Post on 03-Jun-2018
8/12/2019 237exam1 Sp09 Solution
1/4
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 * )$;7 m
w = 5.3 mm
8/12/2019 237exam1 Sp09 Solution
2/4
- 2 -
2. %" %""!en membe*s " 3 x 6 inh )ni#"*m *etan+)$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 #"* #)$$ *e!it.
6:
3:&0
(
(
&0
&0
(;; ( "s &0
( ( sin &0
( 2&00 $b.4x4"
8/12/2019 237exam1 Sp09 Solution
3/4
- 3 -
3. he stee$ #*ame E 200
8/12/2019 237exam1 Sp09 Solution
4/4
- & -
&. his assemb$ "nsists "# an a$)min)m she$$ #)$$ b"n!e! t" a stee$ "*e an! is )nst*esse!.
5ete*mine a the $a*+est han+e in teme*at)*e i# the st*ess in the a$)min)m she$$ 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