Cap 7, Novena Edc_text.pdf

8/17/2019 Cap 7, Novena Edc_text.pdf

1/216

CHAPTER

7


2/216


3/216

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.

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989


4/216

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

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rights reserved. No

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of

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990


5/216

-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.

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Inc.

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rights reserved.

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of

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Manual

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or distributed

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any

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of

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991


6/216

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

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lights

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992


7/216

-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*>


8/216

-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. ©

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part

of

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may

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distributed

in

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form

or by

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means,

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permission

of

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994


9/216

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.

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reserved. A'o /««•/

o//A« Manual

way be displayed,

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or by

any

means,

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995


10/216

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


11/216

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

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rights reserved.

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Manual may

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in

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997


12/216

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

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of

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may

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in any

form

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998


13/216

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

|


14/216

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

<

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MATERIAL. ©

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Manual may

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1000


15/216

-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


16/216

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}


17/216

PROBLEM

7.12 (Continued)

N*-

TF

y

=

0:

V~{\

25

lb) sin

30°

=

V

=

62.5 lb

^30.0°


18/216

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


19/216

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

—


20/216

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


21/216

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

%


22/216

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

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and

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for

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If

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1008


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(


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

-*--


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

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If

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permission.

1011


26/216

.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|


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


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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and

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If

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1014


29/216

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

\


30/216

PROBLEM

7.20 (Continued)

Free body:

Portion

BK

S?2H>

Z6,«ffb

For*

<

8.75

m.(flf


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

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distribution

to teachers andeducators

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If

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student using this

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1017


32/216

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

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limited

distribution to

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and

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coarse

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If

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1018


33/216

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—


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

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distribution to teachers

andeducators


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coarse

preparation.

If

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11120


35/216

\

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

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distribution

to

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individual course

preparation.

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are

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1021


36/216

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

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student using

this

Manual,

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are

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it without

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1022


37/216

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

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or

used

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distribution

to teachers

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If

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1023


38/216

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.


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

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by

McGraw-Hill

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preparation.

If

you

are

a

student

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this

Manual,

you

are using

it without permission.

1024


39/216

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

#)

=


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.

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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

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distribution

to

teachers

and

educators

permittedby

McGraw-Hill

for

their

individual

course

preparation. Ifyou

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a

student

usmg

this

Manual,

you

are

using it

without

permission.

1026


41/216

PROBLEM

7.28

(Continued)

so

M

=

(sin0

+

0cos#-sin0)

or

M=~Wr0

cos

71

dM

2»r, n

—

~~Wr(cos0-ds\n0)

=

O

at


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


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


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


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


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)


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

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Manual

you are

using it

without

permission.

1033


48/216


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


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


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<


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


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


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.


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


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


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


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

are using it

without

permission.

1044


59/216

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


for

their

individual

course

preparation.

If

you are

a

student

using this

Manual,

you

are using it without

permission.

1

045


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|


61/216

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


62/216

:$]-.\

: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


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


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.


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


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

/


66/216

PROBLEM

7.45

(Continued)

For

x

=

9

ft:

V

D

=

4.5

kips,

M

D

= -6.75 kip

•

ft


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


68/216

C

ID

E

-a

>

|

«


69/216

PROBLEM

7.47

(Continued)

For

x-a:

For

x

=

2a:

Vc

4

wa

M

c

wa


70/216

\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


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


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


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


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

Documents

Cap 7, Novena Edc_text.pdf