cap 6, novena edc_text

8/17/2019 Cap 6, Novena Edc_text

1/240

CHAPTER 6


2/240


3/240

SOLUTION

Reactions:

Joint A:

Joint C:

r» 1^. _

•iSc

PROBLEM 6.1

Using the method of joints, determine the force in each member of the trussshown. State whether each member is in tension or compression.

AB

BC

V32

+ l.252

=3.25 m

32 +4 2 =5

£A^ = 0: (84 kN)(3 m) - C(5.25 m) =

O 48 kN±^ZF X =(): Ax -C =

A T =48kN ~—

+]2F y ~0: Ay = 84kN =

A r =84kN f

±^F=Q: 48kN-~ F„=0- 1 **

13

F«=+52kN»#

+? SF„ (); 84kN-— (52k.N)~.F//c =()

/«' =+64.()kN

/^=52kN r


4/240

945 lb

BQ<

1.2 ft

-

PROBLEM 6.2

Using the method of joints, determine the force in each member of the truss

shown. State whether each member is in tension or compression.

3.75 ft

SOLUTION

Free body: Entire truss

jLIF v =0: Bx =0

+)XM B =0: C(l 5 .75 ft) - (945 )b)(l 2 ft) =

C = 720 lb |

+]ZFy

=0: 1^ + 720 lb-9451b =

B =225 lb

WSIb

3.75 ft

Free body: Joint B :

***** fji

&X^ hc ^Z15lb

F Att F ar 2251bAB _ l BC F^=3751b C <

FBC =3001b T <

Free body: Joint C:

ItoJb

£=•720 ih SBACJTftC

Fac^F bc = 120\b9.75 3.75 9

FBC =300\b T (Checks)

/^ c =7801b C ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo/>a// o/r/irs Mwi/a/ way /«? 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 astudent using tins Manual,

you are. using it without permission.

738


5/240

4 in

PROBLEM 6.3


SOLUTION


J^2.FV

=0: Cx =0 C v -0

+)ZM B = 0: (1 .92 kN)(3 m) + C/4.5 m) =

Cy ^-\.28kN C; ,.-].28k.N

+|s^=0: 5-1.92 kN-1.28kN =

B = 3.20kNt

Free body: Joint i?:

3j?of#

Free body: Joint C :

5^ a »£/f,2BkN

/*/? F ftr 3.20 kN

jUx/^0: -— F. r + 2.40fcN=08.5

/V— +2.72 kN

+1 *Fy = —(2.72 kN) - 1 .28 kN = (Checks)8.5

F^=4.00kN C <

V=2.40kN C


6/240

1 u •I. kips PROBLEM 6.4

Al

Dp

1.2 ft-

? 2.4 kips

1.2 ft-

6.4 ft

- : ' : -9 ~

Using the method of joints, determine the force in each

member of the truss shown. State whether each member is

in tension or compression.

SOLUTION

Reactions:

Joint A :

k*f*

Joint D:

r*» i i7.

p = 4.-2. U*K ? i»

\ Wt? 4 W»>* I VeCf-

+)EW i , =0: /';.(24)-(4 + 2.4)( 2)-(l)(24) =

F,, =4.2 kips

LFv

= 0: FA =0

+JSF,, =0: £>-(l + 4 + l + 2.4) + 4.2 =D = 4.2 kips |

£F=0: /^=0

+1 1/^=0: -1-/^,-0

/^=-l.kip

+1^ =0: -1 + 4.2 + ^-/^=0

Fm =-6.8 kips

15..+. ZFX = 0: —(-6.8) + Fm =

F/J/f = +6 kips

^=0 ^

F/)D = 1.000 kip C ^

FBD = 6.80 kips C <

F = 6.00 kips F ^

PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVo part of this Manual may be displayed,

reproduced or distributed in any form or by any means, without the prior written permission of (he publisher, or used beyondthe limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are. a student using thisManual,

you are using it without permission.

740


7/240

PROBLEM 6.4 (Continued)

Joint E:

1?B£

5> e** » >* EF< -4 >•

Z.4 Wtf s

+1^=0: /^ -2.4 =

Fafc . = +2.4 kips Ffl£ .= 2.40 kips T ATruss and loading symmetrical about 1

PROPRtkTARl MA1LRIAL. © 20.10 The McGraw-Hill Companies, Inc. All rights reserved. Afo part of ,lm Manual may be displayed,reproduced or distributed in any farm or by any means, without the prior written permission of the publisher, or used beyond the limiteddtslrtbutwn to teachers and educators permitted by McGraw-Hill for their individual course preparation. Ifvou are a student using ,/iis Manualyou are using if without permission.

741


8/240

JOS kips PROBLEM 6.5

Using the method of joints, determine the force in each

member of the truss shown. State whether each member is in

tension or compression.

SOLUTION

Free body: Truss

XM . =0: D(22.5)~(l 0.8 kips)(22.5)- (10.8 kips)(57.5) =

IO^Ujw IO^Wjm

D = 38.4 kips I

ZFy

= Q: Ay

= 16.8 kips.

Free body: Joints :

Free body: Joint # :

If-^a ^^g ^ 16.8 kips22.5 25.5 12

OS.

w

f*tf*>*5 >

3/


10,8 fcrp

£K=0:

XF3 , = 0:

tBC-Wi

f 3y

^js *•«*^^^ = 108kips

37 35 12

Fnc =3 1.5 kips T (Checks)-3c

F.(/?

=31.5kips T <

FAD -= 35.7 kips C ^

FBC =31.5 kips 71 ^

FBD = 10.80 kips C ^

jF cd - 33.3 kips C <

PROPRIETARY MATERIAL. © 20)0 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 by McGraw-Hill for their individual course preparation. If you are a student using this Manual,


742


9/240

900 N

900 N

2.25

D

r* 3 m

PROBLEM 6.6

Using the method of joints, determine the force in each member of thetruss shown. State whether each member is in tension or compression.

2.25 in

SOLUTION

Free Body; Truss

ISSm

SUSa

We note that AB and BD are zero-force members:

Free body; Joint A :

+)LM E = 0: F(3 m) (900 N)(2.25 m) - (900 N)(4.5 m) =

F = 2025n)

++ ZFX = 0: Es + 900 N + 900 N =

Ex = -1 800 N E x = 1 800 N —+1^=0: £

;, + 2025N =

£),=~2025N E

;, =2025Nj

FAB =F BD =0<

F*r F jn 900 NLw> AC _

i

AD2.25 3.75 3

9ot>N

Free body: Joint Z) :

t T **

VFree body: Joint j? :

3 2.23 3.75

J^£F v = 0: F^-1800N-0

+| IF = 0: FCE - 2025 N =

F^ C =675N T A

F^=1125N C A

FCD = 900 N 71 ^

F^=675N C <

F£F =.1800N r «

FC£ =2025N T <

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of /his Manual may be displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution 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.

743


10/240


11/240

4.5 in

s.4 m

'

'Pr.G'

2.8 in

HA ItN

PROBLEM 6.7


SOLUTION

Free body: Truss

A

ZBm

Free body: Joint A :

. / Eh.


£ c * a,$v**f

AC

+|S/^0: B„=0

+ JZM B = 0: D(4.5 m) + (8.4 kN)(4.5 m) =

£> = -8.4kN D = 8.4kN

-^LF x = 0: £ v -8.4kN-8.4kN-8.4kN =

£ v =+25.2kN B v = 25.2kN

Far Fac 8.4 kN

5.3 4.5 2.8/^=15.90kN C <

FAC =13.50 kN F ^

+t^=0: .1.3.50 kN-||F CD =0

Fa) =+15.90 kN

JLX^ =0: -F BC -$AkN-— (15.90kN) =

Fa) = 15.90 k.N F «

Fm, = -1 6.80 kN FRr = 1 6.80 kN C ^

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Ab ;««-/ o/rtfe Afamra/ «wy Ac displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill far their individual course preparation. If you are a student using this Manual,you are using il without permission.

745


12/240

Free body; Joint D :

We can also write the proportion


FJD= 8.4kN4.5 2.8

^=13.50kN C ^

/> D _ 15.90 kN

4.5 5.3FB/) = 1.3.50 kN C (Checks)

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



746


13/240

12 ft

PROBLEM 6.8

Using the method of joints, determine the force in each memberof the truss shown. State whether each member is in tension orcompression.

SOLUTION

Reactions:

Joint D:

tf*>jrC£>

£

JZft

BCD = Vl2 z +16 =20 ft

£/^=0: Dx =0

+)l,M E = 0: Dy (2l ft) - (693 lb)(5 ft) =

+tlF v =0: 1651b-6931b + £ =

±-ZF x =0: ~F AD+ UDC =:012 f . 3+t^ =0: -/^+ JF /JC+ I651b =

Solving (I) and (2), simultaneously:

^d^ ~260 lb

FDC =+125 ib

£33

D..=1651b

E- 528 lb

(1)

(2)

7^= 260 lb C <

f^c =125 lb T <

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 limiteddistribution 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.

747


14/240

Joint E:


£ce

(3)

(4)

Solving (3) and (4), simultaneously:

/r^=_8321b

Fa; =+400lb

/^.= 832 lb C <

FCI; = 400 \b T <

Joint C:

e«t

4 .,is» &~*&m*° -

Force polygon is a parallelogram (see Fig. 6. 1 1 p. 209)

FAC = 400 lb r^FBC =1251b f ^

Joint v4: _t.SF v =0: —(260 lb) + -(400 lb) + F B =0

.v^*fit.

FAB =-420\b

£/«-„ = : —(260 lb) - -(400 lb) =y

.13 5

= (Cheeks)

7^= 420 lb C A

PROPRIETARY MATERIAL. © 20.10 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 a student using this Manual,


748


15/240

•J.fikN

10.5 feN 10,5 kN

5.7 kN D(\

-3.8 m- -3.2 m- -3.2 ni- -3.8 in-

fo. 7 kN

4 mT/iA

PROBLEM 6,9

Determine the force in each member of the Pratt rooftruss shown. State whether each member is in tensionor compression.

SOLUTION

Free body: Truss

^£>M/o.Skd

10.5 kM

5V? Urt 5.7 WM

£F=0: ^ =0

Due to symmetry of truss and load

A =H=- total load = 21 kNJ

2

Free body: Jo int / :

5.7 k*J

12.1

3 * *« 3S %

F,„ /< ,.. 15.3 kNB _ * ,4C37 35 12

F„, = 47. 1 75 kN F.,. = 44.625 kN


to.Skd47.i75^»i- >r lJ0.5kW

5-:47.175 M |f.

From force polygon: Fm - 47. 1 75 kN, F^ - 1 0.5 kN

F /IR =47.2kN C <

F, r =44.6kN T A

7^= 10.50 kN C A

FBD =47.2kN C<

PROPRIETARY MATERIAL. CO 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 limiteddistribution 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.

749


16/240


Free body: Joint C: +|zf;-0: 1/^-10.5 = ^/> = = 17.50 kN T <

FBc -*/0.SM4 pff«

44.625 kM > + Jf G_+. 2F V = 0: FC7 , +-(17.50) -44.625 =

FCE = 30.625 kN ^ = 30.6kN T ACFree body: Joint is: £>£' is a zero-force member ^ == A&•*

,- , t^* FT

Truss and loading symmetrical about C


distribution to teachers and educatorspermitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,


750


17/240

1.5 m,

.1 .5 id ^1.5 in . .1 .5 m 1 .5 in 1.5 m

'2k.NI l 2 ^ |2kNlkN'

f2 in

M

PROBLEM 6.10

Determine the force in each member of the fan rooftruss shown. State whether each member is in tensionor compression.

SOLUTION

Free body: Truss

EFx = 0: A,=0

From symmetry of truss and loading:

I

I

Total load

15m ISra 15 m 15 m Urn 15m

lkN

tint2feN 1

2kN2kN

k IlkN

,*

(SETi2ra

t^ GA

>(

= 1 = 6kN'


A, = 6kM

F ;I0 /%, 5kNC9.849


LF9.849

(12.31 kN + Fm +F DC ) =

or

or

Add (1) and (2):

Subtract (2) from (1):

^/tf> + **BC -12.31 kN

+tlF,

Fbd ^bc

9.849

-7.386 kN

(12.31 m+ Fm -F BC )-2kN =

2F, BD

2F> BC

-J9.70kN

-4.924 kN

FAB -12.31 kN C ^

F, iC =11.25kN r ^

(2)

F/iD =9.85kN C <

FDC ^2A6kH C <


751


18/240


Free body: Joint D :

p I' *..«&>^

h*-^ ^-JH5

?


19/240

sOO ib

Am

600

—8 ft-

(iOO Ih

I-IL

f300lh fift

¥p I iI i

X 1 6 l't

5.v,v:~. V.l±E G

-*—«)>— —8 ft- —file—

PROBLEM 6.11

Determine the force in each member of the Howe rooftruss shown. State whether each member is in tension orcompression:

SOLUTION

Free body: Truss

300 fb

eotiih

600 lb

3Wlb 6ft

A h-8fi-H*-8i

£F V =0: H v =0

Because of the symmetry of the truss and loading:

Free body: Joints ;

A~H v -~ Total loady2

A = .H= 1200.1b

UAnt-r W6t

f% =r Woo fftc

9eoli>

^ = 7^ = 900jb5 4 3

F(/

, = 1 5001b C

F /jC =1200 lb T

Free body: Joint C:

BC is a zero-force member

ixooib c f ce

F«,=0 F , =1200 lb r

PROPRIETARY MATERIA ,, & 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 limiteddistribution 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.

753


20/240


21/240

600 lb

600 if) I 600

500 II) B%D)

\*~S ft -8 ft-

2 ft 4 in.

MM.) Ibl

6 a

-8 ft-

1/J

-8 ft-

PROBLEM 6.12

Determine the force in each member of the Gambrel rooftruss shown. State whether each member is in tension orcompression.

SOLUTION

Free body: Truss

£/ *

, = 12()0ib(

^^c^OOlb5 4 3

or

or

F„„=0BC

&JO&

Cj £ C|

•-8 ft--4--a ft-4^-8 (t~4—' '4

F. R =1500Ib C


22/240


Multiply (1) by (3), (2) by 4, and add:

100^= -120,000 lb ^no = 1200 lb C <

Multiply (1 ) by 7, (2) by -24, and add:

500F BE = -30,000 lb Fbe = 60.0 lb c <

Free body: Joint D:

60oti>i

jt_LFv

=0:24 24±-(1200 lb) + —/w-0

Fdf =

—(1200 lb)

= -1200 lb

—(-1.200 lb) -25

Pdf

-6001b-

-12001b c <

Fde'- = 72.0 lb Fde = 72,0 lb T <

Because of the symmetry of the truss and loading, we deduce that

F - FEf - r BE Pef = 60.0 lb c <

Fbg=Fce Peg -12001b T <

t FG = t BC Ffg = <

F = FFII

1

AB Ffii= 1500 lb

C MF = F

gii ' ac ^oii -12001b T A

Mote: Compare results with those of Problem 6.9.

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



756


23/240

J2JjkN L2.5kN 2.5kN 2.5kN PROBLEM 6.1

IJti^JtLJ; * Determine the force in each member of the truss shown.

o.

rI /J

SOLUTION

12.5 kN FCD _F DC

F/JC =32.5kN C

Joint G: \XF = 0: Fca =0 4

JhA t Sfkf** a 'l* /2F = 0: Fre =32.5kN C «

Joint C: £.F = f (2.5m) = 1.6667 m = ZBCF = tan ' —-39.8 1 ° /^. i'^,

3 2 %7-fe^+|EF

1,=(): - 12.5 kN-/v 7 , sin /? = () -^

-1 2.5 kN - 7^,, sin 39.81° = ^

FCF = -l 9.526 kN FCF = 1 9.53 kN C <..+. E/ ; ; = 0: 30 k.N - FliC - Fa , cos/? -

30 kN~F /JC -(-19.526 kN)cos39.81° =

FBC = +45.0 kN FfiC - 45.0 kN 71 «

Joinl£: ±»ZF X =0: ~--^-/> /; - -—(32.5 kN)-A' cos^ =6.5 6.5

i-cr.

Pyt

'po~t***»* FEh . = _ 32 .5 k N ~(M 1(1 9.526 kN)cos39.8Ajk^Gf 3?,SK# 6

Jf^ r 7v =-48.75 kN 7v-48.8kN C

+JXFV

=0: F;„, --^-F

Fr --^(32.5kN)-(19.526kN)sin39.81o

=6.5 ' 6.5

2 5/; V -—(-48.75 kN)- 12.5 kN - 1 2.5 kN =

6.5

FSP = +6.25 kN FBp = 6.25 kN 7'

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> ,«»•/ o/'/A/v Mv/»«// ;««v te r/K/j/«v«/,reproduced or distributed in any form or by any means, without the prior written permission of the. publisher, or used bevand the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using if without permission.

151


24/240


Joint B: tanff = ^^-; r=51.34°2m

+| £/


25/240

(> ill

.2kN

a in 6 tn

f .v

2.4


26/240


Multiply (1) by 2.5, (2) by 3, and add:

~F fl0 +—(6.24 kN)- 7.2 kN = 0, F/WJ = -4.16 kN, F/M> =4.16kN C<

Multiply (1) by 5, (2) by -12, and add:

Free body: Joint C:

Bt \ y,83lfr3V&3

45

3.905F,, r + 28.8 kN = 0, FBr = -2.50 kN, FBC = 2.50 kN C <

+1 2F = 0: -5— jr ——(2.50 kN) -1 y 5.831

u>3.905

FCD = 1.867 kN T A

hfsMm F t6 ^ jjp = 0: Fr , - 5.76 kN +^^-(2.50 kN) + (1 .867 kN) =* CL3.905 5.831

Free body: Joint E :

F« F«=UkN

E^a.tkW

Fc/ , = 2.88 kN r

+| £F, = : -^~ F„, + 3.6 kN - 1 .2 kN =

Fnr = -3.75 kN Fn; , = 3.75 kN C ^:>/;

.+_ IF. - 0: - Fr/ , (-3.75 kN) =

FC£ =+2.88kN Fc/ ,=2.88kN T (Checks)

PROPRIETARY MATERIAL. O 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo /wr/Y 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



760


27/240

^Lu-wft-H^wft^H^L

aS1.2 ft

CI K

—18 ft*-J-

—18 ft>f* —IS ft-

ft to] is {> kips

PROBLEM 6.15

Determine t he forc e in each member of the Warren bridgetruss shown. State whether each member is in tension orcompression.

SOLUTION

Free body: Truss ZFX = 0: A.-O

Due to symmetry of truss and loading

Ay =G-~ Total load = 6 kips


Free body: Joint B:

Free body: Joint C:

&t-7.5k^rf*~ c >

St-Siir* +.2F Y =0: FCE -4.5-j(7.5) =

+l ^7-= +9 kips

Truss and loading symmetrical about 1

4»k»fS» (a l«p*

5

- ^c _ 6 kips3 4 ^/i

= 7.50 kips C <

/5* 1/ Fac= 4,50 kips r <


28/240

9 ft

Am

-18 ft- 18 ft-9 ft

C 1

// \v

EJ

•

1 8 ft *\> —1 8 ft —J-* —18ft-v' ¥

() kips ft kips

T12ft

*

JXc.«3«f* 6 Wf*

+tlM„=0: ~(5) + |f c/j -6 =

+t2F,=0: FCE +|(2.5)-|(5)-3 =

Fcn = 2.50 kips T


29/240

Free body: Joint D :

&r^fc >>S— Jsg

f cfi *2>5k; f s

Free bodv: Joint F :

fois^S »oe*

§?'*^ 44

¥^F, F&Free bodv: Joint G :

Also:

if* 4

3 kff

PROBLEM 6,16 (Continued)

+J2^ = 0: -|(2.5)-|^=0

FDP = -2.5 kips

+.ZF x = 0: ^+6-|(2.5)-|(2.5) =

Fo; , = -3 kips

f kf _ fro. = 3 kips5 5 6

/rfct? _ 2 kips

jM2UV*

Peg-

fv*

5 4

/> G = 2.50 kips C (Checks)

.^ = 2.50 kips C <

FDF =3M kips C ^

F£F = 2.50 kips T ^

Fm = 2.50 kips C ^

F£G = 1.500 kips 7 ^

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No par, of , his Manual may be displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manualyou are using it without permission.

763


30/240

5.76 (t

4Do

:

Mt M

6,72 ft

5,76 It

800 lb

¥«

5.76 ft

800 lb

'; /»

1.0.54ft--

5.76 ft

S00 ilt

'/

100 l .i

II

(

;

-12.5ft H

PROBLEM 6.17

Determine the force in member DE and in each of themembers located to the left of DE for the inverted Howeroof truss shown. Slate whether each member is in tension or

compression.

SOLUTION

Free body: Truss :

cL_ «£a cf* d*5.78ft ,.&76ft . 5.78^. 5,76ft.

Angles:

ZF,=0: A v =0

+) L/V/,., - : (400 lb)(4 J) + (800 lb)(3 J) 4- (800 lb)(2 d) + (800 lb)tf - Ay (4c/) =

Av

= 1600 lb]

6.72tan a

tan ^

10.54

6.72

23.04

a - 32.52°

/? = 16.26c

Free body: Joint ,4 :

AKm IbJ fc *7A(f

F ,i« 12001b

sin 57.48° sin 1 06.26° sin 1 6.26c

FAB =3613.8 lb C

F i/ ,=4114.31b T /•-,«= 36 10 lb C, /


31/240


y^ lb t Zb

V/ L- *


+/ZFy

=0: -F BC -(800lb)cosl6.26° =

f bc -768.0 lb Fsc = 768 lb C «

n^EFv

-0: FM) +36l.3.8Ib + (800lb)sinl6.26°==0

FgD - -3837.8 lb Fm = 3840 lb C «Free body: Joint C :

F *. (tfwib

«r«^

+t EF, = 0: -F C£ sin 32.52° + (4 1 1 4.3 lb) sin 32.52° - (768 lb) cos 1 6.26° =

FCE = 2742.9 lb Fc/ ,=2740l.b T<J_ 2F V = 0: Fco - (41 1 4.3 lb) cos 32.52° + (2742.9 lb)cos32.52° - (768 Ib)sin 1 6,26° -

Fc/> = 1371.4 lb /^-13711b r «Free body: Joint F:

f« /%£

V /^ rlWfc St/r

E«,

F,/)/ 2742.9 lb

sin 32.52° s.in73.74 cFm. = 1.536 lb C <


765


32/240

5.76 ft

100 lb

A?

6.72 ft

5.76 ft

800 lb

-.1.0.54 ft-

5.76 ft

800 lb

,-;&V .J :.• .••• •:q.-£

D

5.76 ft

800 i}>

'F

>o- . : ;::. vcr-:

'100 il>

Sll

.1.2.5 ft H

PROBLEM 6.18

Determine the force in each of the members located to the

right of DE for the inverted Howe roof truss shown. Statewhether each member is in tension or compression.

SOLUTION

Free body; Truss

dc

•1054ft- •ULSft-

+ )ZM , =0: H{Ad) - (800 \b)d- (800 b)(2rf) - (800 lh)(3d)- (400 lb)(4rf) =

H =1600 lb

Angles:

Free body: Joint // :

tana = ~ — a^ 32.52°10.54

tan /f=:-^- /? = 16.26 (23.04

F-6tt

FGW =(1200 lb) cot 16.26

F0// = 4114.3 lb F

^/:,

12001b

sin 6.26c

4285.81b

FGH =41 10 lb T A

FF// =42901b C <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. M> 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



766


33/240

Free body Joint F :

f F6 * nzsjBk

Free body: Joint G:

J

f«


+/ XFy = 0: -F ro - (800 lb) cos 16.26° -

FFG = -768.0 lb FTO = 768 lb C ^

n^ ^ =0: ~FDF - 4285.8 lb + (800 lb) sin 1 6.26° =f df ~ ~ 406 - 8 lb ^V = 4060 lb C <

+\ZFy

=0: FOG sin32.52°-(768.01b)cosl6.26° =

7^= +1371.4 lb

F nrt = 13711b T 4C

±^IF x = 0: -F EG +4114.3 lb- (768.0 lb)sinl6.26°- (1371.4 lb) cos 32.52° =

FgG = +2742.9 lb FEG = 2140 lb r <

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 limiteddistribution 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.

767


34/240

i UH

'-\

2kN

IK.

I.

1 kN

/A

c.

2 in 2 in 2 ni 2- in 2 m

/ _L

in

1 m

zTt-l '»

PROBLEM 6.19

Determine the force in each of the members located to the

left of FG for the scissors roof truss shown. State whethereach member is in tension or compression.

SOLUTION

Free Body: Truss

IkN

y am ' 2m

XFv

=0: A v =0

IM A = 0: (I kN)(l2 m) + (2 kN)(l m) + (2 kN)(8 m) + (1 kN)(6 m) - Ay (1.2 m) =

A v =4.50kN

We note that BC is a zero-force member:

Also: i'ac = f ac

Free body: Joint /f :

5»

4s<hc

- ZF^ 0: V5^ + ^^ = °

+1 £/\. =0: U-/^ B + -U/-V +3.50 kN =0

FK =0<

0)

(2)

>/2'^ >/5(3)

ArHoOkH

Multiply (3) by 2 and add (2):

4i/^ B -7k N=0 /^ B =9.90kN 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 the limited



768


35/240



From (I):

Free body: Joint B:

2fcN

>--BE

l

+ IF,.

FAC - 3.50 kN = FAC = 7.826 kN

f ce^ f ac^ 7 - 826 kN

0: -~. w/) + —(9.90 kN) - 2 kN =

FAC = 7.83 kN T <

FCE = 7.83 kN .7* «*

F,, D = -7.07 1 kN f ' n =7.07 kN C <

+* £F V = 0: FBfi + -~(9.90 - 7.07 1)kN =

FRF = -2.000 kN /'«, = 2.00kN C ^

Free body: Joint E:

^i %

r rV *zezbktlJ-CE

Free body: Joint D:

or

or

Add (4) and (5):


+ £F.

+t^;

0: ~y{F eg ~ 7.826 kN) + 2.00 kN =

.F/,c;

= 5.590 kN F£G = 5.59 kN r ^

: f oe —t= (7-826 - 5.590)kN =FnP = 1.000 k'N FnF =1.000 kN T 4

+ SF 0: ^(^df +^w) + ^(7.071kN)

Fry,+F nn =-5.590 kN

2kHi f»F

r7.o7UN

+ IFI

0: -7-(^i>/.- - ^/x; ) + -^(7.07 1 kN) = 2 kN - 1 kN =

^-'V;=^-472

2F D/ ,= -10.062 kN

2F /x; =-U180kN

(4)

(5)

F/w , = 5.03 kN C ^

F/x; = 0.559 kN 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 bevond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using il without permission.

769


36/240

'

m^c

2kN

m

1

i kN

2 in 2 in 2 m 2 m 2 m 2 tu

: :l m1 in

-.Q-.-i^

; I in

\\ in

I, |I

m

PROBLEM 6.20

Determine the force in member FG and in each of themembers located to the right of FG for the scissors rooftruss shown. State whether each member is in tension or

compression.

SOLUTION

Free body: TrussUN

+)I,M A = 0: 1(12 m) - (2 k.N)(2 m) - (2kN)(4 m) - (1 kN)(6 m) =

L- 1.500 kN

Angles:

tanar = l a = 45°

tan/? = - P = 26,57°' 2

Zero-force members :

Examining successively joints K, J, and I, we note that the following members to the right of FG are zero-force members: JK, IJ, and HI,

Tims:

We also note that

and

FGI ~ FIK - FKl

FHJ - FJL

(1)

(2)




770


37/240

Free body: Joint L :

From Eq. (1):

From Eq. (2):

Free body: Joint H:

Free body: Joint F :

h7 \ f fif S,03kNJ*


***• hsooub x nr

FJL F, KL 1.500kNsin 1 16.57° sin 45° sin 18.43°

FJL = 4.2436 kN

f g, = *fc = *k

^/w =^1= 4.2436 kN

E&U f^MUM -Fit

F;/ =4.24kN C<

F^-3.35k N r ^

FG/ =F /je =3.35kN r ^

F/A/ =4.24kN r ^

ia*3r*-

'V//

sin 18.43°

4.2436

sin 53.14°Ffh = 5.03kN C <

sin 108.43°

F G -.lkN + (5.03kN)sin26.57 o ~(-5.03kN)sin26.57° =

FFG = 3,500 kN /i-. G =3.50 kN T ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distnbuted in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student mine this Manualyou are using it without permission.

771


38/240

J2(10 lb

911 1 #D.

(I i't a it

6 It B ft

1(H) Ih

400 ii> I '(00

( )() lb I .,& I 4(50

6 It

F

c^L-t>:s>-.';- -'i o:-- -o s^;o L -

6 ft

3 n

9 ft-*

PROBLEM 6.21

Determine the force in each of the members located to the left

of line FGH for the studio roof truss shown. State whether each>0ti lb member is in tension, or compression.

..¥/.

Oft —

SOLUTION

8ft , 6ft

4*»Jb

400 tb

4001b400 lb

400 ih

Gft

A ye. _

fts' 1 200 lb

H

400 lb

>xm* S^f B t

-f ab

Free body: Truss £/? =0: A v =

Because of symmetry of loading:

I ^/I, = I Fotal load

A, = L = l200lbj

Zero-Force Members . Examining joints C and //, we conclude that BC, EH, and GH are zero-force members. Thus

^IiC- F£H

Also,

Free body: Joint /J

From Eq. (I):

Free body: Joint B

±* £F. = 0:

or

or

^^ / (: (I)F /IB F

s[000 lb

F„, =2236lb C/WJ

F,,, = 2240 lb C <

FAC = 2000 lb T <

FCE =2000 lb r A

+]£F V =0: - r /-';

J^*>+ ^^ +^(2236 lb) =

Fm + FBB = -2236 lb

jj />o-

JjFee + jy

(2236 lb) - 400 lb =

(2)

FBD /^ -1342 lb (3)

PROPRIETARY MATERIAL. © 20 10 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



772


39/240

ffoo/i> l »F

£ De=&oo& 106


2F m = -3578 lbdd (2) and (3):Subtract (3) from ( 1 ) : 2F m = -894 lbFree body: Joint E

7^=1789 lb C ^

/^= 447 lb C ^

±*XF T -0: -~F i?o +-~(4471b)-2000Ib =

F, Y> = 1789 lb 71 ^

+] 2/W+/w.-=-17891bF ' ' DC (4)

]

+1 E/\. - 0: -~=rF or ~^),g + 4r (1789 lb)

Fdf~Fdg= ~2236 lb (5)

+ 600 lb -400 lb =

or

Add (4) and (5): 2F D/ , = -4025 lb Fw, = 20 1 lb C <

Subtract (5) from. (4): 2F 0G = 447 lb FDG = 224 lb T <

PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. AH rights reserved. A'o pro-/ o///ife Manual may he displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution 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.

773


40/240

1,6 in . 1.6 m . 1.6 m 1.6 in . 1.6 m 1.6 m

1.2 k-N

,1 I1UV

ft/-li^.. ** 1 ,

f.c

i^v?

fy=y,«>wr«

-9.39 W ' V0/

Free body: Truss

£F V =0: A v =0

ZMK = : (1 .2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1 .2 k.N)3«

-/L(6a) = A =5.40kNJ

Free body: Joint /J

F,i/} ___

Flf; __

4.20 kN

VJ 2

FAB =9.3915 kN F /4/} =9.39k.N C <

p u , =8.40 kN 71 ^

Free body: Joint i?

-±. 1F X = 0: ~|-F go +-~.F /}C +-|(9.3915) - (1)

+1 ZF V =0: -^ --L.^ c + -L(9.3915)-2.4 = (2)Add (1) and (2):

-j= f bd + -r=(9.39I 5 kN) - 2.4 kN =

Fm = -7.6026 kNMultiply (2) by -2 and add (1):

-~=Ffi

+4.8kN =v2

FBC = -2.2627 kN

FBD = 7.60 kN C <

FIK: = 2.26 kN C <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this 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



lib


41/240

^.MOkrt t «


Free body: Joint C

—^--0: ^F co +-=^+-ir(2.2627)-8.40==0 (3)1

Multiply (4) by -4 and add (1):

(4)

—j=F CD + —(2,2627) - 8.40 =/V ;/) =-0.1278kN 7^ :D =0.I28kN C <

Multiply (1) by 2 and subtract (2):

~j=F CE + -—(2.2627) - 2(8.40) =

/;;.,.= 7.O68 kN Fc;£ - 7.07 kN f ^Free body: Joint D

¥2,1 K*i E-

-t-X/^0; ^dp+- — Fw,+ -W7.6026).524

m'

V5V y

+~^(0.i278) = (5)

+ -W0. 1278) -2.4 =v5

Multiply (5) by 1.15 and add (6):

(6)

3.30 3 30 3 15--j^Fw + -—(7.6026) + ~(0. 1.278) - 2,4 -v5 V5 v5

FDF = -6.098 kN Fn/ , =6.1.0 kN C


i|if ot . _-|(0.,278) + 4.8 =

^ =-2.138 kN FnP =2..14kN C ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Aii rights reserved. M> /«»/ o//Afa AAwiki/ «m>- /« displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution 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.

in


42/240


43/240

G~-~

PROBLEM 6.24

The portion of truss shown represents the upper part of a powertransmission line tower. For the given loading, determine the

force in each of the members located above HJ. State whethereach member is in tension or compression.

SOLUTION

UMJ

WtbXtf

~A0

r


44/240


=t.viun Ice

^gSO.fcOOkN

Free body. Joint C

2.21±~ S/\. = 0: Fr/ , + —(2.29 kN) =

/-^=-2.21k.N FCE

= 2.21kNC ^

0.6+1 ZF V = : -/>.„ - 0.600 kN —(2.29 kN) =1

•

c 2.29

Fai = -1 .200 kN Fc// = 1 .200 kN C <

Free body: Joint E

±* T.F., = 0: 2.2 1 kN - ^-(2.29 kN) - - FFII =2.29 5 ''

FOT =0


45/240

I M)>,

.'IV

N

r ••

S,

,':

jfo.fiO 111

jO.fiO 111

T.20 m

O.fiO »i

O.fiO Ill

.20 »i

_L| O.fiO in

1 0.60 iii

PROBLEM 6.25

For the tower and loading of Problem 6.24 and knowing that

P ai = Fgj = 1.2 kN C and F £ y./ = 0, determine the force inmember HJ and in each of the members located between HJ andNO. State whether each member is in tension or compression.

PROBLEM 6.24 The portion of truss shown represents theupper part of a power transmission line tower. For the givenloading, determine the force in each of the members locatedabove HJ. State whether each member is in tension orcompression.

SOLUTION

G

lA'kN

2ki/OHt r* -±&i^ 3.03

r^n

F v^.-.


46/240


1

,905 RH

*^&L fafw* 83 ^—*r3.03kH

* *F

-IK

£„»I.B«0kNJjk-

Free body: Joint I

2.97JU. LF r = : FIK + -^—(3 .03 kN) =IK

3.03l

Fyjf=-2.97kN

F/A= 2.97 kN

C <0.6

+| XF = : -F w - .800 kN —(3.03 kN) =3.03

F,„=-2.40kNIN FlN = 2.40kN C <

A JUi5*.MkK

Free body: Joints

4 ? 97±.XF=0: —F™ +2.97 kN ——(3.03kN) =

5Kh

3.03

^=0*

+llF v =0: -F, 0.6a/- -; :-«— (3-03 kN) - 1 .800 kN - -(0) =3.03

Fro =-2.40kN F^ = 2.40 kN C


47/240

0.(50 m0.60 in

PROBLEM 6.26

Solve Problem 6.24 assuming that the cables hanging from the

right side of the tower have fallen to the ground.

PROBLEM 6.24 The portion of truss shown represents theupper part of a power transmission line tower. For the givenloading, determine the force in each of the members locatedabove HJ. State whether each member is in tension orcompression.

SOLUTION

f^.

4*

Zero-Force Members .

Considering joint F, we note that DF and EF are zero-force members:

FDF =F EP ^<Considering next joint D, we note that BD and DE are zero-forcemembers:

Fbd=Fde=*<

A&

f Ml•»c

M»

flB

I Aft ' * *

-6c

^ _ ^c 1.2 kN2.29 2.29 1.2

Free body: Joint B

-±*EF=0: -F,2.21

F /}K =2.29kN 7 ^

F jr = 2.29 kN C ^

5BE

2.29(2.29 kN) -

F,}/

, = 2.7625 kN F/yf = 2.76 kN r ^

+| £F V =0: - /> c - -^- (2.29 kN) - - (2.7625 kN) =

FBC

=-2.2575 kN FliC

= 2.26 kN C

PROPRIETARY MATERIAL. © 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 the limiteddistribution 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.

783


48/240


Free body: Joint C

2.21

10.6 i,. p

-+*. XFr = : /w + -^-(2.29 kN) =

0.6

FC£ =2.21kN C ^+ £/v = 0: -/v 7 , - 2.2575 kN —(2.29 kN) =v ui

2.29

Fay = -2.8575 kN Fcn = 2.86 kN C <

Free body: joint ff

Fg^zjurku

-r c


49/240

I Oft—

15 kins

in

ill I

o it

/J>

40 kips f %|

}


50/240


Multiply (1) by 6, (2) by 5, and add:

40,_ /s B +230 kips =

V29

/^= -30.96 kips

FAB=31.0 kips

C <Multiply (1) by 2, (2) by -5, and add:

40-=F w,-110kips =V61

Fm - 2 1 .48 kips FfiD =21.5 kips T <Free body: joint D

+| £F y = : - r F/fD —r (33 .54) + --L (2 1 .48) =v5 v5 v61

f^= 15.09 kips T <

fjp* 33.5V kips +~ £/%. = 0: FDE + 4-0 5.09 - 33 .54) —~(2 1 .48) -v5 v61

F0/; - 22.0 kips T <

A Free body: joint A

-- L//29 V29 >/5

(4)

F/fC - -28.27 kips, F 4C = 28.3 kips C <

Multiply (3) by 2 (4) by 5 and add:

o 2() 1

/^,= 9.50 kips T A


786


51/240


2ff.l7Kt'f*

ft6

,

ft* &%*«**(

Free body: Joint C

From force triangle

i^___ FCG __ 28.27 kips

JfT\~ 8 ~V29

Free body: Joint G

±* £/? = 0:

FC£ = 41.0 kips r ^

FCG = 42.0 kips C <

F, n =0


52/240

4S kN

*—4 m—*+«— 4 in— JSl D

—4 in —*- -*— 4 in—*j

4,5 in

S-

PROBLEM 6.28

Determine the force in each member of the truss shown. Statewhether each member is in tension or compression.

SOLUTION

Free body: Truss

£FV

=0:

+)ZM n = 0:

+|£/V=0:

H,=0

48(16)-

1 92 - 4b

46w k 1C « ^Ah«T a d ^i

G'(4) = G = 192kNJ *£* '-4 *

+ By = £f

'*

Hv = -144k.N H,, =1.44kNj

Zero-Force Members

Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE,DE, DG, FG

Thus, ''liC ~ ^BE ~ ^Dli ~ rjxi ~ **FG ~ '

Also note: p _ p ~ p __ J? / n

Free body: Joint /(: ^_^ ( c_48kN8 ^73 3

8 Jac 13 L>

F }fi = 1 28 kN F^ = 1 28.0 kN T <

FAC = 136.704 kN F/JC = 136.7 kN C M

From Eq. (1): FBD = FDF =F fy ^ \2S.0kN T 4

From Eq. (2): f ce = ^c =136.7kN C ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o /«w7 «/'//zi'.v Manual may be displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,


788


53/240


Free body: Joint .//

Ffm

144 kN

9

144- kN

J, /H

Also

144 kH

f';//

=192.7 kN C

&«

it®*

FFn = .144 kN8 ~ 9

Fni = 1.28.0 kN 7 (Checks)

(/ F&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 the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using (his Manual,you are using it without permission.

789


54/240

r /.

PROBLEM 6.29

Determine whether the trusses of Problems 6.3 la, 632a, and 633aare simple trusses.

-%

i'~ • ::Oi -o :

* / i: n I i. c


55/240

yipi

//c

1 1- ..•::-'%°S : ;; „J7/?»;;-

°c ?%^ .'.; *L

Bo : : /Y. k'% '%N:;/?(. m \

a£(b)

12

A a c D E

//To/

. -P..

z:::::y3:t;:~;~:

•' O :.'.

'V

Jk.. K N O &3 J'

M

J._^

— o

,

i .

?()

(t)

PROBLEM 6.30

Determine whether the trusses of Problems 6.31/?, 6.32/?, and 6.33/?are simple trusses.

SOLUTfON

)~-« » 4 - « » j » . « - «j»-4— «|

Truss of Problem 6.31/? .

Starting with triangle ABC and adding two members at a time, we obtainsuccessively joints E, D, F, G, and //, but cannot go further. Thus, this truss

is not a simple truss A

Truss of Problem 6.32/?.

Starting with triangle CGH and adding two members at a time, we obtainsuccessively joints B, L, F,A,K, J, then H, D, N, I, E, O, and P, thus completingthe truss.

Therefore, this truss is a simple truss

Truss of Problem 6.33/).

Starting with triangle GLMand adding two members at a time, we obtain joints Kand / but cannot continue, starting instead with triangle BCH, we obtain joint Dbut cannot continue, thus, this truss

is not a simple truss M

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 limiteddistribution 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.

791


56/240

J-o:' : o

: '. '/&•• •.' • •'-• I/'

&5¥ K M( )

/4

£ A(i>

PROBLEM 6.31

For the given loading, determine the zero-force members in each

of the two trusses shown.

SOLUTION

Truss (a) FB: Joint B: />. =

FA: Joint C: FCD = Ihq.FB: Joint ./: F„ = o^¥)^NFfi: Joint /: F7i =

^TY YTligFB: Joint JV: FMV =

( fl )

FB: Joint M: FlM =

The zero-force members, therefore, are BC, CD, A/, /I, L M, MV ^

Truss (b) FB: Joint C: F„,- -0

FB: Joint B: FM,

FB: Joint G: FK;

FB: Joint F: Fw,

FB: Joint E: Fm p\B: Joint /: Fu =

(*)

FB: Joint M: F v/A , =

FB: Joint JV: F^v

=

The zero-force members, therefore, are BC, BF, DF, FF, FG, /J , A7V, MV ^

PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill 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-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.

792


57/240

PROBLEM 6.35* (Continued)

Substituting for FCA , ¥CBt ¥CD , and equating to zero the coefficients of i, j, k :

24 ,., 24 ,

J : ~7 f ac ~ 400 Jb = /r = 1 040 lb r «26

Substitute for FAC and Fco inEq. (1):

24 24-—(10.401b)-— (2F BC )-0 FBC --5001b F/iC = FCD = 500 lb C^

Free body: B

(HBD^f /** FbC= (5 °° Ib) ^ = ~ (48 ° Ib)l + G 40 lb)R

^-^ii —= -^-(10j-7k)B BA ]2 .2]Frd ~~ f bd^

XF - 0: F^ + FSD + F //c + (480 lb)i + (200 lb) j =

Substituting for FBA9 FBDt F flC and equaling to zero the coefficients of j and k:

j: ^ F^ +200Ib==0 f ab --244.2 lb /^ 244 lb C *k:

~l^T^- / ^ +140ib =

F/iD =~T22T

(~ 244 2 lb) + ] 40 lb = +28 ° Ib Fm = 28 ° lb T *

From symmetry: Fad^ f ab F- 244 lb C<

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVe port o/'/W* 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. limiteddistribution 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.

797


58/240

0.8 m

i.H m

PROBLEM 6.36*

The truss shown consists of six members and is supported by a

ball and socket at B, a short link at C, and two short links at D.

Determine the force in each of the members for P = (-2 1 84 N)j

and Q = 0.

SOLUTION

Free body: Truss

From symmetry:

Thus

Fr ee body: A

tmty

fi»li

Dx = Bx and Dy = By

ZFx =0: 2B X =0

BX =DX = Q

XFz =0: BZ ~Q

ZMCZ =0: -2£ v (2.8m) + (2.184N)(2m) =

fs-OietH)^

By = 780 N

AB fi AllAB 5.30

(~0.8i-4,8j + 2.1k)

—f=S (2i - 48j)Ar —Fas-^^^ = |t (+0 - 8i - 4 - 8J 2 - lk)SF = 0: F^+F 4C +F /1D -(2184N)j =

B = (780N)y <

PROPRIETARY MATERIAL © 2010 The McGraw-Hill 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



798


59/240

PROBLEM 6.34

Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.

1.6 m I .6 m 1.6 itt , 1,6 in 1.6 m 1,6 in

1.2 kN2,4 kN

2.4 kN1.2 kN

2.4 m 1.8 m

t

0.9 m

48 kN

4 m -*-}-•— 4 in— >f«— 4 m—*(-«— 4

A m A *

1.8 m 1.8 m I.8111

SOLUTION

(*) Truss of Problem 6.23

F#: Joint 1: Fu =0

Fff: Joint J: F6V =0

FB: Joint G: FGH =

Bl Di FCKrt- - —

_

4.5 m

W'

L6m WtnL6m,Wm 1.6m

1.2 kN

t«m

14 m L8ra JL8 m ' lit oi ' L8

14 171

The zero-force members, therefore, are

(*) Truss of Problem 6.28

FB : Joint C: FJC =0

FB : Joint B: F/f£ =0FB: Joint F: FD£ =0

FB : Joint £>: FDG =0

FB : Joint F: FFG =0


GH\GJ,U


60/240


61/240

a¥ B

<

1I&*.

PROBLEM 6.32

For the given loading, determine the zero-force members in eachof the two trusses shown.

A B C D EPf ~~-yQ awes; ;\-: - -:.--c î~~ .-..::::c, s

•FPX CM -01 JfjrIP ' ' l '' ' •' $ ' ' .::B&:y. .. sBssssid p

-#s K Lj ill N O 3fS

(W

SOLUTION

Truss (g) F5: Joint 5: FR/ ~0

FB: Joint D: FD[ =0

F#: Joint £ : F£/ =0

/TC: Joint/: F /(/ =0

W?: Joint F: /^ =0

F£: Joint G: FGK =

F5; Joint /f: Fcx =


Truss j bX FB: Joint K: FFK =

FB: Joint O: FIO ~0

(a)

AI, BJ,CK, DI, El, FK, GK 4


AD other members are either in tension or compression.

FK and 10 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual way be displayed,reproduced or distributed m any form or by any means, without the prior written permission of the. publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manualyou are using it without permission.

793


62/240


63/240


Substituting for FAB , ¥ AC , FAO , and equating to zero the coefficients of i, j, k:

j:

k:

^W- + ^)~^c-2184N«0

(1)

(2)

2.1

5.30

Multiply (I) by ~6 and add (2):

1.6.8

(^~^) = o r AD r AB

5.20

Substitute fori^ c and F JD in (1):

F 4C -2184N = ) F 1C =-676N

0.8

5.302^ +

5.20(-676N) = 0, FAR = -861.25 N

F fr = 676 N C ^

^=^.d=861N C <

Free body: i?

F^ = (861 .25 N)— = -(130 N)i - (780 N)j + (341 .25 N)kAn

FBC ~ FBC2.8/ -2.1*

3.5^(0.81 -0.6k)

Fflfl = -^ k

SF = 0: F^+F sc + FB/J +(780N)j =

Substituting for F^ , FBC , ¥ BD and equating to zero the coefficients of i and k:

i: -1 30 N + 0.8^=0 FBC - +162.5 N

k: 341.25 N - 0.6 FBC - F^ =

FBD =341.25- 0.6(1 62.5) = +243 .75 N FBD = 244 N 71 «

F. r =162.5 N f ^

From symmetry: FCD=^BC FCD = 162.5 N r

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> /*** / ,/,« Mwm«,/ m» te rffcpftwreproduced or distributed m any form or by any means, without the prior written permission of the publisher, or used beyond the Untileddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manualyou are using it without permission.

799


64/240

PROBLEM 6.37

Q A,

0.8 m-

,D --1.S m

/O. ,c2.) in

The truss shown consists of six members and is supported

by a ball and socket at #, a short link, at C, and two short

links at D. Determine the force in each of the members for

P = and Q = (2968 N)i.

SOLUTION

Fr ee body: Truss

From symmetry:

Dx = B, and Dv = B v

2/^=0: 2B X + 2968 N =

/>\. =£) v =- 1.484 N

£A/„- = : - 2By

(2.8 m) - (2968 N)(4.8 m) =

9«(2468Wi

Thus

Free body: A

(me H)k

B v = -2544 N

r All ~'

'AB

AB' AB

= ils_(_o.8i-4.8j + 2.1k)5.30

F — FAD ~ l AD

AC 5.20ADAD

-^-(-0.8i-4.8j-2.1k)5.30

B = -(1 484 N)i - (2544 N) j O

ZF = 0: F /(B + F , r + F m + (2968 N)i =

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this 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



800


65/240


Substituting for F. (g ,F^ c , FAD , and equating to zero the coefficients of i, j, k:

°' 8 ~ ' AH


6(2968 N) = 0, F, (; --5512N VV-5510N C^

Substitute for FAC and F,, D in (2):

^rr I ^,(c 6(29uo > v - »> /,,


66/240

PROBLEM 6.38*

The truss shown consists of nine members and is

supported by a ball and socket at A, two short links at B,

and a short link at C. Determine the force in each of the

members for the given loading.

SOLUTION

Free body: Truss

From symmetry:

From symmetry:

Free body: A

A2 =B z =0

£/?=0: 4 =

1MBC =0: 4,(6ft) + (1600lb)(7.5ft) =

4 =-2000 lb

B, = C

ZFy

= : 2By

- 2000 lb - 1 600 lb =

5Jf

= I8001b

2F = 0: F^+F, c +F^-(20001b)J =

Fn A+* + Ftr ±Z* + F^ (0.61 + 0.8 j) - (2000 lb) j =

A = -(20001b)j


67/240


Substitute for FAD and FAC into(l):

2

4iFAB + 0.6(2500 lb) = 0, YAB = -1 060.7 lb, FAH =F, r =m\\b C A

Free body: B F« = ^*Jj= +0060.7 lb)itS- = (7501bXi + k)

Vbc=-Fbc*

^/)=^(0.8j-0.6k)

(7.5i + 8j-6k)„, =FBE ~ Fbe1

ft/' TapBE —* BE BE 12.5

SF - 0: Fw + IV- + F /w + FM+ (1 800 lb) j =Substituting for FBA , Fyjc , ¥ BD + FBA- and equate to zero the coefficients of i, j, k:

7jTi: 750 lb +

j: 0.SF BD +

12.5

8

12.5

Fbe=0, F„=-12501b

(-1250 lb) + .1800 lb =

FBE -12501b C ^

F„„ = 1250Ib C A

k: 7501b-F BC -0.6(-12501b) —̂(-]2501b)-0

From symmetry:

Free body: DJ>

FflC =21001b T <

F llD ^F C0 =)25Q\b CM

^r^kLF = 0: F /M +F D/? +F„,.+F n ,i =

We now substitute for FDA ,F DB ,F DC and equate to zero the coefficient off. Only ¥ DA contains i and itscoefficient is

-0.6 FAD = -0.6(2500 lb) = -1 500 lb

i: -\50Q\b + FOB =0 FfW =1500 lb T <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limile,distribution to teachers and educators permitted by McGraw-Hill for their individual course, preparation. If you are a student using this Manua,you are using it without permission.

ding this Manual,

803


68/240

o ^•p'ffiL

II A

/ Ii ft

i -

^0.6 111

-^W - 1 -

PROBLEM 6.39*

The truss shown consists of nine members and is supported by a ball

and socket at 5, a short link at C, and two short links at D. (a) Check

that this truss is a simple truss, that it is completely constrained,

and that the reactions at its supports are statically determinate.

(b) Determine the force in each member for P = (-1200 N)j and Q = 0.

0.6 m^X^T £.^ ^ .1.2 m ~^- t '0.75 in

'

SOLUTION

Free body: Truss

EMfl

=0: 1.8ixCj + (i.8i-3k)x(D;j + Dk)

+ (0.6i - 0.75k) x (- 1 200 j) =

-1.8Ck + 1.8D y k-1.8DJ

+3£>„i-720k-900i =

Equate to zero the coefficients of i, j, k:

i: 3D -900 = 0, DV =300N

Pr-(t20dN)^

k:

Dz =0,

1.8C + 1.8(300) -720 =

EF ^ 0: B + 300 j + 1 00 j - 1 200 j =

= (3OON)j ' /«? 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. astudent using this Manual,


804


69/240

Free body C:

Fcftl J fee


XF = 0: F^ + FCB + F CD + Fc/ , + (1 00 N)j = 0, with

,, „ CA F, r ,

^=^c^ = J^(-1.2l + 3J-0.75k)^ 7f = -Q.6QK)i

^,:>=~~F CD k FCff = Faf ^ = -^ r (-l.81-3k)Cc 3,499Substitute and equate to zero the coefficient of j, i, k:

3j:

3.317F, c + 100N-0, F, r =-\ 10.57 N

-3S7 H,a57)-

,60 -3S9^ =0 - '--233.3

k: —2iZ5_(__i io.57) - Frn --l_(-233.3) -3.317

Free body: D3.499

FAC =\IQ.6N C <

Fa ,=233N C <

FCD =225N T <

ZF = 0: FD/I +F /x ,+F M+(300N)j = 0, with

^,-^^ = ^(-1.21 + 3] + 2.25k)f dc = ^ = (225 N)k F/J£ = -F DE i

Substitute and equate to zero the coefficient of j, i, k:

F3

3.937 7Vad + 300N-0, FAD = -393.7 N,

k;

Free body: £

-^-|(-393.7N)~F w.-03.937 J

D/l

2 25 ^• *- (-393.7 N) + 225 N = (Checks)

3.937 J

.^ D -394N C <

/^ = 1200N T <

Member/*/? is the only member at £ which does not lie in the xz plane.Therefore, it is a zero-force member.

F„,-0 «

PROPHi&TARYMAlk*1AL. «P 201.0 The McGraw-Hill Companies, Inc. All rights reserved. Jfo /*„-, tf *fr MW„«, ft, displayedreproduced or distributed in anyjonn or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGmw-HiU for their individual course preparation. ^ou are using it without permission. '

H05


70/240

l'2'H)ll>)k

T//':

/:

i /10.08 It

PROBLEM 6.41*

The truss shown consists of 18 members and is supported by a

ball and socket at A, two short l inks at B, and one short link at G.

(a) Check that this truss is a simple truss, that it is completely

constrained, and that the reactions at its supports are statically

determinate, {b) For the given loading, determine the force in

each of the six members joined at E.

j/^ 0.60 ft

'11.00 ft.

SOLUTION

(a) Check, simple truss .

(1 start with tetrahedron BEFG

(2) Add members BD, ED, GD joining at D.

(3) Add members BA, DA, EA joining at A.

(4) Add members DH, EH, Gil joining at //.

(5) Add members BC, DC, GC joining at C.

Truss has been completed: It is a simple truss

Free body: Truss

Check constraints and reactions :

Six unknown reactions-ok-however supports at A and Bconstrain truss to rotate about. AB and support at Gprevents such a rotation. Thus,

Truss is completely constrained and reactions are statically det erminate

Determination of Reactions :

1MA = 0: 1 l.i x (B y \ + BM) + (1 li - 9.6k) x G\

+ (1 0.08 j - 9.6k) x (27 5i + 240k) =

1 1 By

k - 1 \B y j + 1 lGk + 9.661- (1 0.08)(275)k

+ (1 0.08)(240)i - (9.6)(275) j =

Equate to zero the coefficient oft, j, k:

i: 9.66' + (10.08)(240) = G = -2521b

j: -ll^-(9.6)(275) = S z =-240.ib

k : 1. 15,, + 1 K-252) - (10.08)(275) = 0, By = 504 lb

G = (-252Ib)j


71/240


£F = 0: A + (504 lb) j - (240 Ib)k - (252 lb)+ (2751b)i + (2401b)k =

Zero-force members

The determination of these members will facilitate our solution.

A = -(275lb)i~(2521b)j <

FB: C . Writing

FB: F . Writing

FB:A: Since

FB: H: Writing

£f>0, l.F v =(), £F 2 =0

2^=0, ZFy

= 0, ZFz =0

4 = 0, writing SF Z =

ZFy

^0

Yields F„ Pa, F Y: =()


72/240


Free body: E £F = 0: ¥FB + Fw + (240 lb)k - (252 lb)j =

-i^-(lli-10.08j) + -^-(ni-9.6k) + 240k-252j = ~-(Ztt«ft14.92 14.6

Equate to zero the coefficient of y and k:

j: Jl^W-252^0 FM =373Ib C


73/240

Al

(240 Hi) <

11.00ft.

\c.

fv If \ 10.08 ft

\ /

/, 4^9.(50 it

?x

&S^M

PROBLEM 6.42*

The truss shown consists of 1 8 members and is supported by aball and socket at A, two short links at B, and one short link at G.(a) Check that this truss is a simple truss, that it is completelyconstrained, and that the reactions at its supports are staticallydeterminate, (b) For the given loading, determine the force in each

of the six members joined at G.

SOLUTION

See solution to Problem 6.41 for Part (a) and for reactions and zero-force members.

(b) Force in each of the members joined at G.

We already know that

Free body:/:/ Wx = Yields Fa „ = 275 lb C <Free body: G IF = : FGB + $ 0E + (275 Ib)i - (252 lb) j =

^~H0.08j + 9.6k) + ^H.li + 9.6k) + 275i-252j =u.yz ]4.6

Equate to zero the coefficient of i, j, k:

11

j:

14.6

10.08

13.92

Fw, +275 =

/V, -252 =IK

k:9 - 6

V-348) + f96

13.92 14.6(365) = (Checks)

6=-(?5-?%-

^c; = 365 lb T <

FBG =3481b 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 the limiteddistribution 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.

811


74/240

6.25 It -s^ .12.5 ft 12.5 ft(

12££t 12.5 ft

15 II

m d\ in in J\—o - . ::a ' o. ;o: -'. ' . :... c

K I G

1 2.5 ft 1 1 2.5 ft X 1 2.5 ft .1.2,5 I'lwOf 5(H) th (jtlOOtli

^K

12.5 it

PROBLEM 6.43

A Warren bridge truss is loaded as shown. Determine the force inmembers CE, DE, and OF.

SOLUTION

6.25ft 12,5 ft 12S ft^ teS ft 12.S ft^

bP dY p \ h\ )\

u 8 \ K4

A

c E C /\ —>

12.5 ft. 115ft, Us ft 12.5ft 115ft 3fiOUfllb fiOWlb

6.25 A , 12.5 ft,

OOWlb

Free body: Truss

J_Xf ;.-(): k v =0

+ )ZA^ =0: A/62.5 ft)-(60001b)(12.5 ft)

-(6000lb)(25ft) =

k = k v =3600ibt 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 educatorspe rmitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,


812


75/240

6.25 l> x . 12.5 ft . J 2,5 ft 12.5 It 1.2.5 It

15 ft

i:

B\ D\ F\ III J\—,o;v • :•.:..::, or,-; .;.:_:: ,o •. .. ..— o ~:rr: ' ~c

PROBLEM 6.44

A Warren bridge truss is loaded as shown. Determine the force inmembers EG, FG, and FH.

- J —-^:z^. r ^i:..r/:::z^:- ::y.'d:.:: .6':.-::.-:i; . .:d fcv I e i a

1.2.5 fl A 1 2.5 ft 1 12.5 I'l 1 2.5 It i 2.5 It¥ ?

6000 li 600 ) .

SOLUTION

See solution of Problem 6.43 for free-body diagram of truss and determination of reactions:

A -8400 lb k = 36001b <

We pass a section through members EG, FG, and FH, and use the free body shown .

+) SA/ f = 0: (3600 ib)(3 1 .25 ft) - FK0 (1 5 ft) =

125 ft 125 ft , 6,2? ft

F c * c

12.5 ft 125 ft K^3fc00|b

J^. , = +7500 Jb

+X.F=0: -ii-;

16.25Fw + 36001b =

/r= -3900 lb/y;

+) 2MG = 0: FFH (1 5 ft) + (3600 Ib)(25 ft) =

FFH =-6000 lb

FEG= 7500 lb

T <

Fm = 3900 lb C <

FF/I = 6000 lb C


76/240

35 kN - A PROBLEM 6.451

2.4 in

^^Determine the force in members BD and DE of the truss shown.

__••:;*; ;_ : .;.._:*;.. _ : :. :. . . . ...*•. . . , . . . . . . - - : . - -

Cr

i]

2.4 mj

JL_- 35 kN ( D ^5> Ikp

2.4 m 1 ^H«

J_ t{ r:--.—— —-:-:I 4.5 in '

SOLUTION

Member BD:

+)I.M E =0: FSD (4.5m)-(135kN)(4.8m)-(135kN)(2.4m) =

FBD =+2\6kN FBD =2\6kN T <

Member DE:

++ XFX = 0: 1 35 kN + 135 kN - Fm =

FDJ ,=+270kN /7, £ = 270kN T A

PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. Ail 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 educatorspermitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,


814


77/240

2,4 tn

2.4 m

2.4 m

PROBLEM 6.46

Determine the force in members DG and EG of the truss shown.

r7JL2

4.5 in

SOLUTION

135 kM4% 5 , Member DG:

15i F=0: 1 35 kN + 1 35 kN + 1 35 kN + —Fnr =17

/,G

F/x; =-459kN

-1E&

FDG = 459 kN C <

Member EG:

+) LM D = 0: (135 kN)(4.8 m) + (135 k.N)(2.4 m)

+/- K (4.5m) =

FfiG = -2 .1 6 kN FEG =216 k.N C ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. jVo /mi-/ o///im M*m«// may te displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution 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.

815


78/240

2 cN :) kN 4k\ :.UN 2 kN 2 ? E G? i/f 1L0.4 m

3rr

PROBLEM 6.47

A floor truss is loaded as shown. Determine the force inmembers CF, EF, and EG.

SOLUTION

Free body: Truss

-±-£F=0: /< =0

2kN 4kN 4kN 3kN 2kN 3kN 1 kN

+)I.M A = : Ar y (4.8 m) - (4 kN)(0.8 m)

-(4 kN)(l .6 m) ~~ (3 k.N)(2.4 m)

- (2 k.N)(3.2 m) - (2 kN)(4 m) - (1 kN)(4.8 m) -

ky

= 7.5 kN

Thus: k = 7.5kN|


79/240

2 kN 4 kN ^f


80/240

I kN

0.46 m I

2 £ =

2.16 m 9

A = I^4kN \<

iw zy* f %9.6 m 40

-I m -5 1Z2.4 m 12

0.46 m 0.46 mA«tfcN

940

2,0444 mSlope A/

+)T.M D = 0: FC£ -(1 m) + (1 kN)(2.4 m) - (4 kN)(2.4 m) =

FCi .=+7.20kN FCE =7.20kN T <

i')T.M K = 0: (4 kN)(2.0444 m)-(J k.N)(2.0444 m)

-(2 kN)(4.4444 m) -[ FDE 1(6.8444 m) -v 13 )

F ni: . = -1 .047 kNDh Fnr ,~ 1.047 kN C M E = 0: (1 kN)(4.8 m) + (2kN)(2.4 m) - (4 kN)(4.8 m)/ 40 l

-F Df (1.54 m) =41

/W=-6.39kN /? = 6.39 kN C ^/


818


81/240

1 kN0.46 tu

aa: c

2kX

D)

2 ;N2 1.X

i k-N

2.4 m 2.4 in 2.4 in 2,4 tn

PROBLEM 6.50

|A pitched flat roof truss is loaded as shown. Determine the force

2.62 in in members EG, GH, and HJ.

SOLUTION

Reactions at supports : Because of the symmetry of the loading,

Ax =0

Ay

^7=1 (Total load) = - (8 kN

We pass a section through members EG, GH, and HJ, and use the free body shown.

A = I = 4kNt<

UN

TJ^I^MW

+)£M W =0: (4kN)(2.4m)-(lkN)(2.4m)-^ c (2.08m) =

FiT; = +3.46 i 5 kN FAG = 3.46 kN T ^

+) ZMj = : - FG// (2.4 m) - FEG (2.62 m) =

2.62Fgh

Fgh

2.4

-3.7788 kN

(3.461 5 kN)

±^=0: -F w2.4

2.46

2.46

F^-0

//./

F,//./

2.4

-3.548 kN

F67/ =3.78kN C <

2 46F ^ —:

—(3.4615 kN)2.4FH , = 3.55 kN 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 the limiteddistribution 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.

819


82/240

I .fi kips

1 .6 kipsI

I .(> kipsPROBLEM 6.51

A^^C

1.6 kips . .—I ——p—r- ^ Howe scissors roof truss is loaded as shown. Determine ^> l;_J_

6 the force in members DF, DG, and EG.

k ^1;

8 ft 8ft 8 ft 8ft 8ft 8ft

SOLUTION

Reactions at supports .

Because of symmetry of loading.

Ax = 0, Ay = I =i (Total load) = - (9.60 kips) = 4.80 kips

A = L = 4.80 kips] -^

We pass a section through members DF, DG, and £G, and use the free body shown .

We slide F^ to apply it at F:

)

ZMa = 0: (0.8 kip)(24 ft) + (1 .6 kips)( 1 6 ft) + ( 1 .6 kips)(8 ft)

- (4.8 kips)(24 ft)8/ ^ ,(6ft) =

3, Tft-

3ft

V82

+3.52

FD/ , = -1 0.48 kips, FDF = 1 0.48 kips C A

\-)'LM A =0: - (1 .6 kips)(8 ft) - (1 .6 kips)(l 6 ft)

2

£fm=,(\6 ft)- ,SFdg

(7 ft) =8' +2.5' V8

2 + 2.5 :

F/XJ =-3.35 kips, F^y =3.35 kips C A

+ )LM n = 0: (0.8 kips)(16 ft) + (1 .6 kips)(8 ft) - (4.8 kips)(16 ft)8^g-

/82 + 1.5

2(4ft) =

FEG = + 13 .02 kips, FKG = 1 3.02 kips T A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No pari of this Manual may be displayed,reproduced or distributed hi any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If yon are. a student using this Manual,you are using it without permission.

820


83/240

1 .6 kips

l.fi kips I 1.6 kips

1,6 kipsJ ,.f

I S .6 kips

o.s kip | ,/

1 A// 0 .8 k in

A If C(.

fi>*-$L 4-5 n

K

6 ft

PROBLEM 6.52

A Howe scissors roof truss is loaded as shown. Determinethe force in members GI, HI, and HJ.

8ft 8ft 8ft 8ft 8ft 8ft

SOLUTION

Reactions at supports . Because of symmetry of loading:

Ax = 0, Ay = L - -(Total, load)

= -(9.60 kips)

- 4.80 kips

We pass a section through members GI, HI, and HJ, and use the free body shown .A = L = 4.80 kips

CJJsLh Ukip>0.6 Wo

~e«-^~ ek-^

+ )ZM ri =Q:\6F G1

(4 ft) + (4.8 kips)( 1 6 ft) - (0.8 kip)( 1 6 ft) - (1 .6 kips)(8 ft) =

Fa/ =13.02 kips T <

162 +3 2

FGI =+13.02 kips

+) ZML - : (1 .6 kips)(8 ft) - FHI (1 6 ft) =FHI =+0.800 kips

Fw = 0.800 kips T

We slide VHG to apply it at //.

+ )ZM / =0:8F, HJ


84/240

:H-N PROBLEM 6,533kN

,5 '-A'

3kN

3kN

o:.

'

o'. ..

o•/•

'o:. - :o:

3 m 3 m 3 m 3 »i 3 in 3 m

A Pratt roof trass is loaded as shown. Determine the force inmembers CE, DE, and DF.

SOLUTION


By symmetry:

Total load = 5(3 kN) + 2(1 .5 kN)

= 18kN

^ = £ = _(] 8 kN)

3k#3tov

2kN

Tv¥\ 1 w>k \A = L = 9kN <

Note: Slope of ABDF is

Force in CE:

6.75 = 3 ^3*9.00 4

+ )LA/ /) =0: F(>: —x6.75m3

3kN

3kN , ^

fi** P -

* CI t£r^

3b» 3» 3*>

&?r»

(9 kN)(6 m) + (1 .5 kN)(6 m) + (3 kN)(3 m) =

Fa ,(4.5m)-36k.N-m-0

FCE = +8 kN FC£ =8kN r ^

Force in DE:

i-)2M A = : FDE (6 m) + (3 kN)(6 m) + (3 kN)(3 m) =

FDE = -4.5 kN Fm = 4.5 kN C <


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.

822


85/240


Force in DF:

Sum moments about E where FCE and FDE intersect.

+)I,ME

= 0: (.1 .5 kN)(6 m) - (9 kN)(6 m) + (3 k.N)(3 m) + ~FCE

(2 . ^-X6.75 m

13 ;=

~^(4.5m)-36kN-m

FCE =-10.00 kN FCE = 10.00 kN C <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill 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 limiteddistribution 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.

823


86/240

:.5kN PROBLEM 6.54

3 kN

3kN

D %:1.5 kN

1 #

111 O 111

3kN

€3 kN

:.Q '. . O ' ' . O . ' . . o

:

1.5 kN I/ 1 6.75 ni

fL

3 m 3 in J 111 .5 111

A Pratt roof truss is loaded as shown. Determine the force inmembers FH, FI., and GL

SOLUTION

Free body: Entire truss 3fcV

By symmetry:

Free body: Portion MIL

Slope of FHJL

£F V =0: Ax =0

Total load = 5(3 kN) + 2(1 .5 kN) = 1 8 kN

A =L = -(18) = 9kN> 2

6.15 ^ 3 3fcC9.00 4

FG 6.75 m . r ....tan « = = a = 66.04 c

67 3 m

Mhw

3kN

La?v«vV

3ot 3nj JJto

Force in FH:

+)ZM,=Q: -Fn - x 6.75 m | + (9 kN)(6 m) - (1 .5 kM)(6 m) - (3 kN)(3 m) =

Force in FI:

-F / , // (4.5m) + 36kN-m

FF„ = -10.00 kN

+ ) 1ML = : Fhl sin a(6 m) - (3 kN)(6 m) - (3 kN)(3 m) =

FPJ sin 66.04°(6 m) ~ 27 kN -m

FFI =+4.92k.N

FFH =10.00 kN C A

FFI =4.92 kN T A


824


87/240


Force in GI :

+)ZM H = 0: FGt (6.75 m) + (3 k.N)(3 m) + (3 kN)(6m) + (1.5kN)(9m)-(9kN)(9m) =

/^(6.75m) = +40.5kN-m

FGI = +6.00 k.N FG( = 6.00 k.N T •«


S25


88/240

3l-'.\

-«§ >o

o

;:

c

4—4.5 m-20 kN

O:

/

f

2.4 in

PROBLEM 6.55

Determine the force in members AD, CD, and CE of thetruss shown.

SOLUTION

Reactions:

34 kM

2.4*.

BB

10 VA 2oW*

4*- 4 (4.5) =

Fc/1 - (2.4)- 26.4(4.5) =

FC£ =+56JkN

FAD - 13.5 kN C «

Fa,=0^

FC£ =56.1kN r <

PROPRIETARY MATERIAL. © 20.10 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 limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,


826


89/240

36 k\

4.5 in —*j-> —4.5 »»—*-{-« —4.5 tir1 20 kN I 20 kN

«9 < : „

//

o

1

/-

PROBLEM 6.56

Determine the force in members DG, FG, and FH of the

—|- truss shown.2.4 m

SOLUTION

See the solution to Problem 6.55 for free-body diagram and analysis to determine the reactions at the supportsB and K. n

B = 26.4 kN |; Kx = 36.0 kN -*; K v =13.60 kN t

36 kM A

2CAM£ -2.ZSyv\

4.5 M *{* 4. 5v0 >|fFH

+>M / , =0: 36(1 .2) -26.4(6.75) + 20(2.25) -/


90/240

0.9 Mi.3 kins I

?Ai.S kip

t-JB0.9 kips I

9 ft

2 ft

31.5 it

-«K

/)

-14 ft- -14 It

* *> $J

«/-

8 ft 8 ft

PROBLEM 6.57

A stadium roof truss is loaded as shown. Determine the forcein members AB, AG, and FG.

SOLUTION

We pass asection through members AB, AG, and FG, and use the free body shown .

x_if.•fej-12 ft-4- »-4r» ft**l


91/240

0.9 kins

T9ft

— J&

/' A-V


92/240

6 ft 6 ft , 6 ft 6ft, 6 ft 6 It 6 ft 6 ft

200 lb400 H

400 lb

350 lb

4001b I 3001b,

II300 lb

3001bSO 11

F I)_

Tift

tl 7«-:5^-?K— M

9.6 ft 8.4 ft 6ft 6ft 8.4 ft

4.5 ft

9.6 ft

PROBLEM 6.59

A Polynesian, or duopitch, roof truss is loaded asshown. Determine the force in members DF, EF,

and EG.

SOLUTION

Free body: Truss

£F V =0: Nx =

+)ZM N = 0: (200 lb)(8a) + (400 \b)(la + 6a + 5a)+(350 lb)(4a) + (300 \b)(3a + 2a + a) - A(&a) =

A = 1500 1blr =0

8ft_ 6 ft. 6 a

^ =1300 lb N = 1300 lb I <We pass a section through DF, EF, and EG, and use the free body shown .

(We apply FDF niF)

x 2001b+)2M E ~0: (200 lb)(18 ft) + (400 lb)(12 ft)+(400 lb)(6 ft) -(1500 lb)(18 ft)

*X)lb*» b

(4.5 ft) = ()

FDF =-37111b Fw- =3710 lb C^

+)SM/t

=0: F£/ ,(18 ft) -(400 lb)(6 ft) - (400 lb)(12ft) =

FEF = +400 lb FEr = 400 lb 71

««

+) SAf,, = 0: F£e (4.5 ft) - (1 500 lb)( 1 8 ft)+(200 lb)( 1 8 ft) + (400 lb)( 1 2 ft) + (400 lb)(6 ft) =

FEG = +3600 lb FEG = 3600 lb F ^




830


93/240


94/240

i

3 in . 3 tit . 3 in . 3 iti j

I a bV ei d\ e\:;o,-

-- o . ' o •'::.:. '..:...;q ::.z:X2

IV

a 4 m

I' a o o _ b b /

4 m

J.

fo

PROBLEM 6.61

Determine the force in members AF and FJ of the t

cap 6, novena edc_text

Documents