Mecánica de materiales beer, johnston - 5ed solucionario

Problem Solutions

By
Dean Updike

EL SOLUCIONARIO
DA AT A.

Chapter 1

a sa ind a nd HC ae wi u an ned
Problem 1.1 shown Deniz de grief face sh eee no A
pane od old |

Bg? ECON IMSS mat :
a 7

a= yy MEE
50130 ep

Ag = POST = 44079 me

Gee? o =P

= MP 00543 nee
Equating Sm de 2 Sx
Sots via Pe 2 (sas = 226-4015 P) Pour
a 2 1.2 te Prod. 1.1, knowing that P= 160 AN, determine the average noemal stress at
da
iva fic ay ae wi pe e
Thera he ea ee Pr ah aoa
nennen RT
(a) Red AB. nm mins ae

Paste ba Chemie) 4

Mao AR. BEE nn » Tard
-

F- 160 (21020) =-80 kw Le 80 AN compression.
Auer Ege TE ers

er 1801 MPa «a,

ay nto. nie coi A e e e

ynwwelsolucionario,net

13 Poo slid stint

rs Aa Cte ele ogee Ban dd e
shown Riga ree na us mt os 75

A 150 AC ene ende la alas od

Paina

40 430 = 10W + 7040 N

P= 3010 = 30x10 N

232x108) ge
müs) TIO

159 mm a

wo Red AR.

1.4 sll india mis A a ae welded ee u a ais
how Koi hat

50 mon and, 3m Breage ona wet
Inden of) ed AN he A

Pz No +30 + Tom = TON

Ar date GO + 1.9635 (hme LS IO" e

= 35.7 10% Pa

Ge = 35.7 MPa a

Su 42.9 MPa e

Problem 1.5 toro as Bone 5.0 a ea itn esti to te DO
m ce as an ASA de ss 2
Fondo tht one dr 2m dm

he bine a Ct be a u
inner dinette

35
me

= 222.9 not m

di 7 14.48% 10m

1.6 To sel is e ob eld tas y ese 6 ae
Probie 1.6 _arength steel bolts ttting snugly inside eylindrical Penes spacers. Knowing
‘Non oman ee nx ene 30 sacs ad 150 Mb he

as

KE cack beft location the opper plade is pulled down by the tenside
force Py cf the bolt. At the same time the spacey pushes that
plate upvand with compressive force Pi. In orden do main
eau dibriom

R
For the haft,

PF Sd

For the spacer, AC]

Equating P and Fe,
Teal = EG)

Ar 262 mm a

Problem 1.7 1.7 Each of he fv a 8 36min ion etna cet
‘ole of fgets or he bs sorting op ad D,
Fin Cm

Lo.mo)F — (0.025 + 0.040) (20107)

Foo = 22.5 #10 N Link BD is «tension,

-(0.0%0)F,, - (2.028 )20.0') =

| Fog = 71S Lite CE 13 in compression.
Wet area oF one Bink For tension = (0,008)(0.086 - 0.016 )

For two parallel links, Ang: 320x100
k BD.

= 160 x10 .

Tensile aires in

> Be Se wot 101.6 MPa, |
Gao > Re: 101.56 wo!

Area Fon one Pink in eum pression = (9.008)(0.086)
= 288x 10 mt Far tuo para Mel Lakes, A= 576-0" m“

= ro? 5
E A =

repr Meri. 0208 Te er te pct Mal y ig pn
e mc o serene on aor eas

Problem 1.8 A King tt bs etal pation fhe ok 29 Rac uno resto
exo 60 me, dto e agudo ler Pe wena:

ini prono 20950 MPA

Fer SA
=(80%104Y(200 15 =
donen e AÑ

ers ar 7 \l
200 LA:

Use Free Baty AC for staticr, Se

DEMO (om ior ed) + BE (or 1e) 4.4)

= Plorsi4)= ©

32.10 Peal kN at

Robe 1.9. 1.9. Knowing ln ak DE 28 rm wie nS wm nick, deme

se moma ae nth on por ht kw 10 0,000 = 90°

Use member CEF as a free belly
Demo
Lo Fat 2124 021n8)-(0:4)(240«56) = O

For = (60 sin® 320.0 N

Aue = (0:025)(0:003) = 75/5 m?

be
Ge: EE
@ 6-0: Fa--326 N
32e
See Faut #27 MPa -

N B= 40": Foes -/60N

> 2.13 MPa =|

Problem 1.10 1.10 Lise salam recta scion Sn tick nd a
Determine he axa tre en pon of he i

Body Viagra of Plate

Note thet He tue 26070 Forces
Form à coupte of moment
COPA

arm

DEM, zo: Mann ~ (Fig con 80 \lnzsm) =O
Fe = 6650 w

Bree oF Sint Aus (Smm\(iz mm) = 36 mnt
En, st

re $
a

= iur Ser 185 MPa Al

Sacer etree tre teers
A AA AS

1-11 Ts bar EPC appart bythe tas ten sb. Ko
ine ect oh nm di oi
- ll the cl ese CG

Problem 1.11

Using perkion EFGCB as a free body
Hag ro Gt tyme
Fue = 28 EN

Using beam EFG as à Free body
Dm zer
Fee = Fae = au
Cross sections! ara of member CG
= But = Eco) = 2544 x10 mn”
Norma cheers im C6.

Ges Bee
. x
i I5EN -
Problem 1.12 1.12 Th ig EFG pont ns ju shorn Det

rin e eos sehn wen member A or whi ha tes
rer à 105 MP

Using orion EFGEB as à Fo body
MAR Or he no
Fe = ZEN

pe + 105%

238.8 me

15 EN TS

rnin ute © 2009 The Mean Compa o. A cure opa Maal mdp pa
Sane green etcetera RAT

yoww.elsolucionario.net

Problem 1.13 1.13 Acoople M of marie 150 mis piso ak faci. or
‘he pcton shown, deine a teen Pee old engine se a

um. (0 de vaae manne crecio rod BC: whch aad

?

Use piston, rod, and crane
together as Free body. Adi
wall reaction H and bearing
reactions Ay and Ayı

D 2M- 0:

Lo.230 m)H - 1500 Nin = ©
We san N

e ; Use piston afore as free Ay
body: Note that rod is
5 fhe orgs mentors hevee the direction
N oF Force is known. Draw He force
triangle and sole Fon Par Far by
Ref |p Pree

Fac leo Be E = 203.8 mn
BP. 2e + pe alot
Pa 22 à: Ps m86n0 N
Boa PIAR -
ea, Arne

Rod BC is & compression member. Hs area it USO mar = 450 AJO m
TAS ST

A 450 410€
@) Ss = = 41.4 MPa «a

=-41.4x10° Pa,

in tt 20 Te Gr A Cpa A re pen a dt
than nay omer) DV
ba pwc Ei era pur. ose sing isms ing pr

1.44 An let br posed by meas ofa sale sae nr
std 125. Suerte ts ie wel is DEF

~The atic en tort DD acer ire 9 For
‘Meo shown, dane he eral ase

Disks aa

Problem 1.14

Nur (200 Nat wf) «1462.00 À
Emo:

850% - usofimer.oon) = 0

Re 2654,5 N

24305 N° (come
„as
roots

A

AS

ar mr €

Try Mon oa er ou. Al es nut pa =
ra par lag Kane

yoww.elsolucionario.net

14, Mi made emba an rta oy Pd esc
Problem 1.15 ‘ibe a dear con Apot edge Is

done a th cesa ls US nl She men st Ie br
deine es simio here sre cl

There ave Rue separste arcas that ave gfveds
Each of these areas framsmels one half the
the 1S kN oat. Thor

FAP = HS) 7.5 W = 7500 N

Let 42 Length of one qlued area and W > 1502 0.05 m be its width.

For each glued ares, Ar dw

Average shearing stress -£
The «Movable shearing ss #8: 70x10 Pa

1 E 2500

Sales Pod, tw” aia
Tate? Length Le Ls Arts P= 192.859 6 + 192.85

L= 242 mm 4

ABC m © 142 BS mom

Problem 1.16 1.16 when he fe Pcs AN, the won rien show li soe
tn ein. Domine un
longa tcs ri a ln

Area being sheared
FO mi» mm = 1350 med + 18602106 m"

Fore Pr 8x1? N

Shearing stress Tr EB - B20, = ne PaaS. A a

rep Man co Can ores Neots henna
arr yey men nad eee nse el BS ele ed nenne
poniendo pop hag a eps sr pe

Problem 1.17 1.17 Two wooden planks ah 2a ick and 28 mm sd rejoined ye dey

mort shown. Koons hatte onde aloo ri wie
‘cra shang ures Teter MP die he magnate xa od
‘hci ne ia

Sa moras mus he sheared off
when the giel Haid, Bach d these

areas hes dimensions IGmm * om,

Hs ren being,
A UE) 192 mal HR

HE Haidere the foro E cueried by each of
Fe YA = (8x10%Mi4z x10) = 1536 Ne 1536 kv
Pr Gh =(6X1530) > 224N

Sines Here am tite fur areas

Problem 1.18 1.18 Ala Ps api os el spor esto by ana pe info

cha rm dame Js eos dled’ Ken tht She sheng eco
‘no wot ee 180 MP in heel sed and 7 MP in he abou pe,
eminem get oa ht can be apli ed.

Ls For the stent rod,
T A, = At, = (mo.al2Yo.o10)

Lo
af + RRA
P = (i80x10°\(376.997 15°) = C7.86 40" N

For the afuminum plete,
Age Tdi, = (1X0.040X0,008) = LoDS 3116 m!

En he th
un Re GA

Py = (70x10 (1, 053% 10") = 70.372 n107 N
The Hiniting value For the fied P is the smaller
FP, ond Pre

Pr 67.46% 10 N T=67.9kN e

Problem 1.19

19 Thoacel cela cho ppg eer em shine = 7540
tote banal ova eh fe bea ee Dg es a
Member nat excess) DAP

sonen
=“
sr

E LP aio®
Selving for LE Lo y Konz

178.610” m
[= 178.6 mm -
Problem 1.20 She The ina of ud 22m adhe rt dom 0 nun 2

art enr ate hele: Dr he alist
‘Moat out dane oft eer, knowing el he a aia res mie
‘ea i 5 MPa an hat he serge brn es Keen he under nd
Hin rst esse SM

=
PES Ste) rod: Az FL,
AS ©» 350% Pa
Pe GA = (35:100(330,12 000%
13,395x10* N

Wache Suit Pa

Es

Required bearing avent

Bt Ar Har- 2")
die ‘à
(0.025) +

= oo

d = 63210 de 633 om a

121 A AGAN ail od apple a se wooden pst ha ie super à
covet Foti eng on aonb oi Detemioe () mu tig
“grees he corte ating, (in er af he et, e sic he Sure
tearing rs ai SP

ta) Bearing stress on concrete Fou
P= Yo kN = Yorto' N

Ar (0020) © 1210 man 129107 met

y

ANS +
S-* Fibs 7 3.2840" Pa
2.23 MPa a
(Footing ace. Prost e las kPa Alo Pa
ak 2. fon . o
6-À ne Es HU + 0.2080

Since Me area is square, AZ bt

wo tA > forsee

0.525 m BF S25 mm «a

122. acl fal sangre a or W200 59 cl |
Pen 1-22 sc wenA © 1560 mul dig oa cons onion

By gs lt shown, Keone the erage ral sch ol
‘nn nmol eased 200 MP ad tbe being sess om the conte
Foundation mut os exe 20 MPa, determine the ide a0 he pte at
vel ori the wot ccm nd sf sig.

For the column = = E

a PR GA -(avowelye7560x/8" sr

Yor Me ara phte, 67 20m

Ae EAE 2 0.0756/m!

Since the plete is square A ot

as Ik hotes gnats ™
2275 mm

reger Mr. 200 Te rm Cope je np of a
A D Poe hasan ay sme ny si

Prem na PEARED) éme stags Sivan ena

Th eae Sete body
dragon of AED.

| Sink RED is a
: ¡EA
the renetion at €
is dlicscted deveod perl €, tha ialecpeation
at the Bines of action of the of hes ts

225 mm

forces.

From gesneiny, CE fps
MER «0: HC-Peo

LEP = (2.6N(500)~ 1300

©), E eo ee 2.0 MPa à
©, Moo

Gomes
ey ee ao
SS

a = ad Pe Ge Bit MPa a

1.26 Kung that a ne P of apie 75 N isp ob po shown,

Problem 1.24 ‘eine (ab dimer one pina for whch he arr mi es ins
pin 40 MP.) de cemepondig being ste in he pel 0) Me
responding eg sin ecc Spo wet

Draw Free bed fie f
dagen of AED. coo AF
Since ACO is a a r
Be menden, =

the reaction at © E

ie divested fowaed peiat E, the inderstchion
4 af the Lines ef action of the other tue forces.

From qeemeley, CETTE © 325 mm

mr

: BEC-Pe0 C=26P-1260s0)= 1950 N
on tax He Se FE (A
teas 22 a Wie 107
Br ase N a a
Be NOR Geste

ÓN

25.571 A= ESP mati

125. A 12 odometer se! wo AB ted rod ole mer e
Ripple 125 {Co he wonden member CD. Fr the kn e, tei ah a

Finn vege al te nthe Wd 0) be dto Tor win hw

rage searing ste 620 Ka a the face inte y he as ie,
Ke average em te un ii wea

60 Hazimon average nomad stress in the word.
Aout © (25-12 X18) = 1.184 SA
Puso kW + 45010 m

HS 3.47410" Pa 2.17 MPa <a

2 EE ne" 102 16 m
a AT FE * Wire Cie * 202

be 202mm a

4.50% 10°

Sento’ © 70,9810" Pa 20.8 MPa a
CAT TT)

aca tr 0309 The ho JON Compa sighs Mae page pee
‘Sate bay oe oy my ees ot pe oe pomo ob woe ans donee eer
ri eta or encore Arden Sor o pm

Problem 1.26 126 Tao iil nk. amater syste col e

poston a the fc ars ch Tue had pre Ihn ps
Sonn GN. ain atthe lacas of meer AD fe 1 mm. sete
‘nin (the vege searing ses 12 mm ine intB th
raring ses Bn menden BD.

a Use one Fork ac à Fees body
¿Ls 12m or
be ts) + o
Es sun =
BIR 0
E+B-0 Ber-
CRETE

Margo ® By Gen
a fare Tran
@) Shearing stress in pin ol 6.
à =
ane Fae TBs RIBS te
eB = 69m Ps a
0) Gearing stress «dB,
a mo
5-2. 40-buPa sa

de onto

ón 127 For ce anion rt 17, dee) emerge den
Fonte? As ep BG) the me Ci nl] in mente 80. (9) Me
Sein a Bien 20€ rente sor 10 +
‘or fem mg tn

1, choeur et ks hs an 8» Sum aim
ion nd aco te poe er te De
{Sho athe scape moma se ie Ina come pe
Pe:

Use bar ABC as a Free body.

(0.040), -(0. 085+ 0.040)( 0 10%)
Foo? 82,5% 10° N

(2) Shear pin at B. t= 2% Tor double sheer,
where As Fute Eo.ouŸ = 201.06 410" mi
e925 410% ioe w
: Copier) = 80.8 10" Y = 80.8 MPa ae
(el Bearing: Pink BD. A= dt = (o.016M0.008)= 128x10* m*
Gehe . LENS). caso Ge 127.0 Hand

©) Bearing in ABC at B.

A= dt = (0.016)(0,010) = 160x10"" n°

5: Fie + Sue = 205 «108 6; > 203 MPa =

opr Mai 920 Te OM Coman Ab cm No pere Ma Be ht eed er
el beni en A Ss aig eae

1.28 Li Ato ik 5 mm ces Gen sitos e
bail ba Koga strpos hinds 100
Mia nd th th ee ser ren to th vo ii RM
vino) dente ot he pr ab in ie a

Problem 1.28

Rod AB is in compression,

As bt where

> SO mm und B= Gun

Ar (2.080\(0.006)= 300.0“

mo wo} 200% 1074

P==SA=-
= 4210 N

ACTA and >
sax 10

ss,

20 Te es A Compas u Arten ar Mem apt pa pts e
ERST ne ioe jepon Ronse nahe ehe asic

Problem 1.29 “The SAN om Pi app hy to wooden members of ni
sil lc sat ope shown. De
in ih lol pe

Pe IN 0: 40° ~ 65

RyelensMoots)= 9-375 u

Ge Pere (Era de) ees 20)
A EPS
| E = 04 MPa an
Pango. (60127 M0 Y sin 60"
AA A]
TY = 0288 MPa a

1.0 uo odon member o nr cos
simple see slo son. Rui le maxi
ine gcd splice 5525 Aa etc a ele ra ta an De:
‘Sil supp (Ue cepo eile ass le.

A = (07125 Jl0.075) = CETTE m
9: 4%" ~Go" = 30°

e Pere
£ is

fay

or

aa Shepley nei de ie pee Aa i so gto per

Problem 1.31

"ho wun member fin engl rs eto ne Jr ye
Simple glucosa pic ora King Da PL EN dee ican
eso aves dpi.

@ = Mora Ys"

Pek = tens,

= su Pe = 489 PA |
fro? Yin 10°)
tase)

= mere Pi To 150 PA a

Problem 1.32

2 io cdo members fem sci eme son lant hy he
pk a ea ho ag en bn
inthe gt pis 360 A, dti (0) pe dot Ian
erlebe cran ein msi ee

De gorse age

Ay USO MIS) © HRS? Wome SA vo

A = Geo" Pa

2 12,602) ua

CAE Xeus 46

i.as «10°
Sexo Pa T= 560 KPa, ma

a

Problem 1.33 AS A ce ld appli he granite lock ran. King

at ie sing menden va fe eg ect eck 17 MP
eine (a) th gnats sl Y (0) he relation ue suce on ich
{th mac scan ares vun, () he noma tes exe on eu.
Ps. mima vals e omas ide Dock.

A, 20-1505): 01022 5 gs UT MPa

= 457 For plane oP Tow

Email a IPl= as Zaun <a aorsiund)
= Tes KN =

(b) sin 20 28% 90° O45? -

= wage; Ts
ee E er

Don. E-

Problem 1.34 134 AN had Pis ail e ae ec dom Dsemie
(erg ma vale oy nemal at om dr Ses
Hythe tin otc plo on ish echo thse mm al or,

Ae (0-5) (0015 )2 0.0226 im

mob. compressive stress = 427 MP a
“ann at 6-0

a A Be See”

== 213 MPa e

Fe lors)
+ 0= 45"

Problem 1.35 1.6 Aste pis fam ne ar ifs en am peo

telling lung ls dt ers om pe 6120" witha plan af othe

= trish pe Ksowagtit max ame lle nal al Sng tes

Fath ct rc ml ad angel ote we aca” MP sand

302 deine he apne Paha ae re tan bape
pe

dr om tetas diem
Yes fe 0.200 0.010 = 0.199 m

As = TERNA) = 7(0.200%- 0.1904)
12.25 wo! ww
6 = 20°
Based on 161= 60 M 5 JE costo
AgS „ (2.25. «07 80x 10" 3
Pe Baby EAS aro m
= 30 MPa zT
wit LE x10", Masnio) - 197210? N
Sin 10
Smaller valve is He <ffoueble vale PP P= 933 kN =
nz {Gt Ase peo dob un ie ur I monaco

ele long at ars u age 20 wa plas: pee othe
ais te pips, Rowing tht #300 ail re he gid tei
eine ternal ad senting ee m cios copii mal a
‘argent ot ld

de= 0,400m les kde = 1.209 m

Me York = 0.200-0,010 = 0,190 m

Aa = WOR Wt) = y (0.200% - 0. (90%)
= 12.25 x[0"3 mi

Pote. = cust 20°
Segre Tate

Ario =~ 21.6 HPa a

EN 2300210 sin Yor
BK, 9 aaa
eT a LAA

Problem 1.37 1.37 Aste oop ABCD of eg Lan 10m darter eco shou

ot a Dan diameter alumina sl 40. Cables BB and DF ec of au
Sater steed ing held Kraming a ise ang el
‘serie op a abs 80 MP min hp oo haa be
{oped Cam real foray of iaa

Using joint B asafree body u
and enden symmetry; el
2. - QO Fe Frs

a: $Fa
Using joint A as a tree body E

and considering symmetry) Eom Fig

2: BF > Fe = 0
BEQ-Fuso : Q=4p,
Based on strength of cable BE,
Qu = GA= SEA = loa) Floor) = s4,24xic N
Based on strength of shee? Poop,
Q =$ ÉGA- EG Fa?
= Ecco) F (0.010) = uso" N
Based on strength of rod AC.
OW? Fee EGA = FFI
= §(acoxiot)F 0.024)" = 38.22 «/0* N
Actus? ultimate Loud Op is the simadfest, - ),= 45.24x10° N

Mahl Jot G2 & = te, gens u

015,08 kN et

aa as tpt eb pasao th ets Cnel o
‘Pei I engen hoa ws mara von pa,

1 Menor ABC wich spol apn an ck an te 0,
Problem 1.38 ‘a dente tet some Kanu es ed
Ieee aD ebb Societe aaa wpe ce

Que member ABC as a
Free body, amd note
that nee ta
two- force member.

SEM = 0

LP eos MO M3) HEP aia 40") (0.6) {Fu ces 80° KO, 6d Fae sin 30” 10.4) = 0
1.20493 P = 9.71962 Foo = ©

Fea = 131238 P = (amisii6a 109 =
Fur = 0010 N

Fer. Borat
Fon Zones

2.9014 wo N

ES,

139 Kai ht ne ante cable AD is 10 sa at
sey of} viera cb ee rind, temic mea oe
Ingest rs Pech ca ey pd shew sora

Use member ABC as a
Free body, and note
that member BD is a
Auo- Force member.

DIM = 0:

16.2) + (Pein 90.6) ~ (Fay cos 30°K0.6)~ (Egg ain 30041) = ©
1.80493 P ~ 0.71962 Re = 0
P = 0.5540% Fao

y Es. Won, ,
Alone value of Fax, Fos Dt OM. gras hr
Ry 2 (.SS404)C3,1a5) TS

tae. 20 In MEC Campania Alpe ol Mao mE gt e
om er oe rep pe att hi ne
‘set pic Sci at icone poo Ava ing eal ac nan ema,

Problem 1.40 1.49 The vz ik BC’ 6 mm thik, as à th = 30 mm,

Je made of see ith 490 MPA line erg in sion, What
eco ar say ste tre sm e desired o apor fad
Pen

je Dam-o
(0-5 cos 30") Fag — (0-45 sin 38MM 0) + ©
Far 3426 kN

Auer (0006)(0-03) = 1-Bx10"A m
Gr Fa. En

Ae RSs
ru (ase
Bs + pS at

Problem 1.41 LA, Tie total ne UC ie 6 mon ick nt made o io ith
1450. at see coon, What seu ete wih a the ink
SAN wie

{the ame shown to Be dig upper od
to of sly ed 1037

DEM 0e
PS

Far 277 EN
Ena. Be
Maa tw

= (FS) Foe
À Ga

”

oT

220:8 mm

opin Mata ©2609 Th Me Compa A es opt Mal may pie psa.
‘a nym yoy meat erage UE cel oe et ek
Keen geyan Aakers cea pe

Problem 1.42

“| Mio,

142 Lick AB 1 be made o's es! wich htm not sc 50
Sa Demin theres seco aes AB sich toro es ml
550, Assume ache ear rencor pis ct

FLAS) NEN Fa le
SEM, =o: NS
—(0.8X sin 85 + (0.216 a

HO) =

er aici wy

= CESE
Bee en

rt

&

o
acre N

NC)
Aso rios

Bret 168. mnt

ai 92 Ma Compe eA ire po hs Mam be ihe ep
EEE Hoon lar nda om Pps ac

porn

-

Problem 1.43 1.43 Th wo wooden enters own, wich spe GAN oa anejo y
pigeon oleo elite hong

ELSE antic ence basen Demon mm Delran O
pie eit oo pls ct of sy of? 75 nto ci,

AN,

4 separate artes of gue.

ve aves must fravamit 8 kN

of shear Poa
Ps gu ao
Required ultimate Pond.

Pr CSP (275 Mar10')= 22410" N

Requiced Joogth ol acck give avec.
we Bo o

Be DAs de tog GONG

Length of aplices Le Mr + Woo Ya 0.006 = aueh"

L= 146.8 me a

= 70.440 m

Problem em ky i

1.8 The to wooden membe sho, which pers TN lad vod

rio ceca pudo oe vecs ooo Thulin en

= a ne lucio 2 MPa he caros Etc he mers Brin te
Wes reqs eg eek Pr a af 273s oie

ae >
<> Theve ave 4 separate areas of give.
eR a Fach glue arte must transmit 3 ki
= of shear Joad.

Pe SN = 8x0" N

Lengtle oP opens Le 2246 where À + dog oye ond &
cJeorence. Ao $lL-e) = £@t20-0.006)= 0,087 m
Area À gdoe. Az Durs (0.08 740.125) = 10.875 =10"* m
Ulimade dond, Por GAS (2.5:10%M00.825>107) = 27,1875 x10? N

Factor of saletye ES À

Bui a 30 oe

Problem 1.45 148 Te ee dote co ot hes pte

ede boo Rte pla wl pot a D NL on i ie
Inte ar ses for teste] we 30 Mn, min ren
ire dos

euch bolt, Ar Bd? = BURY + 254.47 mac
= 254.47 «15% m*
A = (284.47 115 Mco nio)
41.609 10? N
ms For He tree belts, Py > (Mco
= 274-83 x10? N

Po

Factor of safely,

Lo. at =
est: Zitate‘ ES=2.50 à

Problem 1.46 166 Ti el ole et es atc ets pe torn io woden

‘Roem Koi tate lt sop 310 Kt le dann
Sts fo es 6 MO MPa nd a o of ayo 8 eed
mie ree tr ol

For each bolt, P= 2 = 36.607 ku

Required Py = (RAP (5.358.667) = 122.88 Ku

-B - R Lom
aR > En - oR

APL fan _ js

d 220.8 mm a

en 020 Te M Gran A Compu Aro er ye ag ei
Si tagcat pact toner hend Sanita
Serine penny cnet dna cons AS

1.47 Ao ipputed a bon by tel pi hata ben sed in shor
Problem 1.47 ‘San meer Ius ln th ing. Te lit reg of he wo se
Binter SB I cher, le lt stes fe ee
[US Ma. Kong no 40 mm, = 5 mm and Em comio

the land Pa or ctr may fi

othe separ of Prob 1-7, oni bs diet Ue pin d= 16
fin and the map o ut 20 AN, mine () he lor a
‘Sy lo in, 0) required vals of and cif he te of sey the
‘occ mer athe am ath rd part tpn

20 kN = 200’ N | ed
Fins As Fat = (0.016) = 201.06 x10 m

Double shea TR =e

Py = 2A% = (2) Roni xic®)(145 10%)
= 59.336x10° N

ps Pe men. oan

Py = 68.388610 N for same FS.

(6) Tension in wood

== 2 4 = 40 mm © m
= Sg were we 90 0.090

= = 8.3360" = 40.3 Jo"
be de. ee = ao” m

be 40.8 mm

Shear in wood Pot 58.336710 N fon same ES.

Double shear; each area is Az we

"ZA awe

E AE IA 97.2 30°
Ce 17 7077777570827 00m

C= 97,2 mm =<

rps Marl ©3008 The Meran Capi need Np fe Noma moy pay psa,
ergy ie ooo
ingen ip

1.48 nd Ps ppt es shown by tpn it as bon cin a set
Soden member taping fom cg Te alme srengh of hed vd
Pa non nd 7 MPa ar, Wie De ine engl ao
145 Pai hc Knowing Ib 40.0 SS mn, aid Em, Samir
Abe end Pitan rl ctr of ay 1 del

Problem 1.48

Based on double shear in pin
Ps = 2A% = 2 Fd°r,

= Fay(o.oey(insn0*) = 32,80 10 0
Based on tension in wood
Bz AS = wlb-d)6,

= (0,010)(0.040- 0.012 (60x10%)

= 67.2 xo N

Based on double shear in the wood
Pus 247, = AweT = (2W0.040)(0.055 (725x108)
= 83.00’ N

Use smallest Py = 32.8x10% N
AMowahde Pe Pe LB . 10.2000" m
10.25 kN -

Problem 1.49 1.49 Fach ofthe wo eel ins CF song vo bol mens AD
nd Fat 10» Amen fr ean cs sot aod sac a el

‘ria la sega enon of 400 MPa wile exch the inst aná Y

the 220 m dir and made last wah test

Ma, Deine he rl nor ofS he en conc

them othe orion onder

Use member EFG as free body,

DEM, >
0.46 Fey (0.65 (ag lo") © ©
Ri, = 34x10" N

Based on tension in Pinks CF
A= (had) t = (0.090 - 0.02)(0.010) = 200 110€ m‘ (one Link)
Fy #26, A= (2(100«10°)(200 10%) = 160.0 «10° N
Baced on dovlde shear in pins
A= Fat = E(oo20)* = 81.16 x 10° m!
Fo = 2TA = (2150 O 314.16 x10) = 74.248 x10" N
Retna Ey is smafler value, ie Fy = 0424810" N

¿E Mar .
Factor ob solely RS. = fe aay * 242 ~

spun Agree No part a muy dpa ee
ream cds pao pei

Problem 1.50 £30 Sale rb Dunn pins Cam ae eco reed is
ia am den

149 ch of he we sei! nk CF comet he o itl bs AD
an Gs ito tanga us en men ch nd Om ws,
made ste with ln seh tomo of DT he pla and
Fest have 20m aan nd e nnd ts wih an int athe
‘erro 150 MPa Denise ve at oa Fre as andi
conecto heat m

Use member EFG as free body.
Fee

Kaum nun

y DEM.

Sa LN: 0 Ey ~ (065\(20x07) = O
Fe SAT EN

244K

Furdore by tension in Finke CR, (2 parade? Pinks)
Neb section ana for | fink Astb-alts
Fy = 206, = (2) (Geox (zur) =160 AM

Fai2uce by double shear in pinsa
Ar Edt = E (ocd) amet mt
Fy = 2A ty = (msi sob) = 99.28 AN
Actua? oPtimate dond is the smaller volve. For 25 bw

Easton of safely: RS.c242 <i

a jo et
Se pay bes e Cl cr papi A el za ip

ynwwelsolucionario,net

1.31 Ea ofthe ch ks A und Ci once ports o member
Ch 25:00 oct pst inst ss Korg he uate
‘Sheng ca 210 MPa a te sehn pos ad lat lat ma
‘este 0 a oo sc wed de ha, dein lowe a a
nc fcr of ay 003.0 idee, (Se tthe Inka re nt rence
sind hen les)

Problem 1.51

Use member BCE as Free body.

Hau 07 03E,-07P =

DIM=O! 05g 04sP =

Both Pinks have the same area, pin diameter and matevial,
Therefore, they have the Same wlhimete Fade

Az Bate Fans

Faidore by pin in single shears
FL = GA = (mtl) » 103-09 kN

ones sui Int

Faidure by tension in Link, A=(b-d) t= (ms-
atone \(axie*) = 147 kN

Rt G@A-=
Ublimade Dead For Link and pin iz the smallers
Brrosor kit

Ablow ble values oF Fig and Fan,

En? BEE = 3636 kw
Allanhte Jard Bor stnuchere in the smaller of 8 Eu mol Eg,
Pe game) Por =

art apne eared om mye ih pat

Problem 1.52 152 A anat din ting onda tp meme CR a Pa
ESA, a ent € wide sp y tn Ia, sac of Gamma cs

dc, casing pins a nd Dn ou he Arco dt al ter
prions rors uch, demi he aloe ha Pt sell
sw of may of 20 deine.

1.81 Fach fhe ste inks 8 nd CD uma sopor ameno
BCE by 25mm ee pm sting in gle she Kooning tet he
rte chewing rs 210 MPs dr I ed wd ns pi ar
ine nor ses 150 MI for the wel ed in ie ak, eii De
‘Moi a ie

srw nd Ps orl tr it) 0720 de
‘ont need woe the pte)

Use menber BCE as free body.
DEMO: #52 P =O

HA AEM

Pe
03 Tao oe © =
5

Fe

Auen of aff pinst Aus Bd? > Flows} 20001
Net sechen arca of fink ABE Anar (b=A) tea =(ont—orte} 01d) ani fo
Net section area oF the 2 Links CD ic the same.

Fayre by pins A end B in singe shear, (Fredo Te Ag
(Fae) = Guosid)(490-?010"*) = 103.09 iu

Faidore by tension in Rink AB. ou 6 Anat
Cid = aoe Gust) a7 ba

Ultimate load For Pink and pins AB is the smaller’ (Fas), =/02:09 kur

Comespanding vitimete Lon, Py = &(Fualy* 66-73 bw

Faidore by pins Cas in double shear (RME 2% Apia
(Eds = 210x040 910 4) = 206-18 EN
Fasore by Tension ia Sinks CDe ae So Aut

Ready = terol ue) = 167 14
UPhimate Bond Por Dials ancl pins CD in thesmadlers (Eos 2/47 AM
Connesponching ultimate Joli Por Flo), > #88 ey
Retrial vttimate Load is the smaller, Py S08 kW

Lo. se

Monito Load Pi p- fe . Porrera ae

189 Inch ts sie sw Gear in ot and 10 mm

Problem 1.53 met ne sens Ma Tee sheng ts 158 Mu
Comes an tian oo ne tenga
eat tied demain le Ines Pat ee cp
‘ar tk D rot sound ee

” Use Free body ABC.
Leona Mes 0:
6200 P - 0.120 Fae = ©
» Ps he m
+92Ma=0:
oo P= mo © = 0
Fe a

Tension on nab section of Ink BD.

Bas

Fast 6A =

ES.
A CE CS

2. N

Shear in pias af Band De

Fan = Ages Ze Bat (EME NT fous) = 3.9270. 10 N
Gmadder value of Fig ie Man N,

Fre 0) Pr E = 693 ke) N

Shese in pin af C

Ce BCA gn = ai Ta ENT ons Ye g.9n74 ur N

$ zs
Ten @) Pe PS) = 212x107
Smaller yedve FP is Maude voue. Pr nano m

Pz 1683 kN a

. 1.84 Save ib 15, sigh the ac ac en dig un 1:
Problems ra sa Ban Dd charg rca made

LS ft ne sin Gian a es a Ca Li
Siam ps ae wu Bao D Ih let ain ar 150 MP ll
‘ones un ie ulm para ao 300 MPa ne BD. non a
Pr a uf ed. be get lol tn ray
‘Nota ak BD mt nord ee pin

D Use free bendy ABC
theme IDEM, 20
0.280 P - 0.120 Fie = 0
a Pire a
HIM = 0:
auto P = 0.120
Prac @

Fos At

pion ist (
TN)

= Honor N

Shear ia pint af Band De

a

Fant Ap E ne N
Saber vole oh Fey is BORO N.
From (dy PGA. Rov 10") = osx N
Shean im pin ad O,

Cr tA» 2B Fat = coy SHOVE ens ZB à
Fa), P

DIRA) Qtr N
Salle. valor A P rs the Moche vase. Pr 2.06xi0 N
Po 206k <a

100 Pa
os fsa of 30s

Slabies 3 Dse ABC as Free body.

a 8
oe y
hrmro: 020 F-08P=0
sp ren
Em 0.20 Fn-03P = a

Based on double shear in pin he —
A= Hat = $(0.08)° = 50.266 10° es

A (A100 JO S0.266 10) à oN
En EA u 19) = 3.351 v10

P= BR = 3.72 «0° N
Based on double shean in pins al B and D.
A= ¥d* = Flo.owy = 13.10 jo" m*
= BMI tome) > agg yo? tn

97 «108 N

$ Fo

Based on compression in Jinks BD.
For one Link A=(0.020X0.008) = 160x100"

Fay = ZEA - (Maso nico) > 96.7 10° w
A Fr = 2.0 ® -

pe
sable value of Pis emallest 2 Pr 3.720 N
P= 3.22 KN -

Te Foo = Most N

1.6 anale dia for Isar Poh 15 apio Vs
Probe 156 ds to be ud at 4 Armani vera col,
‘coe hella ad Pian onl nor fy of dr
1.58 Inte structs don, an mundane pniesed 4,012 ler
san dan and Knows tliat ia ceo LOOP ta
nner and tk lit ars 250 Main each of et

ti Band trie vo la sd Pitan orl ade ofa ie

Stahics 3 Die ABC as Free body.

e.

‘Oa os
Fa
or 0.20 F-omP=0
ler
E

P

: 0.200 Fae 0.88P =O
Based on double shear in pin A, OPE Faw
A= YA? = F(0.010)* = 78.54 wo“ m‘
= BRA . (al 100 OM 78.54 218 0 SEN
Fa = TÍA, (alto «Jos 180 = 5,236210* N

P= BE = 58240 n
Based on double shean in pins ad B and D.
R= Edt = Flo.omY = 113.10 «07° m*
= BEA | @Koomot Mtro) gy vig?
Fo = FEA > ES 1.54 x10° N
P= ff Fa = 3.97 “107 N
Based on compression in Sinks BD.
For one Pink A= (0,020 (0,008) = 160x/0* m“

25. A sos Kiko mis :
ESA „ se «po Xıkonıs o

We Fao = Motos N
NDowable valve of P is amalfest. P= 3.47 0° N

P= 297kN

Problem 1.57 +157 A ph athlon of SO ween ean A, vice

de sopports ss our y a pinata by sender e208 BC wi LAN
te Ia (a Use the Lo und Résa Far een end à
roses fair 6 090 end lol ts, 125 and n= 1.8, demie De

Tepes hat a e ely placed per (2) Wa the coneja
mein aor fay read BC?

DEM 20: (24)ÉP— 24 M -L2W, P- EW +Év,
For dead Loading, Vis WOMEN" 392.4 N

Wis (S0X%21)= 490.5 N

Po = EX 392.4) HE M490.5) = 1.0623 x 10% N
For Dive Roading, Wemg We = 0

P= Emg from which m 2h
Design anterion.
PP + APR = PP
Rp = PP-%R _ (Esojavio)-(.25)L0628x 10")
Te

a
= 5.920410 N
ed Mlowable had. m= au m=262 kg 4
tische Fate abad
Pr Re P= 1.0628 wo +.5.92040” = 6.9183x10” N
w Ast À. Aue ES. 1718 ma

ES

rp Me 20 Te Co
ise yoo ey mens on te rin re

yoww.elsolucionario.net

sabe owe Sha wi
Sethe penn 298 oft rl wig he weg oF et spent
he ta he dsp Bela feast ale a) Ann ween
Kater 08 tl ol ft yo 12 and n= arene et
‘qi ining tite hd ce co. What i seven e
tee sally or seo ee?

p= ERA
?

0.85

= 283616 kg = 278 48
Conventional factor of safely

Pe Pe Pt 272 + 0052 BB à 16 hey = 148 RN,

=~

Problem 1.59 159 Link BD consists ofa sage bar 24 mm wie nd. 12 mu uk
Keung ht each pi 9 man une, dere the main alo of
he crie nomma Me in E BD a) 0 = (8 = 9.

Use bar ABE

as free body. .

o) 020,

19 FM, 2 OF 04S sin 307 146) = (ares 30") Fag = ©
Feo" Br kN. Übension)
(bd) E = (4 912) om

Ares for tension Soadings

ia CEE

3 lattes 20°)U6) + (013 eus 30%) Fag = ©

Fos = ie. compression,
Area For compression Losing Ar bt= (#12) = 216mm
-24000

: Fe
Stress! Ge Fae.
AS ep.

2633 8 a

een Mata. 40 Ta Gr Cpa a ten No pa ny dt e
en A ote a

Problem 1.60 10 Tes anal 20 AN ho ep a emy

AS |
Use joint B az Pree body,
LN
to Eu Fu
FEN.
Law of Sines Force triangle
Fag _ _ AD
ee Fie > EX
Fo = 284 EN Fac = 34:9 kN

Link AB is a tension member Ñ
Minimum section at pis Ang = (0:005-002)(0:012)=300x10m'

(2) Stvess in AB pe =
Link BC ic à compression member
| Cross sectima! area is A =(0045)o1072,

(6) Stress in BC Soe = he -

A

Snes eo rd han or

Problem 1.61 11 rte assembly ad ding of Prob. 1.60, ete a) the a
‘rage sharing rs inthe pin a (th average hang ses at in ee
AC, (he avg eta ssw Ba una BC

1.60. Two havior 204 frees ae apple opin e assembly
sown. Kooning Mur a pin of Amen camer uso ech cometo, de
“ene Ue massa eo the ergo normal es (on Ink A)
Tine te

Use jet B os free body

ews
EPS ies
3

ET

la) Shearing stress iu pie > fe
Ay = Bat > Ebo)? = 3st mn®
Le sr rn eras3 ma 4

(o) Benring stress ot Cin member ec. Gr BE

Az Ed = (220) = 240 mat

o> ee = 144575106 Gs 4449 MPa me
_at Bin member BC, CET

Az2td = 2042160) = 480 met

6 BZ, mama Gert a

ern Mtr 200 Te Meee Compania. Al ph ere a ef i Maal el epee et

1 jie by he
ssl mor rr sc. Kai ta tet il al en ge Anz
resins he aces 62080, tein te lt alo hee! he
‘tse joint ci hend na a o natal P= DAN

Problem 1.62

Seven surfaces carry the teh?
Jard Pr TE ÍA + Fer.
Lt B= 22 mu,

Each gfe area io A= dt

BRETTEN Es
Te Malo?) a

= 1.32404 +10" mm”
EEE en @.2 de 60.2 mn a

repeal 0209 he Me Cepia igh No ner aa my ie mp
‘Scien nay onary me tt Bap alarmas paar gee aed donne cies
nn por y ees niches A terre

Problem 1.63 1.68 Theale elder CF wc pue ctl te poston fed DE, bus
‘ee kein th pion ha, Mer A mm ka omer o
{he ven m hy 3 Sedona ol, Knowing hat P= 2 EN and = 75",

mie o) be serge har srs in the hol (be Bering res Ci
enter BD.

Use member BCD as a Free body, and
mote thet AB is a wo force member.

pes Gens 15)

jee, la ents" ar

Length of member AB.

Ings [Reo us°
205 mm
DEM FO? (285 Fine cos 20°) - GEE Fag (100 sim 20%)

= (Zus ISO IIS tin 20) = (A ain 75° IIS cos 20") = ©

89.1696 Fe 248.662 = 0 Ras 4.424 ki
BER TOF malen Zem O Ce = 0.397 km
#26: ES 15°20 = 5.9732 km
Reaction at CF Cs Jr GF C+ SAB60 kN

(a) Shearing stress in belt (single shear).
Mar Fa Æ(ooo = caerme

pe Ge = SOMERS snow 7 er
Y SACHS amoo Pa 1s 4.1 MPa

(b) ss stress ot C in menter BD.
A, = dt = (0.009 )(e.015) = 135x106 m*

E so Ca Sr 413HPa

ren Mati 18 The Mera Coe Alig ann arf my
ÉLIRE Anse o pe A ng EE

yoww.elsolucionario.net

Problem 1.64 184 The decide Cie para come iron DE has

been locke epson stow Link AB y aun coun cae son
oF 12+ 25 rm and connect 8 nier 2 by an en
Knowing tthe moin love age sing sui ni OPA,
eerie (et ace Pw iy be pied a E OF he
camping eating sesso in Ink AR (crop maman al
‘ofthe nn esi A

Use member BCD as a free body, and
note that AB is a two force member

Pew com

60

Length oF member AB,

Dar [a0 + as"

= 205m
1D EM =O: (BER) 100020") E, Moo ein 20°

=(P 60s 60°)(178 sin 20") = (Poin go" 175 cus 20°) = 0

84.1696 Fig 172.3819 P 30 Fig? 2.0875 P

Pin at A is in single shear,
So RSS m‘

rs he to 10"

Bem Hesse

P= 3.437000 0 TS
() Bearing stress at Bin dal AB. d= Bun, te 12 mm
dt = {0,008 (0.012) = 96x107* wm"
Fre = (2047S (348750) = 10883 vo M
E À > 13.310 Pa Gramm =

= En. 208
Sie pe Rio

©) Maximum normal stress in fake AB. bo 25mm LFO.00R mm
b- AME) = (0.025-0,008X0.012) = 204x107 m*

Sia? À = FERRO ecu suo he Gyr OSM

Pr 148 Tao wooden mener am mio comp os sen
bl 1.65 Joly carpi pulse sow orig era ano
‘Reming deste nn Sti deme ocean ed tat

me cn bey tid

2 = (0.000. tome 77 O ae

= aso y
P= 11.924 mm

Problem 1.66. 140 the OL tod may te mows lo he ba BD 10 ny por

tio en tops End Kuna ng 42 MPs Tr the st wk
ind AD and CD, deme where the ope ct hepa Wie er
{sd ct of ado Be a ge pose

Permitted wanker forces

AB: (Filan = Gur Aue (200 Eee"

pe TATEN
CDT (Rd Ge Ag = tan (E ori)?
= 844
ce Use member BEFD as à free bodys
ES P= oo ky » 9924 EN
4DZM, = 0

ME Fe + up = 0
Sag (05) 1415010)

E E
0:807
| CARTER |
4D3M = 0

ts Fe 2P =O

P 8.834
x am A

eier ati 200 Tin On, Ic Al ra Nope ih Mee y ol, cher
SSR eye ae cS etre erase

yoww.elsolucionario.net

1.67 Ae! plate 10 mn ick embedded. erkront coer lab mie
Problem 1.67 do ancora vera cleo shove. The dtr of the ben he
foe is 4 oun ten erg heel ce 290 MPs, A tite
Ending acaso ple aout 21 MPa Kooga aora

60 ene whan P IBAN, cern (2) herequre wah he pe,
Gen eh o which pt ft wi shuld be embed be
‘Goer (ele ova tees etw comente and elmer end
epi)

IBM (a) Based om tension in He plete.
At @-at
Por GA

gi. Sat
e ?

Solving For a,
ar de ES

ony US ne”)
Gt 28 CE

= 0.042 m ar nn

() Based omahedrlbelwten plate and concrete ofab.
A= perineler » depth = Cas 28) b %= asie Pa
BR: MA: 2% Cost Bi

a (EP. .00 4B v10%)
Sefving for b, O” an crio)

0.25748 m L=257 mm a

regir td © 20 te ec mi Campane Alan wane No ao Me nh pr
e ri or D Im eb eis
pora ne cd con peat Adem rama El pr

Problem 1.68 168 The no pora of member AB we ghd txt
ara un ale th he hoon. Keng int te

. te groin ir 17 Min tuo ad 9 MPa bar date

{' ee ‘of values of Fo which the factor of safety of Uns mem i

A, = 1005)0.02) 20:00 15m"
P = toi BR =(FS)P = 0 eA
Based on tentife stress

S,= Je coto

cto = GA „ (ue Yoools) _
M tone es
cos 8 = 0.922 @ = 228° 03218"
; 2 ae
Based on shearing abres Gr Besides = Fi «20
e TE A
2 Tower
20 > bane 073210 Os 321"
Menca 228° € @ € 321° 27
Problem 160 10 Toco posts ment Ax gd tp an ple
ange lc hu Ks ht ae O

wis (he ed jo
‘rue of for wich

"Mba tess ADS in ae, erie he
of ty of he ae tes he

‘Seren. yl ut de Gt (i: Een the exes ce
"ener otr of sy pa o a ste ase)

Ay = 01050003) 26-0015 m?

At the optimum angle (FS)e =(ES)e
Norm ad! stress
ES), = Be

Sheoring vers Y= Esidaso : Pie

ne ts?

170 À ace Piste at shown tel rnin rats bande
in block ofeume. Dein these gh for wc he al lobe
onal rss iin arcana doses Kaps bern in ms fie har
‘Toth ay allowable none ste. In the See a the ne allowable
fond een hen he nr ande years Ue bet. Neg
iron sees between the carte tie eno Dee)

Problem 1.70

vp che, Az Tél
Pe TwA + Te dt
For tension, À = Fat

Pr GuA = tar
Equeting, utd = Eu Ed

Seloiey dor L,
tur Sad / ore a

repr Mur © 209 The McCraw Cony, A sd pet Moral mene ga
{ite mony nn prepa pla emer en sain een
nec pmb snd cook pr A

PROBLEM LC CLA slid stot ed coming ly esc! clean eat gie e

jen I din en Th eter len Sod d nd m

Ie plied ne cl F vs agitado ln lo ng ne

y ‘ponte Wa mei dvr sona nd ev emi. 2) Wie a

enc par proto dm dare te ieh lem fe. Ue
ers al Me La

souvron
FORCE IN ELEMENT L 7

It 15 the sum of the forces applied
to that element and all lower ones:

hei

ren

AVERAGE STRESS IN ELEMENT

Arca = Aj =4Mdi Ave stess=

PROGRAM QUTPUTS

Element” Strass tune) tenant Serena. (489)

2 stat
> Hi

Saget erento

raosiaat ic 1.62. A 203N oa esp horn member A

sad ie cater Cs Down edo m 101 Se

sournon

es to ts E-B.DIAGRAM OF ABC:
7 2Egh 0EM=0: 2F2p(80-P(AC)=0
À a Fgp= P(AQ/2(BO) (TENSION)
le a25nit— 0.4 m | +72 Mg= 0: 26, (BC)-P(AK)=0
2 Fan Fe¿= P(AB/Z(BC) (Comp)

() Link an ene

Pap Thicknesas ty

Thickness =
Ago=L, (Wi -d) te

do Ace = E, ay

E Cap= Ly
List Yan=t Fan/Ago Ore == Fee Moe
(3) PI B . (4) Pin ©

Ty = ap / (Má) =F, (NA)

(5) BEARNG STRESS ATB | SHEARING STRESS INABC
Tinekness of Member AC= Ac | RENE | )
SigBear B= Fan /(d ty) ir
(6) BEARING STRESS ATC je

ay MES & = afb
$9 Bear C = Foz) (d he) [I She tne the ;
(connie)

PRODLEM CZ CONTINUED, PROGRAM OUTPUTS

INPUT DATA FOR PARTS (4) (9 (9: P= 204N, AB - 025 m BC~ 040,
AC 0.651, TL.» 8 om, WL 36 mm, TAC 10 men, WACH Sonn

d Signa DD Sigma CE ‘Tau B Tau © Sighear D Sienear ©

mas m. 19.58 125.00

6125 20 Ba 87 156

es 2 es Ha

un an Pl 35 2

01.56-22.70 a ale os ae (b)

es oa —

De an E me as

Bm mo ma 18 VS 6
Ode Da Ba
Bon eo ano

38 10192 120.97 46.30
so Sie Luger 4300
29 alan len ae

CO ANSWER: mm O)
CHECK: For d=22 um Tau AC 65 MPa TMS OK,

INPUT DATA YOR PART (4) P=20 KN, AB =0.25 m BC = 0401.
AAC 0.882, TL = Emm, WL 36 mm, TAC =8 ms, WAC + $0 mm

A Sigma DD Signa CE Tay B Tm © Sighear R Giger ©

1000 26.33 -21.70 19.58 150.28
No 83138. “7 8 152005
12:00 ala Dia 55136 HE
Fee 90°33 2170 fes 1008
wo 92133 Sao Ban FR
100 201038 21070 is Pa
1700 dé 2170 als 2188 os
1800 192.88 2170 43686 20.86 seu
1900 mans 2170 97.31 22.04 2
20000 128.08 Slim ds mu
fie aaa 52 1808 Ben
24509 ETS 16a Ties

gout 0% Ba

28:00 HET 2 sas
26-00 Boat > 80110
BR 20.38 92 Srey
oo ges 2 55:80
Pan 22:00 se EN

Ca er 6: atm TAG USINE cé an = (A)

PROBLEM LES 1.09 Tun horizontal 20 EN fons are api pi Bo he sms
slam, Each oft ce is a Bd Cha the same dr a
In bt shear a Wien compte rogram te cls loros nf om
123 mm 0 375 mm. mg 1-25. inert 1) msi vale ot
the average oma ses In somber AB, (2) tbe argo normal se in
ember; Le erage cae sre pin A, the average sbi,
Ji nio enge Dig a ener (OR enge
ering ses at in ment DC.) he average Bering res a hm
1. Check your progeny comparing ths ales td Cor d=
Sty sos given fr Probe. 1.0 apd 1.61. () Use this prgram o ad
the permis waves ‘of the Gameter of th ping. min. Bt the
alle sales of the nurml, ag, und Searing esse for the
esl ese ae ssp, 150 MP, SO MP, and 250 MPa, Sls part
‘suming war ame dei de Being ined in vico the here
nd wih the wo menden ee hand, petals fm 125 o 8 mon
nd om 45 eo 60 mm.

SOLUTION
RÇES_ IN MEMBERS AB AND BE
FREE Gopys PINS

From FORCE TRIANGLE:
Fag _ Fac 2p

+ ECS ñ Sa Foc ¿nus SM Sin 7S*
los, Ne ker B= 2P(sinus/sn 75)
2P he= 2P(sihto/sin?5)
CI) MAX AVE, STRESS IN AB (2) AVE. STRESS IN BC
ee 7e Age = we
Ane = (w-d)t A Ger Mac/ Bac

Fe Cas Fro/Ans
6) Pin A oH) PIN

Ca Upa /D/ 0470) enla lA MAJO
(6) BEnrING STRESS ATA | (6) BEARING STRESS ATC
Sig Bear À = Fro/dt Sig Bear C= Fac /dt
(7) BEARING STRESS AT 8

IN MEMBER BC

Sig Bear B= Fae /2at

(conrnsveD)

INPUT DATA FOR PARTS (CC P-

PROGRAM OUTPUTS

Zobn weasmn tir mm

D Sms sie Ta Tac SG BRGA SEBRGC ' SIGBRGE
om ME MPa m Mi MP Mia MPa
115 Trisi [tez | 1010800 runs 13:60
17S airs 68687 beats Bosne leal (Tiers Prat
zo mozos) -6:587 #09 400 ENS 154544 TATE 0
ITS Tooth SENTI FRITH ALI ra SOI!
68 ERT Meade (493 Ehzer REAR HED
immer; 15mm £ d §27-5nm

a uses je Ts
nm Ma Me MPa
E 128597
38 ish BENT $0.309
6 Meike ~ 2667 zei
Sas [bye PS BET road
ue PG MH FSET ego

roger aut 0299 The Mtr Copan

AUT DATA HOR PART (8) P=2949 ge=bomM Le Bra

Kenn sinne SrühRn
Mn MM mba

356047 ezo 26614

Dros 2777 87M

RPS] 1B 2EY Ow eey 103-032
BET 0547 ups? Pa
17093 EP RNIT ES ORE

(A) hain gps DEAE me a € 4102 mm

Aion raed orate om no edi op

Ses Ds y memo tsp apo se wtp or dee ende

en Hemel ee o erento

EN CS
as 1.08 À LOAN oros P mg an angle wid Se wees pit
a 1 monher AC, wich upon ya pin and bei a Cam
{eae fem an ane 8 wi he oral (7) Kr et the
mae Tn fe cb 00 AN, wie cout pore coast
ah ae yes ae tea sal of e eb fe vans and 3
{fom 01915 lng creme in ad comeing eres m
an tn Chk fay ven il € th asar abe
tie fain el sty à on foc 8 = 3.56% and espe ah. 1) De
termine he ras posa ue the fc of se for fo Ss
‘elles De spon vals of 1 explain he es ened

Ion hun. soumon
(a) Deaw FB, DIAGRAM OF ABC:

AO ge M¿= 0 :(Psino ré m)#(Pcosoÿ(rsomm)
PK o = (Freosp)(216 an) -(Fsin Bonn) =0

276 mn ;
fi 7e. P LESiNX + 30 cas
around TC | 1S coop + 12 Sins
FS. = Fie /F
OUTPUT FoR P= bin AND Fyy= look 1 /
SNORE GRE ge ave bg se aa

Yat rs

1
(6) Wienß = 38.66; Tanß = 0.8 and cable BD is perpendicular
te the leverarm BC.

©) FS.+3.579 for w= 266% P is perpendicular fo fhe
lever arm Ac |
NOTE;

The valve ES.=3,579 is the smallest of thevalves of ES.
corresponding to B= 38.665" and the largest of those
corresponding ta X= 26.6", The point O=26.6; /3= 38.66"
is a ‘saddie point”, or * of the function ES(0,B),

yoww.elsolucionario.net

LS CS A nie ppt show y two wenn mens o ua
sity Go vestngaar cos acm tht re jel a silage sce pie
do) Doran by oy and eich Ue el

Ss in Pr LA) Ney m chf ee en css hat he ai e
mimo fx 45

souunon

Draw the FB.diagram of lower member:
SER a 07 -V4 Pesoz O VE Pees
VDRO: F- Psmax=o F = Psinx
Arca = ab/sinw
Normal stress:

0= E = (Pad) sin
Shearing sires: Tr à = (Pb) sinx coy

ioe

3) FSifor tension (norme) ne
FEN = Gy/0

(4) FS. for shear:
F55 2 WE

CP OveRALL RS:
FS = The smaller of F5N and F55.

(CONTINUED)

PROBLEM LLCS CONTINUED

PROGRAM OUTPUTS

Brag =" 1226 mea

eR AOR oH
5 à ae Ed
cos Pe A)
ë a BA
Fe 38 aa
Em 2
3: Ber
ta, a
er
ALPAA Ste (kPa) Prackel FIN Fes ES
s 460% SCENE 2957 36400 264
25 Hb-69 250.303 8.498 [SE care
Aas 326669 321761 334 ASE 3314
be 49000 Me 3.182 095 3.143
PS baRxeR 55.417 tbn) Ab top 619

opor Mata 0009 The Me Compete rod Nop id Ans dpi ph.
‘pea ao bay mc uo at nem pal cet yen al en
‘lain oral ctu endo come caos Arig uo ec ai eae.

PROBLEM uch
1.06 Member ABC i supported by win an bh a À nd by two
links whi re panne the bt Santo a xe oppor a)
a compre rogram o eel bellow oa fr any given es
0) dee cof e pin a2) he mm deter a ei 3
dD, G) the nt ora oi ol so tuna
‘Sheri tes each th te pi, (I dented oer tor a ey
FES" You progam mon ao init which othe alain tee sess
“he nal ses olas, o hing ses in Be pin st te
ring sas ath pas a Band. (and) ik your pogo y ing Ue
(ba oras 1.5 and 156, pt nd compa e rom ote fr
Pe ió given te tt. (O Use ost pea a determi he loue
ad a we ax nich oi sone etal en e 1S mm ou

[OM fr toni inks, r= 10) Mba fr el pn od ER D

0) Forgiven dı of pina: Fz ell, Jrs DAD, Py= 20m

(2) Foc given dof pins Band D: Fen= 2(6, Fa fe), ES
nate stress ir. fines BD! = 2G /FSN OO OO), P= 222 Fay,
sheaing etre in ping: E y n Lue

> actgctven in pins: F is the smaller of P And À,

©) Ex desirol ayerall ES: Py is the smaller of Band e

IEE < Py, stress is critical im links
IS E, <Pa and Pe Pa, stress fs Critical in pin A
IE Ph <É cana B< Pis stress is critical in pins B and D
PROGRAM OUTPUTS
(6) PR LES T= oan clam %,=250 MPa, Eg = 100MPa, ES=3,0
3.72 AN. in pis A is critical =
53 DATA: dy En d, =|2 00, 0,2 250MPa, Fe WOMPA,FS=3.0
Faz B97KN., Stress in pins Band D is critical 4
@) DATA; 4,=d,> 15 mm,G,,=HO MPa, %,=100 MPa, ES.= 3.2
Pal ST KN, Stress in links is critical =

Chapter 2

ZA Two gage mars wo placed cacy. 290 mm apt on à 12mm dime

Problem 1 Alain ol. Kong at, than il Id LAN acting oo the to de
lance men he page mark 120.1 mm, dete temo lay
‘fe ain ved in he rod

Lx La = 250.18 ="250.00 © 0.18 mm
0.00072,

112.097 mn = 113.097 #10 m"

na 53.052x10* Pa

78.632410" Pa E=78,76Pa

Sf nh 300 au and dance 12 mais sb
lo! King at = $) GPa, determing a) he
Staion ofthe (the eral res te RL

R= Bat Flema ld ot

e seh = ocen8t7
d es ait Pa ln]
Problem 23 5 Qu und te gh a de odes tU i, ci CB
‘mole Snr ny el ld mi rem
B= 48x10 tm, E = 2ooujo* Pa s- fe

Loro lao) a ¿a
ro A

EST. am 42601 mn a

a = 1Gou0* Fe Gew.“
ana TR

ni md ae Wire rc 1 mm when ele toe te pe,
Kong tat 20 N, dote (a the agate of he fee E.
Le Im (Gy br cureendig nol rs the we

Az Yat Globe 28.27 154m
= - AES. (827110) 20010!) (0°00). 6.91 be
ash P- & 7 LE

ES 0x0) Corost)

© o> Es - ES . 244.4 Pa

Problem 2.5 35 Aces

ute edo spp compre Kenting st =
{GPa atthe maximo alla change neg D ten)

‘be maxima mal rs e ab, () he mien ll ns ara Wa
TZ SE ic pe O mn

= E Lost 1600:
-2 geist _ 000015

E = (colo. cos) = 11.25v10° Pa € = 17.25 Ma)

pautet

€
+2
wort

EEE = 41784110 mt Ta
Re RAP) dhs ad AB RE a
der 14,368 mm b= hd

som)

ENTE}
Problem 26 25 control mado yellow han us ttt eh e wen the
terion in de Wie 14 Knowing o À = 108 GPa ad te ia
wine om tess 180 MP, eco o) sats das tanh
Sehe for doe (the coneponding mem teh the

a Bee mot
Bea = Same
A= 5,22 mm at
Pe AES _ (22.1220 llos mot Mamo)
rg Lie ETES
L =
Problem 27

27 Two ge munk a pla ety 250 mm apa o 12
¿umi linn wih 7 and ana sce 10 MP
owing tht do tance been epa mach 5033 mon ar a
Es 13 +10" Pa espe, determin (a) ne nd, he fer et.
$7 280.28 -250.00 = 0.28 mm = ORAR

Lo 250 mm = 250% 107 m

ey oe ker ES - Waele Worse?) |

zur Terio Pa BLE MPa

ars

Problem 2.8 2 An 0-m ong wie of un ameter mn fa tel with 200 GPa ed
an mate ate sc of 400 MP. Wa ctr of stay of 32 à deel,
cine) te iret lovable tencion in the wi 0) ls eee.
‘Hoagie he wi

fay 6, = Hono" Pa As Edt = BSI 19,628 m0

Py GA + (ooo (ia, casio") = 7854 N

Po. ase
Par Pew Taste.
us PE > BS ausa

= BL. isis o
©) 8 fe = a ©

Problem 2.9 29 A block of 250 mm tan 0 4 man co to mppor ce

Compre ad The mural 1 eo a home or which B98 GPa
‘tein he rg ll weich cn Be pp, trang al he nol se
to rot ed 80 MPa and tht he dc ei ots Wack au bol

A Som OSs sg em
G,= 30 MPa + 80x10' Pa ES 95x10" Pa
Consideriva elouuhle stress
esk P= Ae = (2x0*)(80x10") = 160x107 N
Considering abtovehte detorwation t
Se 73 Pe AEE) = Qnio™)Cas%107}(0,0912) = 228x108 W
The smaller valor guverns. 1601 N ACTE
ei 2104 Lg anne men ce ie unbe eel

ie mu sol Mo nel sl Ue DR ah
Mid = deine ed dera dl
Lo LE
rot, dor Pa, E= Toxo’ Pa, Pazo
NS AaB E AS
A AS
£
Delemeliont So BE

ne Bi N = nat à 69.27 mt

Larger vado FN gomme. A TS mm

Reid de [ER JE ATT mn

2.41 Au aloe cont man th q won a EN tense

Problem 2.11 toa ptt Kooning tha og M wad B70 OP
‘sn ets det sss gl wes yee fot
Praim Seam Gy alla
2008
s-P<e, Ai

Acte def OT

sree + eo

> ES _ Co wo \(o002) _ 0907 m
MPG mur TF
Pisin 242 |

sable od. Knowing hal B= DG o and Uat thelial ate |
en 10 MPa, determino (a) he mau aloe Le e is)
dele densi fhe rome aci Ue te

> 120x10% Pe

E = 70»/0* Pa Se Leo" m
as Pb - Sh 1. E85. EL
@ Ste E be oe oso 9.317

ml a LR = ma 238.858 e

ho ef = ana

528 me

Problem 2.13 243, Rod HDi mode of se!
anal compres member ARC The mm foros tht cn be developed
enr MD et 126 MPa ath na
ange in each of BD am encon 001 times te lah AR.
nine he sales dame od Ua un bean or wrt BD.

10 GPa) a so 1 rae the

Fig = 0.02 P = (0.02 MI2O) 2/09 bw

Comsidering stress? &= 126 474

As D tee

DER à 825% mnt

Considering deformation? 831009113400 )=24 mm

- fale, Foolae „ Af ig gps
So Fete As lees Ce rem

Langer area governs.

Arkh d= A. OUT deso me

Problem 2.14 2216 The mm del cale RC ig mde of tee wh 201 GP, Kamine
‘tat he maximum ssn be eble mt pol cet 190 MPa and ar the
enorme marne Pte,

Peer
La = [e’+4® = 7.2 m
Use bar Alb “au a Pros body %

SPP Fe
und OEM, 20 asp = Nats A

Pr 0.1509 Fa Ag
Considering aflounte strest © 190410" Pa

Ar

A= Babs Hoss)" = Inserent
Sri Ar A= Goma) = 2.288410" N

Considering altouabdo elongetions 82 € 4107 m
E 6x2). 3 on ot w

RE 7.200
Smaller valve governs, er 2071410 N
CS CL AN
Problem 2.15

LS A Bambi toos pelayo eine (E = 3 Ga) and a
ight Car ar only pa eh I Me) ar ad 4 sup u
20 rr og ee so A LE 200 GP of € mm von an SEAN
dol a we demi te open € rod AB (the ds
on po) he menge Fam ES A

Red AB: Par3200N Imsansm d=ómm

Amr Fat = Haooo\ 29076 m
Bols (3200 25)
@ Sq- Bales.
DS Re “Boas
We 7 Hinder? de = 005m

-05-(000)2)= 0009,
(dde) © Zloross 01048) = 0000443 mw
sa, P-320N

(2200)(0+03)

1415 1050 nl

sonate

Durs EN” Trio jlo oooads)
(0) DefPection of point B Sa Sie + Sey © one y
(el Stress in yoo! AB E> Bt se ma -

Problem 2.16 scan Sao dane Ses om ei a
Koi at 200 Gi sore) he oad Pn ne bl dee
an 008 wm, he ce purding tem fe etal paren

@ 5-7 - LE
Pr ESERY" A

Lan] dm] Amt | L/A smn! |
nal [38 [are [oo |

gel % | fest asus |

co| » | 26 | set | connie
ad so

Prien Koos) CozssaTy! =ars62 rio Y Pr arm à
Bir. asa

WM SF € ER em) 8 -

Problem 2.17 17T sl onl od ji) Ban ate shown. Ro An

dealt! (22200 Gs amd ml D of ta (E105 GP). Drm ()

Se {hotel tonto of he compost rd ARC (9) to dono plat

Rod AB: Fg -P= - 20x00 N
| La = 9.250m Tre = 200410" GPa
u = loo) = 706.85 mt = 706.854 100
(30x10?) (0.250)
Be ON
= 53.082415" m

| mx

En

Rod BCE Fy = orto 7 70 kN + 7oxtoN
La + 0.300m = 406 x 10° Pa
Kec = 10 = 1.9638 100 TREE FERTIG

> False - __(lexto?)(0.200
ee een)

= = 101.864 #10

tor Total deformation? Spa” Sagt Sec * ISA = ~ 0.1514 me at

() Deflection oF point B, Saz See Sa" 0.1019 me Y

‘i 218 Fort compas 10 of Prob 2.17, deems ath lad Gr whch he
Bieblem 2:18 ‘deforms of 03 man () decoro date of ot
a 2217 wo sll cle ed ji at B ad load ws shown. Rol Be
‘ae ofl (E20 GP) nd a of ese (= 108 OF). Demi 0)
{eto demain the sompent od AD) Be decian api B

Be 200x191 Pa,

AS

NN Geo ose
= 106301 P

CE Bact = [Pr donot)
lue = 0.300 m Eve > 105% 10" Pa

Au = NASA o BO ant
Sur Bebe. (Pr dori? }(0.300)
Bader (las ¥10*\Ciaeige 10°)
NES IR 10 P = 53, 205107

w
deformation! Spy" Sag + Sm

@ ha
- 106300? P = 14551910"

6.210

Pe 434870 N
aso a. IO") = 58.208 «107%

- 58, 205% 10"

Pro =

Fron (1,
SA 107% m

te) DePfection of point B.

0.1222 mm d à

Sgr Sac Se

Problem 2.19 219 Both peions the ro AAC a de ofan which OR
owing tt the maple ot KN man a) te vl on at
. ct ua, () he responding een o.

tar As = Edm = Foo‘ = 818.16 win“ m

de Au © Feel A

Force in member AB is P tensi

bastion. Syq = Etoo OA
Elus. at EAn " Gemorlan-iems?)

= 72.186 vo“ m

A Ih QUE ein.

de Jia. (Q-Pi(0.5)
Shoes, Ste ER (oras)

= 2.$263 410" (Q-P>
For zero dedection at Ay Sac = San
2.526310 (Q-PY= 727560 Q- Pr 28.8 x10 N

Q= 28.3010 + Yr = 3281 N= 328 kN

3

(6) pt 3a Spt 7220 = 0.0728 mmy -

ropa Mei 020 Te Gran Corpo a, Allegra Van Mae y Eto pd
e Mckee e sI app Au ug areal arg pd

Problem 2.20 220 Tego ABC ise of en amino ch E 70 Pa, King a”
AN a = 22, demi e deco fe) pot, pe

214.16 ES
2.824 x mi

adn Es

Pe rien
PLA ao

SS Lez OS m

Sig be, (6x10? 0.4)
And,” (Sitter one)

= 109.135 v0 m
PST

Bu RE mes ere)
= = 90,947 #10" m
WO Spt Sag + Tec © 107. ICA = 90,947 In 181 m

= 00 man toa

(br Tes See - PO 0.00% mm oe 0.07% mn bd e

eect Mater 0209 Te eal Compan Ie. Al et se Ns per Ma rail
A rm yrs on Sor per plac kel es ane
{ete sere IE u engen Asem vin mal en net yon

Problem 2.21

221 Far the sel wur (E = 200 GPs) and tang dem, de te
Celosa fhe mente AB and AD, hao ande rocio u
{10240 cu and 1800 npc

Stables? Reactions are HA KW upward of AC
Member BD iz a zero farce menben,

Las = mur m

5 285 5
Use joint Aas à free body. # A 50s + BE
En 215.10 KM
Fo
Esto)
ar RL)

Wa ember AB! AR @uSdorte*VC4.207)
Menke AB: Sha = TBR "gas nen armer)

2107 ma Spe = 2th et

Eds „ Gao 4.0 Fer inne

Boo =

=
mi
fa

ropa atl 200 Te Me Compa ne Aro pa! Me
SAT ans

‘Sitio penne hr pepe dag md eg woe pe

Problem 2.22

ok

= Ble «Gite on Matan Gasen =
07 Re nc .

FL_ cratondurtarionel Gern

1222 Forde eel us (E 200 OP) ad lending shove, determine
the elote of the members RD ae DR, krming Mat tere
‘Goal tes ae 1250 m a 1875 0 especie.

Tice God: Portion ABC of truss

mak A, OFM,
Fhe fig (es)
cos nées
Posts — (za Mrs

E 2 qua
8 | hsm AE Es tits bar

Free Body: Portion ABEC of tons

ein, a BZ E=0

mota reten 2
3 Ge = tate kn
To

Enz "AE" 815 mnélGecooo MPa)

Fresco Mario 0 Te Mei Compa Alig ace No o i Me yb dl pco

AR A mo ea ot

ee Sins or incl cea panne Au ig ds mp Re pri

Problem 2.23,

223 Member AM and CD ae henner sc más, and men
er ae AD are 2 mn ame el od When ete gh
cod. the dispel oer AC putos, Keogh — 209 GPa,

‘Seve hr ges able tn a Ct a efron ee
‘Svan sad Dar exes mn.

Tre = So + mm h=42m

Peat A" = Flo:o3 7 =70b-4n m
Feoke

So he

F, = EAE
eo Les
| ENTE EN
Use jeit € as a free body Re
DEE 203 Rig Herero à PAL
Fie get

Problem 2.24

aden

(a) Staties

From simidar triangles
He - fe. fe

CE:
Fe = LE
o = Fu Force Tangle iS
For equal deformations
= Sei Beh. febo: ba
Mi ee Pa hone Pre
Equeting expressions for Fis
+ As hi. Aie, Ede de
we Fee Bo Re” Fat de
Beige tt El
E er b = 01m
= Po) 2 227m -
LE Sehing Sins See + Imm
See = Fab Fa = Ebo Se _ 7) E (0-022) (0:00 1)
e ERC “Ur 7 i.
= PES EN
Elges) Us EN
Freu the farce triangde
Pos Fe tf Fat Fae = MR EN =

224 Fat ser in Prob 223,
the defomtion in members
Srependig int a mets ACS

NAN ds

CCD.

229 Members AB and CD ar 30 diameter sc! ra, nd mem
os Cand AD 072 man ener ste ros, Wie

Feat in citer Ad und CD d no exc mm,

Use joint B

neering tera ole eee nrc
A EEE

ran, (the

in (a he.
Ana

as u Free body

Problem 2.25 2.25 Each oft veto ink oonectingte o horizontal memes ae
‘lamina (79 Gal ax tom recta crn to of 0 40
speeding sn cis tin a RO ot À

Stakie

Free ea ere.

pen =0

Fes suo
Area one Binz 5
É 4DEM, so!
= (1240) » HOO mu 2
“ Bee (av0\(2 Fie) = (ESO MAN = ©
ES lw so

Length 3 LE 300 mm = 0.300m

= Bel. INCA rro
A PE en + - 80.357010

Ser + Gb. N aan

ER © amooo]

sm a

@) Deblecdion oP paint E, Sg * Saal

(0 Deflection pit F, Ser See
Geometry choose.
Let © be the ii Y

Small charge Se
in dopo angle. n

209 pm y

= Get Se , B026mI0"+ 208,9» 5 je
as at 9-93 23.220 melas

©) Deflection À pint G Sur Set due ©

Set Spt lo = tam + or X718.¿2m10)
= 384.73 x10 m $e = 39 amt À

Problem 2.25 226 tf the os a Cis made ol eel (= 200 Ha amd

Ps fr clan ers sesion of 6 % 26 mar Detemine te gst
land hc co be sponded om pit he let of Ina
‘sed 025 ur

Free tly BCE,

Feo
el E

„rer p

DEM, +0: sus 09180 = ©

Area Links ee
Tea sli mn +92 Me + OF 045 F, -obSP = ©
Fen = 25 P

Leng fh? 290 mm
Le

LE Pen .
Deformations. = Fah , (Ls Pre = 1047 y
z SER a © NT YF

2sP\len .
se Bt = DEA à enr te

Gone) bent)

Se 5m EPA SE

Ses taser eis? o

SE

108 we
RA ne“ Pp

Be = Spt Lee + ITEM Pe NIP) = $2027 IG PL
Limiting value oP Get Sp = 0252000
Limiting value PP! 51027 mot pP as?
Pr 43e N Peto EN me

Problem 2.27 2.27 Bach of ik AD and Ci md of sharin (1 =75 Ga) nda
‘rnsatctinal sno 12mm Keogh they oppor he ih ene BC,
ere tha delicia of poet

ann Use member BC ua
Free body.
[sao Fi
Dem = (0.64) Fay + (OMY SO =O Fe = 34818 m0 N
D'ZMa-o: (0. Fy - (2e )-0 Fo 1.5625 lo" N

For Saks AB and CD, Ar 125 mm = 125 «IS m‘
Sa: (3.196.106? Yo.e) |
ne Tso aso) ~

g, > Hele. (Ses me )(0:36 ) e
> fale. : mom = Be
EA” morro) oo iS

RES

122.00 om = Sa

| De Oh
| = NAS x10 vd
|

Ser &+ 10

= 000 + (0.44 Xi. 10)
= 109.540 m 523 0.1095 mb me

‘rey Na. Te Mere Coman e lg Rat y hr
‘Sse mortem eat reno fae mc ol
pony Hoban rs Chonan pop. A ed aig nmel nam

Problem 2.26 2228 The length ofthe easter te wire CD ha ba adjusted o ha
‘Sith old ape, a ap of 1S mm exes Bcn he nd Por gi cum

CD nd cml pot King at = 200 Of, deine were aA.

‘tock oa be laced onthe Da in ode to suso cnc beeen Bad E

Rad rol ACE robots through angle © Pme gap:

AS
er SR - 876 4107 wud
Port © moves desmonte

5.7 0.08 © = (0.0 (3.76 x10") + 800 #10 m

Be Soon

dt Fae SING te smic et

E- 200¥10" Pa Sa: ee |
LE (oon OY. 16 m1 X some) „ |

Fur Ehe ss = 755.98 N

Use beam ACG cs a Foe badly. We mg» GONE 176.2 N

Ref p

1D ZM,= 0% 0.08 Ry -(0.m- mW = ©
Alone suis NS, sn

eZ
+ 0.0926 m = IÓ mn

Forcontecty Re éme A

regeln Mater 208 Te Mera Companies or pa Any ta ot
ESTERS oie gh arama dete wantin teed en

Problem 2.29

far

& Total weight.
EAS .
Fo EAS.

229 bomoqencouscableof et Lund ion rat section suspen on
nel (a) Detling y pte dei asp nt whee) o o caba by E
ds medio felts, determine coma 07 he cable de os om
ig. (8) Show hte sane clado eu be bined ¡CD abe mero
tonal a re nd 1 al pt we ape tence

For element at point ¡detified by corilimbe y
P = weight oF portion helos" He peind

= pg Alley)
as = Ex. along. PUN 4,
5. Sega Ban,
Aus) se
W = pgAL
FAT PRE à dog aL F=4w <1

yoww.elsolucionario.net

Problem 2.30 230 A serial fad P i applied at the eter À ofthe upper scion oa
homogen umm of Gear cate of ea A, mien eas a, an

sien rl. Dora by hemos of ey a he mae sed

leg te lc is wel, determino he defecto of pit 4

Extend the clout sidee

OF the cone do meet i
stn point O and | a
place the ovigin Py

fre Coordinat® ystems
hove a”

ON.

2 (dy , P pi
e b-a _ Plb-a - 2h
TA ab | ae a TEab i

opty ae 9209 Te Men Compa. Al er o peo i ad yb ple pc o
Sembee ct ee ont ng nance
re ETES pun

Problem 2.31 231 Dating by the “ening ani ten spines dw tat be te
Sonate nll)

= (i+ E) > tise)

Ente (ive) a

La 5

& = RE = tn
Thos

Problem 2.32 232. Te some of a til specimen if comia cont wile plate
Germain u, bial dime Oe meinen hou at wen De
Arie heine wing

IF the vole is constr, FFL = Tg
L. de y
Liar à y
es dh In (By ESTA Ze)

ry ti 920 Te CAN Compre AN No perl ra my edie ap
ne A vel rad ir a
BE ee eb we pr A cg Dese ab Roa pom

Pr 22 À 250m or of 15.» Jn retragaor cos section come a o
palmas an yes, Sam tick, taped 1 8 ener eya of thee

‘abject e cm fono uf magne P30 AN, ad
"0 Gand 105 GPa termi the molto (yin

For each Jayer,

A= (0M5)> 150 a= [$0 10°F me”

Let P + Toad on cach allminan layeo
P= food on brese Sayer

We

LER

STO Pa SA

BE. Pa G=85.7 MPA a

Problem 2.34 234 Deere the defoml fe creer Pob 2 1 sab
mi forces mapritde P= SAN,

23 A 28m Ba of 1 amm renin cots sis const o mo
Sania layers, Eme th, bad to 3 Conor ben cr of the ane

chos. IF I miete 1 conte feces ul gado P30 KN, u
I Racine tha Ea 70 Ge wl Bo. 105 GPa tein he nomal stes (a) in
the lemas, (Oi ds bs ayer

oe For each foyer,
As os 150 mb 2 Soria e
Let Re Jond on ench afin Layos

Por Head on bora ayers
Se - Lon >
Re En - GER LER
= 3.5 Pa BaP

ADAL) «ye nfo mn
ne "0er

52 "0.306 mu a

235. Camp cen aces a 160 N ae ape tb ot of

Problem 2.35 he sem shown by sans of ig les Kaowing st, 200 Pad
250 Ga cine) normal ies m he Sve und a
‘mina el, De some of he sey

Let B= portion of aia force carril by shel
Ye + portion «Dania Farce carried by core.

Detar P= fée EN ,
hax FAó-di) > 3 (0:062% 01025") 20.002528 m
As Bat = F lovors)*s 0.000441 mt

A A
(ox1e®)(0-002538) +(200x18)(0:000491) E
(a Sr Exe + (oomet)(-58hsnd4) 163 MPa «
Gar ELE = Conc! ESPA O) = - 407 MPa -

MS he = (o-25)(-sBnswe?). 165115 Mm = 01145 rem A

ropa ata 0100 Te Mea Compa at A la ee ue Mon be le ec.
SER Senator emp pe

Problem 2.36 1238 The kin ofthe sembly ders by 0 mm wn aw
dore spied y esa fp ese. emi the nei 1
the oe ose, (0) the oe tess nthe el cle

Let Re porlion Bann farce eneried by bast elo
Re portion Penal fore cavried by steel core.

Aer (0025) = baso m“
Ay Co) (oo25)*= 144 110 be

Gay P= [llasncly(rane')+ (200 md Suto") JA SEPIE = (OL GEN

& Ay ares
WM G-Ee- EB + Boon) 2222 - 100 mp, a

‘opts Mata 0309 The Gran Copa At No pf Maa lg, er
SCOR yyy msn sp pan es

lem 237 Te Ls compas lcd wih sx ac ae, ah wi a een
Problem 2:37 ‘imate. mb ot f= 20a and Fe 2 GPa demie (e nom
free ao el and a he conte mbr 155 Nal som fu Ps

pico por.

al
R= BAS
PL REA BADE

EA BAL

Cas? > aero mnt = 3.6945 tn

4 de + vo

GEL -
Hd - A + Faso) -asmswie'= 155.394 +107 mm)
ie = 68.842410 m’

rn SO WNC nn ee gut
(a5 «107 Woes. io Rae) ~ 9°

ETAPA = SLIM, ae

Ge = EE = (ao mo Vase 31 «10

Gr Ber (aswio asser ion) + 838 x10 Pa = 8.38 MPa e

Campi ehcp ha Mer aye ape po
NR pte maker tan etna

epg Maer. 0200 Tees

2:38 Far tpt of Pub, 2.37, determine the mania cen fre wich canbe
Problem 2.58: plc the alle hama seas 160 Mba il ie an MPa ln he

2.47 The |S come pot is winored wi ix el pr, each ih 8 28-1
‘Gn, Roown at, 200 GPa sad. 725 GPa eerie the nema
Ses in De atl andi the anton 1550 AN lene face Ps
ped pot

Determine Woehte stain in auch mude
red Bye E
Steh: Es” ©

Concrete” E,

Smaller vedue gevems > B= 720410

Ahead covried by concrete.

Pa = portion carnied by six steel roda»
seht, Por BAB. BAF
sE, Per AR = BAe

P=Por Ps = (EA + EADE
Ag = 6 Babs Elan! = 3.070510! mnt

Rat Hd A sensed 155.299 + 10% mnt
ETES
P= Lcaseiontiss.anuto")e (200 ions] (729 #10"

ATA

= 3.28% t0% N 3330 KN =

pq m ah ere anny ec
Soto

2.39 Tee rod = 200 a) spp 364N of Eco the m
Probe 32: tnd CD hes a NO wow ana md tal EF Bu» Ot
Sono Nager denon ld RED, sie) de
{ingen notre BF bes inch os,

Use member BED as a Free body.

ni E JR By syrmotry, or by EM =e
Le We.

Pas + Re Per ~ Pro
Pr 27e + Por

Since Lig = Le
Since monte BED is rigid Be = So Se
Pre bas . «As La 200 . yo
BASE 5 Pas QE Per = E Per
0.256 Pur

Pr 2B # Per = 20.256) By à Poet 1512 Par
Poe = peg = BHO. agotó? N

Pig = Pan = (0.286 aimons
CRE

6.095 4108 N

23.810 «0° Kuno v0”?
3000107 )\Ce2S m1)

ge, = Los Use")
2S Ba aro 4108) C295 107)
Po
Am

(61

240 Thee wies re sd ape th plate shan. Alun wes

FR arc una Aa thro Sm an a a Cl

tr Zum. Kureing tha he lobo a rain = 70

À A a a ae arm = AN Gy 1
i nin od Pa cs be ap.

By symmetry Paz Po, and Sar Se
Als, S.= Sn = Sa = 5
Strain in each wire.

rer, arf = 2h
Determine ahuebfe strai
Sos mue
Eee 2 uses?
Wiel: En =¢ a sois vio
En Er fee 2 one 16?

Allowchle strain Be wire © goveens + Gi abate
ChE Par Ar Pod’ esse”)

> 15597 N
Pa = 15:67N

ATT P= AE = Ho) (asmio®) = 30584 N

For equidibeiom of the plate,
Pe Pet Ps +R = 776 N

rear Mae 0209 e ray Compe nc ligand Nop sen maybe ill a o
en Be clic poro A dl ma

ynwwelsolucionario,net

Problem 2.

&

Gi oe

Ame: P= eR Le BOmm* 0.180 wm
Bi. Maloun) — E

Sr Po Lo rio Re

Ps R- soxpo* L 0.120 m
600 0.20) a
A OL nt pe ans
CRD: PER conto ws Jonas ie
Ry =G0x10*Y(0.100) à e
Sur Fe ee) eco Ry = 80.891010
Pe R,- 100» 10* Lo 100 mm + 0.100 m
PL llanto Yo | wail
Ser = Ex swore Note) Lauras, 184. 785%5

E BR, DUR

Since point E

a) 3.853837, 10" R,- 292,434 =10%

Res R-

WM Sc = Sgt See = LI6267vI0" Ra ~ 26.848:10"

a 2.41 Two cines vols cu ted and ke ober uf Ets e pina Cu
naine Aeizoppod Forth ant tU
200 GPa sad £ 108 Gm dire () the ons A und BO) De

m etc in.

Jp ane: so mom n
Av Hofe

EAs 251327910 N

ERW mt

as mt

ES
Ar Eo = 106.86 met 76860

ehr Ar retoman

cannot move refiefive to A, ere

© RTOLESIMON cas al

lomo?

62.8110% 1000" =-37,2x10 N g7akve al

= G.1¢36%4 10" (62.8314 10°) - 26.848/0%

55 WB pm > <a)

Problem 2.42 2242 Solve Po 241 rin trod AC made trans ed CE made ol
sa
‘Tyna no one ft and ther sie wt Cw
nd med mar nt Por Deol hv and king at
0 Go and 10S GPa, deter (2) he ct at an 5, (0) he
ein of pie €.

A = los #10" Pe
A OY satis mnt Leere
° ER = 181.947 RON
4 > ERE ChE E 2o00x*/0" Pa x
Sn A + CG 2 706.86 we = 706.8615
un Re EA= 141,372 «10°

Home = 0.80

= ao Ry
HD ame RO

= 109,456 415 = $4,567 910%

100 mm + 0.100 m

5 fe pee = 707.345 7 = 42, 44) x10"

EE
100110 Le 100 mue 0.1000

(Ra= 199*10°)(0,100
EAS

Ser Sat Sac + Los So
de poist E cannot move relatives hy Su 7 0

= 707.854 M15" R, ~ 70, 7354105

S.68834 10° Rm 167-743 x10%

oR 4547 N YES Wm mt

BES KIO” Ra = 107.743 nie
Re = Ryn lorie = Hs. PRO foo +10" + - 54a SE

GD Sert ur RICO À, ~ $4. 567 x10

= aro 05,479 #10") ~ 54. 547% 10"

VRB RIO m See 48.3 pme a

Problem 2.43 243 sel he (= 200 GP) wih a 3.0 tr dane a ©
Sm hes plc avi a i aj oUt ee Js ach
Ihe ni of the be wnt Sri a pene on ther. The to or
‘own ae thn ajdt ee. Ar tes foc ae apd th ie e
slt decree the June bocas ju y 02 Deermie he
loros exer hy the eon a A D, th change dog of
the pon BC of thee,

BEN PEN

For the be 30mm
di = d,-2É =0:03-2(0005)= 0024m

A+ Bas - 49 = E(o103*- 00242) = 2545 x 10 "uf
Ans: P= Ry EN
ge LE. alas) _
ER ‘soon wares nie’)

Bt P= Ret s2kN, La0107S m
A ne a isos vist,

oo nen) 54 x?)
cho Pe Rn + BEN, L=0015h

(Rat Booo)lorots) u e
ae ER tete) rs rg Tan

aoısm

bare Rn m

+ 4115 Yio'm

ARDE Ser Ser See + du + 4240 Rat 58:14 500m

Given jaw movement Sig = -010003 m

@) OT Ry 4 Suis = oive0 2
Ra= 585078 RESRÓ AN

Ro = Ra 48000 = 505778) Res 50 BEN +

Wd 5er (IA TES") (58597) + 71S WI Ce 39 2x13 m =
= 342x107 mm

rey tr 020 Ts mans, Inc o ono a a Bs pa
end At es brad eave pom Ate EEE AS

oer see een ap

244 248. Solve a 24%, sui
Do wen lane y 0 m

the vid sj wire te

243 Muscle (= 200 GP) wi à 10 me eter arc nd à
3m chee pad vist ads dr ojos tach
en ofthe tbe witht cating ny pero on em De o Ces
‘wm ar then sped ote tne Ar tee flocs we apo he ve
dd ecco he ane Den nes 02 ma. Deere he
fon ee by Ie vine oni abe at A ed Degen gt
pon Bathe me.

Bon tan
a AB «eg, et
Lo |
BEN 26E4

For the tobe der So mum
a= RE = 003-203): 0024 m &
FE ds) + Foes? 0.0247) IRIS m

P-R Lroe15m
Se = EL. ——Ra (01075)
ER Zeon 10% (ase oH

LAT RA m.

BRC! P» Kesh, Lro07Sm

= DE. _(Rissr000) nos) „|, 4 E

Su TER otro” CITO Re + ATEO N
Ch D: Reisen Lee07S M

Ps
u BL; AAA cast
BE ee eae) à as ka À RAS m

<9 ze
Ade DI Saa? Shy + Dec + Sug © 44100 La + 59-06 KIS

Given jaw movement 5 = 01000 1m
(a Ar4axta? Ko +58 Axio" = oo |

R= 35467 N BBS IT EN e

Re = Ry + Booo = —27967 R297 EN «a

SR uE Em

D Sue hans 1-26 067) SAI
5-83 Ki? mm

opc Mata 0200 Te Mr Compa

aera Na ul aa my be ed q ur
mona x pate seul onde do

Problem 2.45 1245 Liahs AC und DE are hath mado ol el (= 200 Ga)

re 12 men wide an 6 mam tk, Daze (othe force a
AN foco Ps led ori mente AP ao (0)
election of oia

dol whet a
cerning

Let He vigid member ACDE cobote through
smelt angle O cPockwise abot point FE

Then 8.7 Sa =o10m >

So = - Soe = oerOm =

ml

Use menben ACDF as o Free body. *e
EM =O: 02P - olf a 1005 Fog = © I
P= tend Re oe

220 = (me 10516 ¿ere wide = 864 KIO 0

6: 027178 rio” vad 9
o Far (ms 1000-27778 no”) Rea =
Rec ~ (576 «10° }@-37778 x (0) ers a

(0) Deflection at point A.

Sq 028 = (02)(027778 nto)

rp Mater 020 The Grn Compa Ine. As ene No pal Mar yep se.
‘cl ay yey means at pc wien pore per db ened Sie Dace
pen coi Gros come pension A ire io rca Rc pombe

Problem 2.45 2.46 The ii br ABCD in speed fo four nl ir. Deine de
demini ir en y Be an Poa.

Deformations Let O be the notation
oF bar ABCD and Su, Sn, Si and Se
bbe the deformedions of wires A,B,C, and D.

Fen geometiy, 09 Sage
Ber 5 +LO
Set St 2190 = 25e - Sa ar
B= &+3LO = 35-25, 0

Since all wires ave identical, the Forces
in He wines are proportional fe He

deFormations.

Ts Ta an
? Tor 37-27, an
Use bar ABCD as a Free bey.
4DEM.r0! -2LTp - LR + LT = 0 eS
HERO Te Ta TAPA O an
Sobabiduting 00 inde (5) and dividing by L,
St Bae 0 Tr 27 (3)
Substituding UN, (20, and(3') into),
Tran + BU + In PO
TEA OP PO a
Tr GENER TP ~
T= @&P)- AER he ee =

% 240 Te lin sl lly (sE cor ande any I
IREM 247: mal a tongs of 18 € Come sly mul deus.
termine de ste In th alunitn wen cepa reas 98°

Brass cave Let L be the Jens th oP the essen ty
05 GPa
0.9 110"

Free thermal expansion.
T= (95-15 = more

Brass corez (Sp) = Lola rd

Aleminom shell? (Si) = LAT?

Kamel

Met expansion of shell with verpert to the ewe. SL (AX)

e > Let P be the desside Ferca, in the core amd
She comprecsive Farce in the Shel.

Brass core? Ey = 10S>10"Pe
Ar ESP
ph = E

214

har Bl6o%=257) = 2,3366 10 mor = 2.336620’ m"

440.87 mm = 490.370

coor

Py e

Ea 704107 Pa

Momınum :

5 > (Be + Spe

our: Ps Pe
Llao Es Eu = PL

ee Ke a td = : !
Ko EATER,” Tanase?) À Gono Wa cere)

= 25.5100” N"

(Wy IMA rento 20.910 (80) _ aa N
a = ure

ES

Then Pr

sh our | wo
Stress in aluminum. & apes ora = - 8.16x10° Pa

= 3.15 MPa a

rociar Mt. 240 THe Mr Compe o hd Not ie May he
Seu eyo ry yest rm Fan aon rn
O A ÓN

Problem 2.48 248 Save Prob, 247, sen dat he cor it made of el (E= 200 GPa,
119 IO VO) intend ase

2.47 The lim sell silly onde 1 he bras cor and the assembly is

Urmel ale empatar of 15. Condos ely el defeats
‘Stein the sm De sui we he empetneehes198°C

Aluminum shell Lab L be the Length oP the assent.
ET 10 GPa

d=23.6410 4/0 Eee thevined expansion,
AY > 195-15 = 180%

Steed core? (Sp), + Len QT)
Aeminom shell: (SOS = Lo csT)

Vet expansion dl shell with respect to Heuer. 8= Ll du MAT)

?
E Let P be the tensife Force inthe core and the
compressive Force in the shell.
Steeove? Ey = 2orlot Pa
Ag = BGS)" = 490,87 mn > 440.87 510° m"
e ¥ = fe
SER >
A i Ele 4 «¡Pe
Aluminum shelf! Ey 70x10 Pa Ge E

INTA

Se Sp: + Ge

Lía. omar = EE + E > KPL

ee !
A BAT Toro nee) (Toxo NRTA

= 16.248440 Nt

Then P= (da) à (ases M192) arto Nv

where K

Shress in aluminum . = S¢.210° Pa

Gr 86.2 MPa 4

249 LD conato posts rad y for sel La, cach of

Problem 2.49
18 m ameter. owing ut £, = 290 GP, a = 117% 10°C aad
Band a = 89% 107%, determine the oral ces

ed I dhe sec ad the coaciew by a Vrapeaie ie of 2°C

Ay 4 Fd? Won? = ovooloT ie

Rum AA, = 02-000 10114 = 00039 mt

het Po be the tensife in the concrele. Far
enviibrivm with zero fetal force, the compressive

Force in the four steed rads is = Pe.

- 2 PES
ur By + aan de vun

En E ur)
A
rau = - + aan

Ber
GX + Enr =

; ,
(mira rn Pe

= TRAINER)

&- de Xan

Re B--METN.
Stress in steel u 20 «fn -

2
E. Peces
Shess in emenk rer PEO 021 Ma

ro . :
caia peered

al
Problem 2.50 250 terms 209 107°C) fa Poli tetra, 117 |
LO)" Domine imac epee Ht ae in

Let Fa = auf force developed in He steed core
For equilibrion with zero hte) force the
compressive Force in the brass shell is Pa.

Strain: = LR sacar
se: ta

ee eB + an
Matching, Es + Es

a en =~ Be aa
Get AIR = a Am

Aj= (0.0203{0.020) = 400 #19 m“

Ay (0.020X0.020) - (0.020Y0020) = Soorıö“ nt

He mere

B = GA (Sri M4oow lt) = zai N

ER 1 ae Tae COTE SSS” À

(an.ss x10" Wars 10) = (aan KATY

AT TEM Cm

‘een Try ey mc iz PE ri rd SP bd |

cmd cnc dba peu Aedes lol be rpm

Problem 2.51 2.81 A md aeg of o sine pains AB an BC gui tot
end. Potion AB ae of el (E200 GPa "1.7 10" Chad oro.
Is mate ota 105 Gas 209 « IEC). Kio ta he a i

inured demi te compressive fs dl ABC hen Det

796.86 men = 706,86 WIS ont
[9685410 ne à 86352157 wn

as Free theeme? expansions

Spe Wlan) + Lam
6.250 410760) à (0.300 Yye. 101050
= 459.18 x15 m

> 8.223810" P

For zero nel detections Sp

3.223510" P + 454,75» 107

Pr vacio? y Prima a

Mi en id at co

nn lc ne

Problem 2.52 2.82 A rol consi of to epica pono AM and AC re
strained at both ends. Porn AB de made of bros (Ey = 105 GPs
2 209% 1097) and portion Ci made of lomo 72 Gs

iene 22 2393 8°70, King a thet aly wel, er
1 E in the noma! res nai ponian AB amd AC Ey open
| "monte A me Steno pm

| bs a >

| ” Res = Aha = eo)" = 2.8270 110 ma 2
un hmmm Anes Bales FO] #1. 2506 win à LAS IG
| Same SA

Bre Las (AT + Lu (ar)
AY 20-7 MS Xe) + 01.2 Maa. PGR)
= 2.2708 x 03m

Shortening de 4 induced compressiva Force

oe A
Cos ao (same) © (RA 2566 1078
P = 18.0744 10" P

For zero net deflection Sp = Sr
18.074110" P = 2.2705 x10
Pr Raser MIO M

SMA cines
@ Bar Ay Bang: à IA RERREEMENENRS. ont

so = - BER - _ 100.0 MOP = -100.
ee + - He do“ Pa = -100.0 MPa e

N Se = + Elo - aur

El

LUS a ON Wat atest
Se aan RATTEN)

= +600 4/0" m = -0.500 mu
Le. 0.500 mm Y -_

‘epi a 0369 The Mer Campi ne Agha pate Mane my edge pe
‘afr tray mes oo reno pao el ned ees o |
cr in Mc cl lente ren Ale log pal i eager pen

For

cy

de

Problem 2.53 253, Solve Pro 232, acing ta portar AW of he compet we

Isa o sel a potion BC mado of bes

1282. A na eng of 10 édit pámione A ai ev
suined at bh cds. Potion ABE tudo of bras (Es — 108 GPa,
ing = 209% 10) ond portion is made of alominnm (E = 72 GPa,
(2 239% 10-%C). Knowing mt the radi nally unsre, ler
‘ive (a) normal sees ined ot prions A and BC oy tempera
anne of 2°, the corpo deletion of pat 2

E
Bec Edi = Here)“
Ence Hormel expaction Arta
(Sra = Law AT = len
se = Lec (OT) = (aaa x 10 a2) = 1305 010? m
Total SAS + Gee = 846 10

a ere

astro tm

Er

Sustenirg due te ineees compressive Force P.
Plan à na a
o 7 as LS

ei sa "
(Shae? e Teens jo” P

Ga + dee BT 207 a

qero nel defhection Ser Sr

16803:8>10P= 1906 nio” Pensée N
Saf = à tea na Gr mn a
Gs mern Gr 401 MPa a

Car UMSISA LEI) = 022 110m
Sas (igh +5), 1 = 059100? L + agente

a

Se

(Sg. = (36-580 V1 3-16) = 16626 610% m

Sar Sy De (Spb = Merri? à nore wett

Sgt oem Chera

Problem 2.54 254° À scl od ick €, = 200 GPa, = 1.7 % WO wae
Is ut at à amp of LPC Detenmine übe nr ses e a
helps eches 32°C suing thatthe ral (o) re weed ©

dorm a comma ck, (9) se 13 long wi rm pape tien th,

Le 12m
See La(aT)= C2) (UTM) (571) = 000744 m.

2 or
AS

Pr. L
se Eee Eo

(a Sr Se

PS +0 6=-14MPa a

100b Ber" 4 000704 ob S--24MPa. em

[Deren

a A hero a tte nenn
A pay Br rs ens peras Azote oi tomas ap Av me

248 Yost (200 GP 4-117 10%C) ae wed orinercen
Prcbiem2ss fea ar f= 108 GP 209 10 Ch aged rd
IN. When teste tr wee ace, th dae been ee etre
Ink which wre on pis was de 09 am sal ane Seco.
"The ul bar were he pada on ove ores th eng so ey
. ‘wet jo Gente pe. Flowing ction he gett se tors
17m hoped bck toroomenpetine Demi (0) etc ln evpest at
À ste il baum pin (een he br ar ar ne
iced raped on

FR) Reauiren temperature change dor
CE Fabricetion-
$1 0.5 mm = 0.5710 m
Tempertore change required ‘to enpand steed bar by this amount

Sr: Le AT, 05107 =(200 XT KAT, A

CORTE ICS CA)
AT = 21.368 “e ae a

ie) Once ascobded, m temido hueco PT deselaps ie the steel and
o. compressive force Pt develops in the beast, in order be
elongate Ha stee? and contract the brass.

E Doagation of steeds Ag? (CS Yo)» 400 mar = 400 ~/0"% wt
Rise EL. once mo pr
SS

Conbrackion of mes: RN mt = 600 10 mt

Pavo
Sn a Ta ”

Bot (ple + (Spy iscequad fo the ini

31.146 ries”

À amount of mish

Gel (Sp), 7 0.5%10% 6.74615" PT = 0.410

BPs 8.81% m

+ AUOT. 22.080" Pa «22.08 Va
His

= EB 108 Pa + 19.68 MPa,

To these stresses must be adda the stresses due fe Ha 25 KN food

continued

Problem 2.55 continued

For the added food, the additionat deformation is the same for
Leth the ster? and the brasse Let 5' be the additionad
displacement, Also, det Pa and Py be the additionas forces
Developed the steed and brass, respectively
si: BL. BL
AE Ae
Por Auf si = REGAR gr you 5"

270
Re AE r= Asco MASA e amos"
MO ga Se asnos

Teed P= Pr PL = 2er N
Yowlo® 5% + 3/.Sx10° 5" = asxjo* S'= sacs 10" m
Re (Momo Mot) = arce m
Rs Cane sam ere Yt on

Wao y

A

Add stress due de Fabrication. Total stressen.
Gy = 34.47 x10% + 1208 vof : 57.0410" Pa

57.0 MPa

Sit 18.96 «lo cent = Baba = sce MPa =e

recy Mati 20 Metro Compe. Alsi orn. Np arm al per
SRE emery ee ce cepa ee

Problem 2.56 2:6 Determine masa o Pt muy be ali othe ns te of Pro.
255 Uv labio teinte ame 30M nd alimente
seta 2 MPs

285 Twoceibun (E, 200 GPaand y 117 «105°C are ed enc
rara (= 105 Ga, 209 10% C) which i bj a P28
AN. When th so ar ro aed the dance beeen he emt of he
Tae wc ert i ein was made 0 um mal

‘The sel far wer Hen Paced

ssid at ton he pine. Found hören, Ds tempera ts tet rs
ppt back rst epee elemin (0) he nace aterm et
‘rehired ofits asc thi, he tess in the bese ale the
do api wit

See soluhion ho Problem 2.65 to obtain He Fabrication stresses,
652 22.03 MPa Gos 14.08 MPa

Mlawahße sheewes? Gut 30 MPa, Guy" 25 MPa

Auaidahle stress increase From Pood,

& = 30-2208 = 1.07 MPa
Gir ASV 468 = 34.68 MPa
Corres pon

ng available sirens.

f° Rens © 8510
ays Bs et ere vo
Staller valve gwerns © E 288500

Areas: Aus (2 MSX40) = 400 me
Aye US) = 600 mes

400 = 1% +
PS

P.= BAGS = (20010 "Mgooxio* )(o2.85+ 10)

3.1880 Y
Ps BALE = Croce 10" }coo 10 Nan. 8515) = 251010 N

Tete? ablowable additional Force

PR e PLS alero asus: Srovioin
TON =

repay rt 0200 The er Compa Ie. Arte No pato Mar gt pc
‘ion ny oma y tt pon emi cs pea seb Ind is De
‘Sas poms bccn reste pga Aa sing msn a tt peo

05 Ga av 209 1087) and ae al (E> 200
ave the dimension shown aa temper of 20°C

Problem 2.57 257 an br ik

GPa ar 117 10
“The ie od in cold util ts fy na te ik. Te temper ofthe
‘whe sembly Men ado 45°C. Determine () fal ein he ee
od (the in eh teste od

Teil dimensions ot Te 20°C
Final dimensions at T= 45e

AT = 45-20 = ase

“YAS WO2SO) = 130.625 #10 m
Steel sed? (She AM (PAS VAS YO.ASON = ra, 125 #10" m
AE the Fina? Remperadora the difference between Ha Pree Length
of the steel rod and the hass Link is
S = 120x10% + 13,125x10 À 120,626"

Adel equal but oppnite forces Pdo edge

62.5% 10% m

—
= the bwess Pink and contract the steed Rud.
Ea)
Brass Link + = 108 «101 Pa
re As (2(S0YG7.S) © 3750 mm = 3.750x10° m*

E F
L Ptos did
&a= El = Tes) cdt

Steel ned? Ex wor fa Ar Bor 106.26 mo = 706.26 (nt

+ _Plo350)

» Le me
Gr ER," Coria entr

Geet En, = 5 2.4033 «JO P = tro”

Pr 26.006 x10% N

tay Stress in steel mod,

26,0066 tu?)

= 36.810 Pa 6, = -86.8 MPa

th ef steel mud. Le = Gr)
0.250 + 120410 "+ 73.128410 - (1, 7683BK10 7 (26.003 10

ot Be

0-480 147 m bp = 250.107 mm a

Problem 2.58 2.56. nomina 0S gop cis wen he engere I 2,

sein (apt bdo nae the eo il
tesa 75 MP bythe sepan cat ei he alain at,

komp

ar 75 m
ee 5.

(78x08) (IPo0yie') = 135 EN
nome Em heben de be?
Fee UBB y, Pla, Die
ES
€ Lino (039) 4 U350e0)(or05)
Seat Toca) TE Ber
© 78203 00 m.
Aulable efonyetion for Horna? expansion
5, = (noces 4 27x16 €)
Bt 3 aan) ı Lan 2
sphere E)(ATIA lo 05) (2302000 NAT) = (18 x10 DAT
Bene JAT rr? Ar Tole
Tas = Toa AT 0244701 294-100 =
- ~ Pla
oe TER (135000 Mo-45) 3
NAMED guyz gia? m.
a N
kant 2 OSH ODT KID 50-4507 m -

pat 060 The Mei Cupo o ii o pto Mm a pa gn
“hoo ney Lomo on ent ze ren rn pa a oa ed ee em
‘carne por by toon on Ao ug oor Depor

Problem 2.59 289. Datemine ) he conector in the hare ho a ame

ere isnt Cth mög enge eal fe boa ar

tn

Thermal expansion fcc of constraint

= See Li (AT) + Lal)
Wage PE
Le Eile

= leas )bhéxrS HP) +652) (8)

MIER m

a metrained expansion. 57005 nm
Shortening due de induced compressive force P
Sp 47616? — 000005

CL + Be = Ge Be

976 x03 hy

ose as E E
(ia ne ar

CES

SATT P = 0.776 x 18

Paless N

1728 EN

- - Ph
© Sr Lan - Dr

= A - LIME DOSE,

in = 00236 hm a

rept a 08 The Ma HD Copa o A rl pa Maal y nee ps.
Sogn ene leo mer Be sult o

Problem 2.60 260 Mrocestemperstre (29°C) 20.5 mm gap ex een nds eos

nn pn on oats mm ep kne rate
ser
ef AT = Wo - 20 -
1 poe thus ma? sup

lua Sees Hita

sis 2. Ser Lao lATd+ Leo (AT

mr, ge sel (0.200 ami 120) + (0.250Yi7, 3410" Y 120)

Di E28 Mose = 1307010? en

Shortening due de Ph meet convives
Spt Oi = 0.508 = 0.847 10" wm
> Pla, Ple. sad
fe ERY ek CER AP

- 2.400 2260 . wt
rr e ns P

Eqeating; 26M P = 0.88716" Pe ausm N

@ NER MPa,

B - Elo
Be Laun- ER
237.34110° Yo. 300)

= (ooo) ELO co

0.363 mm at

Problem 2.61 261 In end ei ttn lim do mn amet jd o
ton once = 30 RN” Rowing = 038 and = 70 GPa determine

{apie elengaton o eo nan 10 mu gape eg) he change in dia
se sterol

P= ox10¢ N
OY = BISA van = BY ISTO a

EA

MO
FS NB OS
To» 10°
D Gy Le» somo Mor) 205-110 m
ay > Le
0.205 mon, Le

o”

ave = AIM) STINE o

1.55
= 0.00955 mm a

a
© 5 = des + (too "Y-477. Mémo)

2:62 À Animes ral male of expresa ple eto o a

denle reco! magi EN. Keg an apte o 1 mem und

nn damier of 85 mm are oberen IS mn er, detain he
hemo iit an Po ne Rem

Problem 2.62

das on

Let the yraws be along the Length of the vod and the x-axis
be transverse.

A= FY = BiG mat = Silent Pron

-L Eu BE
GTR Treo

= 11.0485« 10" Pa

Ham + 0.043333

ETS
Modulos of elesbicily: Es & = BOSC = 204.6310" Paw
E-205 Mm a

= 00425

ES
a= >

Peissun's diet ©

vs ous -

CET

E
A Ge zb AMSÈUEE = resi nich Pa

G+ 70.3 MPa,

1.63 42.754 tent lad i applied at coupon mad fom 1.6 at sel

plie. 2000Pa, 030) Dteni resale) ine 0 auge

dona (0) in the th pon AR le et cor () te iene af
3. pation AB) the cos section at of potion AB.

Problem 2.63

Nae 19,2 mn
DENTS)

nes vit
> EL VE» (o8oN6.15410*) = = 214.24410
Sue LE, + (Come SHO = 35.0100“ m
0.0858 mm a

m we O.012m Gyr wey (0.012 IIA 84 10%) 2 GIRO wn
0.00258 mm a.
OMA 113) = 2ST" m
= 0.0008487 mm a

@ tr on Ste

GY Am MM EN AC mt (HF Gs ere) Aas wets
BRS AA. tite (ey te, enegfigeide form) = Abe
= Noa co 248410")

EST = 2000828 mat at

Probem 2.64 2261 An emihofanalamimu ipeof260: m sr die an mm vai
chnel wei a à han ola and ea acento wil had of 640 AN.
Keon hr E= 73 GPuand += 023, deeming) he ange ng of he

om ‘ip e age ins oe inet ee ange in wall kes.

dez 240 mm te 10mm dis do-2E + 20mm
Ar BAS )= F(2io*- 220") = 7.2257 10% mm"
= 12257708 m“
Pe Cuomo N
fos Ph. von (2:00)
= AE GS Tao SX 734107)

Hanno’
243 mm =e

= 12023 lot

= ve = -(0.88%X-i2133 x Jo) 400.4 0"

1

BC) Ad,
(o de

de En =(290X 400,4 x10 Y = 0,0% mm

Lane = (1X 40041) = 0 00800 mm

À

or nn rn i eas
AA A AS

ynwwelsolucionario,net

1268 The change ia dancer of og et hot cory messed st
de tigre nosing a = 200 GP and y = 029, dtr he neal
Fort nthe bl the me ho deb Bm

E = Goro

| = BUS cra
E vio 0
ES

Gy = Em = (2000 MIU = 149,43 «10S Pa
A= Hd HoY = 2.827410 me = 2.027810" me
Fe GA (lanar Mamie = 42240 N

= #22 e ~

Problem 2.66

12.46 an anim pa (= 74 Ga nd = 0:9) ish oa cen aia
tea ht ater 0 vol ss 2 Knowing Ot, bef ding, ie pe
21 seid on he pl, dtr the ethene when or 128 Ma

201:

The slope after deformation is tan 8 = 2046)

Tree
& . agent
BTE

ZE + 659 rio

D006ST)
0.001687

pre Ey = ve, = - (0.28% 1.6892 410) = 0.5574 » 10"

= st _

repay at DY Te Men Campe Ie A IA No O a ny gle pao
nm bat ee me ne pater eben nl ne
rn Gon er Peas Xe a sa og aon

A mers is nt em
aaa
ees

Sep mm gro

E HG -"6, - 26.)
‘

Torio

= 384x154

E tamot=cor36)00)= 36-42]

E Ho +6) - 96)

36) 4axrot)#0-dos0)t-4ax8)]

Length subject de stmip Est Le o-sm
e) = Ley mins) ot 2013 <a

(0) 5, > de = to03r ERA 109) =-001451 710 m = 00146 pom ew

oc ome leia

Problem 2.68 268 forte rod of Prob. 267 deen th ass tase
led othe ens Asal D fe ol e al ot
{ods lorena asthe yet eer, het apt
a | AD ote wd ems chang

287, ‘The alain cod AD ie ite wih joio ai edo py
«cat pesa 2 MPa Sram pt Heo el

Ing ha Rott ad y= 8 termine ah charg te
A3 change dtr ml ld od

Over the pressorized portion BC
| rar 6%
4 yz & oo 4.65 - v6)

= Eau + sy)
@) ña = © 2upt Go
‘Gy tp = - BXaseXarneéy
= 30024 MPa
Ae Eat - Klaas)! share? qm
Es AG; ste?) Cone net) =-34242N
Le. 3423 kN compression al

0) Over onpressorizal portions ABI CO 6 = 6 20
ur Gee E
For no change in Dength
S = Las (yla + Leder + benlEeo = ©
(Las + Leo) Gy ye + LoclE dee = ©
(05-03) & Bapro

wor. (none) _

> on
Pe AG, = Unsorted = — 107878
ie 1029 EN compasión at

90.72 MP,

Problem 2.69 24 A Stic win ina sorte jet 1 ail ing Bat
Tone in nama esis oy 120 MPa a 160 Ma. Kao I he
ropes of the fahre can be primado a 8 = 87 Ga ans = 0.3,

3 Amie he tage ner oan sde 4, (ide Digan AC

ro,

6, = Rowot Pa, = 0, G = Konto‘ Pa
A

-¥G - v6)
= ig mor - Goan Xicono*?]

= 754,02 «10

"El - 8 + 5)» aise] sao) + 160 10°]

1.3701 w10*
@ Sig= (AB dE, = (100 nm d(rs9.02x10%) > 0.0754 mm -
(br Se CRE, = (75 mm LLBTONZ I )= 0.1028 mn ~

Lake sides «Fight triangle ABC ac a,b, ande.

8 :
Obtain differentials by cadesdos.
¢

2e de = 2ada+ 2b db
de> da + Bab
Bit, 02100 mm, bt 100mm, cordon TE) = 125 mm

0.0754 mm dbe Sq = 0.1370 mu

Le (0.0754) + (0.1028) = 0.1220 mm =

28
5 us

Problem 2.70 22:0 The lock hou made of magnesia alloy fer whieh = 45 Ga sad»
= 035. Kiwi Bat oy 180 M, cine (a) the mayne fo ar

‘whch ie chape ae up ofthe Bock lb oro, 9) corses

‘hang i be aren off Fc MCD) corresponding cope the eames

En ‘ofthe ack

o 70 6.0

HS - vO - 16,0

VG, > (0.3530 «10%

“63410 Pa Gye OS MR
(6, - 96, - 6) = - BG, + 67) = O ga

Save

E (6, - v6;

6) A, = Lala

152.95 210%
ETS

6)

as

As Ge) Ll e) = bl (se +e eee)
BAS AA = Ll (Eto, +8) Lits (Gate)
AÁ= (100 mm LOS mn SIEB) BAZ ~ 13,50 mnt 28
© oe bey
Us LDL, gy) Ly OT A]
Lille (14 EL 4 E, ee + Bey + GE, + GE E
AV VV, = LLL, (EHE, 48 à smal dems?

AD = (ao eo as (> 2.510157 + 0-191") AVE Stowe A

reply at 200 Te Gr

rm te
ESSER eaten tact senaamema™

Problem 2.71 271 The homogenous plate ABCD is subte oo xia nding shown. I
{saan ht eye hal he change neath of pe e accio.
tat he Yor ut =A Dering the oda sic un»

CI Foot, determine) he ore magi of (0) ei 07.

20, & 0

a

»
ar
A
a
x
RM
a
me
x
a

0

Ap ni © rd tent rf er a St ac

Problem 2.72

dormia a mo 8° wih e ai al bo fond es de al sb
(0) compaing the Hype ofthe Calo shown in Fig, 234, which
Sepa readvely an clanent bn a lr female, (0) wing the
len af the comesponding sees” and a alone fa Tip 140, andthe
‘ecard Hooke’ awe

e » “
has
Far)
fi Ele
‘ al Ge)
T TE
a Before defirmadics After defacaation

(need = qe vs ven
ale a RE) 1428 shal zu eue
Nea Ze 2e Bh ave + ve"

Neylect square os sual?

1 ny NC PE
e ld ° |

a Nop ti Nam Dpto nee
aman eons nn eet rl lo cea

yoww.elsolucionario.net

= 127 many sition ibn ht sora so nen recone
Robin 2.73} Zen, or le = 0 hue o the in plas o. Fe aoe

Khe pone ae. hve at Ds and have Den din
“ Epinal, ne an espesos ad flows

HAUT

autos

li
EGG) 6 gi A
MuPhip ying CQ) by» and adding te

Eva EEG e 6 = Sele ve? =
Mod pdying CY by vu and adeting ta)

eevee, =H eigen =
es ve ng) PE (Rens oe ve

+ BED (gets ld =

Problem 274 37 a ny Sons pal esis pot ain om cg I

Si decian, fr ecemle & = ithe cae shown, whee kal
‘oro of the long pam de proven at every pein lame seins
venir othe tga xs oui plane an he tame distance apr
‘hw tat for his stun, ich eho a plone ei, can apes 0

vera)

1
A A]

us GA 6) or 6

ECG - 26, - 6) = ¿[6,-08 - 6,0)
¿la-de, - streng] -

by El var ne EL HG, +6 616,0]
ep [0-25 - 2G eG] -

Problem 2.75 2.75, The plc block shown I Bond à od bse and a ha
tal gl plats i wc a force e ap. Kean th

ee pte
‘ed G ~ 388 MPa determine the dels of te ple when?

Sow.
mé crie zu
b plastic block
ah Te shearing,
| Force carried
EN]
The area ix Ar (0084 Vena) rr m*
BD. u.
Shearing stress. Te À BET, - 320675 ube,
x, ans
Shearing stvein, Y= Es APE > 0000453
Bot Pe Bs Schr: (Noro) = 05 mm =

276 À vivio bon wi conse o to blocks of har rbb
onde wo lt AB dto id soporta aca. For he se und grade of
rier eig = 154 MPa aed G > 126 MPa Keoning th à cent
‘ert fro of magia PES EN mt ese à 25 my seca de

th ofthe pit AB. determine the malt hab msn nd Bo
be oct

Consider the vabher block on the right, I
carries a Shearing force equal to 4P

The shearing stress is > AE

oe rapid A= Be 2 ESO nou n°
Bu A=Gorb

Hence b

‚5628 m

des Bun = 563 mm om

Use besssmn and Te Italy

Shearing sheain, > Er BE = 0.12228
2 28

#| 24 7-5

Heu, a7 B=

sam a

Problem 2.77 1277 us pi block sown eb tag spp oa vera pito
sea ADA laa Papp ig a or he plese wed G 1050
Po: ermine he den oh le.

COTE B.Com? = EMO! mt

= Pr atom’ W

MÍ RAE m |
EE |
Er Fe Bee en |

di L hr SOmm + 0.08% m |

5 + hY= (0.050\23.210 +10?) |

= IO KIT” m LOR nd |

Problem 2 78 casas P should be applied to the plate of Prob, 2.77 to produce a S|

2.77 the plas eek me onal id spp ani veria ple o
Shen a KN lol Ps app. Knowing at De laste wed G 1050
Nh seine he defen of pnt

Ba LS mme LED m

| ree |
ER pe ÉLUS sonic
h G = 10s0»10% Pa
e Ee GV oso o

= 31.5» 10% Pa
As Gao W0 + EOS ma = nero

Pe LAs lso Macao) zero — saz kN) e

NOTE: In problema 275 theeogh 282, 2.87, 2.28, and 2.132
it is assumed thet Y is the tangent of the angle change. Thit
makes the relative displacement proportioned 4 the Force.

Shearing sean

Sharing stress

229 Two blocks of eer with a mods lcd G = 12 MPa
‘are Bond to igi spot ad 1 pte AB. King at ¢ — 10 un
a? = 20 RN, eter the sas alot dern a Bu
Biss me sing res he uhr a nen LM he
Mecon of pe eto be at Te Sam

(an Kuss, orten 4m. =

LIT

Problem 2.80

Shearing stress

Shearing cteain

230 two bock of uber wth a mele o di = 10 MPa
‘are bn igi apro ant à plate A Knowing that b= 200 um
01 125 ws die the get love Pee sais a
esl ch th lcs the sheng se th ror al
‘hes SS a th elena of te pe oh tee O

Je
MEA
”

«42
iat
Pr Aber =e)(0-eMornsyleswe') = 75 EN -

Couette 20004 m = tomm at

21 An carie bearing (G09 Ma) i wed 1 spr arg ide as
‘own to provide Hein ring cas. Me bea ans ot dace
‘ore an O ie when 322 AN er ll pide du. Kong at
marin a ab sag ses e430 Ka, derive () I eet
ae dimension (8) he sled Wicks a

Sheating force P= 270 N
Shearing stress %- 420 vio" Pa
2B Ae Be AUS eos wilt

= 52.381 «(0% me

(200 mm bY

A. 523
bris a
AS
E im u
T ke ae

9110

232 Fer the homer Pen in Prob, 281 with 220 mm nd 30 mm,
ein obra modes Can he shea tess or
dond VON and mann pire = m.

Shearing fare P= 14x10" u
Arca À = (200 mn (220 mn) + Ho" mn“
= MIO nt

Mu 451.31 lof Pa
rn = 431 kPa om

1.080 «10° Pa

= 1.080 MP, we

NOTE: In problems 2.75 theovgh 222,287, 2.88, and 2.132
it is assumed thet Y is the tangent of the angle change. This
ouai te the force.

maker the relative displacement propor

+28 A 150 dint cl el ph ere io cen

rth res 0a tet 3 as et the na). Kon.

200 GP ud y= 0.30, rain oh de deren er

the care vto the spb) he percent cre
te deny ofthe sere

DFA = EO sy = Werne

Problem 2.83

For a solid sphere
erg Ge =p = Somm
=:

”
PE E 16) Gp = AURA LO ARS

200x107
foexlo

Likewise 6 > Ee =

er Bt +8 = - Boorıo

(0 -Ad = -d by 2 = (01S (100 mo“) = 15 #10 a
(AV e (TUN -300 not) = S30K 1077 m =
1 het m= mass of sphere à Im = constata
ma BU = PU = PAIE)
Bi LH =| A = deed
Pe Po Tore) m Te
=(sese*-e74.-)- 1 = -eret-etr..
= -e = 300 vo“
Paßwioo% = (30 #10 100%) = 0,03% -

etry til 28 Te Mera Conan E on pt hl ee be pl rea

Problem 2.84 1284 () Forte ox looting show, demie tho change in ight an be
tant hate of bso lider shoe (0 Salve prt sing Dat

Jn abra. a= 0,= 90 MPa

net 135 mm + 0.135 om
As die Eas ¥ = Sons do mt» FE ot
Vat Aghy® 766.060 mi 766.06 "10m?
(a) 6,20, SMS Pa, = 0
es = S
yr bas CRE

A ara

Mes = (135 md 52.38.10) = - 0.0746 vu

an >
es Ur) (ONES
e Wege ENS

= 187.810

AT = We = (70.060? met WC 187.8110"

MR: = 700 Pe
2.3

5

Bye Elo - vo)»

IIA aro

SITE

Bh = hy by = (12S mn VE 226.670 €) = = 0.0306 mm -
(esse 6) CANE Alan) à eso rot
= SU mm? ~

165.06 10% mm i gor

SENET ell liken ft Ue bdr papain Ata ng emcee lp

Ren Determine te dation «a the hun in vol of he Men Ih
Problem 2.85 shown it) thera mat a se! with f= 200 CP el y 030.1)
ed met amine wit À = 70 OP and y= 035

CET mono P Gr ©

&

go) Bot. vo = -v&
en eg AS
Volme Ur AL = (830.18 mur (2000017 76,026 v10° ma?
Aw: Ve
e zu vn“ -
AU + (10. 026/08 L242 #10") = 18.40 mm -
simios =
LF > (76.086 108 MEMO) = 39.4 mnt =

Problem 2.86 26 eine te change in volume fe Sm gag eng ent An,
‘Prob. 263) y coping the ln of ths mater 6) y aci he
‘ginal volume pun AB oma Anal la

260 A254 elo is eptod ats copos male fom men at tee
ple £= 2000, 030) Demmin ere anya Om ago
Fog (in he with of poco AB o le copan, () a a cts oF
‘aro A (dah rn ein en of pion AB.

AE desdicha

ONE ETS 19.2010" m

Volume Vor LA, = (SONIA) = 960 mm?

¿Lo rn «
ee LUE ao Pa, 5-5 -0

(6,06, 20) Beuel. so

By = Es 6 UE + (os) ao
erre” 2métenot
AV= Me = (Moll 250460) = 0.275 mm?

@ Fron the sodubion de Problem 2.62

Sur 008581 wm $F = 0.002578 mm 5

0.000487 me
The dimensions when onder He 2.75 10 Bnd ane?

Bongth Ls Let Se SOF 0.03581 $0.0568{ mm

width wr eS) 12- 0.002582 11.997422 mm

thickness Er ty 4 Sp = MG = ©,0008Y87 = 1.576568

wbome Ur Lwt = ($0.03581K 1, 447000. 5146568)" 960.275 mut

AV + YY = 960,275 - Ho 0.275 mm Zi

epa ne 0386 The Meet Compro do UM O Mu nu gle ph.
RO

Problem 2.87

24 A vien io upon cousins of rod 4 ra ~ 10 mm ed
be of tne alas 25 tan tnd a Rng els rer
‘lake wil smote of ity 12 a eme he rg owe
Ph can sapped I m Ai elias ot ssa Du

Let be à radis

coordinate. Over the hallo

vubbee cylinder RE VER
P Shearing stress Y achng
on o eyfindnicad surface
of tadiuer is

The shearing strain

r=É-

© meno

Shearing deformation over vadiad Jensth dv |
as
a Y

a dr
35: Paes e
TAS debormatí

Dela: Re

G = 1ario* Pa same
p = GB «160.080 a at
A On lets 00 at LE

CRETE

0.025 m ,

he 300 0080.

Problem 2.68 “2.88 à baton dalton support amis fe rod of ay nd a tbe
of ier ml Re Bnded toa Semone hollow mbr ender wih a
Sodas of pty Go 109° Mi. Teen the reed sof he aio

Ra LOU force Piet comes 7. Sefton nd A

Let be à radia? coordinate. Over the hosen
vubber cylinder Rix WSR

Shearing stress Y achug

e
on a eyPindnicad surface
i of redliurr is
ws

= «E

b ore x TR CA

Ey The sheceing shrain ie
Br

I ef
E
fon over vadiad Donyth de

Shearing deformed

as.

dr” Y

a8: fades
Total deFormetion

“
s- 4/95
LL 2
cree
La - 7mEhS . CAmM10.73 x18, 080N 0.002) _ | oy
AG = FHSS ase = hones

3.00

e

Problem 2.89 12.9 A corpo ato wih ann sds the pats shawn nd with
las polymer Hic led ly nen. Te cab a comica ast

csi inthe pad ic

in de» deen Da

sen (he so

tod ba talon and of EN |
= hag in th og of the abe the |

a o
@ eo
e pS)

The constraint condition are 5-0 and Emo.
Doing (2) et (3) with the constraint conch tins gives
Bo we,

5% E
= Ge ES, = ES 0
da Sy ~ ee = SE Ex or Gy - 0.426, = 0.077216 6%
| Eli ha = 216 on -0M8G + 6 = 0.077210 6,
| Solving simoltanecos La, Sy = 6, > 0.189993 6%

Using GV md GV) Es O E - BEL
Ex = & | i- @.2svyo.1se095) - (0.259 Xo. 1349093) ] 6,

A = (Ho)(4o) = 1600 mm“ = 1600 v10°° m“
Gee Be BBR ocaso Pa

continved

Problem 2.99 zonhmued

gg RE) age pid

So «10%

DV Se LyEs > (HO mn (756 72%10*1= 0.0308 mm -

BI Gr 40.casrioPa = 40.6 MPa =
5, = 6.134993 4o.625K(08) > 5.48 x10°P

= 5.48 MPa =

Problem 2.90 230 The compost cut of Pr 289 conse aie dtematon ne
2 dio ad longs nh kn by DNS zum def a eile aa in
thes dsm Determine the sees aa (0) he change ln e
ri nthe y deco,
Reisen aaa
Eon cas

@
wo
[3

Constraint con.

Load condition , 6-0

Fron equation, 07 -2 G+ PS

i+ Maag, = (0.2MUE2 „0.077216 6,
continued:

Problem 2. %corinued

From equation (1) with 6y =O,
ee Ee - > Eu

= Le 026,1] = Alı- Grass Yo. 077201] Ss

= 213082 G,

5
Se = 0.98037
Bot, ee Fe = 20 = gs 10“
immo mess) , PELE |
@ LOs WIRST Pa 19,6 MPa |

6, = (0.017216 Kaneaswio®) = 3.446 vio* Pa = 3.45 MPa a

- Za 46 -
From (2), ES + Ey al 6
AA 625 x10") y y LOMB GO
Sorıor Have
= - 323.73 x10*

WS 3 1y8= Gown 32378710“) + - 0.0120 m

opc Mata 0300 FD Mer Comps Ls A igh sch Np Men o e ped ce
‘Sie nny aa ay rs oat raja aie jet ie ce
0 pomor rior Spore Aces eel loli pi

Problem 2.91 29 The muaa cass EG und war etd by Bs (2.3) and 245)
‘Show tat ay one of hse const may Le pr ns af ny ois to

conte Foren, don a ( 2 CAN anal) = Ok SO
=
a
sem “6
ams
k=,E > RES . _2E6
30-21] 526-2116 mBe-ce
era =
k- 20m
OG = za
3k- Guy = 26426»
BK-26 = 26+ek
2k - 26 =
kr 26
Problem 2.92 1292 She at fr ay ve rh te mis GE o modas ii
ee emo f chi ny en but more an à it Rar
toy CAS 213]
Ben or = 20190
Arson

230 Fer atmast ol medennts ond v<t Pr a positive
bulk modul.

Appdy ing the bourds, 28 E < 2641323

Taking the eeciproceds, 42

Problem 2.93

2390, Kring tht = 40 EN, domine she masimum ess when
(ore mn ihre 1 mm

Pe 4okN D+ 125mm
ge BE aoe

As dt =Lorb62)(o-0l8)= MG m™

d > bam

CRE Sr vos

From Fig. 2.64 b Kr 104

Gar KE. Con)

LEA CT = CEE ahs -
CAES K = 8e
se. -
Problem 2.04

2.98 Know at forthe it shown, te lowe tosis OM,

rie he warn lowe sla of Pa) ¢ = 10 my 8) P=
mm

6... = 10 MPa

D-nsmm de brmm Den

at -(o6ézMoro1P) = Lane” m

so
form £

E eee
a6 Kr ed
ke
a Le
pe ee Gene cp by, =
18
CE Sd cost En
Au „ LEI) bs k
X 27 SEN, =

repr Math 20 Te Mera Compan An Ne pati nt pa dd
(ted om ay =

Dre matiere ten

= 295 For = Née th sion pl ceed
Preble 2.98 it ewe see D MR

Yt lame ds 55 24 231

At the hole
ho Fe
Foon Fig2éta K= 226
Gus KE + KP : £- EL
met he hE © LT de
zes) 2 5410704 m 330: um

LE esnciesno®
M He Set Dem dus tome

fat à = toma Redox + Be
From. Fig 2.646 - Ke . ke

K= ur sr SE. KE
XP. (7025000) jo
te KEL „Aromen. g.oam = Iran
u” (004) osu) EM

The Danger value sis He requived mini mom plate thickes
braoamn e

Problem 2.06 296. Knowing hh ole ho eter of Hmm, deine (a) o

ud y ls E or wich ano im sess oca te rie

md he itt 9 he cmespending anima loc aod Pi the
‘howe eee 108 MPa

For the circular hole

d= (60-10 = 40m
And = dt * (0-04)

Fem Fig 2640 Kate

= Kup
Ale „de Kr 2
> Aut Gna» (010204) (10800) _ 55. =
wp: ABS: an 233150
(ar Fe Hd Detoonmadrermn Peer ed
Ann = dt * (over) (01) = 0100062 ml
Ann Sue „ (or0001 CATA
EN Keane As « ES
-

From Fig 2046 MB SS

1297 A bein be died ne plac 4 mc o tetas vale
{oi the hte ang om 121 24 mon m men noes.) Dee De
vote dF the angst at an De se if le ha at ai io
x at ds il. (0) I the allowable sts in the pte 145 Mn ht
Be sumespondingalenahle bal PP

Problem 2.97

Ae the Elle, Ve mm de TS mm RT ds

Dense E Es
Le Leon fem Fy 20h K=21o |
Bain = S302) = 700 mm = 900 mm" OO m“
= Kae. 2 > Aux. (900.
Eur KEY Go: Per Ayla,

SELON > Gan
3 Aus [De anlt, Fe gt
where DEMAS wm va valin oP ciel ram
Kis Haken From Fig 2.1 a.
Eu KË- 6e à Pas A

K
lode diam] | a=D-26{ w/a | K | Anat! Re,

MM the

Yam | ede wag | easel age nein | size
15 mm | Sat 97.5 mn | 0,077 | 2.75 ©1.7x10®
18 mm | Toma] 24.5 mm [0,075 | 2.71 Co.7» 10
Bi mm | 1050=| FS mm fours | 2.67 Sel
_ A4 mm po 88.5 mm [0.136 | 2.62 |iocaxiot | 58. & «io

@) Bit diametes, Le 12 mm
© Allende lod. CARRE)

ron nr 2009 Te Come Copain, Al ee rere. op any be poe
ei tetaksaacna are

Problem 298 Pelee erent eer ana a

Maximum stress at hole
Dar Fig. 264 à Far lues of K

on rey -
Lez «0.0577, Ke 2.80

Aut = WAYNE IRV = 106 we! = 1206 210 mt

Sue KE > eK) era

anim stress at FD
Dee Fig. 2.64 b
a
Fe on,
IDAS + 900 mt + 40710 m™

L - Goya) . os
k£ Ru 135.310" Pa

+ 1.80, Ke zo

4

(0) With pude and Fflets Sone = 134.7 MPa

(e) Without hole Gone = 135.3 MPa

ry

Hp ey uc tases pr a Et ells In ls
Ana 8 Me lr bdo come ron AU a ewe prin

Problem 2.99

2299 Te aline tet gechnn she e object oto equal and epost
entre ai occ o magstde Po) Know that E70 GP e200
MP, traine the mima alone value of P aad the csp a
congo af he sexe. (8) Se part amin tthe pecan se
replaced y a mim ho he me og iin 00 15
‘eer ere wen

Guns 2008108 Pe Er TOS fa
A A TOO wnt = 100416 ut

cor Ted spent Dr Sam, de Don, O
Bet
BeB-ws 8
E Kr Las =«E
‘om Fig, BONY Kr Las E
Loan Ma O , neue

Pras no —æ

Ps AG, = 00010’ Ka02 10") = 18010 Priso.ow a
PL. dsone®) (0.600) | es
S* RE 7 Morton 7 trete SLT om A

rept Ma © 20 Te Meee
Sh my ovr enn oa
A Auen anna ace

in A hcl Napa ao ee tne
ie pins Sorolla Polea o

1210 Forte tat spain of Pub, 293, Bein he mum valu fhe
Probie 2100) ‘orm ses cmp oa wil congo 010.75 mm

299 Te mim pee shown is subjected 1 1 ead oposite
eats an forces of magia Pa) Keng tf 70 Pa a on” 200
N dio the maximum lowe vue a Fan te cosprdig al
oa u the peca, (0) Sole prt, sing tht he specimen as
E Be rc y lote ts ofthe same lrg ene 40 15
ocn coc cn

P
hz

ER

Be
= by 7 ISO me + SO

> La Bonn = 0,500 m

Ait Au © (Sms

RS met = 12S 1

Ar (Gomes) + 900 mu? = 200x107

lso, 0.20
aso? * Geo. 10

Loose) | ge
Goo = 82.5410" N
concentration, Dr 7S mm,
fee
27%
fig. 254 Kees
EL. Wasyan.szie”) oa
E-- Masas! vero Pa Geiste MPa al

Mote thet Gun © Sue

erlag Mat 30 The Mera Compania. Agree pu me ie at.
ESE aay one ny mn ee pre oh one ed eed Sea ek chen
Icaro Meow ea ei pon Ag ana Se

Problem 2.101

seb

2.101 Rod Ai mad of ste nt am be spate wit B=
200 GP ud 0 - 345 MPa Al ed ls ac ache the ii ve CD,
Jus fond ed Ci Game. Avera ce Qis en ppd Cu
{hi pol ns ove open C= Deena true apie fs the
“elton he ever stomp back 1 tal pa Q soma,

PS
Since rod AB is de Le strich? permanoitty,
Fra Bray = (68.677040 Vans 10)
24.948 410" NO
DEMO: 11Q-07Ey=0

Qu Pe

sae:

2,2

(21.448 707) = 18.967710? N
ren 4
CRT EN
O SÓ

es Zu 3. 9804 10 mel
Se 110 = 3.897107 m 29m

2102 Solve Pr 2 10! assuming Dic pol oft mil el 20 MPa

1 Rd AR sde o ds md tb his wih =
200 musa. MS a Aer em lo adds dl oa CO,
and mo A veni as end Caral
"as pis mora ono: Dateien eed peu al
deco ibe ever ap ack fria poster Qee
Baa = El = 68.617 med © 68.617 110" ot
Since rod AB is to be stretched permanent,

Four AG = (63.617 015X250 x10")

= 15.4043 410" N

HMO 11 Q-0.7 hy <0

anu? 22 (157) 10.12 «107 m
dl 10.12. kN -

N Seagate tm

© Gero VES. a
= 2.2321 21% nd

HO m 246 mm a

2.103 2105 Yh Sse spa ter AB len 22:0: is made ofa mise
Robe Pt acme nh np À 2200 GP and 15 MP. Aloe
Pied othe bar union he moved du y ant Dene
Ihe mern vals a be fro P and ie piment at fe ur fee ce
hasbeen ove knowing th (yw (0) Ben

Bone
CBS LION > 2.715410" = 3.195 m

viet

200 mm
LE, &
sie: bee ©

IS > 5 Pu = AG + (200010 Xa4sviot) = 810. mm
ee se arme

2) hs Pas SOS N = 80.5 EN =~
Spon + Sn 3.76 um OO mm -

fb) Sur Bee > Sy Pur Bio.8 oN = 310.5 kW =
Sp ED = BTR ee = AROS mn -
lanzada Satis EU Eee TEITENTETEETTT

‘hate amd D lp wih E = 200 Ga ad wy +34 Ma À fe
Pi apre tthe bar tad üben teed 1 io À à permuta
ete de maximum se of he fre Pa the aia teh By
ti the ar able seth the dened ale of (a) 38 am) 64

A = o)(32} - 900 mm = Tooxıo“

> Ley» LS. Zotac
Bre hey = Bee an
When Sn exceeds Sy, Hes producing a permenent
stretch af Sp, Ha mastmun Fate vb

Sarasa IU m= 4.3125 mm

Pa = AG? = (900 #10 Maugrlot): SI0.SuION > 810.5 4N e
Sp 5-5 = 5-5. Sn > Sp + Se
8.5 mm Sm? BS mnt HSE mn > PSE mm

fb) 8,7 6.5 mm Sar Gent 4812S me 7 1081 mn =

72105 Rod ABC const of two sli puros A an OC hs ae of
Problem 2.106 ‘este! bat nc Re ls wi 20 GPa a

À Rae sapien ho edad then remove ee pounce
an Deter

+ 706.86 wel Bosse
Ase = Fu Lasa 2107 weed > ASES wt

Par? Ranke = Done Yasorıor)= Mensen
Pee TEN oat

Lang Pilea. (RTE Vas) AO
or Eh al

Bono 70.86.6102) * Goosio® E

El
E

= LEITET = LEUTE m

Sue 0 Gus 5 Bo Rel Bees En? 3.94 me a

Fed ABC cons of tan eal pion A nd BC ls ade of
‘ht el ht and be pete Wah E 200 GPa ey 280 MPa
[hice Pepi lo ho il nd I mod dm by amo =

Smee e musa tao oC ese und th ernannt et he
13 he the fee hs teen eed

An leo!

> 700.86 mnt + 706.86 10m
Ager Boy = Lasa mo LASER

Fae? Ana» (706.80910 AS04182 NGN
Pas 1767 60 a
= Pilas à Plus „ NGN LES NON
Ehre” © Pa” Ria MAT) * Goon 0" Khascetsto7)
Lerma my = 194375 mm

Se - BIG mm G28 a
|
A co teens” |

Problem 2.107 22407 Rod AB consists of two cna potions AC end BC cach wih a
owatonna 1780 mn" Porn A hema of i el wih =

300 GPa ad y= 250 Mi and portion (A me o el teni cl with

r 2200 GPa any = 345 MPa. A lod P ieppled nt Cat dame. Asuniig
im: y o ch ote chape, emi othe sium deco of Cif Pt
rally ice rn o 915 UN and th eed tk fo or, 8) he

al ‘maxim stress in each portion ofthe rod, (c) the permanent deflection of.
Displacement at C te cause yielding of AC
i" En Gis „0. MoYasaue) „ 287510 m

ie: ec = SE 29010

Corresponding fowce Fes À Ga = 07500“ Masonic’)

= 4875 vies N

Fin ro BASS „ euro KO) gon

9.190

m For equidbrion of element at O

€ Fe-CRsBleO Rs FE. Fa= 85x10" N

EstP
Since applied Sood PEÓN > ES,
portion AC yiebele.

Fog = Fie P = 87,510 97510 N == - 837.5 XI W
0.29179 1107 m

» - Abe. ARO!
(a) 8. > - BRS gore rio a car 2
Ge) Haximon sh Sr Spas = 280 MPa -

-307 MPa a

- 307.19v10* Pa =

Sus Fe puta
_and Forces For _foading .

Faélae . - A ts
Re € RE
VIS w1o* = Pim Pag 2 Pe Pa = 487.5400" Y

$! = [BL 0.100)
(00x10 rs ot

+ 026404 x Je m

Sp= Se" = 0.1979 - 0.264618 = 0.027510 n
= 0.027 mm =
ie nt

ton wana atte echo

Problem 2.108 2.108 Fur the epost roof ro. 2.107, Pi sadly nea oz
{inthe deco of poi! (reaches arme val 0 a = DOS men end
{hen Goes bck 1 aero, die, a) dr aii ale af P 0) Me

a ‘ati ses in ch pot lol (bp deflection of Cer
M: Y on rome)
ae 2307 Rod AB soie of an clic pod AC and BC, ex wäh»

‘ons etn nn of 1250 ma” Pot AC mad af a ml el vi E

+ 1200 GPa 0 230 MPa ad orton Cr made o Mae el win

2006 soy = SMa. lend ape at Car sb Asi

ton a cs tects, eerie (he mam decis of CPS

‘telly eme fom sto o 915 KN an en eed tack ar (Be
8 an in cach orion thc me (0) he permanent dein ai

Displcement EC is Bus 0.80 mm. The corresponding Shine ore

= En = 980m> » grap nic

Lee

= Ge ZO 50? Cyiehhing)

| Eu |. set ro .
| RE RS

Fret AG + (i780 «10° Xasonot) = asso" wm

EAE, = (200x10°)(17S0 x10 (= 1.578910" )e -552.6 00? M
For equidilion oF element at €, Fe- Feg- P= 0
Po uo Ro = 132 MLO 56204107 = 930.14 10 N= 99010) e

lor Stresses? ACT Gr Gene = 250 MPa =

BRE = sexo Pas 316 MPa

ud Forces For unlouling.
EM PEA

17] = Pas Fae Tr Pa
E AA EN AER N

1, SAS (oo) | ann
een = ous vie 0226874 mn

Gr FAS 0.80 me Ó.24874 mm * 0.081 mn -

Problem 2.109 12109 Esch abe har a cree sectional ren of 100 me? ad ie mate of an
«soii maria vih oy = 345 MPa and = 200 GPa. À fro Qs
lied at Ct Oe si bac ABC di edly esd om 0 1 SON and
‘hn raced er Kad, ut the cb wer aly dermis (0)
‘earn suet a occ nb 17, (0) de masia dicton pon
EUR digne of oi € (Imparte cl CE ent ait)

Elungetion constraimts for tust cables.
Let © + rolas any oP régi bar ARE

a
Equilibrium Ff bar ARC.
ah Fed &
ki OM: 0: Lafont Lu Re = Lu @s 0
=== aa
= Qs Re + PF = Re +d Feo ar

Asame cable CE is yiedled. Eg AG, won NBI

From Gy Tey > 2(@- Fc) = CT
Since Fae € AG = SUSAN, cable BD is huh when A 50 HW.
(Al Mesiniom stresses

Sr fe oo! Pa Gro” 810 MPa a

15071

© Mawimen D dePection oF print ©.

File , __Gnowe a)

AS

Seo En ‘Trono Woo +10) Bn 10
Fu Sur Sia 7 R Say = GARRO ao

Permet elongation’ cable CE: (See Bee) Ser
Cent Goede” Ela = Sim Ge
© eres Glues) ,

708 y 16°

©) Unbending: Cable CE is sfack (Frro) of QO.

Fron (2), Far 2a Fis)? 2(0-0) » 0
Fan Leo

BD remained efastic, Sr “EG? = © =

27510 7m

Since e

Problem 2.110

Each be us reset wen of 100 mm nd is mae of an
sep mara oe wich oy = AS M an £ = 200 GPa A face Qs
plc ato the ipl bar ADC an gell rre Ine 01 SDI nd
then reed zero. Keng hate cale were inl ty eine)
‘heron tens ha scr ta able BD, (th mst Stadion of pot
CO al placement pi UN ln pr cabe CB rt at)

Elongation contrainte.
Le Ox rotation aude af vint bow ABC.

+ See
“ma
Ser LE Ge? dee oy

fp Fe fi Equibibrion A las ABC.
A, DEMO: Linfu + Le Fe - le@ro
a Qs Faster Faris @

AG, (1008 Nas): 30.510 N

Eugen”) = 31.0410" N

Resume cable CE ie probe.
Fra) Fasz 2(Q-Fur)= 2X0
nee Fae € AG = BASAN, cabe BD is ete wher 2504,

(os Man , GeO; = aus
wo decline geist ©,
erro sinter? m
BE = ona SN
6.20 mn bat

omni) (Sed Seo = 2300 = 62»

Defeating. Q's Sono, 8,

ste Fe ES ant 8)

ee ZEN or 109 x15 ae
Fan (2) Qe ed + 4 Fal = 1258008 Se
Equcbing expressions for a 12.8*10°S,' + 50x 0*

Ble ot

©) En disaleccnet Ser m Sr ERO RR

Problem 2.111 ZN To tempered el as, ah 46 men ich, ar onde to à
125mm mis Du. Thi pompa as bj a an ces
te al ad ol rapide P. Both ch ae lala vil E 200
{GPa aed with ek regs eal to 0 MPa and ES M, ripest
{ete tempor and mil sel Te lend Ple rally iced From er

ttl the deformation ofthe bar noch marin val a > 1016 mm
sn ted hen dern sk to er, Determine () he main valo oP)
Pc le asim sus he epee el rt) he perme st tr

The lad roman

For the mil shel A, r(oroizs)lo-eseg ) 635 x 10h
Sq 2 LE Een, ua
ET
For the tempered steel A,=2(oroeré)oet
Sat LS SSE NOONE) «1335 0
al
Total area AZ AJA

et Brea e
Sa < Sn < Sn The mdd steel yields. Tempered steed is elastica
(a) Forces, P, = A, Gn 6655x165) (345 not) = 219-075 EN,

EAS = (200x101) (483: bx15* o-oo ott
Pa Gore Gesell) np Bi EN

Pr PR =445-9 EN -

(0) Stresses & a + Gy = 346 MPa
GER mr = me MPa E

UnPoading giz PL, (Aastor)lovzss)

Fe
- GENS 17,
EN ng OTE THO M

©) Permanent set Spt SS = hol 00187 = 07224 mm

FOL nn ae

Er Ma 309 18 Gr op poi en e
a er A a lg El pr

Problem 2.112 2.112 Forte composta pool Mo. 2.1, i rly nck
rom seso o 436 RNs then dete ac os et) use
mm foin u the Bat. 0) the comm ses he tempera!
{een th pence af te tad eed
23H Too tempert ee har, cash 176 mm ick, ae td 1
22.5 rate, Ti comp a suet shown ow Se
ie al bond of made Hdi ch are Copan wth = 293
ia end wth yd eng epa 90 MPa o 315 MP, pete,
forthe per and mal tel Don i shally ied eno
tlhe detente br ec a ai aloe da =H or
en Jere hack to er, Deter the mein vale (8)
the marin ss the temperd ste ao () the pret et

Mill steel À, =lero125 (0.0508) = CISNE
Tempevedl steeR Ay = 200047 Nousobl=#Pi4nia n>
That: A+ ARA, 2 eK? mn”

arce to yield He mdd chef

À 2 Pee AG bai lsespé )es85:9 Kir

P> Pr, Heels mild steel yields.

Let P,= force carried by mild steel
Pls fone carried by Temper steel

P= AG = (850164) (488104) 221075 N
PoR =P R= PP Asbo00 21907 > 216425 N

A COTE D meiner am
(0) Sat rro TOMER m2 OFF -
BEE, aug. cen um a

DETTES
+ PL, (42600 )(o-255)

Unfoeding ro * EL Beeren
rm x
297 nm
A S&F S.-5"= 0146 0.692 = 6-164mu -

go nn e an ae i rt

Problem 2.113 213 Therpitbar ABC spp y two ak, AD nd E o air 375 +

‘non restan Eros section und made fa mid te al sad 0 be
oi wih À 2 200 Gs andy = 20 M. The magnate of he farce
el st pray mean fom zero w DEN. Koi at a=
‘eam, deine e) be va uf the renal test in each Tink, @) he
‘movi dl of pol

Stehen! IM, > 0: 0.640(2- Pa) 2.04 Pa
572640,

(BREE) 250 = zus m"
OKs as ar vi 5,

= (RENT NR. 0) = 69.88x10" 0
Ses BE = 310.6 x10" 0

Pre = Edge OA

= (46105 Yo.suo 8 = 28.80 10° 8
Cue = ES = 12321076

= + 20 = + 4.125 Py
Q= Past LEE Pa = Pur ans Po

= fa. so nto (4.125 (69.85.0019 = 317.0¢"10% à
Gt viking ob Rink ho
8, 304.89 io
Qr = (317.06 mot Maou.sawo"t) = 255.2 4105 N

lo * Sy 250w0 = 80.6 vio" 8

(a Sine Q+ 269710" > Q,, Bak AD yiedde. Ems 250 MPa e
Pos AS, = (225710 Masonite) > 56. 25110" N
From Statics, Pan” @- 4.125 Pho = 260 wid - (4.125 M5c.25w10* )

Quo? a
> ITALO 29 riot Pa
GE 124,8 VPa a

a) = 621.52 410" m

md m

Problem 2.114 2114 Sole fob 2113, owing that 1.6 at he maga of Oe
fore applied gay re fun ito ESA
2.118 The igi bar ABC supone by ot, AD and BF usen 37.5 |
mm ecangulae cos section and modo fila ht ad lo be
230 The mate fe eco
rally ieee fom sto 2604N Kopin dt =
Gi

stp
Yard
{isto m. deine) the aac nthe man san ac
ci detection of pot

ne EMO: Ta Py =0

Delornakin! Se 2690, Ses 1.78 ©

Elastic Anadysie
A = (87.8 )(6) = 225 mm = 225x107" mt

FB 5, = (Aero 5, = er 5,

BEATE VAL 8) = 68.88 +106 0
Gp = Bee siento 6
Mazda) 5, = 40 Se = (aSHo* )(1.76 9)
See * ae = 352vi0'8

Peet TL Pe = Par + 1.500 Pao
= [78.8 viot + (1.500 Xerssriotd]o = 178.210 O

Oy et yiekiing J Bink BE Ga = Gy = 250410" = 352010" O,
Oy = 70.240
Qy= (178.61x10° Mrio.2arıs‘) = 126.86>10° N

Since Q= BEIN = Qy, nk BE yields Gy 2.67 = 250 MPa a

= AG, + (225 ¥10%)(250 rot) = 56.25 x10* m

Fron Stadics, Pao” 7 (Q- Pye = 52.5 «lo N
Pre SRS x10" riot e a
Sr + Pp SBSH 233.3v10% + 233 MP
From elastic onclysis of AD, Bra = 151.280" red

176 6 = 1 822 10 m Saz 1.322 mn 1 «e

Se

ynwwelsolucionario,net

1.118 Save Prob 2.13, suming ht he mgt ofthe fre Q apt at
Problem 2.115 ji godly eras Im mr o 240 EN and the dcr Mc I em.
Knowing tht a= (640, eerie () he ion sues en ih (the
Teal din o pi B. Asa dat he Anka re acl o te ex |
“Sey Empreses wih Peking

2.13 The igi ar ABC porte by two ink, AD ad BE of wif 37 |
‘rm rectangle cos sti and made fa mid ho that e atando be
remplie wih E 200 Ge ander 250 MPa. The mnie a he (ce
O pri ah pray incre fem act 260 LN, Knowing

6 m, determino (o the alo of the mm ames in each ik,
traci defino polo

See solution to Froblem 2.113 far the norma? stresses in each dink
and the deflection of pant 8 alter deadin

Es = 250 #108 Pa Sue = 124.3 010%

Ba = GAI.SS “10m

The elastic analysis given in He coftion de PROBLEM 2.113 applies
do the union. “4 3 Us

QÁ= 317.06xJ0* 6°

pr Qu asomo
aaa "Boa ver

= 820,08 410"

Ge = $10.6 110" 8 = (310.6x10"X820.03=¡0%)= 254.10:10% Pa
Sie = 128x107 9 = (tasa Marcos) = 104.06 v:10* Pa

Se + 0.60 0° = 524.820

(a) Lerida stresses

Ginn = Go = Say = 2500 254700 4 1010 Pa
caro HR ae

Gara? Gon Cal + US IHM MO
toner ie

62}.63 +107 - $24. 82 +10"
Ben LOT m = 0.0967 mud 2

CEE

Problem 2.116

2.118 A if test rod of resections ea Ai alice gi
Lappots and is unsuessod al operar of °C. Ihe sete am
tote slides with oy = 290 Ma and E = 2006. Krovi fat
EIER WFT, dete the ee heb fo) eI ete
‘uc ied to 160°, (9) aer tee ha ened 7

¡A

Let P be tha conprescive Foren in the vale

Determine demperdore change do come yielding.

8=~ fe + car) = - 4 Later, = 0

(an, = SL. 2800108

Ea © Tama ets

64 ar: (60-7 = 153% > (Am
e vide

Cooking? (ar)
S's G+ 5 = -ÊL. Lau - o
ots E - Baar.

5=-S, > 250m a

153%.

= = (200 ou Bt) (153) = — 358 MPa,
0) Resided stress
Chet SS 7280204 WE KIO” 109 MPH

tom =

Co pr hy dr

Problem 2.117 2117 The sel 10d ABC is tah og spots ed e unse a

amet

RT ihn. (een oli
Eat Lar aussen
m) Rg ont Lar 0.250=

Constraint? Sp + Sr =

3, datenmas AC te care yield in pain 68.
- Plus _ Ble „
r ER Exe © be AT)

avs Pep (he + bem

Leza Re
A yielding, P= Pye Ag Gy = Boon esoo) 1510 N

-_ Lat y ben)

at (Er)
25x10 [os 0280 Vo ape.
CTS) Gas + 2.) = 90.812 °C

Aetua? AT: 150°C- 25°C = 125°C > (AT) — Yiedding occors.
For AT> (AT), P= R = Twin

lo) me

Or Wh ur MPa

Sige 250 MPa =

(A Fon AT > (OT), portion AC remains efastic.
Pilas 4 Lye (AT)
Ehe + Lau

(OO Yo.150) me “e
E OMT NAS = 106 918%

Since point Ais shetionany, 5.7 Baar 106.910 m

A]

Problom 2.118 “2.118 Solve Pub 2.117, asuming hat the semperure fl 10 sd to
180 md en tat 28°C
à sit

2.177 The sel 10d ABC is alto sg spp ad md el à
tempera of 29°C. The le mod stop, wid = 200 GPa ad
fy 250 APS. The tempe ts pts le ed ete lec 190
FC. inmi 17 > 10°C; eine) the st in Beth pte
fhe (he dci pont ©

FM re
Lar 0.250%

Sao

fo cave yielding in porhion CB.
pa (AT)

ira (HS + RE)
Ab yieding, P= Py = AaSy = (200110 Vasoniot

Ty

IN

|
Le a bs) |
Res

[_2-180 N
"To A SS MT | = 9-812 °C

Actual AT: 150°C- 25°C + 125°C > (ATX Yielding ee ars.
Fe AT > (QT), PP = 7540 N

Cooking? (AT! > 1267
pte — Elu din
2oontot)(0. Yee 7 1 weed 108. 285 x10* y
er
Reside force? Rss P

LR 103,288ab* 15910? = 28.235 910% (tension)

(2) Residual stresses. ¿aso? Ge 56.5 MPa e
10.1 MPa a
or Prmnet detection «E po 5.5 0.0424 mn 4

Problem 2.119 12.119 Dar Ah rs sectonal reno 1200 mn andi mud a el a

sed tobe lali wil» 200 GPa and oy 230 MPa. Kring

‘ha ie fins Feros 019520 N and ben deere zr, Serine
(a) e permanent det poil CP) drid es inte br.

A+ 1200 wm? = ROMO me

ice te yield portion AC Pier AS, = (200x150 asomo
= 800 <10%N

For equilihein, Er Peg = Pac = ©

Py = Pu-F= 300410- 520 #0?
- 220 x10" N

(220 o Moto 0, mo
400 10 lavo x10°°)

Saz E. gene. 183.333 10% Pa

= 0.293333 "id" m

pf = Ela. - (roue Moto -o.ne)

87818240 N

as Let 0-40
Pal = Pl F = 378.122=10°- g2080° = 191.313: % 107 N
1. Ri. arman .
Ou. De. PURES = sisi Pa
els ES Lu = ug seg cio" Pa

(378. 1a2)(0.120) sso”
room = un
@) Sp 0.293333 407 - ON BIO KIT? = 0.104210 m
= 0.1042 mm ==
{bY Guess Sy Ss 250 lot - 315.152 O” = -65,2 x10" Pa
65.2 MPa =
Geass * Ga = Seg > 128,30 NB 2 10% 2-65 2210 Pa

ESR MPa a

yoww.elsolucionario.net

Problem 2.120

2:20 Save Poh 2.19, essing tt = Im

1.149 Bar ABs rs cial wes of 120 1m and ie ae of lt
ds sol tobe catala with = 200 Ga aa a = 260 MPa mia
lero X ers rm 1 520 and then dr 1 ou mie
10) de permanent din of po CB) the dal nin the at

1200 um

ses rot m“

Force do yield portion AC! Pu AG

(120010 aso rot)
300 #10% N

, Fr Pa Pan ©
Re or Rea
AAA = 300 x10" S20 10°
= 220 0% m
Pag tee _ (270x109 (0.490 - 0.180)
> - Pate „ (zz0nwr(o, -
Se: E ECT IN
~ Ra. tomo | Sarre
Gr FE A
Untoad ing,
Pathos, Patten , (F- Pills , Meets Fle
fee Re te Coe 222
+. Ela orto? Y(o.440-0. E
Pl Ele A ro
LA = 202770 - Sao» 18 ag rio N
Se + A 0.220455 610 m

Goo 0 *}(1200 7107

Be = HER = 86.061 110" Pa
Gt et. nase?
ar B= Kr 0.208383 510° 0.130460" = 0.00 788 vo m
0.00788 mn =
WW Gum GG 250n10°- 256.061 nJO" =~ 6.06110" Pa
-606 MPa e
+ 198.985 of + 177.273 X10% = = 6.06 «10° Pa
- 6.06 MPa 4

: 2121, fer be compose tr of Yoh 211, semis the a
Posen ziel ti ep Bar dal Ines om co 38
IN atthe ned bck tose.
2484 Two ket ban, ah 476 un hick, hand a
12: an min hr Di empor ar ed a ah
{ic alan of magpie Passe te Scope wh 4 2
ve

ar the deformation of the bar reaches a maximum vale fig = 1016 crm

{Sule Jacl bed vow. Demin) US mima Ten 0.0)

cnn tne ps pomme

> Mifdsteel Ayr Covonts)lorobeB)= 635 w 86 in
Tenpered steel Ay » tano) 492-6410

Total: Ar

Total force de yield the mild steed

PE & 2 Poe AG mbr) GSE) = 295. TITER,
P> Pez therefore mill steel yiedele

Let B= free carried by mill steed
Bie Ru mou by as at

B= AS + COM) 4x0) = 2191015 EN
PEREA, A PR A a wo 217 EN,

= RB - ETA 5
és ee 5 IHN.

oe?
HA, BIS bs

InBonalin ta 2, Bele > LE mp
pote WO at deine
Residual stresses

mit steel Em = GS BUSA SEM Pa,

Fempered cel Gey = 55-65 = 4487-20
> £84 MPa -

loma o and Noi sn u be lr bc
e pc aci

12122 torte compa hr in Prob. 2 1, dae the rea
Prob 2122 sees in the epee si ap Pl rivals inne Hu ce vn
the defeating of te ber ecos sins Ye dy = mm and en
‘reed bck em

2.411 Two empere ste bac ach 476 men tik a dd a
12. am mids Ti compose tr fe suecia 1 à ca
rx alka of mage Hh ci ae elas with E 200
‘Gt wit ied stent egal 0 MPa onl 205 MP aptly,
tt empero ad a to Ths ka Pi rally seed fn po
‘lhe determin of the ha cher «msi vd LONG mon
ande decreed bac sr, Deter the macnn Yl PB)
tho masia sc I the empty) the pomos sat
tte led rome,

For the mild ste? A,

> LS. aie Gen
Sn E” Dee Kio"

For the fempered steed Ay szloersetitacit)sobantn?

Sue LS = (teen
inal Y ET

Tota) areas A= As A, stern
Sn < Se Sp The mida steed yields. Temporal steel is elastic.
Forces P,= AG, = (6ENE(Busni04) = 2141075 EN

a5 Koasos). 635

= bine?

Re ER en (ARANA bn uy

| Streses 6, = Boz Gy = 345 na

276800! _ sau M
e rr = erp te

Unde.

1e aastos
A a AER Mr

Residual stresses
Gres = 5-6! 240 — 43.3 = Deza

Ses = 6-07 892-4 4423 = 119-1mPa =

rp ata 299 Tae Ga Crp nt eee Nye pe al ph ot
ee rennen see

Problem 2.128 2123. A amb of sani i onde othe se of ick te
rie as sho ala, ti ees er Keng ae

Teepe wile oy mico 7 anes eed nF ses o) e
Tight praetor ol eal end ssc, (ha ope
‘es tha wl cot ends os lt lo 409 Pa
‘Aun my = 295% 10 PC to the im ad = 183 % 10 FC fe
‘tev ut ae ht he login el ts = 8 Pe
m = 200 MPa Un Nes al ees the se)

Determine temperature change te cause yielding

Se + LAAT) = Lasam,
© = -Eta-d)(aTh = -6r
ame ates > 412%
(a Tay = THAT) = 214472 > 4300 -

After yiehling

>= Ga Latari= La, (ar)

Cooling

sie FE etary: Lecary
The resident stress is

Ba = Se Bs Gy - Eldar)

2 et 2 444%
ATs za)" Teen rc
6 T= T+ AT = Dit age = 765% -

TP > 165°C He adominom ban will most Rkely

yield in compression.

Problem 2.124 2.24 “The atm tod ARC ( = 70 GPL) wich comi of to

‘indica pois AB ad BC. so be replace win yaa a u
IDE 200 Gh) of the sare nel fh termine the minimum Fo
‘qi it do! the nea ot he on ect na oe
the ton fe am od de the ae al he one
Seen he tes eo a cod 19 Mia

Deformation oF aluminum vod
» lue , Ple. L/lu , Li
Se mee Re ECE
MESITA “eee ) Z
evel (Ea + Eioasgy) =0-eo3nión

Steed vod S =01683x 15%,
A= PL, MoxéMers) | ne
ES aro) DS
z
s
Required area is He Parger vedve Aro LI”

ar EE» [CMI - 0.029m224mw -

ri Nop Mana y e die pc
en nee

yoww.elsolucionario.net

22428 Th tans lp A bo atc eed
rap. Kong ha ven

en np

A de sopor dtemin Gerne in tempera for which pin il
ingen.

w

Problem 2.125

Detar yrogo Ar aa) com © GOK ta
getan B= losmiot Pa

_(a.e\10o\l4.81)
Y

= erre

Problem 2.126 2:28 Two wid cl rod joie a Band noel a show
Rot Ais mate le = INC) ded Al = HS GP.
‘eterno (th tal deformation te compote ot ANC (he ei
tn iE

tion ABI Phat PRA Ll bom

Nr ar Ear} -0-001963 m* Esyozos GPa,

Bab Cord ERROR
BREE ane eng “Am

Portion BET Pas “OBEN Lec s076m., d= 75mm
Bact Fat Feonvosonsen Eur 05604
8818) las)

Tesi )ío: 00442)

Sect e - = e142 x18 mM

CRE

+45 - 02143 = 01509 mm sr 00m e

CRE 142mm a

2427 Link BD male ot edt = ISGP) alla matin
caca 290 i Link CER me of aie (572 GO and mac
chen a of 50 mint Dts te musimam fe P ie ah De
ete eral ot pit À the isc lA m ar ne 030 m

at € Use member ABC
as à free body
> Fe
?
DIM. = 0, 50P-225 Fez 0, Fos 1.5556 P
DZMero PSP no, RuroseP

So = Res fe. PAS = estan
o Ta ee PT
From the deformation diagram
Shope O > E, A

one

Tobie nto" P

5,7 Sat MgO
13233659 P + (0-125) (rota xo") P
saz mod P

Apply displacement mit Gy + Or 6063S = 22-17 SP

LORIE D 16.707 EN. 15:79 EN =
BRETT

rear ert 030 Try Car, I. Al a pot am be plc rc
À Ro pm po prc einge hed gts when
| Re it des odos pros A ron ma ama cg Keen pc

Problem 2.128 {2128 A2Soemtong aluminum u (E = 70 GPa) of 96mm otr ics and

mn aer dual may be cased at hth end y mens aii mates
am dun desen an ro LE plc. Wien cover seemed on gt aso be

od (= 108 GPs) of 2 a dane spas ise he take and de seco
ver seemed on. Sn thro lily ones has tb peeved
thn te cont mathe ced again de md by lig copiar af à ur
ee can De ply shed,” Detrmie (a he serge noma sen I he
take ud al (0) dla De be oe fe fo.

Au = Hdi at) = Fla 28") = 402.12 mr = MOZA IO oh
Bed = BAY EAST = 490,87 we = 410.87 15° nt

PL P(o.2s0) oo

Eh aa TP

PL - Ploas pasoo”? P
Zaku o EA

5° = Chem} LS = 0.278 mm = AO m
Sue = St Sad or Sane = Smt = St

8815 x10" P + 4. 8505x107 P = 375 «10

‘a — + ameogulo” m
air: em)" 47508610

BL. dor. , 210% Pa = e, e
Wo Ser Be > ARE > ana 0% Pa = 67.1 MPa
o A eo? Aue
A Pure =

WW Ese? (8.8815 OK 27808) = 22.5810" = 0.2020 mn

132, 515% m

Su > (4.8505 » 10°" 127. 308 810" 1325 me

‘ir Man. © Ta Men Compe Alm Nop Nm ny hein

12.129 The nom win AB seh eg 2, aac apports
son and vera Pi ap tthe point D. Del by he
‘oss cional rn of he wir an ty he made aay Do lo, fo
D the decir e ign ie

5 e

Use approximation,
5
Sin 9 % tan 6 e 2
Shties HER, Pa md - Peo

Pas = Zone = 2

= Pa PL

AE AES
From the wight triangle
(B+ Se)“ ES
Ste RT + Sun à
= US (14 452) 7 20 Se

“ PP
ABS

Se BE 25

Ss RE : sea -

james near ét mt id md
nal aspen owed eked dards
ion taming area ng ser ere

2.190 The cgi har AD x pore by two sl wie of LA
ine (E > 30 GP anda pra backe 1 Kong a the wi
‘ere tally Gt, erin (the onl tion a ch we when à
SHON nad ap), de comes deci u pi D.

het © be He rotation of bar ABCD
Then 8, = 03

3. = 060
> Dicker
= Fe
Page LA Bec « Goom Howes}(o30)
w Lee 9.25
= 424150
> gs + Le
2
e. > Ele - Goon) Hows) OO
‘al Lee
= 411234 0.

Using Free bady ABCD

DEM =O 0-3 Re tobe -01P = 0

(0-3)-424 15 8) + 001411234 0)= (01) (Tos,
| 409918 0 = 210

02 ATEN vad
Poe = (42408) (1976x 1073)

= 938 ÑN. ~
Pe = (471234) Châtéxto"?) = WEIN. =

(85 704 8 = (OA) (MATE) = TB UT? m
-

= 1-18 mm
rp atin ©1009 Ta Mer pa

ER nr mp a
‘ve por by cles i ot ab

A en docu

Problem 2.131 2.131 The caer pst (125 Ges 9 «104 Cisne with

ste har, cach of 2m dee (209 GPa andy 117 «07°C.

Dati the rome sss Faced inthe sel ae In cti by à
temper seat °C,

Bex 6 HF GE" ann nt

2808100 we
se 2 Ay O 2.280800° = SSSR mo
| SS
Net Fa + teraile Puce deveiped in the comerte
For equidikrion with zero tote Poren, the
compreasive Force in the six stee? rods equals Pe

Strains? £

AH

eo E MN ar ary
latehing! = Re +
Matching? Re B+ UT)

(eat RD = (anar)
Tasman yes) | Be Gant Yas)

= 0.29) 40S Pa + 0.291 MPa wn

Gis Be PAN arme Pa MPA me

oper Sa Oe md

em pl abe ro
ipa tates Soren

yoww.elsolucionario.net

12182. Aviation elton it cuts ah chs had wih
eros A ricos of py 6 2 19 Ms Mode toe A arto ig pta
ds ron! Deng y Pt grid of the ace app o he ptr a
typ 8 the comeponding te, me te ese sing Coma,
Fi atte sytem

Shearing chain iad
Shearing sten Gr ors SE

Force $P = Az = SAS
= 2GA8
pe GAS

Effective spring constant

eyes
= 2 . 264
pa kr E > 264

Data: Ge um

34 ‘id
el ke se] 4 Ar (ous Yor) 2000159

hr 3omm
ZC xref (0-018) _
Wil aluumes) = 14 NJ, =
Problem 2.133 2.20 min Gana” DMPA sami miam ana de

Sen nd
i Ab hee Az ve Hora 10 me

1 d= D-2n o- 20 + some

Aug = dt = (RS) = 1200 me 1.20410

From Fig. 2.642, K= 2.65
RA ao

we Aloe 25mm dr 100-507 SO mm
50 MIS) = 750 mnt = Bows, 4257050, K= 216
10 1iz0%108)

Po he „ (ute:

41.7 «107 N

Morable vale BP inthe smaller P= AUDIO N + MT =e

Problem 2.134 2134 Koad A consis of wo ent enone AC and BC, each wih a
quel ra 950 1 Fons Aa fa lh
GPa au oy" 250 MPa ad pra CB s mud of gare sc wi
E 200GPuande,~ 345 Na. À led Ph applied ase ahown. Amonio
q na stc be esop, demi (he masimor ellos of CAT
aly incre m cto 19 AN ad on oh sk to (Me
Femanent defecto of

Dicplacoment ab C to cause yielding of AC
= Lis „ CHOSE . 0,500 5" m

. 300 #10"

À Gre = (295040 Joso #108)
= 7872.5 nor N

Fig t= BABE à - Goouot ars None), uns

Las eye

Corresponding Force Fir

bee 0.320
Fre
For equidibrion of eJement ot ©
e
Teer Pe CRae RO Re Fi Fes 195 10%N
Since applied Put Pz 1628 LON > SUN,
portion AU yields.
Rig = Fie P = TRE leas JO NS = 987.5 MIO NW
FL . (32.5 40? (0:20) ut
TER gone ie or) > OBE #10
= OBI mm =
Maximum Stresses. Ge? ire = 250 MPa
Sur GE -- MAC = - g00,51910° Pa = -301 MPa
O DeVection and Forces Son unfloading à
et Rélee Pr 2 gb, .
= Bde Re 2 em
= J62S%10" = Pe Pa = 2m Pac = 8125x100" W

gi = eus ie Mo. s201
(200 «10 raso oy

+0.04068x Jo m

Sp = Sue $' 7 0.171810" - ooEsm = 0.080680 Tm

Permanent debfeclion af C Sp = 0.0407mm à a

21436 Turn 104 Ba ar ar and me of el
Problem 2.138 “sh conte cool to be lapa vio mods of y Ed a
Yt steph sng he tete en so, eed to
That ecto od Col rod cat lowe pra pic
{Siren ha ene ston pla Can] Ca ea or a
‘Slow of PD yt diet fen icon he ec he
Torso uae, dive epson or) eed a a the D
em ora et ang

ra

+ AG

Force deflection diagram for point Cf Beck and
spring system O E

NemgsO Ne mg
AER Or Pa TO Pe ie
IE Week does not move, ne
Hee ur

TPP = omg, then ship at Po Es mg occurs.
W tle Force Pis the removed, the spring returns de its
insti? Length
D Eqvating Pe od Fu,
AG- amg m> AGe -

44

0 Enuating slopes

21 Aro contig of comes, ech of which ie homogenea und of
‘ifr ro section subjected othe lng sow, Te lemon
I emma yt ral te by deal ete By and
‘holon pp at cu y Py de rapide Po thi Ton ba
‘sauna o e postive Ps diet oe ig and negative tke (o)
‘Write compat program ht ean be dt demas Io arc or
tras teach mar, the demain of each clement and the u
termine.) Use hs program tote ee 220364 126

SOLUTION

FOR EAC ELEMENT, ENTER
Li, Ai, Fe

COMPUTE DEFORMATION

UPPATE AXIAL Lom P
COMPUTE FOR EACH ELEMENT
= P/M
8 = PLA Es
TOTAL DEFORMATION:
UPDATE THROUGH ELEMENTS

loner
PROGRAM OUTPUT

À

Probler 2.20
lomont "Stress (Mo) Dotoreation (m)

2 un

lement Stress (Nay Detormation (am)

130.6775 00450

rp Mae 20 Te Mr ona Ae pl Ma bp erh,
hs ip pulp el boo ne lr nates
e peated col pte ops Ai pr

PROBLEM 2.2 2.2 Rot A ooo wih bot ende nds Remmi meets coh

ie omogencss ed of noms cos von and tected oie
Iondngshm. The len ofelmen i denny Ls eros cien rs
(ou sms el y. Fado spe ti ih 8 by
Pa gate of san ig assed tbe postive, iin
{hcg and neve bere (Note that Py D) (a) Wea complet
program wich cn nan determine the ection 0 and D, de args
eral ses ach lament and he defomatan of sh eset. (9) Use
tis program o solve Probe 24 02.0.

soon
WE CONSIDER THE REACTION AT B REDUNDANT
AND RELEASE THE ROD AT B
Compure Sp WITH fy =0
FoR EACH ELEMENT, ENTER
Li, Ary E
VPDATE ANAL Lao
paPrh
COMPUTE For EACH ELEMENT
= PLA:
Go = Plif ME
UPDATE TOTAL DEFORMATION
ee
COMPUTE fy DUE TO UNIT Lean AT 8
quire; = If Re
out à = bef Mie
UPDATE TOTAL WHIT DEFoRAnON
UNIT gy = UNIT J, + UNITS;
gurenssıron
FOR TOTAL DISPLACEMENT AT B= 2580
byt By wur 5,50
SOLVING:
Rae = def NT Se

THEN:
Ra = EPitRs

CONTINUED.

-PROULEM2.C2 CONTINUED,

© =O; + Rp un
La & + Ryu fi

PROGRAM OUTPUT

2.018 oo
PR 02.
228 ma
a Kr

cos

een Mtr © 89 Te Gran Campi nA hare Mp ie y ba |
ie mn D RD
‘dan pay Med lo pi ze tio phe

yoww.elsolucionario.net

PROBLEM 2.3

2. Ro AN consi fn lets, exch of whichis omngeons a of
fem res ton. Fa ie, wile aly ther op bee
Ed B ua he Ans veel sue on de igh. The hgh men Fi
Ente I exemvesetonal ea by Ay modif let hy ant
Cosi ofthe expson by a Aer he ampere of te od as
‘eon ceed by AT, We op BI card ad theres ue eet
‘ula opps forces on rol. () Wate acompuler prog wich cat
[ao mins te gil o recon at nd th normal
‘resin eich lees end the deformation of each espa. Use is
ogro solve Pa 251,239 and 20

souuTion
WE COMPUTE THE DISPLACEMENTS AT B
ASSUMING THERE 1S NO SUIPORT AT BE
ENTER Li, Ai, Ets
ENTER TEMPERATURE CHANGE T
COMMUTE FoR FACH ELEMENT

8 = 0 LT
UPOATE TOTAL PEFORMATION
da Sar &

COMPUTE gp DUE TO UNIT LOAD AT B
unit $ = Li/ Ac Ee
UPDATE TOTAL UNIT DEFORMATION

UNIT Gq = UNIT fy + UT 6
CompurE REACTIONS
FROM SUPER rTIen
Ro = (ig fume E

PRODLEM 2. CONTINUED.
PROGRAM OUTPUT
Flemmt Stress (088)

2

Problem 2.58
Ra aces a

2 En)

ren es (nes
2 20.887

Doors. mieren)

toca. tmberom

PROBLEM 2.08 2.04 Mar AB beng La s made of two dient mea of given

Cross-sectional an, eof lic, ad yield srngh The bar le
jc esa tos it tice rally ie foe et wl he
arm uth ber Js cached meinem vale Sy aad tee
eed bck taco) Wis a Sanur progam which, or sen o 25
Stuer af ely col ever a range xc om ovalo ea
{Dove ofthe dehnen ening bath mol o il, can be ed w-
eine U mesma vos Pe ofthe, he masta soma ro in
Sch maria te porn rman dy of We ba ar he sia se

(By Uns he rogram ok Pb 2 111082112,

sounon
nore: THE FOLLOWING ASSUMES (65) <(4,),

MENT INCREMENT
dm = 0.05 (ey), Lf
DISPLACEMENTS AT vire
fe 5
FOR EACH DISPLACEMENT
IE me:
2 Sm ef
Em ELE
ps (SALMA,
15 by < bm< OBE
(oy),
= dm Ealt
Pip = AG + (nf) AE
IF dm? Oe?
GeO), el
Pme AT, TAO
FORMATIONS y RESIDUAL, STRESSES

SLOPE OF First (ELASTIC) SEGMENT
o SLOPE = (4,8, 4 Aa Es )/L
Sp" En = Pm/ 5988 )

Er

HT
d
|
Y
se
Pa

+46)

4 i (Joes © 5 = (Er Pm/ tt store)
ct 5 (eg = GE Po [le store)
Au] —

PROBLEM 2.04 CONTINUED
PROGRAM QUTPUT

dm Pu sane) Sema) DP Siceu) Sick
‘am fy WMO EE ly SS
o o o o o o o

00369 235511 2eme eh © o o

O76 436090 Su S000 WERTET 010g -4u09z 69.5P0 «Palm

ole MSPS Besen SIE 0134 rio NAS «PIO

ny Hfbero 34068 Toon 0543 Meg Mor

2.05 ‘The plate ea ole center cs the wih he com
re ren afr a or al ac oi th a cet hi

re + se ao he ol nd i the WA ofthe ti Wie a come
Ptr ogra triste allowable Ic or te en stn D,
Me os u he ne le sss ofthe mal Ki
Ing trom, = Sa aora = 10 So ei ovale

SDT cao om Si 18mm wg 3 men Verena
SOLUTION
ENTER
12s ty on
COMPUTE K
RD = 2.0 7/0

K = 3.00 - 3.13 RD + 3.66 RD
COMPUTE AVERAGE STRESS
Soe = Guy K
ALLOWABLE won
Pa = San (U-20F |

PROGRAM OUTPUT

rating Allowable Lead
st ee

repr Mata 260 The Mra Compania ace No pol rl my poe rp

Sa penta Mela er pea A Role rag lc pr

ynwwelsolucionario,net

PROBLEM 2.C6

2.04 A soit nested cae subject 10 aa Kes Pa town. The
act argon PHARE By plasma Ue ce bn crea ner
Sul ticks, wre a compatr prom th cn ett eal de
Elongation the est cor. Wot fe the passage eet ese
aio To he pram wg) 6 (R= Le

SOLUTION

FOR (=I Ton:
bp = (i eos) (1/n)
hy = 2e- eC L/L)
AREA:
Ast,
DISPLACEMENT:
g: &+ Plbjorftne)
EXACT DISPLACEMENT:
Seine = PL/t2-07 eB)
PERCENTAGE ERROR?
PERCENT = 100 ( §~ Geyer) Feier

Approximate Exact Pereane
A Bas 0.108 4000
a Bilge ONE ie
E De Eee

bist ee
‘hea emis con a nda soak pps A eet ig Be hc

Chapter 3

24 or oni a nm, demie de med hi as |
yl mp 72 SIN me
bao TE '
we oe Fe

AT, , DS , A
dues BE = A = 89.682 vo

Toa = 597 MPa a

A ze
Tur D Te Fe

Ts Fete = Flo.02a) (0x0)

= 1338.) Nem Te 198.84 Nm a

tac
nd és por y Mr Wer er ina ce pepe À et weg maa ng ent pen

29, Kacuing the the inet diameter of e Hoe shal shown à
à = 72 om, cermin be mania Scar sess cused by a tng of

Soest f= SON

Ca = bey = (440) 220mm
= ka = IGE) 01 mm
Te Hess Hat"
Ang = TE - „Bodlaon

Csmozm

= 228328 mnt

ÓN =

Problem 3.4 3.4 Knowing that = 30 nn decano que Fat cames mas
ina sho sr 82 MPa in he alow sa sow

© sam

qe Ll ANA arm og

© a

si 3.69 Demis tg nt cn De ell # i af cam
roblem 3.5 dice itt ecc ane ing tes of Mb. I ae
‘ut. ein Satta free ced Ino sah te
ee oes ont strane dase ome
ino.
(a) Solid shaft: c= dtd = 40.020) = 0.010 m
Te Fol = Eloi)" = (S.208e%/077 mi
p= View, UsTonomo (sexo) |
ee 2.010

(0) Hollow shaft: Same arta as solid shah,

As mé el) = nle-Geÿ]: gues = me

2] Te 125.7 Noms

30.010) = 0.015470 m
0,0057795 m

J= Flat") = Z(0.0115470'- 0.0087735°)= 26.180 110” mt
Te Ed LOBEN) a TG Mm

9.0057

ren er. 0309 Tee Compan. ID opera Mana ae il pad
‘Sed ey ene mens wa eng of ol e on oad nhac nen
cad arce piel Male lees ec pres At wg Samana ue hel pro.

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36 A torque T= 3 KN m is applied lo the solid bronze eyinder shown,

Fées Desemino (0) the masi sewing tes, (2) he sheng sus a point D
‘ich Les on 15 mun dis re dv on De nd of he older (2) te
Ga mm ent of th re ea y De jan Gein within de >a

E

m fe Some = Sons me
Je Fete Eons)” A

Te 300 gun

Te _(an10*(s0n10"

Pa 7 70-736 *10° Pa
Im = 707 MPa a

(py IS ISO

nr Samp. =e

er

6.25% =

‘rps Mahal ©2089 The Me Ht Cong, tesa ptf mal eae ge rt
‘rooney mee aspera mete epee om en eher
‘Sesame pan He Bs od san: pro sat wig tne aot rn

Problem 3.7 37 The slit spine A mad. sol whan allowable snag

scat of 5 Mba ad nee CD e mad o ha wi nent se

ng ars 0130 MP. Dato a) ke age logo T na car he aod

Ue alot scan sarc pt ne xen in see CD. he
Spine AB

coord oie e de meer

Sleeve CO: cher ECTS) = 375 mK
OS

E (ete!) - Floors" sone"

Tes

Sexo ene

CANA ron.

wer
| re Tzzoin a
WM Solid spindfe AB: T> 2080 Kim,
JE - PRE - cn sacam

de Forma

38 Ts sod pe AB has x dancer» Wm ae made a
to ia one sat re o MIA wl shee CD ira ot
‘bras wih an le eng res € O MPa Dace the at
ee he lit

Sad spindle AB: ee hd, = 4a9)= [4mm
I = Eel = Eon 204-7 vie"
- Le
Han
Tog Zr. CONTE) 906 qm

oat

Sleeve CD: ¢,=4dy = 475) = 57S0m
CEE EE Bho mn

4 (0315* - 03x") = 1. tan
T= EEd- ef) = Host cout) = 156 oe
+ Le „ LiseuS oud) - sop um

Tri wor Beton!

Allewahle volve torque T is the smeffer, Tr 2 oP Kin ae

Problem 3.9 39 The targus shown ae exento on ples an Keng that ech das
sb, dame he sna searing se (2 ins 48,0) BC:

Ts TS 2 BT, ana)
J We Ras)

56.588 vio" Pa = EE MPa e

5 Tae 300 + 400 = 700 Nom

20,046m, C= m TE „AT, -2xmo
dr 0.0%6m, C= 0.028 Lo FER mo

= 36.026 10% Pa 36.6 MPa «e

Problem 3.10 2410 Thetongo shown are exe nls which ne tchat 080
cpr sho dt and RC. Ton 1 dec te toa ms of the ar
‘Stern tn allo devela Chr wiih gt sear she

nba nea

5 Shatt AB: Tyg = 300 Mm, d=0.030m, C= 0.010 m

Te . 21 - Los)
Ta os

u = 56.588 “lot Pa = 56.6 MPa
ShoPk BC: Tee = 300 +400 = 700 Nem
d= O.04em, c= 0.23 ‘say > ES = 21. Qin
2 ” to A

> 36.626 tota = 36.6 MPa
The Dargest stress (56.588x10° Pa) occur in portion AB
Reduce the diameter oP BC provide He same stress
Tac = 700 Nem ste. 27
y. = 700 N- Troe = EE = 2

2. OS oars
OF es, © O = On

Es 10875407 m d= 20: 39.79 tom Em

= {MI Under nora open codons, the etc meto ent arg 02.
‚problem Seil AN on sat A Knowing Bt ach Sala termine te aan
searing ein (ah a (sh shat.

ES 28mm > 0.088 m

an Le 27. ame .
tar Tee fe A orto MA

LR Tar LU bem > LM IG! Mm, crol Alan 0.084 m
ZE O |

tas Fee = GER = CANTO Pa act OLS HPA a
(©) Shaft CO: OS Nm, Crhde Dons 0,028 m
Los FR = AGE = 28.08 «10° Les 73,0 MPa e
12 inorder ce wal mo te my of rb 1 à wo sgn
Problera 9:12 ‘hin condo ct aa eo Dn

he smile dime lt BC or vih De mein vale ofthe song,
ein emy wil tbe eae

See the sion fe Prublen ZI far Figure od fur maximon sheowine rester
in shePts AB, BC, ard CD. The Aumest shearing occurs in ah? AB. Die

magnitude is 81,2 MPa. Rdjuct He diameter of shaft BE su Het i
maximum sheoving Stress ie 81,2 MPa,

Tre? 8210" Pa

shot Nom

tee = 2L
wT We

SEE YT at = ratte
RL = rr 7 RR -

de 2e WE

Problom 3.13 213 Tor shomner sean pliz, and Keown that ot ll
ool, demie th mann easing coa) sha 4, (sha BC:

(a) Shot ae:

= 76 los Po
us 15,5 Mi
(6) ShePt BC: T= 200 Nom
c= 4d = 0.010m

mo. AL. ANB) à 62.7610
ur SE - AUS = 2.72108 Pa
Tue = 62.7 MPa «a

Fri ais 3.14 Teesiñ of ple sem dhownasetobe redesign Koni tht he
en Sel alone edo ur in tach sa 0 Mh tein he sna owe
‘Sameer oÙ (a) Sha AB, (0) Sah BC.

N Gay Shaft ABE T= 400 flew

Tom GO MPa, = GOxto* Pa.
sto te: zz

Fes

- JT . es
Ve es
= 16.9 me 1619 me
dez 2€ = 32.4

(6) Shaft BC: T= 800 Nem
Come = GO MPa = coxjo* Pa,
EROS
= 20.40 1107 m = 20.40 mu
dies 2c= YO.Bmm A

repr Mal 0288 The Rin Copan A gs nt pt Mac y dn en.
ee erg

2.18 Tele tad ess 100 MP in the set A and MPA the
Frome 1e ras rod AC. Reon a torque ef marie T= DON m apd at ad
remonte ru arf)

Shabk AB: Lu = 100 MPa > 10010 Pa

DE = names» mans

Bar Re = 25,8 me oe
| Shaft BC: T= GO MPa> Como” Pa,
[aa | adhe

AGE = AMA o 2121 mm

de = Les 424 om a

ce

3.16 The allowable searing stress 100 MPa ah 36mm oA
nd 40 MPa in the At dame cod BC. Negecing the ett of se
‘tacts, Jets he lus rg a ca apd 14.

wel, Ir Tine
Shaft AB: Tu? 100 MPa = 1096104 Pa,
07 hdr Ka) = 12 mu = QO
= Flooxio 0.01 = 716 Nem
Shaft BC: ur COMP = GOO Pa
crade bo) = 20m = 0.020m
Tr Elcomo Mo, = 1754 Nem
The allorable torque is the smaller of Tan ol Tac
Te 4 Nm a

opt aa. Te Mera Copas a tir a pr Ma ay die
on iy Be yc elo pl pao eet een eh
‘carat pod Meroe et wba cone pune Asc Sead rc rin

yoww.elsolucionario.net

3.17 Tsd ha shown orme fa bas for vii the llowable hing

Probien 317, Am SS MP egin the ci! of sora cession de e
Slt ann nr whch elon shag ds e
pans

miam "Ving © 55 MPa + SS Pa

Kr « TS 5 TC = JE:
tear Tes Er PR
mm] Skt REE Tar 1200-#00 = 800 Nom

= [REA = arrondis
e a = 21.000 + 21.0 m

minimo de = 2e = 42,0 mn @

Tag? 400 Nem

= 1.0 > 16.67 me
minimor dec = Ze = 33.3mm e

UT

218 Sone Prob. 317 ering at donnee.

2.7 The ai sha een famed oa Bas fr wih slow hing
ss i 58 MPA. Nain the let of soa nca, econ De
‘Spat ame yo dc lor ich de alo sharing sh to

Problem 3.18

THON" Note Md He divedhion TE han been
‘eeversect in the Figure do the Acht,
Tat SE MPa = ESvio" Pa
te. Ag. a

T’ me Ten

Te >
Shatt AB? Tia = 12004400 = 1600 Nm
ce JEU 2 acer tom > 2646 mn

WEF minimum Aha = Ze 7 S24 mm 4

= lem = 16.7 amy
minimum due = Le = 38.8 mm 4

ern nr 200 Te Meo Cpt eA gr pe a ep
‘hae by er pay can natn hep pr be pata Re aa
RnB enna pepe Alcan De dot ohne

13:19 The atlowabe sus i 50 MPa in he hs od AB and 25 MPa in he
amina 108 BC, Kola tata true of magie J 125 N mi ped
Sc deine he require meer a) 08 A.) 108

Problem 3.19

& = Het ar
Toe TE Te Fe e.
(21250. e
Rod A8: c= AS
C= ASASHIT = 2ST mn dies Be = 50.3 mm -
0 2 0. GXns) _ oe?
Rod OC: ch = use = 3, sar xu
= BLES m = 31.68 mm de? 20 684mm =~

nade fa lina foe
1 rs 25 MPa. Rod AB e follo an ha a
‘made o Ens lr which he loo ag
(he agence ase 610148 ch
‘he air of y Te he se foreach ol. (0) the pot nie thet at e
ape 4

eli rs Br: v= Te Zee
Lans 25:10" Pa, ds 0.015 m
= E Le = Floasyassıo®) = 132.536 Nem

Moliow vod 48: Tu = 50% Pa Tay = 132.636 Nm
0.0125 m as?
ce)
+ _ (21132.53 (0.025) «
ons" - QU 2 20777255
PS men de 2c, = 1618 mm A

Ten B25 Vem a

he en Compe A oi Np aa iin pan
LA pop rm nennen

Problem 3.21 221 à tre of mad 7 = 800m
King a hate bag st SD M
A ag dae (tft AE () sat CO.

fes doin
lay, .teottoo) . ae
|

610306 in = Zo16mm

du

DE g = PAS = er0225 me 2 275 bm

Ar Ber dm

(aa Shaft ABE

2e bh2mm a

Problem 9,22 322. tre af mail T= 900 Nm ape Das sown
sowing te he Snmeter sf shit AM le re nd ott cer of
Shot CD is 45 unm mine dhe an shang se (shalt,
man co,

(i Shut AB: Cda = Somm

tee BE
0 ae
To a Lo A = Sh0S Ha La) To ES OS MÍ —
W St CD! ee hd, = 2216 mn
AT. AO Le so.
Tome Fes © bona Sor3 Mem LD Tony + 5078 Ma na

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Mecánica de materiales beer, johnston - 5ed solucionario

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