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CHAPTER
7
-
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-
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2
in.
8
in.
4
in.
B
~n\».
§ik
4
in.
1t~
G
,/|
l/i
r
—
fi
m.
—
*•
60fb
PROBLEM
7.1
Determine
the internal
forces
(axial
force, shearing force,
and bending
moment)
at Point
J
of the
structure
indicated.
Frame
and
loading
ofProblem
6.79.
SOLUTION
From
Problem
6.79: On JD
FBD
ofJD:
D
=
80lb,
V*
3
So
lb
I
o
~i
t^;-o
ZR.
=
+
)ZMj
=0
K-80lb
=
Af-(80lb)(6in.) =
V
=
80.0
lb
f
^
F
=
«
M.
=
480 lb-
in.
^
A
PROPRIETARY
MATERIAL.
©
2010
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
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limited
distribution
to teachers
andeducatorspermitted
by
McGraw-Hill
for
their
individual
coursepreparation.
If
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are
a student
using
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Manual,
yon
are using it
withoutpermission.
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2
in.
fj ill.
'I in.
12
hi.
4
in.
\.E
3
'.'.}
la)
7
I
,
r
*
—
6
in.
—
.a
PROBLEM
7.2
Determine
the
internal
forces (axial
force,
shearing force,
and
bending
moment)
at Point
J
ofthe
structure indicated.
Frame and
loading of
Problem
6.80.
(ii.i iii-
SOLUTION
From
Problem 6.80:
OnJD
FBI) of
JD:
D
=
401bJ
rf
y
*
3
HO
lb
(&
1
A .
|SF
V
=0:
K-401b
=
—*£/>;
=
0: F
=
+)XM.
;
=0:
M
-
(40
lb)(6
in.)
-
V
=
40.0
lb
|
<
M
=
240 lb
-in/)
^
PROPRIETARY
MATERIAL.
CO
2010
The
McGraw-Hill
Companies, Inc. AH
rights reserved. No
pari
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
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limited
distribution to
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and
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for
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individual course
preparation.
If
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-300
mm-
1.4 kN
200
mm
iOti
mm
PROBLEM
7.3
Determine
the
internal forces
at Point
Jwhen
#=
90°.
200 mm
400
mm
-
SOLUTION
Free
body:
Entire
bracket
Free
body: JCD
j*4-krt
-^)EM
D
=
0:
(1.4
kN)(0.6
m)-
.4(0.175 m)
-
,4 =+4.8kN
A
=
4.8kN
—
-±~IF
v
=0:
Z)
v
-
4.8
=
D
v
=4.8kN
—
+|SF
;
-0:
.D
?
,~1..4
=
D
v
=1.4kNt
-
-*-^
c
~
K
V
*D
i.4W{
4,£krt
-±.LF
X
=0:
4.8kN-F
=
F
=
4.80kN
—
<
+\lF,=Q:
1.4kN-K
=
V
=
1.400kNH
+)XMj
=0:
(4.8.kN)(0.2m)
+ (].4kN)(03m)-M=0
M
=+L38kN-m
M
=
1.380
kN-m
)<
PROPRIETARY
MATERIAL.
€>
2010
The
McGraw-Hill Companies,
Inc.
All
rights reserved.
No
pari
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
byMcGraw-Hill
for
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individual
course
preparation.
If
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k«-300
mm-*j
J.
-UN
\\
;
M
3 S
200
mm
400
mm™
200 mm
J
o mm
PROBLEM 7.4
Determine the
internal
forces
at
Point ./when a-Q.
SOLUTION
Free body:
Entire
bracket
Free body:
J
CD
I.4WU
+|
/>;,= 0:
D
J;
=0
D
;
,
=0
I,M
A
-
:
(1
.4 kN)(0.375
m)
-
D
x
(0.
i 75
m)
=
D =
+3 kN
D
v
=
3 kN
—
F
_j
|«-©.3i*-*|
tA
*v
c
T
±_E^=0:
3k.N-/'
=
+|£F
V
=0:
-K
=
h)
XM.,
=
0:
(3
kN)(0.2
m)
-
M
=
M
=+0.6kN-m
F
=
3.00kN*-^
v
=
o
<
M
=
0.600
kN-mjM
PROPRIETARY MATERIAL.
&
2010 The
McGraw-Hill
Companies,
Inc. All
lights
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
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limited
distribution
to
teachers and
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course preparation.
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-240
mm
225
mm
I
1 //
225
irun
A
(o
i
60
mm
t
60
mm
135
mm
400
N
SOLUTION
FBD
ABC:
FBD
CJ:
/2*>
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-240
mm
W-
-®j
m
PROBLEM
7.6
Determine
the
internal
forces
at
Point
J?
of
the
structure
shown.
225 iniii
V
/B
W~
—
m
Tm
mm
•..•J
I
60 mm
135
mm
400
N
SOLUTION
FBD
AK:
E
£F=0: K
=
M
v^o
<
y<
J2F
V
=0:
F~400N
=
{1M
K
=0: (0.135
m)(400N)-M
=
F
=
400n(^
M
=
54.0N-mjH
PROPRIETARY
MATERIAL. ©
2010 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
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distribution to
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educators
permittedby
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individual
course
preparation.
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I20N
I
*&S
180 mm
|
gQo
Irav
PROBLEM
7.7
A
semicircular rod
is loaded
as
shown.
Determine
the
internal forces
at Point J.
SOLUTION
FBI)
Rod:
UO
N
Ax
A
FBD AJ:
tso
(2.0
M
{XM
B
^Q:
A
x
(2r)
=
Q
/
L/y
=
:
V
-
(1
20 N) cos
60°
=
V
=
60.0N/
<
\Z/y
=
0: F
+
(120N)sin60°
=
F
=
-1
03.923 N
Q
EM, =0:
M
-
[(0.
1
80
m)sin
60°](1
20
N)
=
M =18.7061
F
=
103.9
N\
,<
M
=
18.71
J)
^
PROPRIETARY
MATERIAL.
©
2010
'Hie McGraw-Hill Companies,
Inc. All
rights
reserved. A'o /««•/
o//A« Manual
way 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 teachersand
educatorspermitted
by
McGraw-Hill
far
their individual
course
preparation.
If
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J20N
PROBLEM
7.8
180
mm
180 mm
60°
r
A
I
A
semicircular
rod
is
loaded as
shown.
Determine the
internal forces
at
Point
K.
SOLUTION
FBD
Rod:
s
-
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Sin.
-2-4
ii
PROBLEM
7.9
An
archer
aiming
at a target
is pulling
with
a
45-lb
force
on the
bowstring.
Assuming
that
the
shape
of the
bow
can
be approximated
by
a
parabola,
determine
the
internal
forces
at
Point,/.
SOLUTION
FBD
Point A:
By symmetry
7;
=7;
£F=0: 2
'3
^
45
lb
45
lb
=
T,
=7;
=37.5
lb
Curve CJB
is
parabolic:
x
~
ay
1
FBI) BJ:
At B:
x
=
8in.
y-32'm.
8
in. 1
At
J:
So
(32
in.)
2
128 in.
128
Slope of
parabola
=
tan
&
=
^
=
—
—
=
~-
dy
128
64
16
4-i
%
'
•
1
t*s^
=
tan
64
-
14.036
c
4
a
-
tan '
—
1
4.036° =
39.094°
3
/s/v
=0:
V-
(37.5
lb)cos(39.094°)
=
M
V
=
29.1 lb/
<
PROPRIETARY
MATERIAL
©
2010
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,
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beyond the
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distribution to teachers
and
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permitted
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coursepreparation.
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PROBLEM 7.9
(Continued)
\
I/y
=0:
F
+
(37.5
lb)
sin
(39.094°)
=
F
=
-23.647
3
(^ZMj=0:
M
+ (X6'm.) (37.5
lb)
+
[(8
-2)
in/
4
(37.5
ib)
F
=
23.61b\^
M
=
540.ib-in.^H
PROPRIETARY
MATERIAL
©
2010 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
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distribution to teachers and
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individualcoursepreparation.
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8in.
-24iii.-
miii.
16 in.
32
in.
PROBLEM
7.10
For
the
bow
of Problem
7.9,
determine
the magnitude
and location
of the
maximum
(a) axial force,
(b) shearing
force, (c) bending
moment.
PROBLEM
7.9 An archer aiming
at a target
is
pulling
with
a 45-Ib
force
on
the
bowstring.
Assuming
that
the
shape
of
the
bow
can
be
approximated
by
a parabola, determine
the internal
forces at
Point
J.
SOLUTION
Free
body: Point A
±^2F=0:
2
T\-45\b
=
T
=
37.5
lb
O
Free body:
Portion
of
bow
BC
ZZflb
ft.
Equation
of
parabola
At B:
Therefore,
equation
is
F
c
=
30
Lb
|
-
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PROBLEM 7.10
(Continued)
(*)
Maximum
axial
force
+\ 1F
V
=
0:
-F
+
(30
lb) cos
-
(22.5
lb)
sin
=
y**\
\
3
Free bodv: Portion
bow
CST
frsoS
F
=
3Ocos0-
22,5 sin
F is
largest
at
C(0
-
0)
F
M
=
30.0 lb
atC
<
(b)
Maximum shearing
force
+/
TF
V
=
0:
-V
+
(30
lb)sin
+
(22.5
lb)eosc?
=
V
=
3Qsin0
+
22.5
cos9
V
is largest
at
B
(and D)
Where
=
maK
-
Ian
-
'
(-)
=
26.56°
F„,
=
30
sin
26.56°
+ 22.5 cos
26.56°
P;,
=33.5
lb
at
5
and
D^
(c)
Maximum
bending
moment
+)£M
X
=0:
M~960lbin.+
(30lb)x
+
(22.51b)^
=
)
M
=
960
-30*-
22.5
j>
M
is
largest
at C, where
-v
=
y
-
0.
M,„=9601b-in. at C
<
PROPRIETARY
MATERIAL. ©
2010 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
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-Gil
•3
in.-
•3
in.-
-6
in.
—
•3
in.*j
3
in.
3 in.
PROBLEM
7.11
Two
members,
each consisting
of
a
straight
and
a
quarter-circular
portion of rod,
are
connected as
shown
and
support a 75-lb
load
at A. Determine the
internal
forces
at
Point J.
SOLUTION
75
lb
IO0tt>
Free
body:
Entire
frame
+)1M
C
=
0: (75 lb)(l 2
in.)
-
F(9
in.)
=
+*£F
r
=0:
C
x
=0
+\
ZF
y
=
0:
C
y
-
75 lb
-
1
00
lb
-
C
v
=
+
I75ib
Free body:
Member
BEDF
+)ZM
B
=
0:
D(X2
in.)
-
(1
00
ib)(l 5
in.)
=
-±*.ZF
A
.
=0:
5
V
=
+JLF
3
,.
=0: i?
v
+
125
lb-1001b-0
/i
v
=-251b
Free
body: BJ
4EF
v
=():
F
(25
lb)
sin
30°
=
+\
If
v
=
0: r
~
(25
lb)
cos
30°
=
F
=
100 lb <
C
=
1751b
-
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6 in.
1-3
in.-
k}
in,
(
'»
I
.
-6
in.
s-3
iii.-*-J
7K
PROBLEM
7.12
Two
members,
each consisting ofa
straight
and a
quarter-
3
U
circular portion
of
rod,
are
connected as
shown
and
—J-
support
a
75-lb load at A.
Determine the internal
forces at
3
1,.
Point K.
SOLUTION
Jft
Free
body:
Entire frame
6lrt
* T?l
/25/i,
at
f
3d
to.
lO0ft>
+)
1,M
C
=
:
(75
lb)(1 2
in.)
-
F(9
in.)
=
±-XF
x
^0;
C,
=0
+)
SF
3
,
=
0: C
r
-
75
lb
-
1
00
lb
=
C
y
=
+ 175
lb
Free body:
Member
BEDF
+)ZM
B
=0: D(12in.)-(1001b)(15in.)
=
+\ZF
y
=
0:
S
v
+1251b-1001b
=
B
=-25
lb
F
=
1001bl <
C
=
1751b}
-
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17/216
PROBLEM
7.12 (Continued)
N*-
TF
y
=
0:
V~{\
25
lb) sin
30°
=
V
=
62.5 lb
^30.0°
-
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2KON
PROBLEM
7,13
j-«
— 160 i«
m
-
-*
—
.160
nmi—
-i
_
tit
1-iJ
A semicircular rod
is loaded
as shown.
Determine
the
internal
forces
at
Point
J
knowing that
#=30°.
1.60
inn
120 mm
j
D
SOLUTION
FBD
AB:
C
SM
,<
f4
^
=
0:
/*
-C
15
)
+
H
—
15
,
~2r(280N)
=
C
=
400
N
/
=
0:
-/4
V
+-(400N)
=
£x
\S
I
r
/
^.
A
V
=320N
—
K
=
0:
^,+-(400N)-280N
=
FBD AJ:
A
v
=40.0N|
40
M
j
2>lO
hi
\
f
k*
/'
\S/V=0:
/X/y=0:
F
-
(320
N)sin
30°
-
(40.0
N) cos
30°
=
F
=
l 94.64IN
F
=
194.6 N^
60.0°
A
V
-
(320
N)cos
30°
+
(40
N) sin
30°
=
V
-
257.13 N
\
V
=
257
N
-
8/17/2019 Cap 7, Novena Edc_text.pdf
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2-WN
1.60 mm
120
mm
PROBLEM
7,14
A semicircular
rod is
loaded
as shown.
Determine the
magnitude
and
location
of the
maximum bending moment
in
the
rod.
SOLUTION
2S0N
160 mm , 160 mm
Free body;
Rod
A.CB
+ )1.M
A
=0:
l/-
a)
j(().16m)
+
f|^,,
|(0.16m)
-(280N)(0.32m)
=
+~
ZF
X
=
0: A
x
+-(400
N)
=
./(
V
=~320N
+
ZF
y
=
0: A
Y
+~
(400
N)
-
280
N
=
4,=
+40.0
N
Free
body: AJ
(For <
90°)
F
a)
=400N/ <
A =320N
—
-
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PROBLEM 7.14
(Continued)
Free
body: BJ (For
>
90°)
2&o
ti
+)
2M ,
=
:
M-
(280
N)(0.
1
6 m)(I
-
cos
)
=
M
=
(44.8
N-m)(l-
cos
4)
Largest value
occurs for
-
8/17/2019 Cap 7, Novena Edc_text.pdf
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K~
0.6
in
—
>h
—
0.6 m
—
J
h
50(1
n
k
,*;
(
' i
0.9 in
0.9 ni
PROBLEM
7.15
Knowing
that
the radius of
each
pulley
is
1
50
mm,
that,
a-
20°,
and
neglecting
friction, determine
the
internal forces
at (a) Point
J,
(b) Point
K.
SOLUTION
Tension in
cable
-
500
N.
Replace cable
tension
by
forces
at pins A
and
B. Radius
does
not enter
computations: (cf. Problem
6.90)
(a) Free
body:
A
J
SOOft
i
soon
,
/
H
J
_
F
J.
M
*
—
0>»
4-
i.ZF
x
=
0:
500N-F
=
F
=
500 N
+t
1/^=0: K-500N^0
F
=
500N
+)SAf
-|
„
fl
-
[
SCO//
Sao//
B
K
%
-
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PROBLEM
7.15 (Continued)
+f
IF
y
=
0: -500N-(500 N)cos20°
+
F
=
F
=
969.8
N F
=
970NJ^
+)I,M
K
=0:
(500N)(1.2m)-(500N)sm20°(0.9m)-M
=
M
=
446.1.
N-m
M
=
446 N
•
m
)
<
PROPRIETARY
MATERIAL. ©
2010 The McGraw-Hitt 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
permittedby McGraw-Hill
for
their
individual coursepreparation.
If
you
are a student using
this
Manual,
you are using
it
withoutpermission.
1008
-
8/17/2019 Cap 7, Novena Edc_text.pdf
23/216
«-
0.6 m
»
1
<
0.6 m
50 ) N
/
B
0.9
m
0.9
in
PROBLEM
7.16
Knowing
that
the
radius
of each pulley
is
150 mm,
that a~
30°,
and
neglecting friction,
determine the
internal forces at
(a) Point
,/,
(b)
Point
K.
SOLUTION
Tension in cable
=
500
N.
Replace
cable
tension by forces
at pins A
and
B. Radius
does not enter
computations:
(cf. Problem
6.90)
j
Soon
.
(a) Free
body:
AJ:
/ft
J
f
-±~
2/^=0:
500N-F
=
'
y
F
=
500N F
=
500N
—
<
+\xF
y
=Q:
K~500N=0
K
=
500N V
=
50()n(
-
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24/216
-
360 N
=
A
y
+
B
y
=
360
N
+)
SM
fi
=
:
-4,
(2.4
m)
-
(360
N)(l .6 m)
=
Av =-240
N
y
J?
y
=360N
+
240N
^,=+600N
Free body: BJ
We recall that the forces applied to a pulley may be
applied
directly
to
its
axle.
iwtf
\f£F=0:
-(600N)
+
-(520N)
-360
N
—
(360N)-F
=
F
=
+200N
B
=
520
N
-*--
-
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25/216
PROBLEM
7.17 (Continued)
+/1F
X
=
0:
-(600N)-
-(520N)-i(360N)+K
=
=
V
=
=
+120.0
N
V
«
120.0
N/
^
+)ZMj
=
0:
(520N)(1.2
m)
-
(600
N)(l
.6
m) +
(360
N)(0.6
m)
+
M
==
M
=
+
.120.0
N-m
M
=
120.0
N m
)
<
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill Companies,
Inc. All rights reserved. /Ve;
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 andeducatorspermitted
by
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for
their
individual
course
preparation.
If
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are
a student using
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you are
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1011
-
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.1.8
m
if''
,-360N
=
4,+.#
v
=
360N
+
)
2A4
-
0; -A (2A m)
-
(360 N)(l .6
m)
=
^=-240N
#,,=360N +
240N
B
=+600
N
y
B.
=
5201M
—
<
A =520N
—
<
(i)
A
;
=240N|
-
8/17/2019 Cap 7, Novena Edc_text.pdf
27/216
r= 55.0 in.
B
I
C
—
.1
2
in
.
H
-
6.75 in
.-»-
9 in
PROBLEM
7.19
A
S-in.
-diameter
pipe
is
supported
every
9
ft
by
a
smalt
frame
consisting
of
two
members
as
shown. Knowing
that
the
combined
weight
of the
pipe and its
contents
is
10 lb/ft and neglecting
the
effect of
friction,
determine
the magnitude
and
location
of
the
maximum
bending
moment
in member
AC.
SOLUTION
%ft>
e=5*t&
Free
body:
10-ft
section
of
pi
pe
+/
XF
X
=
0:
D
--(90
lb)
=
\J-SF
v
=0:
£--(90lb)
=
Free body:
Fram
e
+)ZM
H
= 0:
-A
v
{\
8.75 in.)
+
(72
lb)(2.5
in.)
+(541b)(8.75in.)
=
D
=
72lb/ <
E
=
541b\
<
L/c
From(t):
4,
=+34.8
lb
A,, =34.81bj
-
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28/216
tf*»A
2t*Mlb
PROBLEM
7,19
(Continued)
Free body:
Portion ^4./
For*
<
12.5 in. (AJ
s=
,4,0)
:
+)I,Mj
=
0:
(26.41b)-x-(34.8Ib)-x
+
M-0
M^12x
A/
=
1
50
lb
•
in. for
x
=
12.5
in.
•Zk4jt>
M
max
=I50.01b-in.atD
<
Forx>12.5m.(^J>/JO):
+)XMj
-0:
(26.4.ib)-A--(34.8lb)-x+
(72Ib)(x-12.5)
+
M
=
M
=
900-60x
M„_
=
1
50 lb
•
in. for x
=
12.5
in.
Thus:
M
mm
=150.0
Ibin.atZ)
^
PROPRIETARY
MATERIAL. © 20J0
The
McGraw-Hill
Companies, Inc. All rights
reserved. M>
/wwY o//A« 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
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educatorspermittedby
McGraw-Hill
for
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preparation.
If
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PROBLEM 7.20
n
•*
12 in.
~
&./
E
r
* ().ioiiir
9 in.
For
the
frame
of Problem
7.19,
determine
the
magnitude and
location
of
the maximum
bending
moment
in member BC.
PROBLEM
7.19 A
5-in.-diameter pipe
is
supported every 9 ft
by
a
small
frame
consisting
of
two
members as
shown..
Knowing
that
the
combined
weight
of
the
pipe
and
its
contents
is
1
lb/ft
and
neglecting
the
effect
of
friction, determine
the
magnitude
and
location of
the maximum
bending
moment
in
member AC.
SOLUTION
D=72i,
C
¥**
Free body:
10-ft section of
pipe
+/
ZF
X
=0:
D
~
-
(90
lb)
=
\
f
T.F
r
=
:
E-
-(90
lb)
=
Free
body: Frame
+)XM
B
=0:
-4,(18.75 in.)
*
(72
lb)(2.5 in.)
+ (54lb)(8.75in.)
=
D
=
721b/
<
E
=
54
lb
\
-
8/17/2019 Cap 7, Novena Edc_text.pdf
30/216
PROBLEM
7.20 (Continued)
Free body:
Portion
BK
S?2H>
Z6,«ffb
For*
<
8.75
m.(flf
-
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31/216
PROBLEM
7.21
A force P
is
applied to
a
bent
rod that
is
supported by
a roller
and
a
pin
and
bracket.
For
each
of
the
three
cases
shown,
determine
the
internal
forces
at Point,/.
-a—
*- *—
a
—
r
'1
,
• /
{*)
if
3j
:
X
4*
(b)
a
—
*->-*—
«—
ft
(c)
D
1
*
SOLUTION
(«)
FBD
Rod:
a-
¥
a
J
J
+
£
—
-
ZF=0:
/,
=0
XM„
=
0: aP-2aA.=0
A.
P
FBD
AJ:
M
*-;*>
j-
EF=0:
—
-F
=
J
2
—
2/^=0:
V
=
^
F-*|«
2*
(ZM,
=0:
M
=
^
(/;)
FBI)
Rod:
i
-
t
r
^
S4
2^=0:
2a
-D\
+
2a D\-aP~0
-ZF=0:
v4
r
-~
P
=
-x
5
14
j£F=0: A-P
+
-—
/>
=
()
'
J
5
14
D
,1
A.
5P_
14
-
?
UP
14
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc.
All
rights
reserved.
No
part
of
ibis
Manual may
he 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
permittedby
McGraw-Hill
for
their individual
coursepreparation.
If
you
are
a
student using this
Manual,
you
are
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PROBLEM
7.21
(Continued)
—
-
1/\.
=0: -P-V
=
7
1
J
1.4
V
2P
7
14
*
7
y
(c)
FBD Rod:
B
.
°~
'
'
a
A,
1
52
*
3
5P
\
y
y
5 2
D
5f
2
4T
=
2P
3
2
XF
v
=0:
2P-K=0
5P
£F
=0:
2
F
=
(lM
;
=0: a(2P)-M
=
V
=
2P^~4
2
*
M
=
2tfPJH
PROPRIETARY MATERIAL.
© 2010
The
McGraw-Hill
Companies,
Inc.
AH 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 permittedby
McGraw-Hill
for
their
individual
coarse
preparation.
If
you
are
a
student using
this
Manual,
you
are using it without
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PROBLEM 7.22
A force
P is applied
to a
bent
rod
that is
supported
by
a roller
and
a pin
and
bracket.
For
each
of the
three
cases
shown,
determine
the
internal
forces at
Point./.
-
a
—*-[-*-
a-
T
]
1
A
(«)
iiw
H—
-
8/17/2019 Cap 7, Novena Edc_text.pdf
34/216
FBD
AJ:
F
V
r
s~p/z~
(c)
FBD
Rod:
&
P
$£
dS-»
<
r
ct*
J>
Di*
Z#-~
^
PROBLEM
7.22
(Continued)
3
5P
-K
=
2
1
>
5
2
-F^O
2
y
CLW
D
=
0: fl/
,
-2a[-^]-
,[
4
>
=
14
—
E/v
=
0:
K-
{SX4)
14
|SF
V
=0:
4
5P
5
14
7
*
C£AO=0:
M~~a
f35P
\
{5\4J
=
14
y
PROPRIETARY
MATERIAL.
C
2010
The
McGraw-Hill
Companies, Inc.
All rights reserved. A'o
/>«>*/
o//Aw
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
permittedby McGraw-Hill
for
their
individual
coarse
preparation.
If
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a
student using
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Manual,
you are
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without
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11120
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\
A
,„-.
B
PROBLEM
7.23
A
semicircular
rod
of
weight
W and
uniform cross
section
is supported
as
shown.
Determine
the
bending
moment at Pointy
when
0-
60°,
SOLUTION
B
W
K
w
/l/V=0:
F +
—
sin
60
W
3
k
F
=
~0A29521V
cos
60°
=
FBI)
BJ:
XM
n
=0:
r\F-~)
+
—(—\+M
=
7t)
2it\-h
M~Wr 0.1.
2952 +
1
1
it
2k
,
On&/
0.28868W-
M,
=0.289^/^
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc. All
rights
reserved.
JVo
/*»/
o/rtfc Mutira/
may
/*?
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
educatorspermitted
by
McGraw-Hill
for
their
individual course
preparation.
If
you
are
a student
using
this
Manual,
you
are
using it
without
permission.
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PROBLEM
7.24
A
semicircular rod
of
weight
W
and
uniform
cross
section is
supported
as
shown.
Determine
the
bending
moment at
Point
J
when
$
=
1.50°,
SOLUTION
FBD Rod:
\lF
'=0:
A
y
~-W
=
A
y
=w\
XM
fi
=0:
-^W-2rA
x
=
71
A.
W
7T
FBD
AJ:
W W
£/V=0:
—
cos30°
+—
sin30°-F
=
F
=
0.69233W
\
7t
6
CzA/
*
0:
0.25587r(
—
+
r F
-—
I
-
M
-
M
=
^r
H« l
+
o.69233-I
6
ic
M
=
Wr(0A\66)
On
AJ
M
=
0A\7Wr)<
PROPRIETARY
MATERIAL.
©
2010 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
(heir
individual
course
preparation.
If
you are
a
student using
this
Manual,
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are
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it without
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iL
a
m
PROBLEM
7.25
A
quarter-circular
rod
of weight Jfand
uniform
cross
section is
supported
as
shown.
Determine
the bending
moment
at Point
J when
- 30°.
SOLUTION
FBD
Rod:
£F=0:
A -0
2r
71
A,
1W
it
FBD
AJ:
a
=
1
5
°
s
weight
of
segment
=
W
30°
W
90°
3
r
=
—sin
a
-
—sin 1
5°
=
0.9886r
12
2fT
W
AW
W
£F,=0:
cos30°—
-
-cos30°-F
=
'
^
3
Q£M
()
=
M
+
r|
F
F
2Ef
W
WS(2
1
2
U
3
/
+
rcosl5°—
=
3
M
=
0.0557
Wr )-*
PROPRIETARY
MATERIAL.
©
2010 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
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using
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Manual,
you are
using
it withoutpermission.
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IM
/•J
PROBLEM
7,26
A
quarter-circular
rod
of
weight W
and
uniform
cross
section
is
supported
as
shown.
Determine the
bending
moment at PointJ
when
6
-
30°.
SOLUTION
FBD
Rod:
**
A
CxM
/f
-0:
rB~W
=
Q
B
2PV
K
FBD BJ:
«
=
15°
71
12
F
=
—
sin
15°
=
0.98862/-
Weight of
segment
=W
fLF.
=
0: /
30°
W
90°
3
,
W
w
SM
=
0:
rF-(Fcos 5°) M
=
2W
cos30°-—
sin30°
=
3
^
/
\
F
s/|
J_
v
6
*/
w
/
M~rW
V
6
K
J
0.98862^^-
Iff/-
M
=
0.289
»>}
^
PROPRIETARY
MATERIAL. ©
2010 The
McGraw-Hill
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
permission
of
the
publisher, or used
beyond
the
limited
distribution
to
teachers and
educators
permitted
by
McGraw-Hill
for
their
individualcourse
preparation.
If
you
are
a
student
using
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Manual,
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w
am
PROBLEM
7.27
For
the
rod
of
Problem
7.25, determine
the
magnitude
and
location
of the
maximum
bending
moment,
PROBLEM
7.25
A
quarter-circular
rod
of
weight
W
and
uniform
cross
section
is
supported
as
shown.
Determine
the
bending
moment
at Point
J
when.
^30°.
SOLUTION
FBD
Rod:
+
XF
x
=0:
A
X
=Q
a
2W
71
r
.
-sma
a
Weight
of segment.
=W—
=
-~W
SF,=0:
4a
2W
-F
W
cos
2a +
cos
2a
-
It
K
r,
21V
2W
F
=
(l-2ar)cos2«
=
(1-0)
cos,
ft
ft.
FBDAJ:
{l,M =Q:
M +
(f
2W
2W
)
,_
x
Aa
\r
+
(r
cos
a)
—
fV-0
ft
J
71
M
—
(l
+
0cos0-cos0)r
sin
a
cos
aft
K
a
But,
sin or
cos
a
=
—sin
2a
-
—sin
2
2
so
for
2/
71
dM.
2rW
M
=
^-W(\--cos0+0cos0-sm0)
71
'
de k
(I-0)sin0 =
O
dM
(sin
(9
-
si
n +
cos
#
~
cos
#)
=
-
8/17/2019 Cap 7, Novena Edc_text.pdf
40/216
PROBLEM
7.28
For
the
rod
of
Problem 7.26,
determine
the
magnitude
and
location
of the
maximum
bending
moment.
PROBLEM
7.26 A
quarter-circular
rod
of
weight W
and
uniform
cross
section is
supported
as
shown.
Determine
the
bending
moment
at
PointJ
when 0=
30°.
SOLUTION
FBD
Bar:
(ZM
A
=0;
rB
—
-W
=
v
A
n
B
1W
71
A
a
2
_.
v
.
r
-
—
sin or
a
so
Q^-a<
K
Weight
of
segment
=
W
2a
Aa
n
W
An
2W
'IF.=Q:
F-—Wcos2a
sin2or
=
2W
F
-
(sin
2or
+
2orcos2a)
n
=—
(sin#+
0cos6>)
K
FBD BJ:
But,
An
ZM
Q
=
:
rF
-
(r
cosa)
—
W~M
=
M
=
~Wr(sm0
+ 0cos0)
TV
sin
or
cos
a
-
—sin
2a
=—sin
2
2
r
•
i
4a
w
—sin
or cos
or
—
W
a
)
7T
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies, Inc.
Ail
rights
reserved.
No
part oj
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
permittedby
McGraw-Hill
for
their
individual
course
preparation. Ifyou
are
a
student
usmg
this
Manual,
you
are
using it
without
permission.
1026
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PROBLEM
7.28
(Continued)
so
M
=
(sin0
+
0cos#-sin0)
or
M=~Wr0
cos
71
dM
2»r, n
—
~~Wr(cos0-ds\n0)
=
O
at
-
8/17/2019 Cap 7, Novena Edc_text.pdf
42/216
1'
c
\
n
PROBLEM
7,29
For
the
beam
and
loading
shown,
(a)
draw
the
shear
and
bending-moment
diagrams,
(b)
determine
the
maximum
absolute
values
of the
shear
and
bending
moment.
SOLUTION
(a)
t£
aL
(X- CX-
?*
i
FromA
to B:
i£
I-
V
+)lM,=0:
A^-P,
From
/? to
C:
+|LF
v
=0:
-P-P-K-0
K
=
-2/>
+)£M
2
=0:
P
Y
+
P(x-tf)+M=0
A/*=-2P,
+
P
a
V
-*P
ZZ
%
-3?*-
(b)
\V\
=
2P;
I
Ml
I
'
I
mas
'
I'
3P
PROPRIFTARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc.
All
rights
reserved.
/Vo
/«irf
o/
/Afa
Mm™/
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
usmg
tins
Manual,
you
are
using it
without
permission.
1028
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43/216
PROBLEM
7.30
For
the
beam
and
loading
shown,
(a)
draw
the
shear
and
bending-
moment
diagrams,
(b)
determine
the
maximum
absolute
values
of
the
shear
and
bending
moment.
SOLUTION
(a)
Reactions
:
From
A
to
B
;
*.**
)A.
r
3
.
—
x
-*y
From
B
to
C:
2P
3
C
4
?/
J
3
+)SM,=0:
M
=
+—
+ sF
v
-0:
K
=
~~
P
*£.
2L
3
^
C
tf(
4.?
(t-v;
h^
b
+;ZA/
2
=0:
Af
=
+—
(L~x)
0)
l^ m
1(
x=—
;
|A/|
m
„
=
—
9
PROPRIETARY
MATERIAL.
&
2010
Tl.e McGraw-Hill
Companies,
Inc.
All
rights reserved.
No
par,
of
this
Manual
,»ay be
displayed
reproduced
or
distributed
m any
jonn
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^/Iill
for
their
indivk/uai
course
preparation.
Ifvou
a^
you
are
using
it
without
permission.
'
*
1
029
-
8/17/2019 Cap 7, Novena Edc_text.pdf
44/216
1
1
I
T
1
c
L
L
4
PROBLEM
7.31
For the
beam
and
loading
shown, (a)
draw
the
shear
and
bending-moment
diagrams,
(b)
determine
the
maximum
absolute
values
of
the
shear
and
bending
moment.
SOLUTION
FB'D
beam:
(a) By
symmetry:
Along
AB:
£
fi
^~x J
^
H
j.
2
v
/
2
4
I
3'
A
4
(SAf,
~0:
M-x—
=
V
J
4
j^/
=
—
x
(straight)
4
M
Along
BC:
wL
fLF,=0:
—
-m^-J^O
F
=
wx,
4
B
ss|
)H
X.
^V
Straight
with
V
=
at
x,
4
A/
w
Y
r2
8 2
Parabola
with
Section
CD by
symmetry
(/?)
From
diagrams:
\
3
,-,
I
M
—
—
wL
at
x,
=
—
32
4
|J/|
max
=—
on^fg
and
CD
^
3w/r
^
A^L«
=
at
center
<
'max
s\r%
PROPRIFTARY
MATERIAL.
€>
2010 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
individualcourse
preparation.
If
you
are
a
student
using
this
Manual,
you
are using it
without
permission.
1030
-
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45/216
AD
PROBLEM
7.32
For
the
beam
and
loading
shown,
(a) draw
the
shear
and
bending-
moment
diagrams,
(b)
determine
the
maximum
absolute
values of
the
shear
and
bending
moment.
SOLUTION
(a)
Along
AB:
Straight
with
Parabola
with
Along
BC:
\.
IA I H
8
«
(2^=0:
-~wx~V^0
V
=
-wx
r/
wL
L
2
2
XM
y
=0:
M
+
~wx
=
M
=
-~wx
2
M~
at
a--
—
8
2
ummm
M
\
a.
~Wi-
X
/
r
2
2
EM,
=0:
Af
+
U
+
-L- =
4
J
2
A/
w£
+
*,
Straight
with
(£)
From
diagrams:
M
=
—
wl
2
at
x,
-~
\V\
mi>
=~
on BC
^
l^l»ax=-r-atC
^
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc.
All rights
reserved,
to
/«/-/
„/,/,& AW*/
mop
te
dfe/rfoyerf
reproduced
or distributed
in
anyjorm
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
mine
this Manual
you
are
using
it without
permission.
1031
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46/216
PROBLEM
7,33
For
the
beam and
loading
shown,
(a)
draw
the
shear
and
bending-
moment
diagrams, (b)
determine
the
maximum
absolute
values
of
the
shear and
bending
moment.
SOLUTION
Free
body:
Portion AJ
V
=
-P
<
M
=
P(L-x)
-
8/17/2019 Cap 7, Novena Edc_text.pdf
47/216
M
f
r
PROBLEM
7.34
For
the
beam
and
loading
shown,
(a) draw
the
shear and
bending-moment
diagrams,
(6)
determine
the
maximum
absolute
values
of
the shear
and
bending
moment.
SOLUTION
(a)
FBD
Beam:
Along
AB:
x.
)
M
Straight
with
Along
BC:
/}
4
«:
£A/
C
=0:
LA
v
-M
o
=0
A
=^
EF
;
,
-0: -A
y
+C
=
C=
K
L
Z.
M
^ati?
M
ZF
y
=Q:
~^-~
y
=
Q
v
M
Mi
L
Straight
with
(/?)
From diagrams:
-H./L.
*
z.
ZM
A
,
-0:
A/
+
j-^—
Af
o
=0
M
=
A/
(l-~
A/=
—
9.
at
5 A/
=
0atC
2
l^lmax
~P
everywhere
A
IROPRlklARY
MATERIAL
a
2010 The
McGraw-Hill
Companies, Inc.
All
rights reserved.
No
part
of
this
Manual
may be
displayed,
reproduced
or
distributed
in any
jorm or
by any
means,
without
the prior
written
permission
of
the
publisher,
or
used bevond
the
limited
dmnbutton
to teachers
andeducators
permitted
by McGraw-Hill
for
their
individual
course
preparation.
Ifvou are a
student
using
this
Manual
you are
using it
without
permission.
1033
-
8/17/2019 Cap 7, Novena Edc_text.pdf
48/216
-
8/17/2019 Cap 7, Novena Edc_text.pdf
49/216
40 UN
32 W
W
V
]
=
50.5
kN
+t
LF
y
=
: 50.5
kN
-
40
kN
-
V
2
=
+)£M
2
=
0:
M
2
-
(50.5 kN)(0.6
m)
=
+|
2/^=0:
50.5-40-32~K
3
=0
+)XM
3
=0:
M
3
-
(50.5)(1
-5)
+
(40)(0.9)
=
M\
=
<
F
2
=+10.5kN
<
Af,
=
+30.3 kN
•
m
-
8/17/2019 Cap 7, Novena Edc_text.pdf
50/216
PROBLEM
7.36
(Continued)
Just
to
the
right
ofB
:
-ifi«
-0,1ft
B
W,
'37.5A.N
+|£F
v
=0:
V
4
+
37.5
=
+)LM
4
=(): ~M
4
+
(37.5)(0.2)
-0
K
4
=-37.5kN rn)
39.8
J.
50
A
C
>
£-B
jMj
max
=39.8kN-m
4
PROPRIETARY
MATERIAL.
©
2010
The
McGraw-Hill
Companies,
Inc.
Alt
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-Hillfor
their
individual
course
preparation.
If
you
are a
student
using
this Manual,
you
are
using
it
without
permission.
1036
-
8/17/2019 Cap 7, Novena Edc_text.pdf
51/216
6 kips
).2k
-i\r<
4.5
kips
C
.
/)
2ft
2 It
I:
2 ft
'
2 ft
PROBLEM
7.37
For
the
beam and
loading
shown,
(a) draw
the
shear
and
bending-moment
diagrams,
(/?)
determine
the maximum
absolute values
of the
shear
and
bending
moment.
SOLUTION
Free
bodv: Entire
beam
6
faps 12)dp*
ISfdps
+)
1,M
A
=
:
E
(6
ft)
-
(6
kips)(2
ft)
-
(1
2
kips)(4
ft)
-
(4.5
kips)(8
ft)
=
£
=
+.16 kips
+t
2/^
=
: A
y
+
1
6
kips
-
6
kips
-
1
2
kips
-
4.5 k
ips
-
/4
V
=
+6.50
kips
(a)
Shear
and bending
moment
Just to the
right
ofA
:
h
||
^h,
V\
=
+6.50
kips
M
i =
Just
to
the
right
ofC
:
6ty\ K
+JLF
v
,=0:
6.50
kips
-6
kips -K
2
=
a^i>*
totyHfl^
+>A/
2
=0:
M
2
-(6.50k.ips)(2ft) =
E
=
16
kips
M
A
=
6.50 kips
t<
-
8/17/2019 Cap 7, Novena Edc_text.pdf
52/216
PROBLEM
7.37
(Continued)
Just
to
the
right ofE:
At
B:
+\lF
y
=0: V
4
-
4.5-0
+)XM
4
=0:
-M
4
-
(4.5)2
=
^=^=0
V
A
=
+4.5
kips
-
8/17/2019 Cap 7, Novena Edc_text.pdf
53/216
US
:
300
11)
120
Hi
f
C /.)?
E
f
Mi
10 in.
5 in.
20 in, loin.
PROBLEM
7.38
For
the
beam
and
loading
shown,
(a) draw
the
shear
and
bending-
moment
diagrams,
(6)
determine
the
maximum
absolute values
of
the
shear
and bending
moment.
SOLUTION
Free body:
Entire
beam
1201b
3001b 1501b
iOfru
25iJt SOto
'
15in-
+)ZA*
C
=
0:
(1
20
Ib)(10
in.)
-
(300
lb)(25
in.)
+
£(45
in.)
-
(120
ib)(60
in.)
=
£
=
+300
lb
XF
x
=0:
C
x
=0
+t^;=0:
C
v
+3001b-120Jb-3001b-1201b
=
C
=+240
lb
E
=
3001bf <
(a)
Shear
and bending
moment
Just to
the right
of
A :
+|LF
v
=0:
-1201b-^=0
4
Just to
the
right
of
C:
iciA
|0ir>.
^?antl
V
t
'jZMfb
+t
£/*;
=
: 240
lb
-
1
20
lb
-
V
2
=
+)SM
c
=0:
M
2
+(1201b)(10in.)
=
Just to
the
right
of
D
:
2.0,0 fc
10
' '^W
+t
SFj,
=
0:
240-120-300-^=0
+)£M
3
=
0: M
3
+
(1
20)(35)
-
(240)(25)
-
0,
C
=
240
lb
f
<
^i
=
-120
lb,
M.-0
-
8/17/2019 Cap 7, Novena Edc_text.pdf
54/216
PROBLEM
7.38
(Continued)
Just
to
the
right ofE :
AiB:
+XF,,=0: F
4
-I201b
=
+)%M
4
=Q:
~M
4
-(1201b)(15m.)
=
K
4
=+ 20ib
<
Af
4
=-1800lb-in.
-
8/17/2019 Cap 7, Novena Edc_text.pdf
55/216
60
kN
A
CW D
25
kN/m
-2
m-
1
in
-2
m-
PROBLEM
7.39
For
the
beam and
loading
shown,
(a) draw
the
shear
and bending-
moment
diagrams,
(b)
determine
the
maximum
absolute
values of the
shear
and
bending
moment.
SOLUTION
Free
body: Entire
beam
+)IM
A
=
0:
B(5
m)
-
(60
kN)(2
m)
-
(50 kN)(4
m)
=
5
=
+64.0kN
XF
x
=0:
A
x
=0
+t
2F
y
=0:
A
+
64.0
kN
-
6.0 kN
-
50 kN
=
A
=
+46.0
kN
(a)
Shear
and
bending-moment
diagrams
.
From
A to C:
From
C
to
D:
555=^
For
For
+jX/1
=0:
M~46x
=
+tlF
v
=0:
46-60-K
=
+)
£M
7
=
0: M
-
46x
+
60(x
-
2)
=
A/
=
(l20-I4jc)kN-m
i
=
2m:
A/
c
=
+92.0 kN
•
m
z
=
3m: M
n
=+78.0kN-m
B
=
64.0 kN <
A
=
46.0
kN
f
<
K==+46kN
<
M
=
(46-v)kN
•
m
-
8/17/2019 Cap 7, Novena Edc_text.pdf
56/216
PROBLEM
7.39
(Continued)
From
D to B:
+|
XF
y
=0:
V
+ 64
-
25//
=
K
=
(25//~64)kN
For
+)2My=0:
64//
-(25//)
//^0:
F
B
=-64kN
/*
M
=
M
=
(64//-12.5//
2
)kN-m
M
B
=
<
(*)
I^Lx=64.0kN4
-|*t.O
-6M.0
rAfkM.irtl
«.D
f^ftftBOLfl
[ML,,
=
92.0kN-m
-
8/17/2019 Cap 7, Novena Edc_text.pdf
57/216
30 fcN/m
80
1
v>kriA«i
ezxi^
From
C
to
D:
K
V
{GO
kt4
80
kM
}•*-
2m
>j<
3
tv\
*-2vn
->J
&
C
fe
+)SM
/;
=
0:
(60
JcN)(6
m)
-
C(5 m) +
(80
kN)(2
m)
-
C
=
+104kN
C
=
104kN
J
+tsF„,
=0: 104-60-80
+ 5
=
B
=
36kNf
+\?F
V
=0:
-30x-K
=
K=-30x
+JXM,=0:
(30x)
+ A/
=
M
=
~I5x'
U
^4
-
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58/216
PROBLEM
7.40
(Continued)
From
D
to
B:
UW—
4l-
^v'>
36kb4
+JZF
;
=0: K
=
-36k.N
O^M,
=0: (36)(7-x)-M
=
M
=
~36x
+
252
V^krvj )
44
5.
M(k^-wi)
GO
(b)
|KU=60.0kN
iWU=72.0kN.m.
«
PROPRIETARY
MATERIAL.
© 2010
The
McGraw-Hill
Companies, Inc.
All rights
reserved.
Afa
/«»/
/ //?«
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
permittedby
McGraw-Hill
for
their
individual course
preparation.
If
you
are a
student
using
this
Manual,
you
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4
kins/ft
art
f
8
kips
C
D
<
5ft
w
PROBLEM
7.41
For the
beam
and
loading
shown,
(a)
draw the
shear and bending-
moment
diagrams,
(h) determine
the
maximum
absolute
values of
the
shear and
bending
moment.
2 it
SOLUTION
(a)
By
symmetry:
fi*
—
A
y
=B^8
kips
+ -(4
kips)(5
ft) A
v
=
B
=
1 8 kips
j
Along
AC:
A
J
&Ui
r*ft*
ZSXT3
l$Ufr
PC
M
(Wi^ft
Along
CD:
&
6
2ft
/P
* ,
A
_
£
4
2(r
w
H
)ZF
V
=0:
18kips-P
=
K
=
l8kips
IMj
=
0:
M
-
x(1.8
kips)
M
=
(1
8
kips)x
M
=
36kip-ftatC(x
=
2.ft)
|
E/-;
=
: J
8 kips
-
8
kips
-
(4
kips/ft)*,
-
V
=
K=10kips-(4kips/ft)x,
K
-
at
x,
=
2.5
ft (at
center)
ZM
K
=
0: M
+
^
(4 kips/ft)x,
+
(8
kips)*,
-
(2
ft + x
}
)(1
8 kips)
=
M
=
36
kip
•
ft
+
(1
kips/ft)x,
-
(2
kips/ft)*,
2
M
=
48.5
kip
•
ft
at
x,
=
2.5 ft
V,:s
-/jf
Complete diagram
by
symmetry
(b)
From
diagrams;
|
Kj
Biax
=
1
8.00 kips
on
/*C
and /;>/?
|M|
max
-
48.5
kip
•
ft at
center
PROPRIETARY
MATERIAL
©
2010 The
McGraw-Hill
Companies, Inc.
All
rights reserved.
No pail
of
this
Manual
may be displayed,
reproduced
or distributed
in
any
form
or
by
any
means, without
the prior written
permission
of
the
publisher,
or used beyond
the
limited
distribution
to teachers
and
educators
permittedby McGraw-Hill
for
their
individual
course
preparation.
If
you are
a
student
using this
Manual,
you
are using it without
permission.
1
045
-
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60/216
2.5
kips/it
*
f
t
t
.?
jljl1L_
_.
|^
6 ft
H-— 4 ft
—
f
12
kips
PROBLEM
7,42
For the
beam and
loading
shown,
(a)
draw
the
shear
and
bending-
moment diagrams, (b)
determine
the
maximum,
absolute values
of
the
shear
and
bending
moment.
SOLUTION
Free
body: Entire
beam
-3/H-
,5
k
'>
5
H*.
^
12faps
v)HM
A
=
0:
i?(l
ft)
-
(1
5
kips)(3
ft)
-
(12
kips)(6
ft)
=
B
=
+11
.70
kips
LF
X
=Q:
A
X
=Q
+\-LF
y
=Q: ^,-15-12
+ 11.70
=
4,
=+15.30
kips
(a)
Shear and
bending-rooment
diagrams
B
=
11
.70
kips
t
<
A
=
15.30
kips|
-
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From
C to B:
-Cl«BB»
^A
11,70
fci
P
1
PROBLEM
7.42
(Continued)
f|]£/V=0:
K +
l
1.70
=
+)SM,=0:
11.70//-
A/
=
M
=
(11.
70//)
kip
-ft
F
=
-11.70 kips
(A)
A/
c
=
+46.8 kip -ft
^
M
;
,=0
-
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:$]-.\
:UN
PROBLEM
7.43
H
.\:
Assuming the
upward
reaction
of the
ground on beam
AB to
be
uniformly
distributed and
knowing that
a
-
0.3 m,
(a)
draw the
shear
and
bending-
moment
diagrams,
(b)
determine the
maximum
absolute values
of the
shear
and bending
moment.
SOLUTION
(a)
FBD
Beam:
Along
AC:
A
t
mz
)
M
'Lf;
=
0: M
-
8/17/2019 Cap 7, Novena Edc_text.pdf
63/216
;UN
:ikH
n-J
\\H
1.5
id
>
PROBLEM
7,44
Solve
Problem
7.43 knowing
that
a
-
0.5
m.
PROBLEM
7.43
Assuming
the
upward reaction
of
the
ground
on h&amAB
to
be uniformly
distributed
and
knowing
that
a
~
0.3
m,
(a)
draw
the
shear
and
bending-moment
diagrams,
(h)
determine
the
maximum
absolute values
of the
shear
and bending
moment.
SOLUTION
Free
body:
Entire
beam
+|
ZF
y
=
0:
w
K
(1
.5
m) -
3
k.N
-
3
kN
=
(a)
Shear
and
bending
moment
From
A
to
C:
For
jc-0:
For
x
=
0.5
m:
From
C
to
£>:
For
x-0.5m:
For
x
=
0.75
m:
For
x
=
1 .0
m
:
+\ZF
v
=0:
4x-K
=
K
=
(4x)kN
+
)£M,=0:
M-(4x)~
=
0,
M
=
(2x
2
)kNm
K
c
=
2
kN,
+\ZF
y
=0:
4x-3kN-K
=
K
=
(4x-3)kN
+)
ZM
7
=
0:
M
+
(3
kN)(x
-
0.5)
-
{Ax)-
=
2
M
=
(2x
2
-3x
+ 1..5)kNm
w
|r
=4kN/m
<
CW«)a:S
y
A
=M
A
=Q<
M
c
=
0.500kNm
-
8/17/2019 Cap 7, Novena Edc_text.pdf
64/216
PROBLEM
7.44
(Continued)
From D
to B:
^
\ChkNfm)i
For
//
=
0.5m:
+\lF
v
=0:
F +
4//
=
F=-(4//)kN
0XM,=O:
(4//)£-Af
=
0,
A/-2//
2
K
n
=~2kN,
My-
0.500
kN-m
VCM
2,00
(/>)
in
max
=2.00kN
^
-*
0,375
C
$
£>
£~
|M|
nm
=
0.500
kN-m
^
PROPRIETARY
MATERIAL
© 2010
The
McGraw-Hill
Companies, Inc.
All rights reserved.
Ato 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.
1050
-
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65/216
•')
kips/ft
-pl^-Cft.
o
kijWll
1
B
3
it
PROBLEM
7.45
Assuming
the
upward reaction
of
the ground
on
beam AB to
be
uniformly
distributed,
(a) draw
the
shear
and
bending-moment
diagrams,
(/>)
determine
the
maximum
absolute
values
of
the
shear
and
bending
moment.
SOLUTION
Free body:
Entire
beam
Skip/it
3ft
.
—
•
3ft
+\ZF
y
=
0:
w
g
(12ft)-(3kips/fl)(6ft) =
(
a
)
Shear
and
bending-moment
diagram
s.
Prom
A
to
C
:
**
t
+1^=0:
\.5x~3x~V
=
K
=
(-1.5*)
kips
h
/
-
8/17/2019 Cap 7, Novena Edc_text.pdf
66/216
PROBLEM
7.45
(Continued)
For
x
=
9
ft:
V
D
=
4.5
kips,
M
D
= -6.75 kip
•
ft
-
8/17/2019 Cap 7, Novena Edc_text.pdf
67/216
3
KipS /lt
n
n
*-3
J't-'-h 6
ft -h-3
ft-*
PROBLEM
7,46
Assuming
the
upward
reaction
of the
ground
on
beam AB
to be
uniformly
distributed,
(a)
draw
the
shear
and
bending-moment
diagrams,
(b)
determine
the
maximum
absolute
values
of
the
shear
and
bending
moment.
SOLUTION
3
Idps*
(\SbpMi
^
(3fej»Wft-ya3*
(tstip*M)x
i
2
Free
body:
Entire
beam
+1^=0:
Wg
(}2
ft)
-(3
kips/ft)(6
ft)
=
(a)
Shear
and
bending-moment
diagrams
from
A to
C:
+)S^.
=
0: 1.5x-K
=
K
=
(1.5x)kips
(b)
\\>.
1 .5 kips/ft
<
+j£M,=0:
M-{\.5x)-
M
=
(0.75x
2
)kip-ft
M
c
=
6.75 kip
-ft
O
Forx
=
0:
For
x
=
3
ft:
K
c
=4.5
kips,
From
C to
ZX
+|lF
(
,
=0:
|.5x-3(x-3)~F
=
K
=
(9~1.5*)kips
+)
ZMj
=
0: M
+ 3(jc
-
3)—
-
(.1
.5*)-
=
A/
=
[0.75x
2
-I.5(x-3)
2
].kip-ft
For
x
=
3
ft: V
c
=
4.5 kips,
M
c
=
6.75
kip
•
ft
-
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68/216
C
ID
E
-a
>
|
«
-
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69/216
PROBLEM
7.47
(Continued)
For
x-a:
For
x
=
2a:
Vc
4
wa
M
c
wa
-
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\C
TO
E
t-fl-*-H-«-
PROBLEM
7.48
Solve
Problem
7.47
knowing
that P
=
3wa.
PROBLEM
7.47 Assuming
the
upward
reaction of the
ground
on
beam
AB
to be
uniformly
distributed and
knowing
that
P
-
wa, (a)
draw
the
shear
and
bending-moment
diagrams,
(b)
determine
the
maximum absolute
values
of
the
shear
and bending
moment.
SOLUTION
*i
[*~
a
-J\*-a
~+\*-
a
-*f«-
a
~*1
**T*
I
:a>x
Free body:
Entire
beam
+]
LF
V
-
: w
g
(4a)
-
2
war
-
3
wa
-
(a)
Shear
and
bending-moment
diagrams
From
A to C
:
5
+lxF
=0:
—wx-wx-V-Q
y
4
K
=
H
—wx
4
x
f
5
1
x
For
*
=
0:
For
x
=
a:
From
C
to D:
K~
=h
—
wa
c
4
+{ ZF„
=
:
-
wx
-wa-V-0
f
y
4
w.
=—
w
-
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71/216
PROBLEM
7.48
(Continued)
Z
__
__»
__.
X>
£ &
•^«JA
c
•;
-.
f^-
*~
;
.ftfa
_
(6)
For
x
=
o:
For
jc-2u:
K
c
=
+
—
M'a, M
r
=
+
wa
2
-
8/17/2019 Cap 7, Novena Edc_text.pdf
72/216
*l
PROBLEM
7.49
Draw the
shear
and
bending-moment
diagrams
for
the
beam
AB,
and
determine
the
shear
and
bending
moment
(a)
just
to the
left
of
C,
(b)
just
to the
right
of
C.
Free body:
Entire
beam
+)ZM
A
=0:
5(0.6
m)-
(600
N)(0.2m)
=
£
=
+200N
IF-0:
A=0
+\lF
y
=
0:
4,-600N
+ 200N
=
^=+400N
B
=
200N(<
A
=
400Nt
-
8/17/2019 Cap 7, Novena Edc_text.pdf
73/216
10.5
m
-1.5
in
-
i
*\.
m-»4«-l
in-*t*—
1-5
m
PROBLEM
7.50
Two
small
channel
sections
DF
and
EH
have
been
welded
to
the
uniform
beam AB
of
weight
W= 3
kN
to
form
the
rigid
structural
member
shown.
This
member
is
being
lifted
by
two
cables
attached
at D
and
E.
Knowing
that -
30°
and
neglecting
the
weight
of the
channel
sections, {a)
draw
the
shear
and
bending-moment
diagrams
for
beam.
AB,
(b) determine
the maximum
absolute
values
of the
shear
and
bending
moment
in the
beam.
SOLUTION
FBD
Beam
+ channels:
(a) By
symmetry: T,=T
2
=
T
\zF
y
=0:
2rsin60°-3kN =
v5
T,
x
2V3
7]
=
~k.N
-'
2
FBD
Beam:
M
=
(0.5m)^kN
2V3
=
0.433
kN
•
m
With
cable
force
replaced
by
equivalent
force-couple
system
at F and
G
Shear
Diagram:
V is
piecewise
linear
dV
dx
0.6kN/m
with
1.5
kN
/.S Aa/
/
f
£~kq
discontinuities
atF and
H.
Vp.
=
-(0.6
kN/m)(l
.5 m)
=
0.9
kN
^increases
by
1.5
kN
to
+0.6
kN
at
F*
V
G
=
0.6
kN
-
(0.6 kN/m)(l
m)
=
Finish
by invoking
symmetry
/Sh
~»/i»t*/M>*-
fS*,
->
PROPRIETARY
MATERIAL.
©
2010 The
McGraw-Hill
Companies,
Inc.
All tights
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
if without
permission.
1(159
-
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74/216
PROBLEM
7,50
(Continued)
Moment
diagram:
M
is
piecewise
parabolic
dx
decreasing
with V
with
discontinuities
of
.433
k.N
at
F
and
//.
M,,.
-(0.9kN)(1.5m)
-
-0.675