Chapter 6 Statics

Short Description

Statics chapter 6 solution...

Description

PROBLEM 6.2 Using the method of joints, detennine the force in each member of the truss shown. State whethereach memberis in tension or compression.

SOLUTION Joint FBDs:

Joint B:

-rF'x= 0: B

1

4

.fi FAB- "5 FBC= 0

1 3 .fi FAB+"5FBc- 4.2 kN = 0 so

7 "5 FBc= 4.2 kN

FBc= 3.00 kN C

~

Joint C:

-

4 12 rF'x=0: -(3.00 kN) - -13 FAC=0 5 13 FAC=-5 kN

FAC= 2.60 kN T ~

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

796

PROBLEM 6.10

:2kips

I

2 "ips

2 k.ips

Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.

SOLUTION

-

FBD Truss:

LF'x

= 0:

By symmetry: Ay ~

H

Hx = 0

= H y = 4 kips

by inspectionofjoints C and G:

t

FAC=FCE and Foc = 0 ~ FEG=FGN and FFG=0 ~

AJ<

Joint FBDs: 'I'

FAD= FAC = 3 kips 5 4 3

Joint A: IN

/P ~ .]"

1/

so

f"l3

~

FAD= 5.00 kips C ~ FAC= 4.00 kips T

file.

"I J'f's

and, from above,

Joint B:

and

-

LF'x=0:

y

5k'f~ Joint E:

FFH

so

~

= 5.00 kips C ~

FCE= FEG= FGN= 4.00 kips T

~

4 . 4 10 - (5 kips)- - FOE- ~ FOD=0 5 5 ~109 3. 3 3 - (5 kips) - 2 - ~ FOD+ - FOE= 0 5 ~109 5 FOD=3.9772 kips, FOE= 0.23810 kips or

FOD= 3.98 kips C ~ FOE= 0.238 kips C ~

and, from above,

FDF= 3.98kips C ~ FEF= 0.238 kips C ~

t LF'y= 0:

~

FDE- 2 5 (0.23810 kips)

=0 FDE = 0.286 kips T ~

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

804

j

24mT2,:':)L2"~::;"T24mII2T PROBLEM 6.23 1.",1' G '~ For the roof truss shown, detennine the force in each of the members Jl~ B ~ ~ "', LLH'I ~m loca~edto the left ,of member GH. State whether each member is in

r

L

f

InoA _"

G

G

F

?f

/I ""'-

} ~~.,

.~

I.

,;rj

tensIOnor compreSSIOn.

SOLUTION FBD Truss:

-~=o: (12m)(My-l

kN)

- (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8m)(1.5 kN) = 0

t ~=O: Joint FBDs:

Ay

= 5.05kN

Ay

= 4.45 kN t

- 2(1 kN) - 5(1.5 kN) + My = 0

5 445 kN - 1kN - - FAB = 0

Joint A:

t

My

.

13'

FAB=8.97 kN C

~

FCE= 8.97 kN T

~

12 FAC-13(8.97 kN) = 0, .ill

Joint c:

-~=o:

t ~=O:

12 13 FCE -

8.28 kN = 0,

5 13(8.97 kN) - FBC= 0,

FBc= 3.45 kN C ~

1111"

,I,

5 5 -(8.97 kN) -1.5 kN + 3.45 kN --FBD= 0 13 13

Joint B:

FBD=14.04 kN II. '1

" I' 111' I

I"'i, 1

-~=o:

FBD=14.04 kN C

~

12 12 -(8.97 kN) - -(14.04 kN) + FBE=0 13 13

'

I' 'W

FBE= 4.68 kN

FBE= 4.68kN T ~

,Ii'

PROPRIETARY MATERIAL t!:>2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduc, or distributed in any fonn or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teachers Oi educators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

826

PROBLEM 6.23 CONTINUED Joint E:

-

6

Ll\= 0:

r;:;;;FEH -

v37

12

4.68kN- -(8.97 kN) = 0 13

FEH=13.1388kN 5

FDE-_13(8.97

kN) -

or 1 v37

r;:;;;(13.1388)

FDE=5.6100kN Joint D:

-

Ll\

= 0:

12 12 -(14.04kN)--FDG13 13

or

FEH=13.14kN T.....

=0 FDE=5.61kN T.....

1 r;:;FDH=O v2

5 5 -(14.04kN) - -FDG-1.5 kN - 5.61kN 13 13 1 + .J2 FDH= 0

Solving:

FDG=8.60 kN C....

PROPRIETARY MATERIAL to 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in anyform or by any means, without the prior written pe/mission of the publisher, or lISedbeyond the limited distribution to teache~sand educatorspermitted by McGraw-Hillfor their individual coursepreparation.lfyou are a student using this Manual,you are !LYingit withoutpermIssIOn. 827

I'

PROBLEM 6.30 For the given loading, detennine the zero-force members in the truss shown.

Q

SOLUTION

H/ l

"1/

"t"

"1/

,L.

»

By inspection of joint D,

FD1= 0 ...

By inspection of joint E,

FEl= 0'"

Then, by inspection of joint I,

FA!= 0'"

b 1::'1

By inspection of joint F,

FFJ(=0 ...

By inspection of joint G,

FGK=0'"

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

837

PROBLEM 6.42 A floor truss is loaded as shown. Detennine the force in members CF, EF, and EG.

SOLUTION FBD Truss:

- LF::= 0: ( WA= 0:

6a(l\v- 1251b)- 5a(250Ib) - 40(250lb)- 30(375Ib) -20(500 lb) - 0(500 lb) = 0 Ky

t ~v= 0:

= 937.5 lb t

Ay- 3(250 lb) - 2(500 lb) -3751b -1251b + 937.51b = 0 Ay=1312.5lb

t

t I I :., I" ii n, .I ~' I' I I , j"

FBD Section ABEC: (2 ft) FCF+ (4 ft)(5oo lb)

!.--E.:&.

B

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

,""-

-2-""'" t:~,

It

~,.- - - - - -,..

=0 FCF= 3.25 kips T ....

FCF= 3250 Ib,

1..312.\$-lb

1312.5Ib- 250 lb - 2(500 lb) FEE'=62.s.J5 lb,

-

LF::= 0:

Js

FEE'= 0

FEE'=139.8Ib T....

3250 lb + .Js (62.s.J5 lb) - FEG= 0 FEG = 3375 lb,

FEG= 3.38 kips C ....

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

"

1, ,,'

855 II

il'11 ,;,

-

i il: I'

PROBL~M 6.51

~ J en ~ 15m ~

1.500115m

1.500 1.5...

,""

~ 1111

IOk!\1 :;k\A

B . ~1.8m

I)

,

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

.

FG,andEG.

'

,_

.

_~

1./))I}

I.8m

J J3k!\ ~ "'-.j~

,

~

1.1;m

...

2.400

.

I.l'im-l

SOLUTION

rflx

= 0:

By symmetry,

Ax

= 1.5 m

=0

Ay = Ky = 18kN

t

FBD section GHK:

(

rMF

= 0:

- (1.5 m)(6 kN) - (2.4 m)FEG = 0

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

\ WI

I

I il

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

FEG = 16.875kN,

ffi.. 2.¥N, 4

FEG= 16.88kN T....

.

r

! I

"il.

,:

f/

'

1-._

!

3

(

rMK

I 2n = 16

constrained

but indeterminate

"'11II

Companies, Inc. AII rights reserved. No part of this Manual may be displayed, reproduced written

permission

of the publisher.

or used beyond

the limited

distribution

to teachers

and

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

892

PROBLEM 6.78 III

For the frame and loading shown, detennine the components of all forces acting on member DECF.

I'FA 250 mm

L

J -

E

"-l:.-:'---

)i

-~

F

200 mm

L

~ J J 250mm-j 500 111In

250nit

I'

SOLUTION

FBD Frame:

(

LMA

= 0: (0.25 m)Dx - (0.95m)(480 N) = 0

FBD member DF:

-

Ex= 1824N

-

Note that BE is a two-force member, Ex = E"

- ~=O:

--

= 1824N

Dx

7=

-1824N+Ex=0,

....

....

i :

~

so

Ey = 1824Nt....

,I

/%2'11'1

I

,'"

V

-- (;I.s;, lot

.

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

~y

t rFy = 0:

=0

C

= 1824 N

Dy

= 480 N

~

....

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

~

....

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

896

I:

r

PROBLEM 6.86

6 in.

Detennine the components of the reactions at A and E when a counterclockwisecouple of magnitude 192 lb.in. is applied to the ftame

!.A

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

:r

"

SOLUTION I d

(a) FBD AC: A

Note: CE is a two-force member "

(

rMA

= 0:

-(8 in.)(

)(

i'

FCE

~

FCE

) -(2 in.)(

~

= 19.2J2 lb,

FCE

)

+ 192 lb. in.

=0

Ex = 19.20lb -

Ey = 19.20lb

-

~ = 0:

t~

= 0:

Ax Ay

19.2lb = 0,

....

!

Ax = 19.20lb -

Ay = 19.20lb

- 19.2lb = 0,

....

....

t

....

Note: AC is a two-force member (b) FBD CE:

FAE

= -12.8ffi lb,

4 Ax = r.;; FAE, ..,17

c Ey

Ax = 51.2lb -

Ay = 12.80lb

-~

= 0:

Ex - 51.2lb = 0,

t ~,

= 0:

Ey

- 12.80lb = 0,

....

!

....

Ex = 51.2lb Ey = 12.80lb

....

t

....

.-

, II 'I" if I PROPRIETARY MATERIAL. II:>2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any fonn or by any means, without the prior written pennission of the publisher. or used beyond the limited distribution to teachers and educators pennilled by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission

904

,.

.,~500li5001

PROBLEM

6.98

..... For

the

frame

and

acting on member

: .! i

, I I

Dimensions

shown,

detennine

the components

of all forces

ABD.

in rom

j

j; I; .

f I'

SOLUTION FBD Frame: ~y A

B

(0.625m) F - (0.75m)(4 kN)- (1.25m)(3kN) = 0 Fx= 10.8kN-

c

-~=o:

F

F '-)<

Ax -

fLFy= 0:

FBD BF: (I)

Ay

10.8kN = 0,

Ax= 10.80kN -....

- 4 kN - 3 kN = 0,

FBD ABD: (II) '1.y

-4B o,z.sr.t

II a.

0,2slN'l

* )< b.37M

y

F /tJ\$

AN

I: (!Me=

0:

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

Bx = 16.2kN,

II: (!MD=

0:

(0.25m)(I0.8 kN + 16.2kN) + (0.5 m) By- (1.00m)(7.0 kN) = 0,

Bx= 16.20kN-....

By = 0.5 kN,

-10.8 kN -16.20 kN + Dx= 0,

Dx = 27 kN,

7.0 kN - 0.5 kN - Dy= 0,

Dy = 6.5 kN,

Dx=27.0kN-.... Dy = 6.50 kN! ....

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

918

I

I

'i

PROBLEM 6.104

PI-12f'-):12f' t~ .

7,2ftThe

B

~,6ft

''f-L'

axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P = 14 kips and Q = 21 kips, detennine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.

1:=2fl

t

I

I;

'I

!II Ii 1'1

I. I

SOLUTION

~

t"

Members FBDs: 21k~

i

I I

/1. f't:

"'" -2/jR-

I

II '

I 1

-

rr

1:( WA=O:

(12.8 ft)Bx- (32 ft)By- (20 ft)(14 kips) = 0

II:( Wc= 0:

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

=0

Solving: Bx = 27.5 kips, By= 2.25kips,

1:-

~=O:

t LFy= 0:

Ax- 27.5 kips = 0, Ay

Ax = 27.5 kips,

(a)

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

(b)

-

~

Ay= 16.25kips

t~

Ax= 27.5kips

Bx= 27.5 kips --~

l

By = 2.25 kips ~

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

925

.

";~.

-r 7,2. It:.

I

S' ", i.

"

~;-.

"O!i'

PROBLEM 6.123

\E

The double-toggle latching mechanism shown is used to hold member G against the support. Knowing that a = 60°, determine the force exerted onG.

\\

~... \'.

\"

\ L'\!0.-: ,..~ :?\

c",;\

(

~.~'//~~

"!IL ~,. 1.5in.

I.:;in.

I

SOLUTION Member FBDs:

()

.,

= tan-I

(:BC+ m)sin60° ..

AF + (AB + BC - CD)cos60°

. I,,,.

= tan-I

(30.-9),'

~.~ ___ --- ---

./

'"

C

.f.,.,

/

2.51~~"' I

.

40.

,'"0.

(2.5 in. + 1 in.)sin60° 4.5 in. + (1.5in. + 2.5 in. - 1in.)cos60°

£'1(

() = 26.8020

£1

~{!..

----- -- ~~

~lh, A

c.

1.5 in.

From FBD CDE:

(

= 0:

We

(7.5 in.)(20 lb) - (1in.)FDFcos(30°- 26.802°)= 0, FDF = 150.234 lb C

-- LEx = 0:

(150.234Ib)cos(26.802°)- (20 Ib)sin60° - Cx = 0, I I

Cx = 116.7741b

t

'fFy

= 0:

(150.234Ib)sin26.802°- (20 Ib)cos60°- Cy = 0 Cv

= 57.742 lb

From FBD ABC:

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

JiBx

= 0,

+ B.v = 385.37 lb

(I) continued

PROPRIETARY MATERIAL. ~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced or distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual coursepreparation. lfyou are a student using this Manual. you are using it without permission. 947

"

PROBLEM 6.123 CONTINUED

Ii

"

I I: I'

From FBD BFG:

(

rMG = 0:

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

- [(1.5 in.)cos300]Bx- [6 in. + (1.5 in.)sin300]By = 0,

Solving (I) and (2):

--

IF'x = 0:

t rEy

= 0:

Bx

= 243.32 Ib,

J3Bx - 9By = 96.7751b

(2)

By = - 36.075 Ib

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

Gx = 109.2261b-

- (150.2341b)sin26.802°+ 36.0751b = 0,

Gy = 31.6671b

Gy

t

G = H3.71b "" 16.17°....

OnG,

"

'. .

j'\ "

1.1'

I(, 11,11

'IIII .

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

948 II

PROBLEM 6.128 :.f~~~l.~ A.couple M o~magnitude6 N.m is appliedto the.input li~ of the four-?ar "j.[f"J_ =-:""... slIder mechamsm shown. For each of the two gIven pOSItIOns, determme the force P requiredto hold the system in equilibrium.

SOLUTION

(a) FBD BC:

(

=0:

WB

(0.045m)FCDsin45°-6.00 N.m = 0

~y e 1',

('

FBD Joint D:

b

FCD

a

-1 38.8 mm

= tan

92.2 mm

= 188.562N C

= 22.8230

-1 37.5 mm - 18.8 mm

= 25.7320

f3 = tan

38.8mm

/

(188.562N)cos(45° -25.732°)

LFx'= 0:

-FDEcos(22.823° + 25.732°) = 0,

FBD E:

rEx = 0:

FDE= 268.92 N C

(268.92N)cos(22.823°)-P = 0,

P = 248N -

...

_135.4 mm = 51.8570

(b) FBD BC:

8 = tan 27.8mm

f3= tan-I 27.8 mm - 24.3 mm = 7.55200 26.4 mm (WB

= 0:

(0.045m)FCDsin(90°-51.857° -7.5510°) - 6.00 N'm = 0 FCD = 262.00 N C

continued PROPRIETARY

MATERIAL

~kN

FCD=7.682 kN

;p"

'~.

,

-'

(~

J ~ '"

(LMG= 0: (2.5 m) (2.4525kN) 1 1 + (0.2 m) J2 FGH-(0.6 m) J2 FGH=0,

0,1>10\

i

lPy

-

F'..

,, - -,

FFH=21.677kN,

d,z.>1

t

FFH=21.7 kN C....

{i"

F _FIf

PROPRIETARY MATERIAL.