Baixe Resolução Hibbe...- Statics 12ªed - hibbeler - statics 12?ed - cap.7 (3) e outras Provas em PDF para Engenharia Mecânica, somente na Docsity! 7-1, The column is fixed to thé floor and is subjected to
the loads shówn. Determine the internal nórmal force,
shear force, ând.moment at points 4 and B.
Fres body Didgram : The support reaction necd not be computed in
this case.
é bt A]
Internal Forces : Apolying equations of equilibriura to the top segment Bona | dsoum
secrioned through point À, we have 150 amor too mm T
SER: n=0 ans
+T2Ã=0 M-6-6=0 Me I20kN ams wa Bta
A +ómm
Cem =O GOIS-SOID=M,=0 M=0 Aus A
Applying eqjuatipns of equilibrium to the tóp segement sectioned durough.
point, we have %
+ Ma
DER |=0, W=0 Ans
i Ng
+TEG=OG No -6-6-820 Ny =200kN Aus
+EMGI=O: 6(0.15)—6(0.15) -8(0.15)4+M; = O
' 1.20kN-m Ans
7” Theródis subjected to the forces shówn. Determine
the internal.normal force at points-A, B,and C.
Free body Diagram: Tho support reaction nec not be computed
dm tis case | SE 550 50h
internal Fórees.: Appiying équations oé equilibrar to die top seg
sectioncd thrr intA, we have
esmo dão po 150% Licor OM | epa
+TEE=0 NM -550=0 NM =550b ans
! A
Applying equations of equilibriura to the top degment sectioned through 4 E 356ib
poim 8, we have. -
+ TE =0; Ng -550+150+150=0 A
Na 250% Ans
Applying equations of equilibrium so the top segment seclone durough [
point; we have
No
Í
+TiEG=0 No -550+150+150--350-350=0
Nç =9501b Am:
7-3, The-forces act on the shaft shown, Détermine tie
internal hormal force at points A, B, and €.
Internal Fórces:: Appiying the equation of oquilibrium to the left segment
sevtionod indoúgh pointA, we ave
DEEa0O NM-5=0 NM, =500EN Ans
Applying ihe equation of equilibrium 10 the Fight segmeit sectionod through.
point 8; we have
DER =0 4-Nç=0 No=400kN Ans
Appiying thé equation of equilibrium tó the right segment sectibned through
point C, welbave
DEE SO No+4-7=0" Ny =300kN Aus
*7=4, The|shaft is supported by the two smooth bearings
A and 8 The four pulleys attached to the shaft are used
to transmit power to adjacent machinery. If.the torques
applied toj the pulleys-are as shown, determine :the
internal torques at points C, D, and E.
Internal Forces : Applying the equation of eguilibrium to the left segmait
secstonid iirpugh poiit €, we have
EM=0, 40>T=0 T=400:R Ans
Applying the equazioh of equilibrium 40 the-lef segraent sectioned through
point D, we have
EM 20 40+15-D=0 T=550h:f Aos
Applying thojequation of equilíbeium to e right segment soctionod iirough
point E, wetave
10-%=0 F=100b-f Ans
SAN
SN Mk
cm
Ne — 4
kN
No ! 448
I
“7a, Detéimino the normal: force, shear force, and
moment àt q section Passing' throubh point €. Assume the
Smpportat À can be approximated by a pin and Z as a roller. sEM, = 0; “ISA = 80) + 28/0294 1969) = 0 ta
f , 'Ozhyp
5, = ITikp tok |--—, Pkir
a i + =0; = x A
7 su skip DER «0; 4=0 9 8
[HOT HO N | +tzg = 4 "10-24 1721-800 sete info ap
É A = 201 Mp
As 6R Seg = O; Ne=0 Ans “tar
1
Pe doo By
+TES =0 W-96+171-8=0 Nest! y
+ [1
Ve =05Hp Ang me
L+EM = O Me = 9.5(6) + 17.112) — (18) = O sta
Mo =36Hpf Ams
79, Detefmine the normal force, shear force, and Tomient
at a section passing through point D. Take w = 150 N/m
CG = cisto + 2Rah = O
: Be= I000N
| ÍB=O. 4 =St1000) = 0
| 4.= 800
+fz7.=0; 4 = 18048)“ 342000) 0
: 4 = G0N
| SER «6 No = -800N
| sfzg =0; 800 — 1504) «Wo = 0 a
bao às demo
i CHEM = 0; om s0OÇA) A ISU(A 2) Aut, O
Mo 1200 Nem = 120kN-m Ans
78 The -beam AB wil-fail df the maximum amtemal
moment at D' reaches 800 N*m or the normal force in
member BC becomes 1500'N, Determine the largest load w ; : ud.
| |
| man
it can support:
f CAÇÃO Mod) =0
800 = 4w(2)
we = 100 Nm
L+EM, = 0 8004) + R(068) « O
w Be = S6IN<IMON (OX)
| | w = 100 Nfm
A , e e
> SO
É gm 4m 3m
| e |
Ea
Td. Determine'the shear force and moment acting at
a section passing thrôugh point C in the beam.
A (9) +2H6 =0
A = Skip
4=0
“NOEHDAM=O
Ho = 48HpR An
9-3-W=0
=
i Ye.= Skip Ans
*7-12. “The boom DF of the jib crane and the column
DE have à uniform weight of 50 Ib/it. TF the hoist and dps
Joad weigh 300 lb, determine the normal force, shear SE O N=0 As Na Aus
force, and moment in the crane at 'sections passing fa n “ap
through poinísiA, 8 and €, +fZZ=0 4 -480=0 Y =450 Aus t Sroth
D.
EU po CHEM O My = I5MLS)- 300) =00; Mm LiZS IR Ars
Ye
e sf 34 SE -O M=0 Ans Ng ff sa Sé
sf NF
+TEB=0; Mo - 3580-3000; Y = B50)h Ans Sent
A e 300% CHEM = 0; My — S50(5.5) — 300011) = 0; "Mp = 6325 br Aus
, eso
1a SE =0 4=0 Ass ch) es
| +TEZ=0" No — 650=300-280= 0; No < 1200 Ans mi sh gosl
Er . C+EMÇ = OM; — GSO(GS) - 300013) = 0; Mo = BS IA Ane Ne
7413. Detérmine the internal normal force; shear force,
ind moment acting at point C and'at point D, wiich is 300 1byft
ocated just-to the right of the roller support at B.
Support Reactions : From FBD (9),
Geri =O: 8, (8) +800(2) 240064) -800(10) =
i 8, =2000%
ROO) = Benth 3008) sadootb
Imternot Forces : Apolying the equaions of equilibrium to segraent: ED Z00i4)=Booi
TFBD (b)!, we have
Sig = No=0 Ans
Ae o À
+TEE=0; W-800=0 W=B0lb Ans 28] 36 Todo [art
den Mo = 800(2) = O Ag mo
EM = Mp -
ç o “Mg 216001. i= <1.6Dkip-f Ans ZaDA)=B0o |
Applving theiequatoms of equilibcium to segment EC [FED (6)] , we have &) H
>
SIE = Ne=0 Ans
ne : HOMERO 00h
+TEE=0 Vo+2000-1200-80020 - Wz0 Ans Zoofo 80:
Gem=o; 20004) 1200(2) = 800(6)— 47 = O «<
I Me = 800 bn Ars
717, Detetimine the normial force, shear force, and
moment acting. at a section passing through point €.
701 L+EM, = 0; —800 (3)-.7046: 00830) — 600 cos30º(6 cog30e + 320630")
+ 6008In30º(3 ginãde) + B(6c0530º + 6 cos30º) = O
B, =9a74ib Too
mou y st tooab
+ 3 E
Szg «0; 800 sin30º — 600 sin30º- A, = ç
x
A, = 100% 1 t
' 8 %
TER =o A, — 800 c0830º — 700 — 60060830º 4. 9274 ="
4, = 98811
ARE * 0, Ne — 100c0830º + 985,]xin30º = 6
Ne = 406 Ans 15% Ss” Me
i le = -
: doca SY
100 8in30º + 985.1609309- Wo = O f
i esa
| Ve = 903% An
i (+EMp = O; =BBS.1(1.5c0830P3:= “200(1.5 81030) + Mp = O
; “Me = 195S bh e 135 Hp Ang
718, Determine the normal force, 'shear force, and
moment actihj at a section Passihg through point D.
| 700 lb
: C+EM, = 0; = 80KB) — 706 co630º) — 600 c0530º(6 c0830º + 3ohi30")
| + 6005n30º(3.83n30%) + B, (6cos30? + 6 0830") = O
; oo
: B, =4tb doou 38 ALA sp doou
me 34
; SIE =0 800 sin30º — 600 sina02 4, = O f
! As qem
' A = 1001 f í
| dp
| +TEg=0 Ay = 800 co530º = 700'- GO0cos30S + 927.4 = O
| A; = 985tp
' ER = 0: No + 927.48n30º + O
No = —46&b Ans.
IE = O Vo — 600 +.927.4 cos30' = 0
i W= Ans
Mp = 6001) + 927:4(4.00530) = O
Mp = 2612 bh = 261 ip Ans
í
719. Deterniine .the normal force, shear force, and
moment at a: section passing through point C. Take
Pabx, +EMyo= 0; -I%0.6) + 84225) = 0 ai T
ti son dn
Ê Ba Eee]
i DEE = 6 A, =308N
aim (Sm
+15 = 0, 4, = BkN
Em = O -Ne-30=0
Nç = —305N Aus
+t2g W+8=0 my
Ni Efe so
fem mam due . em da
! L+Bk =0 -M+ 8019
i Mo = 6RN-m An
: =” : : jon of
*7-20. The cáble will fail when subjected to a tension 9
2kN, Determine the largest vertical Joad P the'frame will
support and: calculate the-internal normal force, shear
force; and mokent at a section passing through; point C
for this loading,
; c A
É e
E ogim— A 03sm—— 035 m—
Pp :
2hN
i (HEM = O -XO06/+ M225) = 0 Pu
dx
| P = QS334N Ans Tee
I ! às
Dex = 0 A =288
+Tz; =0; A, = 05334
-Ner2=0 He e
tz Zku
Ne = 28 Aus o-sagha
o -Ve + 0533=0
Ve = 0533 4N axis
—Me + 0534075) = O
: Me = 0400 kN.m Ama
|
ta,
Determine the interna! normai force, shear force,
and be;
nding moment in the besm at point B.
Free body. Diagram & The support meaciions at À end nôt be computed.
Internal Forces - Appiying the equarions of equilibrium to. segment CB,
wehave
E =0; M=0 Ans
+tig=a W-288=0 -W=2884p Ans
CG =0 -288(9-M =0
| Mg =-1tSkip-ft ans
722, Determine the ratio of a/b for which the shear
force will bé zero at the midpoint C of the. beam:
Support Reaptiôns :: From EBD (a),
; i t
Gem Fm bu[st-a]-A, (=0
lerda)
Internak Forcás : This problem Fequires Vs =, Sumersing f
verticaliy (EBD (b)], we have
+ÍEg jo; membro fas KM o
FU (b=a) = Keane)
1
ã Ans
Mp Elenco
!
727. Deterímine the normal force, shear force, and
moment at a séction passing through point D of the twvo- o, ;
=izodçã) = s00(4) + Bet 0 * e =
member frâme. tri = Re
i 400 Nim 4
Rin Be = 2600N Tr
SERRO O A c= Ste) = 2400N
+TEg = 4, = 1200 = 600 + Stogoo) no
A, * 800N. .
SEE «0 No =2400N = 240EN Ams
+fz7 = 0 8009-600 - 150 - W = 0
| Y =.50N As
| (HEMo. = O; =BOK3) + 601.5) + 150(1) + 34 = O
Mo = 1380 Nm « 1.35KNem Ans
moment at secticus passing through points E and F.
Member BC js pinned at B and there is a smooth slot in
itat C. The pih at C is fixed to member CD.
— 120% — 500 Bins + G(S) = 0
tow Sopa
Sob G = 0281 u «Dik
4 4
: Bs =0
EB f 00 gos60e ey &
na era E B =250% Mo car
Lilbiabosa NE +TEg =0 B, = 120 = 500 sin60º + á07.8=-0
E
i B = 452% mk
| . Ne Efe aço
| SEE = 0; Ng = 280=0 a
: Ne = <250 db Ams
+TEE = 0; Ve= AS Ans
(+EM = O; Mg = 25.47) = 0
Me = 4 it Ans
, N=0 Ans
: MMS W = O RO de
Rk .
i W= 3081 Ans
: (EM a O 307.84) + dg = 0
Me = 1230 )b-f a 1.23 kiphr Atis
7:29. Diteemie the internal normal force, slicar force,
and the moment at points C and D.
Support Reactions : FED (4).
Cm, = By (6+ 6c0s 45º) - 12.0(3+ 6c08 480) = 0
; B,=84854N
+ÍEg =0; A, +8485-120=0 A, =3.5I5kN
Sm =o A
g the equations of equilibtiara to segment AC.
Internal Forces : Applyin
EFBD (8); wehavé
Pe ER =0; 3SlScos 48 -W =) G=249:N Ans
Rozkis0 351Ssin 4P=No=0 No=249kN Am
CH=0 M-Isiscos seta) =o
ME =4S7kN.m Ans
Appiying che fquations “Fequilibriurm o segment 27) [FBD (c)] , we liavé Más ha Ne €>
Ans (b>
Mp =0
sa
+Tig=0; W+8485-600=0 W =-249kN Ans
CEM =0; s48s(3) 65)-M =0
; Mo =165kN-m
[7-38, - Determine the normal force, shear force, and mio-
ment acting at sections passing throúgh points 8 and €
jon the curved rod,
É MARE = 0 4006n30 —30Dcos MY + A; =
Mg = 598 tb ans
FAN EF =0 Va + 400c08 30" + 3008in 30º = 0
Vy = ag 1d Ans
Gem = Ma +400(2sin 30”)
+ 30062 — 2005304 =0
Mo = —480 Ipe Ang
— SS BUD +04 My = 0
My = 480 lbat Ans
As =400 Hb
4=300b
Ma 30044) =0
Ma == 1200 bat
Ne + 40D sin 45º + 300coç as! = (h
Í Ne = 495 lb Ans
400 lb
300 tp
0 Uesmp
1200 br
300 1%
A +BR,=0 Ve — Meus AS + 30Osinds? = O
WS Ans
= Me = 1200 — AB sin 45”)
+ 30002 — 2c0s45") = 0
Me = 1590 beto 159 tip Ans
Also,
4(+EMo
495.062) + 300(2) + Mic = 0)
Mg =: 1590 Ibeftos 1.59 kipil Ang
731. “The cantilevered rack is used to support each.enid
SÊ a smooth pipe that has a total weight of 300 1b. Deter-
mine the normal: force, shear forêe, and moment that act
ih che arm ar its fixed support A along a vertical section.
eps
dtEr =;
No cos = 1500
No = 173.205 Ro
= Ny + 1732058030 = 0
Ny = 866 Ip As
Va — 173.205c0830º = 0
Va = 50 lb Ans
Ma — 173.205(10,3923) = 0
Ma = 1800 Ibiin, Ang
1501b
Me
A
1773.2056
so”
10.3923 ig,
q
M,
Gifs 4 Sor
7 om = to9a3im O Ma
Va
7:34, - Determine ihe intérnal normal force, shear forçe,
and bendinig moment at points É and £ of the frame.
Support Reaerions : Members HD and HG are tva force members.
Using method -of joint ÍFBD (a), we have
DEE =0 Erocis 2687 Bpcos 36.818 <0
to = ho sF
+TEÉ =0 2Psn265T-A0=
| Ho = Bro = F=89443N
From EBD tb.
Gm ao G. (Bens 26.57") +G, (2sin 26.879) 894.431) =0 (1)
From FBD.(6),-
Gm =, BSAAID Cos 26.57)+G (2sin 26.57) 0 123
Solving Eqs 12) and (2) ietds,
G=0 Ge=s0N
Internal. Forces : Applying the equanionis of equilibrium to segment DE
LEBD (d)]; ve havé
+ 26.=0; K=0 Ans
+ERi=0; B9443-M=0 Ne =B94N Ans
Go; Mç=0 Ans
Applying he equiations of equilibrium to segmest CFIPBD (e)] , weitive
7, =0;. Vr+500506 26.579. 894.43 =0
W=447N Ans
i
NES EO: Mo 500sin 26.570 = Nr=224N, Abs
QPIM EO Mp+E94430.8) - S0Gcos 26.878(i,5) = O
! Mp = 24 Nm Ans
2C9$ 28:57
ve AM
Í
i
77-35, Determine the distance q as à fiaction of the
beam's length £ for locating the roller support so that
the moment in the beam at B is zero.
!
*7-36. The semicircular arch is subjected to a unifórm
distributed load along its axis.of tn per unit length.
Determine ithe internal normal force, shear force, and
moment in fhe arch at 6 = 43º,
Resulianes of distribuso load:
[o Evetras sing = Pmtcenofy » rwóll-cos8)
By mil (ras) con = PwolsiaojS = ruo(sinó)
737. Solvel Prob, 7-36 for.8 = 120º.
i amd wa
nIa (rd)
i wetrdg
. a
Mg = Luotrdo)r = 12 mç0 As
| Ms
AO = as 075
M/ERafl VefcsO - Rino = O RutaBim,
V = 0.2929r wa con4s? — 0,70% wo sináse NM
i Va -0.83rm, Aus É
*RES = 6 Nóf, cod + Ressin6 = O “ +
Í Mes Fep= men
Nom = 0.707 wesinas? = 0,2929 we cos4so
N= 017 rw Aus
1 "8
Gim so mer m(E) + 07077 mo = 0
M= OW Ass
: Resuloinos of distibuna load ;
Fes m ( inte dO) sinO = rmetecond)E = rwo(I-co88)
= Entra) coso =. rwo(sinajo = rwo(sing)
Mm Emorrámr a imo
AG a 120º,
Fes memo (t-00s 120º) 1.5 my
Fay = r wosia 1208 = 0,8660831 wo
STE =O N+1.5 riscos 30º -0.86603 > wysia 30º 0
N=-0866rm Amy
ANIS 20; V+ 1,5 rwpsia 30º +0.866037 macas 30º = O
V=-LSrwg Aus
Eimno, =Meusto(EE CO nt6 rmjrab
Marim. Am
41; Determine the x,y, z components of force and
moment aí point. Cin the pipe assembly. Neglect the weight
of the"pipe. Take F, = (3501-4005) db and E, =
(-300j+ 150k| Id.
Free body Diagram : The support resctioiis need not De computed.
Internal, Forces : Appiying the equations of equilibrium to segment BC,
we have !
Em
25+0
280
EM, =0;
2640
Nç+350=0 No=-350db Ane
(Wc), -400-300=0 (Hr), = 700 ans
(Vo), +150=0. (Yo), =-150b Ana
(Me), +400(3) = O
(Meda =-1200D-R=120kip-h Ans
(Me), +350(3) = 150(2) = 0
iso), =-750 4-8 Any
(Me), —300(2) =400(2) «O
(Me, =1400b-A=L40kipih As
i50 tb
00h
Mad
|
Tas, Draw the shear and moment diagrarms for the
beam (a) in terms of the parameters Shown; (b) ser P=
600-1b, fb=71t.
x
* v
deb
r
=
Hm Pb
(emo fasaroo m x Y
pb y
M= E das a
Sp x
Fr acss(arb ZEM
: n Bha A+
sig =0 SE-P-vso MM
: x
i Ps
veDE am sa)
Po sro
(emu = 0; dire P(- a) M=0 A, rop
ab =2L50
: Pa va Cape) 1750
M=Pa- |, ams
; z+5
: «4
|
0) ForP = G0blb au 58h = 78
743. Draw the shear and moment diagrams for the “
cantilevered beam,
800Jb-f
| Monti
i Sole
i > a
tap E p too
v Pio left
(ste)
gn
For Osz<sh: vey
CA TES 0 100-V=0; Ve tdo Am
CÓEM=0 M-100+I8O=0; = 1007-1800 Ame CBI
=
For S<xsI0A:
metb-fes
+*TEE = 0 100-V=0; Va 100 An
(Meme = O M-100r+1000=0; 4 = 1905 1000
T-44. The suspender bar supports the 600-Ib engine,
Draw thé shéar and moment diagrams for thg bar.
Í toolb
Por OSx<15ft:
: mn use mn
+Tir =0 -s0-ved
x E ;
Ve 30 Am sa ' Booih
M+300x=0
veio)
M=-30x ms Ja p
For LiSfiezs 3%: xd
; UR “Jos
+TEh = 0 600 - 300 - V = 0 mé gos
V=30 Amas
: “tra
Gus o M+300z- 6006 -1:)=0 my
| M=300x-900 Ans
27-48. Draw the shear and moment diagrams for the 1.5 kNfm
bem.
hupport Reactions:
(HEM =O, Ga ISCN=O C,= 1 354N
L5tr- kN
TES, Aj>15+135=0 A, =0350kN
: m Ty] ”
Shear and Moment Functioiis: For Bb < x <2m [FBD (MI, z % 3
i oh) do
(HIER=O 0250-V=00 V=02508N ans ou Y DISuN a
i [o (db)
A rEM = M 02500 =0 M= (0.250x] kNm
' VN
Ans
| h 025
For 2m < x:< 3m [FBO (bl 2.167
: im
+1E5,=0 0235-154-)-V=0
0.521 a
: Vo [325 E50x) KM Ams Ma NM os ct3s
A tem =0 das st -n(E
)- a
í M = (-0.750x) + 3,25% — 3.00] kNem Ans “(m)
TH49, Draw the shear'ind bending-tioment diageams for sony
the beim, ] I I I 1 200b'f
Sáppurt Reactious: am F
AÍ +EMo=0: 30000) -200- 4,20) =0 A, = 90h k mr of
Shear and Moment Functions: For O < x< 21% [FED (al. Ss M e jm
' | |
M.
TES SO 40-50 -v=0 |
y
º x
V=[890-500rbib ans ago v
(a)
emm=u cm+so(5) sor =o
i o o 0
f a am
M= (40 2508) bh Ans sm
For 20 Mt < x < 30 ft FED (bj,
i St
+t25 V=0 An
Á+EM=0: -200-M=0 = 200 b& Ans dio 201
i
7.50, Drahr theskicar and morhent diagrams for the beam.
Suppdrt Reaciions : From FBD (9),
y wL/3L 3wL
Beco gr EfB)io edi
i JwL WE »L
irTrg=0 A+ -4=0 A= 7
Sheur end Moment Functions > For 0Gx< trBD tb,
: wL vel,
: =0 “vo ak
sta o val
Que 0; ME ao Mo Es
Fr fersz LFBD (9),
i wL
+TEE =0; me vL-9=0
”
: v=Eçe
| alt=)
IwL L-x
+EM=0; TF t-d-mtt-n[E7E) m=o
i M=P(cErste-a?)
Ans
Ans
Ans
«<,
tá
751, Draw the shear and moment diagrams for the beam.
M= 25(100x - 5 — 6)
Sovoth
som t+ + S00KI0) — 150 +-8, (20) = 0 roncf-oos
Faia ni
Um sem qm SR |
A Jo. meça 1
Es O) ea us p A
Boba] mA Eolb-n
+TER = 0; 4 Saad. ep,
: E
4, = 250% asa
vem
For Osr< 204 as tok
+tmg =o 2500 250r — v = 0 º X
-A80p
V=250M10-5). Ang
: Pobtajy
(HEM = BOM) +10 +2s0m(5j + Mao O a—
-Bo
Ang
47.82. Dravytho shear and mioimentdingrams for the bean»
Hora on
13375 = 40x - V=0 SA PLTI Sit
iV= 133,75 - 40x ans USA Bobastos
É E E
(EM = +40) mB = OM Í |
E x
: Pp. pa
M= 133.75x — 204º Ane Z
É SINE, Ás
s<xsit mea [
| 2386 I
+tE7 =6. V-0=0 od ' .
Lo
v=20 Ans Ta E o
C+EM=0" M+SMI-) + 150=0
lá = 20%-370 Am
755. Draw the shear and moment diagrams for the
compovd beam; The beam is pinconnected'at £ and F.
Support Reactions : From FRD (b),
, do) WE wi. fo
memo s()- So p=nt (o) RE
Pies BA Mo gal REMO
tmg-O sepeç=o gem A
| 4 :
From FED (à). As
, eo so
: L
(emtso; paed(s)
3
From FBp (.
| AWLAL wLfL ML
rg =; SEE) =o 4
dig=o e ME mi O mL
3 6"? B=g
Shear and Moment Funcitons : Far 0Sx< LIFBD (d)];
+TER =o; Tê -mev=o
»
Ve (IL
Te Rx)
Geim=o; Mem) Talçao
»
i M= a (Te ge)
For LSx 4 2L[FBD (6),
+ tzs =; IwL | 1O0wL
AE to ta=V=o
v=SGL-2 Ans
, 3) Tolo 10wL ç uuxt
Grito; mv) Es Le nao a a Tê
í a” 3002-0948 Zu ne
; Me (Le -200 9) ans 18 Va ua
: a: E I— x
| SJ go N
For 2L <Y 5 3LIFBD (DJ, a
un Sae 8
7
+TEF =0, ve RE wgLo9=0
»
Va ate 18) ans
Génio: Tlm wtat-n(MEE) age
Me a(s sor?) ams
*7-56. Draw the shear and moment diagrams for the
iagrams for th
, am Z.
Support Reactiont «From FBD (a),
(a =o SM(m-A,(9=0 4 =300H8
Shear and Momeiit Functions : For 0Sx56 m (FBD(DI,
, X
+TE5=06 300-2-V=0
v-fpsn-Dhex Ans (ã)
|
“The maxirhum morent occurs when. = 0, then
' 2
0=300-T x=3464m
| Nx
“EM = Mei [= |-20020 :
€ (gm sd E
í me (roma Tbim-m ADS Ls e
Thus, |
p
Mas =300(2464) - LEO 690 EN -m VCkN)
,
ST ME L=I8N, the beom will fail when the
maximum shear force is Vais = 800 1b Or the maximum
imoment is Max = 1200 Io-ft. Determine the largest
lintensity 10 of the distributed loading it will support.
+1ES,
rem
ml,
Vau =
som = UÚ)
u = 880 yr
we lar
— 1290 = =
wu =222 Ans
*7-60: | Draw-the shear and beriding-moment diagrams
for'the ibeam;
Support Reactions : From FBD (9).
Qua A (raso -12000)=0 “A, = 650N
Shedr and Moment Functions : For 0Sx<3m. [EBD (b)].
i
+TEg=0 -650-5000. v=0
Va [-650-50,0) N Ans
i , BE -
emma me (some )(5)+ ss =0
i M=4-650r- 1672; Nom Ans
For3jia <x 71 [FBD (c)],
P 4 TEG=0, V-300(-2)=0
V=(2H0-300) N Ans
Gem=o -00-n(T52)-si=0
M=[-180(1-5) N im Ans
'
TOXINAS 20514) =]2000
761. | Drawthe shear and moment diágrams for the beam.
Ai) =0 Vet
lgJeaoso em
“Gs
i
Shear and Moment Functions.: For 0 <xs L(FBD (b)].
i bL ow Tyw
CeTmgeo Dodsildaa-veo
Í w 2 x
V= (soe). ms
“The niaximim moment óecurs wben V = 0, then
|
| O=4U=6lr-3? r=0.8275L
E) tero
= tatiosbdo
i M= (ts 3184) ans
=* [ul z >
EA (Ostist) - 31(0.5275L) —(0.8275L)*]
=0,0940wL?
corpus
| For ac<xaL-a
7-62. Draw the shear and moment diagrams for the bean
fa) in terims.of the parameters shown: (b) set P = 800 lb,
a=3t L=i2ht.
É Por Q<e <a
+IER=G Ve? Ans
: Q+IM=0 M=Pr Ans
Er pr tim
Fotb<yssf
Ve = 1500 b = 1.5 kip Ans
Vi=a Ans
Vi, = 8004 — 9) db As
M, — BOA — 3) (2) =0
Ma = 4DONE vp btt Ans
M, + 150002) = 0
My = 3000 Ibt= -Tkipft Ans
Me + 15008 — v) = 0
i M, = —1500(4 — y) ibft Ans
| +1ER ans
HEM =0, Prrpie-a+M=0
2 x pe doa
M=Pa Ans sã
Fo )
: vo sob to o yM
i f y s00 tb
| Ass TS Em
i VI |
í - -n= r Mr
| MAPL-0=0 ES : , nos
i Ma PL) ans >
Í a ! 2091
GS p=, a=5t L=12ft ” mm:
jfr 0xexón o “
Lo +tER V=80 As 800 ih Soo Ih
í A +EM =: M = 8007 bt
for Sieg <7h noob | so ih
= TRT SR
V=o Ams Valhr |
—H00r + 8006 = 5) 4 M = O “o | |
o
| M = 4000 bt Ais º 1 *
; ; 800
For TA<r sit bd
+1E5,=0 Ven Ane
; o x
Po +EM=0 -mesodz-s=0 «dy
M = (960) — 800) Ib.ft Aus
|
763, Express the x, y, : components of internal loading :
in the rod as a function of y, where 0 < y <:4 ft. EN
Abs
1.500
b Soo
767, Draw the shear and moment diagrams for the
beam ABCDE. AN pulleys have a:radius of 1 ft. Néglect
the weight of the beam and pulley arrangement. The load
weighs 500 1b.
Support Resctions : Prom FBD (a),
Gem =e 59-00) -s00(3)=0" "E =3333b
Ce TEE=O A,+33333-500=0 A = 16661
y aDembe
Jd Moment. Diagramis : The loná on the pulley
Caidty equivalent foroe and couple moment at D as shown on EBD (b).
vá
| te?
| te [li xd)
| f = E
| at L
e | 333
ST -833
-33333/b
50 bo & 33333,
br)
1000 soolb Mb
1
|
f
*148. Dray the shcar and moment diagrams for the , f iate
beam, :
am
769, Draw the shear an
id moment diagrams for the
beam,
Support Resqtions :
! 3
Germano; (go -s0€) -soorn=o E=625N
+TE8=6; Avens(S) -soo-sogmg 4,=625N
i Z5002)25:
º OR) = S00N Erecsa
4 ol RL
| | í . SOON
i 1 | jm 2 |
770. Draw jthe shear and moment diagrâms for the
beam,
20Hp | 20kip
4 tipift
B a
3% 158:
20% Es ao
mis |
Le
mm Ri
77, Drajy the shear and moment diagrams for the
beam,
Support Recctions :
CEM =0; .B,(8)-320(4)--20(11) + 150=0
º B, =206.25:N
+z5=0; 4,+20625-320-20=0 A, =133.75kN
t
Socg)= 320 k4 20W
iso lm
em Em 150 to
= 20625 kN
133:78 dl Eye tobas
TT, Drjw the :shear and moment diagrams for the
beam A 8aN SN
Suppor Resetions ;
Gem =o: D; (3) -8(1) = 8(2) » 15.0(3,5)=2020
i D, =38167kN
+TEG=0 32.167-8-8-150-4, =0
A = LISTKN
| Ve)
ôm BM das oRO R
: 1 2
: 2 , ; ;
| = + E— X(m)
i fi mai
i ne Tom im (oa 95
dasm Deseiorta
the maximum moment is
“ftor the maxi um shear is Vos = 8
Determine the largest distributed load w the beám will Ma = 40; Bady
support. i
Í W =2kp/r
: ”
- j Maa= 6; 30 by
—tur
7:79, The beam consists of two segments pin connected
at B. Draw the'shear and moment diagrams forthe.beam.
i Tot sou,
imoio tri
i Wo der
RE Si E
vt
781, me beam consists of two segments pin-connected
at B; Djaw thé shear and moment diagrami for the beam.
i
Súppart Regctions : From FBD (a), ta)
(+EM=0 G(6-060()=0 G =0200%p
*TEG=O 8, +020-060-0 3, =U400kp
Frorá PBD (6),
(rm =06 My -0:700(8)-0.400(12) = 0
My = IOstipf
tTEG O 4, -0700-040050
Ay = 110kip
Shear qud Moment Diggrams : The Peak valuc of the moment for
segment|HC: can be ovaluad using lhe mmetiod:vf sections. The raaximuim
moment pecurs when V =, Fróm FBD (6)
nTrg=0;
i
Gra ox (EL) aco
M=0200- E
=0.200x - —
Tô
0200-;(2)e=0 2=2/35h
Thus, ci
: 2/3) .
Mu ac = 0200(2/3) tda =0462 bip.
700 Ib
are) =ogodo = nt Rup
==—3
l
RT FR
Byogokip Gozo tip N/0.kip bp oso “ip
Ág=i10 Rip
af ar
Mod pf a,
vit)
Ho
786. Drawthe shear and moment diagrams for the beam.
Supphrt Reacsions « From FBD (a),
CrIM=0: 8 0+150(0)+15
Í = 50.0(5) = 15.0(12) + 15=0'
; B,=400kip
[a TEEsO 4,+400-150-500-150=0
: A, = 400kip
Sheariand Moment Diagrami : The value of the moment at supports
4 and Pci bo evaluaed using the method of sections [FBD-(6)].
| +EM=O; M+ÍSOM+ISSO M=-450kp-h
HOM) =D ok FUN CMDORP Let SOM
4 =400 ip by 400 ip
TO <Bbkip
Víkip)
250.
787. Draw the shear and moment diagrams for the 3 iprft 3 kipyR
team. Es *
| e E [ |
: A RISE Te EE»
: . E sk — st
Sipport reactions: Shown on EBD (a)
From FBD (0) Vexip)
We I6=0 ç
| Var= 9 kip A atm
i Me + OAG = 0 E
: Mg =—36 kipiht
i Muito
TS do
-36,
i
*7-88. Draw the shear and moment diagrams for the 2 ipi
beam.
Siitar and Moment Functions: For O < x« IS R
ir 33/18-V=0
Ans
Veste sfsIN
Ma ceps(G)- ttaa=o
M= 2/2 -*!/45] Nem àns
xt
75
*7-89, | Determine the tension in each segment of the
cable and the cable's total length.
Equartons f Equilibrium : Applying method of joints, we have
Joint
netos o-BilAl=o ul
| ic ua
i 7
+TZ5=0; aula) teme-s-o [3]
Joimic
[SEE=0 Rotsf-heos Os0 (31
[+TEE=0 Eos 0% Eepsin à - 100=0 14]
Geomgiry :
y 5
sinó= 008 8= ao
' ir [92425
| 3+y 3 100 Hb
singa
=== cos fa ae
(46515 Re óy+I8
|
Substinie the above results into Edy. [1], [2], [3) and [4] ind solve. We
have
| fe=67b Ba =830b Eo=881b -Ans
y=2679h
The totaiength of the cable is
Ia VETA (SEI + seio
=202f Ans
*792 The, cable supports the three
Determine the magnitude: of;
Ja = 8£t, Also find the sag yp.
loads shown,
P if P=30b and
| P; |
Lam siena
Me i % E
5 H cet — Esta =
SEE = o, mete qo
i te O
1 - tis - 300 = 0 Srs
+TE; EO; plc + E Te
: vas e dook
Tia = 9833
: as Te
| Tc = 85421 ê E agito
i : i
AC 5
Í =» Top = 0 o
+ . 42) + meto
Demo» O; go (8s4a) ST
, 6 My cr «»
io A + Fo
EGO o gasta + ER
D i
a Ser a a Tem e Top = 0
om = O (Evo + lda fla yol + 285
| M- vp = 300= 0
4% yp o leo — 300
Vero atas O flap + 225
te
: To = MO/BSTA naD om
” Foro = 108 E
Soo
ubgtitite Into Eq. (1) :
8 | Jo = 644% Ans
i Top = 9161 b
&= esp An
793, The cable supports the loading shown. Determine
the distance + the force at poiút B ácts from A. Set
P=40b
BS n=0
a
40 leg —
a 405 dm + 6 a
5 8
=0 Ts — fe =0 so
+TEg =o 5 fra da * dou,
Ê g
=
13x, +15 bs
mo q) %
VOS Diem E
AC 3 3
. 4 xo
= =(30) De =0
E nO 00 tdi ção isto
8 2 3
=0; Toc ——==Ttp — (80). = 6
+TEr = 0; JE na” qu 5
Dr mes 10 (3 dc
Vim + a SÉ:
ade
1ôxg = 15200
30% O
Za E 4I6R Ans
Solving Egs. (1) & (2)
E
7-9, The cable supports the loading shown. Determine
the magnitude of the horizontal force P so that xg = 6 ft.
T;
13
. 8 2 3
= == he — =» 00 =
+18 =0 Ya he es
i 18
he = 102 [62]
qaie
Solving Ego: DE
ts ms us m A,
Method àf Joints:
Join 8
Fac = IOMEN Pago IES RN
Joint C
4
1528 = Feo= 1031 [ )=o Fen = HIDAN
7
ATER = toa ( -P=0 M=250kN An
Sbimt D
328
AEE =D; For (=) -25=0 pl
Sélviog Es. [7 and [2] vielas
= 625N Ans
Fa = [179 kN
Tifus. the maximum tension in the cuble is
Fix = Fan = 125kN Ang
bem
Fog
Yo
799, The cable AB is subjected to a uniform loading ot
200 Nim. Té the weight of the cable is neglected and the
slope anigles at points A and B are 30º and 60%, | 0 M
Tespectively, determine the cúrve that defines the cable O PO
shape ànd te maximum tension developed in the cable, , ER RERRRE [a
Í yu GUN ++) ae!
B, (2002 +€,)
i =—— td:
»=6" G=0
e
E
k
v
»
z ="Wn30, 6» Bjtanzo
»= attom + random
4a Atx = 15m, Ee tanéo, E =B98N
ve (38 sr) m Ans
*7100, Thecáble su A girder which weighs 850 Ib/ft.
Determine thê tension in the cable at points A, 5, and €.
100%
Ay EK, par
sas
wo =
%
Ay =s0R, x =(100-49
= 425(100-2)3
“= &
EP = (6 = Doart+ 100?
6)! + 207 - 10 =0
ve aja DO | doa
& = 3649 ip
MA,
D. -
= Utme, 1.365
8 = sr
159
TS moço O ssa SL
n=SL7kp Aus
AB,
D=H =365kp Anê
AC,
dy Xa2s)x!
> = tm = HE =
E e been 086
Be = 440º
B 36459.
% = E. = S$4o
E cor Toaaade * 0683
% = 507 kp Aus
TON, The cable is subjected to the triangular loading. IE
the slope -of thê cable at point O is zera, determine the
equation of the: curve y = f(x) which defines the cable
shape OB, and the maximum tension developed in the cable,
x
2º idntaránã
1 500:
-s fe ada
1
= aloe + Ca
= a *Cx+C)
Ax=55%, y=88
>= 237010? Ans
Ba =
FA Ema Keig "Lisa
Ou tar T(1.6) = 57.990
Ha
Le * =
Tou = 442 ip Ans
H= 04
2344
BO ca GOSTO
3
8
dl
“E
TE ip
*7104; Determine the-maximum tension developed in
the cable if it-is subjected to a tniform load of 609 N/m.
The Equation of The Cable :
1
ve 5 [ecinasãs
(Eetscsra) 1
Elreste) ta
7
Boundary Conditions :
»=0 at x=0, ten ftom Eq.fi 9= Elo €G=0
É tn 10 ais =, den from Eg 3) mm 1a L€,) Ci =Htan 10º
d& Fa
vos
Thus, vs mera Ox 13
| 2
y=20matá = 100 nm, then from Eg (3)
600,
! 20 =+55(100") +ean10M(100) By=1267265,47N
| a
Sta
and atento
so
“Tay Tag + e OE
=0,4735(10) x + en 10º
i $ = Bus.atx = 100 m and he maximum tension ocouis when 8'= ju.
+ - 3 2
tanfa E =0.4735( 1073) (100) + 10º
| Bexi = 1261º
The rima emósion the cabe is
Fr o. 1267265,47
Cosôga cos 12,61
1298579.00N = 1.30MN . Ams
MOS. “A coble has a weight of 5 If Et cam span 300
ft and has a tag 0F15 ft, determine the length of the cable.
The ends of thg cable are supported at the same elevation.
: 106. Shdw, that. the deflection curve of the cable
* discussed in' Example 7-15 reduces to Eq. (4) in Example
7-4 when.thé Ayperbolic cosine function is expanded in
* termsofa séries ând only the first two tétims are retáined.
(The ansvwef indicates that the catenary may be replaced
by a parabola ih the analysis of problems in which the sag
is small. In fhis case, the cable weight is assumed to be
uniformly distributed along the horizontal.)
wo = SIR 3
From Example 7-15,
He
= É, to
>= ql cash (ala) y
Atx= 1509 .y=I5t
15wo (e)
e = cost emb: — 1
& Br
Hg = 37621
&
se E
Wo
sh)
s= 50%
L=2s=302n Ane
Using .Eg. (3).in Example 7-14,
volt
Fgm ME
Wo ya n? qr»
7107. A Uniform cord is suspended between two points,
having the same elevation. Determine the sag-to-span
ratio so that the maximum tension in the cord equais the
cord's total! weight.
K
1º emiço) =):
z = = uni (0.8) = 0.5493
asc t, | wear Lo yuh
2
E = tuto e mt) Bolo» Bfea(2t) 1]
! Sm al
cosbui FED 2a Am ia ds - | 01347 (E)
Tu = BL
obras QISM7L = 0.5493
L 2%
wo(ts) Hgicosh (E)
(8 lar :
i L = Q.I4i Aná
arrom( e) à Eros (=)
xi 7
!
s*7108. (A cable-has a Weight of 2 Ib/fL..lf it can span.
100 ft andihas a sag of 12 ft, deterímine the length of the
cable, Theiends of the cable are supported from the same
elevation.;
From Eq. (5) of Example 7-15;
e Sfo(s9-]
= Fi ue)
ne Fon) - 1]
aenfoa(i)-]
Fy = 222
From Eq (3) of Example 7- 18:
|
)
711. A telephone line (cable). stretches between two
points which are 150 ft apart and at the same elevation. w= OS biA
The line sags 5 ft and the cable has a weight of 0,3 Ib/ft. . 3
Determine the length of the cable and the maximum Prom-Examplo7- is,
tension in the cable. N / sk
a
sa Euução Ls
»
4
t>
É?
MES TR p=58, w 03h
+
2
pq EerosaçÃÃe —
ES 5 = Sp feçÃ) = 7
| E
' a
! oa E = 690%
: Pa
a
Í o k Il. lindaa = siniçãs)
a 7 delmar E luisa
Um Arara 7509).
ma = (sinta) = 7.606º
; 169
= o 8
| Lo = Casé cosTada = 170 Ans
' 169,0 14:03 z
“= 25 Bla] = 7522
L=2s=150R Ans
m*7-112, Thejcable has a mass of OS kgfm and is 25,
long. Determine the vertical and horizontal components ot
force it exerts on'the top-of the tower,
!
Performing ho integration yiedets :
za asftm [pr cosnercs e a) ty
Fran Eq T-13
Eeplnd
E A a908:4€,)
K
Ats 0;
Atx Sm;
By trial and error
At point,
tinê, =
dr
f 1 sponsor
4.305
dr
4
ce eia + tan30º 2)
a pI
Applying bóundary conditicos at x = O; 5 = Oto Eq.[t] and using he result
Cy'= Eytan30º ylekas C, »-sinhr *(tan30º) . Hénce
E
fm acesas apso nor (83
s=2m FromEq
ia |
Ta teses +xmmcon]
B=354N
4=25m PomBgi2)
dj 50429
+tanzoo
7394
o, =s850º
dilsezsm
E), = HrtanO, = 7394t0065.90º = 165 N dns
(Ha =H=89N Ais
1
A7-1I3, A S0-ft cable is suspended between two points
a distance of 15 ft apart and at the same elevation. If the
minimum tension in the cable 5s:200 tb, determine the
total weight: of the cable and the maximum tension
developed!in the cable.
Tm» Ey = 200b
From Example 7- 15:
Ba. Wo x
lar DM cio [0%
“E to (5º)
2 2ea(5 (8)
Fr 200 a
Deo asi "GR daao POL Am
7. tb A 100-Ib'cable is attached between two points
ata distance 50 ft apart having. equal elevations. Jf the.
- maximim tension developed in the cable is 75.15,
determine the length of the-cable and. the sag.
From Example 7515,
Ta «mem
E
DOS Gui = E
= Ame
1 he
For ; of cable,
100
we «30
s F
una. = et, E
Ex
Thos,
VP = = ss0p
ea fra fr)
s=7.8R
so
Wa Ba L5ODAR
Tonlkomh =25= 556% Aus
+ [eo (22) +) Ef (12200) À ]
= 1068 An
7117, Détermine the distance a between the supports
in terms óf the beam's length £ so that the moment in
the symmietric beaih is zero at the beam's center.
i
Support Reactions :. From FBD (a).
wo a: w
frim=o; ZU+a(5)-8,69=0 B=F(Lra
Free bády Diagram : The FBDfor SEgment AC secdoned throiigh point
C isdeiion,
i :
Internal, Forces : This problem fequires M, = 0; Sumrairg moments
about poi C[FBD (b)], we have
Creu =o; TG)ria-aliaen)]
» a
| -qUro(s)=o
i 20! +20, 1250
i a=0.366L Ans
T4I8. Drivfthe hear and momens digrârei tor the bear.
FU): Diaw thé shear and moment di
Raid apd, lagrams for the.
Support Redetioas : The 6 EN load can be seplacde by an equivalent force
and couple mdment at B as shown on FBD (5).
Grs Fepsin 4S(6)-6(3)-900=0 p= 6364 kN
+TES HO; A +6365n4Sº-6=0 A, =1.50kN
Shear and Moment Functions : For 0Sx<3 m [FBD (b)],
«tigão 150-V=0 V=1.50:N Ané
Crua M-1505=0 M= (1502) Nom Ane
For 3 mexk6 m(FBD(O],
ATER ÃO V+636MNMS=0 -V=-450LN Am
GrEM=O 6364in4SM(6-x)-M=0
; M= (2710-4500) kN-m Ans
bx M
Aero a
z DJ ai
(e) N
Manda
nv
Espeto 364
| em
dote ( EELITI TIO sm
“a poem
vg
Cd