chapter 6 statics
DESCRIPTION
Statics chapter 6 solutionTRANSCRIPT
SOLUTION
Joint FBDs:
Joint B:
B
Joint C:
PROBLEM 6.2
Using the method of joints, detennine the force in each member of thetruss shown.State whethereach memberis in tension or compression.
-rF'x= 0:1 4
.fi FAB- "5 FBC= 0
1 3
.fi FAB+"5FBc- 4.2 kN = 0
so7"5 FBc= 4.2 kN
FBc= 3.00 kN C ~
4 12- rF'x=0: -(3.00 kN) - - FAC=05 13
13FAC=- kN5
FAC= 2.60 kN T ~
PROPRIETARY MATERIAL. \C 2007 The McGraw-Hili Companies, Inc. All 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 tetWhers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
796
'I'
:2kips2 "ips I 2 k.ips
SOLUTION
FBD Truss:
PROBLEM 6.10
Determine the force in each member of the Gambrel roof truss shown.State whether each member is in tension or compression.
- LF'x =0: Hx =0
By symmetry: Ay =H y = 4 kips t~ by inspectionofjoints C and G:
HAJ<
Joint FBDs:
Joint A:
IN
~/P ~ f"l3
.]"
1/ file.
"I J'f's
Joint B:
y5k'f~
JointE:
FAC=FCE and Foc = 0 ~
FEG=FGN and FFG=0 ~
FAD= FAC = 3 kips5 4 3
and, from above,
FAD= 5.00 kips C ~
FAC=4.00 kips T ~
FFH = 5.00 kips C ~
and FCE= FEG= FGN= 4.00 kips T ~
so
- LF'x=0:4 . 4 10- (5 kips)- - FOE- ~ FOD=05 5 ~109
3. 3 3- (5 kips) - 2 - ~ FOD+ - FOE= 05 ~109 5
so FOD=3.9772 kips, FOE= 0.23810 kips
or FOD= 3.98 kips C ~
FOE= 0.238 kips C ~
and, from above, FDF= 3.98kips C ~
FEF= 0.238kips C ~
tLF'y= 0: FDE- 2 ~(0.23810 kips) =05
FDE=0.286kips T ~
PROPRIETARY MATERIAL iD 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means. without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.
804
j
r24mT2,:':)L2"~::;"T24mII2T PROBLEM 6.23
1.",1' G -
?f'~ For the roof truss shown, detennine the force in each of the members
L fJl~ B ~ ~ "', LLH'I
.~~m loca~edto the left ,of member GH. State whether each member is in
InoA _" /I""'- } ~~., ,;rj tensIOnor compreSSIOn.G G F I.
SOLUTION
FBD Truss: -~=o:
t~=O:
Joint FBDs:
JointA:
.ill
Joint c: -~=o:
1111"
t~=O:
,I, Joint B:
II. '1" I'1111'I' I"'i,I' 'W
,Ii'
-~=o:
(12m)(My-l kN)
- (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8m)(1.5 kN) =0
My =5.05kNtAy - 2(1 kN) - 5(1.5 kN) + My = 0
Ay =4.45 kNt5
445 kN - 1kN - - FAB =0. 13' FAB=8.97 kN C ~
12FAC-13(8.97 kN) = 0,
1213 FCE - 8.28 kN = 0,
513(8.97 kN) - FBC= 0,
FCE= 8.97 kN T ~
FBc= 3.45 kN C ~
5 5-(8.97 kN) -1.5 kN + 3.45 kN --FBD= 013 13
FBD=14.04 kN FBD=14.04 kN C ~
12 12-(8.97 kN) - -(14.04 kN) + FBE=013 13
FBE= 4.68 kN FBE= 4.68kN T ~
PROPRIETARY MATERIAL t!:>2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduc,or distributed in any fonn or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teachers Oieducators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.
826
Joint E:
Joint D:
PROBLEM 6.23 CONTINUED
- Ll\= 0:
- Ll\ = 0:
6 12r;:;;;FEH - 4.68kN- -(8.97 kN)=0
v37 13
FEH=13.1388kN or FEH=13.14kN T.....
5 1FDE-_3(8.97 kN) - r;:;;;(13.1388) = 01 v37
FDE=5.6100kN or FDE=5.61kN T.....
12 12 1-(14.04kN)--FDG- r;:;FDH=O13 13 v2
5 5-(14.04kN) - -FDG-1.5 kN - 5.61kN13 13
1
+ .J2 FDH= 0
Solving: FDG=8.60 kN C....
PROPRIETARY MATERIAL to 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in anyform or by any means, without the prior writtenpe/mission of the publisher, or lISedbeyond the limited distribution to teache~sandeducatorspermitted by McGraw-Hillfor their individual coursepreparation.lfyou are a student using this Manual,you are !LYingit withoutpermIssIOn.
827
I'
SOLUTION
PROBLEM 6.30
For the given loading, detennine the zero-force members in the trussshown.
Q
PROPRIETARY MATERIAL () 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written pelmission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hil/for their individual course preparation. !fyou are a student using this Manual, you are using it without permission.
837
By inspection of joint D, FD1= 0 ...
H/ "1/"t" "1/ ,L By inspection of joint E, FEl= 0'"l . »
b1::'1
Then, by inspection of joint I, FA!= 0'"
By inspection of joint F, FFJ(=0 ...
By inspection of joint G, FGK=0'"
PROBLEM 6.42
A floor truss is loaded as shown. Detennine the force in members CF,EF, and EG.
SOLUTION
FBD Truss:
FBD Section ABEC:
B
It
!.--E.:&.,""-
-2-""'" t:~,
~,.- - - - - -,..1..312.$-lb
- LF::=0:
( WA= 0: 6a(l\v- 1251b)- 5a(250Ib)
- 40(250lb)- 30(375Ib)
-20(500 lb) - 0(500 lb) =0
Ky =937.5lb tt~v= 0: Ay- 3(250 lb) - 2(500 lb)
-3751b -1251b + 937.51b=0
Ay=1312.5lb t
(2 ft) FCF+ (4 ft)(5oo lb)
+ (8 ft)(250 lb -1312.5lb) =0
FCF= 3250 Ib, FCF= 3.25 kips T ....
1312.5Ib- 250 lb - 2(500 lb) - Js FEE'= 0
FEE'=62.s.J5 lb, FEE'=139.8Ib T....
- LF::=0:3250 lb + .Js (62.s.J5 lb) - FEG= 0
FEG = 3375 lb, FEG= 3.38 kips C ....
PROPRIETARY MATERIAL. () 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manua/may be displayed. reproducedor distributed in any fonn or by any means. without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manllal, you are using it without permission.
-
855
t
II
I:.,"ii
n,
I ~. 'I'II
, j"
'ItJ
"
1,,,'
II
il'11,;,
i
il:I'
15m ~ 1.500115m 1.500 1.5... PROBL~M 6.51~J ,"" en.
A Fink roof truss is loaded as shown. Detennine the force in members FH,
IOk!\1 I) '~~ FG,andEG.:;k\ B J J3k!\ 2.400A
.,_ _~ "'-.j~
...~.
1,
1.
1,
1~~1.8m 1./))I} I.8m 1.1;m I.l'im-l
SOLUTION
FBDTruss:
FBD section GHK:
Iil r
"il.,:
-i'-- -,--~ 1.(. .I \ WI. ffi..
2.¥N, 4! f/I '
1-._ 3
! E'I>
,)
j
:IIi I
.~
j
' ; !" ,I: !,I
tll:: ,jI!,,\I
Distance between loads = 1.5 m
rflx = 0: Ax = 0
By symmetry, Ay = Ky = 18kNt
( rMF = 0: (4.5m)(18 kN - 3 kN) - (3 m)(6 kN)
- (1.5 m)(6 kN) - (2.4 m)FEG = 0
FEG=16.875kN, FEG= 16.88kN T....
I<.
( rMK =0: (1.5m)(6kN)+ (3m)(6kN)- (3.6m{ In FFG) =0
FFG=8.0100kN, FFG= 8.01kN T....
-+ rflx =0:15 3
17FFH - .j73(8.01oo kN) -16.875 kN = 0
FFH = 22.3 kN C....
j8\ PROPRIETARY MATERIAL. It>2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.
ill 866
B D ,..
= ci EI G
LtHips f) kips ~ 12ki!"
j8ft~--8ft~8ft~8fi
I
SOLUTION
FBD Truss:
..~ff
A
FBD Section ABC:
IJ[ Bt.-
FBD Section FGH:
PROBLEM 6.67
The diagonal members in the center panels of the truss shown are veryslender and can act only in tension; such members are known as counters.
Detennine the force in member DE and in the counters which are actingunder the given loading.
( IMA =0: (32 £1)Hy - (24 £1)(12kips) - (16 £1)(9kips)
-(8 £1)(6kips) = 0, Hy = 15kipstt IFy =0: Ay - 6 kips - 9 kips - 12kips + 15kips = 0
Ay =12kipst
Since only BE can provide the downward force necessary for equilibrium,it must be in tension, so CD is slack, FCD=0
t IFy = 0: 12 kips - 6 kips - ~FRE = 0
FRE = 10.00 kips T <III
Since only EF can provide the downward force necessary for equilibrium,it must be in tension, so DG is slack, FDG= 0
t IFy = 0: 15 kips -12 kips - ~FEF = 05
FEF =5.00 kips T <III
Knowing that FCD=FDF = 0, inspectionof joint D gives
H
I1<;"(t * See note before Problem 6.64.
II-p.,I ~
PROPRIETARY MATERIAL tC 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed. reproducedor di,vtributed in any fonn or by any means. without the prior written pennission of the publisher. or used beyond the limited distribution to teachers andeducators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.
884
101
SOLUTION
Structure (a):
~.
Structure (b):
PROBLEM 6.74
11-I
IIi'.'rti
I"
I
ii,.I,ll, i
Classify each of the structures shown as completely, partially, orimproperly constrained; if completely constrained, further classify it asstatically detenninate or indetenninate. (All members can act both intensionand in compression.) ,
.1I, ,
I
IJ
'II
II
Then ABCDGF is a simple truss and all forces can be detennined.
This example is completely constrained and detenninate. ...I'
No. of members ill!
I
m = 12
No. of joints n = 8
No. of react. comps. r = 3
m + r = 15< 2n = 16
unks < eqns
partially constrained ...
Note: Quadrilateral DEHG can collapse with joint D moving downward:in (a) the roller at F prevents this action.
,
continued Ii
PROPRIETARY MATERIAL e> 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teacher:>andeducators permitted by McGraw-Hili for their individual course preparation. Jfyou are a student using this Manual. you are using it without permission.
891 Ii
ili
p No. of members m = 12
ENo. of joints n=8 m + r = 16= 2n
No. of react. comps. r=4 unks = eqns
Structure (c):
PROBLEM 6.74 CONTINUED
No. of members m = 13
} ~No. ofjoints n =8No. of react. comps. r =4
m + r = 17 > 2n = 16
unks > eqns
completely constrained but indeterminate "'11II
PROPRIETARY MATERIAL. () 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducotors permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.
892
PROBLEM 6.78III
I'FA250 mm
L J E-
~)i J"-l:.-:'---
J-~
L
F
250mm-j500 111In 250nit 200 mm
For the frame and loading shown, detennine the components of all forcesacting on member DECF.
I'
SOLUTION
FBD Frame:
( LMA = 0: (0.25 m)Dx - (0.95m)(480 N) = 0
7= D = 1824N - ....x
FBD member DF: Note that BE is a two-force member, Ex = E"
- ~=O: -1824N+Ex=0, Ex= 1824N- ....
~
--i:
,I
/%2'11'1 I,'" -- (;I.s;, lotV .
~y
so Ey = 1824Nt....
(0.50 m)(1824 N) - (0.75 m)C + (0.95 m)(480 N) = 0
C = 1824 N ~ ....
t rFy = 0: -Dy + 1824N -1824 N + 480 N = 0
Dy =480 N ~ ....
PROPRIETARY MATERIAL () 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manllal. you are using it without permission.
896
I:
:r
"
r 6 in.
!.-A
SOLUTION
(a) FBD AC:Id
(b) FBD CE:
.-
, II
'I" if
I
c
PROBLEM 6.86
Detennine the components of the reactions at A and E when acounterclockwisecouple of magnitude 192 lb.in. is applied to the ftame(a) at B, (b) at D.
Note: CE is a two-force member
( rMA = 0: -(8 in.)( ~ FCE ) -(2 in.)( ~ FCE )+ 192 lb. in. = 0
FCE = 19.2J2 lb, Ex = 19.20lb - ....
Ey
Ey = 19.20lb !....
- ~ =0: Ax - 19.2lb= 0,
t~ =0: Ay - 19.2lb = 0,
Ax = 19.20lb - ....
Ay = 19.20lb t....
Note: AC is a two-force member
FAE = -12.8ffi lb,4
Ax = r.;; FAE,..,17
Ax = 51.2lb - ....
Ay = 12.80lb !....
- ~ =0: Ex - 51.2lb = 0,
t~, =0: Ey - 12.80lb = 0,
Ex = 51.2lb - ....
Ey = 12.80lb t....
PROPRIETARY MATERIAL. II:>2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any fonn or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teachers andeducators pennilled by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission
904
A "
)(
i'
,.
.,~500li5001.....
PROBLEM 6.98
For the frame and loading shown, detennine the components of all forces
acting on member ABD.
Dimensions in rom
SOLUTION
FBD Frame:
~yA
c
F -~=o:
(0.625m) F - (0.75m)(4 kN)- (1.25m)(3kN)=0
Fx= 10.8kN-
Ax - 10.8kN=0, Ax= 10.80kN -....
B
F'-)< fLFy= 0: Ay - 4 kN - 3 kN = 0,
FBD ABD: (II)
0,2slN'l
I: (!Me= 0: (0.375m)(I0.8 kN) - (0.25m) Bx= 0, Bx = 16.2kN, Bx= 16.20kN-....
II: (!MD= 0: (0.25m)(I0.8 kN + 16.2kN) + (0.5 m) By- (1.00m)(7.0 kN) = 0,
By = 0.5 kN,
-10.8 kN -16.20 kN + Dx= 0, Dx = 27 kN, Dx=27.0kN-....
Dy = 6.50 kN! ....7.0 kN - 0.5 kN - Dy= 0, Dy = 6.5 kN,
PROPRIETARY MATERIAL. iO 2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
918
:, I
.! Iij
j; f I'I;.
FBD BF: (I)
'1.y
-4Bo,z.sr.t
*II a.
b.37M y
)<
F
/tJ$ AN
I
t
I
'i
PI-.
12f'-):12f't~B
PROBLEM 6.104 I I;
'I
''f-L'
1:=2fl
7,2ftThe axis of the three-hinge arch ABC is a parabola with vertex at B.~,6ft Knowing that P = 14 kips and Q =21 kips, detennine (a) the
components of the reaction at A, (b) the components of the force exertedat B on segment AB.
!IIIi
1'1
SOLUTION
I. ~t"I
"'" -2/jR-
21k~ i/1. f't:
-r It-I 7,2.
II
I S'I ",'I i.
. ";~.
Members FBDs:
:1
. ~;-."
I- rr
1:( WA=O:
II:( Wc= 0:
(12.8ft)Bx- (32 ft)By- (20 ft)(14 kips) = 0
(7.2 ft)Bx+ (24 ft) By- (12 ft)(21 kips) =0
Solving: Bx=27.5 kips, By= 2.25kips,
1:- ~=O:
tLFy= 0:
Ax- 27.5 kips = 0, Ax = 27.5 kips, (a) Ax= 27.5kips- ~
Ay= 16.25kips t~
Bx= 27.5 kips --~
By = 2.25 kips l~
Ay - 14 kips - 2.25 kips = 0, Ay = 16.25 kips,
(b)
PROPRIETARY MATERIAL C>2007 Tbe McGraw-Hili Companies, Inc. AIl 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 teachers and
educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.
925
~...
"O!i'
\E
\\\'.
\"c",;\\L'\!0.-:,..~ :?\
~.~'//~~"!IL ~,.1.5in. I.:;in. I
PROBLEM 6.123
The double-toggle latching mechanism shown is used to hold member Gagainst the support. Knowing that a = 60°, determine the force exertedonG.
. -I (2.5 in. + 1in.)sin60°I,,,. = tan(30.-9),' 4.5 in. + (1.5in. + 2.5 in. - 1in.)cos60°~.~ - 40. £'1(
'" C
___ --- --- .f.,., / ,'"0. () = 26.8020./ 2.51~~"'
. I £1~{!..
- - - - - - - - ~h,~~ A l 1.5 in.
SOLUTION
Member FBDs:
c.
From FBD ABC:
() =tan-I (:BC+ m)sin60°AF + (AB + BC - CD)cos60°
From FBD CDE:
( We =0: (7.5 in.)(20 lb) - (1in.)FDFcos(30°- 26.802°)= 0,
FDF = 150.234 lb C
-- LEx=0: (150.234Ib)cos(26.802°)- (20 Ib)sin60°- Cx = 0,
Cx = 116.7741b
t 'fFy = 0: (150.234Ib)sin26.802°- (20 Ib)cos60°- Cy = 0
Cv = 57.742 lb
+[(4 in.)cos600](57.742Ib) = 0, JiBx + B.v =385.37 lb (I)
continued
PROPRIETARY MATERIAL. ~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written permission of thepublisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual coursepreparation. lfyou are a student using this Manual. you are using it without permission.
947
(
.,
..
II
"
II
PROBLEM 6.123 CONTINUED
From FBD BFG:
( rMG=0: -(1.5 in.)[(150.2341b)sin26.802°]+ [(1.5 in.)cos300][(150.2341b)cos26.802°]
- [(1.5 in.)cos300]Bx- [6 in. + (1.5 in.)sin300]By = 0, J3Bx- 9By =96.7751b (2)
Solving (I) and (2): Bx = 243.32 Ib, By = - 36.075 Ib
-- IF'x = 0: 243.321b- (150.234Ib)cos26.802°- Gx= 0, Gx = 109.2261b-
Gy = 31.6671btt rEy = 0: Gy - (150.2341b)sin26.802°+ 36.0751b = 0,
OnG, G = H3.71b "" 16.17°....
PROPRIETARY MATERIAL ~ 2007 The McGraw-Hili Companies. Inc. All rights reserved, No part of this Manual may be displayed. reproducedor disttibuted in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.
948
"
Ii" II:I'
'. j. '\
"
1.1'
I(,
11,11
'IIII.
PROBLEM 6.128
:.f~~~l.~ A.couple M o~magnitude6 N.m is appliedto the.input li~ of the four-?ar"j.[f"J_ =-:""... slIder mechamsmshown. For each of the two gIvenpOSItIOns,determme
the force P requiredto hold the systemin equilibrium.
SOLUTION
(a) FBD BC:
('
( WB =0: (0.045m)FCDsin45°-6.00 N.m =0
~y
e 1', b FCD =188.562N C
FBD Joint D:
FBD E:
(b) FBD BC:
-1 38.8 mm = 22.8230a =tan 92.2 mm
-1 37.5 mm - 18.8 mm = 25.7320f3= tan 38.8mm
/ LFx'= 0: (188.562N)cos(45° -25.732°)
-FDEcos(22.823° + 25.732°) = 0, FDE= 268.92 N C
rEx = 0: (268.92N)cos(22.823°)-P = 0,
P = 248N - ...
_135.4 mm = 51.85708 = tan 27.8mm
f3= tan-I 27.8 mm - 24.3 mm = 7.5520026.4 mm
(WB = 0: (0.045m)FCDsin(90°-51.857° -7.5510°) - 6.00 N'm = 0
FCD = 262.00 N C
continuedi
PROPRIETARY MATERIAL <C2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced
or distributed in any form or by any means, 1,1,oithoutthe prior writtenpermission of thepublisher. or used beyond the limited distribution to teachers andeducators permilled by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
955
I'Ii
PROBLEM 6.128 CONTINUED
FBD Joint D:-1 20 nun + 35.4 nun - 26.4 nun = 42.3630
r =tan 24.3 nun + 7.5 nun
a =tan-I 20 nun + 35.4 nun -26.4 nun 16.8584092.2 nun + 27.8 nun - 24.3 nun
F--- Cc -(262.00N)sin(90°- 7.5520°- 42.363°)= 0,
FDE = 196.366 N C
FBDE:
- r.F'x= 0: (196.366 N)(cosI6.8584°) - P = 0,
P=187.9N -- ~
:1I
'il PROPRIETARY MATERIAL. i!:i2007 The McGraw-Hill Companies, Inc. All 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 teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
956
Zin.
SOLUTION
FBD top handle:
FBD Joint D:
'10II, PROBLEM 6.139
A hand-operated hydraulic cylinder has been designed for use wherespace is severely limited. Determinethe magnitude of the force exertedon the piston at D whentwo 90-lb forces are appliedas shown.
I~I
Note CD and DE are two-force members
(4 in.) ..k FCD-(1.5in.) k FCD-(13.2in.) (90 lb) = 0
FCD=n.J61lb
FDE = FCD = 12 .J611bBy synnnetry:il
D= no lb ....
.1
PROPRIETARY MATERIAL (02007 The McGraw-Hili Companies, Inc. All 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 teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using thi.rManual, you are using it without permission.
-
967
II
I
DhlU."n~;ons in mill
~y (~FBDEntire linkage:
65,.2
,
,5-
,
""- -'
J,I) 0,1>10\,.. i
;p" lPy d,z.>1
l '~. - t
~-F'..
,, -- '" {i"1. q!>~kN -, F_FIf
PROBLEM 6.150
A 500-kg concrete slab is supportedby a chain and sling attached to thebucket of the ITont-endloader shown. The action of the bucket iscontrol1edby two identical mechanisms, only one of which is shown.Knowing that the mechanism shown supports half of the 500-kg slab,determinethe force (a) in cylinder CD, (b) in cylinderFR.
( LMD= 0: (0.8 m)(2.4525 kN) - (0.5 m) FAD= 0
FAD = 3.924 kN
w = (250 kg) (9.81 N/kg) = 2452.5 N = 2.4525 kN
(LME= 0: (0.68 m)(3.924 kN) - (0.54 m)G~ FCD)
+ (0.35 m) U~ FCD)= 0, FCD=7.682kN
FCD=7.68 kN C ....
(LMG= 0: (2.5 m) (2.4525kN)1 1
+ (0.2 m) J2 FGH-(0.6 m) J2 FGH=0,
FFH=21.677kN, FFH=21.7 kN C....
PROPRIETARY MATERIAL. <02007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.
978
SOLUTION
FBD Bucket (one side):',II I
If.Afj.(a)
fr
! , t'!>.-"",
to.r1:,
Dx"\.0
I
III
ii', Ii
,ilI FBD link BE:
. il0, I
3.'2.¥ ktJ B.I,
--r0.II H1
_J.it I,
E J IJ. r;v 111.]
€,llI -I i