PROBLEM 6.2 Using the method of joints, detennine the force in each member of the truss shown. State whethereach memberis in tension or compression.
SOLUTION Joint FBDs:
Joint B:
-rF'x= 0: B
1
4
.fi FAB- "5 FBC= 0
1 3 .fi FAB+"5FBc- 4.2 kN = 0 so
7 "5 FBc= 4.2 kN
FBc= 3.00 kN C
~
Joint C:
-
4 12 rF'x=0: -(3.00 kN) - -13 FAC=0 5 13 FAC=-5 kN
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 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.
796
PROBLEM 6.10
:2kips
I
2 "ips
2 k.ips
Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.
SOLUTION
-
FBD Truss:
LF'x
= 0:
By symmetry: Ay ~
H
Hx = 0
= H y = 4 kips
by inspectionofjoints C and G:
t
FAC=FCE and Foc = 0 ~ FEG=FGN and FFG=0 ~
AJ<
Joint FBDs: 'I'
FAD= FAC = 3 kips 5 4 3
Joint A: IN
/P ~ .]"
1/
so
f"l3
~
FAD= 5.00 kips C ~ FAC= 4.00 kips T
file.
"I J'f's
and, from above,
Joint B:
and
-
LF'x=0:
y
5k'f~ Joint E:
FFH
so
~
= 5.00 kips C ~
FCE= FEG= FGN= 4.00 kips T
~
4 . 4 10 - (5 kips)- - FOE- ~ FOD=0 5 5 ~109 3. 3 3 - (5 kips) - 2 - ~ FOD+ - FOE= 0 5 ~109 5 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= 0.238 kips C ~
t LF'y= 0:
~
FDE- 2 5 (0.23810 kips)
=0 FDE = 0.286 kips T ~
PROPRIETARY MATERIAL iD 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 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.
804
j
24mT2,:':)L2"~::;"T24mII2T PROBLEM 6.23 1.",1' G '~ For the roof truss shown, detennine the force in each of the members Jl~ B ~ ~ "', LLH'I ~m loca~edto the left ,of member GH. State whether each member is in
r
L
f
InoA _"
G
G
F
?f
/I ""'-
} ~~.,
.~
I.
,;rj
tensIOnor compreSSIOn.
SOLUTION FBD Truss:
-~=o: (12m)(My-l
kN)
- (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8m)(1.5 kN) = 0
t ~=O: Joint FBDs:
Ay
= 5.05kN
Ay
= 4.45 kN t
- 2(1 kN) - 5(1.5 kN) + My = 0
5 445 kN - 1kN - - FAB = 0
Joint A:
t
My
.
13'
FAB=8.97 kN C
~
FCE= 8.97 kN T
~
12 FAC-13(8.97 kN) = 0, .ill
Joint c:
-~=o:
t ~=O:
12 13 FCE -
8.28 kN = 0,
5 13(8.97 kN) - FBC= 0,
FBc= 3.45 kN C ~
1111"
,I,
5 5 -(8.97 kN) -1.5 kN + 3.45 kN --FBD= 0 13 13
Joint B:
FBD=14.04 kN II. '1
" I' 111' I
I"'i, 1
-~=o:
FBD=14.04 kN C
~
12 12 -(8.97 kN) - -(14.04 kN) + FBE=0 13 13
'
I' 'W
FBE= 4.68 kN
FBE= 4.68kN T ~
,Ii'
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 Oi educators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.
826
PROBLEM 6.23 CONTINUED Joint E:
-
6
Ll\= 0:
r;:;;;FEH -
v37
12
4.68kN- -(8.97 kN) = 0 13
FEH=13.1388kN 5
FDE-_13(8.97
kN) -
or 1 v37
r;:;;;(13.1388)
FDE=5.6100kN Joint D:
-
Ll\
= 0:
12 12 -(14.04kN)--FDG13 13
or
FEH=13.14kN T.....
=0 FDE=5.61kN T.....
1 r;:;FDH=O v2
5 5 -(14.04kN) - -FDG-1.5 kN - 5.61kN 13 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, reproduced or distributed in anyform or by any means, without the prior written pe/mission of the publisher, or lISedbeyond the limited distribution to teache~sand educatorspermitted by McGraw-Hillfor their individual coursepreparation.lfyou are a student using this Manual,you are !LYingit withoutpermIssIOn. 827
I'
PROBLEM 6.30 For the given loading, detennine the zero-force members in the truss shown.
Q
SOLUTION
H/ l
"1/
"t"
"1/
,L.
»
By inspection of joint D,
FD1= 0 ...
By inspection of joint E,
FEl= 0'"
Then, by inspection of joint I,
FA!= 0'"
b 1::'1
By inspection of joint F,
FFJ(=0 ...
By inspection of joint G,
FGK=0'"
PROPRIETARY MATERIAL () 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 pelmission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hil/for their individual course preparation. !fyou are a student using this Manual, you are using it without permission.
837
PROBLEM 6.42 A floor truss is loaded as shown. Detennine the force in members CF, EF, and EG.
SOLUTION FBD Truss:
- LF::= 0: ( WA= 0:
6a(l\v- 1251b)- 5a(250Ib) - 40(250lb)- 30(375Ib) -20(500 lb) - 0(500 lb) = 0 Ky
t ~v= 0:
= 937.5 lb t
Ay- 3(250 lb) - 2(500 lb) -3751b -1251b + 937.51b = 0 Ay=1312.5lb
t
t I I :., I" ii n, .I ~' I' I I , j"
FBD Section ABEC: (2 ft) FCF+ (4 ft)(5oo lb)
!.--E.:&.
B
+ (8 ft)(250 lb -1312.5lb)
,""-
-2-""'" t:~,
It
~,.- - - - - -,..
=0 FCF= 3.25 kips T ....
FCF= 3250 Ib,
1..312.$-lb
1312.5Ib- 250 lb - 2(500 lb) FEE'=62.s.J5 lb,
-
LF::= 0:
Js
FEE'= 0
FEE'=139.8Ib T....
3250 lb + .Js (62.s.J5 lb) - FEG= 0 FEG = 3375 lb,
FEG= 3.38 kips C ....
'I tJ PROPRIETARY MATERIAL. () 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manua/may be displayed. reproduced or distributed in any fonn or by any means. without the prior written permission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manllal, you are using it without permission.
"
1, ,,'
855 II
il'11 ,;,
-
i il: I'
PROBL~M 6.51
~ J en ~ 15m ~
1.500115m
1.500 1.5...
,""
~ 1111
IOk!\1 :;k\A
B . ~1.8m
I)
,
A Fink roof truss is loaded as shown. Detennine the force in members FH,
.
FG,andEG.
'
,_
.
_~
1./))I}
I.8m
J J3k!\ ~ "'-.j~
,
~
1.1;m
...
2.400
.
I.l'im-l
SOLUTION
FBDTruss: Distance between loads
rflx
= 0:
By symmetry,
Ax
= 1.5 m
=0
Ay = Ky = 18kN
t
FBD section GHK:
(
rMF
= 0:
- (1.5 m)(6 kN) - (2.4 m)FEG = 0
-i'-- -,--~ 1.(. .
\ WI
I
I il
(4.5 m)(18 kN - 3 kN) - (3 m)(6 kN)
FEG = 16.875kN,
ffi.. 2.¥N, 4
FEG= 16.88kN T....
.
r
! I
"il.
,:
f/
'
1-._
!
3
(
rMK
I<.
= 0:
(1.5m)(6kN) + (3m)(6kN) - (3.6m{
FFG = 8.0100kN,
E'I>
-+
rflx
= 0:
15
FFG= 8.01kN T.... 3
17FFH - .j73(8.01oo kN) -16.875 kN FFH
,)
:
IIi
I
~
' ;
"
In FFG) = 0 =0 = 22.3 kN
C....
!,
j!
j
. I:
,: tll: ,j I
I !,,\ I
j8\
ill
PROPRIETARY MATERIAL. It>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 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.
866
B
=
ci
t Hips
D
EI
,..
PROBLEM 6.67 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Detennine the force in member DE and in the counters which are acting under the given loading.
G
f) kips ~ 12ki!"
L 8ft~--8ft~8ft~8fi j I
SOLUTION FBD Truss:
(
IMA
= 0:
(32 £1)Hy - (24 £1)(12kips) - (16 £1)(9kips) -(8 £1)(6kips) = 0,
.. ~ff
t IFy =0:
A
Hy = 15kips
t
Ay - 6 kips - 9 kips - 12kips + 15kips = 0 Ay
= 12kipst
FBD Section ABC: 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: IJ
12 kips
- 6 kips - ~FRE = 0
[ Bt.FRE
= 10.00 kips T
Since only EF can provide the downward force necessary for equilibrium, it must be in tension, so DG is slack, FDG= 0
t IFy
FBD Section FGH:
= 0:
15 kips -12 kips - ~FEF 5
=0
FEF
Knowing that FCD= FDF
= 5.00 kips T
= 0, inspectionof joint D gives
H
I 1<;"(
II-p., I~
t
* See note before Problem 6.64.
PROPRIETARY MATERIAL tC 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed. reproduced or 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 and educators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.
884
11I
IIi'
'rti I .
PROBLEM 6.74
"
Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify it as statically detenninate or indetenninate. (All members can act both in tension and in compression.)
101
ii
,.
I,ll,I
,I,
.1 ,
I
SOLUTION
Structure (a): p E ~.
No. of members
m
= 12
No. of joints
n=8
m + r = 16 = 2n
No. of react. comps.
r=4
unks = eqns
I J
'I I
I I Then ABCDGF is a simple truss and all forces can be detennined. This example is completely constrained and detenninate. ...
Structure (b):
= 12 n =8 r =3
m
No. of members
No. of joints No. of react. comps.
I'
ill!
m+ r
= 15 <
2n
= 16
I
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. reproduced or distributed in any form or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teacher:>and educators 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
i
PROBLEM 6.74 CONTINUED Structure (c):
m = 13
No. of members
}
~ No. ofjoints
n No. of react. comps. r
m+ r
=8 =4
unks > eqns completely
PROPRIETARY or distributed
MATERIAL. in any form
() 2007 The McGraw-Hili
or by any means,
without
the prior
= 17 > 2n = 16
constrained
but indeterminate
"'11II
Companies, Inc. AII rights reserved. No part of this Manual may be displayed, reproduced written
permission
of the publisher.
or used beyond
the limited
distribution
to teachers
and
educotors 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 6.78 III
For the frame and loading shown, detennine the components of all forces acting on member DECF.
I'FA 250 mm
L
J -
E
"-l:.-:'---
)i
-~
F
200 mm
L
~ J J 250mm-j 500 111In
250nit
I'
SOLUTION
FBD Frame:
(
LMA
= 0: (0.25 m)Dx - (0.95m)(480 N) = 0
FBD member DF:
-
Ex= 1824N
-
Note that BE is a two-force member, Ex = E"
- ~=O:
--
= 1824N
Dx
7=
-1824N+Ex=0,
....
....
i :
~
so
Ey = 1824Nt....
,I
/%2'11'1
I
,'"
V
-- (;I.s;, lot
.
(0.50 m)(1824 N) - (0.75 m)C + (0.95 m)(480 N)
~y
t rFy = 0:
=0
C
= 1824 N
Dy
= 480 N
~
....
-Dy + 1824N -1824 N + 480 N = 0
~
....
PROPRIETARY MATERIAL () 2007 The McGraw-Hili Companies, Inc. AII 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 Manllal. you are using it without permission.
896
I:
r
PROBLEM 6.86
6 in.
Detennine the components of the reactions at A and E when a counterclockwisecouple of magnitude 192 lb.in. is applied to the ftame
!.A
(a) at B, (b) at D.
:r
"
SOLUTION I d
(a) FBD AC: A
Note: CE is a two-force member "
(
rMA
= 0:
-(8 in.)(
)(
i'
FCE
~
FCE
) -(2 in.)(
~
= 19.2J2 lb,
FCE
)
+ 192 lb. in.
=0
Ex = 19.20lb -
Ey = 19.20lb
-
~ = 0:
t~
= 0:
Ax Ay
19.2lb = 0,
....
!
Ax = 19.20lb -
Ay = 19.20lb
- 19.2lb = 0,
....
....
t
....
Note: AC is a two-force member (b) FBD CE:
FAE
= -12.8ffi lb,
4 Ax = r.;; FAE, ..,17
c Ey
Ax = 51.2lb -
Ay = 12.80lb
-~
= 0:
Ex - 51.2lb = 0,
t ~,
= 0:
Ey
- 12.80lb = 0,
....
!
....
Ex = 51.2lb Ey = 12.80lb
....
t
....
.-
, II 'I" if I PROPRIETARY MATERIAL. II:>2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced 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 and educators pennilled by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission
904
,.
.,~500li5001
PROBLEM
6.98
..... For
the
frame
and
acting on member
: .! i
, I I
Dimensions
loading
shown,
detennine
the components
of all forces
ABD.
in rom
j
j; I; .
f I'
SOLUTION FBD Frame: ~y A
B
(0.625m) F - (0.75m)(4 kN)- (1.25m)(3kN) = 0 Fx= 10.8kN-
c
-~=o:
F
F '-)<
Ax -
fLFy= 0:
FBD BF: (I)
Ay
10.8kN = 0,
Ax= 10.80kN -....
- 4 kN - 3 kN = 0,
FBD ABD: (II) '1.y
-4B o,z.sr.t
II a.
0,2slN'l
* )< b.37M
y
F /tJ$
AN
I: (!Me=
0:
(0.375m)(I0.8 kN) - (0.25m) Bx= 0,
Bx = 16.2kN,
II: (!MD=
0:
(0.25m)(I0.8 kN + 16.2kN) + (0.5 m) By- (1.00m)(7.0 kN) = 0,
Bx= 16.20kN-....
By = 0.5 kN,
-10.8 kN -16.20 kN + Dx= 0,
Dx = 27 kN,
7.0 kN - 0.5 kN - Dy= 0,
Dy = 6.5 kN,
Dx=27.0kN-.... Dy = 6.50 kN! ....
PROPRIETARY MATERIAL. iO 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 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
I
'i
PROBLEM 6.104
PI-12f'-):12f' t~ .
7,2ftThe
B
~,6ft
''f-L'
axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P = 14 kips and Q = 21 kips, detennine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.
1:=2fl
t
I
I;
'I
!II Ii 1'1
I. I
SOLUTION
~
t"
Members FBDs: 21k~
i
I I
/1. f't:
"'" -2/jR-
I
II '
I 1
-
rr
1:( WA=O:
(12.8 ft)Bx- (32 ft)By- (20 ft)(14 kips) = 0
II:( Wc= 0:
(7.2 ft)Bx+ (24 ft) By- (12 ft)(21 kips)
=0
Solving: Bx = 27.5 kips, By= 2.25kips,
1:-
~=O:
t LFy= 0:
Ax- 27.5 kips = 0, Ay
Ax = 27.5 kips,
(a)
- 14 kips - 2.25 kips = 0, Ay = 16.25 kips,
(b)
-
~
Ay= 16.25kips
t~
Ax= 27.5kips
Bx= 27.5 kips --~
l
By = 2.25 kips ~
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
.
";~.
-r 7,2. It:.
I
S' ", i.
"
~;-.
"O!i'
PROBLEM 6.123
\E
The double-toggle latching mechanism shown is used to hold member G against the support. Knowing that a = 60°, determine the force exerted onG.
\\
~... \'.
\"
\ L'\!0.-: ,..~ :?\
c",;\
(
~.~'//~~
"!IL ~,. 1.5in.
I.:;in.
I
SOLUTION Member FBDs:
()
.,
= tan-I
(:BC+ m)sin60° ..
AF + (AB + BC - CD)cos60°
. I,,,.
= tan-I
(30.-9),'
~.~ ___ --- ---
./
'"
C
.f.,.,
/
2.51~~"' I
.
40.
,'"0.
(2.5 in. + 1 in.)sin60° 4.5 in. + (1.5in. + 2.5 in. - 1in.)cos60°
£'1(
() = 26.8020
£1
~{!..
----- -- ~~
~lh, A
c.
1.5 in.
From FBD CDE:
(
= 0:
We
(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, I I
Cx = 116.7741b
t
'fFy
= 0:
(150.234Ib)sin26.802°- (20 Ib)cos60°- Cy = 0 Cv
= 57.742 lb
From FBD ABC:
+[(4 in.)cos600](57.742Ib)
JiBx
= 0,
+ 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. 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-Hill for their individual coursepreparation. lfyou are a student using this Manual. you are using it without permission. 947
"
PROBLEM 6.123 CONTINUED
Ii
"
I I: I'
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,
Solving (I) and (2):
--
IF'x = 0:
t rEy
= 0:
Bx
= 243.32 Ib,
J3Bx - 9By = 96.7751b
(2)
By = - 36.075 Ib
243.321b- (150.234Ib)cos26.802°- Gx= 0,
Gx = 109.2261b-
- (150.2341b)sin26.802°+ 36.0751b = 0,
Gy = 31.6671b
Gy
t
G = H3.71b "" 16.17°....
OnG,
"
'. .
j'\ "
1.1'
I(, 11,11
'IIII .
PROPRIETARY MATERIAL ~ 2007 The McGraw-Hili Companies. Inc. All rights reserved, No part of this Manual may be displayed. reproduced or disttibuted 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.
948 II
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 mechamsm shown. For each of the two gIven pOSItIOns, determme the force P requiredto hold the system in equilibrium.
SOLUTION
(a) FBD BC:
(
=0:
WB
(0.045m)FCDsin45°-6.00 N.m = 0
~y e 1',
('
FBD Joint D:
b
FCD
a
-1 38.8 mm
= tan
92.2 mm
= 188.562N C
= 22.8230
-1 37.5 mm - 18.8 mm
= 25.7320
f3 = tan
38.8mm
/
(188.562N)cos(45° -25.732°)
LFx'= 0:
-FDEcos(22.823° + 25.732°) = 0,
FBD E:
rEx = 0:
FDE= 268.92 N C
(268.92N)cos(22.823°)-P = 0,
P = 248N -
...
_135.4 mm = 51.8570
(b) FBD BC:
8 = tan 27.8mm
f3= tan-I 27.8 mm - 24.3 mm = 7.55200 26.4 mm (WB
= 0:
(0.045m)FCDsin(90°-51.857° -7.5510°) - 6.00 N'm = 0 FCD = 262.00 N C
continued PROPRIETARY
MATERIAL
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or distributed in any form or by any means, 1,1,oithoutthe prior written permission of thepublisher. or used beyond the limited distribution to teachers and educators
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for their individual
course preparation.
If you are a student
955
using this Manual,
you are using it without permission.
i
PROBLEM 6.128 CONTINUED
I' Ii
-1 20 nun + 35.4 nun - 26.4 nun
r = tan
FBD Joint D:
24.3 nun + 7.5 nun
a = tan-I 20 nun + 35.4 nun -26.4 nun
92.2 nun + 27.8 nun - 24.3 nun
F --- Cc
= 42.3630
16.85840
-(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
:1 I '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 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.
956
'10II,
PROBLEM 6.139 A hand-operated hydraulic cylinder has been designed for use where space is severely limited. Determinethe magnitude of the force exerted on the piston at D when two 90-lb forces are appliedas shown. I~ I
Zin.
SOLUTION FBD top handle:
Note CD and DE are two-force members
(4 in.) ..k
FCD-(1.5in.) k
FCD-(13.2in.) (90 lb) = 0
FCD=n.J61lb By synnnetry:
FDE = FCD = 12 .J611b
FBD Joint D:
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 and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using thi.rManual, you are using it without permission.
967
-
I I
I
PROBLEM 6.150 A 500-kg concrete slab is supportedby a chain and sling attached to the bucket of the ITont-end loader shown. The action of the bucket is control1edby 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. DhlU."n~;ons
in mill
SOLUTION I
',II
FBD Bucket (one side):
f
,
!
If.Afj.
r
(a)
(
t'!>.-"",
FAD
Dx
w
"\.0 I
II
LMD= 0: (0.8 m)(2.4525 kN) - (0.5 m) FAD= 0
= 3.924 kN
= (250 kg) (9.81 N/kg) = 2452.5 N = 2.4525 kN
to.r1:,
I
,ii'Ii I FBD link BE:
,il il
. 0,
I
.I,
3.'2.¥
B
ktJ
--r it
_J.
I,
0. II
(LME= 0: (0.68 m)(3.924 kN) - (0.54 m)G~ FCD) H1
+ (0.35 m) U~ FCD)= 0,
E I
FCD=7.68 kN C ....
J
]
IJ.
r;v 111.
€,ll
-I
i
~y FBDEntire linkage: .2,5-
""-
65,
,I)
,.. ,
l
1. q!>~kN
FCD=7.682 kN
;p"
'~.
,
-'
(~
J ~ '"
(LMG= 0: (2.5 m) (2.4525kN) 1 1 + (0.2 m) J2 FGH-(0.6 m) J2 FGH=0,
0,1>10\
i
lPy
-
F'..
,, - -,
FFH=21.677kN,
d,z.>1
t
FFH=21.7 kN C....
{i"
F _FIf
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 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.
978
