8/12/2019 237exam1 Sp09 Solution

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MECH 237-002 Name ______________________________ Exam 1, Feb. 16, 2009

1. Link BC is 6 mm thik an! is ma!e "# stee$ %ith a &'0 M(a )$timate st*en+th in tensi"n.

hat sh")$! be its %i!th w i# the st*)t)*e sh"%n is bein+ !esi+ne! t" s)"*t a 20 kN $"a!

%ith a #at"* "# sa#et "# 3.0/

h"% a$$ %"*k #"* #)$$ *e!it. he FE Han!b""k ma be )se! as ")* *es")*e.

600 mm

&0 mm

90

w

( 20 kN

4 B

C

5600

&0

'

&3

s$"e

( 20 kN

5

C

4

Plan:Ultimate strength given.

Determine the Allowable Stress = P/AWhere P = internal force along CB.Where A = (w(t of member CB.

!iven "S = #.$ = %ltimate / allowable

Allowable = %ltimate / "Sallowable = %ltimate / "S = &'$ Pa / #.$

= &'$ Pa / #.$ = )'$ * )$+,/m- = P / A

Use Statics analsis to etermine the val%e for

P in member CB. Call it 0 for tension.

1 abo%t 2t. A = $ = 0(&3$ 4 P(+$$0 ( &3$ = 5$ 6, ( +$$ 0 = 5$ 6, (+$$ / &3$0 = 5' 6, = 5' * )$7 ,.

,ow anal8e member CB

= )'$ * )$+

lb/in- = P / AWhere P = internal tension9 0 = 5' 6,.

A = 0 / = 5' * )$7 , / )'$ * )$+,/m-

A = )++.++ * )$;+m- = w t

w = A / t = )++.++ * )$;+ m-

$.$$+ m

w = 5.3 * )$;7 m

w = 5.3 mm

8/12/2019 237exam1 Sp09 Solution

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2. %" %""!en membe*s " 3 x 6 inh )ni#"*m *etan+)$a* *"ss seti"n a*e 8"ine! b a

sim$e +$)e! s$ie. n"%in+ that ( 2&00 $b, !ete*mine the n"*ma$ an! shea*in+

st*esses in the +$)e! s$ie.

0he cross;sectional area = (#$ 2si

Shear stress = = P// / Ao = 5&$$ cos &$(sin &$ / )3 in- = +'.+'& 2si

h"% a$$ %"*k #"* #)$$ *e!it.

6:

3:&0

(

(

&0

&0

(;; ( "s &0

( ( sin &0

( 2&00 $b.4x4"

8/12/2019 237exam1 Sp09 Solution

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3. he stee$ #*ame E 200

8/12/2019 237exam1 Sp09 Solution

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&. his assemb$ "nsists "# an a$)min)m she$$ #)$$ b"n!e! t" a stee$ "*e an! is )nst*esse!.

5ete*mine a the $a*+est han+e in teme*at)*e i# the st*ess in the a$)min)m she$$ is n"t t"

exee! 6 ksi, an! b the "**es"n!in+ han+e in $en+th "# the assemb$.

4$)min)m> E 10.6 x 106si, 12.9 x 10-6; F, 1.2': ")tsi!e !iamete*

tee$> E 29 x 106

si, 6.' x 10-6

; F, 0.7': !iamete*

0otal eformation of al%min%m = total eformation of steel

@ 0 /; P@ / A = @ 0 /; P@ / A

Al%min%m: A = /& ).5'- ; $.'- = $.3'& in-

Steel: A = /& $.'- = $.&&)3 in-

Al%min%m: A = ( $.3'& in- ()$.+ * )$+lb/in- = 3.#5' * )$+lb.Steel: A = ( $.&&)3 in- ( 5> * )$+lb/in- = )5.3)5 * )$+lb.

ach term has the length9 therefore o% can cancel the @Gs in all terms.

0 ; 0 = P / A steel ; P / A al%m

( )5.> 4 +.' )$;+0 = P ) / A ) / A

Ho% have info abo%t the al%min%m shell: + * )$7 lb / in- = P / A

P (al%min%m = A = ( + * )$7 lb/in- ( $.3'& in- = &.)5& * )$7 lb = P

0 = &.)5& * )$7 lb / +.& * )$;+ ) / A ) / A

0 = #+.#)5' * )$+ ( ) / )5.3)5 * )$+ ( ) / 3.#5' * )$+

0 = )&'.> bac6 s%bstit%te this into either sie of the e%ation

Deformation of Al%min%m = eformation of Steel = )$.'# * )$;#in = $.$)$'# in =

h"% a$$ %"*k #"* #)$$ *e!it.

in

4$)min)m he$$

tee$ C"*e

0.7' in

1.2' in

al%m > steel

al%m com2resses against

the steel

0hin6 IAction = JeactionP al%m = ; P steel

Bone I

al%m = steel