Chapter 6 Statics
Short Description
Statics chapter 6 solution...
Description
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 ~
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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 McGrawHili 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 McGrawHili 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)(Myl
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 FAC13(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 McGrawHilI 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 McGrawHili 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)  FDG1.5 kN  5.61kN 13 13 1 + .J2 FDH= 0
Solving:
FDG=8.60 kN C....
PROPRIETARY MATERIAL to 2007 The McGrawHili 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 McGrawHillfor 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 zeroforce 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 McGrawHili 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 McGrawHil/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 McGrawHili 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 McGrawHili 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'iml
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 2n = 16
constrained
but indeterminate
"'11II
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permission
of the publisher.
or used beyond
the limited
distribution
to teachers
and
educotors permitted by McGrawHill 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 250mmj 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 twoforce 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 McGrawHili 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 McGrawHili 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 twoforce 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 twoforce 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 McGrawHili 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 McGrawHill 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 McGrawHilI 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 McGrawHill 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
PI12f'):12f' t~ .
7,2ftThe
B
~,6ft
''fL'
axis of the threehinge 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 McGrawHili 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 McGrawHili 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 doubletoggle 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:
()
.,
= tanI
(:BC+ m)sin60° ..
AF + (AB + BC  CD)cos60°
. I,,,.
= tanI
(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 McGrawHili 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 McGrawHill 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 McGrawHili 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 McGrawHili 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= tanI 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
~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.
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