o$ $v e r y s m a l l t h i c k n e s s t .$
	Commented by BHOOPENDRA last updated on 02/Aug/24

	$T h a n k y o u$
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