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Unformatted text preview: H5 filiZa’US _ I ‘__ HW #14 Key TahluGmup I 6.3 Member AH nfthe [mas is subjecled to a IﬂﬂU—1b tensile force. Determine the weight W and the axial force
in mcmhcrxtﬂ‘. r 1"" x ‘ ___ __L__ “A! I Analif A _I'
1 195$ '1. . 'H _..'——___'
i H. ’ LL60 in—ﬁjnﬁl} in a . I‘ +
hulln “cal #21:; 2: O 2— —Mdam}zgg:
: 1 ’ o
i.B»'' ' ;‘(/'§ . ‘by/ EACH} 4.:
U W
ER T5 if = = "fUOO whims? 7
“ A; 9:44‘?
' W
ﬂ
9% AC =ﬂl2.é»4*.cr: 1269£k¢lb WMFF. W : 447.; {g 6.18 The lengths oftlu: members of the lruss arc slmwn. The mass urlhc suspended crutc is UGO kg. Determine [he axial] forum in the mcmhcrﬁ.
__________.. 6&2 WWW(9 rem/“32’ch EMM— (chﬂ 0L1.wa un‘f’ f5 :14 ha), “Si/“p.10 : ﬁ(l$mzz§3) uﬁﬂﬁfdm) _
“ ‘fﬁﬂff. 3’ Egg}: :3 Hmﬁw) _,_ "B:  L60 m )mLjf Cl; *rF'fiF 55’ '— “EvCLng’DH Cbmglio FEE—pa x
#17553 5“; '17.}?
9:: E: (15% w}
'\ “3/ +1£5=0= ﬁdeM‘DFCDﬁLL‘F‘D
51:. r‘  4o":
/ iL—Li ﬂintE
C.
p cp= ﬁlm; : inn'1’; w «mph
Mi 1;.
_______.
.19 .
ﬂqub Fit =C‘) 1:. ’015[3 5M (I‘D " 5:14.55
“two i:' dgb C455 2?
an}
/ WW w M
6 (21%:
p I
4.21;; = 0 = H [1.60 Hasn Les #0
+9.75 mm: sue: 56.1.2153)
f. O :3 0 Chad;
Am‘dtb. a, E Fx=0=Hlbo+l.sl «um27.3.3 # no.2; mam “.[aﬂ 46a —; 2F]:O:grg3+hmmn}8 «913.2«440 1m Matlab Ex. 1. See the ﬁle "Matlab 2" in the Matlab Documents folder. This time ttse the approach we
discussed itt class in which you draw ALL the FBDs for all thejoints ﬁrst. showing all unknown member tbrces
as tensile and showing the unknown estemal reactions in the directiotts assumed on the FBI") in the documettt.
Use either of the 3 Matiab methods discussed in tltc document for solving a set oflittcar equations. Print out
your code attd your results from the cotnpttter. Make sure you get the sattte results as for tltejoint—hy—joint approaelt we ttsed in the LEX itt class. First draw all the FBDs for all the joints at once assunting all intemal forces are tensile and any directions you
wislt tor tlte estcntal reactions. Then organize all your equilibrium equations itt matrix form and solve for all
the unknowns. A negative sign will simply indicate the assented directions are wrong. and speciﬁcally for the
member forces. a negative sign will indicate the member is irt compression. ln lvl'l‘l [SC sou. toast of you will do [some already ltavc done) a tniss analysis in which you solved a large set
of linear simultaneous equations using lvtatlab. The method there is tnorc complex than our simple equilibrium
approach. There you assign a “stiffness” to each member that measures how much force it takes to stretch or
eontpress it a cenain distance (actually a gag distance}. You then minimise the total potential energy in all the
members of the structure and due to the external loadittg. The minimization process creates for you the set of
goventing equations rather than doittg as we do itt ME 2:31 by applying Newton‘s Laws directly. The main
difference is that. ifyour minimisation process is done with respect to small changes in the positions of the
joints {as the members stretch or compress), you will ﬁrst solve for these movements {deﬂections} and back
substitute into member stiffnesses to find the member forces. I believe this is the approach used itt your MT] [SC sou course. You will also use this approach lit a simpler way itt ME Elli artdr'or 3'36 and you will call
it lC‘astigliano‘s Principle. if your truss is statically determinate. your resulting forces will be identical to those
we would ﬁttd using Newton‘s Laws directly. But with the energy minimization approach. you canjust as
easily solve statically indeterminate trusses and in these cases. the goveating equations will account [hr the
relative "stiffness" of the different members. in such stmctures. you will find that sonte member forces change
its you change the stiffness ofsonte mentber relative to others. It. the structure is “estemally determinate", these
members with changed stiffness must be those that cause the truss to become internally statically indeterminate.
As the stiffness of such a member increases. its force will also increase. Conversely. as that member’s stiffness
decreases. its force will decrease. If all members that cause the truss to be internally statically indeterminate go
to zero. the truss revens back to a statically determinate truss. However. if you let a member stiffness go to zero
that is not one of those that are causing the truss to be statically indetemtinate. the truss will collapse and your
solution will indicate that the governing equations are “singular”. i.c. cart get a solution. As we know itt ME
201. this is also saying that the static truss is "unstable" or has entered ittto a dynamic condition and its collapse
would need to be modeled by a dynamic analysis method. 0a approaclt to such a dynamic formulation is to
tttiriimixc tltc total mechanical energy. i.c. the sum of the potential energy and the kinetic energy oftlte system.
We will actually do this type of formulation later in ME 20] . but not for such complex assemblies as a
collapsing truss. For tltosc ofyou who find these topics interesting and want to estcod your abilities to solve more comples static
and dynamic problems. c.g. to solve problems wltcrc we do not have to make so many simplifying assumptions
as we do in truss analysis, you will want to consider taking ME 413 (Finite Element Analysis in ME Design)
as one of your technical electives. This will build on lvtF. 2m. ME Sill. ME 31116. M't'ltSC see and all your
olltcr math cottrscs to allow you to solve very realistic problems using computer methods. There you will be
able to tnodel a tmss but not make the assumption that thejoints are pinned as we have do in these 3tltlIcvel
courses so that you could investigate the effects that realistie welded or bolted connections at the joints have on
the beltavior of the truss including the deflections of thejoints. the member forces. and the strength of the truss.
You cant even investigate what happens when a connection fails or a tnentber fails. Does the truss collapse (the
recent disaster in Minnesota is but one oftnatty examples  check out the site
litigation.wiitipedia.orgrfwiki.”[—?l5"l.'vr Mississi i River bridge} or does the trussjust readjust its internal forces. increase its deflections. and “hang in there" until some other
failures occur. Hopefully the beneﬁts of statieal ittdetemtinacy are becoming evident to you. Example 2 .513 a hmncwork problem due on Friday. afar. solve the Imss problem. EXAMPLE 5.? using
MHI'LAH. Th} 11:11 use [he suppnrl mucliunx gix'cn in the example. Instead apply [11¢ Inclhﬂd ul'juinls ElT. each 01'
[hr I'uur pins and win: I'm' all uighl unknuwm uh'im:I MATIAH. 115:; Ilu: tuuurnf'unkncmrm :15: linlluws: fr,”
My. .13, .H). BI]. HE ('D. If]. (1.}, when: [he dircclinnsnfthc suppm‘l rcnctinns an: deﬁned in the FED and lhe bar Ihrcer'. are Elﬁﬁun'li‘d IL] he in tension. . )9 Ag,
.5 emf A : é
——_ —_"5
c}, T A”
Pl 1 _ a;
3’ —a “CA m 1?— Fy” a = A ﬂ’rﬂﬁxéﬁ'mij (E) J'JVG'": A 51' “D 3::
3D
A? Afr x" F__q_ @
PF? :65 E:ﬂ :(ﬂ : aﬁﬁcﬁszSJS fﬁpméﬂiﬂﬁi‘m: f:— F/ QEEEDGQ ﬂQﬁJIEJFfﬁggrw m‘ﬂx c.11153al} $59.6» GIN—49 mxifﬂ—TW Wtﬂﬂr” Hullm
III“. _H«:2 a—wﬂﬂr 5w “La. MﬁTLﬁrbﬁ.opacJ f§1
A1, A65, AD, 6515,, M3, cal)ch (2/ ow?wa lbw NIH4'3 t): KH‘E. CD .QDA’f? Ha) : O
(3;) Afrua‘ M: :0 +IQDEZI> :0 6’) "‘g% ﬂ'SED : 406
C9  Er; FCF =0
@ —CJ:> '57 : :33 “AD misﬁts =~£poo
CS? .3635 +d> = O '{a I O D C) If) D g
l A? *9 0 0 0 c» 0 ﬁre 0
g "C; ﬂ *6: I o o 0 H9 0
_. 0 .'z? o D 0 0 " 40D
{3' D E) o _I C, _I 0 6c '— D N
O O O a ‘3’ "i 0 ﬂ: 51> c:
O b  #:50 D a O 0 C“ EEG?
O a o .E o I o 0 6/ O Here is the IIIﬁle used: %Input E equations as shown an previeus page to solve for the eight
$unknewns Ay, AB, AD, ED, EC, CD, Cx, Cy. The E unknowns are stored in
$urder in the vector x. Nate the directions assumed for the external
%reaetinns are as shown on the FED and all her ferees have been assumed to
%be tensile. Therefore, positive bar forces indicate tension. negative
%indjcate compression. All units in N. ' a=[EI.61IDDDIDI;1 .BDOﬂGCID;D.I5IJElDDD;D.ﬁlﬂ.EGDDD;D
GUUlﬂlU;UEUGﬂlﬂ1:UDl.EDUUU:UUU.EGIUU]
h:[U;ﬂ;ﬂ;4UU;D;ﬂ}Eﬂﬂ;ﬂ] suinutal'h x=aﬁb x=rref{[a bl] Here an: the results from MA'I'IAB:
a _ Fnlumns 1 thruugh ﬁ [1 {}_t5{}[} I .[lElEIfI E] [J [I
I .ﬂﬂﬂ'ﬂ ﬂﬁﬂﬂﬂ 11} i] [II [I
D —{}.ﬁﬁﬂﬂ ﬂ ﬂﬂﬂﬂﬂ I .Uﬂﬂﬂ G
U U.Eﬂﬂﬂ ﬂ —U.Eﬂﬂﬂ ﬂ 0
L} U {1 D —l .ﬂﬂﬂﬂ (II
U H H H H ].UﬂUﬂ
II} [I 1.ﬂﬂﬂﬂ ﬂ.ﬁﬂﬂﬂ l] 0
If} [1' Cl' 0.3000 U l .ﬂﬂﬂﬂ Columns 7 through E U
U
'3' 1']
ﬂ
[1
f} I .ﬂﬂﬂﬂ
ﬂ
0 MHMJUUD
T5E}.UUUU
45ﬂﬂﬂﬂﬂ
ESUﬂﬂﬂﬂ ﬁ[}U.ﬂﬂﬂﬂ
—?.U{J.Uﬂﬂﬂ
em} . [Jenn lﬂﬁﬂﬂﬁﬁ x: ﬁﬂﬂﬂﬂﬂﬂ —T5ﬂ.ﬂﬂﬂﬂ dﬁﬂﬂﬂﬂﬂ Eﬁﬂﬂﬂﬂﬂ EJUI.ﬂﬂf]If} 2ﬂﬂ.ﬂf}f}f} ﬁﬂﬂﬂﬂﬂﬂ Eﬂﬂ'ﬂﬂﬁﬂ Jr:
Iﬂﬂﬂﬂﬂﬂﬂﬁﬂﬂ
UlﬂﬂﬂUUUTSU
Uﬂlﬂﬂﬂﬂﬂdﬁﬂ
U ﬂ {3 1 [l [1 1'] CI 25!?
UHﬂﬂlﬂﬂﬂ—ﬁﬂﬂ'
ﬁﬂﬂlﬂﬂlﬂﬂZUU
Uﬂﬂﬂﬂﬂiﬂ'ﬁﬂﬂ
[J U ﬂ l] E] [II D l2ﬂﬂ Se we see the results are the same for all 3 methods and they also check with the results
obtained by hand in Example 5.2 all yuur text. 6.4] Tho Pratt hridgo truss supports. five I'orcos F L CtrHi kit. The dimension L — ‘3 In. Use Lho method ot‘sootions
to dotomiino the axial toroo in mombor JK and also .IK and U. 'I'hon douhio tho hoight oftho tl'llSS to height 
2L and rosolvo to: .IK. L1, and Al. Conunont on the effect of tho increased depth of the truss. Now let the height of the truss bu. doubled. f ,9 2M5: 5): F0.) +5Lﬁ2‘9Kmkt.) i= is if i= r F": {33 M333 = IJCLL) nfsroﬂ.) up.
ZEWS 1:3“: 4'15“ w mm, .7‘)
easeT 4' 135
m J;
A: Jrze’. F2: 1 *3 :4Lh‘v—A’I
a! i J‘F'zf far: : 413 ILA] matG“
F5 6.54 The truss supports Inads at N. P. and R. Detennine 1he axial forces in members IL and KM and I'M. 1m .111: 2m 2m 2m "7‘5" 1m _ . : J_ . : 'L Ar
. h. +'. a, Ikxlkxlkx EL ‘L (L L Mm =0 = {mm #0 )(L) ~(aX4) We») IL: — “5’ 1‘”? PM aﬁmFr‘ f1 III ...
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 Spring '08
 Biggers
 Statics

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