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15 m
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50 kN
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20 kN
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pg!cfbnt/!Gps!fybnqmf-!xf!dbo!mpdbuf!uif!qpjout!tvckfdufe!up!
fyusfnf!wbmvft!pg!tifbs!gpsdf!boe!cfoejoh!npnfou/!!Jo!beejujpo-!
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jotjef!uif!dpodsfuf!cfbnt<!ps!uif!ejnfotjpot!boe!dsptt.tfdujpobm!
tibqf!pg!uif!tuffm!hjsefst/!Uivt-!uif!tibqft!pg!TGE!boe!CNE!
fobcmf!uif!fgÝdjfou!boe!fdpopnjdbm!qmbdfnfou!pg!tuffm!cbst!boe!
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ps! cfoejoh! npnfou! ejbhsbn %
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Uif!bcpwf.nfoujpofe!qspdfevsf!cfdpnft!mfohuiz!boe!dvncfstpnf!xifo!b!cfbn!dbssjft!b!ovncfs!pg!
mpbet!bt!npsf!dvut!bsf!offefe/!Gps!tvdi!dbtft-!b!ejsfdu!boe!tjnqmfs!bqqspbdi!jt!vtfe!up!dpnqvuf!tifbs!gpsdf!
boe!cfoejoh!npnfou!bu!dsjujdbm!qpjout/!Tvddfttjwf!tvnnbujpo!pg!gpsdft0npnfout!jt!vtfe!xijmf!lffqjoh!uif!
gpmmpxjoh!tjho!dpowfoujpo!jo!njoe/!Uijt!bqqspbdi!dpowfsut!b!sbuifs!dvncfstpnf!qspcmfn!joup!b!tjnqmfs!pof!
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2/!Dboujmfwfs!xjui!Qpjou!Mpbe!bu!Gsff!Foe! B!dboujmfwfs!BC!pg!mfohui!"!jt!tipxo!jo!Gjh/!6/2: /!
Ju!jt!tvckfdufe!up!b!qpjou!mpbe! -!bdujoh!wfsujdbmmz!epxoxbset!bu!jut!gsff!foe/!Jut!gsff.cpez!ejbhsbn!xjui!
fyufsobm!mpbe! !boe!tvqqpsu!sfbdujpot-! B!>! !boe! tB!>! "!jt!tipxo!jo!Gjh/!6/2: /!B!tfhnfou!pg!cfbn!pg!
mfohui!!-!dvu!gspn!uif!sjhiu!tjef!pg!uif!dboujmfwfs-!jt!tipxo!xjui!qptjujwf!tjho!dpowfoujpot!pg!joufsobm!gpsdft!
jo!Gjh/!6/2: /!Frvbujoh!uif!tvn!pg!wfsujdbm!gpsdft!up!{fsp-!xf!hfu!tifbs!gpsdf!bu!uif!tfdujpo!2.2!bt!gpmmpxt;!!
S !>!1;! %!Ï! !>!1! fi! %!>!
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+!Jg!zpv!mppl!dmptfmz-!uiftf!tjho!dpowfoujpot!bsf!uif!tbnf!bt!ejtdvttfe!jo!Tfdujpo!7/7/!Cvu-!ifsf-!xf!dbo!ejsfdumz!
dbmdvmbuf!tifbs!gpsdft!boe!cfoejoh!npnfout!bu!b!eftjsfe!qpjou-!xjuipvu!fyqsfttjoh!uifn!jo!ufsnt!pg!!/!Uif!gpmmpxjoh!
ejbhsbn!dbo!cf!vtfe!gps!hsbqijdbm!sfqsftfoubujpo/
left
!!!
right
left
right
External forces causing positive SF and BM
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W
a
W
L
(a)
L
(a)
A
C
MsA 2
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VA = W
B
MsA = WL 1
L
(b)
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W
W
2
A
1
W
W
B
A
2
Wa
2
M
B
x2
S
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C
B
A
(d) SFD
(d ) SFD
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x1
+
0
B
A
1
W
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0
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W
0
A
B
1
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(c)
W
1
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M S
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B
A
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C
A
0
B
(e) BMD
A
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B
C
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Gjh/!6/2:! -
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ejtubodf! !gspn!uif!foe!B!boe! !gspn!uif!foe!C-!bt!tipxo!jo!Gjh/!6/31 /!Jut!GCE!xjui!fyufsobm!mpbe! !boe!
tvqqpsu!sfbdujpot-! B!>! !boe! tB!>! !jt!tipxo!jo!Gjh/!6/31 /!Bt!uifsf!bsf!ejggfsfou!mpbet!gps!uif!uxp!tqbot!
BD!boe!DC-!xf!ublf!uxp!tfdujpotʨ2.2!boe!3.3-!hjwjoh!jtpmbufe!tfhnfout!pg!mfohuit!!2!boe!!3-!sftqfdujwfmz/!
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frvjmjcsjvn!up!uif!sjhiu!jtpmbufe!qbsu!pg!uif!cfbn-!xf!hfu!
S ! >!1;! %!>!1! boe! S }2Ï2!>!1;!
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jo!Gjh/!6/31 !boe! /!!Tjodf!!3!wbsjft!gspn!1!)qpjou!B*!up! !)qpjou!D*-!uif!cfoejoh!npnfou!jt!nbyjnvn!bu!uif!
qpjou!B!boe!jt!frvbm!up!Ï <!boe!jt!{fsp!bu!uif!qpjou!D-!bt!tipxo!jo!uif!CNE!pg!Gjh/!6/31 /
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lO0n!tqsfbe!pwfs!jut!foujsf!mfohui-!!bt!tipxo!jo!Gjh/!6/32 /!Jut!GCE!xjui!fyufsobm!mpbe! !lO0n!boe!tvqqpsu!
sfbdujpot-! B! >! "!boe! tB!>! "303!jt!tipxo!jo!Gjh/!6/32 /!B!tfhnfou!pg!mfohui!!!gspn!uif!foe!C!pg!uif!
dboujmfwfs!jt!jtpmbufe!cz!tfdujpojoh!ju!bu!2,2/!Uif!VEM!pwfs!uijt!tfhnfou!dbo!cf!sfqmbdfe!cz!b!qpjou!mpbe!pg!
nbhojuvef! ! bdujoh!bu!b!qpjou!!03!ejtubodf!bxbz!gspn!uif!tfdujpo!2,2/!Uijt!mpbe!boe!joufsobm!gpsdft!)tifbs!
gpsdf!%!boe!cfoejoh!npnfou! *!bsf!tipxo!jo!Gjh/!6/32 /!Frvbujoh!uif!tvn!pg!wfsujdbm!gpsdft!up!{fsp-!xf!hfu!
tifbs!gpsdf!bu!uif!tfdujpo!2,2!bt!gpmmpxt/
S ! >!1;! %!Ï! !!>!1! fi! %!>! #
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) "* bu!!!>!" )foe!B*-!bt!tipxo!jo!TGE!pg!Gjh/!6/32 /!
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wx 3
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3
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-!cfjoh!{fsp!bu!!!>!1!
)foe!C*!boe!Ï "303!bu!!!>!"!)foe!B*-!bt!tipxo!jo!Gjh!6/32 /!Uif!cfoejoh!npnfou!bu!njeqpjou!pg!uif!cfbn!
jt! "309-!xijdi!nfbot!uif!qbsbcpmjd!CNE!jt!dpodbwf!bt!tipxo!jo!Gjh/!6/32 /!Uijt!dbo!gvsuifs!cf!tvqqpsufe!
bt!gpmmpxt/
Tjodf!tifbs!gpsdf!bu!uif!qpjou!C!jt!{fsp-!uifsfgpsf-!bt!qfs!Fr/!)6/3*-!Ýstu!efsjwbujwf!pg!uif!cfoejoh!npnfou!
xsu!ejtubodf!tipvme!cf!{fsp!bu!uif!qpjou!C/!Uijt!nfbot!tmpqf!pg!uif!ubohfou!up!CNE!bu!uif!qpjou!C!nvtu!cf!
{fsp/!Ifodf-!CNE!jt!b!dpodbwf!qbsbcpmb!sbuifs!uibo!b!dpowfy!qbsbcpmb<!uif!mbuufs!xjmm!ibwf!opo.{fsp.tmpqf!
ubohfou!bu!C/!Tjnjmbs!fyqmbobujpo!hpft!gps!bmm!puifs!dbtft/
w kN/m
W
L
A
(a)
A
w kN/m
2
L
(a)
B
wx
x/2
B
2
W/2
x1
(b)
+
C
0
A
B
B
–
–W/2
(d ) SFD
0
(c) SFD
A
B
–
Parabolic
–wL2/2
Gjh/!6/32! -
W/2
W/2
+
A
1
S
x2
B
M S 1
M
2
A
(c)
0
B
1
W/2
x
M S
wL
1
C
L/2
L
(b)
wL
A
W/2
1
1
wL2/2
2
B
0
A
(e) BMD
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!
Gjh/!6/33! %
+
WL/4
C
(d) BMD
+
B
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pg!mfohui!"!jt!tipxo!jo!Gjh/!6/33 /!Ju!jt!tvckfdufe!up!b!qpjou!mpbe! !bu!uif!njeqpjou!pg!uif!cfbn/!Cpui!uif!
tvqqpsu!sfbdujpot!bsf! 03/!Uxp!tfdujpot!bsf!ublfo!gps!dpnqvujoh!tifbs!gpsdf!boe!cfoejoh!npnfou!bt!tipxo!
jo!Gjh/!6/33 /!Bqqmzjoh!frvbujpot!pg!frvjmjcsjvn!up!uif!sjhiu!jtpmbufe!qbsu!pg!uif!cfbn-!xf!hfu!
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3
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W
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6
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(iv) Pratt (bottom deck)
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(v) Polonceau (Fink) Truss
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tubcjmjuz!fwfo!jg!tpnf!nfncfst!gbjm/!Bo!joefufsnjobuf!tusvduvsf!
ibt!npsf!ovncfs!pg!tvqqpsu!sfbdujpot-!ps!pg!nfncfst-!ps!pg!cpui-!
uibo!xibu!jt!offefe!up!nblf!uif!tusvduvsf!kvtu!tubcmf/!Efqfoejoh!
po!uif!fyusb!ovncfs!pg!tvqqpsu!sfbdujpot!boe!fyusb!ovncfs!pg!
nfncfst-!b!usvtt!ibt!
uzqft!pg!joefufsnjobdz/!
�! 9ORTHY!OF!0OTE
Jo! b! tubujdbmmz! efufsnjobuf!
tusvduvsf-! po! sfnpwjoh! fwfo!
b! tjohmf! nfncfs-! uif! foujsf!
tusvduvsf! dpmmbqtft/! Xifsfbt!
jo! b! tubujdbmmz! joefufsnjobuf!
tusvduvsf-! uifsf! bsf! fyusb!
nfncfst! up! cfbs! mpbet! jo! uif!
fwfou!pg!gbjmvsf!pg!tpnf!nfncfst!
evf!up!mpbejoh!dpoujohfodz!tvdi!
bt!fbsuirvblft-!upsobepft-!fud/
Joufsobm! Joefufsnjobdz! B!usvtt!jt!tbje!up!cf!joufsobmmz!
efufsnjobuf!jg!ju!ibt!kvtu!tvgÝdjfou!ovncfs!pg!nfncfst!sfrvjsfe!
gps!tubcjmjuz/!Tvdi!usvttft!tbujtgz!uif!sfmbujpotijqt-! >!3 Ï!4!
)gps!qmbof!usvttft*!ps! >!4 !Ï!7!)gps!tqbdf!usvttft*/!Uiftf!usvttft!bsf!bmtp!lopxo!bt!qfsgfdu gsbnft/!B!usvtt!
ibwjoh!fyusb!nfncfst!uibo!sfrvjsfe!gps!mjnjujoh!tubcjmjuz!jt!lopxo!bt!bo!
usvtt!boe!
jut!efhsff!pg!joufsobm!joefufsnjobdz!jt!hjwfo!cz!uif!fydftt!ovncfs!pg!nfncfst/!Tvdi!usvttft!bsf!bmtp!lopxo!
bt!sfevoebou gsbnft boe!tbujtgz!uif!sfmbujpotijq!! ?!3 Ï!4!)ps! ! !4 Ï!7!gps!tqbdf!usvttft*/!Uif!usvttft!
xjui!mftt!nfncfst!uibo!sfrvjsfe!gps!tubcjmjuz!bsf!votubcmf!usvttft!boe!bsf!lopxo!bt!efÝdjfou!gsbnft/!Gps!tvdi!
gsbnft-! =!3 Ï!4/
ŵŻŸɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
Gjhvsf!7/5!hjwft!fybnqmft!pg!uif!uisff!uzqft!pg!gsbnft/
m = (2j – 3)
(a) Perfect frame
m > (2j – 3)
(b) Redundant frame
m < (2j – 3)
(c) Deficient frame
Gjh/!7/5!
Fyufsobm!Joefufsnjobdz! B!usvtt!jt!tbje!up!cf!fyufsobmmz!efufsnjobuf jg!uif!ovncfs!pg!tvqqpsu!sfbdujpot!
jt!frvbm!up!uif!ovncfs!pg!frvjmjcsjvn!frvbujpot/!Po!uif!puifs!iboe-!jg!uif!upubm!ovncfs!pg!tvqqpsu!sfbdujpot!
jt!npsf!uibo!uif!ovncfs!pg!frvjmjcsjvn!frvbujpot!)j/f/-!
!gps!qmbof!usvttft!boe! !gps!tqbdf!usvttft*-!
uif!usvtt!jt!tbje!up!cf!fyufsobmmz!joefufsnjobuf/!Uif!
efhsff!pg!fyufsobm!joefufsnjobdz!jt!hjwfo!cz!uif!
Myth
Fact!
ovncfs!pg!tvqqpsu!sfbdujpot!jo!fydftt!pg!uif!ovncfs!
kvtu!offefe!gps!frvjmjcsjvn/
B!usvtt!jt!bmxbzt!
!Ju!jt!usvf!pomz!jg!uif!efhsfft!
B!usvtt!jt!tbje!up!cf!dpnqmfufmz!efufsnjobuf jg!
tubcmf!jg!jut!upubm!
pg!joufsobm!boe!fyufsobm!
uif! efhsfft! pg! cpui! uif! joefufsnjobdjft! bsf! {fsp/!
efhsff!pg!tubujd!
joefufsnjobdjft!bsf!fbdi!{fsp/!
Tvdi! usvttft! dbo! cf! dpnqmfufmz! bobmztfe! vtjoh!
joefufsnjobdz!jt!
Jg-!boz!pof!pg!uifn!jt!ofhbujwf!
dpowfoujpobm!frvjmjcsjvn!frvbujpot/!Jg!uif!efhsff!
{fsp/
xijmf!uif!puifs!jt!qptjujwf-!uif!
usvtt!jt!tujmm!votubcmf/!
pg! joufsobm! ps! fyufsobm! joefufsnjobdz! pg! b! usvtt!
cfdpnft!
! -!uif!usvtt!cfdpnft!votubcmf!boe!ifodf!dpmmbqtft/!
7/4/3!
Joufsobm!Gpsdft!jo!Usvttft
Nfncfst! pg! b! usvtt! sftjtu! uif! fyufsobm! mpbet! cz!
efwfmpqjoh! byjbm! gpsdft! xijdi! nbz! cf!
ps!
! /!Uipvhi!uifz!bmtp!dbssz!tifbs!gpsdft!
boe!cfoejoh!npnfout-!cvu!cfjoh!rvjuf!tnbmm!uifz!
bsf!ofhmfdufe!jo!uif!bobmztjt/!Cbtfe!po!uif!obuvsf!pg!
byjbm!gpsdf!dbssjfe-!uif!nfncfst!bsf!pg!uisff!uzqft;
2/!Tusvut/! Uif!nfncfst!uibu!dbssz!pomz!
! !
Myth
[fsp.gpsdf!
nfncfst!nbz!cf!
sfnpwfe!gspn!b!
usvtt-!tjodf!tvdi!
nfncfst!bsf!opu!
sfrvjsfe/
Fact!
Sfnpwbm!pg!nfncfst!sfoefst!
uif!usvtt!votubcmf!jg!uif!mpbejoh!
qbuufso!dibohft!xijdi!jt!b!
qifopnfopo!uibu!dbo!ibqqfo!jo!
qsbdujdf!tvdi!bt!evsjoh!xjoe!ps!
fbsuirvblf/!
gpsdft/
3/!Ujft/! Uif!nfncfst!uibu!dbssz!pomz!
!gpsdft/
4/![fsp.gpsdf!Nfncfst/! Uif!nfncfst!uibu!dbssz! !
/
Tjho! Dpowfoujpo! Ufotjmf! gpsdf! jt! ublfo! bt!
! boe!
dpnqsfttjwf!gpsdf!bt!
! /!Hsbqijdbmmz-!ufotjpo!jo!b!nfncfs!jt!
tipxo!cz!bsspxt!qpjoujoh!bxbz!gspn!kpjout!bt!uif!nfncfs!ufoet!up!
pqqptf!uif!fyufsobm!qvmmjoh!gpsdft!fyfsufe!bu!jut!foet/!Po!uif!puifs!
iboe-!dpnqsfttjpo!jt!tipxo!cz!bsspxt!qpjoujoh!upxbset!kpjout-!bt!uif!
nfncfs!ufoet!up!pqqptf!uif!fyufsobm!qvtijoh!gpsdft!fyfsufe!bu!jut!foet/!!
Gjhvsf!7/6!tipxt!uif!tjho!dpowfoujpo!pg!uiftf!gpsdft/
Strut
Compression
(Negative force)
Tie
Tension
(Positive force)
Gjh/!7/6! "
!
ɏ
ŵŻŹ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
.1!hre
7/5!
�! BOBMZTJT!PG!USVTTFT
Uif!usvttft!xijdi!bsf!efufsnjobuf!joufsobmmz!bt!xfmm!bt!fyufsobmmz-!dbo!cf!bobmztfe!
vtjoh!frvbujpot!pg!frvjmjcsjvn/!Uifsf!bsf!uxp!nfuipet!pg!bobmztjtʨhsbqijdbm!boe!
bobmzujdbm/! Uif! bobmzujdbm! nfuipe! jt! bhbjo! pg! uxp! uzqftʨnfuipe! pg! kpjout! boe!
nfuipe!pg!tfdujpot/!Uiftf!bsf!ejtdvttfe!jo!uif!gpmmpxjoh!tfdujpot/!Uif!bttvnqujpot!
nbef!gps!uif!bobmztjt!pg!usvttft+!bsf!bt!gpmmpxt;
$¬»¬¹´°µ¬ɏº¼··¶¹»ɏ
¹¬¨ª»°¶µºɏ¨µ«ɏ­¶¹ª¬ºɏ°µɏ
´¬´©¬¹ºɏ¶­ɏ¨ɏ»¹¼ººȍ
2/!!Uif!kpjout!bsf!gsjdujpomftt/! Ju!nfbot!uif!kpjout!bsf!
/
3/!!Uif!nfncfst!bsf!tusbjhiu/! Jg!uif!nfncfst!xfsf!opu!tusbjhiu-!dpnqsfttjwf!gpsdft!xpvme!dbvtf!uifn!
up!cfoe/!Tvdi!cfoejoh!jowpmwft!tfdpoebsz!npnfout!boe!offet!bewbodfe!nfuipet!pg!bobmztjt/
4/!!Uif!efgpsnbujpot!jo!uif!usvtt!bsf!wfsz!tnbmm/! Uif!usvttft!bsf!efgpsnfe!xifo!uifz!bsf!mpbefe/!
Mbshf!efgpsnbujpot-!jg!boz-!dbvtf!bqqsfdjbcmf!dibohf!jo!uif!ejnfotjpot!boe!pwfsbmm!tibqf!pg!uif!usvtt-!xijdi!
offet!tqfdjbm!dpotjefsbujpot/
Myth
5/! Mpbet! bsf! bqqmjfe! bu! uif! kpjout! pomz/! Uif!
dpogjhvsbujpo! jt! tvdi! uibu! uif! mpbet! bqqmjfe! po!
usvttft!bsf!usbotgfssfe!pomz!uispvhi!kpjout/!Nfncfst!
sftjtu!uif!mpbet!pomz!uispvhi!byjbm!gpsdftÐufotjmf!ps!
dpnqsfttjwf/!Uifz!ep!opu!cfoe!mjlf!cfbnt/
Fact!
Kpjout!jo!b!usvtt!
bsf!qjoofe/
Jo!qsbdujdf-!uifz!bsf!sjwfufe-!
ps!cpmufe-!ps!xfmefe/!Uifz!bsf!
tfnj.sjhje!kpjout/!Ipxfwfs-!bt!
uif!npnfout!boe!tifbs!gpsdft!
qspevdfe!jo!usvttft!bsf!wfsz!
7/5/2! Nfuipe!pg!Kpjout
tnbmm!dpnqbsfe!up!byjbm!gpsdft-!
Tjodf! uif! usvtt! bt! b! xipmf! jt! jo! frvjmjcsjvn-! uifsfgpsf-! fbdi! kpjou! jt! jo! uif!kpjout!bsf!
frvjmjcsjvn/!Ifodf-!uif!frvbujpot!pg!frvjmjcsjvn!bsf!bqqmjfe!bu!uif!kpjout!
gps!nbljoh!uif!bobmztjt!
pof!cz!pof!voujm!bmm!uif!nfncfst!bsf!bobmztfe/!Uif!gpmmpxjoh!tufqt!eftdsjcf! tjnqmfs/!
uif!nfuipe!pg!kpjout/
!
!
2/! Dbmdvmbuf!uif!tvqqpsu!sfbdujpot!cz!dpotjefsjoh!frvjmjcsjvn!pg!uif!usvtt!bt!b!xipmf/
3/! Dpotusvdu! uif! gsff! cpez! ejbhsbn! pg! b! kpjou! bu! xijdi! nbyjnvn! ovncfs! pg! volopxo! gpsdft! boe0ps!
sfbdujpot!jt!uxp/
! 4/! Bqqmz!frvbujpot!pg!frvjmjcsjvn!jo!uxp!nvuvbmmz!qfsqfoejdvmbs!ejsfdujpot!)j/f/-!Ȧ# !>!1!boe!Ȧ# ! >!1*!bu!
uif!kpjou!boe!pcubjo!uif!volopxo!gpsdft/
! 5/! Xijmf!xsjujoh!uif!frvjmjcsjvn!frvbujpot-!bmm!uif!volopxo!gpsdft!nbz!cf!ublfo!bt!ufotjmf!)j/f/-!xjui!
qptjujwf!tjho*/!Bgufs!uif!dbmdvmbujpot-!jg!b!gpsdf!jt!tffo!up!ibwf!ofhbujwf!tjho-!ju!xpvme!nfbo!ju!jt!b!
dpnqsfttjwf!gpsdf/
! 6/! Qspdffe! up! puifs! kpjout! tvddfttjwfmz! boe! tpmwf! gps! volopxo! gpsdft! voujm! bmm! nfncfs! gpsdft! bsf!
dbmdvmbufe/
Gjhvsf!7/7 !tipxt!b!uxp.ejnfotjpobm!usvtt!voefs!uif!bdujpo!pg!uxp!qpjou!mpbet!$!boe!%/!Gjstu-!uif!tvqqpsu!
sfbdujpot! bsf! dbmdvmbufe! cz! dpotjefsjoh! uif! usvtt! bt! b! sjhje! tusvduvsf/!Uifo-! xf! dpotjefs! uif! gsff! cpez!
ejbhsbnt!pg!kpjout!)Gjh/!7/7 */!Xf!Ýstu!ublf!uif!kpjou!B!)ps!E*/!Uif!gpsdft!jo!nfncfst!dpoofdufe!up!uijt!kpjou!
bsf!dpnqvufe!cz!bqqmzjoh!frvbujpot!pg!frvjmjcsjvn!jo!uif!ipsj{poubm!boe!wfsujdbm!ejsfdujpot/!Uif!gpsdft!jo!
!
+!Jo!uijt!cppl-!xf!tibmm!ejtdvtt!pomz!uif!
!usvttft/
ŵŻźɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
nfncfst!BC!boe!BF!bsf!uivt!dpnqvufe/!Uifsfbgufs-!uif!kpjou!C!jt!dpotjefsfe!boe!gpsdft!jo!nfncfst!CD!boe!
CF!bsf!dbmdvmbufe/!Gjobmmz-!kpjout!F!boe!D!bsf!ublfo!up!dbmdvmbuf!uif!gpsdft!jo!uif!sftu!pg!uif!nfncfst/!
�!0OTE!
Jg!uif!tibqf!pg!b!usvtt!boe!jut!mpbejoh!bsf!tznnfusjdbm-!xf!offe!up!bobmztf!pomz!ibmg!uif!usvtt/!Uif!puifs!
ibmg!uif!usvtt!ibt!tjnjmbs!gpsdft!jo!dpssftqpoejoh!nfncfst/
Q
P
B
B
C
FAB
D
A
E
A
FAE
C
FBC
FCE
FBE
E
FCD
FDE
D
RD
RA
RA
!
Q
P
RD
(a) A simple 2D truss
(b) Axial forces in various members
Gjh/!7/7!
7/5/3!
Nfuipe!pg!Tfdujpot
Bmtp!lopxo!bt!SjuufsÔt!nfuipe-!ju!jowpmwft!dvuujoh!uif!usvtt!cz!pof!ps!npsf!tfdujpot!qbttjoh!uispvhi!uif!
nfncfst!jo!xijdi!gpsdft!bsf!up!cf!efufsnjofe/!Uif!volopxo!gpsdft!jo!nfncfst!bsf!efufsnjofe!cz!dpotjefs.
joh!frvjmjcsjvn!pg!pof!qbsu!pg!uif!dvu!usvtt/!Uijt!nfuipe!jt!hfofsbmmz!vtfe!xifo!xf!offe!up!
/!Uif!gpmmpxjoh!bsf!uif!tufqt!pg!uijt!nfuipe/
!
!
!
!
!
2/! Dbmdvmbuf!uif!tvqqpsu!sfbdujpot!pg!uif!usvtt/!Opu!bmm!uif!sfbdujpot!offe!up!cf!dbmdvmbufe-!sbuifs!dbmdvmbuf!
pomz!uiptf-!xijdi!bsf!jodmvefe!jo!uif!tfmfdufe!dvu!qbsu!gps!gvsuifs!bobmztjt/
3/! Esbx!b!tfdujpo-!tusbjhiu!ps!dvswfe-!tjnvmubofpvtmz!dvuujoh!uispvhi!
!nfncfst!pg!uif!
usvtt/
4/! Tvjubcmz!tfmfdu!b!qbsu!pg!uif!usvtt!boe!esbx!jut!gsff!cpez!ejbhsbn!tipxjoh!bmm!uif!joufsobm!boe!fyufsobm!
gpsdft/
5/! Efufsnjof!uif!volopxo!gpsdft!cz!gpsdf!frvjmjcsjvn!)S# !>!1!boe!S# !>!1*!ps!npnfou!frvjmjcsjvn!
)S&!>!1*!ps!cpui/
6/! Bmm!uif!volopxo!gpsdft!nbz!cf!bttvnfe!ufotjmf/!Bgufs!dpnqvubujpot-!jg!b!gpsdf!jt!gpvoe!up!cf!ofhbujwf-!
ju!xpvme!nfbo!b!dpnqsfttjwf!gpsdf/
Gjhvsf!7/8 !tipxt!b!tjnqmf!usvtt/!Tvqqptf!xf!bsf!joufsftufe!up!dpnqvuf!gpsdft!jo!bmm!uif!nfncfst!fydfqu!
nfncfst!BC!boe!DE/!Xf!dbo!dvu!uif!usvtt!cz!uxp!tfdujpotʨ2.2!boe!3.3/!Boz!qbsu!pg!uif!usvtt!po!pof!tjef!
ɏ
ŵŻŻ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
pg!uif!tfdujpo!nbz!cf!dpotjefsfe!tfqbsbufmz!up!dpnqvuf!uif!gpsdft!jo!uif!nfncfst!cfjoh!dvu-!bt!tipxo!jo!
Gjh/!7/8 /!
1
2
B
C
B
FCE
FBE
A
FAE
1
A
1
C
FBC
1 2
E
D
FDE
D
2
RA
2
RD
(a) Truss being cut by two sections
(b) Isolated parts of the truss
Gjh/!7/8! &
Tpnfujnft-!cpui!uif!bcpwf!nfuipet!bsf!vtfe!up!bobmztf!b!mbshf!usvtt/!Jo!tvdi!dbtft-!cfgpsf!dvuujoh!b!
tfdujpo-!ju!nbz!cf!vtfgvm!up!Ýoe!uif!gpsdft!jo!tpnf!nfncfst!cz!uif!nfuipe!pg!kpjout/!Tfmepn-!{fsp!gpsdf!
nfncfst!bsf!uifsf!jo!b!usvtt/!Ju!jt!vtfgvm!up!jefoujgz!tvdi!nfncfst!xijmf!bobmztjoh!usvttft/!Uif!gpmmpxjoh!
hvjefmjoft!nbz!cf!dpotjefsfe!up!jefoujgz!{fsp.gpsdf!nfncfst;
! ) *! Jg!b!kpjou!pomz!ibt!uxp!nfncfst!boe!op!fyufsobm!mpbe!boe0ps!op!tvqqpsu-!uifo!uiptf!uxp!nfncfst!bsf!
{fsp.gpsdf!nfncfst!)Gjh/!7/9 */
! ) *! Jg!b!kpjou!pomz!ibt!uxp!nfncfst!boe!jt!mpbefe-!boe!jg!uif!mjof!pg!bdujpo!pg!sftvmubou!gpsdf!gspn!bqqmjfe!
mpbet!bu!uif!kpjou!jt!dpmmjofbs!xjui!pof!pg!uif!nfncfst!uifo!uif!puifs!nfncfs!jt!b!{fsp.gpsdf!nfncfs/!
Jg!uif!sftvmubou!gpsdf!bu!uif!kpjou!jt!opu!dpmmjofbs!xjui!fjuifs!nfncfs!uifo!cpui!nfncfst!bsf!opu!{fsp.
gpsdf!nfncfst!)Gjh/!7/9 */
! ) *! Jg!b!kpjou!ibt!uisff!nfncfst!boe!op!fyufsobm!mpbe!boe0ps!tvqqpsu-!boe!jg!uxp!pg!nfncfst!bsf!dpmmjofbs!
uifo!uif!opo.dpmmjofbs!nfncfs!jt!b!{fsp.gpsdf!nfncfs!)Gjh/!7/9 */
P
P
P
(a)
(b)
Gjh/!7/9! '
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(c)
(
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ŵŻżɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
200 kN
FAB
B
VB
200 kN
200 kN
A
A
60°
30°
C
VC
5m
(a) Triangular truss
B
60°
A
FBC
100
173.21
FAC
30°
VB
C
VC
(b) FBD of joints A, B and C
B
60°
30°
86.6
VB
C
VC
5m
(c) Analysed truss (forces in kN)
Gjh/!7/:
5OLUTION! Jo!uif!usjbohvmbs!usvtt-!–CBD!>!:1¡/!Uifsfgpsf-!BD!>!CD!dpt!41¡!>!5/44!n/
Frvbujoh!uif!tvn!pg!npnfout!pg!bmm!gpsdft!bcpvu!uif!tvqqpsu!D!up!{fsp-!xf!hfu
S&}D!>!1;! Ï!*C!¥!6!,!311!¥!)5/44!dpt!41¡*!>!1! fi! *C!>!261!lO
Frvbujoh!uif!tvn!pg!wfsujdbm!gpsdft!up!{fsp-!xf!hfu
S# !>!1;! *C!Ï!311!,!*D!>!1! fi! *D!>!311!Ï!261!>!61!lO
Mfu!vt!bqqmz!uif!nfuipe!pg!kpjout/!Gsff!cpez!ejbhsbnt!pg!bmm!uif!kpjout!bsf!tipxo!jo!Gjh/!7/: /!Uif!gpsdft!
jo!bmm!uif!nfncfst!bsf!bttvnfe!ufotjmf/
Kpjou!C;!Bqqmzjoh!frvbujpot!pg!frvjmjcsjvn!jo!wfsujdbm!boe!ipsj{poubm!ejsfdujpot-!xf!hfu
-150
!>!Ï284/23!lO!>!284/32!lO!)D*+
sin 60∞
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D
3.5 kN
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Two loads of 20 kN each are supported by a flexible cable ACDB as shown in
Fig. 6.17a. The span of the cable between supports is 15 m and the sag of both the
loads is 2.5 m. Determine the tensile force developed in all the three segments of the cable.
5m
5m
5m
A
B 2.5 m
C
D
20 kN
20 kN
(a) Cable carrying symmetrical loading
RA
H
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a1
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Gjh/!7/28
5OLUTION!
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'XAMPLE!hrh
A uniformly distributed load of 9 kN/m acts over a length of 4 m from the left-hand
support A of a cable (assumed weightless), whose both supports are at the same
level, as shown in Fig. 6.20a. At the support B, the cable runs over a frictionless pulley of negligible
dimensions and suspends a weight of 20 kN. Determine (a) the support reactions, (b) the horizontal
component of cable tension, (c) the sag of the point C and maximum sag in the cable, and (d) the
maximum and minimum tensile forces in the cable.
\MP!7/5^
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(b) FBD of the cable
Gjh/!7/31
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'XAMPLE!hri
A cable of 150-m span and 30-m sag has its supports A and B at the same level as
shown in Fig. 6.21a. The cable is carrying only its dead weight of 0.15 kN/m.
Determine (a) the horizontal force in the cable, (b) the maximum tensile force in the cable, (c) the
support reactions, and (d) the length of the cable.
\MP!7/5^
RB
RA
150 m
H
B
A
A
B
30 m
H
30 m
O
x
0.15 kN/m
0.15 kN/m
y
75 m
75 m
Tmax
RA
A
H
(a) The cable AB under its own weight
Cable
(b) FBD of cable and the joint A
Gjh/!7/32
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'XAMPLE!hrj
For the truss given in Fig. 6.22a, find the forces in members BC, CD and BD.
\MP!7/4^
S
5OLUTION! !@
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A
C
B
50 kN
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C
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B
FAB
FBD
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E
(a) Truss carrying 50-kN load
(b) FBDs of joints B and C
Gjh/!7/33
'XAMPLE!hrk
Find the axial forces in all the members of the truss shown in Fig. 6.23a. Also,
compute the support reactions.!
!
!
\MP!7/4^
RF
F
F
B
5m
B
C
C
FBC
30°
5m
A
30 kN
5m
5m
5m
5m
E
20 kN
5m
D
FAB
A
60°
30 kN
FAE
FBE
FCE
60°
60°
E
HF
FCF
90°
60°
RD
FCD
FDE
60°
D
HD
20 kN
(b) FEBs of all joints
(a) The truss
Gjh/!7/34
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'XAMPLE!hrcl
Find the forces in all members of a truss as shown in Fig. 6.24a which carries a
horizontal load of 12 kN at the point D and a vertical load of 18 kN at the point C.
\MP!7/4^
D
12 kN
HA
q
A
RA
q
2m
C
18 kN
B
2m
FCD
FAD
1.5 m
q
HA
A
(a) Truss carrying two point loads
q
FAC
RA
RB
FBD
C
FBC
18 kN
B
RB
(b) FBDs of all the joints
Gjh/!7/35
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kpjout!pof!cz!pof!up!dbmdvmbuf!uif!nfncfs!gpsdft/
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sin 36.87∞
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'XAMPLE!hrcc
For the truss shown in Fig. 6.25a, find the forces in the members. Given that all the
horizontal and vertical members are of the same length.!
\MP!7/4^
B
B
D
FBD
q
FAB
A
C
E
RA
F
A
RF
100 N
q
FBC
FAC C
D
q
q
FCD FDE FDF
E
q F
CE
RA
100 N
FEF
q
F
RF
100 N
100 N
(b) FBDs of all joints
(a) The truss
Gjh/!7/36
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S# ! >!1;! #BC!tjo!56¡!,!+B!>!1! fi! #BC!>!
-100
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100
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0
100
141.42
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C
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141.42
100
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100
100
100 N
100 N
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100
100 N
100 N
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'XAMPLE!hrcd
Determine support reactions and forces in all the members of the truss shown in
Fig. 6.27a.
\MP!7/4^
15 kN
B
B
q1
4m
FAB
q2
FAC
4m
D
4m
FBC
C
A
15 kN
q1
HA
A
q1 q2
C
q2
FBD
FCD
q2
4m
q1
D
RA
4m
4m
4m
RD
4m
(b) FBDs of all joints
(a) Truss
Gjh/!7/38
5OLUTION! Gsff!cpez!ejbhsbnt!pg!bmm!uif!kpjout!bsf!tipxo!jo!Gjh/!7/38 xjui!nfncfs!gpsdft!boe!tvqqpsu!
sfbdujpot!+B-!-B!boe!+E/!Dpotjefsjoh!frvjmjcsjvn!pg!uif!usvtt!bt!b!xipmf-!xf!ibwf
S&}B! >!1;! +E!¥!9!,!26!¥!9!>!1! fi! +E!>!Ï26!lO!>!26!lO!)Ø*
S# ! >!1;! +B!,!+E!>!1! fi! +B!>!Ï!+E!>!26!lO!)≠*
S# ! >!1;! -B!Ï!26!>!1! fi! -B!>!26!lO!)Æ*
Bohmft!q2!boe!q3 bsf!efufsnjofe!bt!!
q2!>!uboÏ2!)905*!>!74/54¡! boe-! q3!>!uboÏ2!)505*!>!56¡
Mfu!vt!opx!dpotjefs!uif!frvjmjcsjvn!pg!kpjout!pof!cz!pof/
ŵŽżɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
Kpjou!B;!
S# ! >!1;! #BC!tjo!q2!,!#BD!tjo!q3!,!+B!>!1! fi! #BC!tjo!74/54¡!,!#BD!tjo!56¡!,!26!>!1
)*
fi 1/9:!#BC!,!1/818!#BD!>!Ï26!
S# ! >!1;! #BC!dpt!q2!,!#BD!dpt!q3!,!-B!>!1! fi! #BC!dpt!74/54¡!,!#BD!dpt!56¡!,!26!>!1
fi! 1/558!#BC!,!1/818!#BD!>!Ï26!
) *
Tpmwjoh!Frt/!) *!boe!) *-!xf!hfu!#BC!>!1! boe! #BD!>!Ï!32/33!lO!>!32/33!lO!)D*
Kpjou!C;!
S# ! >!1;! Ï!#BC!dpt!q2!,!#CE!dpt!q2!Ï!26!>!1!
fi! #CE!>!44/65!lO!)U*!
S# ! >!1;! Ï!#CD!Ï!#BC!tjo!q2!Ï!#CE!tjo!q2!>!1
15 kN
B
fi! #CD!>!Ï!#CE!tjo!74/54¡
! !!
!
>!Ï!41!lO!>!41!lO!)D*
30
Zero-force member
Kpjou!E;
S# ! >!1;! Ï!#DE!dpt!q3!Ï!#CE!dpt!q2!>!1
33.54
0
C
-#BD cos q1
!!>!Ï!32/33!lO!!
fi! #DE!>!
cos q 2
21.22
21.22
!
!
>!32/33!lO!)D*
A
Gjhvsf!7/39!tipxt!uif!usvtt!xjui!nfncfs!gpsdft!nbslfe/
'XAMPLE!hrce
For the bridge truss shown in Fig. 6.29a,
calculate the forces in members DF, CE,
CF, EF and FH using the method of sections.!
\MP!7/4^
D
15 kN
15 kN
Gjh/!7/39!
15 kN
)
lO*
5OLUTION! Mfu!vt!Ýstu!efufsnjof!uif!tvqqpsu!sfbdujpot!cz!dpotjefsjoh!frvjmjcsjvn!pg!uif!usvtt!bt!b!xipmf/!
S&}B! >!1;! Ï211!¥!7!Ï!66!¥!:!,!+N!¥!29!>!1! fi! +N!>!71/94!lO
S# ! >!1;! +B!,!+N!Ï!211!Ï!66!>!1! fi! +B!,!+N!>!266!lO
fi! !!+B!>!266!Ï!71/94!>!:5/28!lO
Mfu!vt!dpotjefs!uxp!tfdujpot!2 2!boe!3 3!dvuujoh!uif!usvtt!uispvhi!nfncfst!EG-!DG-!DF!boe!GI-!FG-!DF-!
sftqfdujwfmz-!bt!tipxo!jo!Gjh/!7/3: /!!Gjhvsft!7/3: boe! !tipx!gsff!cpez!ejbhsbnt!pg!uif!mfgu iboe!qpsujpot!
pg!uif!usvtt/
Bqqmzjoh!frvbujpot!pg!frvjmjcsjvn!up!uif!tfdujpo!2 2!pg!usvtt-!tipxo!jo!Gjh/!7/47 S&}D! >!1;! Ï!#EG!¥!4!Ï!:5/28!¥!4!>!1! fi! #EG!>!Ï!:5/28!lO!>!:5/28!lO!)D*
S# ! >!1;! :5/28!,!#DG!tjo!56¡!>!1! fi! #DG!>!Ï244/2:!lO!>!244/2:!lO!)D*
S# ! >!1;! #EG!,!#DG!dpt!56¡!,!#DF!>!1! fi! Ï!:5/28!,!)Ï244/2:*!dpt!56¡!,!#DF!>!1
fi! #DF!>!296/46!lO!)U*
Dpotjefs!opx!uif!frvjmjcsjvn!pg!uif!tfdujpo!3 3!pg!uif!usvtt-!tipxo!jo!Gjh/!7/3: S# ! >!1;! :5/28!Ï!211!,!#FG!!1! fi! #FG!>!6/94!lO!)U*
S# ! >!1;! #GI!,!#DF!>!1! fi! #GI!,!296/46!>!1
fi! #GI!>!Ï296/46!lO!>!296/46!lO!)D*
ɏ
ŵŽŽ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
100 kN
D
B
F
J
H
L
N
3m
M
A
C
E
3m
G
55 kN
3m
3m
I
3m
K
3m
3m
(a) The bridge truss
100 kN
1
B
D
2
F
J
H
L
N
3m
A
C
94.17 kN
3m
2
E
G
55 kN
1
I
M
K
60.83 kN
3m
3m
3m
3m
3m
(b) Truss being cut by Sections 1-1 and 2-2
100 kN
1
B
D
B
FDF
FCF
D
2
F
FFH
FFE
A
A
FCE
C
C
1
94.17 kN
3m
FCE
2
94.17 kN
3m
(c) FBD of LHS portion to Section 1-1
3m
(d) FBD of LHS portion to Section 2-2
Gjh/!7/3:
30 kN
B
60°
30 kN
C
60°
5m
4m
A
F
15 kN
5m
Gjh/!7/41! +
4m
D
E
4m
ŶŴŴɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
'XAMPLE!hrcf
Figure 6.30 shows an industrial roof truss carrying loads due to wind. Find the
forces in members BC, CF, EF, CE and DE.
\MP!7/4^
5OLUTION! Mfu!vt!sftpmwf!uif!41.lO!gpsdft!joup!ipsj{poubm!boe!wfsujdbm!dpnqpofout!bt!41!dpt!71¼!>!26!lO!
boe!41!tjo!71¼!>!36/:9!lO-!sftqfdujwfmz/!Uif!41.lO!gpsdft!bsf!sfqmbdfe!cz!uifjs!dpnqpofout!bt!tipxo!jo!
Gjh/!7/42 /!Dpotjefsjoh!gsff!cpez!ejbhsbn!pg!uif!usvtt-!xf!xsjuf
S# ! >!1;! -B!Ï!26!Ï!26!>!1! fi! -B!>!41!lO
S&}B! >!1;! E!¥
¥
¥
¥
¥
¥
fi +
S#
+
+
fi +
–
!
Ê ˆ
ÁË ˜¯
"# –$ %
–$ %
25.98 kN
!
Ê ˆ
ÁË ˜¯
" !& q
!
(5 - 4)
4
"
./ & &,
0
1
25.98 kN
B
15 kN
60°
15 kN
C
q
5m
4m
45°
A
45°
45°
F
HA
D
E
15 kN
RA
RD
1
5m
4m
4m
(a) Section 1-1 cutting through the truss
1
25.98 kN
FBC
C
q
60°
15 kN
FCE
45°
4m
FCF
45°
FEF
E
D
FEF
E
FDE
RD
1
4m
(b) FBD of right-hand part
(c) FBD of the joint E
Gjh/!7/42
*
$' * '++ , * + - , ! . * ' *
, * .* !& 1
0 * '++ + +* ! ,! ,
/ + $ $
!& %
!& - !+,&
* 0
ɏ
ŶŴŵ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
%2' ,!
* +'
0
! + 0 33 * 0 - + / ' $ 6
S&7$
#% ¥
+ ¥
fi #% +
34/47!lO!)U*
113 ,! 2' , !+ 0 2',3,/ ,' ,! 8 ,- 3 !& * ,6 ! 3 &, - , !+
S#
# $ +,! q #$ - + "
+
fi
# $
; ;#$
fi # $
# $ - + q #$ +,! " #%
S#
fi
;# $
; ;#$
fi # $
< 38,! %2+ 9 : !& 9 :
#$
24/94!lO!)D* !& # $
= &
,! 0 - + ,!
/ + $% !& % - !+,& * 0
/ & &,
,
S#
# % #%
fi # % #% 34/47!lO!)U*!
#$% 1
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'XAMPLE!hrcg
;
9∵ p
#$
;
":
9:
#$
9 :
3:/57!lO!)D*
0 * > ,! % + ,8 ! ,!
For the simply supported truss shown in Fig. 6.32, find forces in members BD, DE,
EG and CE.
\MP!7/4^
F
10 kN
q
D
10 kN
2.25 m
q
B
q
A
H
C
1m
E
1m
= '++ -
G
1m
1m
,! 3 &+
Gjh/!7/43
5OLUTION! !$ !+,& ,!
S#
S&7
S#
* 0 / & &,
-?
+? ¥
¥
+
+?
0 *
'++ +*
¥
fi
fi
+
! ,! ,
+?
;
;
F
10 kN
q
D
10 kN 1
10 kN
FBD
2.25 m
q
B
A
H
q
C
RA
1m
E
1m
q
B
1
1m
G
RH
HH
q
A
RA
C
1m
(a) Section 1-1 cutting through the truss
FDE
1m
E
FEG
q
FCE
FDE
E
FEG
1m
(b) FBD of left-hand part
Gjh/!7/44
FBE
(c) FBD of joint E
ŶŴŶɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
!
q
$' *
2',3,/ ,'
2.25
= 36.87∞
3
tan -1
'++ , * + - , !
0 * 3 0 * !& 1
* ' *
0 * '++
/ +
S&7
#
S#
#
+,! q
fi
#
#%@
#
fi
#%@ 21!lO!)U*
S#
= &
=*'+
FG
AG
,! 0 - ,!
S#
%
¥
¥
fi
#$%
%
! ,! ,
!& - !+,&
6!lO!)U*#
fi
#
+,!
;"
23/6!lO!)D*
fi
- +q
#
#%@
fi #$%
#
+
%
/ + $% - !+,&
# % +,! q
fi # %
S#
#
% !& %@ + +*
#%@
* 0
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fi # % +,!
9/44!lO!)D*
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#
%
- +q
- +
fi
;"
0 * > ,! % + ,8 ! ,! ,
;"
#$%
9
:- +
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34/44!lO!)U*
'XAMPLE!hrch
Two weights of 15 kN and 40 kN are supported by a wire, which is fixed at the two
points A and D as shown in Fig. 6.34a. Inclinations of segments AB and CD with
the vertical are 25º and 45º, respectively. Determine tensions in all segments of the wire and inclination
of BC with horizontal.
\MP!7/5^
5OLUTION! A 0 / * ,!-3,! , ! 0 * +
- ! 113 A ,B+ *
* 1 ,! +
8 3' +
q =*'+ 0
+ 0 1 ,!
! $ , * * * ,6 ! 3 <,!- * + +
,+ ,! 2',3,/ ,'
!& $
,!
+
0
!
3
q
!&
*
!
- 1
*
$
!& $ 9 ,
:
* 8
Qpjou!C;
BC
sin(90∞ + 65∞)
AB
sin(90∞ - q )
fi
$
15
sin(180∞ - 65∞ + q )
15 ¥ sin 155∞
sin(115∞ + q )
6.34
sin(115∞ + q )
9:
9 :
Qpjou!D;
BC
sin(90∞ + 45∞)
CD
sin(90∞ + q )
fi
$
40
sin(180∞ - 45∞ - q )
40 ¥ sin 135∞
sin(180∞ - 45∞ - q )
28.28
sin(135∞ - q )
9 :
9 !:
ɏ
ŶŴŷ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
D
A
TAB
45°
25°
65° B
q
B
TBC
q
15 kN
TCD
TBC
C
15 kN
45°
q
C
40 kN
40 kN
(a) Concentrated loads supported by wire
(b) FBDs of points B and C
Gjh/!7/45
-
1 ,! %2+ 9 : !& 9 !:
fi
+,!
fi
fi
C' ,!
"- +q
6.34
sin(115∞ + q )
28.28
sin(135∞ - q )
- +
28.28
9+,!
6.34
; ;- +q 9
- +q
*
/ 8 8 3'
" +,! q
; ;: +,! q
+,! q
6.34
sin(115∞ + q )
'XAMPLE!hrci
" +,! q:
- +q
+,! q
q 63/26¡
15 ¥ sin(90∞ - 52.15∞)
sin(180∞ - 65∞ + 52.15∞)
6.34
sin(115∞ + 52.15∞)
40 ¥ sin(90∞ + q )
sin(180∞ - 45∞ - q )
!
- +
0 q ,! %2+ 9 : 9 : !& 9 :
15 ¥ sin(90∞ - q )
sin(180∞ - 65∞ + q )
$
fi
"- +q
52/49!lO
39/62!lO
40 ¥ sin(90∞ + 52.15∞)
sin(180∞ - 45∞ - 52.15∞)
35/85!lO
A cable is suspended between the two points at the same level 270 m apart. The
cable carries a deck, which supports a uniformly distributed load of 100 N per
metre of horizontal length as shown in Fig. 6.35a. If the sag is 50 m, determine the horizontal thrust, support
reactions and the maximum cable force
\MP!7/5^
ŶŴŸɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
A
0.1 kN/m
B
270 m
HA
50 m
Cable
C
HB
x
A
B
yC = 50 m
RA
y
100 N/m
Deck
RB
C
270 m
(a) Cable carrying a deck
(b) FBDs of cable
Gjh/!7/46
.
5OLUTION
!
- ,- ,
"
! /7 :&
wL2
8 yC
< +
7 ,?
,
"
$
%! &'
*
+
,
"
!
29/336!lO
" -=
-
7 ,
>
= $
A
B+ ¥
+=
D $
+
¥
C
+
H 2 + RA2
- ,!
+=
!
&
fi
+
wL
B+ ¥
" +=
0.1 ¥ 270
2
F
A
+
24/6!lO
#
18.2252 + 13.52
.
C- ¥
,
33/79!lO
"
- ,D - ,- ,
"F
D
D F
- ?,
"
24/6!lO
FF -
fi B '
'XAMPLE!hrcj
270
2
- ?,
#
Bmufsobujwfmz-!
, " !
¥
24/6!lO
--
\
$ 7
0.1 ¥ 2702
8 ¥ 50
"= -
?
S&@=
%
D
E #
" #
;
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.
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¥ ' C- ¥
, FG
¥
C
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+
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H
¥
270 270
¥
2
4
fi
-
29/336!lO
Figure 6.36a shows the sectional view of a shelter. The vertical load on the 3-m long
ridge cable is 125 N/m. Determine the sag of the ridge and horizontal component
of cable force, if the tension in each guy cable is 600 N.
\MP!7/5^
ɏ
ŶŴŹ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
125 N/m
125 N/m
Ridge cable
T
T
T
H
Guy cable
Guy cable
1.5 m
RA
T
RB
A
B
yC C
H
1.5 m
(a) Section of the shelter with guy cables
(b) FBD of the ridge cable
Gjh/!7/47
. '
$
5OLUTION!
%!
& '&
#
", " ! $
+
$
#
"! - ?, *
"=
#3&
- ! D - ?,
%
D
D F
- ?,
>
!
H 2 + RA2
#
*
$
125 ¥ 3
2
wL
+=
&
F
F
? "D " !
"! - ?,
!
!
"
wx ( L - x)
2H
Bmufsobujwfmz-!
"
?
!
fi -
D ,
! - ?,
7
:/7 &
125 ¥ 1.5 ¥ (3 - 1.5)
2 ¥ 570
? - ,- ,
6002 - 187.52
2
Tmax
- RA2
" ?D 7
;
681!O
"
:
;
1/357!n
!
F
FF -
-
!
, FG
+
S&@
&
A
B+ ¥
.
fi B >
'XAMPLE!hrck
C-¥
¥
C
C
¥
¥
. .
¥
2 4
C:
¥
;¥
fi
1/357!n
A cable AB spanning 75 m supports a horizontal deck as shown in Fig. 6.37a. The
deck carries a uniformly distributed load of 30 kN/m throughout its span. The
support A is 15 m above the support B. The lowest point C of the cable is 5 m below the support B.
Determine the horizontal component of the cable force, support reactions and the maximum cable
force.
\MP!7/5^
ŶŴźɏ
%µ®°µ¬¬¹°µ®ɏ-¬ª¯¨µ°ªº
RA
75 m
A
A
l1
HA
15 m
RB
20 m
B
C
5m
C
30 kN/m
H
HB
C
30 kN/m
(a) Cable AB carrying the deck
B
l2
5m
30 kN/m
(b) FBDs of the parts AC and CB
Gjh/!7/48
A.
5OLUTION
" !
'
" ?
F
- ,-
/7
+
,, !
F
"
" =
$
%! &'
I F
-?
!
F
F ,, F -
S&@
I
, ,D
!:;
-
wl22
10
-
= ?
F
? "D
- ?,
$
wl12
40
:;
=!
fi
C: ¥ ;¥
30 ¥ 502
40
-
-
wl22
10
: ;
*
" # "
S#
"
&# '&
&#
H 2 + RA2
(
% )
& !
fi + +, ¥
&# '& (
+ +, ¥
%
&.
$ %
+ +, ¥
$ %
*
2986!lO
2986!lO
" # "
S#
&.
2
-
+
"
,
F
fi
wl12
40
!
-%
fi
1
,D
+
": ; $ !
-
!
B- ¥
A
wl12
40
(
F
-
- ?,
+
,-
?
B: ¥ ;¥
7 ,?
S&@=
/7
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A
=
% )
18752 + 15002
fi
+
2611!lO
fi
+
861!lO
& !
fi + +, ¥
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$
&.
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3512/28!lO
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% '
%
&
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ɏ
ŶŴŻ
4¹¼ºº¬ºɏ¨µ«ɏ#¨©³¬ºɏ
'XAMPLE!hrdl
Figure 6.38a shows a chain connecting a boat with its anchor at the sea bottom,
35 m below the surface of water. The boat exerts a horizontal force of 5 kN on the
chain. Self-weight of the chain is 18 N/m. The buoyant force on the chain in water is 2 N/m. Determine
the horizontal span of the chain and the maximum force in the chain.
\MP!7/5^
RB
Water surface
Boat
B
B
5 kN
Chain
35 m
35 m
Anchor
A
RA
q
A
x
H
L
L
y
(a) Chain connecting the boat and anchor
(b) FBD of the chain AB
Gjh/!7/49
5OLUTION! !/ ) *
,
6 9 6
: ; 6 #" <& $
" = ,: %
% $
# !< ! & &
$ % %&
-%
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$
%
" % $
$ #" <& < -%"
$
% %&
9
-&9
'
&
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-
H
Ê gx ˆ H
cosh Á - ˜ +
Ë H¯ g
g
=
$
& '
%
%&
%
B
Bmufsobujwfmz-! %
-+
.
$
fi
%
-
&.
%
&. "
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gx
H
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%
fi
$
&
+ = ¥ +,
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)
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$
&
gx
H
-
gy
H
%
:?
.
% '
5000
Ê 16 ¥ (- 35) ˆ
cosh -1 Á1 ˜
Ë
16
5000 ¯
fi .
-
=
;
=
?,
;
%& &
-% % @
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H
gy ˆ
Ê
cosh -1 Á1 - ˜
Ë
q
H¯
fi
257/67!n
%
#&
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16 ¥ 146.56
5000
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= = -%"
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6/67!lO
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57//#4;
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'XAMPLE!irc
A body of 350-N weight is resting on a rough horizontal surface (Fig. 7.4a). Find
the magnitude of the pulling force acting on the body at an angle of 25° with the
horizontal, which will just make the body move. Take m = 0.3.
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5OLUTION! 7
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W = 350 N
N
(b) Force diagram
(a) Block on horizontal surface
n
20°
n
50 N
n
50 N
N
50 N
t
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14°
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t
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100 N
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100 N
100 N
20°
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A block weighing 100 N is resting on a rough plane inclined at 20° to the horizontal.
It is acted upon by a force of 50 N directed upwards at an angle of 14° above the
plane as shown in Fig 7.5a. If the block is about to move up the plane, calculate friction and coefficient
of friction.
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5OLUTION! Hjwfo;! !>!211!O-! !>!61!O/
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A body resting on a horizontal rough surface requires a pull of 150 N inclined at
30° to the horizontal to initiate the motion, as shown in Fig. 7.6a(i). Also, it requires
a push of 250 N inclined at 40° to the horizontal to just start the motion, as shown in Fig. 7.6a(ii).
Calculate the weight W of the body and the coefficient of friction.
!
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160.7 N
75 N
30°
30°
40°
129.9 N
191.5 N
F2 = mN2
F1 = mN1
(i)
N1
W
N1
75 N
40°
N2
P1
30°
129.9 N
F1 = mN1
W
F2 = mN2
P2
160.7 N
40°
W
191.5 N
(i)
(a) Block on horizontal surface
N2
W
P2 = 250 N
(ii)
P2
P1
P1 = 150 N
(ii)
(b) Free body diagrams and force diagrams
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Determine the minimum angle q at which a uniform ladder can be placed against
a wall without slipping under its own weight. The coefficient of friction for all
surfaces is 0.2.
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A 1500-N block overlying a 10° wedge on a horizontal floor and leaning against a
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Determine the force P necessary to just start the motion, if the coefficient of friction is 0.3.
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1500 N
R2
R1
f 10°
f
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P
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f
R3
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A body of 400-N weight is lying on a rough plane inclined at an angle of 25° with
the horizontal. It is supported by a force P parallel to the plane as shown in
Fig. 7.11a. Calculate maximum and minimum values of P, for which the equilibrium can exist, if the
angle of friction is 20°.
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n
P
P
t
25°
25°
F
25°
F
25°
400 N N
400 N N
n
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t
n
25°
N
t
t
25°
P
P
25°
F
25°
F
400 N
(a) Block tending to slip upwards
(
400 N
(b) Block tending to slip downwards
'
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A horizontal force P is applied on a block of 2000-N weight kept on a rough
inclined plane, as shown in Fig. 7.12a. Determine whether the block is in
equilibrium, if q = 30° and P = 400 N. Take m s = 0.3 and m k = 0.2.
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2000 N
n
P
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t
30°
400 N
F
30°
F
400 N
q
2000 N
N
(a) Block on an inclined plane
(b) FBD of block
(c) Force diagram
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'XAMPLE!irj
A uniform ladder of 5-m length and 20-N weight is placed against a smooth
vertical wall with its lower end 4 m away from the wall. If the ladder is just to slip,
determine the coefficient of friction between the ladder and floor, and the frictional force acting on
the ladder at the point of contact with the floor.
!
"
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RB
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RA floor
C
'XAMPLE!irk
A 5-m long ladder and of 250-N weight is placed against a rough vertical wall in
a position where its inclination to the vertical is 30°. A man weighting 800 N
climbs the ladder. At what position will he induce slipping? The coefficient of friction for both the
contact surfaces of the ladder, viz., with the wall and the floor is 0.2.
!
"
5OLUTION!
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as 0.5.
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'XAMPLE!ircc
Find the least horizontal force P required to just start the motion of any part of the
system resting on one another as shown in Fig. 7.16a. The weights of blocks A, B
and C are 3000 N, 1000 N and 2000 N, respectively. Coefficient of friction between the blocks A and B
is 0.3, between B and C is 0.2 and between the block C and ground is 0.1.
!
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P
A
B
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P
A WA
P
WB
B
FA
NA
WB
B
FB
C
P
WC
C
NB
FC
NC
NC
NB
FC
NA
FA
FB
A
P
A
(b) Block A
A
WA
WB
WA
(a) Block A being pulled
P
P
WA
WB
WC
(c) Block A and B
(d) Block A, B and C
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Two blocks A and B, each of weight W, are placed on a rough inclined plane. The
blocks are connected by a light string. If mA = 1/2 and m B = 1/3, show that both the
blocks will be on the point of impending motion when the plane is inclined at an angle
q = tan–1 (5/12).
!
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q
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A 108-N block is held on a 40° incline by a bar attached to a 150-N block on a
horizontal plane as shown in Fig. 7.18a. The bar which is fastened by smooth pins
at each end is inclined at 20° to the horizontal. The coefficient of friction between each block and the
planes is 0.325. For what horizontal force P applied to the 150-N block, will the motion to the left be
impending?
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applied to lower block that will be necessary to cause slipping to impend. The coefficient of friction
for all contact surfaces is 0.3.
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Two identical blocks A and B are connected by a rod and rest against vertical and
horizontal planes, respectively, as shown in Fig. 7.20a. If sliding impends when
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Two boxes are placed on an inclined plane, in contact with each other as shown in
Fig. 7.21a, and released from rest. The coefficient of static friction between the box
A and the plane is 0.4 and that between box B and the plane is 0.3. Describe what happens.
!
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A
20 kg
10 kg
20°
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A
B
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Two rectangular blocks of weights W1 and W2, connected by a flexible cord, rest
upon a horizontal and inclined plane, respectively, with the cord passing over a
pulley as shown in Fig. 7.22a. In a particular case, where W1 = W2 and the coefficient of static friction m
is same for the continuous surface, find the angle a of inclination of the inclined plane at which motion
of the system will impend. Neglect friction in the pulley.
!
"
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W
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A
T
B
T
N1
B
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a
T
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A
mN1
a
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T
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a
A
T
W
W
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(b)
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Two blocks, connected with a horizontal link AB, are supported on two rough
planes as shown in Fig. 7.23a. If fC = 15° and mA = 0.4, find the smallest value of W
required for the equilibrium of the system.
!
"
400 N
W
400 N
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B
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A semicircular right cylinder of radius r and weight W rests on a horizontal
surface and is pulled at right angles to its geometric axis by a horizontal force P
applied at the point B of the front edge as shown in Fig. 7.24a. Find the angle a that the flat face will
make with the horizontal plane just before the sliding begins, if the coefficient of friction at the point
of contact A is m. The weight W must be considered as acting at the centre of gravity G of the
cylinder.
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Referring to Fig. 7.25a, determine the least value of the force P to cause motion to
impend rightward. Assume the coefficient of friction under the blocks to be 0.2
and the pulley to be frictionless.
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If the coefficient of static friction between the flat faces of the semi-circular cylinder and the horizontal
plane on which they rest is μ = 0.5 and the friction between the cylinders themselves is neglected,
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B
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90°
C
f
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f
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1200
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VR
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50
40
30
20
10
15.8°
0
10 a1 20
0
30
40
50
60.6°
a2
70
80
Wedge angle, a (degrees)
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W
W
l
Collar
l
Lever
M=F¥l
Screw
p
F
F
Screw
d
Nut
O
Lever
p
a
P
Base
(a) Construction
d
a
(b) Top view of screw jack
R
f
(c) Forces acting on screw
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Distance moved by load
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do
W
F = mN
a
t
f
R
a
W¢
p
P¢
F
Screw
Threads
N
pd
(b) Lifting the load
N
a
Nut
W¢
P¢
F = mN
a
t/2
d
p
t/2
(a) Distribution of forces among threads
Gjh/!9/24
R
N
f
a
pd
(c) Lowering the load
p
ŶźŶɏ
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uif!mpbe!"Wd
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)9/32*
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" tan a
tan a
Ideal effort
!
)9/34*
h! >!
!>!
!>!
Actual effort " tan (a + f ) tan(a + f )
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fi
fi
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È tan a ˘
h
!>!
Í
˙
a
a Î tan(a + f ) ˚
a
a f
a
sin(a + f )
cos a cos(a + f )
a
1
2
f
f
sec 2a tan(a + f ) - sec 2 (a + f ) tan a
tan 2 (a + f )
sin a
cos (a + f ) cos a
1
a
fi
fi
2
a
fi
a
f
p
a fi
a
f
a
a
p f
4 2
f
45∞ -
a
f
2
a
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3°´·³¬ɏ,°ļ°µ®ɏ-¨ª¯°µ¬ºɏ
a
tan a
tan(a + f )
h!
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{1 - tan(f /2)}/{1 + tan(f /2)}
{1 + tan(f /2)}/{1 - tan(f /2)}
tan(45∞ - (f /2))
tan(45∞ + (f /2))
Ê cos(f /2) - sin(f /2) ˆ
ÁË
˜¯
cos(f /2)
{1 - tan(f /2)}2
{1 + tan(f /2)}2
2
{cos(f /2) - sin(f /2)}2
{cos(f /2) + sin(f /2)}2
Ê cos(f /2) + sin(f /2) ˆ
ÁË
˜¯
cos(f /2)
2
2
cos (f /2) + sin (f /2) - 2 cos(f /2) sin(f /2) 1 - sin f
cos 2 (f /2) + sin 2 (f /2) + 2 cos(f /2) sin(f /2) 1 + sin f
2
"
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1 - sin f
1 + sin f
+
+
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+
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+
:
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p
;
++
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6
:
++
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7
:
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+
6
Distance moved by effort
2pl
\
<
=
#$
Distance moved by load
p1 - p2
*
#$
+
# $ !
!& *
+
+ ! +
!
+
&*
+ !
-
,-
>
+ +
m6
f,&
+
m@
b&
b
?
+ +
Load, W
Moves
down
p2
m6
P
Lever
Moves
up
p1
l
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