Solucionario resistencia de materiales (beer johnston, 3rd ed)

i CHAPTER |

Normal stress

PROBLEM A

= SOLUTION

u (a) rod AB
Area: As
Norma? stress?

(b) cod BC

Fore! P =
Area: A =

LL Tao só inl rod AB and BC are welded ober at and loue as
showy Row Mat,» 30 am ad.» SO me. Ed de overage aaa en
emi seo ff AB, (0) 08 8

Force! Pz 60x10" N tension

Fa

+
Sue E 2

a
1

= 706.86 110° m

ee = ameno Pa
Get 343 MPa 4
Goxo® - (2)(125 x10") - 190x107 u

Hal + Elson)! caso? m*
= 96.77 «10% Pa
Cae? - 16.8 MPa 4

PROBLEM 12

SOLUTION

| rod ag

rod 8C

Stress à

ay

Force P =
Gece -150%10* Pa
Si -

as: E?

1.2 To tesina rod AB and BC are wide toto at Bm odds
sou Rating that te average norma ess mua record 1S0 MPSA er
‘od, deen he ais! alle aso te dames d ad 4,

4

—o |
Force? P = 60x10 N Stress! Ge ISor10“ Pa |
As ta"
Re A
Fa |
de SE = ME on aust

di 4

- 90x10" N

Asa

Goxio* - (aXi2s #10) =
Areas
mono) _ x.
m iso 10°)

40.16 x10” m a

1.612810" m’

= 402mm

PROBLEM 13

1.3 Two si phil 168 A and AC re welded toga at 8a ones at
‘om Koma tac» 125 and d= 075 in ed he somal sues athe
moi fa) 10448, (8 08 BC

souunox
(a) rod AB
Pa 12+10 = 22 kim
As Bat = (25) 1.2272 int
.2 5 = csi
er À - ón > 17.93 ke -
& rod BC
P= 10 kips
A = Pal = lors) = 0.412 int
Lo A
Gre À = Re ki

LA Two oi yin rod AB and BC ar ld opt at Bad lote»

PA shown. Kaowing thatthe normal te must not exceed 25 kai either od, determine.
‘ebm oval aso be dar un 4
im souwion
aa ä rod AB:
| Pr 12+ lo = 22 Kips
F Es Cre = 25 kei Aw =a,”
+ à ou À ts
el 242 - 00D _ za
at = Bes DER = hizo in
sabe di 1.059 in =
pod BC: Pr 10 kipa Gar 25h Age = Fdo
de; = Duo . :
af = E + WG 015088 in 45 0.714 in

PROBLEM LS

1: Asin pa ened us Conte sre of bone 48 nites th
soa ca bone 330 MPa wen an sut 10120.
ona. Astamn e cou ain bone nC ro e arar one
(out dane 25 ri, dci e er ier ofthe dore cos Scion

SOLUTION

=

(4412009
(3.80 210°

= 222.9 10 mi

a? = (asno? Y

dy = 14.93 107 we 14.93 nn

mins our dir oe pacers ich ils eo D an al
dan
SOLUTION

A4 each bolt Jecation He upper plate it pulled clown by the tensile
force Pa of the both At the same time He spacer pushes

that plate upuard with a compressive Force Pe. In onder to
maintain equidihrium
R= P

For the bolt CE = ELA or PF d
5

R= FGA)

Po ap

For the spacer 63 ran *

| Erving Pa and Re
Feat = FS la - ay)
aes dis Bai + (+ Bay
de= (Ir SY = 0.160667 im”
ds = 0.408 im, -

PROBLEM 17 12 Link AD soma ingle bt 30 nu wie end 12 nick. Keone hat
ei ar cm Same een he mais ave of reg nom
se HD 10) O =. 9990

soLurion

Use bar ABE 290
as free body

D
Fao
Ar
=M,=0
N (a) @=0° (0.450 sin 80° X20x/0) ~ (9-320 cos 30) Fan = O
>» Fee > 17-32 "10° N

1] (br @ 90" [0.460 cas 30°X(20010") - (0.800 css 30*) Re ©

E

Fa = - 30 210 N

Ar (0.030- 0.010)(0.012) = 240 10m

At (0.090)(0.012)= 360 x/c“ m

ot

i „Fe 17.32%/0% „ wie e
er 6+ JR - 7aavio 72.2 MPa

w © Fe. Zeno + aso” = 83.2 MPa a

sks us 1. 36: om una rectangular ss
ofthe opine he 16 in dama, Dene te asin
tl te es) pe Bans pou

PROBLEM 18

SOLUTION

Use bar ABC aso Free body.

ao

EM, 20 (0.010) Fie - (0.025 +0.010)(20%10") = ©
Fao = 32.500 N Link BD is in tensvon
= (0.040) Fig ~ (0.025 )(20 wc) = ©
Fer
Net anea of one Sink For tension = (0,008X0.036 -0.016 )

RSS N Link CE is in compression

= 10x10" For two parole fiaks Aug: 32040 m!
Tensile stress in dink BD
5 mio? 7
@ Sp Be LHS 10156210" = 106 HPa 4

Area fon one Pink in compression = (0,008 )(0.036)
= 288x 10% mi Fon tuo para Mel Links A = SIEM mt

> fe. aso | e
Gig E = - Alto vI0* or ~2L7 MPa

19 Two borro! Sp foren ae applied 1 pin Hof he ase shown |
Romain ta eps of Om deter i ied a each commen, detente
mn vale fe serge normal erst (a tink 48, DE AC

SOLUTION

Use joint 8 a: Tree body

10 pa

LANG
Bf NE

D ke
Force triangle

Law di Sines
Fe - Ea =
mir 7 mts | Sn a
Fag + 7.3205 kips Fou = 8.9658 ke

Link AS is a tension member
Minimum section at pin And = (.8-0.80.5) = 0.5 in

(6) Stress in AG Og = a = ZEN 14.64 ksi -

Link BC is a compression member
Gross sectional area is Ar (1310-5) = 0.9 in?

(2) Stress in BC See

- | z% =
Co] sm

2250 ds, e

750 de

I minimum
section

EUG Fee) - (4001590,
© (XE Ree) = (1S) Dy

of our wooden werben ABC, DEF, BE and CE
cows ech ember as an rezan ros sect and at eh pin
basa in diameter tee the cc alo le mea normal rs a)
in sen BE. (a member CF

SOLUTION
Add support reactions te Figure

mn as shown,
Using entire Frame as Free body

zum =o 40 D, - WrsoXum) = ©
Dur 900 A

FF R=o
an, -$0,=0
Dy = #D, = 1200 db,

Fee = - 2250 Ab.
Fer 750 4

Stress in compression member BE
Area Az Zink tin = Bin
(2) Cae Fat A asp

Stress in tension member CF

Minimum section area occurs at pin,
Buen = (QUHO-O.5) = 7.0 in?

&) 107.1 pss =

|
|

LT Foes Peat ge tse and log show, deme he avenge norma!

END ess in member AF. knowirg that the cross-sectional area of tat member is $.87 in
or sorumon
| K Use entire truss as Free body
A ZM=0
ate 4 (aXe) + (181680) + (ar Iso) = 86 Ay = O
in a ee

slg whe be

Use ion of truss to the Pet Æ à section

Fe cutting. members BD, BE, and CE.

+t Zu = 0
r y Fie = SO bips
Fee
8.52 wi =
no ke 30 kips
PROBLEM LI nn ee dene owen et

Pia menda whch lil! he mos economia aná sae design Assume at
Ba co ofthe member wit be azar rre

soLunox
52 entire truss as Free body
’ DEM,» 0
T Te Mer Cia Mo) Hanke) - sch = ©
CU ES A ote .
= 120 ki
AAN ea Pee
e fo Dre partion oP truss te the Det of à section
_— cutting members BD, BE, and CE
Nm Dzm 20
€ A A
A Lo
} Fa Gun Er
Ace
Wo Fps 30 kips

a
O

PROBLEM 143 113 À col mare SOON mi ao othe naka anengte For
(ie poston town. dear) he fos Rreqaed fo hae engine em ia
‘ira. (0 ie aeg carl a e Gone 8 Be cas 3450:
souuriex
Use piston, red, and crank
together as free cosy hob
wal reaction H and bearing
reactions Ay and Ay.
OZM=-o
(0.230 m)H - 1500 Nm = ©
He S.3571x10 N

P

ley piston aline ae free 4

+ Note that rod is
© fuoforce member; hevet_ the divestion
ot Force Far is known, Draw the force
triangle and solve Re ? and Fae by
‘eel lp propecia
er L = [rots 607 = 208.81 mm

Lo we
H E]

CS
Fae BB fus acta lo w

Rod BC is & compression member. Fs area it USO ma $50 7/07 mt

= la. 218.698 x/0* __ <
rt 18-618 210". 41.4 x10" Pa

@ Se 7-44 MPa a

116 Tao dre esse usd cole een chers um ABC

owing econo osc! a And Deen Ps à 2m ace! nd

> O)
Ma Als member DO

PROBLEM 1.14

Soo y

1

E] DEMO (100) HF (0.600 Mao) =O Fes din
1 :
A Area ff rod in member AE in A= Fd*= Fonic”) = 81x10 m*
7 ly AE _ — ajos

= Stress in rod AE? Gig? BE = OS = 12.73 108 Pa

Par, @ Er = 1273 Mr €

Use con bined members ABC and BFD as free body.

a DIMp= 0 |
E (o.150XÉ Fre) — (0.200 (4 Fo) |
= (1.050 -0,550)(8%9) = 9 Foe = -ISeo N

: | Aven ved DS m AZ + Fzono)* 3140 m‘

Stress in red DEE Epa * > AS
[0] M ST MPa a

—

PROWLEM LIS IS The orden member dan Y set een non spice pits wich

ly lec on e srt one par o th deg of he jt a
Krone tt se cleanse bean the fs ol De mentir ts Dm, ire
al ane gi ne svete hang rs ele Smt exceed

ra

sowvtion

There are Four separate areas oP qhue. Each
orea must transmit half of the 248N Loud.

Theron Fat BN = ON
Shearing stress in qe ts 800 x/0% Pa

ee Rs ke Be Bes sowie ot

Let + Jongh d qhe area and ws width = 100 mn + Où m

As dws pe Ae Je ist = so
Le Qe gap = (AMIsod+ 8 = 308 mm -

1.16. Detemine te unter of res naar ol wich can be puna into
1 he pape atk, non ht he rs mr byte pur 45
{Nandita a CS MPa average monte sr i equce ao be Mae Bi

PROBLEM 116
soLuriox

Ar mat fon cydimbeicad Par cortas
CO 8 Acs

Eau AY mat - E
«io? i 16°
String on di dh mirra” 43.410 en

d= 43.4 rm =

L..

I

r

eS. ca

PROBLEM LAT 1.17 To wood pass cach ei thik and Sin wid, rey the ed
mor shown Reg tt heat will when he average he es |
e a vences 0 pe; eres be sre lobe op ls ofthe
io anhand a od of ade P= 1300

SOLUTION

pra Seven surfaces carey the teh?
haut Pr 1200 Lo.

Li on }
st [epee Ae (MG) d= Ba

Ae E Bq = ne d= 1.633 jn ~

AA Aion Pape sett gupore s honey a tu pe mo
ch un ame he ar Dm ced Koa ha he man res mst
‘ov ecos 1 kt nthe el fd and 10 fe ina pe, Stemi the
Ergo Pa ma Se appli oh ro

PROBLEM 1.18

quel

au sonriox
T ose es
| 7 For steel A,= mat = loo
| Bane be
i a É 2 Pr AT © (0750190)
= 13.57 kips
For aluminum A, 3 Wd t © mULENO2S) > 1.2566 int
ASE + Pe At = WasccKlo) = 12.57 Kips

Limiting valve ch Pis the smattervalye + P= 1257 Kips €

[=a]

a]

[=]

1.19. The aval re nthe colin spp he ier beam Son ig = 7 EN
Date the mat enable length of eb ple pc sus
‘he timbers oc 1 exceed 3.9 MPa |

Sn
s-i- 5 à
+ pec
Solving Re LE Le ESC)
128.610 m

173.5 mm -

en

Bearing plete =

Ara

120. Ana lou Pimp a ho W250» 67 column of ross seen
ca A= 3580 rn cd à url Lo orte ound by à square pia
(Down. Kong ha ie ses normal rs ithe cola mus a exceed 150
Pa ad eng ce on e corte ondaa mat o! exe 12 SMPs,
aa e side ofthe pate wich el pone he ant come and safe
on

soLunoN
Area # cofomn Ar 680mm = 2680 10° m“
Normal ‘stress in colomn? 6 = ISO x10% Pa
o= £2 Pe AG= (essoró “Uso 0“)

= 12871108 N
CE Ë and Ayz at for square plate. ¡

1.237 X 10%
1.5100

= 32110 ms. ln

a

ronca 121 121 Tyee woode park are need together by à seres of Bt fom à

laa The ance fetch bls in ad tens mtr o as water a
1. wc sight rer ar e amer oF Sle nthe parks Deen he
‘inet alone ote dee ofthe waters Erving ta ne verge coal
Stress sn he bls $ sa ut the beg sess betwen te wasters 08 he
Pas must ot exceed 1.2,

SOLUTION

Bott? Ans Pa + HU oa int

67 Loi Tensile Force in bolt P= 6, A = (SW 0.19638) = 0.9315 kip:

A
Washer: inside diameter = di = $ in, outside diameter = de

Bearing area Aus Fld b+ AP) ond A = &
Equating Has - di) = &

dir dia cg + (o1820) . 1.4323 int
de = 1.197 in -

inc 49 filth = Zin and ken = sues o sopor te end
fanaa Kaowing tl he rage norm sro ne = 208 and
Shear ara each of uo pis 12H, etre (y e

PROBLEM 122

souumox

Rod AB is in compresein.
Az bE where br Zin and bain
P=-6A =-(20X2X4)= 10 kips

and Ap=Po*

1
to

1.8 Each fe Sur vera as Hg an 8% 36 om form vectagu tos
an ado apra as em dete

128 For ry non oF Prob 13. eine (0) De
sia pa D, 8) e average Dang esa nk BD, Ce)
‘rng ses a Fa member AB Roig that hs mee a
om rca eos eon.

PROBLEM (23

sorerion

Use bar ABC as a free body

EM. =O (o.molf, =Lo.0r5+ 0.0020 108) = O

(ar

(0)

Kc)

Fes” 32.5% 107 N

Fae fon .
Fe For double chen

where Az Late Eloy = 261.06 «10 mi

Shear pin at B ve

Len = 80.8 wo‘ 20.3MPa =
Bearing Pink BD A= dt = (0.018)(0.008)= 128 x10" m*
See. Laie). i6.9sriot 17.0 Ma

128 x 10
Bearing in ABC ot B
A= dt = (0.o16Mo.o10)= 160x107

Ge o Fe > sit 203 MPa a

18 Eh of te ur ver ats aa 8 36. m un ecg cost ]
RE AA
„ Fee

Use bar ABC as a Free body

ams 00 —

IMgs0 -(0040)F, - (0.02520 xo") = © Res = 129408
(a) Shear in pin at © A=Ed*= Flo.o6Y= 201.060 m
Double shean 2 FE = RS «0? = ateo

ZA * (Sono)
311 MPa -

(D) Bearing in dink CE at C

A= dt = (0,016 )lo.00%) = 123 #4

+ dE . tasado - una 488 MPa a

©) Bearing in ABC at C
A= dt = (0.016Xo-010) = I6omo* mi

<2 Fe MEM og) spot MP od
az. RS 784 io 78.1 MPa,

PROBLEM 128 19 Toter Ship res apldto pn Botte sien sou. Kong
ones pen ian dancer ned tach comen dete he more ve
‘ine menge nenne) ni a) met"
AS Torte sven ae arg oP 1% omiso) he enge sein
asin, O) he ave bang ren A ment AB
sorcriox

Use joint Bas Free bad.

Pur
ae
KG

10 Kips
Force triangle

Law of Sines

Fe, Fe Ie eis
Sin dg nr me Fig = 7.3208 kips
(a) Shearing stress in pin at A re ze
pres Aye Edt F(o8) = 0.5026 int
ay ‘si a
Tr aa 78 1.28 k

(6) Bearing stress at A in member AB
A: tds (.5¥0.8) = 0.4 m‘

Fis _ 13206 sj a
= 18. 18.30 ksi
oO Be: E 18.30 18.30

] PROBLEM 126 13 Two boron Si rs spp pia Bo ese own Kaowng
stata pnt fon Care a escocia, deter e mac se

ofthe average nel ses (o) LA cy nk I

126 Forte end a ocn of Prob, dete) he average teca

rein the pia € (8 the aveo ding stes Cin member (teaver

asia sc a Bin member BC

sono
Use joint B es free body
10 Kips
a sun
haw of Sines a
= ae Fac = 8.9658 Kips
(a) Shearing stress in pin at © tE
Ap = Hd? > F(o.8)* = 0.5026 in*
+ 3.9658 PA
. Ct as) 7 2 rw |
= (0) Bearing stress at C in member BC 6 E
0 Az td = (050817 04 int
gl 5 E -
E ©) Bearing stress at B in member BC
ü! A=2td = 20.5 0.8) = 0.8 it
Tee 6: 22% - 1.2 MA bo

127 Koi hat = 40° acd P= 9 RN, determine () he sie
ameter of he pn at ite wege seg cs Pr ut io
MP (the covrexpodig erat Dun ares in aber AB a: 8.) fhe
B (tmepondegwrrge brag rein uch fhe supp back a

PROBLEM 127

sorumon
Geometry: Triangle ABC is

Bs un isosetes triangle with
1 angles shown here.
e

Law of Sines applied
+ Force triangte

Es ¡Ps En
sin20" sin |10°

Fro Él > A
once > E site
triangle Fer ao"

Sin 20°

(2) Aifowable pin diamehn

ve Fao, E hee Figs 24.75 1108 #

TS

ate fee MT ao wt

A= 11.95 »10'm 1.95 mn a

Kb) Bearing stress in AB of À.

Az tals (0.016 Yi4s mot) = 183.26 wo” m*

y Fe, RIGO _ ape
Sit FE rot = 134-90 134.9 MPa, a

(©) Beariny stress in support brackets at B
A= td = (0.012 nas wot) = 187.410 mt

= Glsin lo’. 24.730

A (

PROMLEM 128 1124 Oeernseteergestlon ich Depp Ae 9 6D Anos

dat Me tags dar esa 1 Gare pina Bm o enc 120
7 San rag sng ac omer 48 adi cies
ied 90 NP

SOLUTION

Geometry? Team de ABC is
an isesetes triangle with
angles shown here

vu: joint À as Free be
as A haw of sines applied to

Forze twangle
120°
Non he da ar
Es E : arg 0.57785 Fi
he ETS
Pe

IF shearing stress in pin at B is cribleed
Ae = Fate Hood! = 78.590 mt

Fra RAT = (ares Roma) = 18.850710 N

IF bearing stress in member AG at bracket at A is erihiend
Ag tel + (0,016 )(0,010) + 16010" m*
Fie = AS = Como" X40 x10°) = 14.40% N

IF beasing stress in the bracket at B is critical
A: 244 = @)o.02\%0.010)= auorıo“ m*

Fre = ALS, + (240 x10) 90410") = 21.6x10? N
Abou ble Fra is the smallest, ne 1.402108 N
Then, Bron Staties Patas ? (0.57785 109.40 107)
= 8.31 w10* N 33)4N <a

LU as a ás

rose 129 The 6 XN load Pia apport by two wooden member 0075 = 125 ms
lor vecanglr aoe sehr whch jie the ple gue Sat sce
“Shown” Decree coal ans Deng sess n he Bee spe

q sounox
= Pe 6x0 N 6 = %0°- 10° = 20°
Ay = (0.075 (0,128) = 4,375.10" mt
to LO ot
soso es
E => SES kPa
n e. 2 e l6xioN sin dot _ wot
Ce nao + 2060
7 t= 206 kPa 4
~ PROBLEM 1.30 1.30 Two wooden members of 75 * 125- men uniform rectangular cross seston are

À de
1 eo hela eo serene
ee en
Lise soumox
wef \ L
ROA A = uns ours = as m6 wt
8 = 90°-70° = 20° E = Soo% 10 Pa
u o-oo
ee BS, = (eue NSO). agro
3% z

CES
Ps 5.31 kN ~

- (5:208S<10) sin 407. wo!
Aaa EAA HIO

C= 122.0 kPa —

PROBLEM LS

121 Two wooden member 3 in rif eng cos sone joined
Wiesn pel at spice sown Keowee acme swe heey
ses nthe i ice 9 ps, Senin) eager ead Phan he le,
pie. (besorgen reine spice

soveTiON
D = Io" - so"
À = GE) = 18 ist

Th sine
re BEG = EXD sere

P= 3290 th —
Sige SOS 155 CFS pi a

PROBLEM 132

1.38 Two wooden members o 6. im nor rc ers section eine
‘byte spe gue serpin shawn. Kong at 240 temic ora
A enn Sess: nthe ued pce
SoLtTiON
@ = mor Ho"
Ao = (GES 18

= Beste (2100) sha?
o- foro 2% ss.

AS

-2 = (Bie )si0 100° à
Ee gg 0920 = je Er

ES

Ü

=

tE

PROBLEM 133

1.33 A camion Papi anios own Koi hat he rer
ram abe fhe a erro te RC kn deere (ye nape
SEP. Ch tc seta of anon which be eve tin ren cto,

ips ss ed arc, (eu ae ofthe soma cs
beac

sournon

ME) = 36 int Lans 25 kei
45° Fon plane O

Dons à lets aA Zane = AMBAS
"2 180 Kips -
©) sine] 2990" @ 245° -
A ee ae ia
ee bts sks
D Gus f= HE: Shi -

PROBLEM

134 Naples ick own, Dm eng
stn de ra o oe eng os. Se de cta
panos ach ech fae ms es oes
souerion
Ao > G6) = 36 int
6 = Beate = ave - rn

(o) mon tensife stress + O ot 9 90

mobs compressive stress = 6.67 hor a
at O20
he LE y og
wr da a neo
4 0= 45"

1.38 te pipe of 300 cuter diet bated lm 6 nn ck pla by
vaig longa ric wn fon an angst? witha line perpen he aus
Ute pe. Kong ha 50-4 sl ce Pia oe pe determine te
rama ang ro ciao peces managen wei

PROBLEM 135

souurion
de = 0.300m fe hd * DISD m
Tes Yo-t = 0.150-0.006 = 0.144 m
An Ten Y) = 1 (0.150% - 0.144?)
= 554 xjo$ mt
8 = 25°
S> eso

37.1 010% - 37.1 MPa, e
= = =250x/0* sin So”
T= 7h sin 20 = SO Ne sin SO

(NEGO
7.28 10° C= 17.28 MPa ®

1.36 A rl pipe 00. me ove mer Die om 6 mm ic lt.
eig long ae nich oma ange 29" va lr papa 2
Se po. Krona ta the mama nase som nd eg ses a
(rec respective nora andere the sd me 9 = 30 MPa and + = 10
MPa, duran th rade Pf ergs asa te that an be apd tie
pre

PROBLEM 126

sorwrion
dy = 0.30 m =kde> aisom
r= fo=É = 0.150 ~ 0,006 = 0,144 m
A. = TAO) = (0.150 0.1444)
= 5590 mt

0 = 25°
Based en 151= som: = À costo

ps AS 2 sq en) = ga7xio*

cea 76
Based on IZ1= Zo MPa t= Fs 20
p: 2 = @Yss4x10%)20x109 434 x0"
28 Gin SO

Smaller valve is the cMouable value PP & P=337kN =

PROBLEM 137 1.99 Lisk BC 86 mm ick as awn = 2 um and made oa sel
A: MPa rate sei son, Wal was sy etn eed ithe cre
‘tos as cres 1 apor à EN Ton P?

em sonumon

Use bar AC D as a free ad
mechs Bo ae A Fe
force member > 5
IM 0
(480) Fae - (600)P = © ft
Fe = LP „Leoexiewiot = 20%108
UHimdte load for member BC Ey = GA
Fo = (30x10 )(0,006X0.025) = 72.10% N
Factor lade FE = eco -
paca adit 138 Lk AC en ik ande ts wih 50 Matin seran

in tenon. Wat soul be it ith wi te Sutton Di do 10
por 4 204N ond Ph fo ley D

sen
sorumon
5
Use bar ACD as a Free body ri
and note that member BL ic =
© two-Force members
e
ZM,=0 .
3930 Fac + 600 P = © Ls
sep (20410%) . 25m
Fac” = Lege: 250 N

For à factor of salety es = 3, the ultimate fond of member BC
Fo = (FS) (Fa) = (8125x10%) = 75% 107 N

5x0" _

Bt = GA: Az E = 166.674 15 m°
For a rectangular sadion none ar who gro

w=22.8x0 mor 218m9 e

hoe 1.39 Menter8C, this supped bye pin an bach
saved o suport the Pp ad Paz mau. Ras
“ible HD 35 ipa dern th trey wh espe eae ar

SOLUTION

Use member ABC aa ln

Free body and note Fa
i thet mentes BD is a
Ñ Ans force member.
1 a

ufr À EM, =O
(P cos 49") (ZOin) + (P sinter JS in) = [Fan cos WS)
Foo sin 32 im) = ©

en, Lo .
Foe BE LE. ars Kips

7 Rector of safely for calle 80 RS = Fr

1.40 Koni: lio od cabe BD 325 hips an ta cto of sey
32 with pes to cable flere segue, mire the magne ofthe Lg
face which can te sale apple a shown o menda ABC.

3

q soumon .
mm pee member ABC asa
fi Freu bedy and nee Fo
Hat mebber 2D à a 3
r | quo Force mem ben. A
a ZM = 0
p! (Pens 10130) + (P sin do MIS im) (Fan cos 86 NTI
01 = Fag sin Bora in) = 0
Pleno po
Ë P= ETS Fey + 0.58216 Fe
E Alfowable Pad For member BD is Fay = on FE rans ki
ni Allowable dead P =(0.582I0M7.B125) = SS kips =<

fi pre

| 1 | PROBLEM Lat 1.43 Mertbers AB and AC of the tus shown consist of bars of square cross section

LE ee oft same al. Ru known ai a 2 m gat br ef De sane ay was

| tested fre ad at an ke uf 120 N vas recorded tar AD as 5

tune sue clots scion demie (2) te cor af ay for ta AB, (8) the

Simone ehe roar chem oar AC ove he sane act ol ay as ar
#

Length oF member AB
Ana = Jo15t+ 64° = 0.85 m
Use entire test as a Free body As f Ay

. DEM¿=0 14 Au -Gas2)= 0 Aer 15 kN
i ZA =0 A-2220 PRET ia
7 Use joint A as Free body Er
u ES en 97
> a th Zo OB Re - Ay =o ce
i Fig = ACI
OS
Fre HZF,=0 | Ay~ Fe Ela = 0

= Fes 2 LA = un
For the test bar A= (0.020) = 000% mt P,= 12010" N

0 A A
r A en s- fe > SA, (BoonotYloaıs)!
1 ©, Brei ES Fe Fa 17 10°

= 3.47 -

‘or bar B&B. GA Ga
(or For bar AC RS = gt an > Ge

Ñ at (ESE > (Mom) „264.7 810°“ mt
: © 300 #10€

Q= 16.2910” m 16.22 mm oe

1.42 Members A nd Cf tus sow consi ars o square cos secion
nad ofthe sum oy, is known that 0-3 cut Dr ofthe sae ay wat
testeo fle and hat an ta nad 120 as records. We tr ley
9132415 be ated bots, dre aque dena he st
Sesion oo) ar AB 0) ar AC

PROBLEM 142

souumion
Length «P menber AB

Ina = LOITTORE = om

Use entire truss as a Free body Ar
DIM =O NUN 20 Aya SKN
+EF,= 0 Ay- eo Ayez kn e
Use joint A as Free body kw
Lo 2 .
SER r0 BH Fla Au so on
EASY) -
Fig O = 17 kv
HER =0 A-Fe- Paso
Fre = 28 - 10402). - 20 kN
For the test bar À = (0,020)*= Yooxso® m“ = 20x10" N
Fer the material Ge Pes HOME = aeoxiot Pa
(a) For member AB R Fe *
2. ESF , (82117100
= u 00 x 10 +
@ = 18.47 10m 13,47 mon 20

(6) For member AC ES. 2 BE SA. St

Epa al) o

be cio en

|

13 The ru wooden members sum, which pec à 204 and, ied dy

PROBLEM L43 pod apes ll ued on be sarees cert The ite hear ares à
{he gue? MPa ed he leven tee te mess 3 mn Dee be
gas nm {Seto fy, nomina ah eg of nh spices 200.
NS

SOLUTION
There ave 4 separate arcas of give. Each
give area must transmit 10 WH oF shear

al Toad.

Pe ox m
Length f splice Ls 28ec where Le fengthéahe and c=
cheérance. $(L-c) = 4(0.200-0.003) = 0.016 m.

Area of ge A= Bw = (0.096Mo.120) = 11-52 #16 mt
Ultimate bad Po= WA'= (2.3x10* (11.52 0107 )= 32.256 v10* N

Factor Bande Ss Le = ER = 3.28 «

1.43 Theo wooden members town, wich pps 2040 had, a ind y
od pes Uy pueden be sc cost. The ie seo a
lle le 2 Ma mé ue crece Smee o menden Rn.
E de Fa ns og oP 1, eo ma
4 sie ifs sor fsa of 8 810 be achieves
“SN,

PROBLEM LA

> sum
There ore # separate areas uf glue. Each

aloe area must transmit 10 kN oP shear
an dead.

P = 1onlo! N
Required ultimate foad Pot (ESP) = (a.,5\loriot) = 35x10" M
Required dength À oP each’ ghve area

Pe paseo _. u
Pr GAS kw 2s Be ts

Length of splice L= 20s = (2104.17 x107) + 0.008
= 216.310 m 216 mm =

HL.

In
u
=
(El

a

PROBLEM LAS 1.4 Tre fin its tro at ee down
toa woocen set Keon a pte ml apor 25 oda ae

that sr o baca dean ete sal da
ei

soumon
For each bolt AsEdt. HS) = 0.4418 int
Po = Ay = (0.4918)(52)= 22.97 lps
Per bolt P= ds 8 Kips

he
rss Be 497 5 287 -
a 4 ge md cl ntc inde

Xabi tl hepa por plo the sme hang pr
Be nenn Fi md ht hoe sty 90997 e se, ern fe
‘moved er oft bl

soumox

For each bolt p= St = 8 ki
Required Py =(ES.1P = (3.278) 26.96 kips

0.5139 im

= 0.8125 in. -

a

I

1

cS EN ds

1.7 Aout Pia spponed as Dong ate pin which asbeeninsetdina sor
coder enter hargng fom be cling The tte seg old wad ed
E Main enson 2875 MPa car, be te urate suengih afte sel 150
Nevin shew Rowing tl e deer o de pn ed = 16 mm ata be
grue ofihelndisP = 201, determi () facto ey o pin (8)
required aus of and ifthe factor o ly o wood meter bee
sia una ar a re pa

PROBLEM La?

souunox
Ps 2ouN= 200 N | a

@ Pin: A= Fat = F (0.016) = 201.06 x10 m
Double shear Ti Ye 2

Py = 2AT = (GMrontexo® )(150 x10%)
= 60.319 *10" N
Po, cos
ES.= = Gans 7 302 -
(6) Tension in wood = Py = GO.819 0 N fon same FS.
she Y a -
= Tu where we dO mm = 0.040m
= Po. A so
be d+-—- 0.016 + see easy = Bodom
pam
Shear in wood P, = 60.819 «10° N for same ES,
Double shears each area is A= we
u: =
SE ZA awe
= Be - 60.3910? 100.5 210°
Tu 7 BKowvoyn.swior) de
C= 100.5 mm -

ct Cr Eee

| fl

147 Amar
PROBLEM LAS wenden ete
Stuur

A8 ae agpon of Pb 17 wig tt 400.0 $5 mm d= 12
som deren Nawal Pits vel co E UP

apport sou Wy ee pin which bs ben retina short
om decano, The wheat seo i od weds
ise Whedbee sen th a8 1

sorunos
Based on double shear in pin
Po = 2A%,= 2Fdte,
= FGYo.or2}(sou0*) = 33,43 10% N
Based on tension in wood
PB: AS = wlb-d)6,
+ (0.040)(0.040- 0.012)(60x10*)
> 67.2 x10*N

Based on double shear in the wood
Pur 24% = 2wcT, = (2X0.040)(0.085 (7.5108)

= 83.0 10° N
Use smallest Pi = 33.010" N
Alomahde Pr Gb DO . 10.2100 u

10-3) kN -

ei
u

=

PROBLEM 149 149 nthe suture sn, an & mn: etes pin und at A ad 12 mme

meer pir ae wees a and D Kaowing hat (ee sheng us 100
are otha th Utena ress 250 MPa mech be o
Us jong and, decido slow ad Pian oneal ais a sly of

row Dowden
iam soon
pee SS Statics 3 Use ABC as free body.
2 ©
I Feo
ua E Peer

ZM=0 0.20 Fyp-0.88P = à

Em

Based on double shear in pin A
A=4d? = Flo.008)* = 50.266 «10S m‘

F, = LA, Glisomo Unes") 39912108 N

P= BR = am N

Based on double shear in pins al Band D

A= Hd? = Kom © 118.10 wo“ mt

= AGA _ @dllcox/0°)(113.10x10"¢)
PRESS 3.0

15410’ N

P= BF + 2.97 108 N

Bosed on compression in Sinks BD
For one Pink Az (0.020 X0,008) = 160 #0 m*

= REA _ (aV(aso wong‘) - wos
Fao = 25:4 . as = 26.7 “10° W

P= Pep = 1.04 x10 N
Movable value of Pis amaflect Pz 3.72 ro N
3.72 KN -

PROBLEM 1.30 at sucios shown, an 8 dame in is wad a, ad 12 me
Samet: pis sewed Band Keown ht de limale hear rs 100
Malone an he inte nar sss 250 Pain ah ofthe
ine jimng and D, determino one aad Pian gveral eto of ey oF
Eee
a mare desig ode area Prob 149, pin 0-maimter
seta, Assumag bla be species ae urcanged, etre
sala lo Pian nel ctor oy oF 0 desd

Torten

porn nn

5 um sorunon
A — Stahes 3 Use ABC as free body

8

» TL zmso oxF-omP=0
A un Prey

Based on double shear in pin A
A= Hd? = F(o.010)' = 18.5% PR

Fl =2%A, (L100 xjo°)( 7854 io“) „ 5.236610: N
= E 3.0

P= Pre = $.82 x10" N
Based on double shear in pins af Band D
B= Bat = Klon = 118.10 wo“ m“

AT SD

Fs.
P= rp = 3.97 x09 N

Based om compression in Sinks BD
For one Pink A = (0.020%0.008) = 160110 °m*

ESO AE
Poo = 1.04 Jo N
NDowable value of Pis smallest 4 Pr 3.1770 N
3.97 KN =

1.51 Each oft note A a CDi comes
by à natos sess inginsigle sea Ki

sis ish oh sel usen is nd ht he tne roma ess 60
foc De tel edi tents, demic he loud oud Pif ar ove factor a
‘say 1328 des (Note th he ak re ot ied acu ep en)

soerion

Bn i | Use BCE as Free body à
en ZMs=o

Br.- PO
Pe: She

ZMe tO 8F -1P=0 P= dhe
Both Pnks have He same area and same pin diameter ; hence,
being A the same material, they will have the same ultimate toad,
Based on pin in single shear

As Hat = EG) à 0.19635 int

Fu = TA = (24M0.19685) = 4.7124 Kips
Based on tension in Link

Aslb-d)t = (= NE) = 0.125 int

For SA = (60)(0.125) = 7.50 Kips
Ultimate load fon Sink is smadfest: Fy = 7124 hips.
Ablowabile Load Fr Dink Pres Aue = 1.4726 kips

Allowehle ad For shuchre P= BF = 0.589 Kips

Fo

F = seq de ~

Use menber BCE as Fret body

The ShytoPo Präm
Zero 3Fa-12P=0 Pe$Fe
Batud on pin A in single shen
Ar Ede BENT asi
Foz WA = (avile.ness)= 4.7124 Ko
Based on tension in Pink AB
Wet AYR) = 0.125 in
For GA = GoXous) = 250kips
Vlfinde Road Jor Sink, AB iis Smallest ie ig M1124 hips

Corresponding uPtinate load for shrucvre: Re
4 Cond D in double shear
FAY = 0.19685 mt

GXAN(o. mess) = 9.4243 Kips

Based on tension in Finke BC

As (b-d)E = UENO) + 0.0625 i Cone Zink)

Fo= 2654 = (aXeo)(0.0625) = 7.50 Kips Chotal, both finks )
UMinate out for Hinks EC is smallest, de Fy 7.50 hips
Corresponding vWhimate Loud Br «hrwtore Pas R= 3.00 hips
Actual ulimate Load is smatiest, ie P, = 3,00 kips
Alfowable load for strocture Pe Bt = B22 = 0.108 kip

Pe En. +

BR = 80416 np

PROBLEM 133 1,59 Ech ofthe wo veni Es con he wo horton meer AD
snd RO as 10 0. mm num esangilr os econ nd iad el vi

ns ae srnih eno 0 MPa, wach ehe pies ur ad Fasa.
om ‘tm deer ad ad of eel wii ite strength ashe of 159 MP
oN eine be rl cr nyo is Fe ps ei Demo

eh menes

soLunoN
Use member EFG as tee body.
Re

I are feas

DEM =0
0.90 Fr (0.65 aa wo!) = ©
WS Ry 84x10 N
Based on tension in Pinks CF
A= (b-d)t = (0.040 - 0,02)(0.010) = 200 110 m (one Pink)
Fy = 26, A+ (2{uoox108)(200 x10") = 160.0 «10° N

UN

Based on dov de shear in pins
A= Fat = Floor = 814.16 x10 m°
Fo = 2UA = QYiS0x10* (314.16 "10 )= 74.298 x10° N
Actos) Es is smaller vole, Le Fy = 9929810" N

= fe, o?
Factor oP solely Ss pe = SRE « zu ~

oS cr

un „19 Etat vaine conecte e o ra mente 40

Kw Ban use in cad pos og
mn put cee

Se ano hp Fe ered pit
Sips

souunox
Use member EFG as free body.
Far Fee

loto — 025

DEM, = 0 Pr
OHO Fer - (0.65 K24 xis) = ©
Fes = 3910" N

Based on tension in dinks CF
Az b-dlt = (0.000- 0.080)o.00)= 100 em" Cone Auk)
Fo: 28, = (2)(Woowlo®)(100 x10") = 80.0%10° N
Based on dovble shear in pins
As Fat = F(o.080)* = 706.86 «70° m
Fy + 2GA = @Xısorıo* K 706.86 410") = 212.06 « 10" N
Actual Fy is smaller value ic. F, = 90.0¥10°N

Factor of safety FS.+ E = 4980 > 2.05 -

=> ©

N
J

Se ©

D 0

PROBLEM 15 155 A me pie in thick à embeds hora conce sh adit und
tanec ahighsteng veia cil atom, Te amer fe oleate pe
2 lsd in, eat rer fhe sel wed is 3 ks, ar e ult ondes

{ab [Negima setae betwee te sont ed e or mdf hate)

souurios
2.5 Kips Based on tension in plate
A s(a-d)t
SA
Le - Sa-Nt
ah Br b
as IP. 3, Nas)
cas * Gee)
a= 1.550 in -
Basedon shear between plate and concrete sfah
Az perimeter depth = 2(a+2)b % = 0,300 ksi
R= TA we Fs. =P
olving for = {SP [ERSTER RB]
Selving for b 2lartv% ” (AN.sso- 80.300)
br 8.05 in -

oo Ly
O

a

PROBLEM 1.56

188 Ases lt a cis embeds na orina ori i ana is wed
rocaille nn. Th th hs
D ein gh te a ir o
rom betwee tan cont i 300

156 Bene re oc
og that a= 2 ande

ob 155 when Pa

souvnos
3 bips Based on tension in plate
A=lo-d)t
2- EGE) = 0.3906 in!
P= GA
(86 (0.3906) = 14.06 Lips

RSs Be. 12% = 4.09

Based on shear between plate and concrete slab
A= perimeter » depth = aCavt)b = 2(a+£)(75)

Ar 34.69 in“

Xu = 0.300 ksi

Ps TA = (0.%00)(8467) = 10.41 kips
ps. = Eo. Mil > 3.47
Actual Factor of safety is He smaller value ES.= 347 =

PROBLEM LA 157 À 0. plaide end a SO wooe beam AB, hich
signed oy dy apna A aby ender ro BC ih LAN mie
lod! o) Using he Lad ar ste Pactos Design meld wäh risa ato

+ ke > 030 and led facts yo 125d y, 1, demi te gs oa which cn
T Ke ay place ne platform (2) Wat ise cospondn comento ter of
satay rod BC?
SOLUTION

| h >
¿3 j = Da
For dead Soading W 3 WURST 392.4 N
= (0X81) = 490.5N
= (FX 392.4) (E4908) + 1.0623 x lof N
For Live Joading Wem W= 0
P= mg o from which m=$
Design ceterion

RP + MPA OR
R= P= TR - (0.90)(iaio") - (.25 1. 0628x10"*)
3 3

u

“ip

= $.920%10* N
oe
Adlowalde toad me PENAL + se ty -
Conventionad Factor sabety
Pr Re P= 10628 wo + 5.920%107 = 698310" N
2 BR, wo
RS tS + CeasnoF ie -

DEM. = 0 (AMB zum -L2W, = Po Swi EV,

PROBLEM SE

1.58 The Lan nt Rice Fete Dig ef isto be se tose he uo.
cater which lee and lower fom sopping we window washers. The
hen wig 160 and ach the iow washers asueto wih 195
Bt ciment Sos thee varie are foso move ont pio 75% ofthez
ai and ofthe wight De een wb ose asthe dsg velado
‘Senco a) Asset rem air 0m 04S ard lo torse 1 208
PAS demie the ried mostra al of oe cable (8) Wha the
seven tr of ly fre ste able?

SOLUTION
Ta Po + YA = PR

p= ARANA

©
A tó NO

0.35
621 fb. -

Conventional factor of safety

Pe RoR

Le
P

= dos orsxaxias = 3725 dh
= Be = veu -

I.

By isim iar Fring
Fo. 180,
ds? 3

Joint D
zs

Feo

Fr

he a 08 odo

me ri us ed ong own, d de weg o ul nn

th coset ee ofthat en 08 a
souuniox

Using method af jointe to Find member forces

Joint 8 : AB and BD ave zero force members.

Joint At A E

180

N 130 kn
le LA EN Re
Force
Figs 225 de Tang te

(compression )

Es By similar triangbes
Fg - 25
> dl pe zas * 378

hr = 135 AN (comp)

135x109 y
Area? Ay = 2500 mm = 2500 10 m“
aso = e -
Zoo cit 7 “Sto Pa = -S4.0 MPa

Shres Se = -

1.60 Link AC has uniform care xo son a, ik an st
soma Determine termal res in he oral porton of ha ak

SOLUTION

e,
Use the plate together Be
est bre polleras a, 34
Free badır Mole thes re
the cable tension ite

casses af 1200 fhein
hacker couple toad |
Fe

on the body.
DEM = 0

=(12 +4 Fog cos 36°) + (101 Fc sin 30°) - 1200 = O

Fe = mer 7135.50 Me.
Aen fi Bink ACH Ar tiny Sine OURS int
Stress in Bink AC: Get = Esa = 1084 qui © LOS ksi =e

PROBLEM 1 at We ihe fe Pech Nh wonde peimen shows filed in shear
“ nthe nos nested y cabo ie Denn de average eg ee
ight auch ine ae.

Area being sheared
Re GO wid» 16mm = 1350 mnt = 1850910 m*

Force Pr 8x10" N

. 2,80
Shearing stress ve E

5.98 x(0* Pa =6.98 MPa at

1.62 Tuvo wooden nec 12s hk an 28 me vie are one he dy

PROBLEM LE suc hows Kr lar nod cet arf ag tg ne he
ewe being uc ares SMF, Seeman he mde? of al oad

SOLUTION SESSA So o
Six areas, each Komme (2m are En

4 sheared. =
Area: A (SUECIA) = M2 mm

= 152 "10° m*

«2

zer

Pe LAs (Bis) AH N = 7.22 KW -

PROBLEM 1.09 1.63 The au ee CF, wich pany coto the postion 06104 DE, es

Beeniocedin he posan shown, Manor AD sin tick tnd iscomecte o
mn ‘eric rodby x 2-in- meer bo. Determis (he average sheng ses in
cag Ub bol (bt beings u member BD.

SOLUTION

Use member BCD as a free body, and
om mole that AB is a two force member.

ESTE

= 8.2 in

Wo cos 75°
YOO sin 79%

DEM =o (des NE Fig) - (Hein 20° EE Fag)
= (7 00320" (400 aia 75°) = (7 sin20" (400 es 75°) = 0,

3.80678 Fig - 2787.35 - © = Fes 828.47 2,
“Zr so Ehh 4 Cy 4 900076 = ©

as Lasa = 400 cas 75° > 78.84 fh,
LIZA < 0 ce + - 400 sin 5° =D

Cy = MI à oo sin 28° = 110% 05 Mh

CHG Gy = 197.2 te

Shearing stress in the bolt: P= 1197.2 4h

As Far = E(B) = onoss int

Te Re RR 2 oo + |

A? oies
N Beaning stress at C in member BCD : Pe 17.20
Aye dt = (BME) = 0.234825 in
«Li e ce oa 1
GER ame” Url SM -

=]

Pat

“ 1.66 A tin dumeter eel rod AB do à round ole ren Cote
a ‘wooden meer CD For the ada sows, deerme (a) the maman average
‘ora ores the weed (1) the ice DD wich he oca earn sr 3
Ken no keit te bed en, (0) etre eng où

SOLUTION

(0) Maximum normal stress in the wood

Ant = 3(3-4) = 1.875 int
a 100:
Gebr 1892 = saa pes -
2 leo. a -
NS à
1200. ,
Estey © 2667 mi -

1.66 Tuo ple, ach Jam dic, ae usd to ce a lie sp as shown,
Hong teat shearing cs fe onda Devin oe uae 0
Pa Grange tro stay wth rept ot wa DISCO

PROBLEM LES

sounox
Bond area: (See figure)
Az à (Go)@o)1s}lee

1500 ma = 1500 10m
Py = 2AT, = (Alison Macoxto*) = 2700 y

Fs. B+ 2e. 1.800 =

1:66 Two wooden menbs of 3.5 + 55a wife rectangle con sion are
Soins y e Sop ue sat pce shows Kran a mima alo
‘Sewing ares ao pued pices 7s pl deere res a wh can
‘esa ppd

PROBLEM 1.66

sownon
As: (3.55.5)
@+ 1-20 = 70°
pe Pe sneme = Ron 2e

25 int

Zax, (mana - s
pe ZA. BUN > yuan j= dut kn -

167 A tellop ABCD ofl 1 2 nd 1m eter laced ns town
around» DE maria shear of 40. ales BE end DF ach of m

PROBLEM 167

soLurioN

Using joint B asafree bed; a
and considering Symmetry £

2a -Q=o LA Fis
Qs $fn
met Ll
2: 3Fu- Fe 0 Fin
PEOQ-Fu-o 2 Qs dR,
Based on strength à cable BE
Qu GA= Gaz foret) F (0.012) = 59.200 N
Based on strength of steel Loop
CLR CIE E
= ÉCasomot)F (0010) = usager n
Based on strength of tod AC
CESSE LUE CE CE
= Eaowot)F (0.024)* = 98.22 x10° N
Actor? ultimate Load Oy is the smadfest = Qy= 4S.24uio" N

Movable Sod = B - iin, is 0840 à
= 15.03 kN =

norm engl cos soon and is
ra res. Tis sensed up 14
otr pi, wie meme BCD corne 1
rapto ba amet pi Af beis cen Single send are made
fe mel wit 28 tte rare ses Koni a en real tr of
T3 280 dee, rm gc fad Pw canelo ap
[Reet AC a at enc rod pi

1. ink AC his a unto dx
ide of el with a 6

PROBLEM 1.68

sournos
DEM¿ +0 GKER)-I0 Pro
Fm = 2.0833 P P= 0.480 Fu
E izo Bráfi=o
n° B = SF, -(Va0mP)= 1.25 P
ES er WEA +0 8,+35.-P=0
8, N B+ P- ¿osasp) = - 0.66667 P
BAJE TB: LH&7P, PromsseB
Bosed on strength of Sink AC: Gas 60 ler

Ant = (EH) = 0.08/25 in) Fuge = GAut >(60X0.08195 )= 1,225 kp,
Po = (0.480 )(1-875) = 0.900 ki.

Based on strength of pin at ©:
E Fig, = LA

Po = (0.480 (2.761) >
Based on strength pin at B
By = To Apu = (25X0.07670
R
Actucd Po is the smaddest >
Afowable vole For P: ®

Anz Fate FB) = 2 MoN int
= (25\0.11045) > 2.761 kip
1.825 ke

Im Fal? FGF) =0.076%0

1.95 lip.

de

(o:rosas\ii.ar7s) = 1.3535 kip

Pos 0.900 kip

0.00 | :
GAP = 0.277 kip > 277 4h «a

e

FS

8 169, The oo prion of member AB ar hund together longa plane fring a
PROBLEM LO ne Oi Monroe Keown u out ee or the cd oot 17
[iP in eon apd in ea, dera te range o aer of Of wich
Ber ay ol menber lent 3

souuriox

(0.080) 0.050) = 1.50x10* mt

P = lox N PR =(RS)P = Sonic N
Based on tesife stress
Sr Beate
cate = SA „ (I7H10%)(iSonG) |
“ Pr Bios =
cos D = 0.92195 @ = 2.2 93 22.79"
Based on shearing stress U = 2 sinB e © = A sin 26
simzo= Zt 2 GXiso mo KIKI) o,
Pe TT Gute
29 * 64.16" 8 = 32.03” 9 < 32,03"
Henca 22.797 © € 8208" =
wont ue 2.70, Te wo orion lento Al e pd pear dang se Gs |
Le pes en ag tonic sete cn
mv SP na nd ar dano (bev lo nh e or
1 ol safe al de member in mim, () e crrespoing valve ofthe cto of
Sa Uline Eo decesos seed era ml a
Som sree ee)
SoLUTION

A, = (0.030)0.050) > .Soxto” m*
At the ophimum angle (FS)r *(FS)e

SAL Vel chee 9 Late Rye BAS

B
SS = Salts
Shearing des! C= Psindase : Rye - des
Be. LA
N Stee
e EA A

Peas Pbro?
hing EI > tne E ea, Are me
The, Pr Sie à Mira \sone*) , 2.05.10

Es > Po O x

i=)

me EX |

e)

ac

PROBLEM LCI

tt

AVERAGE STRESS IN ELEMENT L:

C1 A solid te ro coming of elek elements welded 10.

KERN
an nee dci dorado te snd gue
steno) ae oan eh et
HA CE man ane ico del

O tees mpm Pe ma

souvmox
FORCE IN ELEMENT L
IE 15 the sum of the forces applied
fo that element and all fower ones:)

ein

Area = A;
PROGRAM OUTPUTS

Problem 1.1
Element Stress (MPa)

1
2

ea Ma: Ave stess=

Problem 1.3
Element Stress (ksi)

à 22.635
2 17.927

PROBLEM 1.2 1.02 A 2040 force is wp shown tthe haiona member ABC
Member ABCha 10 % 50 mono rrtanular cos sion and p>

the nicks of member ABC has been reduced fom 10.10 8 mu.

soutien
EE JA LINKS FB.Dieesam OF ABC:

264 HIM =O: 2 Fap(89-P(AL)=0
BT E Fop= PIAD/2 (BO (TENSION)
Host 0.4 m ==] +92 Mg= 0: 2£,(80)-P(AB)=0
2Fao Fep= PUAB)/2(8C) (COMP)

esse (2) LINK CE
Ie ness = ty | Le Thickness = ty

D
Agy> À, (0% -d)
ig ate > A Les tum
pete | et De = Force

hs a
G)E PING |
= Fe / (0/4)

| (5) BEARING STRESS ATB HEARING STRESS IN AR
‘Thickness ef Member ACS “tac Whe) oho
Er
\ Si Bear BE Fan / (dl tye) x
(6) BEARING STRESS ATC! Y
The

= ES the Cine /2)

Fate Fer/ldge) | LE e
> ‘comm

oa

E eS

PROBLEM 1.C2 CONTINUED PROGRAM OUTPUTS

INPUT DATA FOR PARTS (00), (@ P= 2040, AB 025 m, EC=040m,
AC 05m, TL + 8mm, WL 36 men, TAC = 10mm, WAC = 50 mm

4 Sigea BO Signa CE Tau B Tau € Sigsoor B Sigkear €
10.0 m -21.70 79.58 125.00
no D “it te sn uE]
we Bega 2m Far ie
20 86.33 2m => Sens
un “arte [ge] Ki 85:25
29 Seas Arne el 385% Fe
36100 101:56 21:70 HOW? 31.08 213 e (b)
Rn = =
1800 IS cn O6 Ai HE
1.00 15.43 22 e
20100 12605 Fu aris 6280
fee nie ieee die 58:82
Zoe 13.0 hi az Sea
Ba 1 1230 sus
20 Fer 1352 52.08
25.00 3233 Don 520
22.00 7 men ales
EX 12 120.97 eo
25.00 1113 110.07 Ana
35:00 Sie Ne 43:20
EXT 4 8 der

ANSWER: 16 mm s d'< 22 mm (€)

CHECK: For d 22 mm, Tus AC = 65 MPa <S0MPa_ OK

INPUT DATA FOR PART (0) P=20 uN, AB = 025 m, BC 040,
AC=0.:5 1, TL=8 mn, WL 36 mm, TAC = mn, WAC = 50 men

4 Sigma ED Signa CE Tau B Tax C Sighear B sigeear €

39.00 79:19 -n.7 B} 79.50
me a An En
oo fra Ce Sue
oo fes A7 fes
F0 sem ale 2
delos 201.56 Same NE Stas

1700 ingle: 21.78 Muss 2704
18:00 1288 21170 Sie alee
19:00 ses 2170 Sagi 22.08
20100 26.95 21199 Sm 136

21.00 12 217 dé 18.0
2700 148003 2170 AS 10004
23:00 am ju iso
200 m 2 De
28:00 im ue 1m
25:00 Hum a am
2500 Bm 2838 1092
32.00 am 3638 ide
25:00 ar 20 5:46
32:00 E ame su

OASSUER 15m «de
CHECK: For dm Ta de

156.25
Fa
108197
9766
us
82.24

m $ (d)

1.03 Two iso Sp forces ae applied 1 in 8 of he ate
sown, Each of the te pin at A.B, and Chas he same diameter ad
In denke set () We computer program to Gifs vals of fom
00 0 130 m. ing On Iren, (I) e mai vale of de ar
age mal ses la member AR, (2) be average nomal ars m
CE meine hearing ars In pi A, (te ase she sc
Be CS) serge bearing sus Ain mener AB, the vera Dear
Jpg ses Cin tear DE, (7) average Wearing sex at Bin mer
E) Cesk pu program y compañia the ales ted ford = Lin
‘rise an gion fr Probe 19, 128, ad 120. (6) Us this program 10
Fi de pme wales of the diner ofthe pins, Loing ab a
lina al ofthe nor, hen. ad Dearing stress fr the acl wed
‘oe respec, 22 bs 13 o and 36 el) Sle prc, sui at à
‘hw di being testa vos De cine nd wi fe two
‘ember ar changed epee, rom OS 1 03 a. an fom LA m. to
EE

SOLUTION

AE AND B

Bar 2P (sin as/sin 75°)
Fac = 2P(SiN60/Sin 75)

(DMAXAVE, STRESS INAB | (2) AVE. STRESS IN BC

A
Be Age = we
SR, Ge FBe/Age

2 Fe

NA (4) Pine
Caz Fae /2/014/0) ARA)
(5) BEARING STRESS ATA | (6) BEARING STRESS ATC
Sig Bear À = Foe /dt Sig Bear C= Fac fat
(7) BzARıng 5 AT

N MEMBER BC

(conneueD)

FROM FORCE TRIANGLE!

3

[=]

=

in

eer

PROBLEM 1.C3 CONTINUED

PROGRAM OUTPUTS

OOO Pci a
et cs i
ge ae at ee
Le ue deb
ae ym EE EE Un
LE sE Er
HBL Ce ia A ass dese dcr
Seelen
Erbe
ere cee a I
eee u 5
ee Eee
See
TEE
Teuer
DEE
eins be
PE ae ae I
JGR ER Eee LE
SES eee le
PEE BEER GE La
SRE TR Rg IN
JE EN NS En st
RE Sn

KÖANSWER. 070m cds 1 10in 4

INPUT DATA FOR PART (4) P= Skips w= 24 in, t= 03 in
‘TAUC SIGBRGA SIGBROC SISBROR

8.500
Pen
&soe
9.880
3.700
00880
21300
81580
31280
1350
2408
Lise
21506

stone sine

ket

308
190
785
252
73
m
EN

ke

212.482
5452
Dis
us

121452

ees

DE
EE
REA
5104
20560
1227
Het
259
2098
227
on

Rei

e
50
32
72
ns
ses
653
ER
21512
Parts
31539

Kar

Bee

Base

301503,
26.708
22.183
BAS
20.238
1807
15.078
50
15029

kel

%
60

25.980

22.

388

HET
Ber]
13.924

es

ms
ss

KOANSWER 085in sds 125 (A)

(0)

©

1,04, A ¿ip ce Pfoming an angle awit the vetas apli ws
sown lo meer ABC, whch ls apply pin ad racket u Cand by
Cl BD Teran ans wi th orzo a) Kewing that de li
ae ad of te eae 2 ip, et compar pgm t cost ble
fe ls of De factor ol ay ofthe ele fo values of u ad fom 0

SS ng lee fe and omespnding to 0.1 neremens in
in a and tn 0) Check at fray pee val of de maxim ale
ihe ocr fat one foe = 38.66" and expan why. () Dr
‘Bie the sal polo vale of he factor of safety for f= 38.66, weil
4e comen vale of, nd expli ve res aie

= wenn sopa

(a) Deaw FB. DiAGRAN or ABC:

FAYLM.= 0: (Prinay{is in) + ( Pcosa){ 30%)
~ (Fcosp)( 19 in.) -(Fsin p)[1Zin,)=0
Ca Fr pP [sind + 30 sx

of
fl
IS)

AL von

rag 15 cosp +12 sinß
im FS. = F/F
OUTPUT FoR P= 4 kips AND Fuy=20kips

Bade ae a eee oe

16)
(8) Wien = 38,66; fans = 0.8 and cable BD is perpendicular
fe the lever arm BC.
©) F.S.= 3,579 for w= 26.6% s P is perpendicular fo the
lever arm AC
NOTE:
The valve RS = 3.579 is the smallest of thevalves of FS.
corresponding to B= 38.66° and the largest of those

correspending to Xz 266°, The point X=26.6; [3238 66°
is a “saddle point’, or ‘minimax’ of the function RSC PB).

4.05 A lead is sopor a shown by uo wooden members of si
oc rectangular cos schon which ue ted by a imple ld sur pe.
(a) Dening by oy and ro espese, be unse tag of te jon fo

dension and in her, wre 3 compute progam which for sven values of
1,0, Pon zp express eier Stor US. etry ns and or va
{sf rom 8 to 89° 8° neral, can be used o cc CD De nomal

Draw the FB.diagram of lower member:
GER 207 -ViPosx=0 VE Peeso
ZE < 0: F- Psinx=0 F = Psinx
Area = ab/sinw
Normal stress +

0= E = (Pad) sino

Area
Shearing stress; © = = (Bab) sinx coso
(3) FS: For tension (normal Stresses)

pe

Fsn = 0,/0
(4) Es. for shear;
F55 = C/E

C) OVERALL RS:
FS = The smaller of FSN and FSS.

ccownnurn)

PROBLEM LCS CONTINUED. PROGRAM OUTPUTS

For PO.L24 P= RN
| 423125 mu, b=75mm, X= 70, Gr 126 MPa, %,= 1.50MPa,

ALPHA SIG(MPa) TAU(MPa) REN rss Fs

5.0000 0.0049 0.0556 259.1782 26.9942 26.9942

10.0000 0.0193 0.1098 65.2905 13.7053 13.7053

12.0000 0.0428 013600 29.3899 9.3750 3.3750

9000 0.0749 0.2057 16.0301 7.2925 7.2925 |
2510000 0.1143 02482 11.0229 6.119 6.1191
30.0000 0.1600 o27n 7.9750 5.4127 5.4127
32.0000 0.2106 0.3007 5.9942 4.9883 4.9883 a
10.0000 0.2644 0.3151 4.7649 4.7598 4,7598
18.0000 0.3200 0.3200 3.9375 4.6875 3.9375 (0)
So:0000 olanse 3181 3.3549 4.7598 3.2549 D
55.0000 0.4294 0.3007 2.9340 4.9883 2.9340
60.0000 0.4800 0.2771 2 5.4127 2.6250 is
$5.0000 0 gas 2 ell1sı 2.3068

3 dao 22286 u (4)
33-0009 0 z
80.0000 016207 2.0300 13.7083 2.0300 Ñ
85.0000 0.6352 0.0556 1.9838 26.9842 2.9838
For Prob 1.22: Ps 249016 Ol
in, in =, = si, ©, = 214 po,

Q=6in, bz 3in,, X= 40%, Gy = 150 psi, T= 214 poi, =
APRA SIG(psi) TAUPE) SN m FS.

‚0000 1.0120 11.5765 248.2028 18.4857 28.4857 T

10.0000 4.0205 22.8013 37.3089 9.3854 9.3054
ood 8.9316 33.3333 16.7942 6.4200 6.4200

20000 15.5970 42.8525 9.6172 4.9939 4.9938

350000 23.8142 51 6.2998 4.1904 4.1904

30.0000 33.3333 57. 415000 3.7066 3.7066

35.0000 43.8653 62 34196 3.4160 3.4160

“0.9000 55.0901 69.6534 27008 9.2595 9 7270 4/0) y
480000 88.8607 &8.6857

ee ON

| 5910000 89.4880 62.6462 1.6766 2.4160 1.6766

| 08.9000 190.0000 87.7350 1.5000 3.7066 1.5000
63/0000 109.5192 51.0696 1.2696 4.1904 1.2696

93.0000 117.7363 42.8525 1.2720 4.9939 1.2740

79/0000 124.017 32.3353 1.2058 6.4200 1.2058

80.0000 129.2126 22.9013 1.1600 9.3854 1.1600

| 8910000 132.3205 11.5765 1.1336 18.4857 1,1336

PROBLEM 1.C6
1.06 Mene ABC in supported by a and ace: 4 a y vo a

share pionnier and a ed appur (9 We

computer program cat Be alowabe od Pfr any sve es 1

Te Ence othe inl 4, (2) th common diameter of te par at and D,

(healt oral sus ; a uch fio ak, (4) telnet eins

sm rs, nec of re pis (5) he rd oneal tor of aay ES. Your
Tope progre hu ao itouch o falar vos ses e ria the

sou o 90 a, res ik he being enn th ps or ala
pt ti in be pins at Band D. (band) Check your program by ang the data of Proba.

1149 nd 150, respete, and comparing le aos ti o Pe he
Ep ration (Lie your pogamıodeemietheulembiekadP, anch
TES ET suis of te mess lia, when d= y= 13 mm, 0 = 110 MPa Br

‘Seni nk, = 100 MPa forse ps en AS 232,

N © sono

0) Forgiven dy of pina F=2@,/rs\irdiy), Ps 20 ra

2) bax given d, of. pins Band D: Fan” 2 (6, JFO(PAZJO,
6) Far ultimate stress io links BD: 7, = 2(Z/F5)(0,02N0,008), p=200

for ultimate = 60” 2 fr) 0000), p= 202 Fa,

(4) Koc ull sheering cts in pins: Bis the smaller of Pana zo
2

(E) Fix desited overall ES: Py is the smaller of Band P,

i Be Py, stress is critical in links j
4 <P and P< Pa, stress is ceitieal in Pin A
UF PL SB Ana Bec, Shresc Is critical in pier B and D

OGRAM OUTPUTS
(6) Erab. LEZ. DATA: j= me d,=lènm,G,=250 MPa, 25 = IMMPa,

Fay 3.72 KN. Stress in pin A is critical a
©) Pros.1.50.DATA: A, =109m dy =}2mm, Os: 250MPa, 72 OP, ES = 20
Bus S:97 kN; Stress in pine Band D is Critical 4
(A) DATA; A =d,> 150m,Q,=110 MPa, 92100 MPa, ES-= 3.2

Pal S.T7 KN, Stress in links is critical

CHAPTER 2 |

IL. Lu a

2.1 A sel eis 22m long and must ot ech re tha 12m when $51
] À rrosiemzs load i applied tt. Knowing that + 200 GPa determine 2) he sms ant
Fed ao be sl (Pe oresponding ora! ses cca bythe ka

sorunox
2 A= BE N. ran mt L
cr 7a“ m
| Acht: d= JRE EE 00
= 276 mm a
7 AT <
} fm @ E = 1071 MPa =
Ben 22 à gi E me ec wie y os o 50

Tense load. Kzomin dat E =29= 10a, determine (jte clon ofthe wire,
(6 be ceresponding orales.

Le des 576 in Achte HN =42087 110"
0 gs PE. MER. 0.3710? m + 00908 à

L SES)
ee E = ERBE pei 15:28 ksi - |

I PROBLEM 23 23 Two gogo mas ae placed ec 10 aces apar oa a Sinner |

] alo od wit 10.51 pea ia ag of Ia Koma |

‘Bathe dance benne ie gage mars Is 10.09 ln
SOLUTION ‘eee (te ses in va) te of

er eed pe,

(a) $S= 10.009 - 10.000 = 0.009 in.
O CEA

E 10
sa a
-

W Fs "ee en |

24 A contol rat mde of ein bss mut ot ch ce an 3 wan wen > |

a tension in the wire is 4 KN. Knowing that £ = 105 GPa and that the maximum
een e ios mite ol em Sree
A EE Secon cn act |
] mE e LS Ge |
@ o-Ps Ach = M = 22.222%00

Zu | |

fl As Hots Jef . [OR . 5
e 5.32 mm et

Pe. 2 AE 2. 222 10 (or 10") (3 x10" )
ei se Bb à Le AEE, Camano lucena

Rio

1.150 m -

25 A lengua ia wie so o ue ina ange. Mis nt
dad ares mm whens tel face Psp Kant 200
{hs deen) maps oft cP) scope Roma sees
sorwrion neu.

ta) A= Ea? = Eo.coc)’ = 28,274 010° mi

e= Ph: pe AES, ero Xrooroliano?)
E 3

PROBLEMAS

SE

Re
= Eon = MSI UN =
MONOS «
0] A = 400 *10° Pa = 400 MPa ~a
ai 26 AS: alain pie hat archos Sn when ahead
tose ot Rowing Mt = 01 10 a and al de alow ea
ni 1, ce ana a ap op (De
souunox Fa ae ane LD
= Ph à Le EAS ES - U0.Imowlo.os _
a Sehr SAS 2 ES - Leta) = 36.1 à -
[Sir Ze AE? LIENS au ae -
> 2, Arten ad otero. Kang 31 Pa
PROBLEM: LR RS ee TP cio de dinge
souris ead a nts Ora
&
a &
gx à
Ar =<
error + 863010" Pa = 303 MPa -
ua 28 à un we sued por maire in, pot at 19
À ta aa ia args ra 10.023 et, seras
Leman coma ai he ae) mn wal ise Resto
sorumox Ic ue Gur of ne
@ SRE = 0.0095
© = ES - omo ours) = 26 Of psi © 25K me
BL too _ at
oy = & + Se = octo i

20 Me ais di
t=4(d.-d:) © £(Q0- 1.7847) = 0.1077 in. -

29 A ak of 10 gh a LX cs sen 1 ppt ene
À mon, Concilio tote aca ra el a
mos eg sd inc bp Tao ll De ale
acuer tened Iisa hat des Leo de och tl m1?
pero foal eng
Considering allowable stress 6 = 18 ls = 18710% psi

A=G-8G-6) = 2.880 2
Pr GA = U8so% (2.88) 51.8 wo! de
Considering abfowabte deformition Pz JE = 0.0012
se: Pe AGB = (298) (io (o.002)= 43.4 ro Le
Smaller valve governs Pr 4840" th = 48.4 kips -

200 ASI ricos Sm
ha “Seem ten Sate
a Sc
soumox ER
Considering altowable stress S = 150 xlo* Pa
= E = Bl que
orks ae bs oe
Considering Movable elongition 8 = 25x10" m
g- Pls p= PL, (Suso
AE ES rd von),

hanger aren governs = 90 «tot

Axdt df (oe 10.10 110% m

hotel ic wit = 200
be uae. Eros ht te
een being ehe wie

PROBLEM 210

couo* m

To no“ mt

= 10.70 mm -

A The meer io DC made of ste wih = 200 GP. Koon
‘PROBLEM 2 ,
A momia aros ios ce mat cd 90 MPa >

pnl e ee must nc excl 6m, Bd e main oad Pl cn
‘pple shown

soLunon

Les POOR + TRI m

Use ber AB as a free body LE =
DEM,2°0 35P - (6 elt ©
Pre 0.1508 Far Ar
Con sid ening abou ete stress FF 190410" Pa la,
A = Bat = E(o.0o)" = 12.566 xo" m“
cr: Rye As Otomo Minscents*) = 2.520010 N

Considering «Movable elongation 37 € 1107 m

5e Elus pe A

Smafer value governs Re 2041410 N
P= 0.9509 Fre (pas omo) 1.983x10°N © 1.958 40 <a

momie Tt Rod BD ea = 92 pa we ce e iy
ae a am incendios et BD
Sean a no Tbh sl Uni can a EE
pom Sole, Tin doin DC éme le send
anna.
soon
Feo * 0.02 P = (o.0R 130) = 2.6 Mips = 2.610% db
Considering stress = IB ksi = 18%10" quí
Feo
oe Ba As Be BE orme
Considering deformation Som? 0.144 in.
“Fale, À: Folie. ero?)
see ES

Larges aren governs AT OM it
NES = [RO = 0.429 in ~

225 À gl axial ond of gre P= SHAN salad en Coe bard
PROBLEM 213 ABC. Kong hat = 108 GPs determine th ameter do portion BC er which
ite decia one Cv e mm

‘ SOLUTION
5. = EZ tilo. Ella, el
1 AE" El Ae” Au
ssl
RNA 1-2 psg y
ONS PR = 2788000 w
28. = 214.2810m*

TIS ISS

> 3 DE FR : [QUEDAO . 6.5200 m
+ > 16.52 mm. =

2.24 Bot poro of the rol 48C ae modo an alanis fr which £=73 GPa

PROBLEM 2.14 ‘Keown at the rete parton BC = 29 rm determine le ang ro P
Sm ane applied fo, 160 MP ande comerpendin dsc stat Canal
] souumiox

St Ar Floor)» 70600"
“4 Au = HCono)*= 314.16 x10" m°

Considering «loable stress = 160 %10* Pa

SR: Pras
Portion AB — Ps (706.8605 \isoxio“) = 13.110? N

Portim BE Pr(ammiexio*)Ueoxo") » $0.3x10* À
Considering er election & + 410% m
A)
Pe ES, (+ + bey" ao (A
= 68.8 x10" N
Smablest value For P governs Pr 50.33 N - 60.84N e

E
0

=

ju cry

25 pin don ma der ST
PROBLEM 2.15 1.$-in-oaler-dinmeter sleeves bonded to oda nen Sooo
dete) he ad Pm ate ean SB Be
En seman fe cnr pan

SOLUTION

re An i
@ she. Bei
P=ESEBY As Bay
Liml din | Ajint | L/A., im
sepa [is Pure] wise

Be] 3 fio | 0.754] 3.8197
cop 2 [us | tm] 11318
6.033 + sum

P (aero X0.002 6.083)" = asirio” th, = RSS Lips «a
WW Su= Pl "Ep E a). 125907 in =
ans 216 Dago oe nd AC fn RE CGR

Knowing ae magie of Pi 4 IN. denne (2) Be Val ar 0 aos
def 14 0) the comesponding defection o

SOLUTION

(a Ag = Fog = Flore) = git.ic vio mt
Rec © Feed = F(0.0¢0)" = 2.8274 010 mt
Force in member AB is P tension %
ae = Plu . (4x107)(o.4)
had yg > EES Sera
= 72.756 wo” m

Force in member BC in Q=P coupressi

Shoes Sip le Bea
= 2.8263 10? (Q-P)
For zero dellection at A Sue = Due
2.826310" (@-P) = ISA OT Q- P= 28.840 N
Q= 28.3%/0" + quo = 32310 N= 32.8 kN -
(61 ms Sur Ser TIA = 0.0728 mm _

))

4 PROBLEM ET

47 The rod ABC made an luni or which £70. Kooning a?

EAN Q 421N,detemine he den a) pan 4) pont

SOLUTION

T a her Fi = Ho.) = série
An Au Fast = o.oo)" = 2.8274 «10° mt
| A Pa = Pr Grin
i Pac = P-Q= 64h HO niche non
03m Lio = 0.4m bac? OS m
nn dune
Ser Bala. (rien
Bab” (Eco
= 109.185 10% m
= '~36 410% )(o.S
Sue as Lio v0")
= = 90,947 * 10% m
Sa Bag + Spe © 107. 18016" 20.947 015m + 18.19 «10S m
> 0.01819 mm
W Se = See t= Maritim = = 0.0904 mn

e

218 The Moma ee ABC anda ra rod CDof bo am ane we
Joltedas pit Ct forthe? Sara CD. Fo eo om ae gc
the weit rl determine de decian of) pol.) Pla
SOLUTION

Ar

Fat Hose)! = 1.01787 #10" m

Pi] R TE: | & | Pil: AB

A8 [iso ui | 2m [200 Ge. | 1.474 x10"* mn
BC | took | 3m |200 Gfa| 1.474» IO” m
CD | 100 kN | ZSm| 10s GPa| 2.334,10? m

Sue + Ste = LANG à LATA IO
= 21480 %m = 2.95 mm «a

We) Sur Set Se * 2.748710 7 à 2.999 ro?
2870 n= S29 mn a

PROBLEN 210

219 Th rss be 48 (E15 » 10" pa oeste are 022 ane
[edwithapugat 4. True isatuchedu Dto ii a ah
ai Cie the bovomefan alain (= 1045 Up wil st
Ares 04017. Thecytinder ben Roma spprta D. In rt lose e
ner, plug mas move den ough À in, Deen
te pid othe inde

Shortening af bras tube AS
Le IS+Á > 15.047 in Ag 0.22 in‘
Dé ISO pui

> Ela „ Püson) , vo“
Sa a a ss‘ À

Lengthening of aluninum cylinder CD

Los IS, Aus O40 nS, Eier 10.910" qui

Le POS

fa = LE PO 0%
Sue amor Eee) 3605810 P

Total deblechion Sp = Sue + See

2

a

(4.8597 116% 4 3.6058 10°) P

?

5.74 x10* Jb
SET Kes mt

220 À 12.m sean of lain pipe o ro sen ase 110 um es ona
PROBLEM 220 Fed upportat The 1m dame rl nd Changs fome mp br
r Omi fie pipe 8. Knowing to mods fey 200 Ga arsch
a Orto onza dot esr po be ON ete

applied Co

soumon
Rod BC get dim, Es = 200 107 Pa

Reet Fd* = Food) > 126,715 410" m“
So» Pine , (Goro! Ma)

EsAu | (200 x10) 76. SHIT)

> 3.585710" m

Pipe ABE Lag? 2m, Eyes 12:10" Pa, Ang? 1100 ma OO mt

= Plu , (6ox10*)(1.2)

Sem > nda 7 Vario Woo) 7 1071 KIO mi
Se = Sens Suns 908110 + 3565-10" = 4.47 NOM HAT A
225 ae (200 svn age nt
uamumazaı 1920 mm’. Determine the largest allowable load P if the change in length ofmember
a
» Ir SOLUTION
Seis m, Ag 1420 mma 19202105 mo
pS Lu STE = 7810 m, Enr 200 210" ña
Sa = Fla

8 me" Etam . Dieses

E E EAS. omo linda

i Enf - a

IH sn = 78.67% 10° N

Use jomt 8 as a free body? SEF, = ©

ies Fa - Pe
N r Pen» Osma
e 1.810
Fi! Fae = 50.4x10" N = 50.4 KN «

222 Forte stat ns E =200 Gr) dls sown din comatios
PROBLEM 222 ‘mente AB nd AD, koa el hero cs ass ae 2400! a
1800 mu espe

EN
> EN Reactions are 14 KM upvard at 44 €.
4 Sl
Monde ED is à zero farce mentor
ml Lag sat rast = mur m
Use joint Aas a ee body: HERO A
Faw = 215.Jo kN
- 5 E -
Fo O Ferne + o
se = Ms = 182,4 kN
whew ende AB! Sun = Babes > GiSiemet}(#.717)
Mae, BaP ee anaes)
zar dm € Zen -_
> Elu, (mm gn © 2.08 mn =

wien CORTE

223 Mente AB and AC are made of sa (= 29 « 10 pa) wil som sent
aca of 080 in and 041, pete, Forte lndng shown, deerme de
login of) member 48,0) member SC.

OS

pg
¡mr sum .
se (a) Le ATREA =100R.= 93.72 m
Use joint A as a Free body
ai Fe
id IF 20 ks Fe- 2820 à
Fe = MT kip = 837 À Fe
Figs „ (sue Wa8,72) _ V2 kip
"ERC taxa se) = 91767 im

(6) Use jent Bas à free body
Fe SZ. tO Fe -

Tin Fa = ©
AR en 33.60 kip = 33.00 x10” A

EZ) = 0.1804 im, <
rn en

een 224 Members AB ad CD we ini sol ros, and members BC and AD
ace -in-dameterstocola. When temburlintighenal he diagonal member

yu le ACspitensen Kzonng t= 29 1998 04-48 domino Berg
‘lobe rin AC wo De lemas bes 4
Sen
à À souunon
| Sie or 0.04 in he APL Hin = Les
4 0 Bess Hat = Plis] 0.99402 int

Feat
lla Se: Fe
Foo = Eheier Lam (0.99402)(0.04)

Leo 48
= Momw0 Jb
Vie joint C as à fe bony Fu c
PEF, Remo Fer af |
Fu = Flmomsıo)= 3.0.10 À ruf
0.0 ke =

=}

224 Members AB and CD ar ini sc ds, and meres BC and AD
ue inter iso. When mulet o igonalmenber
Acisputintenden. minga = 29» 10 pend = 4 deemio de lag
‘Slowabl eon fa AC 0 at he deere in members 48 and CD do ct
nr

225 Forth mueren Poh 0224, determine () be ista at the
‘teenage in mente AB, BC, CD und AD a eal 00.0, (he
ordis tein ln mener AC

soLunioN

(a) Statics? Vee joint B as a free body

8
From sinidar bangles

Fe 2 Fe o feo %

CC

II

Fe = Y Fue Force una D
For equal deformations
> Set Fab. fi 2b Aw
mess Sr Rein
Equaking expressions for Fin
+ b
AA
h. de Ble
5 de Wa 7
he $b (6) = 2268 = 468 in -
(6) Selling Se» Se + 0-08 in :
See = bh pue > Efe. amo) ER) (0.04)
HER “ee EJ

= 1287640 Lo
Fra? bre Flare) 24.902 do? A
From the force Triangle

Foo Fe la Fag 91.6 rio! A, =<

DzH

ron mE

20

226 Members A0C snd DEF we ned with el ik E = 200 GP. Tach of he
Tika fap of 25035 om inten of)
apart 25cm pen, Dacre tgs ego)

soummiox
S(T ER, Use member ABC as à Free body
ell DEM,
Fee (0.2601(13x10") ~ (0.130) Fie = ©
A MESA
. Fog = end N

(4.840 Mis > 10°) + (0.180) Faz =0

onu) |
D: me
WII.) | se
Gaon AE T) © "20-2410 m + -0,0802 mu a
= Lane ne)
ASOMO KLIS O PY

Area for Dink made
= Ht M of to phates

A = (20.028 0.085 }+ 1.2513"

7. 88¥15m = 0.0188 mor =

227 Bac fie 48086 CD made fein (= 75 GPs) ed bas
second aes of 125 mn Koi da hey sepron the gel ment

PROBLEM 227

‘Eiomie he een ao
P soLurion
oe Ba Fo Use member BC ara
Free body
Og a
res Sun
DEM -=0 -(0.64)F, +(0:49 Sed =o AS

DzZMso (our, = (0.20X6x10*)=0 Ro 18625 wo m

For Pinks AB and CD Az 12S mm > SAI m‘
Su: ale „ (3.9826 sto* YC 0,96)
m “ER” lonas lo)

= Bela, (SG mo )(0.5€) | gases
E a] A E

132.00 dom = Sa

6 + sty 0+ x. eens!
| 2% 212.5 to nd
ole Sez S.+ 4.0

Deformation gran + acon’ (os mo‘)

"ORSAI m = 0.1095 mm

229 Link PD nade of trae (E = 15 10" a) nd hs rs ena ae of
‘ab ins Link CEs made oli = 116 = IF pa) an co seal
rauf I, Dtemin he mann free Phat nb pled vor
fn the election fo otto excel OA

soLunoN
A Foo € Use memcer ABC
as à bree bod:
+
Re
MPA Fo O0, Fo LS P

DZMso SP-Iuro, Fu + 0.5056 P

En > iur Faber . (15556 Plao) je
ee tt
Set Bes Erler. LOSE or À

Exe” ÜoAntorY0.50)
From the deformation diagram
> Sat& | 2.9708 10S P
pe 0 et ae
een A = 03308 «10 P
Deformation Diagram
“ 5” + Mae

= 2983340" P + (5X0. 3805 0")?

3.1353 #10“ P

Needy displacement Puit Se + 0.0/4 in = S.ansB nto" P

p= ost + So tb

2.29 A omogencss cable oleh and ui ros seins suspended from

or end (2) Dent by othe deny (ass per ni vole) ofthe cable aby &

a modus o sic, determine the longue oe cable ae o own wei

sournox (0) Aasuning mon the ate lo be ran, determine a fre ar shoul De
lid uo af thecal toch the same soni an pa

PRODLEM22S

(a) For element at point identified by esorilinate y
P = weight oF portion below tle por)

1
y
? + ren
+ AE Le, guy)
| as» Ee. Al, polly) à,

Ly y
= 5. (AED > AF (y 40,

Bw gy + -

dol Pr se PR SAS AE glo dw =

PROBLEM 230 a ee er

kr sownon
T L Let ere
+ NOTO ct the tess ow
Ed =
element a cn wilt
pe] eee
kee zh

Vodone oP portion above element Vs ¿rey ¿mty?
Pe y - ah Azmet ERY
Se, ce Ce ce ae ça
A m me
. x
5 p »
> saz)" - ey -

231 Tas volume ofa tee specimen à cuenta souriant wile pac
PROBLEM 231 demain occur he na ame of he pcan a a when he
amater tros sain sg» 2nd /6,

SOLUTION

IE the volome is constant Fs r= Tai,
Le af. (ay
a

E
a= dk DY - 29 -
PROBLEM 2.32 meen as in a tensile specimen, show that the true

sovcTion
es ME = tS = (is 8) = lise)

Thos EME -

233 oo ay mn lod

Do rROBLEM2SS late. Demi (2) the nol rin sé

an Dace (0) be oo are ba el be mn
‘deformation ofthe assembly. © _—

sm 2 Sma SOLUTION

et Let PL portion of axial force canied by brass shelf
EC en 1 Re portion db axiad Ponce carried by steel core
Ru 2: HAS
cg PM s - RE à Ef
FOR Rs
Po Res (BAY BADE
dei
E BATEA

Ag = (0.020M0.020)= Homo m!

Ay > (0.080 X0.080) - (0.200.020) = 500 x10°° mo

60 x10%

= 452.88 «10
Tor” 52.88 x10

LE see
Sse

(ar Si > Eje = (105x107 (452.8310) = u7.5 10% Pa
= NTS MM -

WM 8 = Le = (2sonto*Maszgg ict) = 113.2 210°“ m
= 0.132 sof m
= 0.1182 mm -

2.34 Teen fe ae decreases b 0.15 man when an axial ac app

PROLEM 234 "oy mens of gd ed lat, Dein the guide fhe pies fre, 0)
te coreo xr bo el ee.

Sam Sas

un Zum SOLUTION
Se Let P= portion cf aval Force envi by bones shell

selec > portion of arial force cavvied by steed core.
Po ES a t'a
«pe + EAS.
s- PE Ron

E P= RR EA EA) Ë

As = (0.020%0.020) = Hoë #0"* m“
Ay = (0.020X0.032 )- (0.020X0,020) = $0010" mt

(a) P= [osmioriksoone*)+ (200.10 Yoo vo] LLE

e

= SO N = 75.9 kN =
A AN
> 120 mr =|

SH eo

e

Li “LJ

oia 238 The 4: cert pos sens wth ix mel br, uch with in.

et, comio that 29 1p and,» 42% 10 pa, deere the sco
sein tec azi he concrete when 504 cei ce sapped

SOLUTION

Let Pe = portion dPavia® Force carried by concrete
Pa = pectin camied by the six stec? rods

: ÉL A
eo mäß
ih 5
ER Be BAS
Pe Ra R EA EAE
ete EA BA,
As= GEA + GS = 5.964 in
E E SE IAS
FAS Fe SH in

= = 350 vit

A
Varo an. Ari ar) 7 AS

Ge = ESE = (270% 287.67x10%)= 8.390” pri » 8.34 loi à

Gr Res (URREA) rm 1208rlO pora 1.208 40 a

236 Anas cea fre of mgitde P= ASIN aide compos Heck

PROBLEM 236 nny mano ideale. Keown ut = Dan, demi te seal
= es) debas (0) ds men pa
EN SOLUTION
Let A= pain Find For card by
mass Core
Paz portion camied by tue alominon
Pe É

+ BL - LAS
cl

P= Re R= (GAS ADE
770 "SOHN NEE
LT Apr Eh
Ay = (6040) = 2400 mu = 140010 mt
Aa= (2¥(60Xt0)= 1200 mm” = 1200 107%

= E 1.3393 io”

"ARM MOS Nico roy
@) Gr Re = Úasu1o? 1.439310") = 140.6x10% Pa > 140,6 MPa 6
) G+ Euer Cromo Marc) co Pa > 98.75 MPa a

E

O oo

237 Fothe compe Hock sown in rob. 236 determi (he vale Cie
rion ofthe fod came by ie aluinum pa i al he prion ofthe td |
tid by he ras ee 8) he a ond ees ine bas 30 MPa

SOLUTION

gt
ee Let Re portion of ant Foren cari
by brass eave
Paz portion carried by the too
nino plates

= AL - BAS
3: Be RSA
@) Giver Pr AR
EAS EAS
ESE
An =tBe As

Ay = Wolle) = 2400 mé = 440010 mi

A, BR 2400 « 180 mn‘ = Qncorh
> ,
he BES IS mn -
&-8
(er À

Po > ALS, = (osx BO tot) + 192 wo? N
Pas APR TION
Pr R+P,s 288x107 N = 288 UN =

=] =

[=]

235 The 4: conce pn reinfred wi ae bars, each mia in.
mer Kooning Sat, 29 10 mind E, 4427 0 pi, Stern th oma
sea nd e coc we à 330. a centre Pop

Wirren.

138 Forte pt of rob. 238, determine the maxime ene force which maybe
oli ala oral sees 20a be sel and 2.4 ate const

PROBLEM 238

Determine alfowable strain in each mutenal
Sr eye Se Bah carer eet

IT

Concrete? E.= =
Smabher valve guess e+ B= 571.48 110%
Let PLA pectin oP Dood cavried by concrete

Pa = portion carried by six Steel pode
SeÈE, Per BAR? EAE
Ar, Per BAP + BAC

Pa Pet Pe > (A+ EAE
Ag = Edd. Bist sat int

Ae = Baia = Bley - S964 > 248.5 m

P= [tumor Xx 248.594 (armo ene

= Gas «10% + 69S Kips ~

PROBLEM 239

= Pale Paleo Peg Lor
E >

139 Theses (E 200 Ga) sponta 36 nd. Echo rot AB ans
CO as 330 mm rss seca es a ro EF ha a 625+ ma conocen
sea Demi the) ar m eg fro FF.) te sues cao

SOLUTION

Use member BED as a Free body
if P ir By symmetry, or by Zero
= Ro + Pa

E E
Past Bot Pr Poo

Pr 2Pa + Per
Li

Since Lig = le ond Ag Aco, Se = Seo

Since points A,C, and & ave fired Sg =

San, 85> Eno ; 3e * Ser

Since member BED is mgid Be = ber de
Pepi Lolo: pes Am te p, = 200. 2 p,
= 0.256 Po
Pr 2B + Per © (2MOLSO Pos Par? SI Per
Per = bg = BHO - asso wot N
Po = Pes = (o.2scMas mod) + 6.095 vit N

AS CTN MES
ey, EN

seu

= 0.0%2 mm «e

4.or5xto? )(S00 ot) | ee
SÓ

fe goss = 80.840 Ph = 30.5. MPa

U O MA ae

n
IL

eee AD Abrus bol, 15510 pi) wa in darter ied inate tbe
(6229 tt psy wa jie cue meer and fin wall ihe. Aer et

ua oe sly ii bend ne quan o al um. Kai ute bl
leed wh. pith determine oral ess (9) ot, () 8
oido

2.—— SOLUTION

The movement of the nut aforg the bolt after à quarter torn is equal
te E Y pitch.

S =($(0.1) = 0.025 in

Ales = S = Sum + Snte where Sum: elongation of tle beit
and Shute * shortening oF the tube

Let Putt = axial tensile Force in the boft
Pie = axial comprersive force in the tube

For equidibriom af each end plate Pan = Prue = P
Aunt Hd = TRY 0.1105 it
Base > Cd = dit) = FCG -Y) + 0.29452 int
San = ye . ss = 2248140 P

po BE Aal pe y xt
Ses ee roma? [Hosen P

0.028 = 7.298) IOP + 1.4080-10*P = B.c48ix10" P

P= 2.8908 «10° de

À + Acad pui + GA kei -

9.3210 pei +-7.82ks -

241 Two cylin rods, CO made of el (E = 29 = 10 a) and AC wade of
tain ( = 104 1" are joda and enced ld spp a
D. Determine) be actions a and D) elem of pot ©

PROBLEM 241

SOLUTION

AB: PER, Last
Au = Haas = Fans + 0.99402 int

MON
PESTE CTD]

= 0.177386 x10 Ry

BC: Pz Ry 18xi0, Lolo in, A+ 0.99402 int
= PL, (Ra 12x10" Xo) se -s
MER atmo Kose) = NIEHS RA = nA
CD: Pe R= ao = 14x10” = Ry 32m
Leloim As Edd = F(le2s) = 2.0729 in
Son BE + MOI 016027 00% Ra - S821 x10"°

See" ER * aaa y

SI ER
Since paint D cant move relative te A Sp = ©
Kad 1.9075 10“ Ry - 22.13810*= 0 Rena ae

Ry = 820 - Ry ‘20.080 Mb + ~

(0) Se Sip + Sig
Faro Ry - 12.412 10

RUN RO MAS = 3.58 008 in me
or 8, = Bale .(2008M XI). . sage in -

Esha * Gitar iCzorany

aia a]
bearers a ge cer
ee EM
Being

swan Be eae ean eterna,
Daoudi Gone

an Em

For the tube dis dé -2E
32-210) + Hmm

Aide 92) = E(sz*- 24%)
= 351.86 mm + 351,86 10m

= 0.080 m

«PL, Balo.ose) 1808 ci" Ra
Sat ER" oceano = 1188 ie

BC: P= Rt Yario?, Le 0.080%

RR aca + 47 Nero
EA * (2000 MESES vie)

Cd: P=Resiario?, L 20.080
= Rar 1210 Moos)
¿004107 XES-B ri

r
i

11368 IÓ RA 13.642710 €

Total: Spo = ag + Sat Sep © 34104710" À, + 61.328810

Given jaw movement Buy r-0.Amn = 0200 m

OR" 340410 R, + 6138810 à Ru = 76.6 vo N

276.6 kN -
Rye Ry + Retos Ro <= 64.6 710 N
Ehen =
(6) Spe = (1136810 VIGA) + 47.746 HIS 39.9 010"
-0.0394 mm mt

= =. A te |

PROBLEM 24 24 À ele (E = 200 GPs) vita 32m oe der and 40 ces |
ie placed Ina thar eed no Jl mh e ena ake
‘edt exerting any rr u dem. Ta o es ln re en ep the
SOLUTION The Aer these ose appli he ise aed to ase the dance
Between dj by 02 men. Deter (the fires ced bythe vs onthe ube
AS

f 24% Sn Pb 2 ming tal te res een vis

[ R RE 9 a,

un TES

+

A For the tube dis -2t
SEI > Hem
ñ As Fla -di)= Flazt- zu)

L = 351.56 m

ES

A8: PER, Leon
L Sas LL. Melone) =
ER " Gecuotnmaet
BC: Pr Rs mario, Ls 0.080
4 Seq = Le. Armor o R à nist
] fae = so Re ner
EB Pa on L = 0.080
= Lr avi (0,080)
METTRE TT

1.1368 10"? Ra

Sr L138 IRL + 13.642=10°

E

THE Bro + Sue Sat Seo © BM IO4M IÓ? R + 61.388870"

Due the movement of He jowe Bor O. mm? = GRO m

- trazado? N
42340 =e

=O. KIO = 310410 RA + Gl.88m0 Re

PORTS
BR SN —

Ro Ras tano =

= FA E

Su #868 Mt e 97.790 «to
ook =

a
L

=

Lx |

rRosent2s4 2.04 Te a co ep pt sows, Ann wes we wed
Janda hs Saro in mb id wee
E oiga be nr i ain 14 po 4
Sl ate lab (910 plo WH, dee
A
soLunoN

Sy symmetry Pa
Also, Se: Sa Se = 5

Pa, and Sur Se

Strain in each wire
eur, as - 2e
Determine «flute strain

Ate ae 3462 10%
Ben B= 26924 wo”

-& . 0 . coe

€ fe = = per © 0.620710

Enr Er 4» loss
Movable strain Br wire © gevems 4 6 + 18x10" pai
KRG Par Abt" Bb) Cono (ose)

+ 12261 Ib
Par Janer wh

ATA Pee ALS = E) Cis «i0*) = 78,17 4,
For quil ium Pf He plate
Pe Pr Pe eRe 177414 -

on mS

+

245 The ii AD is suponed by two ste wires of dame (= 29
AU pu and pin and basket aD. Knowing dat de wires war nelly up,
semi (6) e aban ce in ech ne when 20.0 oad Pi spp ai
‘Dud the comenpordes else pi D.

PROBLEMAS.

SOLUTION

Let © be tHe rotetion of bar ABCD

etal Then 8,2: 120

» a y & = 240
+ 43

- EÂBe o ana ere)
= H8.6%x106

Using Free body ABCD
DEMO Ae +R. -&P » 0
(2X106.77:10' 8) +(24M118.68>10* 0) -(86 (220) = O
4288x108 © = (BeKaro)

@ = 1985107 pad
(a) Pag = (l06.77 103 Kı.asswıo”) = 204.3 4 -
P= (18.63 colmo? ) = 227.6 th -

te) $7 36 0 = (2% Mins 10*)= CHI KH

= 0.0001

Lea]

0
N
N

246 The sel rod BE tnd CD tae diameter off in (= 2910p). Toe
nds ae eds wa ich an. Knowing ae bing ugly fe
318 üghenedene fl ar, Sterne) esi 08 CD. De len
pain Co te pd member 480.

PROBLEM 246

SOLUTION

Let © be te rotation of bar ABC as shown

Then, Sez 60 ond S.= 100

Bt Sq = Sum - ele
ks Pos = (Ese Au MS pon = Se / Loe
\ Le = ESF + 90 my Shin 7 On! im

ole hue Ha EGY
Pre » (2108 Ko.3068)( 0.
en 10

= 7.886 x10" = STB. ©
= Bele 2 Pr EAR
a a

Lent oh > IZ in, Aus 0,3068 in

CELO 8068) (10 8
“ 72

= 1.23972 70° ©
Dzu,-o SP -10 Py = 0

RE

(6)( 4.886 rio - 538.16 10* 8) ~ (1011.23572»10%) 0 = 0

$4,316 #10" - 15.916 * 10° © =O 0» 3. 726810"
LR Pag = (123572 10° Ka.ncgriot) + 4.61 » 10%
= 9.61 kips ~

(md S¿= 10 6 = UoWs.718x107)+ 37.3 ¥ 107 in
= 0.0873 in -

cco om co ce

El

2.47 The igid rod ABCD irene fom the ios fo sr mei. The
tres setae of he wires qua to a oh eus arcs of
‘he wires dan Determine the lesion cach wie cated by e oad

SOLUTION

DIM +0 e+e

put
TA

Hno AR -LRIÉLPO
CELL

Ler Las ne tin OF THE es

2.48 Tb gd bar ABCD is mapendad om fur et ie. Detemnioth F
PROBLEM 248 ‘zion in ach wire expe ya ln

SOLUTION >)

Let © be the supe cf bar ARCO after E

deformation |
Sie Sie H
Se 5,+ 2L0
So - & + 3LO
= EA
BR: Bs,

{ 4 + EAs,» Es, , AL
Per SA Se + Shs, + Ebo

- ER + ZEAL 7
ia ae N
Ps GAS, + EBS, + te

ze: Pas P RP Pr 0
HAS, + SER Lo =P
4s +c6lo Ho Di
DIM=-o LR + ALR + SUP, - ALP = 0
SEALS, à MEAL O > zur

6+ te = 8
Solving (1) end (2) simudtanewst, Lo = 5 22

bh
Soe

| a: BER - Le =

o PRE E -

Po PE Ar -

Por EB sh - 2e =

( rr “snm at ea Ge need en om |
| (2) See ant + (emo We + HE He
I Se, Er Egg
[ B= se, © AOS sone LS
sure

[o [mo san < am ns + ane

or ut. sro won =

C

250 The am sl fly bonded othe base curs, nd te assembly is
PROBLE anses al a tener 078°. mining ony ata déformation eme
‘he ses when the temperature eh 180" (a) In the Bas sore) he

bent oluminanı shell
a, man
4 Donne AT= 180 -78= war
ERED PONT 7 be the torso Force developed
ul in the tore

Fon equidibeiom with zero tete) force, the compressive force
es a Fe E

Strains Es a AN, ee E Haar)

Matching Eu = tn BD + ar) = - Br, + (AT)
(Art ER) Pe = (0 MAT)
Als Edt= FUP = 0.7854 in!
Acs HOS ~ ast) = F(RS*-Lo")> 4.1232 int
E IAS
, m
Tagen manana] e + Cesu (ice)
Ps 1.2805 x10* 4

Ge Re RUE er sr

Ge À A

a 251 Th rusa, «209» 104°C) fly ond tothe ste core (611.7

ERSTE) Decl eons tee ng Forte

Le Nm SOLUTION

u => Let Py = anal force developed in He steel core

rs For equilibrion vith zero ted Force the
compressive force in the brass shel ia Po.

=. Shrine nr E sogar)

m
ay + AT)

N Matching Es + Es
(AT) «= & + oylar)

ER CERA = aa AT?

Ag = (0,020%0.020) = 400 #10 m“

Avs (.080X0.080) - (0.020Xa020) + Sao “16 mt
m-asr 4.2 "10 /2C
Be GA 7 (So doo xs) + 22710 N

Tose ci à 1
Ay’ ELA © oo mor Goon) * Coso” X500x10*)

= (mamo NATY
Are 75.4 e -

EST

131.55 10" raro?

252 The cone paa (£.=25 la anda, = 99 107°C) res with alk
tel bra, each of 22m dater (E, = 200 Gea and a, = 11 I
termine the normal sas lud ia the sel and I the comerlo by à
tempe roof °C.

SOLUTION

PROBLEM 282

Age @ FA = GE (2a) 2280810 m= 22808416 m
Ace 20 Ay = 290" 2230840" = 55.2200) mnt

= S532 210" mt
Let Re = tensile Face devehpal in the emerck.
For equilibrium with zero fetal Force, tHe
compressive Force in the six steel pods is Pe

HAAN, &r à + AT)
Hatching? en & & 3 (AT) = = a + ar)

(CRE ALERTE
\ ı
aaa?) * Croma] À» (mis Was)

Fe = 21.6/xJ0 N
a A

A a + Os MPa

Goo A + ro Pa = m

O

2.53 A rod conning of wo clinical pains AB and AC heine ota

PROBLEM 253 ads. Porion AB ls made af sel (2, = 29 10 pe = 63% 109"P) ad porn

Ba made ttm (= 1510p, 104 X ICO) Kong are de
Aero demi (4) e tra ra duel apa AD
by temper ri 0145) De conepondng ain pie

soLunoN
Au = Eda = aso 1.2272 int
Aes Yan = Beas} = 3.976) int
Free thermal expansion

Sy > Lag ls (AT) + La (AT)
= U2X6.x15* (ES) HOS MIO eo) Es)

Shortening due + induced compressive Farce P

= Ble + Blas
Se En A An

PE Ps xo”
Gaara sra © S88 SANS UP

For zero net deflection Sp” Sr
(628.674 )07)P = 18.21% 10"
Pe 25.86 «10% Ub

- M à rio? paro = i

BERLE a rio pa 21.1 he -
Saum! = - C50 no pi = -6.50 kei -

= + Ple _

= + ER Loe (ard

2 + RE. 8108K12) 10° Yes“ +3,64 710° im
RN + (AXE SxI0* Yes) á 3.410 im À

8.644108 in +
0.0036 ia 4

254 A rod coming of wo clindrcalparone AB and BC sets at both
PROBLEM 24 es, Pon AB made brass (E108 GPa, = 209» 10°C) and pone BO

‘Sime flumisum (6, = 72 GPa, 239% 10°C), Kaowing that he eds
{nl unswesseddtenine () De nörma trees induced pomor AB and PO
bya temperate rie of 42°C (0) he orespondia decimo point

SOLUTION
Ne dia lo = 2.8274 110" mnt
eet Fda > Hlio)* = 1. 2566 10h
Free Herma expansion

Set LagOhAT + Lect Car)
AX 20.9 OX) + (1.2 Nas PD
= 22705 x 107% m

SAS HUG on

Shortening due + induced conpressivo Force
Pla , Pla
Haut ER
MES a sp
Rs”) G2 Kw M25 o

P "Tor
= 18.0744 10" P

For zero net deflection Sp
18.074v167 P = 2.2705 «10
Pr (26.62 xl0* N

Sr

A 2 yy yes Re —
CN Gus vie SB 444 do Pas 44, MPa et

Sue E RO

100.0 MPa, e
ee 1.2006 v0" Pa

(o) Sa=+ En = Les (AT?
AE NY OE

Cos it 2.8271)

= +600 4/07 m = -0.500 mn

28mm6r |

£. 0.500 mm y -

uy 2s 285 The sembly a consi fan henna aa
PROBLEM 255 129 1099 ally bonded tom mel core "29 10
inte. Dern de gs alow ang pei eset
inthe alnlaum sel nt enced GE (9) e sumepending hang in

ES (Oye samespondingcbnge in en

soLimox

Since 04> 0%, the shelf
poutive temperature

het @ =
As EG dd) + FCi2s*- 07s") © 0,7884 int
As = PA Fos) + 0.44194 int
P= -GAL= GA where Pis He Hunde Farce in the steel core |
Ge = - da, (Eno rést acero psi

As 0.4174

in compression fon a.

6 bei = -6 10% pui

es Dear) = $ ram

Es
(an = & - €
re ANOS PTA 1e
(6.4010 (AT) = bla 4 SIE = 0. meso
(a AT = Mar -
= A (ge - ist
(a) E = nor ESAS) = 13168 ve
or + E aMSN = 8163 o?
S= Le = Goa) Aor in. -

253 A rod oasis of wo clinical portions AB
‘xcs. Pron AB em of tel, = 29° 10 pa,
BC amado lbs (E, = 15» 10 pda, 104 1

AC cine bota
65 x 109") wd porion

Kong ht Oe rod a
il unsresed,deiomine() tal tence induced spoon A an BC
by anger SF.) compos deletion pit Be

1256 Fec trod of Prob 23, semis be maximum allowable temperate change
‘tthe srs in he tt ren Ab nt exceed IN and res athe bss
pion CB isnot to exce Th.

SOLUTION

Alesuhle Force in euch portion

AB + Garn textos, Ag? Eola >» Flas)! + 1.2272 m*
P= Sub = (- 18108). 2272) = - 2.070 x10% de

Tiei = -7mto'psi, Ame Eder > Maas‘ = 3.9761 in‘
P = Qu Ân = C7e1o*Xa?7e) = - 27.858 “10% A.

Smaller absolute value governs 2 Pr 22.090 x10° Sb.

VeFommation due + P

> Pla o Pla . (MN row Xis)
A SN TOS)

= = 13.00#x10°° in
Free thermal expansion
Sr = Le ATI+ La AT) > (12 ME SUS XAT) + (is Mio. qui KATY
= (294 «10 (AT)
Titel defarmation is zero
Sr+ Sp > Castriot WAT )- Ia soul = ©

avs 55.6°F -

53

PROBLEM 27

a

Shortening due to i.
Sp = 17280?

2.57 Deine) e compressive usen ars shown aer temperature se
‘51 96°C, (te coespontin change in eg of De bone Br

souvron
CaPosite Free Heemall expansion
Sp = L¿0 (dr) + Lada dr

= (0.35 Wanéme*Xne) + (095X23.2.16%X96)
= 728% 107% m
Constrained expansion

Sr OS mm = 0.500715"

duced compressive Force P

= 0.500x 10° = 1,228x10*m

But, in terms u i

- Ply Br

5 Ret = Ce + JP
paa

AN a
| = 56496 wo’ P
(a) Enueting 56640" P = 12280 Pe 217.46 o? N

an a N

CEE hour) Ela

= (0.35 lane =

728.76 «10°F

ae (gine «10? Mo. 85)
SN REST

TES TRES
20-2925 mm =<

a 238 Kwame nese poe 21 ac
] rom a wena nesses be SS
5 e ET

5 Arosa sn SOLUTION

1 Gus Foor As momo nt
2m ps ,
m o Pe - GAL = (040% (1800-10
Brie See
LU Shortening Wve te P

DEPOT

- Pla, Pl
P e #7 BAT EAL

2 i „(lea Yo.ss) „ _Lie2xio: Yo.ys)
(CINC)
= MAT MIO m = 0.91974 mn
Le fength Fon thermal expansion

Sp = OLS mm 0.19 mm © LENZ mm = INTA RIS m

l But Sy > LACT )+ Lowry

(0.35 Xa1.6 to) AT + (0,49X23.2 rio) AT
18.00 xJ0" (ar)

=

Equation 18.000 (AT) = Lua? AT = 786 te
e) Tra > Tou + AT
J = 204786 + 800 _

1
j 4 5, = Luür- Ee

FC = aaa) Samer os)

n = 820.510 = 554.79u10° = 266.78 710 m

kant = bet Se Om 2678800 m

= 0.050266 m = 450.0266 mm -

E

5]

25) Atroun temperatur (20°) 4 0.02-n. ap exis heveen the eds e be rode

PE ‘shown. Ata later time when the temperoture has resched 320°F, determine (a) the
un tox cease sana od te Gage nego sein fox
& { SOLUTION
AT = 320-707 250°F
| Be Free thermal expansion
ETS Spe Le (AT d+ Llar)
SE 28.8 «10 JAS)» (109.6 « 10"* Kaso)
= 63,9107" in = 0.0689 in.
> P Shortening due to P de meet construint
— Sp = 00634 - 0.02 = 0.0439 in
= Pa, PL (la, Le
$ o ke

(a)

er

da is e
es es)? = Benno" P
Equting 702.7110 P = 0.093 P= ceso e

2x0
2.3

= 22.09 o pai
= = 22.04 ksi _
%= Laan- Ple

(a\ea.sitorKaso) - LE u

84.90 ot SO et

= Dom =

PROBLEMA 240 br ik (E = 15 «10% pa, = 104 107") ad el rod, = 29
PROBLEM 260 Hei. = 63 NOE ave incio doma emp leo ds
sel ad cold cf ay int the Ink. The tempera old whee
‘ semi is then raed 1 100°. Determine (a) the sl normal ste the el

mue (3m, hhh nal ng othe edad

F3 souvtion
su ar ed with diffe bei Find
M associated with difference bebveen
ota {Livin} 25 dando and initial dimensions
=> i»
Lo asa AT = 100 - 45 = 35*F

Free thermaß expansion of each part
Brass Link (Sr); = OLATHE) = Coin" Mas Mo) = 3.04 210° in
Steel red — CSr)s = Oy CAT XL) (6.5100 JC Kio) = 2.275 #10" in
Bt He Find tempersture the Free Length of the stee? vod

0.008 62.295 WIS? = SHO"? = lo in
Donger than the brass Pink

Add equal bt opposite Forces Pto
ePongate the bears Pink and contract

&3 the Shee? rod.
Ea oa teat

(Gye Pee
rie’ RE © Sas)
PS = Heth xo Pp
- = PO —_
Stel ro (he EE = EG

= 70.94 «10 P

(Sp HS), = 2.685 10

(392.101 *)P = 3.0250” Ps 9.2708 «0 th
(a) Final stress in stef rod & À - “RE

= = 7SSMOŸ psi + = 7.56 lai =e
QW) Fined Length, oP steed vod
Ly + 10.000 + 0,005 + (Srl (Sp la

= 10.006 2225 #16? -(280. 34 = 10" )(9, 2705 10)
= 10.00467 in. ~

Ga ed = 1.9» 109°C) ar eed rare a
PROBLEM 241 teassbr (= 105 GP 2 -209 = 10%) ic bu sion P= 2808

‘When th se! bars were fbi, he distance been Re otis fhe oles
which veto ton pos ws made rm alr thane made, Teel
as were en place man on incase ciego Ma hy would ju fot
the pis. Following bricano, he temperate no tel ar oped back to
‘oom tempera. Determine (o) de incre temperatures vas equ o
‘hee ars the pins, (9) rs the ae ae te led is pi

SOLUTION

(@) Required temperature change For
Fabrication =
Sr OS mm 7 O5 110m

la) Temperature change required to éxpandl ste? ber by this amour

Sr La, AT — 0.810" =(200K NO XAT), AT =

0.5x10° (MMS AT)
AT= 2l.368 "0 ae

(b) Once assembled, a tenido force PY develope in the tee? and
o. compressive force Pr develops in the brass, in order de
elongate the sheed and contract the brass.

EPingation of steeP: Ay» (a\(sX¥0® 400 mnt € 400/07 we

EL Pa pana
A Ke o = 4707 P

Contraction of brass : Ay = (60105) + 600 mm = 600 * 4
Favor op
CT meristp

Bot (Sols + (Spy is-equal do the initial amomt of misfit
(Sp) 4(Sp),= 0.500, SEES’ P* = 0.410"
Ph = 28h N
Stresses due to fabrication
Pe) &. Sano?
2

Regul 22.03x10% Pa = 22.03 MPa,
: E IA AN
Sr 468108 Pa + 19.68 MPa

To these shvesses must be added the sivesios due to Ha 25 KN dood

continued

Problem 2.61 continued

For the added toad, the additional deformation is the same for
both the steed and the brass. Let S' be the addiHonat
displacement. Also, det Py and Py be the additionnd forces

developed in she steel and har, won pectivedye
u BL, BL
E Ae ME
A+ ME 5e Wong, ojo st
AS

Teed PEPA 2540 N

Howto SU + 315° 5" = Sue S'e arco
NES

Ar (aereas \r

1.486 mo

Ho 108 N

Be à nano Pa
ects .
gee = 18.36 ¥ 10% Pa

Add stress due te Fabrication
Gy = 349710 + 2203x108 = SONO Pa = 57.0 MPa

Sr 18.86 rlo%= mesnpte acsriot Fa = aCe MPa e

dm

242 Two mesitas (2,200 GPa nd > 117° 107°C or seo intern
Basar (= 108 GPa, = 209 109°) mich ab Id P=25IN
‘When th el brs wee fabricate, eds Denen the eet oft Bales
hich were to fon pits was made 0S men slr han the need. Teel
us verle plc nan rento nes ng tal hy noni ju Eon
the pins, Following fabrication, he tempe in he el tre Goppe bck To
‘oom lenpratre, Deere) reise empese hu was eed Et
‘Qe ta Br one pis, (ih es In a brs ar a ead epi o

romana

242 Detrai te maximum and Pd be paid 0 the bras bar of Prob.
261 ie allomable sess In se br 30 MS nd he albert
‘as bara 25 MPa

soLunox
See Solution to PROBLEM 3.61 to obtain He fabrication stresses
63 22.03 MPa Sr 1468 MPa

Alfoualle stresses? Gus 30 MPa, Say" 25 MPa
Available stress increase From bond

& = 80-2208 = 7.97 MPa

Gz AS+IRES » 8.68 MPa
Corresponding available strains

re PUS. mario“

MEETS 4
ar FREE. ao

Shabler valve governs 3 82.8510

Aves Aus (AMBX00) r 400 mut = 400% O m“
Au= (S990) * 600 mm = Go0r10'* mt

Ps = BASE = (200x107 Koo xi (89.8510) = 3.138 vit y
Par Eine = Coss 10%)(c00 «10 )(a9.asxis*) = 2silwio" N
Total «Movable additional Force

Pe Pt PL 3.80 + 2 suelo? = S.70¥10'N
= 5710 KN =

See
a

2463 na sand eset so of in. dancer subjected tension
Tore of 17 kis. Kooning tht v= 0. and # = 29°10 a, determi (a) he
again ofthe od nan i. gape eng () Le cans dam of ed.

— SOLUTION
- Tact A. Ar Bats FG) + 0.60132 in"

N E. 28.27x10°
FER = marys en Ss BEDS

= 974.9 x10°°
Lee = eae Ie) = Tom + 0.00780 in, =e
E A

= (Fe aeswist) rasero” ine noise. at

acia 264 A sanded ein ft is set detras ce pops fan experimental

ve

G

L Be Aa Ed = Elsi

Pla. The tat specimen ea 8 m.dmeter m and ted to 3.5 EN
fens force. Koowing dut e logon of us and a eco dime of
D a oberen dam gage ls, toi be muaa of eat,
(be mods strip, and Poisons ati te mates

souunon

SENS A IETS eI
P= 26x08 N

ase A
SRA + Iori! Pa

Es E ¿LL acer mo

RS
= © Samen > cr joe =
Er E » Berek = ziert a + 216 MPa me
= 0.62 0m

El AA . 9 ugg -

E Caro

E UA quo Pax
REND) 7 o PLE Oz

45 MPa ~

e

286 A Zum length of en alumiur pie 0240 ute dater and 10 wal

PROBLEM 265 cine 5 ied u 3 or clame ad care à ee al oad of 60 IN
ming at = 73 GPa 033, determino a) chang in eng he pip,
mes (0) change in outer mr, () ang nal tikes,
F souunox

do 240 mm Ex10 mm dis d-2te 220mm
À = BOS dat) Flavet-a20!) + 72257108 mm
= LEST 10% me
P= Gore N
ng = Pb. TER]
q NS ETS)

2421 10m
-243 mm oe

AN u
es ee 1.2133 fot

Ege = VE = -(0.33)-12189»/0%) = 400.4 810%
&) Ade = dear =(210Kge0. qui") + 0.0961 mm ~
@ Ab = Heu = (loXtondate® ) = 0,0400 mm -

266 The change in meer of are sea otis creta messe athe nr
PROBLEM 266 "ptes Knowing at = 200 Pa and s= 023, determine the tea fe athe
tel fe inne in observed deco y D

sorurion
Sy = -1Bx Sm de colo
ey Be Bae en ae c7 wet
ur Auot. zur 13 or
G = Es = (2000 19.310) = 149.43 10% Pa
Ar Bats Fo = 2.827010" mm 2 227010" mt
F= GA = (149.43 x108 Marry aros N

= #2 ka =~

247 An minu pt (= 74 GP, y = 035) te sujet conc el
Jo wich aus anormal sess. Knowing before edi, ino lope 21
‘seed on he pte determine he ape of he ine when = 125 MPa

SOLUTION

a

tt
3
x

The slope after deformation in tan 9 = 20260

Tree
& uo F
er E BE + near

Ey + VE, = ~(0:88X 16892015") = 0.55744 10

= Se ©

A) e
ACI- 0.000687) | a
[5 tne EN = 09551
A A0 as gpl se op mad hu in pe
omer LE = 29 x 10 psi, v= 0.30). Determine the resulting change (a) in the 2.00-in. gage
[ A rte ie a cine
‘SOLUTION AB in consi arena ia
u Dre
ACNE) = 0.03125 i oun EAS u
Beni à
Se Kama" ar Ja
LE and | ge met
Ñ bet Se A cea. ors
Ka) Se = LE, = (ROMER OO) + 1.324 It in. —

Bytes VE, > -lo.soeczorio*)

E mr wir > Dec im. -
Mer Sanur de = E o in. =.
Do la Ar wer wordt)
je > Wot, (1s ey + fe + Eee
u AA = AA, wot, (ey +8. +68)
= AM YC 198,02 110" = 498.0241 + neyligiMe tem)
= 12.41 10% in Eu

ome

29 A Lin aque eed on the ale ofa age sel premrevemel. A

msn ‘pressurization the biaxial stress condition at the square is a shown. Knowing that E
Pen "pep 00 ne aan gal,
os
AMIA
E somos
EL te BGG) Hig flare ose Neue]
Tn > 85172 «10

> EG 2605 ag [od (aro ao]
= 2.7% "10"
(0) Se + (AB hs = (00 Mas1s72 10%) = 35.7 #10

4

A

We) Sa, ~(BE Ey = (00 M82.76x10") = 82.810"
lo GE 1= fie” + BOF = GB, + Su) Cie, + Sy)"
SS See
= [Os asa Ya (1122840)

= hase
A = Wa HAL). = 307x10% in. =

or use catevdus as Fous:
8

Leble sides osing a,b, ande as shown
Pea Obtam diMeratial 2e de = Zadar 2bde
Frouvhich des & do + bac
Bt ar h0om, be leon, ez da in
das Sq = 35.0% in, b= Sa = 82810 in
Se” de Elan wie" 4 (82,82 10°)

= 07x10 in. ~

x 269 A Lin square is cod on he de of a impo tel pee ve. Ales
PROBLEM 270 presario he biel tes coin esque eso Knowing tnt

1229 10 pa and v=030 determinee change intel of) ide dB, (she BC,
(agent aC.

170 For the aque of Prob, 249, determine the pot change in de spe of
(insel DB uso de preston ofthe ves,

= souvnon

8 LabeP sides O and b as shown. in]
/À The slope is oz “
The change in of J

ET ar TEE Fre diérea)

cubes dos
ds + ade . ds. a odb-bda . db_ da a
o > sv a" b a

To change in slope = ¿Ex 100%

de = E = FS, - 28) [ma - (0.206 11087] Ey
= 351.7210 DT,

de, - Eu)» zelnen] N:
= 82.76 niet i

de.

E = 351-1210 BL IO 268.96 107%

To change in aper 268.2600" %
”- 0.0267 % _ E

Ra
e)

71 Abi nl asc inated sacri bete able area

PROBLEM 27. da es = ID MPa a d= 1S MP. Konig tt proprio
cheri can te approximated Ga and v= 03%, deerme de ange st
erg) she, (ide BC, (ago aC.

SOLUTION

6, > Rowot Pa, G= 0, 6 = lost Pa

= 46, - ug - ve)

Lores - @anticono®)]
= 754.02 x10"*

El ve 45) par] -Civliaount) + 160 moe]

= 1.3701 vio
(a) B= AB Jey = (100 mm (754.0215) = 0,0754 mm -
(bd St CRC )Es = (TS mm )CLATONR 10%) = 0.1028 mm -

Lake? sides of wight triangle ABC as a,b, ande
chs ats bt
Obtain differentials by ea feudus
2e de = 2a da + 2b db

de: Sda+ Bae

Bot = 100 mm, bo 100mm cr TOO TET) = 125 mn

dar Sig + 0.0754 mm db» Su * 0.1370 mu

Su» de » 48 (vorne 15 (0.1028) » 0.1220 nm a

272 The bras rod AD i ited wit a jacket ui ed to apply an row
presu 06 Pt the 250-mm pron Cafe ro. Kaowing hat E= LUS OPA,
fed 033 eemine() Ine change Inte tot gi AD, () the cage in

PROBLEM 272

amer of prt BC te od
SOLUTION
G2 6 2 -p-2#vot Pa, 6/20

= $6. - gy - 16.)
Isar - 0.300 -(0.3018x105)]

aco

= 306.29 v8

Pl ve +6 - 7G)

= adage ante) + © are]

>

= somo”
(0) Change in dength: Ondy portion BE is strained. L> 240 mu
Sy * Ley = (QUO X-39L.7IxJ0%) = - 0.0724 mn -

N Change in diameter? de SO mm
Sur 5 > des + SOX 306.29 10") + -0,0/531 mm -

1273 Toe homogeness ps ABCD subject o basal dig aso. 1

PROBLEM 273 town tat = ada age in ergeht pe nthe diction mort

A zer, ls 0. Denoting Eb modulos els aby Pulse sale,
trie (0) the equi mude of , (pb rio 2

5%, Gro, ero
ee 10227 227952 20.2278)

la) E, = 26 =
[3
"ze

MN En El-26 -28 +6): ¿(ve -o.+6

274 For mente der axial loading. eps the cal ir 2 Ina dicción
feng an angie of 5" with ait of load ine thea an by)
‘ermparng Ip e Be gl shown ni. 254, whic eset
| Piel an element bie and er efit (0) ang the als of he

eC
1 ve] Rie
T a
@ Before deformation After deformation

[Re] = ans Oe
ala REEDS ia ze veh al = Ques votes

Mesa ts 2Ex +E une + WES
Neglect squares as smab? Hel = 28 -2v&
es Lee, ~

275 In many sions shown tat he noma rs na given disco nm,
PROBLEM 275 for samale 4 = Oi the cae ode lapa sham. or caso wich aot

plane re, sow Baie an gt have ben ra oprima,
Een eres q, and Klo!

eee ne ne
ir

ar)

« SOLUTION

CE
er) M0 grrr a
Mkipbying a) by » and adding to 0)

Eye VE = es or rer we) -
Mobtiphing GV by Y and adding to(2)
Bye Go 6 = ER G+ ve.) ~

Be El-¥6,- 2G) = -2
= - E (gag)

ER

(ELA Ey + Ey + Ve)

we) -

Phys cote prevent omega gen

nie case shown, er login more oe
«eve pe. Baro scene pemendic to the
‘optical u ei plane and be same tance apc, Sta tat fore is

PROBLEM 2.76

shunt, whichis Known plane ai, we can expres and as
0 0(0,+0,)

1
= rene) le + De]
5 SOLUTION

Hit Bros bl Ur E) or Gene) =

Bee HCG - 0, - 2 6,)= 41-25 - »*(5,+ 60]
=E[0-2)6, - (1,0 6,1

Elan) 2-06, +6, - 26 )]
lO 209006, ] -

277 Two Dock ofr, cach of width w= 6 me, ae bonded ii spp
‘PROBLEM 27 “ndo morabepte AB. Koowing at re of magne P= IDEN cases
aci de emis ada fy of te rbber ee.

SOLUTION

Consider upper block of rubber. The Force

sree the a seer:
= The kan
Al ee

tr ek
u
where A = (80m MGOm + 10.80 mn =
= 17108 Ri
depot 9.97968 x10" Pa
ys in BP, = onsen
|
Be arent „ set pate =
Ge SSIS” > 10.260 Pa = 10.26 MPa
ligado G = 75 MPa rbd trig spread
momia À moval nea Ecos ae wa ech cken en,

seco ig on Pd of pe
sounos
Consider the upper block of rubber, The

force carried is PR
u | fd dea

cit
a
A pe u
Ps 24%
The sheuving sirein in T= À fan diet 8h

Effechive spring cmt k= B+ 28%

Noting Hut Te Gr

k= 286 + ETATS ETS 3 Emo NI

eno kN/n me

279 ‘The pat block sown is bond to ie as ando arizona pie
lo white free Pi api. Knowing fr pl used = 3 a tri
Ibe deci of it win >= kp

PROBLEM 279

Consider He
Plastic block
The shearing
force carried L
is Pr gro.

The area is Ar (ASNESI 19.25 in!

Shearing stress Tr Lo FIL ver ca psi

LSE + 0.008506 |

But Y= Be Ss hrs (22 M0.0085006) = 0.0187 in. -

Shearing strein Y

2.80 A viral nt nos of vo Dock of hard ber bonded le
A cd tong spam um. Forth ype and rade a woah 230 i
{land 10 pt Knowing tart vera ee magninde P= 3 gs

‘at ae à in. vere decian of ee pue AR, deine the aci -
onde tension e 28 bof he eck

souunon -
Consider the rubber block on the vight. Tt i
carries a Shearing Force equal to 4P- !

The shearing she is t= dE

on raid A A gaa int
Bd A= Gorb 1
Hence b= À: 2.42 im. ~ |

Use br242i ond + 220 psi

Shearing strain Y 0, 12222,
; pr ar v-5
ll $ i Se Ot -
LE lence E ama, © po
+

La

PROBLEM 281 221 An castomai eng (0 = 09 MP) i wed 10 suport aide ide as

‘own to provide Nini ring sens. The eam must ot dpi more
{han am wena 22 RN dis cpl shows. Denne (a ese
owabledmenond, (0) desmalleareqursdiicinss abe mirame
‘honing tess 2023

soLumoN
Shearing Force P= 22410 N

Showing stress T- 420 v10" Pa

CB Ae Be Bae 2 oz. ss 11S

T 5 a = 82.381 110" mm
Es A = Goom Cb)
a RE -
? RS
Bt Th as 4.225, = 21.4 mm -

Ha 282 forthe clore hera in Prob. 231 with b= 20 mm anda = 30 mm,
tenio sean ous Cand the har uns fora mamon ae oad
Pa AN and a mum deplcenea! de 12 nm.

SOLUTION
Shearing force P= 14x10 N

Area A= 200 (220 mn) = galo mn“
= Boe ot

aL. Maoh. 431.510 10 Pa
A is > 481 kPa —

Shearing strain Ys £ + Ban = 0.400

ETS
Shearing modulus

e .
ot 1.080 x0" Pa

= 1.080 MPa =e

“2.89 Determine the Jai nd te cane in lu ofthe 200 mun eg of
{teed shown fo) e red sod of sel wi B 20 a and 91.30, 0) be

Fe einen th O en v= 03,
PS EL sorumon
Hama | Az at Ear} = 380-13 m + 380.18 40m"
Pr iu N G= 7 + taoivlot Pa &=&G-0
Es 6-25 vr) $ EE re > VE

er Besse: lo). A

Volume Y = AL = (330.13 wm )(200 mu Y
AY: Ve

Ged stent: e = OOO yz ion

10 HOT
DY + (76.026 110°) 242 #1074) = 18.40 mm®

Ue) afaninum: + (0.302101 x108) |
jo ro!

AV + (76.016 rio? MSM) = 39.4 vant

76.024 wo! mn

4

so“

da

From

4

(a)

53 “2.44 Detemino te cng in one of he 2. pg ent segment 4B in Pot
man 281 o) y comping th maton f mal, 1) y strain de cgi
‘ele pari A era inal ne

PROBLEM 2,68

pa

A= (Ai) = 0.08125 int

Name: Uz = Abe =(0.03/25)(2.00) = 0.0625 in

(0)

LE 00 game” a
ai am? Mare pi 5:6

EUG 16-16, )= Gee RHE + ce2.07 mos
Ey = Ez = VEL = - (80 V(ce2.07m/0%) = - 198,62 10°

er EL te RES aso”

AY = Vie =(0.0625 Xas0 230%) = 16.55 010 -
From He solution te PROBLEM 2.08
Se = 1 BH IT in, ST — 99. S41 in zarte in

The dimensions when under à 600 4b pende toad ave?
Jength
width w=WwasS,= L- 9a.gxio%%
thickness B= byt Sy dm IO

2.001324 in.

Lars, + 24 1.324 HO"
0.4999007 in.

= 6.06249754 in
volume Ve Lwk = 0.062516539 in?
AU = V-U% = 2.062516539 - 0.005 = 16.84K 10" ine

‘SLES A Gin ameter sali st! sphere omord int the ocean 1 pin were

PROULEM 285 te pes
pi and yo
icio nome ote sphere, (te premiere he day te spe

Fore solid sphere VS 4? = E (6.00) = 113.097 100

Ge = G+ Op -p = riet pi

(abst 3 miles below te sures}. Keowing that 229 10
determine he dereaein im fe sphere te dense

Be BGO) = = Amp = LOM ae) ana

HET
Likewise 8,28 = 97.98 200%

Er HE +8 = 293.79 110%
(a) -Ad = des, > -(6.001( 97.981) = 588 107 in
(e) -AU re > -(118.0470(-293.24% 10°F) = 33.210 in
@) het m= mass of sphere ma constants
m= RU = PU = PAIE)
LB: Por Le es Fe
Po B®

“Gorey | RE"

lle lr -eret ety.

= -e = 293.79«10"
PAaßrıoot = (293.700 (100%) = 0.0294 %

4

4

==]

7.86 (a) Fort ail long own, Gaerne do change ii nd te change
PROBLEM 246 in voları oe ren er show (1) Solv pat mung ae ong

ET

Pre ah a qe = TON
SOLUTION
hye 18S mm + 0.135 m

ql B= scr

er Aye dr Baste SAR = see

Vaz Behe? 166.064 10 md = 766.06 41% wo?
(a) 620, G*-S8xK Pa, & = 0
ECM + -15)= S

À Bl
> ot
Dhs be, > (135 mm C~ $52.38 10°) = - 0.0746 mm _
> lee LE) E - (O ES 10)
er Hé ea = CBD - (ene eo
= 17.810
AY © Me = (766.060 mt VC 197.8116" Sa
= & + -70x10%Pa 6; + 6 + 6,2 - 21010" Pa
HE + Sy - 9S) > LE G
= Gsm) | —
ee = 16.67 HE
Bh = he Ey = (RS mmdC226.6710%) = = 0.0806 mu _
Y METETE of
e BY (5 + 5y+ &) = Tos rior = -c30 mt
AY = de = (16.06x10% mt -680n10*)= S2 mm? -

rine A Sn de ee ar

Let à be à radial zonrlinahe. Over the hables
vobher cylinder Ré VER
e Shearing stress T achm,
on à eyFindnicnd surface
rad is

ae gobs
1 TR za

EU ie chemins ehren ie
1: de
ng debormatin over radial font dt
ger
ds. aes he
Tate deformation
5: (as

EN
wer à
“ons, RT Ne in,

arg, Br 0

PROD à sen hr anche =

Let be à dias covedinate
votes cylinder

e

Re ver,

édit is

Pe

ng cru

FRE
Inés + 200$. Use Mo)
ar 27008

exp (none) Gull

r A
H te TX “ami
tule The she

REA O2 LE Dama ed voca has ace

+ One Ha halos

Sheering stress Y achng
on à eydindnicnd surface

©: mon

19 compost cube with 40mm eds and the pops sown e made with
gs polymer be aig inthe» direcion. The ee I conse guint
deforma in th yan eos de bed stale lado 6s KN nthe
“direction, Determine o) de chang eth length of he tenth eon)
Fesses ae,
Esc m0
BIG pce
ERC neo

do -shrain equations aye

ca 0 o)
. . @ o
a Ya ta
e ae e
The emsbraint combtions are 520 ad &=0.
sing (2) end (3) with tHe constraint conch Hens
LE,- = de
ES ha e a
“ES ESE)

By WHR GL = SUS, or G,- 0.4286, = 0.077216 6%
Bo e da = BEG, om -0M8G + E = 0.074 6
Solving simodtaneous by Sy = 6, > 0.134993 &
Using CY mt CV in ES BG - Be

E. 2 & | i- (o.259Xo.134918) - (o. 254 (o.134998)] 6

> 2.93142 &

A = (4040) = 1600 mm“ = 1600 x10°“ m’
ES

continued

Problem 2.89 conhinued

2. LAMAS got

E
GS Ly ee > (YO mm 7561810 1= 0.0308 mm =<
©) Gy =4o.615:10 "Pa = 40.6 MPa =<

Sy > ©, = 0.134993 J(40.c25%10%) » 5.48 x10" Pa

= 5.48 MPa ~

PRODLEM 290

1290 The compost cube ofPrb.2.89 contained ag deformation ia he z
‘rection an elongated the ot by 0.035 mn acta eee ia he +
esi Deine) hese gy de, (9) cage n de menos

Remon ma
ach nam
Buen oa

souvmox
ay @
“aaa o
eee (a) gx ©
Consteaict condition 8e 0
Load condition Se

From equation (3)

S®

Ya Ez
Es

oros

Continued

m
Ot
-|
Problem 290 corhinued
oi
O
From equation (1) with 6y=0 N:
at Der 26 - r
Ber ES - 26 ES we. 0
= Ele o.2516,] > £li- (0.190. 077210] 64 “|
2.93081 6, 7
Su = Es |
wt en $= S88" so i
m 6, = (Somtor ars 10") > 44.098 wich Pa» 44.6 MPa e n
5-0 - ll
6, = (0.077216 Yun.easuio‘) = 3,096 10% Pa = 3,45 MPa a 7
a
.--2 be. « fa
um rt Es y
= (0.25%) (44 c2swio) y o LOMB MAME 10)
Soo rer
= - 323.73 xı0*
lg: Womens) om = |
A
y
|
fi
n
1

pee 231 Show st forany given mil, be ao Go he mods ogy over
he modulos of easy salas os a $ but mere dan [ir Refer to E.

(243) an Se 213]
SOLUTION

ELE E.
cn + E = 2010)

Assume 720 for almost al matenints and 7 <t fara potitive
bulk module

Apply ing the baunde 24 Eve 2001403
Taking the peciprocih 42-275

ow teed 7

292 The mat rn Gk and var relates by Et. (23) and (29)

PROBLEM 292

‘Show hat anyone fe con ma be crec a tema lay che vo
const Fo example Bow bt) f= GEASG-36) 180) ve GE. 2G)6E>
souunon =
FD)
wet
ae
eT = 286
30-117 Siac - 28 1G " ee ce
- ES Br
4-68
a Lo 20m
67 30-20
8k- Gkv = 26+26%
8k-26 = 26+ck
Sk - 26 -
Yee ae

Bae Noise ease er ore
SOLUTION
(ar Athole À reich
a= 3-k = 25 à
Aa = dt = (250K) = Nas

din

ás -
CRE

Leds oio rom Fig 4ëta Ke 2.70

Gus Kö = (LIE + 14.04 dei -

(ey A bole B re dlis)s O95, d= 3-52 1.5
Ant dE = USKEY = 0.98 int Gus EE + 8.667 asi

3:28, m Fig 2a Kr
$228.05 Fun Fy acta Kr Zio

Suar? Kö (210X867) 7 18.2 Wei -
iaa 294 Kong at a = 16 kl, em Ds marian allowable also de

‘cot e P
SOLUTION
Ab hole À ve ti

d= 3-4 = 250 in
Ant = dt © so KE) = Ian"

=0.0 Fron Fig2cYa, K=2.70

Fe

ous KP : Pr Age. La) © ur kim
At hole Bred (S)= Om, de 31.5 = Sin
= dt = US UE) = 0.18 u".

5 LE 05 muy 2cta Kerze

Pr As. GO 5.11 sips
Smelter value for P controds Pe ST Kos -

=)

2.95 Koowing da, or pn own, ee tr 25 MPa determine
the maximum sllvable value of? en (o)r= mm Ö)pe ibm.

PROBLEM 2.95,

SOLUTION

A = GOXIS) = 900 net » 400%

tm Gr Be eo

Fon ig 2.06 ci ae Sue KE
= Au. (200 yo Viasno*) | s
Ps Ans. a = S86 motw
= 68.8 kN -

Em +080, fon Fig 26h Ke as
(santo Xiaçuio)
oe

= UU N = CLE KN a

1296 Knowing that = AN, dete be maximum res when (2) = 10
PROBLEM 296 win o)r= 10mm,
us r= locam (d= 1mm

souvnioy
A= KoXıs) = 900 mt » 900 X16 m*

2== = 200

(ar Prion + des nor

From Fig ae Ke zos Sy KE
Guen GONE) _ grow Pa = #70 MPa a

Ho vio

rel Ee ln à or

Fron Fig 26h Ke 1.78
, se) a

Cg? A O Pa = name A

Orit $2 BR à om

Frm Fig 2048 Kx LIS
u AS 75 910 Pa © 73.3 MPa

noO

1297 Kooi a ava dencia in, deine (tras he

ono fies vi same mu es ona tele and le.)
Pr een Slows o elle b Si
4 sourrion
Pr For the circuler hoe ee ME )S Ours à
{ de 4-92 30s Fs GRE
] Du meute Gene 1357
J From Fig 2640 Kun = 282
5 = er
E a Bat
> Ant „ (ISI „ =
W Pe A = CES + 7.23 Kips

(a) Fe Rd Da din, drasn Fs EB co
Au. = der as x= 0.9875 m

Sue = Kent P Kain Sure à LOST)

Ren u: u

From Fig 2.64b 1% 0.174 =(0.1712.5)7 0.43 in 4

Kee 1.945

Pe 200 Bech 2 AS Ke, dermis imam pt ihn ehe

sounox
At He botes Me kin dr RO = LA in
ee oan

From Figzeta Kr 222
A KE

owe KE = ste
te
Aa the Set D=22m, de
neg ows Bs 085. 0.030

Fron Fig 2.646 E
. KP (4.3089
as" des

The Jemer value isthe required wini mum plate Pieles
tom -

0

Sa

Fron Fig 2.646
Aus Ede UsWgorr Tran © mao tnt

Wo u KE pe Age. Weng ent). io W
x Tomy

CETTE

PPA,

Le sou Ye E A vos = 15 m
Ta dano > EE) r00% + 110% -

oe”) eg ne rom

Mike luk: rede dr 9001+ am
roms Fon Rg acta Ke 2.65
Bap À de OS MIDE 1.08 1 orte
Sue AE

eso iss 10
Ec

Pe 256710" N = Ss WW

ne Bit Ds nom, de gun, Boe

Mom FB sam ad Ke 20

= 1.283

Rant Bd A 1 + 2600 mt
ge
uf

pe Any. MaS co à ga» 0 w

Smaller sabes for P controle Pes kN -

| SOLUTION

|

2.101 The 30 nm aque br AB ss eng = 23m his ofa mer
PROBLEM 2101 E Al

pos same 1 de liplsie wih 5 = 200 GPs nd ey = 345 MPa. A face Pt
plied othe bar pl end as moved down by an mona 6. Demis te
aim vaa of the fre P und permanent e e beer the Te ee
‘been removed knowing that () a+. an (0) de 8

À x (80X(80) © 400 mm = 00 KO

eue = AS LEO aaa" aan

2 TR Sn 2 Sy Pu = AG + (R00 «15 X34svi0") = Slo. SHIN

r Ubi Sr Be Gh gy gras mn
5

@) Sat 4San> Sy Pur BOSTON + 810.5 LN =
Span = 48mm = BIS mn + OS mn =
D Re kane Sy Per Blog NA Bios kw =
E ae + 4208 mn -
210% The JO qe ua ar Ys ong = 25m mec at
FPE Ma ost ih = 30 Ghat à = She Are Pi
pal th baad an move 1 give amet 4 Dome
TT mu eof ce ade misc by wh nb sed

@) 37 3.5 mm Sm? SS mms BIS RA > EN mm

(b) $) > 6.5 mu Sur Sent RIAS me = 10.81 mn

|
L Is When Su exceeds Sy , thus

then desd ae of is (2) 33am, 0) Sm,
SOLUTION

A =(ol(ao) = 900 mm © Goo 10"
Sales LE CANO e said à ans mm

rodveing a permanent
stretch oF Sp, He animan force da Y

Pm = AS, = [900 wid" Mado") © $10,5¥I0N » Blo.s LU me
Sue erh

4

i

PROBLEM 2103

8
Ll

22109 Rod AB made of ld cl divas etape with =29
AG Maud» 36 XL Afr he ud hs be atch ai lever CD. m
und tat end Cs zi. 100 gh. vera farce Qi then apple Cu dis
Point hos moved o pion C'. Dotmins th regu meni of Q nd the
‘eon dite eters to snap ck o a oral aan a Q sema.

SOLUTION

Since He rod AB is do be stretched permanently,
the peak force inthe vod is P = Pr, wha

Py» Aby= FG) (ac) = 3.976 kgs
Rederning the Free body diagram of fever CD

IM= 0 23Q-MP=0

Qs Eps Haga. Les wise
Doing ondeading, He apringheck at B is

Sa” Last r Be = GONE) 0.0746 in
From the deformation diag vam

She 6= $85 fe SES: 0.017

72403 Rod ABs ode oa el thas nando be elogia wid E=29
2 10 al ná 9 » 38. Aor the rod has bon stacked roer CD, iit
found at end Cis in to high À erica Tare Q then applied at Col his
pain ae monod to poso €. Deine the requ magne of Q sad the
‘lee di he ever so sup bak han pal ale Qi removed

2.104 Soto rob 2.103, suming a he il pin of hemi sts sede 30.
SOLUTION

Since the vod AB is do be stretched permanently,
the peak Force ia the vod is Po Pr, where

Ps Asp = HaY(S0)= 5.522 kin
Referring te the Free bedy olingrum of deen en

FM =o #a-21P-0

a: Er = SUE). seek
Doring unfoading, the springhack at B is
Lin

Sa = buety= ba. Leo God) 1034 in.
From the deformation diagrum
Spe 07 B= Hs $= Boyz 0.1552 a

ROBLES 2.105 2105 Rods 48 and BC us mad of mid tt same ode elastepatic
in = 200 GPa and = 348 MPa. The ods are sche uni end has move
oa mm. Nogal sees concentrations, determin (a) de maximum ae of
‘efor (0) he permanent ct case at ols dater he fr ls been

mowed
SOLUTION
E E E SS
EN Pre = Ama Sy = (062. 1910 395 10%)
» Danse = 332 KN =e
asm [sand (6) Spring back Sn EE Elim. à)
EST La
x ce a? a
" + pes = 2030
Be point A Spt Bu Sr Amm- Zinn = CBT mm we

1

At pent Bio Mo yielding in BC; bee Sp ©

‘2105 Rods AB an BC rade of aid sc ut sed 1 beastie
Ai ~200 Gand = 245 MPa. The od we sh vai ends moved
‘down am. Neg ers concert, denis (o) o pim vale of
one, (0) e pemanen se menued pois 4 and 8 wer he fre een
removed,

PROBLEM 2106

2106 Solve Pros 205, anomia il poof he ml sel weds 250
mie ant a.

sorumox
Ans F (0.025) = aezirio tn Au = Moos) 159040"
@ Poux = Amin & = (962.1 10" )(250x 10%)

= 2005300 N = 241 UN sn:
a ‘x Da „Ble „ Pins
(6) Springback 5'= Eye + Ele Ge =)

gi = 210,53%107(_0.8 12
ENE

= 1.908 x 10m = 1.908 mm
Ab point A Sp> Sy-8"
A pont B, no yielding in Bt

Anne LICR mue 7.0% mm

$, =0 -

2.107 Each fe tire Goierri ade fu elegante
‘eal wc 0-39 MPa nd B= 200 GPa A frei pli o rió
Br ABC ul he barbs moved dd diam 0 Zum. Rearing at De
‘ables were intl nat, dicen (a) te maxima sl of, (Émis

tcs ht or in cele AD () he Gl pace of he a dr do

‘mod (np cable BE iret tit)

sorwrion
For tach cable A= HloonŸ = 2821900" mt
Strain of

ey E. Sere,

Strain in cables AD and CFE E? Eu
Strain in cable BES dues Be An = 250w0

Le
Since EE Ey, Sin = Em © (Z00m07N( 1.25410") = 2S0x10 Pa
Since Eee 7 Er, See

Forces’ Pag = Per = A Gin = (28.274«10°*)(250x10") = 7.0685 vic” N
Pes = AGae = (28.274 x10 M24sx10%) = 9. 754Swio® N

ATA

For equidibniom of bar ABC Past Peet Ra -P = o

@) P= Pao + Part Pu = (7.0685 + 7.7845 7.0625 )x 10” m
= 23,9 elo? N - 23.7 UN =

WG = 2S0xO Pa = 250 MPa ~

After ondoading P=0
Cable BE is not tust
By syumetez Pho Ree

For equihbaum Par Re = O

| ©) Fin? displacemed 5 is comtroffe# by the Final tengths
oF cubes AD and CF. Since thistcables were never
permanently deformed, the Final displacement is

S> Ses Se * © -

A107 Fah ttre 6 diameter sel als ls made fa copie
‘Bal be which = 45 MPa md ES BOO. As à plo
ADO nl th bt ta moved vetar dre dd Ronin Ut
SE sere AI tu One (2) he masimun vl of 0 mie
rest cr in ale AD. (0) he al places bar ae the oa
Res ir input abe BE wet a)

“2108 Sve Po 2.107, ming hat he cables ae reply rods ofthe same
ac aca andar arbre are ake no
LT ETES

souvron
For euch vod A» F(0.006)" + 28.274 10% mt
Strain at initial yielding Ep? Le PENSE sao?

Strain in rods AD and CF:

Las ao

Strain in rod BE: &

Since Em < Er,

Ep = (Room M2S=10*) + 250410" Pa

Since Ese > Er, Gar > Sy = 345 #10" Po

Forces: Pos Pr = AGo = (28.240 Mason): 70685 xd N
Pan = ASH 7 (28. 2749100 (BUS HO) = 1.7545 IO? N

For equilibriom of bar ABC Pat Pre + Po = P = 0
Ce) P 2 Bet Pat Pe = (Roce a 9LTSIS + 20665 Ju N = 23 à

() Gy = 2504105 Pa + 250 MPa =

Let S'= change in displacement during undonding
pas + EA yr La gr o ssspunt 5! +R,

1600 105° =
nag = GA se ae. ocaso 5°
For equidibriom P's Pal + Par + PS 7 14.137 010% 5"
Bt P- Pio Pie Ps 23.870 m
28.39 +10?

Serós 7 ton m
Permanent displacement oF bar

Sent = SS 2x10 - LEON IS 0.310 116% m

0.310 mm ~

nos ‘sectional area of 1750 mn”. Portion AC is made of a mild steel with = 200 GPa and
Eu ndpren je me of rat = MO ras
ACES M As Psp Ca ont, Asa bh sa oe
Hd sopla decis) ds ositos off Pec ceed
Fa 1 109751 an rc ma es ch
soma Perlen othe pest dees
+ soLerIoN
El Displacement «HC fo came yielding of AC
Tb Sete, = me. Gone. Green m
Correepending Face Fie = Aga = (1260010 aso mio")
= unsern
=. EAS. „ _ ooo" XnsomoXo.221949*) | ;
fa BAB pe Ben
Fac
For equihbeim of element at C
€ Tue Pertiemo Ra RFs sismo N
Since applied Saad Pa MEIN > Sch,
portion AC? yiehden
Fa = Fie P = 487,51 - 97S 10°N-= - $37.5 WI" W
= - Gales, (627.540 (0.190) wo
car 8+ a RS oo
= 0.292 mm -
(6) Maximum stresses O7 Shap = 250 MPa -

2109 Ros 48 condes of two eric ponlo AC nd BC, ch wi a eos

Su: GA. use! - = go7.mvi0* Pa a 307 MPa a

"so o
©) Deflection and Forces for mfouting

Bel . - Ral 2 Age q.
os Se Be nent.
Pie mex = Pi - Py = am Pro = 487.5 x10" N

gi = UM race m

oo oso TEE]

Se= Sy = Sr 0.111770" - 0.204618” = 0.02715 10 m

= 0.027 m

-

o 2109 Rod AD consists of wo yl portions AC nd BC ech wi nee.

TF

+

L

‘eco tea 750 ma. Porton AC made fa id is with = 20 GPs
1977250 MP, prin Be cmo of high strength ssl with = 200 GPa td
= 345 MPA lod Ps spl at Ca shawn Assuming both ef be
opi, din (the maximus defloen of HP 1 ray etes
fro 01975 EN and thea due back o.) de mas mum are Inc
orton ofthe rd, () Be permanent detection af ©

2.110 Porte compte od of Prob 2.109, Pi uly increased tom ero
‘mt the deflection a pol C rss a murio vaa of A = O. om ad ee
seed bck 1 eo, die, (th mai vale o) e maximum
Ses in ec nn e e) Be pm econ el Car eo

SOLUTION

Displacement AC ig Su 0.40mm. The corresponding straint are

= Sa 280m ergo?

Le? Tao ne
bout - BE = Some. 1. s789 ni
Strains at intial yiofling
Byes Site = BO nass? (yielling)
A)
fa) Forces? Fig = AGy = (som NRO nO) = ot m
Fa = EAE4 = (200K10")(1780 x10 1.5780" de -552.6 1107 m

Fer equilibion oP element at C Fin Ra- P20

Pr

I Streses: AC Gr Sune

Pr
si

Fac Fen = 482.5 10 4 552.6410"

190.1 ION 390 kN e

250 MPa -

ce um EEE à 1610 Par 816 MP a

ion and Forces for undauling

Belle Pelo. PL

“L
EA ER Fe Te

La >

AS E 2 AE N
(IS.0s x10") (0,190)

= 74 x 107 m = 0.26874 mu
(ATI

0.30 mn + 0.26874 mm " 0.081 mm -

(a) Forces

(0) Stresses
& =

UnPoading

PROBLEM 2.1

| Sa< Bn < Sa The ml steel yields. Tenpered steed is elastic)

(©) Permanent set 3p7 S,-S' = 0.04 - 0.08094

ZINN Two tempéré br, cach fi. Bic ar bonded à ca milde
ter, This composite sujetos a shown oa cent ala lad magnitude
Bothselsareclstoplasi wih 5 =29 10 pe and with yu stengis cal 100
Jai and 50 a respectively, o be temporal sel, The lod spray
increta! ror zo un the defeat of he bu rss a mim vl =

OH andthe ere back zer, Demi (othe maximum valo o, (9)

gu Rerum mar br) pane ard od

SOLUTION

For the mdd eteed A, 2X2) + oo
Sn = LE à (SU _ 6 oeyisg à
E in.

me ®
For the tempered steed A, = alAYa)= 0.753

Sn = Ah. WISE) _ à
Sn > Se rd). 0.098276 in

Total area! A AHA, = NIE

P + A6 = C.co)(Son*) = Soxio* Fi.

PR, ES x (eu Xo22X0.0n N

Pe Pe Rs M2 MO = 2.1 hips -
Gs 2. Sp = Sox? psi > SD ksi

B= RA = agar pai à 8286 ki A

0.08094 in

giz PL, (sola
ER Gre 1.75 Y

= 0.0096 in, —

2111 Pwo tere br, cacho thick ae bonded 1. fin. mic
bat, This compost bar subjected as showa tow ani ai ld regis
Bathe arce wi. E +29 10 pelan mid olmeca 1
. ‘ond Ol spray, fr o tempere dll sa

PROBLEM 2112

ie E ip enden deren ac 1 zo, deine Ines fora ol

fa Me temanimam sues ithe emp el (the pemmanent eee

22112 Forte composite br of Prob 21116 gradually res fm soto
the eed demand.

ie
souriox
Areas! Milá steed A,
Tempered stee7 Arz AC (2)
Tilt REA SAL = Ent

er Total force to yield He mild steed

Gn= À à Ree AG SEO) 87.500
P> Pr, Heeb mild steed yields.

het Pis force carried by mild steel
o eut à Raped leer

P= AG = U.00(soxio) = Some? dh

Pez Pl Ps PRs aude sonic? = usr A

cs Sar BE OC donne à =
ty 6 E - cst 6 ei -
Unbouting 5° + ee SEM) = 0.02708 in

© Sy S,-5'= 003090 - 0.02708 = 0.00287 in =

En 2.113 Bar AB bas crosscctiral rn of 200 nf ands mado of lat à
sin be eaepaste with £ 200 GPa and 230 MPa Koomeng ta Be
Tore ieee fom 0 o 520 AN and ten deceso I er, deena) Be

E A 4 Permanent dez ef pa, the eel etre nd ar

soumox
== Am I A= 1200 mé = ooo mt
Face te yield portion AC? Pree AG, = 1200 15" (50 x08)
Phe Re = 300 xj08N
ee For equilibrio Fr Pog - Pa © ©

Ras Pur Fe 80048-52040?

= - oxo" N
= - Pole „ (20%05)(owo -0,120 3
= - Ese. Rl = 0.203338 ic” m
k ER" (20d "107 Wario) MASAI
= La, mou x
Cg = Rs TARR = 7188. 988 410% Pa
Unfoading
"= Pole Pal (E-
Bale = Palo «
% ER EA EA
(tu ka). Ela
Fre (& * ER)” ER
a (520 “10% )( 0.440 -0.120) 3
Pu = Eh. - ay = 878.1826" u
Pas 3 Pa F = 378.1840(0"~S20xto® = ~141. 318 810° N

Si Qu. et. A
Pat.

ET
E = O = users Pa

E nn

+. (378.192)C0.120) 2
Some < 0.189091 x10° m
) Bp $e- Str 0.213824
(od Gurs = Sy - Sir 250 lot - 318.152 wo“

0.138010 = 0.1042 115% m
= 0.1042 mm =

“65.2 «10° Pa

pa -65.2 HPa _ .

Gear * Ga ~ Sig = 193.200 18.112710 = 65210“ Pa

=- 65.2 MPa 4

—_ 2.119 Bar AB hac arsenal re of 1200 um! and ie made of asia
rooms sama tobe sioplie wilh E= 20 GPs and & = 250 MPa. Kann du he
fare F acess for 010520 AN and len decease I zei. dete the

em! din poi (9) resi res in her
Pa] 2414 Sole Prob 2.11, amg tt 140 nm,
Pood
al SOLUTION
B= 1200 mt [200 410% m‘

Force te yield portion AC: Pac AG

1200.10" Kaso rio)

300 10% N

lee à For equilibrium — Er Pag = Pre = ©
Par Re = F = Boni. Son

F - 220107 y

Pauken „ (220%109(0.440 - 0.180) „

> - Bates , (220 n109(0.490- 0.120) _
Ber TER 7 (zoomo™ Ciao wiert) = & 79898810 m
Son Ba --2000" 193.888 010 me

À 1200 xo
Unteling
Pal Paste , (F- Paddles | ” > Ele
a en ea De

ER

>Yoswo- one) |
Ro

Rés Pl-F + 30227340 Sao » 10% - 212.7271 N
+, (307.2710 \(0.180)

te Ele, (row

e 307.278 x10" 0

= 0.280455 ¥15* mm

&oosi0 ")Ci200 #107
Bal = 2022788007 aora Pa
Ri. se = = 177.273 710°

0.98 04USSx1® = 0.00 788 x18 m
= 0.00788 mm =e
WY Gum* GOT 250x10°- 256.061710° 7 -6.0640* Pa
= -6.06 MPa
Eure = Sig Sig + VIA RB viot à 177,778 10 = 6.06 #16 Pa
= - 606 Mee ae

(a) 8,7 Ss one

[==

O El

2.111 Two temperate! bas, cach in ho, ae Bonded ta fin leet
PROD us Pur, Ths compose aris bete show ential od of agate P
Bot clore shop wal 29» 10 pe on wshyildsenglanegel 10
A nd 50 ka respectively, fore ener ued mi se

ES Fo te composi a of Pot 211, determine the réal ins nthe
i remparts Prada ieee bee 20198 ipa en eee
nm ter

souumos
Areas : Midd steed Ar (K)la) 1.00 u*
Tempered teed Ay» OXÍXA) = 0.754"
Totodi A= ALA, > 1.7
Todas force te yield the mild steed
Me Ree AGH = GrsXsond)? 87.500" Dh
P> Pes He

Hare mild steed yields

het Ps re canted by mild steel
Fur Dre carried by temperecl ste?

P= AS = (o0MSoro) > somo” de
RAP, Par P-P = rd Korn’ = 4800 de

ae à
AL

= 64x10" psi

Unbending Eis À = Br! sevie qu
Residual stresses
mild sts? Eng = 5-8 > 508 Seo = -6 x10 par
a
tempered del Qu = 5-5, = 64x10*- Strict
OS =

2.111 Twotemperdstel brs cach fic, ar onda 4 in ile
a. This composi bas bete soto cic cl de magie.
olmo caso wih 29 1 pend with al sent ao 100
sant SO especie, rte a ml se,

PROBLEM 2116

2.116 Forth composi aia Prob, 2.11, determined red reses inthe
rene ele pad cessed fo eo unl fe deformation ote
de Perecer minor vale dy = 0.04 in and then deemed beck Io zero

SOLUTION

For the mild steed A, =X) = 1.00 in

Sn LE = Maa. 0.024188 in

F——

For the tempered stee? A, = 22 = 0-75 in“
= Ee. Aer 9, 948276 in

a EOS
Tota) eat A> A4 Aye int
Sn < Sn < Sn The mA steed yields. Tempered steel is elastic..
Forces P, = AS =(ooXsox10*)= Sorte’ de

9: EAS re) = Gera

Stresses Be En Sono pi
Sao? > 52.36 m10" pai

Unde > 64,08 rlo* par

Residual stresses
Gure = 6-6": Sox” -E9.0Bxo" = - 14.0890" pri
= 19.08 Ksi
18.78 10° pri
= 18.78 ko 0

Suns = SÓ 7 82.8617 ~ 6.0310?

m 2417 A vaio set od of oe anal ae is che tig supports and

SSunutenedatatmperstareof8°C. Tes aim elotes vi 6

280 MPa and G= 200 GPa Knowing ata 117 107°C, deals tra
inte bar) whe he leprae ned to 165%, (0) ale tempera De
repel 08°C.

L dE sonmiox

Determine temperature change to cause yiehding
s= -RE +Lalam = - EE + Lair =0

- Se 250 x10* - E
ro IÓ

Bot ATs 1658 = 157*0

© 2-6) = - 2 MPa =

A AE 22
£ = - £a QT)

= (20000 1.710 C17) © - 267.38 «10° Pa

| Sig = - 6 ROMO + 367.88 110% = 117.38 «10% Pa

= 117.4 MPa -

Ei Au fa steak pl

PROBLEM 2118 dw. al. ADO ah ono er, Roig Deere
stl don dd rs ato dario de
ines ft dnd ance) og ds
‘ite renal a Y sani eso Anz 6
Een ed CRD C10 ke dena Fre asin
Bean wi AA
Be

sorurion
Determine tempenstune change te cause yielding
$ = & + LAT), = Lostany,
ES = © = -Elóta- AT = - Sr

ot - were. =

MA aman Ic
(ad Tay = 7, HAT) = 20+ 120.04 = 140.087 € de

After yielding

5 = Ss Losar) = Loar)
Cooling

e: EE = y

Bis ig tbalaT) = Las(AT)

The rescdonl stress da
Gus SF Sy El du)(AT)
st Sr -6
-&% = Gy - Eldu- al MAT)

= RÉ. GKporo) = "
AT = aa)” one Ka nT Tay = 44.17€

@ Tir T+ AT = 204 or = 260.1°C =

TP Ty > 260.3°C , He alominom bar wild most Like

yield in compression.

ao

> oS

PROBLEM 2.118

2119 The sel rod ABC is wach vo ii supports and la unsre a à
tenues OFF. The el anomalie, with y= 361 and 229
A pn. The temperate of bt pores ofthe 18 fen raed 290 “E.

2 amp awe tate" 63 IR, demas de men ones AGO) Be
‘Sut ote
soumon
kun Son = Suns + Say 70 Coonstrnint)

Determine AT do cause yielding in AC.

Eur “KE - Ele _ Pla + Li CAT)
oo rn Eau” ER sat e
apa cP [te a ta
(aT) = alte + te)
A+ yield P = ALS,
1, = An Se (lu „La > (0.70¥ 36x0%? a SE
The SE (Re E “ao asar) año * TS
= 152.785*F
Achued AT = 250- 88 = 212%F > (ATI, 2 yieddiny occur
Gre =~ Gy = -86 lei -
P= GA = (séxiorXo.15)> 25.2 »10* dh
= - Bla _ı y)
Se = Sem EGE ~ Le lal
CNN iia
> BEETS - 14 esno ean
= 0.012176 - 0.019292 =-0.007116 in

Se 0.00712 ine met

2119 The sel 108 ANC i atacod origi supports and is anses
PROBLEM 2.120 fence oP. The sl ame cli, wth y = kind E

SA pi Te tempers of a portions o he rod is thea ated o 250
Ram tat a = 5 IDE, determine (2) the sess portion AC (9) he
se ola ©

"2120 Sole 96.21

suming ht the temperature of he od sie 250

Post amado 38
ese souvnon
n= Same + =0 Costes
e OS
+ Determine AT te cause yiráding in AC
Ble E Pla à 2
LR = ae lualar à
=P (ie à Le P (28
rta (at +32) = amar Gs
= 6.062440 P Ab yielding Pe * Gr Ave = (36210 lor) 25.210 A
UT) = (6.0629 110 Nasa ro )= 152.785 °F
Aetued AT = 280-38 = ar >(AT)y 2 Yielling occurs

HOGS KR)

TE

= 2.007116 in
N en +, AT, ar
Cooking ATS ai °F E
pre A „u TT ‘à
* Goes GoemriOS 7 34. 167 «10!
(0) Residual stress in AC
u agen! qi
= es her -
, Pin, er
Be Eko La alan
> guess Ra, ns“
Ea. quest Ya)
= o.01¢881 + 0.019292 = 0.002411 in
Sep = Se 48! =~ 0.00706 + 0.00240 = ~ 0.00471 in

0.00471 in

2121 Tao righ ta AC apport ik, AD and, wir 37.565
‘om rectal eros secon and male of «mild sel tht i tsuned 1 be
«lapa with = 200 GPa and 20 ME. The mapritdo a te force Q
plc at i duly oras fom zer 0200 RN. Kuowing at 0.60 m,
emit oes oral neh ik eam een
pos

PROBLEM 2.21

sourmon
Stehen? IM =0 0.Wla-Ad- 264 Re 20
Deformation? 572.640, Sa = 40 = 0.640 9
Edastic Analysis

A (375 NO) = 225 mw = USO ES
— Pho = GA 5, = Pc 26.47 105,

= (26.4710 VZ.e4 0) = 69.8810‘ O

© $10.6 x10" 9

sLE7 Par = EBs, = (acoso Maso) ns = 45x10‘ Sy

Garde
= (46x10 Vlo.s00 8) = 28.80 «10° O
Ger Ge = 128x107 8
From Statics Q= Pret BE Po = Pet 4.125 Pro

= [28.2000% (1.125 V(o%.88%10]8 = 317.0610 O

Oye yielding ob fink AD Gig > Sy? 250M" 10.6410" O
A, = 304.89 vo"
Qy = (217.06 x10* Ma04.29910"*) = 255.2 810: N

Sime Q more? > Q,, Lick AD ill Bios 250 MPa e
Bus AGS, = (225110 asomo) 56.2510" N
From Stutics Paez Q- 4125 Pao = 260 x10 (4125 50.25 10”)

Pues 27.97 "10° N Sue” Pu = BATH? > agent Pa
A Asa EN a

Blue „ (aran Lo
So: Pele. (mam raro m
ER" (do m ~

22 Te gd br ABC sppond ok, AD und BE, of wim 756
rar retngua cos esto und tate of a mid sold sto] De
laca wth = 200 GPa and o, = 250 MPa. The mage of te foe à
‘pple a gradually near fom zer 1 240 KN. Koowing U om 0 m
emi () o alc ofthe oral ses in each ak, () de misa dete
poa.

zn Seve Pro 2.121, owing that a 1.76 a tha he iodo af the
fare Q apo al is ray erated tom ares SIN

sorumon.
Stehes! EM=0 1L76(Q- Pal, co
Deformebion! = 2610, 591.78 ©

Ta Elashic Analysis

6 A = (87.5)(6) = 225 ma" = 22500" mt

E sa Pr EA 5, = ONO ao 5, |

G-47xIo* 204 8) = 67-88 x10* ©
Ge: B= socio e
Tre = EAs, A aso = (5010 0)

= 742x108 8 Su he = 524010
From Statics Q@ = Pact PER, = Pag + 1.500,Pro

= [73.8108 +800 Yer 88 »10)]O = 178.240" ©

Oy ot yielding S Bink BE Gag * Gy = 250410" = 352110 ©,
Oy = 710. 29%107°
Qr = (178.620 Maro.23410%) = 126,86 x10" N

Sie Qu 185x10'N > Qy, Bink BE yields Gyy>.O, 250 MPa =
Gee = AG, + (z25*10")(250 rot) = 56.25 x10* m

From Statics Pro" m (Q- Pag) = 52Sxl0* N

> De SRE Ht? toe x 5
So” De = SASH. 233.810 233 MP.

Fron elastic ancl aie of AD 15120 rad

Sar 120 = 1380 m + 1322 mm -

[>

oo

Y Theda ABC nen by i AD a Ba ir 355
PROBLEM2.23 un romp at sto de ee a e ios Ve

Aa win 20 Pa md 4250 Mb, To mente ete Q
Soria et sonda IAN. Ko ata OO
mn (voi se ek) e ml
oom Eu

"2123 Solve Prob 2121, sung that the magnitude fre Q aplicó is
(al increased fom er to 2001 an a dosent ba oar, Ko,
Hin 610m, denne (4) seis rss ven) nl deacon
‘tpt 8 the esd res im euch ink. Aare a the ink ae braced tht
They eu cy compres (res witht sting

IH

See soluhion te PROBLEM 2.121 for the normal shresses in each fink
and the deflection of pont B ateo Loading

SOLUTION

Ey + 250 410° Pa Sec = 124.3 x10" Pa

Sa = 64.53 «10m

The elastic analysis given in the solution te PROBLEM 2.121 applies
Trans Bere bi
Q= 3m.o6x0* 8

A ag.
3i7.d6 «15% "Bios wee

820.08 ¥10"*

Ge * 310.6 110" 0 = (310.6 %10")(820,08 x10) = 254,10x10* Pa
Sie = 128 “10° e = (imBrior(a20.oasio“) = 109.96%10* Pa
Se

0.640 6" 521.32 110 m

(a) Residual stresses

Gros = Gin Go = ROMO 254.70 410° = 47010 Pa
= = 4.70 MPa =

Garn = Sanz Gar = 124.810 104.96 x10" = 19:54:10 Pa
“ = 184 MPa me

21,53 10" ~ 524, 82010"

Sap * Sn
FC. 11 #10 m = 0.0967 mm -

PROBLEM 2124

er

2.124 Te air ABC(E = 101» 10 pa) wich const of clic
pions AB and BC tobe replace à mél as od DEE = 2 1
i oft same oval lg, Dei he minima quid amet dof the

Bie elds vet deforma erro exceed the domain of th alanonm
Fed ane he same an ande lovable sts ine rl ods oto Cerda
pa

SOLUTION

Déomation af alominon vod
à > Fla, Ble . {le , Lee
Her RE ECG + 2)

CTA Yoo >
fiver (ase + Ease) = 0.054576

Steed vod $ = 0.081376 in

> PL. (zoo) 5 .

ses À Een = ann
As = BE nier int

Required area is He Parger vadve As 11667:

dE. [ED . am. -

I

> ot

ca

AS 250m logalomison ibe (E = 70GPa) of 36mm ote dante and 28
mm ir muy be closed bo end by beans of pe threaded ren.
‘ove of | Sey pitch, Wi oe cover reed on ih oli bo 2 103
‘Gayo 2m amer lea inet thant he second coer se om
‘Siete 101 y tegen the tbe, saved that th ever ma be
faced api the od by ati ior of à um her I an be hy
‘esd Determine () he serge nora ares tb and ea (0)
“mation fhe be and oe ol.

PROBLEM 2.125

SOLUTION

A
A

sae HAAN = Esc ~ 284) = deze mt
wt FAS = art = oz
unbe Please) 2 TO

WOR nt

Sake” ER * Code Vion aso) ~

@

wy

PL = Plone) oy scogw io" P

E Io

S' = htm iS mm = 0.228 mm = BIS m
Site = S74 Smt or Sete = Sur = St

B.BBIS KIO" P + 4.8505KI0P = 375 4107
oss to!

Pose»

N Ser Re RSS = 624 wor Pa = 67 MPa e
Gun =~ O ae

She > (0.8815 ID K27B0 BE) = 242.510 mn = 0.220 mm a

Sa > — (4.8805 » 10° (27, 808 «10° ) =-182. 510 m

1325 mm a

2.125 25D alo lamina (E = 70GPa of 36 mun our dmaer 28

PROBLEM 2.126 mm mer ek may be led ba nd y mon le end er
Sen el} Smith. Without cover stone oa ag slid ess ol 2 10

ee (Gra 25 mm diste paces tie aban te second ner toned

Sine he eds alight ona an he be ti ced a he Con mot be

(ir agaist he ed by ta. ons f en befor cae be tights

cd, Deemine (he vera ermal test ts ihe dl be ta) e

5 eirmations of hee and he ee

ET smn 1 242618 Prob 212, demie argo mama re in he tbe le

‘suming lnea was 15" Cube hc at wee say ited no de
Fiat 53°C. (For lamina > 236 104°C; hebra e203

SOLUTION rey
Anse = q (AS?) Fast 28

Aou à Bat» East > 990.87 met = 490.2741" me

= 402.12 mm = 402.12 x10" mt

AT= 55-15 = Hore
P (0.250)

» PL © + 50) 1540)
SN O

= Ben TP + 236 «10%

as Macano
Lal > ¿EGO + learn")
= 4.8500 lo "Ps 20410

"or dm LS mu = 0,878 mm = BIS 10 m
Bite = Smt + St
8.3815 OT P + 286 10° = - Ha oSxIOP à 20710 + 375 ANOS

12.782410 P= 3480” P = 25,94 010% N

= Bo. 25.34 10%
Site ju aos

= 6200" fa = 630 MPa me

Smut Ey eus = SEXO Pa = -S.6MPa «e

nr

©)

2127 The beck hen sma o mugre alloy which 20 10%

pi sad
PROBLEM 2.127 2035. Knowing 20 ka, drin (2) magie of hr whch
à (Shane nthe height of lock willbe an, (he coriponing ge then

ae ce ABCD, (0) he comespening carge inthe vole eft Beek
SOLUTION

go
+ ECG - 26) =0

©) 62 6% = (0.25 X2ox109

Tai Ti a
rm - 2G)
o, Lu ue) _ vor
222 es = 1.4838 vo
à = =20%0* - (0.55 K-7w10%)
Ext CSV) = CET
= -27 0”?

Ag AA = (IED CIE) Lil, 48 EL + En)
Bt Az Lila
DA > LlCe + E, + ELE)
= Go 0 1.4538 107 2,710 4 small or Y

MO int = 0,0098 int -

Sie Ly is constant

AV = Ly (AA Y= (1.375 14880107) = 6.35 lo it
= -0,00685 int me

2.128 The in ede AB and BC are me of el nd ar loaded shown.
PROBLEM 2.128 owing tat E = 29 x 10 pl, derive the maple and diesen of the

O
Ki,

fe sournon
ee e],

a a de Poe * Poo 0 = (2507 em 227 = 23.12 210 Ab
4 Ses Belge = EEE | 0.0446 in
ook 8 Paps Poin =( 28010" ) sin 227 4.565010 Dh,

corte Se Sq = fight EN 6 5019 in
ont | pe 85" m =~

Se [Coma = 0.0955 in

ae 2.129 Keoniog ttf = 29 = WF pi, dete (a) he vale of Or wich e
Gein of pit demand tll lng lie orina anangle 36 its
es ‘Se oi) oespndng mars ofthe econ of
fe
«es q soumox
Es MO Ses Bier
| p, = EAeeSm . (910 (0.8) $ cos 36"
ee ES
os = sono? 5
al Sa? Seine
Eu Su, (20100295 sin 36°
i Pe eT 25
a fe = 318,200” $
al METATETI E e a
tan » BEROHTS 190107 0= 68.0
Pr 2840 = ]OIT0F or 5° + Un Pot S TT = 918.280 S
g= 2% 0.0272 in. -

q

Oo

[>

a]

[==

oS

PROBLEM 2150

Elongation
Debfection

e

Shhés TF0 — 2Pigsin@- Peo
Peri. LL

= el. oe

5

2:30 The aora wre ABC, of red eg I, cha 1 e sopor
sown an vel o Pi pplid he point Dori by 4 be ous.
‘cles há y Pm (uo ow et Boch do
‘estos the main

SOLUTION

Use approximation

‘ 3
Mme rime e 2

Remo

From the wight triangle
(B+ Se) a st
= A+ Sue + Sad 27
= 2288 (14 £38) 20 Se

~ PE
ABS

y E
Se E

%

‘PROBLEM 2.191

2191 The eel bart BE ad AD cach have a 6m cos con, Krona at
52200 Ga deci the duce of pot 4.0, an CF te igi ABC.

SOLUTION
Use wigid bar ABC os a Free body
DEM¿=0 (15) Pao -(S00 3.2) = 0
Pio= 128 kw
BEE O - Pr 6824, o
Pee = 16 MN
Deformations

Az (6X18) > 1037 = 10310 m“

= 8,25. = Folie - (28x10? Yeo xo)
5,7 Sho EA Goo (01 (108 #10"

= 237.09 (Dm = 0.237 mmo _

>» Bale. 6 110* oo”)
ER” Gooxi Klosriot)

= 96.30 10m = 0.2% mn ee

gr Sr2Se , (937.04. + 290.80 10%)

he 75x10"
= Wiz xo?

+ Sa See

S = Se + Lae
= 296.80 "10% (800 410" 7.124107?

= RABID Um = 248 mm «

ont

motes 2.132

2131 The el bers BS and AD each have a8. eos section. Knowing at
E 200 GP, deere the defections o pun A snd € ofi ig br abe

og 2432 In Poh 219, 03240 ec cased point Cto dele the ight. Using =

ZEIT 101°C seine a) over hunge in tempers tht mos pot

Cte tum ois ig ps, 0) e coespondi al def pad

soumos

Use vigid ABC as a Free body

DEMeso 15 Pa 003.2) = ©
Pa = 12.8 kW

HDR +0 Pet Bat Poze
Por = IE kn

Deformations:

sce
| © 8p = Spo + Luclär)
a
Ja

Fre

= (2.8210 (Hoos) nS uses AT
CT a
= 237.04 107° 4 4.6810" (AT)

+5, Sw* Lola?

Pas!
yoo")
= Menton Y"
a + oo» OTE)
Ape = 2600 + 4.68157)

3000

Se See
“Se LES, > Las Sa

A > 0.375 ©

= (232.08 10 + 4.08415 Ar

125 [296.30 15 + 4.68 STAT]

+ AT=-57.68# €
= -527C me

10.680 tar) = Go7.HIS O

282.0410 = (HB SNE) = ~82.92K10"F m
Sen 4

Casio STERN) = + 26,347 10" m
Sa * O.0263 mm —+

296.301

2.133 A pol ob dled inde pit. Te diam ofthe is valle te

PROBEN, drill the hole range from Y to 29 mra in0-mm inczements.(a) Determine the diameter
ot pan or 1 la ut ok ali snes
Ha a Ai (1 de wae ses ee LS MP watt
ei ‘roping alot os Pt ve
saque u den
Ar Be Bee, ne Ian Y an
Denes Br Es PS
Le Leon Fi Fy2éb A 20
Bain = (SYR) = 900 mm? = 900 mm OO m“
Sone = KEM En 2 o x08)
ELIO” N = GIN
At the hole: Aut = (D- 2r)t, fe +
where D> 112.5 mm ee radis of circle Leiten
Kis token from Fig 264 a
Some? KEL Ou : Pur AS
Woke diam] y | a=D-20) w/t | K | Ane | Pe
A mm | 45 mm] 103,509 | 0.0435) 2.87) 1200406 nt | 62.7107 NW
15 mm [TS ma| 975mm ]0.077 | 2.25 | ens nt | 617 SN
Am ES [0.15 | 2.67 ESA
mm |13.5mm] 35,5 me | 0.152 AS RN
Largest hole with Pau > GREW is the Tam diamater hole,

AYovalle

Sora

A -

o>

oo oS eS

TBA FPN om a tm an
PROBLEM 2.134 ‘shown. Led. ”

SOLUTION

Macimum stress at hale

Use Fig. 2.640 Fer values of K

rs fon 7
Leg * 0087, Kram

Aut = GANS IR 1206 mt = 1206 10 mt

er KE > Ga) - urn

Maximum stress of Hide

Dee Figo 264 &

Lions
F:2=0.,
A

nin © MZ MIS) + 900 met +

2.10

Sur KE ee). 195,3210% Pa

Bain SS
QD With bode ond Flete Eu = 134.7 MPa -
(e) Without hole Eu + 135.3 MPa -

29» al 9, SOLD Lodel
Fe au de mama learn «= 0003, e) Neglecting the ees
Shim les onchange canal de sposien, demi the sag overall
ang ab ofthe me ac he ode moved. (0) Falling he renova ok
‘icon par a compe oe le ep wall he mim compres
Janse on Daten ne eating owe gh afer tea

PROBLEM 2195

sovurion SE
= & Qu 9 cores
WERE! E Fa à OU

Eme © 0.0025 > Ee Yielding occurs in portion BE
Geer Ey = Sono psi

Permanent strain in BC

Ent = Exe E * 0.0028 - 0.001724 = 0.000776

Sue * Lac Ee * WKO.000776) = 0.00510 in. -

In reversed Doudins, A point E
on stress stein plot

£ > - 0.0020
as given. Doving removal of the

rental tout, Phe change in stein
GE + 0.00124.

The permanent strain in BC is
Em" - 0.0020 + 0.001724 0. 000176
Sue = Lees WMX-0.00027) = - 0.001104 im. e

Note Hat portions AB and CD are always elastic, thes Her
deformations during Lond ing and untoading do not contribute fo

PROBLEM LCL

2.01. Aredcoesisig of elements. echo wc ic homogeacss ed
ni crs ion, abject the loading shown. Te ag e

nent derby fits cretion ae by Aye modulo of ls
DIO Ev and he Toad plied woe rg end ny de mentado Po is
‘ial being sumed bs pose 1, rested o ek and gave oth

Sie (a) We à computer rogram tha cn be sed o eerie tea
SE mare in ah leet the deformation of ech element, andthe
{Sul ¿formato e he od (0) Use ns progra t se Prob, 217 and

soruron
FOR EACH ELEMENT, ENTER
bi, Ae, Ei
COMPUTE DEFORMATION
UPDATE AXIAL LORD P= P+,
COMPUTE FOR EACH ELEMENT
as F/M
G = Plif Ei
TOTAL DEFORMATION
UPDATE THROUGH n ELEMENTS

$2645
PROGRAM OUTPUT

problem 2.17
‘lenant stress (MPa) Deformation tam)
20 name es:
Total Defornation = .0162 m

probien 2.18
Element’ Stress (MPa)

Deformation (em)

1 pese st
os an
a ln An
Totaı Deformation = 5.2068 se

PROBLEM 202

2.02 Kos AB soil pt ents Fed omis 9 el
met cach which mp nd ram essen, se =
Jo la sown, The eng of elem dent by Ls ro
feral area by A, is modules of elasticity y E ad the led applied wis
‘ph end by Py he ago Po! this ond Bein sumed o De pte it
Pi did gaie the (Not tat, 0.0) Wie
‘cute prog a can be med o deermine Ihe cios aA nd B
‘te average formal ses seach leer an de deormsion of each ce
‘Rea bh Us In progam sole Pod 24

SOLUTION

WE CONSIDER THE REACTION AT B REDOMPANT
AND RELEASE THE ROD AT B

COMPUTE Sy WITH Ry =O
FoR EACH ELEMENT, ENTER

Li, Ai) Es

UPDATE ANAL LoñD
p=P+P;

COMPUTE FOR EACH ELEMENT
a = P/ai
6; = PLL Mi Es

UPDATE TOTAL DEFORMATION
er

COMPUTE Sa DUE TO UNIT Lond AT 8
amt oy = If Re
umT & = bef MBE
UPDATE TOTAL UNIT DEFORMATION
UNIT gg = UNIT Jy UNIT Sy
SUPERPOSITION
POR TOTAL DISPLACEMENT AT B= 2580
Gt Ra wit Tu» 0

CONTINUED

In
|
|

fo]

PROBLEM 2.C2 CONTINUED,

FOR EACH ELEMENT

€

E 1 Run fe

Ge + Rg ur G

PROGRAM OUTPUT

pronten 2

2.909 ase

Ro > "20/001 Ripe
monene Steves (ken)

13.887 conse

PROBLEM 2.3

Route a

9.2.03

2.02 Rod AB conics an cements. ech of whic i homagencos und
aio crs ection End

SOLUTION

WE COMPUTE THE DISPLACEMENTS AT B
ASSUMING THERE 15 NO SUPPORT AT Br
ENTER Li) Aj, Ec, a;
ENTER TEMPERATURE CHANGE T
CONFUTE FoR EACH ELEMENT

€ = q LT
UPDATE TOTAL DEFORMATION
$e = ber 5;
COMPUTE dy DUE TO UNIT Lose AT à
UNIT & = Life;

UPDATE TOTAL UNIT DEFOR MATION
UNIT Eg = UNIT f+ uur

COMPUTE REACTIONS
FROM SuPEPrOSITION
Bo = (4-8 fur &
THEW
Bn

FOR EACH ELEMENT

Rofhe

= lit + Bali hi Ej

Ri

a

coNnINvED

PROBLEM 2.¢3 CONTINUED

PROGF AM OUTPUT

ES

Element Stress ut)

2 En)
rates 2,58
Element Stress (Ha)
1 ET
2 ERA
les a
Element

etes 2:98
Es pe
Flament Stress (kai)

1 22.002
2 THe

atom. 10°-3 in.)

EU

befor. imteren)

500.106

242.504

etoca. (10-2 an.)

PROBLEM 2C4

Nenn ee

y

204 Bar AB hu leg Land
hes eme aa, modal a ic nd sd sent The a
bes sen à ko Puh pr cence fe

te deforoin oft br har rach main wale By arden een
Seok to er. (a) We a conput:pengram th a sc 2 valo of 6
ly anal ve à rn nern frm a à ae equal 1 120% othe
rin casing bc aerial to ii a be as trans Ue ma
ms vale Pa fhe hd, de nani normal sess in ach ner
mane deacon he bar a the ek ae im eae
(@)Use ds progra oe 2108, EL and 2

sournon
NOTE : THE FOLLOWING ASSUMES (5) < (9),
DISPLACEMENT INCREMENT

Em = 0.05 (0), L/E:
DISPLACEMENT a7 rame
GeO L/Er S25), L/ Fe

FoR EACH DISPLACEMEN

IE mai
G = Sm Eft
Sy Eft

Bee (SALE, + ALE)

IF En ime 68:
el),
= dm Eafe
Pas AG (mf Arey

IF Sm > Oe?

Gr) alte
Pm= AT + ALT,
PERMANENT DEFORMATIONS , RESIDUAL STRESSES
SLOPE OF Fikst (ELASTIC) SEGMENT
SLOPE = (AE, + Ba Ea )/L
Spe 6m Pn / S002)
(res = G - (Es Pm/ UE store)

(neg = SG (Es Po [le 0088)

CONTINUED,

E

LL

J
J

=

0

=
9

==]

PROBLEM2.C4 CONTINUED.
PROGRAM GUTPUT
proble 2.209
ox mm son
un De
‚000 00.00
1 6097 acaso
mE ee Suse
foie tetas 270
Ses 2150 6:00
Be 201.08 seite
MS Sr 169.800
AMOR dés 12070
1510 463.000 138.000
diner 5431979 131280
162.075 643.780 172.900
lue GLS 188.780
e] 207.000
on HE
125 21:00
2 280.000
200 380.000
589 250.000
u 250.000
Eier] 20.000
327/380 350.000
34017 250-000
305 250.000
ven 280.000
399.309 1041.20 250.000
Problems 2.111 and 2.112
om Py sim
sowed an, pe ees
zie 08
fes 172800
Tai 26200
12.069 43.750
deu sos
Hé enr
19:00 30.000
Ron 78.20
99.000
102.500
106.250
119.000
117.500
121.280
128.000
0.620 128.000
EE

sion)

„000
38.500.
83.000
56.250
183.800

345.00
38:00

sa
kai

130.009

1003 in.

673
Bi
000
‘000
1.9
cl

900
Bid
000
1200

900
.000
1009
2000
2000,
2000

„375

280

Lee

26800

247.500

2000

2000

‘900
a

sim

000
298
1009
2000
209
1900
308
“000
08
00
as
47500

stat

eit
20:00

2.08 A sil trance cone is subjected so an axel force P as
one à computer progr ht canbe use Lo tin an aproxima:
ti of de eogain ofthe cone y epa By ciel Hide of
al ces ad ld equal the eam ais fe pain o cons
‘ty rpc Kooi the De ea! vale ofthe iongain a he one 1
(PLE 708) and ws fr PL. and ales of your shoe, ermine
fhe pecemge eme nes when the progam is ssed Wah (a)
irren

SOLUTION

FOR fel Ton:
Lp = (1+05)(1/n)
rn = 2c-c(Li/L)
AREA: ,
herr;
DISPLACEMENT:
be b+ Plejnd/ine)
EXACT DISPLACEMENT:
Ena * Pijiz.oreE)
PERCENTAGE ERROR:
PERCENT = 100 [8 Egeo Feuer

RAM OUTPUT

mo ponte act,
5 0.15952 gases
Fi Pe rer

ES SUN Susa

CHAPTER 3

A1 Detemise te tongue T it une a ima sheng stress of TO MPa ia

PROBLEM 3.1 te seeds
PROBLEM 32 242 Paton the mannan sing rescued bya torque of mage P=

oN

souvon
Lu = E cc
Tome Le Te He!
Pron A
Fes © Hop = 510 Pa

87.2 MPa -

PROBLEM IS

iS

al

33 Knowing hat the itera ameter fe halle af shown i = 09 m.
Sein th nasi rg stm causal hy tore of mae FO Kip

SOLUTION

by = (Ye) + 0.8 in C= 0.8 im
€, = kd = 10.9) = 0.46 in
T= Hoït-ctls Plos"-0.95*)= 0,5770 int

= Te. Ces - A
Tu = FE = USE = 12.44 ks -

PROBLEM 34

234 owing thet d 1.2 in, determine the orgie wich ea à reia
si set 07, kn the ow at shown

SOLUTION
Cas hd, ((LE= 0.8 à C= OB im
Gmbh = (EI(02)= 0.6 in

J=H(c*-c!)= $(0.8'-0.6") = 0.4398 in’

Come =

Ta So LIO 12 kiprin a

mas 28 (o) Fr te how aan ing sow determi the mama sai
ns in e) Dame te cer of az sa or wich he nm hr
Rs he ane a parta.

pd mm
a e, = Hd, =(4X0.040) = 0.020
| 0, = kd, (F)(0.060) = (00m = 0,080 m
Tele (0.030° - 0,020)"
= 1.0210 210" m*

& = Te. (2000 X0.03) „ sise pe
Un TE = GRR) = 70,52 rio Pa

TS MPa

. 15 Be’ oe
wm or, Tre in
2-27 10 ve

Cy 27.88 110% m A= 2Cy* SEO SEE mm at

PROBLEM 36 34 (o) Determine rg wich y be apd a od sa of 9 ter
diameter wit cata sn abuse tearing erent O75 MPa. (8) Sole pat a,
Sig atthe bd sat reps ol sa of ese ms ad of 0

soLenion Shee

(a) Fer the solid shot = $d = {0.090 = 0.045 m

Zs Het = Focus) = 143.14 wu
teas Ten Ts TaD 0 Mp 10!) = 10.74 x 10 Nem

10.74 Km <a

10) Hollow shaft = $d, = (4)0,090) = 0.045 m
For equa? masses the cross sectional areas most be enved

Aztech s west) m faire
Ca = fos ts 0,045' + 0,0636896 m

Te H(o 6") = 19.3237 210 m"
E ES x ateo. 323710“) = 227740 Nem

22.8 UN met

27 (Forte Singer sb ender and lung shown, determine the
mcm sharing str (0) Determine be ne meer of tower, oF
ER er meer hich ie mun tes te sae par

SOLUTION

É (a) Solid shit ew dds Hao) = LS in

= Tes 20 aX. ;
4 Thay? Ges BE ern
(6) Hollow shaft che =4(90)= 20 in
ato). T

=
TR . zor. DRA _ ind
a A IA
C, = 1.74395 in d= 2c,= 3,49 in -
rons 2s) ee nn he ie a foren

nei vna cea a able asomo mc o 10 ki. (8) Solve pra.
Sing tht te sod shat ac brn epic by a hello a ofthe am cr
SOLUTION ‘Sel rn an than ber rte oque 1 a cute dame.

(a) Sedid shaft:
Js Be" = Hlo.37sY

Te In

(6) Hollow shaft
For He same arta as the solid shaft
A= r(et-c?) = mlei-(aÿ]s dran = me?
Cy > fo = Fy (0.875) = 0.433018 in

= (M075) = 0.225 in
0.031063 int Lans Vo Mos

0.828 kiprin or 828 dein a

es ka + 0.210506
T= Fletes)» E(o.ustors! - 0, e506") + 0.051772 in"
= Gad „ (0.051772) - i con
Ts Se MO.081972) = 1,196 kiprin or 1196 inh <a

9 Theo shown at exc on pal 4 and B Koning ha ech at à
sob, ermine nam Seng es (o) in ta 48.0) m Bu BC

(vemo

Shaft AB: Tu = 30m, d=0.030m, er ausm

PROBLEM 39

sms 5 Tar IE. BT, (alten
és T Fes Has
on = 5658810" Pa = S66 MPa at
OD.

Skat BC: Tac= 200 + 400 = 700 Nem

# m, Ez m = Te .2L +
doom, es 0.02% m Tur TE = BE

= 2.62610" Pa

PROBLEM 3.10, 210 The torques shown te reed o leg 4 and which tacho sai

colt sat AB a BC. In oder to reduce the ttl mass ofthe ase,
eme the malt diet of al BC fr wich pest shearing ses inthe
erly soot ced

SOLUTION

Shaft AB: Tyg? 200 Nem, d= 0.080m, = 0.015" m

2Te . 21. (jo)
Ta T er (0.015 y
= 56.588 rio Pa = 566 MPa

Shaft BC: Tee = 300 +400 = 700 Num

da 00m ex 0.8m Tage TE = 27. Qro)
a e a es
= 36.626 »0° Pa. See Hm

The Dacqest stress (S6.S88x10* Pa) occuns in portion AB
Reduce the diameter of BC th provide He same stress.

= 700 thm TS , aE
Tag = 700 N tome FE = 3%
2. 17 , Qro wot et
ee Whey, > TS.“ Teen
es 19.898 0167 m d= 20: 39m 40m Em

O S © oo

=

1

[=

It Knowing hu each portion fhe tat AD const of ld clar 04,

mon mins) the potion ul au nich he mai sng Arc oa,
De maga of at Sess
un souunos

Shaft AG + T= 400 Abs in
Nu 07 EEF 0.80 in
u La TS »2L
a TS

au eh 481 pi

Shaft BC: T=-40041200 = 800 dim

erde nem tur ls FLA > aaa

\

ne

Vu 060

Shot CD: T= -400 #1200 + 500 Les Pb-in
A „Te - 27 QM) -
edle 0.46 Toy = 21 =e ES © 2082 psi
Hrsuerst (a) shaft BC (6) 9.66 kei -
PROBLEM AIR 12 Ku tr 0304 tte ben dled ogc porn of

shat AD, rea (potion in ch immer
seu (he gad of Wa sree

souros
mn Hole: c=dd,> 0.5 in
S Shaft AB: 400 Ain

me "eso

cpt hd, = 0.30 à
SS T= #(@'-c' )= Flo.30%- 0.15%)

ALLER
fay 2 Ta > (400 K0.30

Toner PS + LO ei à
She BC: T = -400 +1200 = 800 Ms in Cabos = 0.875 in

T= Eed-c)+ Fons ous") = 0.030268 in
Tees @olosm _ A

Tes Te = poi

Shaft CDI T = -400 +1200 +500 = 1300 Min us hdp" os in

Ir EGG") = Eos - 0.15%) = 0.068617 int

= Te, Golem) |
Tray = Tee + Gaseous) > 9196 pi

Answers! (a) shePt AB (b) 10.06 ksi =

313 Under normal opel conto, here motor exe torque 24
Koma. Koowing th cach a sul, dr de mem ear ess
(natal AD, (aba BC, (on a CD.

PROBLEM 313

SOLUTION

Shaft AB: Tag = 2440 Nm, cake = 0.027m
Vas TE. ZE. 240

E+ FE DAH rasen TLEMPA 4

ShePt BCE Tae 24 Um = La Kim = LAN, Cael > 0.028 m

Tes Fe I = OU gue) cores a C2.8 MPa a
Shah CD: Teg? OJO Nm Cake = 0,023 m

Ta = Te = BE + DOME) zo MPA

214 Under normal operas combos, the ceric motor ee one of 24

PROBLEM 3.14 Nim at 4. In oder fo edie mas ofthe men, deine ais
lamer of salt BC fr wich be let ering ei hr am I
‘ema

SOLUTION

See solution ta problem 8.14 For figure and For mai mom cheat
stresses in portions AB, BC, and CD ob the shaft. The lanes?
Value de tay = 77.62% it Pa occuring m AR.

Adivet diameter of BC te obtain the same value of sinest

„Te.
E. 2

32 2L - Qíaxion — O
OF re = reason] = PES HIC
= ads I mde de 42.3800 m 422 on 8

y = =
PROBLEM 315° eee od Dan in the
once determine the

T= Bez.
Shar AB: Tu sisi a nkdr arsin
T= Floss) = 9.94 kip-in

She BCE Gus Shi CH Kd = 0.90%
T = Elosoj'(a)= FIG Kiprin
The abDowable turque is the smaller value Te 7.6 pon me
aa 316 The aoa sess hl inte ste rd Ba ia bas od

Kong Uta tng 710 Mp appl A, determino the rid
‘Smtr oa) 4, Dr BC,

Fa. SOLUTION
L. a. 12 sek,

"I

ShePt AB: Te lo pin Zum IS hei

3, QUe, ns
oe ek à

© = 0.7518 d=20 7 1.503 in -
Shaft BC: T=lokpin Toy = BUS

60.967 in A> Ze 71,883 in -

ao

a C3

3.47 The wid od 48 ls a meter dy = Mn. Th pipe CD Jas an outer
PROMEM ST dam 090 and al et King at oi rod and
[phe are made far wich i alowabe sharing tres 5 Ma eterno
Be neat tn Y wich ay be paid

soumon
Tay = Rs mot Pa Tar = Tau
Rod AB: C= dd = 0.080m, Se Ec”
Ten = FS Lay = Llosa IO
= 3.181 410% Nom
Pipe LD: Cz, = FA ~O.04Sm — C= Cy-E *0,045+0.296 = 0.089 m
T= E(e- ct) = F(0.095°- 0.033") = 2.8073 «157m
Tu = LOMO 175 HO) = 4.67940" Nom

9.045
Alfowable torque is the smedJer Value (3.1810 Nom) 3,13 KNom et

228 Te sll wd 48 us deter d= 60 maná ls mud fel o wich
tie anale hey ses 85 Pa, Thine tasa rt deter 09 mn
à al ices of 6m made ofa lenin fr which he lowe
‘Henig ares 54MPa, Determine the get torque T which maybe appli 4

PROBLEM 318

souvniox
Rod AB: Caps ESxIO RR, c=fda 0.030 m
Tey = ie Er

= EO ty
= E(o.030\(35x10*) = 3.605410" Nem
Pipe CD: Lys 440 Pa = bd = 0.045 m
C= Gob = 0.050.006 = 0.03% m
T= B(ct- 7) = F(o.o4s"-0.089*) = 2.8073 vis" m"
ny = Lu RTS MS NME) O Den

Ce 0-088

Aliowohße torque ie smaller vale Tap? 3.26110 Nom

3.37 Nom -

| T |
PROBLEM 3.19 7 3,19 The allowable stress is SO MPa in the brass rod 44 and 25 MPa in the aluminum. |

pes en e o ON tna dto ed
Ts TS T= He" A ar
Rod AB:

CASI mM = 25.15 mm
E AS

Eso)
| C= 316948 tm = 81.64 mw dur Zee CH mm, =
PROBLEM 320 120 The slid 04 BCs dtr oO mandas nun wich

bel ss sv 625 MPa Rod AB hallow nd Js an ue diameter fé
2 eae à made à bas a wich albo ewig ares 30 MPa.

sn ‘Determine (a) he largest inner diameter of rod AL for which the factor of safety is the L
ea ee bon os ty om ed
/ n
Y souuron i
POON Solid vod BC: C= Te I=Fe ,
= dh On

Tar 25108 fa = Ed = 0.015 m
Tae = Fe? Le = Fos aso") = 192,536 Nem N

Woltow vod AB: Tay = Sowie Pa Turm 182,536 Nm |
Ce = 44, = 40.025) = 0.0125 m a 1
Ta = be = ga GB

+ _ Z Tte
a

3.3208 vi mt

CG 2ST 4107 m = 7ST mm

2c = 15.18 mm a

Adley able torque Tan = 132.5 Nom -

321 Atonpeof pine P= 1000 tn soles Ds. Kom

PRORLEM an sp a

thence mal 1 6 nt 1 dar dl CO wm des
Dn sary en) a 0) Sal CP.

soLunoN

Teo = 1000 Nm
Ta + FE Ten = 122 (1000)= 2500 Nem
Shalt ABE Cedd > 0.028 m
„IE. 21 „(aXasoo) , .
te Te: - UE). somo T5 MPa —

Shaft Bc:

= 0.020 m

= Te. 27 &Kiooe) LE
Ce ir cr G8.7 MPa ==
PROMLEM 3.22 322 A tue of spite 71000 Na ab at Paso, Kring it

he uw shearing res Pain uch sha, trae be eel der
of) sa AB, (all CD.
SOLUTION

Teo = 1000 Net
Te * E To = 122 (1000) = 2500 Nem

Shaft AB: Tanz SoxIo Pa
« 27 3,27. @lasoo) _ aya
VE o zus

CF 22.820 = 29.82 mm d= 2c = 51.6 mm -
Shaft CD: Tue = Gox10* Pa
¿Te 2 2T (2.0 _ BR
ES A E [O-CIPRIO” M

C= 2197 x 10% m = 21,17 mm d= ac: 43.9 mm -

Solucionario resistencia de materiales (beer johnston, 3rd ed)

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Solucionario resistencia de materiales (beer johnston, 3rd ed)

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