Chapter 6 Statics

(1)

SOLUTION

Joint FBDs:

Joint B:

B

Joint C:

PROBLEM 6.2

Using the method of joints, detennine the force in each member of the

truss shown. State whether each memberis in tension or compression.

-rF'x=

0:

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

~

4

12 -

rF'x=0: -(3.00 kN) - -

₅

₁₃

FAC=0

13 FAC=- kN

₅

FAC= 2.60 kN T ~

PROPRIETARY MATERIAL. \C 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher. orusedbeyond the limited distribution to tetWhers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

(2)

'I' :2kips 2 "ips

I

2 k.ips

SOLUTION

FBD Truss:

PROBLEM 6.10

Determine the force in each member of the Gambrel roof truss shown.

State whether each member is in tension or compression.

-

LF'x

=

0:

Hx

=

0 By symmetry: Ay

=

H

y

=

4 kips

t

~

by inspectionofjoints C and G:

H

AJ<

Joint FBDs:

Joint A:

IN

~

/P ~

f"l3

.]"

1/

file.

"I J'f's

Joint B:

y

5k'f~

Joint E:

FAC=FCE and Foc = 0 ~

FEG=FGN and FFG=0 ~

FAD= FAC = 3 kips

5

4

3 and, from above,

FAD= 5.00 kips C

~

FAC= 4.00 kips T

~

FFH

=

5.00 kips C ~

and

FCE= FEG= FGN= 4.00 kips T

~

so

-

LF'x =0:

_{- (5 kips) - - FOE- ~}

4 .

4

10 _FOD=0

5

5 ~109

3.

3

3 - (5 kips) - 2 -

~

FOD+ - FOE

=

0

5 ~109

5 so

FOD=3.9772 kips, FOE= 0.23810 kips

or

FOD= 3.98 kips C ~

FOE= 0.238 kips C ~

and, from above,

FDF= 3.98kips C ~

FEF= 0.238 kips C ~

t

LF'y= 0:

FDE- 2

~

₅

(0.23810 kips)

=

0 FDE

=

0.286 kips T

~

or distributed in any form or by any means. without the prior written permission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.

804

(3)

r

24mT2,:':)L2"~::;"T24mII2T PROBLEM 6.23

1.",1' G

-?f

'~

For the roof truss shown, detennine the force in each of the members

L

f

Jl~ B

~

~ "',

LLH'I

.~

~m

loca~edto the left ,of member GH. State whether each member is in

InoA _"

_G

/I ""'-

} ~~.,

,;rj

tensIOnor compreSSIOn.

G F I.

SOLUTION

FBD Truss:

_-~=o:

t

~=O:

Joint FBDs:

Joint A:

.ill

Joint c:

-~=o:

1111"

t

~=O:

,I,

_{Joint B:}

II. '1

" I'

1

11'

I

' I "'i, I' 'W ,Ii'

-~=o:

(12m)(My-l

kN)

- (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8m)(1.5 kN)

=

0 My

=

5.05kN

t

Ay

- 2(1 kN) - 5(1.5 kN) + My

=

0 Ay

=

4.45 kN

t

5 445 kN - 1kN - - F

AB

=

0 .

13'

FAB=8.97 kN C

~

12 FAC-13(8.97 kN)

= 0,

12 13 FCE

- 8.28 kN = 0,

5 13(8.97 kN) - FBC= 0,

FCE= 8.97 kN T

~

FBc= 3.45 kN C ~

5

5 -(8.97 kN) -1.5 kN + 3.45 kN --FBD=

₁₃

0 FBD=14.04 kN

FBD=14.04 kN C

~

12

12 -(8.97

₁₃

kN) - -(14.04 kN) + FBE=0

₁₃

FBE= 4.68 kN

FBE= 4.68kN T ~

or distributed in any fonn or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teachers Oi educators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

(4)

Joint E:

Joint D:

PROBLEM 6.23 CONTINUED

-

Ll\= 0:

-

Ll\

=

0:

6

12 r;:;;;FEH

- 4.68kN - -(8.97 kN)

=

0 v37

13 FEH=13.1388kN

or

FEH=13.14kN T...

5

1 FDE-

_

₃

(8.97 kN) -

r;:;;;(13.1388)

=

0

1 v37

FDE=5.6100kN

or

FDE=5.61kN T...

12

1 -(14.04kN)--FDG-

r;:;FDH=O

13

13 v2

5

5 -(14.04kN)

₁₃

- -FDG-1.5

₁₃

kN - 5.61kN

1 + .J2 FDH= 0

Solving:

FDG=8.60 kN C....

PROPRIETARY MATERIAL

to 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced

or distributed in anyform or by any means, without the prior written pe/mission of the publisher, or lISedbeyond the limited distribution to teache~sand

educatorspermitted by McGraw-Hillfor their individual coursepreparation.lfyou are a student using this Manual,you are !LYingit without

permIssIOn.

(5)

I'

SOLUTION

PROBLEM 6.30

For the given loading, detennine the zero-force members in the truss

shown.

Q

PROPRIETARY MATERIAL () 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced or distributed in any form or by any means, without the prior written pelmission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hil/for their individual course preparation. !fyou are a student using this Manual, you are using it without permission.

837 By inspection of joint D,

FD1= 0 ...

H/ "1/

"t"

"1/

,L

By inspection of joint E,

FEl= 0'"

l

.

»

b

1::'1

Then, by inspection of joint I,

FA!= 0'"

By inspection of joint F,

FFJ(=0 ...

(6)

PROBLEM 6.42

A floor truss is loaded as shown. Detennine the force in members CF,

EF, and EG.

SOLUTION

FBD Truss:

FBD Section ABEC:

B

It

!.--E.:&.

,""--2-""'"

t:

~,

~,.- - - -,..

1..312.$-lb

- LF::=

0:

( WA= 0:

6a(l\v- 1251b)- 5a(250Ib)

- 40(250lb)- 30(375Ib)

-20(500 lb) - 0(500 lb)

=

0 Ky

=

937.5 lb

t

~v=

0:

Ay- 3(250 lb) - 2(500 lb)

-3751b -1251b + 937.51b

=

0 Ay=1312.5lb

t

(2 ft) FCF+ (4 ft)(5oo lb)

+ (8 ft)(250 lb -1312.5lb)

=

0 FCF= 3250 Ib,

FCF= 3.25 kips T ....

1312.5Ib- 250 lb - 2(500 lb) -

Js

FEE'= 0

FEE'=62.s.J5 lb,

FEE'=139.8Ib T....

-

LF::= 0:

3250 lb + .Js (62.s.J5 lb) - FEG

=

0 FEG

=

3375 lb,

FEG= 3.38 kips C ....

PROPRIETARY MATERIAL. () 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manua/may be displayed. reproduced or distributed in any fonn or by any means. without the prior written permission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manllal, you are using it without permission.

855 t

I

:.,

"

ii

n,

I ~

. '

I'

I

, j"

'I

tJ "

1,

,,'

II il'11 ,;,

(7)

i

il:

I'

15m ~ 1.500115m 1.500 1.5...

PROBL~M 6.51

~

J

,""

en

_.

A Fink roof truss is loaded as shown. Detennine the force in members FH,

IOk!\1 I) '

~

FG,andEG.

:;k\A B J J3k!\ 2.400 .

,_

_~

"'-.j~

...

~

.

1

,

1

.

1

,

1 ~

~1.8m 1./))I} I.8m 1.1;m I.l'im-l

SOLUTION

FBDTruss:

FBD section GHK:

I

il

r

"il.

,:

-i'-- -,--~ 1.(. .

I

\ WI

.

ffi..

2.¥N, 4

!

f/

I

'

1-._

3

!

E'I>

,)

j

:

IIi

I

.

~ j ' ;

!

" , I:

!

,

I

tll:

:

,j

I

!,,\

I

Distance between loads

=

1.5 m

rflx

=

0:

Ax

=

0 By symmetry,

Ay = Ky = 18kN

t

(

rMF

=

0:

(4.5 m)(18 kN - 3 kN) - (3 m)(6 kN)

- (1.5 m)(6 kN) - (2.4 m)FEG

=

0 FEG

=

16.875kN,

FEG= 16.88kN T....

I<.

(

rMK

=

0:

(1.5m)(6kN) + (3m)(6kN) - (3.6m{

In

FFG)

=

0 FFG

=

8.0100kN,

FFG= 8.01kN T....

-+

rflx

=

0:

15

3 17FFH - .j73(8.01oo kN) -16.875 kN

=

0

FFH

=

22.3 kN C....

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.

(8)

B

D

,..

=

ci

EI

G

L

8ft~--8ft~8ft~8fi

t

Hips

f) kips ~ 12ki!"

j

I

SOLUTION

FBD Truss:

..

~ff

A

FBD Section ABC:

IJ

[

Bt.-FBD Section FGH:

PROBLEM 6.67

The diagonal members in the center panels of the truss shown are very

slender and can act only in tension; such members are known as counters.

Detennine the force in member DE and in the counters which are acting

under the given loading.

(

IMA

=

0:

(32 £1)Hy - (24 £1)(12kips) - (16 £1)(9kips)

-(8 £1)(6 kips) = 0,

Hy = 15kips

t

IFy

=

0:

Ay

- 6 kips - 9 kips - 12 kips + 15 kips = 0

Ay

=

12 kipst

Since only BE can provide the downward force necessary for equilibrium,

it must be in tension, so CD is slack, FCD

=

0 t

IFy

=

0:

12 kips

-

6 kips

-

~FRE

=

0

FRE

=

10.00 kips T

<III

Since only EF can provide the downward force necessary for equilibrium,

it must be in tension, so DG is slack, FDG= 0

t

IFy

=

0:

15 kips -12 kips - ~FEF

=

0

5 FEF

=

5.00 kips T

<III

Knowing

that FCD

=

FDF

= 0, inspectionof joint D gives

H

I

1<;"(

t

* See note before Problem 6.64.

II-p.,

I ~

PROPRIETARY MATERIAL tC 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed. reproduced or di,vtributed in any fonn or by any means. without the prior written pennission of the publisher. or used beyond the limited distribution to teachers and educators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.

(9)

101 SOLUTION

Structure (a):

~.

Structure (b):

PROBLEM

6.74 11-

_I

IIi'

.

'rti

I

"

I

ii

,.

I,ll,

i

Classify each of the structures shown as completely, partially, or

improperly constrained; if completely constrained, further classify it as

statically detenninate or indetenninate. (All members can act both in

tension and in compression.)

,

.1

I,

,

I

J

'I

I

Then ABCDGF is a simple truss and all forces can be detennined.

This example is completely constrained and detenninate. ...

I'

No. of members

ill!

I

m

=

12 No. of joints

n

=

8 No. of react. comps. r

=

3 m + r

=

15 < 2n

=

16 unks < eqns

partially constrained ...

Note: Quadrilateral DEHG can collapse with joint D moving downward:

in (a) the roller at F prevents this action.

,

continued

Ii

PROPRIETARY MATERIAL e> 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced or distributed in any form or by any means, without the prior written pennission of the publisher. or used

beyond the limited distribution to teacher:>and

educators permitted by McGraw-Hili for their individual course preparation. Jfyou are a student using this Manual. you are using it without permission.

891 Ii

ili

p

_{No. of members}

m

=

12 E

No. of joints

n=8

m + r = 16 = 2n

(10)

Structure (c):

PROBLEM 6.74 CONTINUED

No. of members

m

=

13 }

~

No. ofjoints

n

=

8 No. of react. comps. r

=

4 m + r

=

17 > 2n

=

16 unks > eqns

completely

constrained

but indeterminate

"'11II

PROPRIETARY MATERIAL. () 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed, 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

educotors permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.

(11)

PROBLEM

6.78

III

I'FA

250 mm

L

J E

-~

)i

J

"-l:.-:'---

J

-~

L

F 250 mm-j 500 111In 250 nit 200 mm

For the frame and loading shown, detennine the components of all forces

acting on member DECF.

I'

SOLUTION

FBD Frame:

(

LMA

=

0:

(0.25 m)Dx - (0.95 m)(480 N)

=

0

7=

_D

_x

=

_1824N

-

_....

FBD member DF:

_{Note that BE is a two-force}

_{member, Ex = E"}

-

~=O:

-1824N+Ex=0,

Ex= 1824N

-

....

~ --i : ,I

/%2'11'1

I

,'" -- (;I.s;, lot

V

.

~y

so

Ey

=

1824Nt....

(0.50 m)(1824 N) - (0.75 m)C + (0.95 m)(480 N)

=

0 C

=

1824 N

~ ....

t rFy = 0:

-Dy + 1824N -1824 N + 480 N = 0

D

y

=

480 N

~ ....

PROPRIETARY MATERIAL () 2007 The McGraw-Hili Companies, Inc. AII rights reserved. No part of this Manual may be displayed. reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manllal. you are using it without permission.

(12)

I:

:r

"

r

6 in.

!.-A

SOLUTION

(a) FBD AC:

I d

(b) FBD CE:

.-, II

'I" if

I

c

PROBLEM 6.86

Detennine the components of the reactions at A and E when a

counterclockwisecouple of magnitude 192 lb.in. is applied to the ftame

(a) at B, (b) at D.

Note: CE is a two-force member

(

rMA

= 0:

-(8 in.)(

~

FCE

) -(2 in.)(

~

FCE

)

+ 192 lb. in.

=

0 FCE

= 19.2J2 lb,

Ex = 19.20lb -

....

E

_y

Ey = 19.20lb

!

....

-

~ =

0:

Ax

- 19.2lb = 0,

t

~ =

0:

Ay

- 19.2lb = 0,

Ax = 19.20lb -

....

Ay = 19.20lb

t

....

Note: AC is a two-force member

FAE

=

-12.8ffi

lb,

Ax =

r.;; FAE,

4 ..,17

Ax = 51.2lb -

....

Ay = 12.80lb

!

....

-

~ =

0: Ex

- 51.2lb = 0,

t

~, =

0:

Ey

- 12.80lb = 0,

Ex = 51.2lb -

....

Ey = 12.80lb

t

....

or distributed in any fonn or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teachers and educators pennilled by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission

904 A

_"

)(

(13)

,.

.,~500li5001 ...

PROBLEM

6.98

For the

frame

and loading shown, detennine the components of all forces

acting on member ABD.

Dimensions in rom

SOLUTION

FBD Frame:

~y

A

c

F

-~=o:

(0.625m) F - (0.75m)(4 kN)- (1.25m)(3 kN)

=

0 Fx=

10.8kN-Ax

- 10.8kN

= 0,

Ax= 10.80kN -....

B

F

'-)<

f

LFy= 0:

Ay

- 4 kN - 3 kN

= 0,

FBD ABD: (II)

0,2slN'l

I:

(!Me=

0:

(0.375m)(I0.8 kN) - (0.25 m) Bx= 0,

Bx = 16.2kN,

Bx= 16.20kN-....

II: (!MD=

0:

(0.25 m)(I0.8 kN + 16.2kN) + (0.5 m) By- (1.00 m)(7.0 kN)

= 0,

By = 0.5 kN,

-10.8 kN -16.20 kN + Dx= 0,

Dx = 27 kN,

Dx=27.0kN-....

Dy = 6.50 kN!

....

7.0 kN - 0.5 kN - Dy= 0,

Dy = 6.5 kN,

or distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

918

: , I .! I i j

j;

f I'

I;

.

FBD BF: (I)

'1.y

-4B

o,z.sr.t

*

II a.

b.37M

y

)<

F

/tJ$

AN

(14)

I

t

I

'i

PI-.

12f'-):12f'

t~

B

PROBLEM 6.104

I I; 'I

''f-L'

1:=2fl

7,2ft

The axis of the three-hinge arch ABC is a parabola with vertex at B.

~,6ft

Knowing that P

=

14 kips and Q

=

21 kips, detennine (a) the

components of the reaction at A, (b) the components of the force exerted

at B on segment AB.

!II

Ii

1'1

SOLUTION

I. ~

t

"

I "'"

-2/jR-21k~

i

/1.

f't:

-r

It-I 7,2. I I I S' I ", '

I

i. .";~.

Members FBDs:

:

1 . ~;-.

_"

I

-

rr

1:( WA=O:

II:( Wc= 0:

(12.8

ft)Bx- (32 ft)By- (20 ft)(14 kips)

=

0 (7.2 ft)Bx+ (24 ft) By- (12 ft)(21 kips)

=

0 Solving: Bx

=

27.5 kips, By= 2.25kips,

1:-

~=O:

t

LFy= 0:

Ax- 27.5 kips = 0,

Ax = 27.5 kips,

(a)

Ax= 27.5kips

-

~

Ay= 16.25kips

t

~

Bx= 27.5 kips --~

By = 2.25 kips

l

~

Ay

- 14 kips - 2.25 kips = 0, Ay = 16.25 kips,

(b)

PROPRIETARY MATERIAL C>2007 Tbe McGraw-Hili Companies, Inc. AIl rights reserved. No part of this Manual may be displayed. reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

(15)

~... "O!i' \E \\ \'.

\"

c",;\

\

L'\!0.-:

,..~ :?\

~.~'//~~

"!IL

1.5in.

I.:;in.

~,.

I

PROBLEM 6.123

The double-toggle latching mechanism shown is used to hold member G

against the support. Knowing that a = 60°, determine the force exerted

onG.

.

-I

(2.5 in. + 1 in.)sin60°

I,,,.

= tan

(30.-9),'

4.5 in. + (1.5in. + 2.5 in. - 1in.)cos60°

~.

_~

-

40. £'1( '" C ___ --- --- .f.,.,

/

,'"0.

() = 26.8020

./ 2.51~~"' . I £1

~{!..

- - - -

~

h,

~~ A

l

_1.5 _in.

SOLUTION

Member FBDs:

c.

From FBD ABC:

()

=

tan-I

(:BC

+ m)sin60°

AF + (AB + BC - CD)cos60°

From FBD CDE:

(

We

=

0:

(7.5 in.)(20 lb) - (1in.)FDFcos(30°- 26.802°)= 0,

FDF = 150.234 lb C

--

LEx

=

0:

(150.234Ib)cos(26.802°)- (20 Ib)sin60° - Cx

= 0,

Cx = 116.7741b

t

'fFy

=

0:

(150.234Ib)sin26.802°- (20 Ib)cos60°- Cy

=

0 Cv

=

57.742 lb

+[(4 in.)cos600](57.742Ib)

= 0,

JiB

x

+ B.v

=

385.37 lb

(I)

continued

No part of this Manual may be displayed. reproduced

or distributed in any

form

or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers and

educators permitted by McGraw-Hill for their individual coursepreparation. lfyou are a student using this Manual. you are using it without permission.

947 ( ., .. I I

(16)

"

II

PROBLEM 6.123 CONTINUED

From FBD BFG:

(

rMG

=

0:

-(1.5 in.)[(150.2341b)sin26.802°]+ [(1.5 in.)cos300][(150.2341b)cos26.802°]

- [(1.5 in.)cos300]Bx- [6 in. + (1.5 in.)sin300]By

= 0,

J3Bx

-

9By

=

96.7751b

(2)

Solving (I) and (2):

Bx

=

243.32 Ib,

By = - 36.075 Ib

--

IF'x = 0:

243.321b- (150.234Ib)cos26.802°- Gx = 0,

Gx =

109.2261b-Gy = 31.6671b

t

rEy

=

0:

Gy

- (150.2341b)sin26.802°

+ 36.0751b = 0,

OnG,

G = H3.71b "" 16.17°....

PROPRIETARY MATERIAL ~ 2007 The McGraw-Hili Companies. Inc. All rights reserved, No part of this Manual may be displayed. reproduced or disttibuted in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.

948 "

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(17)

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:

('

(

WB

=

0:

(0.045m)FCDsin45°-6.00 N.m

=

0 ~y

e

1',

b

FCD

=

188.562N C

FBD Joint D:

FBD E:

(b) FBD BC:

-1 38.8 mm

=

22.8230

a

=

tan 92.2 mm

-1 37.5 mm - 18.8 mm

=

25.7320

f3

=

tan

38.8mm

/

LFx' = 0:

(188.562N)cos(45° -25.732°)

-FDEcos(22.823° + 25.732°) = 0,

FDE= 268.92 N C

rEx = 0:

(268.92N)cos(22.823°)-P = 0,

P = 248N -

...

_135.4 mm = 51.8570

8 = tan 27.8mm

f3= tan-I 27.8 mm - 24.3 mm = 7.55200

_{26.4 mm}

(WB

= 0:

(0.045m)FCDsin(90°-51.857° -7.5510°) - 6.00 N'm = 0

FCD = 262.00 N C

continued

i

or distributed in any form or by any means,

1,1,oithout

the prior written permission of the publisher. or used beyond the limited distribution to teachers and

educators permilled by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

(18)

I'

Ii

PROBLEM 6.128 CONTINUED

FBD Joint D:

_{r =}

_tan

-1 20 nun + 35.4 nun - 26.4 nun

_{24.3 nun + 7.5 nun}

= 42.3630

a

=

tan-I 20 nun + 35.4 nun -26.4 nun

16.85840

92.2 nun + 27.8 nun - 24.3 nun

F

---

Cc

-(262.00N)sin(90°- 7.5520°

- 42.363°)

= 0,

FDE = 196.366 N C

FBDE:

-

r.F'x

=

0:

(196.366 N)(cosI6.8584°) - P

= 0,

P=187.9N

--

~

:1

I

'il

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.

(19)

Zin.

SOLUTION

FBD top handle:

FBD Joint D:

'10II,

PROBLEM 6.139

A hand-operated hydraulic cylinder has been designed for use where

space is severely limited. Determine the magnitude of the force exerted

on the piston at D when two 90-lb forces are appliedas shown.

I~

I

Note CD and DE are two-force members

(4 in.) ..k

FCD-(1.5in.) k

FCD-(13.2 in.) (90 lb) = 0

FCD=n.J61lb

FDE = FCD = 12 .J611b

By synnnetry:

il

D= no lb ....

.1

PROPRIETARY MATERIAL (02007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using thi.rManual, you are using it without permission.

(20)

I

DhlU."n~;ons in mill

~y

(~

FBDEntire linkage:

65,

.2

, ,5-, ""- -'

J

,I) 0,1>10\ ,.. i ;p" lPy d,z.>1

l

'~.

-

t

~

-F'..

,

-

'"

{i"

1. q!>~kN

-,

_{_FIf}

F

PROBLEM 6.150

A 500-kg concrete slab is supportedby a chain and sling attached to the

bucket of the ITont-end loader shown. The action of the bucket is

control1edby two identical mechanisms, only one of which is shown.

Knowing that the mechanism shown supports half of the 500-kg slab,

determinethe force (a) in cylinder CD, (b) in cylinderFR.

(

LMD= 0:

(0.8 m)(2.4525 kN) - (0.5 m) FAD

=

0

FAD

=

3.924 kN

w

=

(250 kg) (9.81 N/kg)

=

2452.5 N

=

2.4525 kN

(LME= 0: (0.68 m)(3.924 kN) - (0.54 m)G~ FCD)

+ (0.35 m) U~ FCD)= 0,

FCD=7.682 kN

FCD=7.68 kN C ....

(LMG= 0: (2.5 m) (2.4525kN)

1

1 + (0.2 m)

J2

FGH- (0.6 m)

J2

FGH=0,

FFH=21.677kN,

FFH=21.7 kN C....

PROPRIETARY MATERIAL. <02007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.