solucionario 2
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Transcript of solucionario 2
Problem Solutions
byDean Updike
Chapter 1
r°1 A
Problem 1.1 1.1 Two solid cylindrical rods AB and Be are welded together at B and loaded as shown. Knowing that di '" 50 mm and d, = 30 mm, find average normal stress al themidsection of(a) rod AB, (b) rod Be.
Roc:-\ ABp:::- lotO + 30 -= 70 ~lV '::' 7D >( I(.)~ N
= 1[112._"t
-)1. 3 'l. o'! ....~ (SO :- /.Cf63$></O...-.",,:: L'1(,~S><'!O """
p:::-
Ae- 70>( 10! :- 35.7><' 10" Pa.
I.Q,.3S.x/O 3
RoJ Be
? -= 30 hJ '=: 30)( 103. l\J
A = !-J'l.'l.;: ~(30Y =- 70(..86 ~"",l.-= 7of;.B'x}o'
30 kN
6ée. -: _.E_A ::
30)1( 1037oG.8G1'lo-C.
" R-:: '+2.4>,/0 a
Ok:- '+2.4 H Pe<. -4D
Problem 1.2 1.2 Two solid cylindrical rods AB and Be are welded together at B and loaded as shown. Knowing that the average normal stress must not excced 140 MPa in eitherrod, determine the smallest allowable values of di and d1·
f{oJ A B.
P :- 40 + 3D ':" 70 k~ -= 70>'10' "J
€S" _...E.:: _E_ '"
4Pdi
Al!> -AAU t: J 'l Vd,1.
f- 't 1
d": 4f ::-(4 )(70 'ji lO'?) :<5_ '). >' 10-3. ¡...,
I lio¡'& lI(l'fo )t io- ') -
d, -= '25". ~ 1M"", 4l
.:!.';::' Ib-52)( lo ~
dt_ ~ Ib.5'Z hlk-\.
PROPRrETARYMATERIAL. © 2006The McGraw-Hill Companies, lnc. AII rights reserved. No part of this Manual may be displayed, reprodueed or distributed in any fonn or by any means, without the prior wriuen permission of the publisher, or used bcyond the lirnited distribution to teachers and cdueators permiued by MeGraw-HiH for their individual course preparation. If you are' srudent using this Manual, you are using it without pennission.
-----------~ "--"- --_ ... _._--. --------_ .._----- -------------_.-
ro
4'
\
Be
Problem 1.3 1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the average normal stress at the midsection of (a) AB, (h) rod Be.
(o.) RoJ AB ...
.p:=- lfO k ,)" ~ ( + I!': '" ~ " c; '" )
A - 7Td~ 1T(~)? '" 3.1"f1(, ~Ag - 't IVI
3 in.
6-Ag' =- _f - Lfo -_ I~.73 K~ IA~B 3.1 '"11'" ---+---- 40 in. ----....j. ¡
Be_
lfD - (Z)(30J -
Ase:. :' 7T (~)2. ~4-
- '2_ 83 ks;
Problem 1.4 1.4 InProbo 1.3, determine the magnitude ofthe force P for which the tensile stress in rod AB has the same magnitude as the compressive stress in rod
1.3 Two sol id cylindrical rods AB and BC are welded together at B and loaded as
A AS -= E.( :z
,<.::- 3.1'-\lb
. 'l shown. Determine the average normal stress at the midsection of(a) rodAB, (b) rodtt J
3.1't l'IV)
Be. 3iu.
::"' 0.31831 P
~ ('6")"<. ~ 7. O~g6 11'7 'l
(;:()(3o) - PAA'!.
(;0 - P
l' ""·10 kíps I
L30 in. ---1---- 40in.----+l.¡
7. o e s c
EqvwJ-;,,~ ~13 ~ 011C.
- O.ILf I '+7 'P
O.31~31 'P::- 8.«7883- O./'iI~7 P
PROPRIETARY MATERIAL. 02006 The McGraw-HilI Companies, Inc. Al! rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, withoul the prior wriuen perrnission ofthe publisher, or used beyond the Iimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation, (fyou are a studenr using this Manual, you are using it without perrnission.
A_ ~- Pb
As rr (d,sZ,-
Problem
2
or
•
d
Problem 1.5 1.5 Two steel pIates are to be held together by means of 16-mm-diameter high strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 200 MPa in the bolts and 130 MPa in thespacers, determine the outer diarneter of the spacers that yields the most economical and safe designo
AJr ~~-:.~ bo}t )OC.CI.+I·O~' +~e. <)pp?"1' 'Pj~+e is pvllec:J cJOWII'\ by +At +evtst"le
fOf'ce PI. of +he ho)t. A+ ~e S::"'iV\e +,'""e. +~e. .spcu.e~ (->v!:,i-tes
thc:vt p)Ct-tc ufwq.roi wi+~ C<. =-r=:« f:o-.r,c.I.~ P.s. In o."JeV' +1.)rn ~',;~, -+ Q.,I· V\ e'l t) ¡ir'!:, f'" :.J~V'_
Pb - Ps
Fo(' +~e. 602+ Oh Pb ~ 't.. Pb
""""'. ._1:' b
iT6cJ2.lf k· b
¡::-or- T-he -r=« 6s :::: Ps :- lf Ps OV' ~ := f6.s (ds"- j~
')
Pb rótn'J Ps
lf s, db'1.
ds =-jl+1.6 A strain gage located at Con the surfaee ofbone AB indicares that the average normal stress in the bone is 3.80 MPa when the bone is subjected to two 1200-Nforces as shown. Assuming the cross section of the bone at e to be annular and
knowing that its outer diameter is 25 mm, determine the inner diameter of'the bone's cross section at C.
120(11\
() -- _AE
_
."~p
A := O
Geow. dV'1 A := l (J, 1- el,.). )
d a. = d2. /.fA d' ,+p2 TI I
; ! 1TrS
J' --( 1.S>l/O
-3
)2- -
(L/)( I z 00 ')
Tr (3. 80 )(10'
- ':<2.::<.'1 -,)1 10
I~O() "_3
z ";: ILf.Cf3 «te l'V\
---,'---._----------_. --_-------.,_ .. .-
+
,
Problem 1.7 1.7 Knowing that the central portion ofthe link BD has a uniform cross-sectional area of800 mm', determine the magnitude ofthe load P for which the normal stress in that portion of BD is 50 MPa.
135 mm"i:-.. 240 mm~......j .... - ,.¡
CaD ~ 50 MPe.. ::- So '¡( lo' 't'a.
A'{I.D ~ "600 01. o _'- <z: 00)( 1
W'l1Y\ 0 0M
F"ao = OSD Aa!) ":' (.so)o( I ol.j (í?00 x 10"( ) :=' lfo)l Id' IV
D...c.<."" -lv"H. boct, J,.~",c,_""" úf bocJ.¡ A'B D.
C:J Me.. = O
(O.~50)(;~~l;c) - 0.135 P : O
'P ~ I.SC'6b Fso
'P-z(1.5b8b )(4-u)< Id')
p f.1..7 )t. lo"!. N
Problem 1.8
12()lh
D2:Me,: O
1.8 Link AC has a uniform rectangular cross section t in. thick and 1 in. wide.
')etermine the normal stress in the central portion ofthe link.
Use the. r.PGde. to~e.+~e'"w i+h two fl)))e'fS "-S ~
f-l'ee 'bocly. Note. +~a1+ke c~bYe +~~"OVlCuvS€S ~+ I~DO .f.'b~í",cRoc..k Io.II'se (.0 IJr)e +o Q.c.t
()V\ +~e hoc:J'I.
- (I~ + ~)( ~"Ccos 30") + (10)( File sin 30") - I ZOO '::- O
I:{OO ":-',3S. So lb.
A z:. I .
.Y ::. ~
O. I;¡_S- ,
In ;j •."
- -_1._3~- 5.._...yC-~o-
c.tis
¡,¡
. Problem 1.9 1.9 Each of the four vertical links has an 8 x 36- mm uniform rectanaular
cross sectron and each of the four pins has a 16- mm diarneter. Determine the rnaximum value ofthe average normal stress in the links connccting (a) points B and D.
(b) points(' and E.
( 0.0'10) FC4jI - (O. Ó;¿S)(1.0 X/03) ::: O
(0.008)(0.0.36 - O.OI~ )
:: 'bOXJO-G t
M. Fer: 1-wo pQ,.o.11eP ji" ks A"q:
TeVl$; J<. s.~~S I'VI -}I\'') k BD
<.a.l Oeo -= hA···!a.·2_.... ·1
&.5 '><'/03
3~o )( 10-(0
t: \01.5"6..,./0 0(' 101. b n Pv." ....
=
(b) 6'CE -
OYl€
0'(' - 2.1.7 M Pa... 1IIl
PROPRlETARY MATERIAL. © 2006 The McGraw-HiUCompanies, lnc. AII rights rescrved. No par! of rhis Manual may be displayed, reproduced cr distriouted in any fono or by any means, without the prior written permission of the publisbcr, or used beyond the limited distribution lo reachers and educators penniued by McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual, you are using it without permission.
------------
ClV'e((. ~s = • q
Problem 1.10 1.10 Two horizontal 5-kip forces are applied to pin B of the assembly shown.Knowing that a pin of O.S-in. diameter is used at eaeh connection, determine themaximum value ofthe average normal stress (a) in link AB, (b) in link Be.
0.5 in.
'0 k;ps
FoV'c~ +('IGl"l~ le.
I~
L¡'f\ k A g t'S Q.. +e", ~iolo'l ~ e~ bf.f\
M¡~t~l,)1'I'I sec:.-+¡oll'\ at.,f p¡'"" JA.l1d
(1:4) S-t~ss '1'1 A 8 OAS ':. FA
"I"
I-
§.t·
=
(/.8-0.8)(0.$) -=- o.S , 1.11'1
L;nk se IS e, eOH\p r-ess ICJII"I ~eW\ber'
t,.."ss S ec+i(.)VI q) ,
A-= (/.8)( 0.5) o . lo
(6) S~es.s . Be 6sc.
- Fec 8.cU~S'8
=- Cf. qr:; ksí
1'" :::
A O.Cf
J'ROPRJETARY MA TERJAL. II:J 2006 The McGraw-Hill Companies, Inc. AII rights reserved. No part of Ihis Manual may be displayed, reproduced or distributed in any fOM Ot by any means, without the prior written pennission of the publisher, or used beyond the lirnited distribution lo teachers and educators permitted by McGtaw-Hill for their individual course preparanon. lfyou are a student using this Manual, you are using il without permission.
-:- ::-
Problem 1.111.11 The rigid bar EFG is supported by the truss system shown. Knowing that the member CG is a so lid círcular rod oro. 75-in. diarneter, determine the normal stress¡neG.
USiVl'a ?o""'¡";o",
+t '2S -= O:
EFGC B t...s G.. fY-~e. 'boJ.y:; FA6 - 3(,00 z: O
F...-'" -= GODO lb .
LJs:n~ be",,,,,,, Er=G- ~ t:.. +-~~ b"J1
~ ~ ~ o : -:(4) ~ ~é' + (4) (~ t:"" ') : OG Fe!;. = FAlO' = bOa::> ib.
E C~ S~ secJio
...J.
Cl.ve.c.... of l"v'Je"",be.r- ce
Aa;. -=- :; J'1.. ~ .~ (O.lS')'" ':' O.4..flfl7"t ¡..'.l...
No"."""J s-tv-e ~$ '11\ cs.Gcc;.. z: I='=- (;;000
/3580Ac<-- o.Ll417'1
13.5~ ks; ~
Problem 1.12 1.12 The rigid bar EFG is supported by the truss system ShOWD.Determine the cross-sectional area of member AE for wh ich the normal stress in the member is 15 ksi.
~·l)::l¡:o---~---Qo...I
3 ft
.Lj~~t=:~~~
L..-l4n+:dUSII'\~ ,?or-holl\
+t 'z Fj = O:
FAE' -= 6000 lb. =
6",e.~-= /5 ks .:
(;.00
IS. t
O.L.f O O '1'\
PROPRIETAR y MATERIAL. 10 2006 The McG,aw-Hill Companies, Inc. AH rights reserved, No part of this Manual rnay be disployed, reproduced o. distributed in any fonn or by any means, without the prior wnnen permissíon of the publisher, o, used beyond the limited distribunon to teachers and educators permitted by McGraw·Hill for !heir individual course preparation. [fyou are a student using this Manual, you are using il without permission,
*
-
Problem 1.13 1.13 Two hydraulic cylinders are used to control the position of the robotic armABe. Knowing that the control rods attached at A and D each have a 20-mmdiameter and happen to be parallel in the position shown, determine the average normal stress in (a) member AE, (b) member DG.
Use VYlel'l1kY- A Beo..~ -t ree b o"ly.
e
fi:.
(o. 6(0)(800) = O FA!?"= Lf)(/O 3
N
A.".~(;(.. óT V'uJ;", ~'i! .... boo A E i~ A == fell...= f (~O"'¡l(JO'3)'l. ':" 3/'+./6 x IO-~
Sfr.~:.s ;1') 'f'oJ Ae-: SAE: íc-- ::'31~.~~~/ó-'- = /2.73 )1/0'- 'Pc:t
(a_) S"é:" I:? •73 VIPo.
D. Use Ca .... bll'led ~e ....h..e""s ABe a,.v¡d BF.D «s -fl"ee bo..-!y,
V ¿:Mf= = B(o. ISa X -t FM)' - (o. ~oo)(
r:~)( r.OSe> - O..550 )(800 '¡ ': o F"os. =- - !Sao N
A= if el?. = f( ~o »to -.! y~.':
6~ ~AFQG. _ - ISoo
3. l'-Ilbx lo-e. =
(6)
-4-.77 >tI ob
Po..
PROPRlETARY MATERlAL. 02006 The McGraw·Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to reachers and educators permitted by McGraw·Hill for their individual course preparation. Ifyou are a studeru using this Manual, you are using it withoul permission.
H
Problem 1.14 1.14 A eouple M ofmagnitude 1500 N . m is applied to the erank of an engine. For the position shown, determine (a) the force P required to hold the engine system inequilibrium, (b) the average normal stress in the connecting rod BC, which has a 450- mm' uniform eross section.
60rnrn
Use ristov\) roc:l" 4.",J cV'~ k+(.)~e,t~ev- o..s ~e. bod1~ AJ..,I w o..iR re"-.d,ollf H CA. '" J b e ~V' .. ~4 "('~o.c:,.,+ioVl~ A" o.V\d At.
~_J.B..O*rnrn D Z MA = o
(o. ~30 w¡) H - ISoo N-W1 '= O
)-l::- .s.3S 7 I .)(I 03 N
p Us~ pistóv¡ ~.p()tt~ Q.S -webody. Not~ -HtC4t "('oc! ¡.sCA. +wo-fo~ce ~eW\ ber; h42ltc.e t-h e JjV'-eGti¡,)~
o-t ~Oi'e~ F~ ís kili O""""'. Dv-a..-v +~~~"K ~
+'('i~.,.,Je Q.>I\ti so;~ -tu". '? t:.I.",j Fu by
p P t" o P o"i+i oVI 'So.
2 -= / ~uo 2- + Go2 - zog.81 ..... '""
-p
-= ;(00 p= 17. 8b #'-.103 N
H So(Q) p::. '7. 8' k~ .....
s+~ss
15 O:. Ci;)"",po"e.ss"CM
6ac. - - F&:..A
fee -= ~O"!.'al
FSc. -= j s..css x/ol N
Me"", ber-. I+s Gtt-e.~ i~
-/8. ~tf3 ){103 :
t.tSO x lO-&'
(6)
PROPRIETARY MATERIAL. 02006 The McGraw·Hill Companies, Ine. AII rights reserved, No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without thc prior wriuen pennission of .he publisher, or used bcyond the lirnited distribution to teachers and educators perrnitted by MeGraw·HiJl for their individual course preparation. Ifyou are a srudent using ihis Manual, you are using ;1 without pennission.
7
Problem 1.15 1.15 When the force P reached 8 kN, the wooden specimen shown failed in shear along the surface indicated by the dashed lineo Determine the average shearing stressalong that surface at the time of failure.
Al"e4. b€.tl"\j s~e"Y'e¿ '1.
A = qO.,.,~ )t 15 I't<M -= t ~S"o ...",..,.;1..::: ! '35'0 )1t (,:>" ~ I'VI
-r ':."p' .¡ Pa =5.93A - ...8 x /_f) '-'~'.. .:- 5.93 l<'{03r- :Z.5'O Yo / O - r~
MPa.. -4
Problem 1.16 1.16 The woodenmembersA and B are to bejoined by plywood splice plates which will be fulIy glued on the surfaces incontact. As part of the design of the joint, and
knowing that the clearance between the ends ofthe members is to be t ín., determine{he smallest allowable length L if the average shearing stress in the glue is not to exceed 120 psi.
Thetre a.v-e -Fov.t' S erc...r-Je C<..~:S ~c...t c..\Nt j.Pue.¿ •
EC(..CJk of He.se. a. I"e. c,....s. p...,.1I\ 5 V"'\ í-Is one ~".Qf +~e
-He s: SJ k.; l' +'::>I'c..e • Tli..,s
F = 1'P z: ~ (S.·s) z: 2. '1 k :p~ = '2croo J~.
L-t.-t j.: )eny~ of o"'e j.P"uP, CtV'e~ a..v» W t: ~ t'n. be. ,'''¡'s w t)flt.
E01/' ed 3.Pue d Gtv-e~ A ~ 1 IN
The . Jjo~e.)J~ she~vt·hj $~¡-e_5S t S
súi",,'l'I') ~,. j ~qoo
(120)(4 ):"' b.0417 in ..
L := .l + (j~f ) + .P =-
f..04.l17 + I + b.0417
:;:- I~. 33 in. -4
2
.
.
\.)b -
Problem 1.17 1.17 A load P is applied lo a steel rod supported as shown by an aluminum plate into which a O.6-ín.-diameter hole has been drilled. Knowing that the shearing stress mus! not exceed 18 ksi in the steel rod and JO ksi in lhe aluminum plate, determine the largest load P that may be applied to the rod.
",y-E..""2. - Al.
A ::
Fo,," stee2 A , '::" rr Jt = 1T(o. (O 1(0. Ion. ~
= O.7S'+O j~
.. P -::A,1:'1 :: (o. 75"40)(
~~.)
- 13.57 k;ps.,.., 4
Problem 1.18
-1 r-16mm
. -ti r16mm
1.18 Two wooden planks, each 12 mm thick and 225 mm wide, are joined by the dry mortise joint shown. Knowing that the wood used shears off along its grain when the average shearing stress reaches 8 MPa, determine the magnitude P ofthe axial load which will cause the joint to fail.
$iy C1.V'ec..S \N>vd be she4.\I'€ol offw~eV\ file jO.'" t -tc...i _h. E (Á,(, \.. of He se
a.l"eAS h",s c1i ...\.e ... ~'()"'s le. 101m )( I~ ...,..,.)
i ts Ct re", be. ih<j
A = (I~ )(1,<.):: Iq7 ",,,.,,1. -= 1'92, >( 16(. Wlz..
A+.. +c.,·1(JV'~ +ke +0 v-(.':" F Cú,I"f','~J by ec...c_~ "f ú",ec<..S ¡..
F z: "tA = ('$?x /O')(ICf-;Z )( 10-6) z: 1536 N::- 1.53b kl-J
P = e F =- lG )( l. 53C.) -:: 9. 1.1.. k ~ ....
r----------------------.--------~-----------,Problem 1.19 1.19 The axial force in the column supporting the timber beam shown is P = 75
kN. Determine the smallest allowable length L of the bearing plate if the bearing stress in the timber is no! to exceed 3.0 MPa ..
SOLVTION
c--...r:....;> 7S')({o':'
6¡.,~. = l3J) Y/(I r¿ )(D.I ~(J )
17 ~. 6 '. 10-3 m
----- ---'_- -_.-----~ _.. _ ..-._-._-~---- ----------------------- ------_._._ ....-.,-_.----_._- -
>
Problem 1.20
P = -10k\
1.20 A 40-kN axial load is applied to a short wooden post that is supported by a concrete footing resting on undisturbed soil. Determine (a) the maximurn bearing stress on the concrete footing, (b) the size of the footing fOI which the average bearing stress in the soil is 145 kPa.
100 m 111 (o..) l3e4 f','V\:r s+r-ess 01" COVlC~e"+e.. foof¡"'j'
p = L/o krv ~ Llo )l/Ol N
2A ::' (1 00 )(J~O) .,. 1;< "1. lo 3
Mm '2. .- ¡')._y/Ó3)'Y¡
6" =- ..E ::'"
40 '1( IO~ ..- 3.33.)(.10(,. Pe.A \':{ lo( IO-!.
3.33 M?c... .4\
(b) Fc:>o+í n j ~H '" •
6-:::. pA
-po: 40 "¡Ol N (5"" 1~s k?G.. =: 't5')(/O'!. '?~
A :" 40)( /0'3-
1'+5'><103
A ~ b2.
= o. SRS Yn
Problem 1.21 1.21 An axial load P is supported by a short W8 x 40 colurnn of cross-sectional area A ~ 11.7 in.2 and is distributed to a concrete foundation by a square plate as shown. Knowing that the average normal stress in the column must not exceed 30 ksi and that the bearing stress on the concrete foundation must not exceed 3.0 ksi, determine the side a ofthe plate that will provide the most economical and safe designo
ov P :: 6A ::: (30)(11.7) ::- 351 k; ~>:.
~of' +he :0.. ')( Ct. pl6-te E)=- 3. O h".
A :: 1:. -:-351
-=- 117~
IY¡6" 3.0
A=o..l.
10. -g")_ ll'\.
PROPRJETARY MATERIAL. e 2006 The McGraw-Hill Cornpanies, Ine. AIl rights reserved, No part of chisManual may be displayed, reproduced o, distributed in any form o, by any means, wirhout lile prior wrirten permission of Ihe publisher, o, used beyond the lirnited distribuuon lo teachers ami educators permiued by McGraw-HiIJfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission.
.• "" .. "'. 1ru -- ",.
v
ProbJem 1.22 1.22 Three wooden planks are fas tened together by a series of bolts to form a colwnn. The diameter of each bolt is t in. and the inner diameter of each washer
is t in., which is slightly larger than the diameter of the holes in the planks.
Td ez-. i':····"I,.'
~._._I..:.~~'.. +"210.
Determine the smallest allowable outer diameter d of'the washers, knowing that the average normal stress in the bolts is 5 ksi and that the bearing stress between the washers and the planks must no! exceed 1.2 ksi.
l_. o'
~ . 1, J,. •
6¿:: :
w~her:
A .TI:. _A' _ ...!L ( .l.)a. _ ~ oIf"'" l"bJt:: '+ "fl:, - 't \,~ - O. 1~6~~ i..,
-Force i(l 60)7- P =. 6b A = ,(5)( 0.""35") :: 6.9817S- kips
B ec<V'; n~ c:tV'"E'(;\ Aw:: f (JD~ - d.(_ 2.
')
Ec¡ud;ns I[(c;io2. - J¿:2) = y
do"z':: el. 2.._. 4 P ': (~)l. +
Aw = !
, ~
A. 11G"b ~ = J~lf3:?3 '"
)1").
Problem 1.23 1.23 A O.12-mm-diameter steel rod AB is fitted lo a round hole near cnd e of the wooden member Cl). For the loading shown, determine (a) the maximum average normal stress in the wood, (b) the distance b for which the average shearing stress ís620 kPa on the surfaces indicated by the dashed lines, (e) the average bearing stress on the wood.
ca..) MCv)t.; ..... v"".·cc.ve.1f"tA."je V1o",""", J .5-t~s 1'" +\.e wood.
A",et ~ (7S·· 1:< Xlg):- l. 13'1 >i Id r.-."","'"" 1./34)t ló..! ~-~
p::- 4.50 kJJ =- 4.$0)1 /0'3 tv
3.Cf7 M?4 ..
(b)
le)p
rJtLL S-o 'l< I o~
(f1)1 ID-3 )(I~ l< 10.1J;:(0. 8 MP~
PROPRJETARY MATERIAL. 102006 The McGraw-HilI Compames, Inc, AlI rights reserved, No part ofthis Manual may be displayed, reproduced or distributed in any fonn or by any rneans, without the prior written pennission of the publisher, or uscd beyond the lirnited distribution ro
teachers and educators permined by McGraw-Bill for rheir individual course prepararion. lfyou are a student using this Manual, you are using it without pennission .
.. -_.- -_ ... _._.---_. -----
Problem 1 .24
1.24 The hydraulic cylinder CF, which partiaJly controls the position ofrod DE, has been locked in the position shown. Member BD is ¡in. thick and is connected to the
vertical rod by a t -in.-diameter bolt. Determine (a) the average shearing stress in
the bolt, (b) tbe bearing stress at C in member BD.
Use MeM be..- B~D O-.S o, +-re e boJ)'.) ~",dVI o+e. th "Jo A 8 t ~ C<. -+"-'0 +D~-;' (¿ M el"'l b e l',
H1.8 in.
_P.46 ::. 1/ s1. ... L 'B '¡.
= 8.:<' ;"l. ·400 ces 7So
c)( =·d ZJ:"y = O
et ::
e -:
88.:2. 1=:13
(4 CbS loo )( b G.s ) - (Lfsl"I ~Do)( ~~~ h",,)."...(. 7 ~zoo)(Lfl)O s i« 1~-Q) - (7 S;"l;(o~)(~oo ~75°) = o
3_3"7g F';! - Z7~9'. 3S =- o :. F"B = 8-:2..8-lf'r lb.
- ~ ~8 -+ ex + 4-00 Co.5 75° -= c>
{v~8,2.s·!-#l1 - 'ioo ,c..o~ 7S"';- vs. 3'1 it_B. ~
A)She<t ...;.V\~ st~ i.; +he boif ~ P t: 11'17_ 2
lb
A:: fol'1. ~ ~(f)2. :r 0.110'15';.., ~
t:> .E.. -= " 'i 7. '2.A D.IIO'tS"
(6) BeC(...¡.·'" s+V'ess cJ- e in 1VI€w.he/- 'BCD:
Ab": dt ~ (~)(f) :: O,~34-37S";~'"
I o. ~c.¡ ks l'
'P:: 11 <=f 7, ':l .Rh.
es;. ~ 'P : 1 J 97. ~Ab O.2'!>lt37S
-= 5. 11 /,(5"
V
«( 23.D MPa..
p
e
-"
.J..
e
Problem 1.25 1.25 A 6-mm-diameter pin is used al connection C of the pedal shown. Knowing thatP'" 500 N, determine (a) the average shearing stress in the pin, (b) the nominalbearing stress in the pedal at C, (e) the nominal bearing stress in cach support bracket at C.
75 mm
]: ¿ i ~ "',_, d
5mm
Dv'C!w +v('€ b,,)y Jlo.."~VCJ."'" of AeD. Si",c~ AC.D 1.." e-,3- +u{'ce ¡""ew-.~e.,. ~4he re.:...c ..t 10",\ 4.4 eis Jiv-ec..te.".¡ to..Jt:.".) \:>01",-1 E, ·/-he i",,1e~e¿t¡o ......r +hE' l,'n-es .r c:..d;oV\ of fJ..e ,,1 hN to..lQ ·tOi'Ce s ,
rr'o"", :'1eo ...c...~tt"y,. CE -::4300'l.-\- 'I:<S~ - 3:{S ""...,.
4-t ~Fj =" o: ~~~C· - F -= o e = A_~ P = (1.~)(5DO)"" 1360""
ca..) .."y'p,'.,
---
-teA,
p..>t
_ ~C-!I._I2.
'1 01'
= (2)(1'300) - 1'1>.o'ilID' pe¡1I ></e>3)'2. -
(b) 6b~ s.. s: =-1'300 (
-::- ~ L/. I )({OAb di ((, ><ló'3)("1'>lIO-3)
(el 6!, -z: te :: ---"13.00
'::' Z1.7-,¡lo(, Pt;, '2.1. 7 M pe¡_. ~f;,. ... ::<dt ('2)(G></o-3)(S)(/D-<i ')
Problem 1.26 1.26 Knowing that a force P of magnitude 750 N is applied lo the pedal shown, determine (a) the diameterofthe pinat Cfor which the average shearing stress in the pin is 40 MPa, (b) the corresponding bearing stress in the pedal at e, (e) thecorresponding bcaring stress in che each support bracket al C.
c. =: 2.. b P ~ (7. {,)(750):::- I <f 5 O N
Yp;" ~ te .tf... el 7;" f{t¿ ::-f7.)(
¡'1S0)5.S7 )1/0 5.'57 .....
(o.)A!'I...,
Ah
~.JrcP''1
z:
1fT"" -
ICf50
iT ( LfD >< {o <:-
::: 1M ""4@
(61 6\, -::- (' -=- di. (!;.57)'.,o-~)(1><IO-?' ) 311. '})" IOr. Pe... 3g_9HP4, 4
(el 6'"). ';;' -2--. -:- e -:-
Ti)(sICf50
-' )(_ 35.0 x/O' Po.. 35',0 MPa.. oCIIIll~ 'J.Jt '
::::-
s)( 1 v-:\ ')
Problem
+ 'D 2.
A
1.27 For the assembly and loading ofProb. 1.9, determine (a) the average shearing stress in the pin at B, (b) the average bearing stress at B in member BD, (e) the average bearing stress at B in member ABC, knowing that this member has a 10 x
50- mm uniform rectangular cross section.
1.9 Each of the four vertical links has an 8 x 36-nun uniform rectangular cross section and each of the four pins has a 16-rnrn diarneter. Determine the maximum value ofthe average normal stress in the links connecting (a) points B and D, (b) points C and E.
I)se bao/' ASe ClS Q. +re~ body
0.025\
~O(o
Io.o+o
A 8 e
Feo Fc:.e
(O.O'fO)F'b -(O.02SofO.OQO)(;;'O>l/03): D
Feo ':' 3~.S)( 103 N
(0.) FI.VI a o e =-FAeD
A:: t ~2. = f (o. 016
)2. e
'i 3~.S X 103_..--,.=- (~)(~" l. Ob ~/o-' )
;(OI.Ob)(/O-6 (.
M
80.S HPe<..
eb) 8 ec..rin'l ; }i~k 3D A :: Jt :: (0.016 )(O.OO&)~
6'~:. t Ft~
--
lO.S)( 3~.SX Jet) :=rzs )( JO-'
(c) 8e.a..~¡t'\, jlfl ABe Ov+ B
A= dt -;: (D. 0/(;)(0.010) = ¡f.O '¡(
-, t10 t'\\
~: F'BD :::-
3"(, S)( 10'3
A 160 ')( 10·"-:: ~03 x 10(. ~o3 MPa. ....
PROPRJETARY MATERIAL. © 2006 The McGraw·HiJICornpanies, lnc. Al! rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior wrinen permission of the publisher, or used beyond tbe Iimited distríbution to teachers and educators permitted by McGraw·Hill for their individual course preparation. lf you are a student using this Manual, you are using it
without permission.
-
O~t
6;,
= 1.
Problem 1.28 1.28 For the assembly and loading ofProb. 1.10, determine (a) the average shearing stress in the pin al e, (h) the average bearing stress at e in member Be, (e) theaverage bearing stress al B in member Be.
0.5 in.1.10 Two horizontal 5-kip forces are applied 10 pin B of the assernbly shown. Knowing that a pin of O.8-in. diameter is used al each connection, determine the maximum value ofthe average normal stress (a) in link AB, (h) in link Be.
10 J<.'psFOi'ee tv-"Cl"'1;~
Feo:. JO
Shect,...\'n~ s+f'eSS iv¡ F"Y! ~t eA 'f
7:.-::
_E 't
d 1.. -
JI. (,'\ <:J)'1- -'t u. o -
(~)( D.SO-;¿6 )
('o) Be~.,·.'r'0 sf"e!lS oJI- C. 11>'\ ....e""" bef' Be
A=- td = (0.5)(0.8) -:: 'loP''I
8.q6580.'1 - 22.4 ~';(.4 I<si .
te') 'BeGt -: VI j s+~ss c:o.+ B ,., WleNlkef' Be
A,-=2tcJ ::2(0.5)(0.8;) .0.8 ''''11. ~ I , l. ~ I ks¡'
PROPRlETARY MATERIAL. 02006 The McGraw-Hill Companies, lnc. Al! righls reserved, No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written pennission of the publisher, or used beyond the limited disttibution lO
teachcrs and educarors pcrnutted by McGraw·Hill for their individual course prepararion. lfyou are a student using this Manual, you are using il without permission.
--- --_ ..._. .-
r
Problem 1.29 1.29 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that P = 11 kN, determine the normal andshearing stresses in the glued splice.
(" "IO->;')c .. ,sZ.45D
0= Ir. Z5 >" ( 0- 3-
-= 48 t¡ y,d p""
(ll>qd' Xs;", 'D°)(~Y:II. ~5")(!
D-"J.
Problem 1.30
(o)
(b)
1.30 Two wooden membcrs of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximurn allowable shearing stress in the glued splice is 620 kPa, determine (a) the largest load P that can be safely applied, (b) the correspondíng tensile stress in the splice.
t: = G~O k p~ = G~O )1/03 P&t
'P 5;11 ZB2. A(>
~A. .'é _ ~Y.II.2Sx,ó3)((;20 >'i03)
51""1:le - $oí V> '10°
:: (l3.ttS'>«O!. )((",;>5 4~~)'l.
(1. Z5){ /O-~
-
p.::--
Problem 1.31 1.31 The lA kip load P is supported by two wooden membcrs of unifonn cross section that are joined by the simple glued scarf spliee shown. Determine the normal and shearing stresses in the glued spliee.
p =- 14-00 110 e= 90t:> - boa -= 30"
1:\",:- (S.O) (3. o ) -e 15 ¡..l.
o
6 = 'P ce-s ~B =
(J'100 )((..<>5 30")1.
A, 15
D =- 7Ó.OP-.L' ....,
T ":: 'PS;v.;(e - (lttoo ~ siV\ 60'0
2Ao (1.>( 15')
'L -=- 40_ 4 psi ~
P'
Problem 1.32 1.32 Two wooden members of uniform cross seetion are joined by the simple searfspliee ShOWIl.K.nowingthat the maxiinum aJlowable tensile stress in the glued splieeis 75 psi, determine (a) the largest load P that can be safely supported, (b) the corresponding stress rn the splice.
:,1:
lo:: (S.O)(3.0)~ 15 "VI
!'/t\~:·.t\\")\~¡
e = 90° - '00
P c.es'l. e-- c--
=- 30°
~;\h;1.;'It
..'. (a)(7S )(15)
1500 lb\,:::,.;.1l'.
(bl T - =- _~!SDO) S.-'" f.:,o"(2)(15')
PROPRlETARY MATERIAL. (:) 2006 The McGraw-Híll Companies, Inc, AlI rights reserved. No part of this Manualmay be dísplayed, reproduced o, distributed in any form m by any mcans, wiihout the prior wriucn permission of the publishcr, or used beyond the limited distribution lo
teachers and cducators permitted by McGraw-Hill for their individual course prcparation. lf you are a studcnt using Ihis Manual, you are using il without permission.
o .... • ......
-- --_ ...._.- -----
2
= (s.
337
~A
Problem 1.33
p
1.33 A steel pipe of 300-rnm outer diameter is fabricated from 6-mm-thick plate by welding along a helix that forms an angle of 25° with a plane perpendicular to the axis of the pipe. Knowing that a 250-kN axial force P is applied to the pipe, determine the normal and shearing stresses in directions respectively normal and tangential lo the weld.
do -= O. '300 IV) ro : -!iJo -=:. D. IS-o M
e: ,. V'o - t = O. '5"D - O.OOb = O. 14 L4 .~
Ao = 1f (ro-a. - n2.):: TI (o. ISo '4 - o. 1'11(';)
= S~Sq x/o-.3 Iv¡ ~
e :: zs"
0= ...P.....--C,,,$"e -Ao
:::-37.1
- ;l.S"O)( (o" ct>s:l lSoS:St ')([0-3
G = - 37_' HPo., --
't = _e_ sI'., le _ - ~50 :x 103 siVl SO'O~A~ ('7.. )(5'.5'+ >(/0-3)
-=:-17.:(S «¡ or;; 1: = 17. ~3 MPa.. 48
Problem 1.34
p
1.34 A steel pipe of 300-rnm outer diameter is fabricated from 6-mm-thick piate by welding along a helix that forms an angle of 25° with a plane perpendicular to the axis ofthe pipe. Knowing that the maximum allowable normal and shearing stresses in the directions respectively normal and tangential to the weld are 0= 50 MPa and r= 30 MPa, determine the magnitude P ofthe largest axial force that can be applied to the pipe.
dI,): 0.300 tI1
r.-.=..r.o -t::'('o: tJo: o. \5"0 M
O.ISo - 0.00.6 = O_I'-Iti ~
A.:J - lT(t<)'"-rt):: rr(O.ISO'l.-O.I'ICja.)-= S.5'1 )('10 -! 1M ~
Bo.seJ o... 1 () \ -= 50 MPa_: (5= p
cos eAop:: A~rs S"'LI )( I o·~J ( So ><' (p6).
é~a 16 eos 1. 2So ,
Bc...st!J ~YI I:"! 1 = .30 M~ ?::. -= ....E._ 'S ,'"" 2e
p:: ;<A .. ~:::
(2.)(s..sq 'XI o-!o )( ~o x 106) -= 4-34 x lo
3
SI'" ~e si", 500
SW\o.lPe""" VD,.}'/( is +he ¿jjl()~~ 4 e. V4-Jv-e of 'P t.. P = 337 kN ~.~
.
3
~A
= _e_ 'f. o _f_ -/~D -:<.5""~s"
Problem 1.35
p
1.35 A 240-kip load Pis applied to the granite block shown. Determine the resulting maximwn value of (a) the normal stress, (b) the shearing stress. Specify the orientation ofthe pIane on which each ofthese maximwn values occurs.
36p ~eAa CúS
. z.In
- 210~!o
~B :::.
M0.,6 +eVl5,-j,¡¿ s.h. -e~s :;: O o,..t 8 = <7'0°
~~. cu......_,p...e..!l S ,. ve, s-tv--e5S = ((;.67 k~,.
~t 6::0- O"
(6) 'Y,.,,~ ::. ~z:
")..'+0 ::
(~ ) (~b )3.~3k~l'
". + e~ 'ISo
Problem 1.36
p
1.36 A centric load P is appIied to the granite block shown. Knowing that.the resulting maximum valué of the shearing stress in the block is 2.5 ksi, determine (a) the magnitude of P, (b) the orientation of the surface on which the máximum shearing stress occurs, (e) the normal stress cxerted on the surface, (d) the maximum value ofthe normal stress in the block.
A,,:, (b) (~) = 36 i\O\ a.
e -= lfSO feo/'_J1:J
rJc:c,i>\ e of 1:~
(ti) ?:..~... - 2.Ao
Ipl=- ~Ao T",~ :::;« )( U. )(:2.5 )
(b) si"
04S
'le
4-5 :: -=Aa ~Ao (~)(3( )
6Y">~ P
Ao>::.
- 1,&0 z:
36- .s- ks¡ ~
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+
Problem 1.37 1.37 Link AB is to be made of a steel for which the ultimate normal stress is 450MPa. Determine the cross-sectional area for AB for which the factor of safety willbe 3.50. Assumc that the link will be adequately reinforced around the pins at A andB.
f':- (l.~ )(8) =- s. ~ HJ
'2 MI> -: O
-(O.~XF~l\5;" 35") + (O.:<')(~.6)
+(().'f')(~o) == O
F~~=- 21.619 klJ = 21. e ¡q )/.1'/ N
AA!!. -: (F.S.) ~G
6"lt
= (3.50)( ')./."¡q )l1()~)1.f50)l LO"
PROPRJETARY MATERIAL. «:> 2006 The McGraw-Hill Companies, Inc. AII rights reserved. No part ofthís Manual may be displayed, reproduced01 distnbured in any form 01 by any means, without ihe prior wriUen permission ofthe publisher, or used beyond the hmited distribution ro teachers andeducators permitted by McGraw·Hill fOI their individual course preparation. Ifyou are a student using this Manual, you are using ji without permission.
Problem 1.38 1.38 The horizontal link BC ís t in. thick. has a width w = 1.25 in., and is made of a steeJ with a 65·ksi ultimate strength in tension. Whai is the factor of safety if the
p structure shown is designed to support a load of P = 10 kips?
= ~It
0.31 AS ,., 7
F. s,AsCo OvH_ -:: (O.3./~S)Cb.s"') ::-
Problem 1.39 1.39 The horizontal link. BC is t in. thick and is made of a steel with a 65-ksi ultimare strength in tension. What should be the width w ofthe link ifthe structure shown is to be designed to support a load P> 8 kips wíth a factor of safety equal to
]> 3?
+:) ~ t--'1(. ~. D
(I~ CI>S 30D)FAt, -(f~ s;",30D)(.g) -= O
FA!! ~ b. '12W2 k:ps
~B
AAf'>
D
F"Btw
(3 \(f.. 9~8:C)
(~\ (bS)
w = J.:2. 7cr In.
PROPRlETAR y MATERIAL. <O 2006 The MeGlaw-Hill Companies, lne. Al! rights rcserved. No pan of this Manual may be displayed, reproduced or distributed in any fonn 01 by any means, wirhoullhe prior written permission of the publisher, 01 used beyond the limited distribution to teachers and educators permiucd by McGraw-Hill for their individual course preparation. lfyou ale a student using this Manual, you are using il without permission.
~-_._--------_.-------
¡
UI I.
:. '.
Problem 1.40 1.40 A steelloop ABCD of Jength 1.2 m and of 10-nun diameter is placed as shown around a 24-mm-diameter alwninwn rod AC. Cables BE and DF, each of 12-nun
Q. Knowing that the ultirnate strength of the
r Q
l240mm E B '40mm
¡BLI I~~::::c=::::::~) e
180rnm
F
Q'
diameter, are used lo apply the loadsteel used for the loop and the cables is 480 MPa, determine the largest load Q that can be applied if an overall factor of safety of 3 is desired.
US;h~ j oi ... + 'B o..s~.fi'H. bo..t~I).~J c.o~",·Jtol"l·"~ 'st""W\~,t"q
2· -t '~IS - Q -= c
& -= ~ F;"
Os;",~ jo;", + A a.s GZ.. ty.ee boJyo.V\J C.Oll\ s i J e.,..;"",~ .5'1 ~W\e..+r'1
2. • 1J;c I=",.c. =- O
J.-[Q FA<:. ~D
.8o..seJ 01"1 sfV'e",,~+h J' c~ le gF
Qu .= ~A ': C5'; Jfd2.. = (Ltaoxlt/') l(O.OI.zt - S'I_7.Cf)l.IO'!. N
B(Ue J OV\ s+i~,-H. 0+Q¡) = ~ F"J\J -=
s+eeJ loop
~ e; A = } Cv lf dl.
': f(lf80'XIO{,,)f(o.olo)2. - 4S.?..4y!o:1 N
B~eJ 01"1 sfv..~,4-~ of '('oJ A eG" ::.~ Fk.J~ -
- .1't.. r::- A z:
~ct
~ JI. cl2..Ot)
= ~ (UD'XIOb)f (0.02't)1.. -=
01) I~ +~~sfNt~JJest :.
88. :<~ x 103 IJ
Qv -= 4-5. 21/ .,¿/o'!. N
'Q,
1=:$. = 15'.08 ~/O ~
~ IS.08 k IJ
PROPRlET ARY MATERIAL. (:¡ 2006 The McGraw-Hill Compsnies, Ine. AII rights reserved. No pan of this Manual may be displayed, reproduced or distributed in any fonn or by any means, withoul the prior written pennission of Ihe publisher, or used beyond the limited distribution lo teachers and educators perrniued by McGraw-HiII for their individual course preparation. Ifyou are a student using Ihis Manual, you are using it without permission.
o.
A
Problem 1 .41 1.41 Members AB and Be of the truss shown are made of the same alloy. It is known that a 20-mm-square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be achieved for both bars, determine the required cross- sectionaI area of(a) bar AB, (b) bar AC.
L et'l3+1.. ol ...e......b..~.1'
A 8
)M6 ': -{O.iI"i. .. O.Lf 1.
:: O.~m
I.f) 2M, :: O
.f.t!r-y = O
U se jOln-f A
A ~ t Ay
lJse en-h'"f'e. t('\)SS o.s.
1. ti Ax - (O.1S")(2~) = O
Ay - ~2 = o
~$ ~re~ body
j.. L f=")( = O O- .7S"rr-,.s -
D.8S
o. fl"~e IocxJf
A)t'~ 1S"/dv
At ':28 l(N •
A )/-- o
)C~t.tl
Fk- l
h = (O.8.5")(I.s) := 17 kNAa Q.7S'"
Ay - FA,- - o~'i1;, = O
F,. = 2& - J.º,~_)(171. _.lOe.
30 kN
For the t-es+ hClV A -= (O.D:<'O)1. = 400 ><"/0-'
So :: 'P\) _ I:(o y 103
=t) A - Ifoo x 10-&
M~ ~u=l:<ox/03N
300 )< I o' Pa...
(C~J ~c)(' ~ ~WI be,.. AS
A"6 ~ (F. S,. ) FK =<5'"u
(6) \="011' rne....b.e-- AC
= (F.S.)Fk
A<.. Q()
6'oAASF"a":' ¡gl.33)(lo-'"",,'l
A,.. ::: I~J. '3
PROPRJETARY MATERIAL. © 2006 Tite McGraw-Hill Companies, Inc. AlI rights reserved, No part of litis Manual may be displayed, reproduced or distributed in any fonn or by any means, without the prior wríuen permission of the publisher, or used beyood the limited distribution to teachers and educators permiued by McGraw-Hill for rheir individual course preparation. If you are a student using this Manual. you are using it without permission.
---_._---- _--._ ..,_._~---_."-- -----~----_._ ..._-~.~_.._. ._---
~-
-
-
Problem 1.42 1.42 Members AB and BC of the truss shown are made of the same alloy. It is known that a 20-mm-square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded. Ifbar AB has a cross-sectional area of225mm', determine (a) the factor ofsafety for bar AB, (b) the cross-sectional area ofbar
1:-~·75m I AC if it is to have the same factor of safety as bar AB.
1.4 m
10.4 m_tB
Lenrt·h 0-) VI1elN'lber A13
.RAa -= -l-O.75 2. + O~t.f--;_ := O. 8S ~
Us e ef'lf,'f'e ffvsS ClS '" ~('~e. body
~~Mc = O \.4 A~ - (o.7S)(:<8) ::::o A;I. z: 15 kN
.f.Z.
~ := O Ay - ;( g :: o A'j = 22 J¡ N
Use jo¡",t AAy
~ ~('ee. b~y
..i>,. Z F;.c -=: o
Fol' tÁe fes+ b~r
5'1) =- '?IIA
120 '1(/0'3
400 'ié'lo-&
(a.) FOil" ba.r AS F.S. -= ..&.he
'300 )(/0" ) (~:<s )l/p·")I 7 )C 10 3
(b)
AAC
F.,.
(3. q7)(~o )(/O~) -300 XIO'
F.s.=3.97
PROPRIET ARY MATERIAL. (:1 2006 The McGraw-Hill Companies, Inc. AII rights reserved. No part ofthis Manual may be displayed, reproduced or distributed in any form or by any rneans, without the prior written pennission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. Ifyou are a student using rhis Manual, you are usíng it wilhout pennission.
Problem 1.43 1.43 Three steel bolts are lo be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a 110 kN load, that the ultimate shearingstress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired, determine the required diameter ofthe bolts.
ReC(<.); Y'ed
"""L. - 'Pu -_ 'Pu
p z: 110 '7' 3b.~b7 kiJ3
(F 5.) P ::-(3.35 )(3~."1) i:
du -
~A
m.rr~t2'i
t J 1() k.\' -= (,+'(n ..'1..~~)(/o'!»«! 360 y/o"')
;;(0. g >lIO-3m
J -= 20_8 1M"", ...
~--------.-----------------.----.--.,,--.-----.----_----------Problem 1.44 1.44 Three 18-rnm-diameler steel bolts are to be used to attach the steel plate shown
to a wooden beam. Knowing that the plate will support a t lO-kN load and that theultimate shearing stress for the steel used is 360 MPa, determine the factor of safety for this designo
.G..'~.·O~
........ OIG.'· .. ' .<, ÍX:.," .
.. v..'!~
~-.........t J I(J k:\
-= -::<5lf.lf7 x/O-'- MZ
Pu :: A -t; =- (;?S'f_'17 x/O-(.)(3Go)( loG)
::::- Cf I ..r¿ Dct ')( lo?' N
Fu "" (3 )(9/ ..'09 ..(03 ~
= ::Z7tt - '83 x 103 N
:{}LI '< (o'!
110 )< I~2.50
PROPRIETARY MATERIAL. (O 2006 Thc McGraw·Hill Companies, lnc. AlI rights reserved, No part of'this Manual may be displayed, reproduced or disuibuted in any fcrm or by any means, without the prior wriuen permission of thc publisher, or uscd beyond the limired distribution ro lcachers and educators pcrmiued by MeGr.w-HiII for their individual course preparation. lf you are a studcnt using Ihis Manual, you are using it without pennission.
-_.- --_ ...----_.- ------_._._----_._----- _ ..-- ----- ...._----
11'\
--
Problem 1.45 1.45 Two plates, each t in. thick, are used to splice a plastic strip as shown.
Knowing that the ultimate shearing stress ofthe bonding between the surfaces is 130 psi, determine the factor of safety with respect to shear when P == 325 lb.
0.15' ',,,,.
(See +i~l)t'e. )
A:: ~(~.tS")(O~75)+(1 ..z5')(O.b~5) z: .<.~S'¡n2.
ES. \.~oo
+----------------_._-------------_._------_ .-..
Problem 1.46 1.46 Two wooden members of3.5 x 5.5·in. unifonn rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum
5.5 in.allowable shearing stress in the glued splice is 75 psi, determine the largest axial loadP that can be safely applied.
L -:::.2- s ,..e c.QS
éA..
A o = (3.S )(S.5)::- I q. 1.5
e ':' c.¡o" - 20° 70"
~-Aeo s . ~ éJ
1.
'>'1
'P =7. A::'
':: (~)( I~.~~)L7§)
--SI"" ~e ..siV\ 140"
44'12. ib -= 4. 41 1<:(15
PROPRJET ARY MATERIAL. © 2006 The McGraw-HilI Cornpanies, !ne. AII rights reserved. No par! of this Manua! may be displayed, reproduced or disrributed in any form or by any means, without the prior wriUen pcrmission of the publisher, or used bcyond the limited distribution to
teachers and educators permitted by McGraw-Hi!! for their individual course preparauon. Ifyou are a student using this Manual, you are using il wuhout permission.
Problem 1.47 1.47 A load P is supported as shown by a steel pin that has been inserted in a shortwooden rnember hanging frorn the ceiling. The ultimate strength ofthe wood usedis 60 MPa in tcnsion and 7.5 MPa in shear, while the ultimate strength ofthe steel is145 MPa in shear. Knowing that b = 40 mm, e = 55 mm, and d = 12 mm, determinethe load P ifan overall factor of safety of3.2 is desired.
13Q.SE'd on Jot.)~'e S ~edt../' ¡VI pi....,
'Po = 1..A 'Lti:: ::l lf dI tu- J¡f(:J..)(O.OIZ.y"(f'tS>l/O&) ~ sz.so-to' N
P¡,J:- A'S() w(b-d)()u
:- (O. O..,0)( O. OL}0- 0.01'<)(60)l( 10(.)
Ba.se d 01"\ ~OIJ b.Pe sh eav- , .... -+Ae wooeJ
?u :.- 2 A '1:"lJ = ~ W t.. 1",33. O >'10'3 Iv
-:: (:< )(0.0'10)(0. O.sS)(7,S~/c6)
Use .s I>l fAlJ esf Fu ., 32.8 )(10J N
AiJ()wr.~bie ? -~
ptJ= 32.9 x 103 3
:' 10. 25 lo N
F. S. 3.'l,. w
10.25 kN
PROPRJETARY MATERIAL. © 2006 The McGraw-HiII Companies, lile. Al! rights reserved. No pan of this Manual may be displayed, reproduced or distributed in any fonn or by any means, without 'he prior wrüten pcrmission of the publisher, or used beyond the límitcd distribution 10 teachcrs and cducators permiued by McGraw-HiIl for their individual course prcparation. Ifyou are a student using this Manual, you are using il without permission.
-
=
Problem 1.481.48 Forthe support ofProb. 1.47, knowing that the diameter ofthe pin is d =]6 mm and that the magnitude of the load is P = 20 kN, determine (a) the factor of safety for the pin, (b) the required vaJues of b and e if the factor of safety for the wooden members is the same as that found in part a for the pino
1.47 A load P is supported as shown by a steel pin that has been inserted in a short wooden mcmber hanging from the ceiling. The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of'the steel is145 MPa in shear, Knowing that b = 40 mm, e = 55 mm, and d = 12 mm, determinethe load P if an overall factor of safety of 3.2 is desired.
p = ;(0 kN _ ~O)lIO~ N ,-
(o.) ?¡I"\: A: :¡,. d lo = ~
(0.01~
G) ::.Q
201.06 >' lo
N_? ,.,_J _ HJc- - ~ ~u - :( A
z A '(1) = (2.)( zc 1./6 ><IO-)~(llf5 )(10' )
5'8.:33"'></0$ N
F. s. =
P'P
u
SS. 336 X /f/'~O)lI03. =- 2,q~
(6) TeV'ls,'oll\ '111
PI)A
w oo d
w( b - 01)
SS •3.3~>I/OJ ~ folt' .S~i"'l e F. S.
b =
She411' iV\ woo,,}
.s~eCit(';
PI)2A
0 ..01(; + s~.33~ >1103
(O.O'fO)( ~O -:,clO&)-
.~ -
PI) ": S~.336)iI03 N -tOL" sC(.~e F.S.
e GI."'~ ~('eC<.. is A = wcPy
:(wc_ §_~.~3-:..l.~~/D:::;_3 _
- (~)(o.OYo)(7. S ~'O~)
PROPRlETARY MATERlAL. © 2006 The McGraw-Hill Companies, Ine. AII rights reserved, No part of this Manual may be displayed, reproduced or distributed in any fonn or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and edueators pennitted by McGraw-HiIIfor their individual course preparauon, Ifyou are a student using mis Manual, you are using it without permission.
r- :pI.
~d
Problem 1.49
l'
1.49 A stecl pIale 16 in. thick is embedded in a horizontal concrete slab and is used
to anchor a high-strength vertical cable as shown. The diameter ofthe hole in the
plate is t in., the ultimare strength of the steel used is 36 ksi, and the ultimate
bonding stress between plate and concrete is 300 psi. Knowing that a factor of'safery of3.60 is desired when P = 2.5 kips, determine (a) the required width a ofthe plate, (b) the mínimum depth b lo whích a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the lower end of the plate.)
1':,,'J leb
" :: ¡3 in
.
'BCt..seJ 01'\ +el"ls¡'~", ,.., r2~+eA ':: (o..-d) t
Pu -= 0",14l! ji
~ ... ..J ..
F..s. ': P ov(a -d) t'P
o~
( F. s.) P::-
(3.60 )(:<..-5 ) (36)( -¡f )
Ba...sea 01'1
(a...)
pY~-+e
a. z: l.S6-'0
allld CDV'lCv--e.+ e sj~h
A -= pe ...,.·Me-fe V' )1 df¿p+h =z: 2Tu(c.u·t)h
2 (O+t) b
F•(
.;l.::"
:~P
0.300 \'¿SI'
b = (ES.) 'P2(a.+t)t'v
(b)
(3.') (2.S)=
(1 )(/.55'0 +;i)(0.300)
b =- 8.0S ;1;\.
rROPRlETARY MATERlAL. I!) 2006 The McGraw-Hifl Companics, Inc. AII rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any mcans, wirhout the prior wriucn permission of the publishcr, oc used beyond thc limited distribution to teachcrs and educarors perrnitted by McGraw-Hill for their individual course prcparation. Ifyou are a studcnt using this Manual, you are using il without pcnnission.
'o
P
l
A
Problem 1.50
p
1.50 Determine the factor of safety for the cable anchor in Probo 1.49 when P = 3 kips, knowing that a = 2 in. and b = 7.5 in.
1.49 A steel plate 16 in. thick is embedded in a horizontal concrete slab and is used
lo anchor a high-strength vertical cable as shown. The diameter of the hole in the
plate is t in., the ultimate strength of the steel used is 36 ksi, and the ultimate
bonding stress between plate and concrete is 300 psi. Knowing that a factor of safety of3.60 is desired when P = 2.5 kips, determine (a) the required width a ofthe piare, (b) the minimum depth b to which a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the Jower end ofthe plate.)
3 KI'rs B 0.$ ed on tev¡ ~,.O'" ,., pPcAe
A :- (o. - d ) t::- (.z - ~ )(;f ) '!' O.3Cfo6 I~
Pu : 6"vA(3' )(0.3906): \Y.o, ~;rs
Lr
'.
S._-
.P&
--
13"'·o~ -- Tt. .
r- Q
8 CA.seJ Dh pPa.te aVld c.ov>cY'eJe sj~~
A =. 2-1J'l
~(Q.+ t) b := 2{~ 1-/~ )(7.5)
"'" ':. 0.300 ks i
Pu = 'tIJA = (0,300) (3,+. (1) IO .. i.f1 k,'p$
F.s. -=: .e, ~ 10. 'i I
-= 3.4-7
AcJu(¡J ~~cA,,~ o-P S~.f~+y 1.$ +Áe oS ~ o)jeV' v ",j..;,,? F.S.=3."f7 .....
PROPRIET ARY MATERIAL. 'O 2006 The Mcflraw-Hill Companies, lnc. Al! rights reserved, No part of this Manual may be displayed, reproduced or distributed in any fonn or by any means, without {heprior written permission of rhe publisher, or used beyond the limired distribution lO
teachers and educators permitted by McGraw·Hill for tbeir individual course preparation. Ifyou are a student using this Manual, you are using it without permission.
A
Problem 1.51 1.51 Link AC is made of a steel with a 65-ksi ultimare normal stress and has a t x
t -in. uniform rectangular cross section. It is connected to a support at A and lo
member BCD al C by i-in.-diarneter pins, while member BCD is connected 10
its support at B by a ft -in.-diarneterpin; al! ofthe pins are made ofa steel with a 25-ksi ultirnate shearing stress and are in single shear. Knowing that a factor of safety of3.25 is desired, determine the largest load P that can be applied at D. Note that linkAC is no! reinforced around the pin holes.
Use -bree bo.J.1 BCD.
p +JMB == O: (G)C~ F"c.J - 10 P .....O
P ~ O.!.f'b ~".._ (t)
B,~By
~~t
---"'¡; .±21=',..:O= B)(- !F,.<-
fp SI(=- I~ F;c -- ,-~p
":: O
-+) Me :: o: G By - 4 P ::: O
B_y:- - ~ P r.e . ~ = ~ 'P ~
B -= lB: + 'By" -: -/1.')$1. -+ (~/. p ... 1.1..11(.(,7 P
Shec.'" '''' p;",s wt A a ......I c' ..
p -= O.7c>S8~ 'B (2)
Fv:o.J"'" (,)
~J:: s.
p= (O.70S"88)(O.S8QQ'1) "'"
o.s-g'1Qq
P # 0.300 k:ps
DI" 'P -: 300 l/-'. ~
PROPRIETARY MATERIAL. © 2006 The MeGraw-HilI Compaoies, Ine. AII righrs reserved. No parl of this Manual may be displayed, reproduced or distributcd in any form or by any mcans, without the prior wriuen permission of Ihe publisher, or used beyond the limited distribution 10
teachers and cducators permined by McGraw-HiII for rheir individual course prcparation. If you are a studem using tbis Manual, you are using il wirhout permission.
-- --- ...._--_._----
)1 - 10 t-,..._ ":.
=-
s
Problem 1.52 1.52 Solve Probo ].51, assuming that the structure has been redesigned lo use ~ -in.
diameter pins at A and C as well as at B and that no other change has been made.A
1.51 Link AC is made of a steel with a 65-ksi ultimate normal stress and has a t x
t in. unifonn rectangular cross section. It is connected to a support at A and lo
member BCD at C by t -in-diameter pins, while member BCD is connected to
its
support at B by a ~ -in.-diameter pin; all ofthe pins are made of a steel with a 25-ksi
ultimate shearing stress and are in single shear. Knowing that a factor of safety of3.25 is desired, determine the largest load P that may be applied al D. Note that linkAC is nOIreinforced around the pin hoJes.
p
U'S e t:1I'.e..e. b~1 r~CD
+) ¿MB -= O: (, )(/0 FA<.) - 10 P == O
B'f 8j
~~C
ep::: 0.4% F,., (1)
t'P +f -¿ Fj ~ O ! 8 G. - .
O
6)( -= I~ ~c. -= I.~S-.P -+)2Mc_. -:;::O: :
\3 = -/8/ + e;
- G By - lf P :: O
';:;)J.?.st. + (i)4 ? ... .'P -- 0.70583 B (2)
5h~a. e- In pins al A 0..",) C ..
r A _ ~ ..Ir p. -=( ~s V 17 ) ( S)~ -r-A<:. =- 'L pi.... - F.. S. 't ~ 3. Z-S"·J\. '+ lf'
T.e-Vl,/t>V\ 01'\ nef sect,'o,," "...4 A 4..V\J er /C:"" A ~ A _( es "(1.)(.1_~) =-1- Ac. 'S 1..,) nef S, ,..J - 3.2.5") <t :2. "
S"", ,..Jle.r \/G.)"e of FAc. is o. .ss=« C, k;~~
s~~Cc~ i"" pi.,.. c;J 'S.
'=- ..vA. - 'tI) Ji. p.. _( '2S )(.Ll)(~)2. -::f... f''''· ES. '-t CI - 3.~S '1 IC.
FNW\ (~)
PROPRIETAR y MATERlAL. ~ 2006 The McGraw-Hill Companies, lnc. Al! rights reserved. No par! of rhis Manual may be displayed, reproduced or distributed in any fonn or by any means, without the prior wriuen permission of the publisher, oc used beyond [he Iimited distribution ro teachers and educators permitted by MeG,a\V-Hillfor their individual course preparation. If you are a student using this Manual. you are using it without pennission.
te : )~,A S
'2~~
' m
Problem 1.53 1.53 In the structure shown, an 8-mm-diameter pin is used at A, and 12-mmdiarneter píos are used at B and D. Knowing that the ultimate shearing stress is 100MPa at all connections and that the ultimate normal stress is 250 MPa in each ofthe two links joining B and D, determine the allowable load P if an overall factor of safety of3.0 is desired.
Top view
¡--200mm-r1SO rnm-j12 mmj_
8mm<1 stc...tI<.s: USe A Be c.s fV"'ee body.I A B e I
LA B e -e _ P B t=o.,0+0.;q p
- 1.-20mm p8 mm-e-
- 1-8 mm...
~¡- D
FA Feo
¿ Me. = e 0.20 FA - O./~ P :: O
Front view12mm-l f-
Side view
?: ~ FA
0.10 Fe!> - 0.38 P ::; o.
8"-sed "" oIo"ble she,,'" ¡", pó"" A ~ !; F.."A :: ~ el 'l = -if (0.002'» 1. -= so. ~'bx 10-' WI 'L
l='A ': = (:lA I00 )(;.~ )(50. ~Íl"~~<?c2-:: 3. '3SI )1 Jo 3 N
P = -tf ~ =- 3.7~ «te' N
Bc..~I?J 011'1 Jovble sheCtV' 111 pn~s C\.i 8 Q.v-.d DA -: lLLf . d'4 -
=
1L (O,OI~)a. :- J 13. lo )1./0
-~ '1.
FSD = :;{~~ :: (a)(IOO)(/OC. )(113.10)1/0-6) -= 7.S,+ »to? NF. s. 3_0
p :: ~ F8D -:::- 3. en )(lo 3 N
BQ.sed 0111 c,e
....p.r-ess" 011'1 In JI,W'\ ks BD _.~ 7_
FoV' one. R¡"k A = (o.o,'{o J(o. (08) : 1 ~o -;'/0 ~
Feo = 2 E>" A
-:::<~l( ~So)(/0' )( 1'0 )(/C-'- ) ::- ~'.7 )/t/O'!:, IV
F-S. 3.0
p = ,I~Feo - ILt.04)1 10$ N
A'~CI..,o..b.Pe vo..lue of P is :5W¡Q,JJest ... 'P= 3.7~)l/O~
N
3_12 ktv ..,
.•_._-_ .._--_._ .. _._-~-.-
+1-
«
Problem 1.54
Top view
1.54 In an altemative design for the structure ofProb. 1.53, a pin of lü-mm-diameter is to be used at A. Assuming that aH other specifications remain unchanged, determine the alIowable load P if an overall factor of safety of 3.0 is desired.
1.53 In the structure shown, an 8-mm-diameter pin is used al A, and ) 2-mmdiameter pins are used at B and D. Knowing that the ultimate shearing stress is 100MPa al all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the alIowable load P if an overall factor of safety of3.0 is desired.
¡-200 mm-r-180 mm"112 mm
Bmrn<:lj__~ ; ~ )-1r! _A B e I stcd-t'c.s: Use ABe c.s fv-ee body.
lA B eA S c.
fO.,0 0;q-- k-20mm P
1- ~8mm FA8mm- 1-
PFsp
D o...,.¡ _,
Front view
( - ~D----""7~
12mm-JSíde view
¿ Me. = e 0.20 FA - O. I~ P = O
p: ~ FA
0.10 Fr~D- 0.'3& P =: O
B«.sed O" dovble sheQ~ ¡", pi", A >z~F"D~A -= ~ J'Z z: l (0.0. 't)~ = lfi.,S"~· xIO-<; ~ ~
l=A = 1.TuA :;;(2l 100)( IO~)( 1S_.stt ~ -= S.l3G)I Jo 3 N¡::: s. '3. O
P = T FA ;:- ,S.a~ ~II)~ N
B~sed OVl Jovb/e sl,eCtV' 111 p'\f\S ,d 8 t:<~d DA -= 11- e(1. '::
lt:. (e. Ol~)~ :- 11 s. lo ></<)
• 6 'L'f 1'>1
F. - ~~A :: (:t)(/oo)t/O')(113.IO)l/o-6) -= 7.S't)(IO~r-Jso - F. S, 3_0
P = !~FSD ~ 3. en x lo 3 N
Sa.see! OVl c.Otrlpr-~SS;OVl In ilf'lks BD
FoV' en e Rif\k A = (o ..ozo )(0. 008) ::
Feo = 2SúA ~ (:¿)(~So )(/o')(¡'OX/c--) _F..S. 3. o
'P = ;)c0r
F.;8D -
_._------.- ._-- ----------------------
rSJ) e F.S.
Problem 1.55
Front view
1.55 In the steel structure shown, a ó-mm-dlarneter pin is used at e and lü-mm diarneter pins are used al B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes.
Use. +~e boJ, A s c .
+'¿:Mc.. :; O:
O.2So P - O.I~O Fs.o -= O
(1)
l'
At- ... ·
Sideview
+,Zf'1e. -==0:
o. I bO P - 0_ r;(O e =- o
P ~ ..~
3 e (~)
She4'" IV> píV\$ J B 4"'1.,1 D.
FIlJ:) -= 6' A~t ": ~. A".f
::- (400 le 10)(6~IO-!.)( /'~ -10 )(lól)3
c. ltO )l/o?' N
r -= ,yA - :- 1".; ][d4. ~(}SO)l./O')(li)(IO"l63y' -= '3_9''),;r¡o: ')(/03 NPI", 't '3
,S..",,)Pe.r I/(AJlJe of Fso t'S 3:c:¡::t7{t¡.10 N.
i=',..,,,,, (1)
S h e a V' , VI P iVI c..,.+ e
c= 2rC Ap;" =-'"1 ~~t=:s.
JI. ,<. .,. (2) (15'0 x(ol. )(JI )(bxlo1)l ~'tcx 3 1-1
(1.') ? = (~l(,<.g:<7'+)(td) ~ 2. r~)( 103 rJ
p =- 1.'~3x/o' N
'
N1
Problem 1.56
Front view
1.56 Solve Probo 1.55, assuming that the structure has been redesigned lo use 12- mm-diarneter pins at B and D and no other change has been made.
1.55 In the steel structure shown, a 6-mm-diarneter pin is used at e and l O-
mm diarneter pins are used at B and D. The ultimare shearing stress is 150 MPa al all connections, and the ultirnate normal stress is 400 MPa in link BD. Knowng that a factor of safety of 3 is desíred, determine the largest load P that many be applied al A. Note that link BD is no! reinforced around the pin holes.
Use F..~. e bool..1 ABC.I
6 mm +.)2M
e, = O -;
A ..···
Sidevicw
o_28oP - D_'~O Fep :: O
p -= ~ Fa.!' (1)
+'L"Ma -= o:O.lbo P - O.IZO e = oP .= SiC (l)
·Shh e- ¡....p;"''' d B CtVlJ
D.
FSD ~ SA.,J -= :~. A.-.'rt:c (400; 10"')(, )1103 )(1 ~ - 1:)((16~)
= 4_'80 '1. r03 N
r- -ro ..,lA. ':" 'f'", jI .1'1. =-( ISO>l/Ú')(' Ji )(I~)(lósy.rSI) t: p.~ F; S. '1 CI 3 't
sMaJiev: vr.,J,Je 01 l;o 105 't- 80 }t lal t-J.
'f"~"" (1)
She c< '" .. 11\ P i1'\ cvl e
S. bSLfC, )l s
e :::Z-1: Ap;" ~ 2 :.~. fc1' -:(').)
(/5'0;10
")(J¿)(b)lló3jt.;:- 2_BZ7y}tlo3 N
F.t'oWl l?.) P = (~)('Z_~2.7'" Y{DS) ~ :2.11.. )I{03 N
--------------- .._-
._---------
Problem 1.57 "1.57 The Load and Resistance Factor Design method is lo be used to select the two cables that will raise and lowcr a platform supporting two window washers. Theplatform weights 160 lb and each ofthe window washers is assumed to weighl195 lb with equipment. Since these workers are free to move on the platform, 75% of their total weight and the weight oftheir equipment will be used as the design Iiveload of each cable. (a) Assuming a resistance factor t/J= 0.85 and load factors Yo =1.2 and Yt. = 1.5, determine (he required minimum ultimate load of one cable. (b)What is the conventional factor of safety for the selected cables?
?u -= "Y"o 'PQ + ~ H.cp
(I.?_'~( ~){ 1(0) + (1.5")( ¡~~)l1~S')
-:=o.as
b'ZCf lb.
37;<.S lb.
PROPRlETARY MATERIAL. <O 2006 The McGraw-Hill Companies, Inc. Al! rights reserved. No part ofthis Manual mal' be displayed, reproduced or distríbured in any form or by any mcans, without the prior wriuen permission of the publisher, or used bcyond the limited dislribution lo reachcrs and cducarors permitted by McGraw-HiII for their individual course preparation. If you are a studcnt using this Manual. you are using it
--------------- .._-
._---------
without pcnuission.
-
i: 3~~ k'J
rProblem 1.58 "'1.58 A 40-kg platform is attached to the end B of a SO-kgwooden beam AB, which
is supported as shown by a pin at A and by a slender steel rod Be with a 12-kN ultimate load. (a) Using the Load and Resistance Factor Design method with aresistance factor f/J = 0.90 and load factors Yo = 1.25 and Y/. = 1.6, determine thelargest load that can be safely placed on the platform. (b) What is the correspondingconventional factor of safety for rod Be?
1.8 m .¡
I ':L···~ >1
~--2.4m-
(2.4)}P- 2.'1 vr; - /.~ WzFo(' d~J .PGlC4J I~~
;Vi. ' (SO)( Cf. g I ') -
:: (uo ) ( ~. 8 1) :- N
l.J,qo.s N
"Pe :"
PL. =
(~X3C:Y2.4) i-(f)(4'lO.S)::
W, = rn~ -w';.=
"1'1) Po + 'Y',- 'PA.. ': <P p~
Ft -= <:pP", - Y'b 'P¡ - ~o.qo}(I~)(~Ol) - (,.:<~-)(I.Ob~8)(/O-!)Y'L.
-- S. 9 ~o)( lo 3
N
APJO'N Ov~;t¿ iOA.J M - ,3 s.er2 )/(IOl"3' Cf. 8'
P ':' 'PD4- 'PL = i. 06"<8 )I/o~ .. S.92.0'/.)0 ~
-~
6.':;&3 >'{O N
F. S. - "PI)::
I~ )i/O~_._-_ ..
-::
t. 7 18
p ,. C"f2.3 )o( I O :3
PROPRIETARY MATERIAL. te 2006 The McGraw-HiII Companies, Inc AH rights reserved. No pan of rhis Manual may be displayed, reprodueed or distributed in any form 01 by any rneans, without the prior written permission of the publisher, or used beyond Ihe limited disrribution lo
teachers and educators permitred by MeGraw-Hill for their individual course preparation. lfyou are a studenr using this Manual, you are using il without permission.
Problem 1.59 1.59 For (he Pratt bridge truss and loading shown, determine the average stress in member BE, knowing that the cross-sectional area of that member is 3750 mm'.
:3(iOk:\ 3(iO k:\ ,1(iOk:--:
B
Problem 1.60
Use e~hV'e.t('t)s,s o.s -h--~~ b~i-+~ ¿-MI-( = o:(3)(3'0).¡ Cb ")(3,"0) ~ (9)(3{.0) - 1:<.~ -= oAt = .[;40 ~....,
Use po,...+t'OV1 o--t 1-V'vss +0 +he .Pc#+ ot- CA. .seoh'o"'l
cu+tiv\~ mel!\1 b~S 8D) BEJ a",d CE.
;·+1 ¿Fy = o
1.60 Knowing that link DE is I in. wide and t in. thick, determine fue normal stress
in the central portion of that link when (a) 0= O, (b) 0= 90°.
.:4,4 ¡n.•,E
u S
eI'vle. ......be... eE F c.-!> o, f..,..ú b ~ y
-t) 2:Mc_ -= o
I~ Fo,;: - (g'(Gb S;"l e) - (1') ('0 c..,.s é) = o
F"é -= - ,+0 si" e - ~ c..:>s e lb
(0.) e =- o: F"E = - 80 )1,.
F
(6) e -=- qDO~ rpt: ':- - 40 .,eb.
- '10GOE ~ O_Us ': - 3.to f '!>;
. • . ·.0 _
. • •0 • •.•• •••• ••• 0
.•0
:"
Ar
d=~ f;{ .3~. 8 kSI'
Problem 1.61
¡-di
1.61 Two wooden planks, each 22 mm thick and 160 mm wide, are joined by the glued mortise joint shown. Knowing that the joint will fail when the average shearing stress in the glue reaches 820 lePa, determine the smallest allowable length d ofthe cuts ifthe joint is to withstand an axial load ofmagnitude P = 7.6 kN.
Se\ler'l sC)V'f'lce~ Go''{'''Y tl,e ~-t.P..Qt)~ P ': 7_b k~ -= 7.~ "!cr
'f-: .z. A ':' .e. -:
Ea.rh ~P...,e C(.n", ís A -: Jt-I.b )1./03
L3:<464 )/ IQ-S 2-
7A 77: (7)('8~o )1 tOS )1.
- 1_ 3Z4 0'1 ~ /03 h')"",
-z: {~3~'ioq)(/O'l -=;('l.
Problem 1.62 1.62 LinkAR, ofwidth b = 2 in. and thickness t=t in., is used to support the end
of a horizontal beam. Knowing that the average normal stress in the link is -20 ksi, and that the average shearing stress in each ofthe two pins is 12 ksi, determine (a) the diarneter d ofthe pins, (b) the average bearing stress in the link.
Rocl AS ¡s In Co~ p'..e.ss ,'?t1.
A -= bt v.I ~ e-l'e b: ~;"" et,.,d t=* ,11'1
P -oA :::-. (- 2O )( ~ )( t ):- 10 K ir:os
Pil1 ; 'Ir - E.. ON\cI Ap IL d z,
( 0.) '::' (4)(/0) /.0.30 1101 • -4
r -rr(I-:t)
tb 1 J:_d..t
10- (l.o30)(O.~5)
PROPRJET ARY MATERIAL. (:)2006 The McGraw-Hill Companies, Ine. AH rights reserved. No part ofthis Manual may be displayed, reprodueed or disuibuted in any fonn or by any means, without the prior wrinen pennission of the publisher, or used beyond the limited distribution to teachers and educators pennitted by McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual, you are using it without permission.
E ,j.' , ni~. \ 1.500.ro?
t.t f"" -= ::.--
''1
3
Problem 1.63 1.63 Two identical linkage-and-hydraulic-cylinder systems control the position of (he forks of a fork-Iift truck. The load supported by the one systern shown is 1500 lb. Knowing that the thickness of member BD is í in., determine (a) the average
shearing stress in the t -in.-diameter pin at S, (h) the bearing stress at B in mernber
BD.
-> ¿ Me, -:: o:
:zy e -(~o)('S(}o) ~ O
E :- t:<SD ..Jb. ~
z, ~ F)I = O
;~ 1::7 E + B" :. O 8)(: - E
ac ': 12SD lt ....
+t 7F1 ::-o:. Bj ISoo = O 8J ~ /5()f;J')1
8 -::--/8/ + By:!' =- ~)").so"t + 1.500"'::- IQS2,S6 n:
A r;" -= Ti J.'- ~(~f o. ¡q, 35 1.
'L. :o- Ji. ::- I'lS~ •
1 .5
•• ..,
...._1
-=c:¡. q'l lo
3
A,:", O./"tb35
(\:'1 BeCl.rl·l'\~ ~..¡.res~ w-l '8.
)< p:, .
6' = B ~ le¡5"~.Sb-= b_::<S'>< lo f~(clt ct )(i")
PROPRlETARY MATERIAL. el 2006 The MeGraw·Hil!Companics, Inc. Al! rights reserved. No part of this Manual may be displayed, reproduced or disrributed in any fonn or by any rneans, without the prior wriuen pcrmission of the publisher, or used beyond the Jimited distribunon to teachers and educarors permittcd by McGraw-Hil1 for their individual course preparation, If you are a student usmg this Manual, you are using it without pcrmission,
Problem 1.64 1.64 Determine the largest load P that can be applied at A when 8 = 60°. knowing that the average shearing stress in the 1O-mm-diameter pin at B must not exceed 120MPa and that the average bearing stress in member AB and in the bracket at B mus!not exceed 90 Ml'a.
". ' ..~.
A Geo~e+r---I ~ T{'Ic:.Vljie A'B(' ¡~
D.VI i soSE:...Je~ tV',oa V'¡~Je wi+\ry
o.h~J~~ Shol.V'n b ev-e
U se j oílfl" A tl..S f r-e e bojy
8 75"0
p ~alA) uf S ¡r\ e s Cc.f pi ..eJ fe,
p -to t-Le +V'ICIr\ ,Je
F~aSi", I~oo
If sheC("";VI~ str-tss P'I piV\ cVt '8 is CW"i hc~l
Ar. -= lcl'l.~ t{O.OIOY- = 78.S'1><JO-' M2.
F~I'3': "J.A't. = (~)(7g.~'1)f)O·';(I';{O'W'J06.¡:: 12.2Soxlos N°
:rF beGtf';"~ s+~.s \Y\ 'M eW"\ bev: AS tJ..+ b~c.,lte+ 0.+ A 1$ G'/',' +,'0.)
AbO: t~:- (0.016)(0.0/0):- 1&0)1-/0·' t"I~
F"B: AbO... :: ( 1(,0)(/0 -, )( Cfo »ro 6) -=- ) Lf. 'fo ><10'$ N
Ir beotV'\'n-, ~$S \~ +~e b~c-kef ~+ 13 is c't'\+ie.~)
Ab ':: 2 t J = (2.)( 0.012 )(0. Ole) '=: ~~o )(/0-6 1'Vl~-
FAB -z: Ah 6k_. ~ ° ( 240 )t'lo"~)( Cfo 'I<':/Ofo) -: ~ l. (. x/o ~ N
A}.PooV "hJe FA6 15 He s~QJ)es.+.; I.e. /4. L/o ~ lO!. tJ
TJ, e~J rt~~ S+JI'C.S 'P..J.I.a.J ~ (O. S7735 )( N. ttox 103)
¡li
Problem 1.65 1.65 The 2000-lb load may be moved along the bearn BD lo any position between stops al E and F. Knowing that 0.11 == 6 ksi for the steel used in rods AB and CD, determine where the stops should be placed ifthe pennitted rnotion of the load is lo
I¡..·-. ---60 il1.---"" be as large as possiblc.
. ,,------i+,!::;"xE~ ~
El: '/. F1
Br-;:==~~~==~DA8 ~ O".p A~""= (6 )( ~ )(1) 1..
'-17'& lo k:ps
e D· (F~ t", -::G.,.u ACf> ~ (,)( ~ Xj) lo
-:- /-~'to7g k:rs
Use. Me-~e<l" BEFO e...s 4. f,..¿~ boofr ..p= ;Zooo lb "" 2.000 k:rs
+:>'2HD= o
- (bo) FAe + (Go - ~e) f' = o
bO F"c,P
35.3'+3
o
(bO)( l. 178 IOJ::2.000
'2'1.7
~p -z:
{;O ~o ...,p
o(bo )(1. s 'Ion)
;(,000
x..p:: SS_,z In. --4
PROPRlET ARY MATERIAL. lO2006 Thc McOraw-HiUCompanics, Ine. AII rights reserved, No part of this Manual may be displayed, reproduced or distributed in any form or by any rneans, without the prior wriuen permission of rhe publisher, or used bcyond the limited distribution lO
teaehers and educators pcrmittcd by Mcflraw-Hill for their individual course prcparation. Ifyou are a student using lhis Manual, you are using it without permission .
..-._-------------
Problem 1.66 1.66 Two wooden members of75 x 125-mm uniform rectangular cross section arejoined by the simple glued joint shown. Knowing that P = 3.6 kN and that theultimate strength ofthe glue is [.1 MPa in tensión and 1.4 MPa in shear, determine(he factor of safety.
p:.- 3. b ktJ ~ 3. b )( fo.:s
t-J
A,;= (7S)(t~ ') 1 t:
(3_')< lO~) C~.s ~ 5<»1-'l_ '!:,7'S >< (O-~
=- . 3.4-9
(3.(, "'lO~ )(s; ...S. Do)
(A)(<7.37Sx {0-3)
F. s. (.1./ )( 10"['-17_0&1 ><[0'3
PROPRlETARY MATERIAL. © 2006 The McGraw·Hill Companies, lnc. AII rights reserved, No par! ofthis Manual may be displayed, reproduced or distributed in any fonn or by any means, wi thOUI rhe prior wrinen perrnission of the publisher, o, used beyond the limited distribution lo teacners aud educators permiucd by McGraw-Hill for thcir individual course preparation. lfyou are a student using this Manual, you are using jI
withour pennission.
f
=- -
Problem 1.67 1.67 Each ofthe two vcrticallínks CF connecting the two horizontal members AD and EG has a 10 x 40-rnrn unifonn rectangular cross section and is made of a steel with an ultimate strength in tensión 01' 400 MPa, while each.of the pins al C and F has a 20-mm diameter and is rnade of a steel with an ultimate strength in shear of l 50MPa. Determine the overall factor of safety for the links CF and the pins connecting
1'250 mn~
,«,~,,,~. ~ I 400 mm
-, ~ 250mm
1
them to the horizontal rnembers.
, Use Me~be,. &'FG a.s ~ee baJy.
Fes FcF
~ O.lfo ~"_O.2S
2H'\ 2'4' kN~¿ ME :: O
O. 'lo Fc,F" - (O .. bS)(~4'1(101) e O
Fe~ ':: 3~)( 10" N
8CLsed OVl te.Ii'IS;~14 rn 2i.,., ks CFA = (b- el) t :: (0.0'10 - O.o~)(O_OIO) -= ';(OD X./o-' yY)" (one);t\k')
Fu -=2Dc,)A= (2~1l00'l.IO(.)('lOO'1C/O-') ':' Ib'O.O « to" N
B I:t~ eci Oh dO') ::;)e S h eC(.V' iY\ t-r»
A -= ][ J~i-f z: "74T-'-
(0_020
)z. = 314.1f, X /0- (. Vl1 a.
Fv :::: 2_ TtI A :- (2)(15D';(/0')(31'1," )CID-e.) = Cflf.;¿48 'I</I.,)! N
A r.+ ,).; i) t,: ,
StTl ",Jjevo VA}Ve .) l.e. Ft) =- Q4/. '2'-18 ~ 10 3 N
FQ.c..-\·(),;- ot S"'.¡:e.ty t=: s. ¡::-¡) s« '248 )l/O~-=
FcF=' gGf -11012. 'f 2 ~
rROPRlETARY !\lATERIAL. :o 200ó The McGraw-HíJl Cornpanies, Inc, AJI rights reserved, No part of this Manual may be displayed, reproduced or distributed in any fonn or by any means, without the prior wrinen pennissíon of the pubhsher, or used beyond the limited distribution 10 teachers .00 educators pcrmined by McGraw-Hill for thcir individual coursc preparation. Ir you are a student using this Manual. you are using it
witbour permission,
--_ .. ._--_ .._ ..- ..--.. U~ • ~ ~ ._ ---- •• -- ~ - --.-- - - Oh -_ ... -~ ----
+
e - ~ -.5 ::
Problem 1.68 1.68 A force P is applied as shown to a steel reinforcing bar that has been embeddedin a block of concrete. Determine the smallest length L for which the full allowablenormal stress in the bar can be developed. Express the result in terms ofthe diameter d ofthe bar, the allowable normal stress o.n in the steel, and the average allowable bond stress T.n between the concrete and the cylindrical surface of the bar. (Neglectthe normal stresses between the concrete and the end ofthe bar.)
A:= TidL
F -:::1:'J/ A = 1:J' .I 1T 01 L
r-v'" e",~ío""'J A -_ ..!'Lt
.~c::;{
Problem 1.69 1.69 The two portions ofmember AB are glued together along a plane forming an angle e with the horizontal. Knowing that the ultimate stress for the glued joint is2.5 ksi in tension and 1.3 ksi in shear, determine (a) the value of e for which tbefactor of safety ofthe member is maximum, (b) the corresponding value ofthe factorof safety. (Hint: Equate the expressions obtained for the factors of safety with respect to normal stress and shear.)
Ao = (2.0)(1.1..S) = -;'.$0 i..1-..
( F. S.Ja- -: (F. s.) t"'R - 6't) .A~
u,,<S" - Cos Le
(b)
feos 'Le~ ":'ces éf)
Pv :::-6"'1) AgC.ús "'8
_ T" Ao- 'Ps: ..e.> cws (f) J..l.. _ 1:'..JTa.v>
::. (!'2~51 (2.$"0)
t:.o~'1 ~7.&,"
L'31
e
-
5'v
Problem 1.70 1.70 The two portions of member AB are glued together along aplane fonning an angle with the horizontal. Knowing that the ultimare stress for the glued joint is2.5 ksi in tension and 1.3 ksi in shear. determine the range of values of e for whichthe factor of safety of the members is at least 3.0.
A" -= ( ':<. O ) ( /. ~5 .) ':" 2.S0 .ivt t
P -:::: e.tt k.p s PI) 1: (F ..S)P =- 7. ¡ k:r~
94.seJ 0\1'1 telll~¡le S1v-es,'Pu . 2.eA. e.es
cos.2e :: 5wA~ ~ (;"2..5 )(:(.So~
O_8b8'o6'Pu 7.2-
CO$ e o.c¡g re t:f e -::- 21 .. '3° e~ 21.30'
B~5~J (:111'1 s h ~(ot",; ... ~ .s~ss t"C) :: 11. si~ e ~s e ::
Ao
Pu
;¿Ao si.,., ~e
SI", 2& ~ 2Ao r" ::: (;()( ':< .. $0) 0.3)"
:: O .. QO?7 gPu 1_':<
2e ~ G'1. sz· .e t: 3~. '3" e ~ 3~.3°
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~
PROBLEM i.ci
Element n
Element 1
l.el A solid steel rod consisting of n cylindricaJ elernents welded together is subjected to the loading shown. The diameter of elernent i is denoted by di and the load applied to its lower end by Pi with the rnagnitude P, of this load being assurned positive ifP, is directed downward as shown and negative otherwise. (a) Write a computer program that can be used with either SI or U.S. custornary units to determine the average stress in each elernent of the rod. (b) Use this program tosolve Probs. 1.1 and 1.3.
SOLUTION
(DRCE Irl f;L_cM'ENT L ..ft 15 t~é'5un) 01 fhe f~r:e5 apphedp¡
AVe R_A. G'E
Arca. =- AL -
I¡
10 +~l{tt e/ement Ond all 1[)~·Vf( OI1~::;:
p~~oe:
t. L-. 'kk = I
i?&06'RA~1 OUT PU T~
Problem 1.1Element Stress (MPa)
Problem 1.3Element Stress (ksi)
1 42.441 1 12.7322 35.651 2 -2.829
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PROBLEM l.C2 l.C2 A 20-kN load is applied as shown to the horizontal member ABe.Mernber ABe has a 10 x 50-mm uniform rectangular cross section and is supported by four verticallinks, each of 8 x 36-mm uniform rectangular cross section. Eachof the four pins al A, B, e, and D has the sarne diarneter d and is in double shear.(a) Write a computer program to calculate for values of d from lOto 30 mm, usingl-mm increments, (1) the maximum value ofthe average normal stress in the links connecting pins B and D, (2) the average normal stress in the links connecting pinse and E, (3) the average shearing stress in pin B, (4) the average shearing stress inpin C, (5) the average bearing stress at B in member ABC, (6) the average bearing stress al e in member ABe. (h) Check your prograrn by comparing the valuesobtained ford= 16 mm with the answers given for Probs. 1. 9 and 1.27. (e) Use this prograrn to find the permissible values of the diarneter d of the pins, knowíng that the allowable values ofthe normal, shearing, and bearing stresses for the steel used are, respectively 150 MPa, 90 MPa, and 230 MPa. (d) Solve part e, assuming that the thickness of member ABC has been reduced from lOto 8 rrun.
SOLUTION
BDTAic.kneS6= 1.'1
JiBJ) '=::. tJ_ (1.1.-[ - d)
CE_[) z: + ~D/AEO
PROGRAM OUTPUTSPROBLEM 1.C2 CONTINUED
18.00 112.85 -21.70 63.86 24.56 180.56 69.4419.00 119.49 -21.70 57.31 22.04 171.05 65.7920.00 126.95 -21.70 51.73 19.89 162.50 62.5021.00 135.42 -21.70 46.92 18.04 154.76 59.5222.00 145.09 -21.70 42.75 16.44 147.73 56.8223.00 ~.... -21.70 39.11 15.04 141.30 54.3524.00 ~/ -21.70 35.92 13.82 135.42 52.0825.00 -21.70 33.10 12.73 130.00 50.0026.00 ~ -21.70 30.61 11.77 125.00 48.0827.00 _,.-21.70 28.3-8 10.92 120.37 46.3028.0029:O O ....z9Y;i.8"- - 21.7O
26.3924 .6O
10.159.46
116.07112 .07
44.6443 .1
3O . OO ..-13 8-: 5-1('" - 21 .7O 2 2 .99 8 . 84 108.33 41.6 7
~. 142.05./~ 130.21
120.19111.61104.17
INPUT DATA FOR PARTS (a), (b), (e); P == 20 kN, AB == 0.25 m, BC = 0.40 m, AC == 0.65 m, TL = 8 mm, WL = 36 mm, TAC = 10 mm, WAC == 50 mm
. -d Sigma BD Sigma CE Tau B Tau C SigBear B SigBear C
11.00 81.25 -21.70 ~'9 65.77 113.6412.00 84.64 -21.70 ~·.6. 55.26 104.1713.00 88.32 -21.70 47.09 96.1514.00 92.33 -21.70 ~~ 40.60 89.2915.00 96.73 -21.70 9y'% 35.37 83.33 (b'\16.00 101.56 -21.70 .82 31.08 203.12 78.13 ~ 'J
:_.¿s~_ . -21.70
(c)ANSWER: 16mm5od5022mm ~ ce)CHECK: For d = 22 mm, Tau AC == 65 MPa < 90 MPa O.K.
11\o11.oDT ATA FOR PART (d): P = 20 leN, AB = 0.25 m, Be = 0.40 m, AC = 0.65 m, TL = 8 mm, WL = 36 mm, TAC = 8 mm, WAC == SO mm
d Sigma BD Sigma CE Tau B Tau e SigBear B SigBear e10.0011.0012.0013.0014.0015.0016.0017.0018.0019.0020.0021.0022.0023.0024.0025.0026.0027.0028.00
78.1381.2584.6488.3292.3396.73
101.56106.91112.85119.49126.95135.42145.09
.2,5-9.. '
• :9-6'..5
-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70-21.70
~
~-80.8271.5963.8657.3151.7346.9242.7539.1135.9233.1030.6128.3826.39
79.5865.7755.2647.0940.6035.3731.0827.5424.5622.0419.8918.0416.4415.0413.8212.7311.7710.9210.15
~2~725.69'213.82203.12193.45184.66176.63169.27162.50156.25150.46145.09
156.25
97.6691.9186.8182.2478.1374.4071.0267.9365.1062.5060.1057.8755.80
29.0030.00
-21.70-21.70
24.6022.99
9.468.84
140.09135.42
53.8852.08
(d) ANSWER: 18 mm $ d s; 22 mm -'<Iif Id)CHECK: For d == 22 mm, Tau AC = 81.25 MPa < 90 MPa OK 1....1 ••
AAB~ (w-d)tFAs O"A E3:: Ffr B / A A
B
PROBLEM l.C3
0.5 in.1.C3 Two horizontal 5-kip forces are applied to pin B of the assembly
shown. Each of the three pins al A, B, and e has the sarne diarneter d and is doubleshear. (a) Write a computer program to calculate for values of d from 0.50 to 1.50 in., using 0.05-in. increments, (1) the maximurn value of the average normal stress in member AB, (2) the average normal stress in member Be, (3) the averageshearing stress in pin A, (4) the average shearing stress in pin e, (5) the averagebearing stress atA in member AB, (6) the average bearing stress al Cin member Be, (7) the average bearing stress at B in mernber Be. (b) Check your program by comparing the values obtained for d= 0.8 in. with the answers given for Probs. 1.10 and 1.28. (e) Use this prograrn to find the permissíble values of the diameter d of the pins, knowing that the allowable values of the normal, shearing, and bearing stresses foc the steel used, are respectively, 22 ksi, 13 ksi, and 36 ksi. (d) Solve parte, assuming that a new design is being investigated, in which the thickness andwidth ofthe two members are changed, respectively, form 0.5 lo 0.3 in. and from1.8 lo 2.4 in.
SOLUTION
FR.D/'1
FAS
Sín4-5°
FORCE.. TR/AN6Lc..=
_ Fec _ Zp- .5 In 601) Síffl7So
( 1) ¡t'\ tiX, A V t. 5 r R E. S ~ rN ~
BI
(2.) AVE. STR~SS IN Be
rA B ~Iiid ih =- ~VJrf¡¡ckne..~s -=t f&~
03c'
ABe z: 1JJt
~c== F.ac/Aec
(3) PI'" AGA z: lfA8/2 )/(7T di/¡)
(~)EEAR/NG STRE55 ATA
51,2 Beai f1 ::- fÁB / d t
(7) B r= A f{ u-J G- S T ': F: S·'~: A T
6IN /vrE¡VIElER Be
5'[3 &ctr B= F8C/2cCC
(4-) Pr('l ecee z: (Ff:< liJ/( TIeljI¡)
(f;) $Ffi f< J N r; 5rn E: 5 5 ~ I eSiS Bear e;:: P.e e/ dt
(CONTINUED)
-------_._-- _._-----_. --- ---- ----------------------_._--- -. --------- o
PROBLEM 1.C3 CONTINUED PROGRAM OUTPUTS
0.500 11.262 -9.962 29.282 35.863<:.550 11.713 -9.962 26.620 32.603
O. 6OO 12.201 -9.962 5 24.402 29.8860.650 12.731 -9.962 11.030 22.525 27.587
0.700 13.310 -9.962 9.511 20.916 25.616
0.750 13.944 -9.962 8.285 10.147 19.521 23.» 9O9
0.800 14.641 -9.962 7.282 8.918 18.301 22.4140.850 15.412 -9.962 6.450 7.906 17.225 21.0960.900 16.268 -9.962 5.754 7.047 16.268 19.924
0.950 17.225 -9.962 5.164 6.324 15.412 18.8751.000 18.301 -9.962 4.660 5.708 14.641 17.9321.050 19.521 -9.962 4.227 5.177 13.944 17.0781.100 20.916 -9.962 3.852 4.717 13.310 16.3011.150 -9.962 3.524 4.316 12.731 15.593
1.200 -9.962 3.236 3.964 12.201 14.943
1.250 -9.962 2.983 3.653 11.713 14.345
1.300 -9.962 2.758 3.377 11.262 13.7931.350 -9.962 2.557 3.132 10.845 13.283
1.400 -9.962 2.378 2.912 10.458 12.8081.450 -9.962 2.217 2.715 10.097 1.2.3671.500 ~
~ .8 . -9.962 2.071 2.537 9.761 11.954
1.450 ~-12.452 2.217 2.7151.500 -12.452 2.071 2.537
4-
~:;~~g :i;:
. .u
INPUT DATA FOR PARTS (a), (b), (e): P> 5 kips, w = 1.8 in., t = 0.5 in.
D SIGAB SlGBC TAUA TAue SIGBRGA SIGBRGC SIGBRGBin. ksi ksi ksi ksi ksi ksi ksi
17.93216.30114.94313.79312.80811 .95411.20710.5489.9629.4388.9668.5398.1517.7967.4717.1736.8976.6416.4046.1835.977
Ce) ANSWER: 0.70 in. ~ d s 1.10 in . -<fiI (e)
INPUT DATA FOR PART (d): P = 5 kips, w = 2.4 in., t = 0.3 in.
D SIGAB SIGBC TAUA TAUC SIGBRGA SIGBRGC SlGBRGSin. ksi ksi ksi ksi ksi ksi ksi
0.500 12.843 -12.452 29.8860.550 13.190 -12.452 27.169C.600 13.556 -12.452 45" 24.9050.650 13 .94
4-12.452 11.030 22.989
.0.700 14.354 -12.452 9.511 21.3470.750 14.789 -12.452 8.285 10.147 19.9240.800 15.251 -12.452 7.282 8.918 1'8.6790.850 15.743 -12.452 6.450 7.900 17.5800.900 16.268 -12.452 5.754 7.047 16.6030.950 16.829 -12.452 5.164 6.324 15.7291.000 17.430 -12.452 4.660 5.708 14.9431.050 18.075 -12.452 4.227 5.177 14.2311.100 18.771 -12.452 3.852 4.717 13.5841.150 19.521 -12.452 3.524 4.319 12.9941.200 20.335 -12.452 3.236 3.964 12.452
2.983 3.653 11.9542.758 3.377 11.495
1.350 -12.452 2.557 3.132 11.0691.400 -12.452 2.378 2.912 10.674
10.30519.924 9.962
(d) ANSWER: 0.85 in. ~ d ~ 1.25 in. (d)
16.699 2.837 3.049 3.227 3.370 3.477 3.553 3.600 3.626 13.633 3.627 3.611
d
PROBLEM l.C4l.C4 A 4-kip force P fonning an angle zr with the vertical is applied as
shown lo member A Be, which is supported by a pin and bracket al e and by a cableBD fonning an angle jJwith the horizontal. (a) Knowing that the ultimare load of the cable is 25 kips, write a computer program to construct atable of'the values of the factor of safety of the cable for values of a and jJ from O to 45°, using increments in «and jJ corresponding 10 0.1 increments in tan zr and tan /]. (b) Check that for any given val ue of «the max imum value of the factor of safety is obtained for {3= 38.66° and explain why. (e) Determine the srnallest possiblc valueof the factor of safety for jJ = 38.66°, as well as the corresponding value of «, andexplain the result obtained.
SOLUTION
(a) ])J<AIA F: '8, DI~GKA¡l1 OF ABe:
A~P1c. ----..;._/5 in,
OUTPUT<
F. 5, =-
/5 sin ex + 3 O (05::x..
/S Ces() + f 2 5 ¡n (3f"ult / F
ALPHi\
VALUES OF FS BETA
o 5.71 11.31 16.70 21.80 26.56 30.96 34.99 ~6 41.99 45.00
0.000 3.125 3.358 3.555 3.712 3.830 3.913 :".956 3.994 ~.O02 3.955 3.9775.711 2.991 3.214 3.402 3.552 3.666 3.745 3.796 3.823 13.830 3.824 3.807
11.310 2.897 3.113 3.295 3.441 3.551 3.628 3.677 3.703 13.710 3.704 3.687
1.801 2.805 3.014 3.190 3.331 3.438 3.512 3.560 3.585 ~_.592 3.586 3.5702·265 2.795 3.004 .3.179 3.320 3.426 3.500 3.547 3.572 ~. 579 3.573
3.558J30.964 2.803 3.013 3.189 3'.-3303.436 3.510 3.558 3.583 .590 3.5tl4
3.5btl34.992 2.826 3.036 3.214 3.356 3.463 3.538 3.586 3.611 .619 3.612 3.596
3 8.660 2.859 3.072· 3.252 3.395 3.503 3.579 3.628 3.653 .661 3.655 3.638
4 1.987 2.899 3.116 3.298 3.444 3.554 3.631 3.680 3.706 .713 3.707 3.6904 5.000 2.946 3.166 3.351 3.499 3.611 3.689 3.739 3.765 .773 3.767 3.750
( D) !Alh e r¡ Ir') -=- 3 P, f.
r,;'le tt« /(>\1(1- Ot'¡¡!
1(01/) -= 0,8 n nt! [(-'¡tJe BDBe,
Ce) F s. ~ 3,579 {('ro CY z: 26.6 (); fh~AC
NOTE.'Tl: e Vn hlc F 5. ;:::3- , :;,/ 7 I:'~ i J,.¡ e
C.crYe.:: J-:>(" Dei i I j (. 1(', (3 :;:-3 8, 66 l
-..)
5(1)(1/1(;) óf fhf'V(7h:é(~ (f T.S.('1 I J ti j!)(:; I a I~S'c~ f ({ f J-¡ () ~ e
('r I J', { :;.( I Ji. i f,
,._•/\.: /•.J (',f· ..
''T/¡f Fe il;! ..j(--2 c. b ~ /3.::- 3[-', bb 6
/ ,
/5 (' r lí) Ji! ! " !) J ft :/ l' (' f t h r ./( '/ )(. 1ir 'iI F. S; (O() (J ),
-..- _-------------- .< .. _._--_._--
PROBLEM l.CS
p
(f) an ce (2)
!JI F
l.CS A load P is supported as shown by two wooden members ofuniform rectangular cross section that are joined by a simple glued scarf splice. (a) Denoting by O"uand Tu, respectively, the ultimate strength ofthe joint in tension and in shear,write a computer program which, for given values of a, b, P, a., and Tu, expressed
in either SI or V.S. customary units, and for values of zrfrorn 5 to 85° al 5° intervals, can be used to calculate (1) the normal stress in the joint, (2) the shearing stress in the joint, (3) the factor of safety relative to failure in tension, (4) the factor of safety relative to failure in shear, (5) the overall factor of safety for the gluedjoint. (b) Apply this program, using the dimensions and loading of the mernbers of Probs.1.29 and 1.31, knowing that 0"(/ = 1.26 MP and t¡J = 1.50 MPa for the glue used inProbo 1.29, and that o"u = ISO psi and Tu = 214 psi for the glue used in Probo 1.31.(e) Verify in each ofthese two cases that the shearing stress is máximum for a=45°.
SOLUTION
Draw ¡he FB.dt·Q<.;<·rJa..lll c,f /Ov./t'.rn¡ernOCr:
~ Z r =- O': ..- V + PcosD<_= O V z: Pcr~'O(J-
..¡.!~ ~ -::O: F - P6 il') ex == o F z: P:; in O(
A re.a_, -= a b/ SIn LV
N (j rraa / S Tre~.5 ;
S he6.rí nj
(3) pf>:-for!-SN -
v::= F (PjCLt)A reo.
ce-
,¿_Sr,... D(
(_ft.)
r.5. for S heó_r:¡t='55 z: Z-u / Z·
OVEt<A L l
pS =-r. s . .,
The smo. //er 01 F55.
(CONTINUED)
5 .007 .085 169.644 17.669 17,66910 .029 .167 42.736 8.971 8.97115 .065 .244 19.237 6.136 6.13620 .114 .314 11.016 4.773 4.77325 .175 .375 7.215 4.005 4.00530 .244 .423 5.155 3.543 3.54335 .322 .459 3.917 3.265 3.26540 .404 .481 3.119 3.116 3.11645 .489 .489 2.577 3.068 2.57750 .574 .481 2.196 3.116 2,19655 .656 .459 1.920 3.265 1.92060 .733 .423 1.718 3.543 1.71865 .803 .375 1.569 4.005 1.56970 .863 .314 1.459 4.773 1. 45975 .912 .244 1.381 6.136 1. 38180 .948 .167 1. 329 8.971 1. 32985 .970 .085 1.298 17.669 1. 298
PROBLEM i.cs eONTINUED
P F. ) 'j-R A M OUTPUTS
Problem 1.29a 150 mmb = 75 mm p = 11 kNSIGO 1.26 MPaTAOU 1. 50 MPa
ALPHA SIG (MPa) TAO (MPa) FSN FSS FS
~ (b»(C)
Problem 1. 31a 5 in.b = 3 in.P = 1400 lbSIGU 150 psiTAUU 214 psi
ALPHA SIG (psi) TAU (psi) FSN FSS FS
5 .709 8.104 211.574 26.408 26.40810 2.814 15.961 53.298 13.408 13.40815 6.252 23.333 23.992 9.171 9.17120 10.918 29.997 13.739 7.134 7.13425 16.670 35.749 8.998 5.986 5.98630 23.333 40.415 6.429 5.295 5.29535 30.706 43.852 4.885 4.880 4.8804045
38.56346.667
45.95846.667
3.8903.214
4.6564.586
3.8903.214 -4 (e)
50 54.770 45.958 2.739 4.656 2.7395560
62.62870.000
43.85240.415
2.3952.143
4.8805.295
2.3952.143 ~ (h)
65 76.663 35.749 1. 957 5.986 1.95770 82.415 29.997 1. 820 7.134 1.82075 87.081 23.333 1.723 9.171 l. 72380 90.519 15.961 1. 657 13.408 1.65785 92.624 8.104 1. 619 26.408 1. 619
PROPRJETARY !\'lATERIAL. 02006 The MeGraw·Hill Companies, Ine. AII rights reserved. No part of this Manual may be displayed, reproduced or distributed in any fonn o, by any rneans, without rhe prior written permission of the publisher, or used beyond the limitcd distribution lo teachers and educators permiued by Mcfiraw-Hill for iheir individual course prepararion. Ifyou are a srudem using this Manual, yeu are using il without permission.
