Answered: A. 5.25 in. 0.5 in. х -x 10 in. 0.5…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

A 20-ft-long column is made of aluminum alloy 2014-T6. If it is pinned at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 13–30.

A.
5.25 in.
0.5 in.
х
-x 10 in.
0.5 in.-
0.5 in.
A10 in.

Expert Solution

Step 1

Now we have

$A = {(b h)}_{f} - {(b h)}_{w} {(b h)}_{f} = b r e a d t h a n d h e i g h t o f f l a n g e {(b h)}_{w} = b r e a d t h a n d h e i g h t o f w e b A = 10 (10 + 0.5 + 0.5) - 10 (10 - 0.5) A = 15 i n^{2} N o w m o m e n t o f i n e r t i a a b o u t x a x i s i s g i v e n b y I_{x} = {(\frac{1}{12} b h^{3})}_{f} - {(\frac{1}{12} b h^{3})}_{w} I_{x} = (\frac{1}{12} 10 {(11)}^{3}) - (\frac{1}{12} 9.5 {(10)}^{3}) I_{x} = 317.5 i n^{4} N o w m o m e n t o f i n e r t i a a b o u t y a x i s i s g i v e n b y I_{y} = {(\frac{1}{12} h b^{3})}_{f} - {(\frac{1}{12} h b^{3})}_{w} I_{y} = (\frac{1}{12} 1 {(10)}^{3}) - (\frac{1}{12} 10 {(0.5)}^{3}) I_{y} = 83.43 i n^{4}$

Step 2

Now radius of gyration is given as

$r_{y} = \sqrt{\frac{I_{y}}{A}} r_{y} = \sqrt{\frac{83.43}{15}} r_{y} = 2.35 i n N o w t h e s l e n d e r n e s s r a t i o c a n b e g i v e n a s {(\frac{K L}{r})}_{y} K = E f f e c t i v e l e n g t h f a c t o r L = l e n g t h o f c o l u m n r = r a d i u s o f g y r a t i o n {(\frac{K L}{r})}_{y} = (\frac{1 x 20 x 12}{2.35}) {(\frac{K L}{r})}_{y} = 101.76 A s {(\frac{K L}{r})}_{y} > 55 H e r e t h e s l e n d e r n e s s r a t i o s a t i s f i e s$

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