-

.... .. 0.__ .. _.... . ... _ ...... _......... .' ___ "_. ____ . ,. .. .- - ----_._. -- ------,~ ... _~_.:---,-~-,.-'''',-_.---.. --"---~- --_.-. -_._-------_ .. _. ---------

1 D, lu !e o D

D o D

1 DI'

o C o D C o o ,(

o o

C'~\APTER \

[J 1 fl IlJ

[J

[]

C

c

D o o [J

o

o 1

1

[]

o

PROBU:M 1.1

SOLUTION

((l' roc! AS

(b) ('oJ BC ~of'c.e : P

Ate()o,. : A =

::-

1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that dl = 30 mm and d1 = 50 mm, find the average normal stress in the mid section of (a) rod AB, (b) rod BC

O.9m 1 1.2m ,1

~oll'c.e: P = 60 X /0' IV fe",s:o ..

i{:-(80)(/O·3/ ': 70,.~&)tIO-' "..,'

E: 60X/03

, A 7Gb. 8' l< /0" ,. B'+. 88 )( ID 1><1

0AS: 84. q M PQ,

r;.ox/o" • (~)(I:(S x10 3) :- - IQo ></0 3

N

=If- Jt

l : ~ (SO I</O',!)''' 1. CJ.,3$ ){/0·5 "..,'

1> • 1 CIO x IIY' .; No""""J stress: G8c ~ A : 1.9(;,3S><'IO.' = - ''16.77><10 Pq,

bac. ':: - 'lb. ~ MP~ .-..

,

f 1

(

!Dr IK ,C ( 'Dl 1 ,

:0 Of

Of 0\ Cl Cl D· D)

10: 0-Q. o Di , .

0\ o

PROBLEM 1.2

SOLUTION

rod A8 Fo""" : A'('e IJ. :

6"~6 '"

~ d ~ 't 1

el ~ " 1

d, ::

roJ BC

- -~ - -- -..,....--,.,..- -" •. ---,,<-~.

1.2 Two so!id cylindrical rads AB and BC are welded together at Band loaded as shown. Knowing that the average normal stress must not exceed 150 lv1Pa in either rad. determine the smallest allowable values of the diameters dl and d,.

d,. .

'. ""~ d.~u.~ 1 1 125 kS 1

'--- 0.9 m •• 1.2 m .1

p :: GO X ID3 N S~S~: 6u = ISO)fIO"'P.:t. if- cl, t A ::

P . A = 1:. A .. €;As

~ .f.. Ei'".

If? (If'l( GD)( IO'l) 3 -, 2

1T"E;Ae := soq. "10 m 1T (ISox/oG )

:n..S6 "/0.3 ... cl, = :(:Z., ,., ... ~

".l:. = .!±..f.. A 1\d2" .:if.. _ (If) (- l<to " loS ') ïrG'e, - 1T t- ISO III 0&)

.. M

'+0.16 x/0-3 .,.,

"$ j 1$>' . 1 . P'

1 .;

T I,.J

C D o

. [., 1 J

1 [J

'ID iD

D o o o o o

ID 1

ID 10

PROBLEM 1.3

PROBLEM 1.4

10 kips

\"0"/ SC: p., & .:œ dl. =- -7TG"Bc.

1.3 Two solid cylindrical rods AB and BC are welded together at B and laaded .s shawn. Knowing that dl = 125 in. and d, = 0.75 in., find the normal stress at the midpoint of (a) rod AB, (b) rod BC

SOLUTION

(Cl.,) ra cl

'P

A = ::'

AS 1:( '1- 10 - 22. Icip\

:tt d,' .. if:' (I.~)~ =-

6i - op AB - A

(bÎ roJ Bt p == 10 k.;ps

A :- if d.l ' ::: ~ (O.7S)1. = 0.'+'+/2

cc- _ P = 10 :..., ", )<SI' 01\8 - A o.'+'i'i .0; ....

, a 1 ..

. .. '"

.....

1.4 Iwo solid cylindrical rods AB and BC are welded together at Band loaded as shown. Knowing that the nonnal stress must not exceed 25 ksi in either rad, detennine the smallest allowable values of the diameters dl and dl'

SOLUTION

1'0 J AS: p :' l~ .. 10 '" 2'2. klps

ÔAI! '" ~s k~,' AI4 : -1f d,L

d, = '.06" ,"1 ... \0 k:r s os~:: ,?S li!.; ABc. ::: If Ja. ....

N'l(IO '> • 0:"50'f3 ln da = 0.7 1 If' i..,. .''''' lT ('lS) -

7 ' fi_ .

r 1

(

r

ne t

["~ ,

~~

0 [J • PROBLEM 1.5

120n ~

[J

'J A \':~. ',l,

D "

1

0 C i

li ~ 1

0 1",

B~~ '0 1200 :\ 1;

,0 ,

0 • '0

1

o

105 A strain gage located at C on the surface ofbone AB indicates that the average normal stress in the bane is 3.80 MPa when the bane is subjected to two 1200~N forces as shown. Assuming the cross section of the bane at Cto be annular and knowing that ilS outer diameter is 25 mm, detennine the ioner diameter of the bone's cross section al C

SOLUTION

~ _01:. V - A • . . Geo ..... J"I d ~ :

2 cl 2.

1

J'" 2 = ( -3 )2-~S"'O -

- n~.9 ><10·'

d, IY."13 "JO .3

':

'"

Cl/)(/ZOO) 11(3.80 )//0'

• m

d, '< 1 t.j .'1'3 ",," ~

n_________ _ ..

[]

o [J

[J

c o r D o rJ

o o o

o o

PROBLEM 1.6

... 1: SOLUTION

1.6 Two steel plates are ta be held together by means of Y4-in,-diameter high­srrength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 30 ksi in the bolts and 18 ksi in the spacers, determine the outer diameter of the 'pacer, which yield, the mo,t econonùcal and ,afe design.

At e6.c.~ boN )O<' ... +,./WI +~e. t)ffe-r pJ ... fe '$ foi"e~ dow", by +A~ +&'5,'1e

fot"ce Plo of +I.e boit. A+ -+he sc.""e +,,,,,e +I.e spD...<.e .. p()$~es

th ~+ pPa.fe UP",q",J \AI i+h 0. eO""(J f'f..S.SI·-n fo~(.e P.s. 1", o."r:Je'" + .. f<\Q.;"'+q,'''' e<flJii,·b ..... v'"'"

Pb = Ps

FO<f' +J" e. boN 6~ =

For He SfD.ce ...

E<t vJ'Vl9 H, alhJ Ps IJ Sb db" :: t <5"s ( oIs\

§d' Os c ::

0'"

. 1 lOI

c

~ .~

1 t

•• J

1

1 }.

1

o

o. '0,

n o o fJ

o iD 1

ID

o 'o· 1 .J

•

•

O···.L·············_"~ 1-- ' . , -- ---~ , ,

PROBLEM 1.7

Cb) ê: "0·

1. 7 Link BD consÎsts of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a 1O-mm diameter, determine the maximum value of the average normal stress in link BD if(a) 8 = 0, (b) 8 = 90

SOLUTION

Us~ bQ,1'" ABC Q,S -tfee 60J1 Co

(o.'1SQ s,,,, 30·)(~o)<I03) - (0.3QO C:IJ$ 30') Fat> ... C

Fel>"" /7.3.< lA/o3 IV

( O.~S"o CDS 3D·)(~0~/o3) - (0.300 C~ Soo ') 'F~.. 0

F~" - 30 ... /0" N

(0.) ie,,:,,'QII\ lo...,.J: ... ,o

(b) c ..... {l..cu,·o ...

.stl"&uec

A:, (0.030- O.O/O)(O.O/~) ~ ':(40 _10-4

... ~

A:, (O.oSO)(O.O/~):' .360 w/o·r. i"\'

(0.) S .. Fu \?a~ )t/oa .. 7~.~ )1/0' n.~ MPe.. ~ - .. A :l'lowfo··

Fac. -30./03 Co - sa.a MPo. .... Ch) s ":: .. - 83.3 "'/0 A

:: 3'0 If IO-C.

[J

n o

C

ID o o o o o o o o o o o

20

PROBLEM 1.8

0.4

-s '2. =ICOxIO M.

Af'eQ. +D'" o .... e .P,,,,,, k

= .eSS x 10-<:' M ":

(b) 6'~ = ~ =

-- ~'.-'~----~----... - ."", -',.-

1.~ Each of the four vertical links has an 8 x 36 p mm uniform rectangular cross section and each of the four pins has a 16- mm diameter. Determine the maximum • value of the average normal stress in the links connecting(a) points B and D, (b) points , CandE.

SOLUTION

Use 'cx.< ... ABC ..., 0- f.-ell bf)J.y.

:10 UI o.~~ ,/'" 0. 040 --"i

A S c

Fse Fe"

&.5.,.,/0 3

3:to )( 10-" ~ IOI.S"G ~/O Co

101. f; 11 Pa.. ...

A ::- 57' >1/0-' ..... a

or - ~/. 7 HP,,- ...

n-l]

lJ • 0 c 0 ' .. ;,

1 ')

D, Ci

0 0 • 0 0

1 0 1

0 0

.Iil!il'

1

f·~ D';;:'

1

1 O~~' ( Ir~ l "!i', , ID

PROBLEM 1.9

0.5 in.

(1:4) Sfress ,., A8

1.9 Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing that a pin of O.8·in. diame~er is used at each connection, detenrune the maximum value of the average normal stress (a) in Iink AB, (b) in link BC

SOLUTION

•

10 I<,'p'

FOtl'cr +"'''-11\' le.

= (/.S - 0.8)(0 • .5') -= O.S III"

7. 5~o'î = 1 if . Qf ksi -O.-.s-

l;",j( Be \s Q. 't.o ..... p r'"es.s '0>"1 ~e",bel" (1.3)(0.5') o. q /1'1 ~ C"ClSS s ec+iClVl oJ Q.l"e~ ,$ A-= -=

(6) S-h'ess '''' SC Gee. :: - F6C. - 8.%.5"8 - '=f. CfÇ; kSI = 0."1 = A

[J

o ID !D' ! '

D o o o o o o

o o o o

10 1

1

PROBLEM 1.10 1.10 The frame shawn consists of four wooden members, ABC, DEF, BE, and CF Knowiog that each member has a 2 x 4-in rectangular cross section and that each pin has a y~. in. diameter, detennine the maximum value of the average normal stress (a) inmember BE, (b) in member CF.

1---- 4IS in, ---tI~B - 301n'l C SOLUTION

40in.

If. D

~3'J Y

AJJ suppo.-t v-eG<d .. "",!> -f>,. -hjUM! o.s s\.., 0"'''.

Us ;"''l e",+,'N. -Pro. "" e tlS ~<. boJy Z-I'-'", = 0 /fa O. - (4S+3D)(4~O) -= 0

D)( '" '100 lb.

LF'- =0

~Pr -~Dlr~O

"0y :: * n.>l' -= 1 ~ 00 Jb.

t>

Lt4F := 0

L. HE '" 0

-(2':lYf Fe/?) - (30+105)'0/ = 0

(30)(-! FcE ) .. CISJ Dr .. 0

FEE := - 2.ZS'o lb.

FeE" ïS'O J4.

:?~so )~ l "

/ /

/ /

'" >mi", .. IY\V'"'

.se:ohoVl

stress '''' CD ""pw-eS'" "'''1 r>oeWlbel' BE

(b)

Areo.. A:: ~j.., l< 'f,.", ... 8i .. 2.

t'I i n ;IM ... ""

-2650 _ 8 -

se c;f. iotll a. .... a.

À ... ;", :: (',i..)(tf.o -D.s~ =

O~ -= F.". 7,so A:...· .. - ~

7.0

- Â,8/ f~'

OCCIJMi tai 7.0 ; .... 1.

/07./ fS"

pl'rI.

c

n [J

[J

0, c:

!e ' 'l

"

0 ! 0: 1 Dl c*

l

D~

1 o~ cr' ..

Q'

çt •

0

•

•

•

PROBLEM 1.11

B D F

.'SO kips ',0 kips 'iU ki.ps

Fer , 1---=-1--;.. ,

PROBLEM Ll2

HO kip$ so kips l;itl kips

8 FGb

FlsE

A C

~ç8

I~o f;ps '0 kirs

1.11 For the Pratt bridge truss and loading shown, detennine the average normal stress in member BE, knowing that the cross-sectional area ofthat member Îs 5.87 in2

SOLUTION

Use e.-Ih ... e. +NSS tU ~"" boJ't

zM~ = 0 (1)(110) + ('~)(30)'" (;17 )(80) - '3' At ... 0

At = 1'20 kirs

Use pot"+,"O\A ~ +V'VS$ +0 +he .P~t of Q. oS ec.:oI-io"l

c.rlti~ l'VIe"" ws 'BD, BE.> ",-.. d CE:.

+t Zry :: 0

I~O - 80 12

FSE = $0 je;l'" - ïi FSE =0 ..

6 SEf ';: F'6" .: 50 8.9. 1,('$ ; .. A s.S? -

1.12 Knowing that the average normal stress in member CE orthe Pratt bridge truss shown must not exceed 21 ksi for the given loàding, determine the cross-sectional area of that member which will yield the mast economical and safe design. Assume that both ends of the rnember will be adequately reinforeed.

SOLUTION

U :: ~ el'1+ ;y".e, +v-..JSS "-S .fv-ee body 9 ~ /'llfI : 0

(o.\(8o)+-(li)(8o) 4(~7Xgo) - .3, A'1 : 0

A, = 1'2.0 l<:ps

Use p"..ti 0... "f frJSS 1» +he .Q~+t o-t ~ .sec.:tio"l evtti~., ~e .... ber-s aDJ 8E:~ "1,,01 Cli

t>~M" = 0

l:t F .. a - (9)(I~o) = 0

Su -F,;,

ACIf!'

A"" FeJr := Cio = - ::-6"c:ti ~l

JJ o [J

o [J

o 10 1.0

C o

o

o

PROBLEM 1.13

p

S+I"CSS

1.13 A couple M of magnitude 1500 N· m is applied to the crank of an engine. For the position shown. determine (a) the force P required to hold the engine system in equilibrium, (b) the average nannal stress În the connecting rad BC, which has a 450-mm2 uniform cross section.

SOLUTIQN

Use Pisto") roclJ "1."..1 ~V'<VIk +Ij:! e..th ell' o..s ft-ee. c ad f' AeJ.J wo..1R l'ea.c:tIOl1 H 0.111) be"4l1'il\1 t~ ... + iOVlc Ai 0.",..1 Ay,

t)ZMA=o

( O. ~ 80 "" ) H - ISoO N'IM ... 0

H.. oS. 3S7 J >( 1 O~ N

U~"- p;5+0~ Clio .. e. <O.S ~ body. No+1t -Htc..+ Nd ioS 0.. t"'O-f"lf'ee .... e""b&"; fo,a"c:A! J-he Jiv-cot/loWI ot: ~ot'<.t Fee. ;5 k .. 0""1'1. DM. ... +he f'o'IC~ +dCl~~Je o.r..J SO)-R, .j;,,,, '? t.I.~..I Fu 11 P f'0p",'M-io Vl s. •

.Q = / -;tua 1- + 60 2 - Zog,81 .... ""

P -= ;/uO • p= 17. 8(; ""'0' IV .-H Go

(Q) p:. n. 8' k~ ~

& : ~n.tl FSc. ':" 1 S.b'f3 "101

'" H '0 ..

(b) 6ec. =- - 41. if MPQ. .....

i

-- -,~ .'

•

[J:

o 1

Cl

0: >

0: .!

D-l

D' 1[1'

-l .' i •.. · . r . 01'"

•

•

PROBLEM 1.\4 1.14 Iwo hydraulic cylinders are used to control the position of the robotic arm ABC. Knowing that the control rods attached at A and D each have a 20-mm diameter and happen to be parallel in the position shown, determine the average nonnal stress in (a) rnernber AE, (h) rnernber DG.

, 1 JOO ....

SOLUTION

Use MeW1 ber A E C ~ oF ('ee. boa/y.

(o. ISo) 1 FP.E - (o. 600 )(8c.>o) = 0

c

A~~t:<. 6f' ru,) j", ... e ..... bu AE i~ A = fd'l.:: f(;),o>'to3)'" =- 311f.16)C 10-:'

SeL ,,_.~ ,'. '" - • Ae-: t::" - FAr - If )(/03

-- - ':l 7"> lI/O" -0 en" -- .,. C>t7 VAe - A - 31 If. /6 )(/0-4 - .;;; ri:<.

(CL') 0,,&:: 1:<.73 t"I Po.

.. Use Co~ bine.,) ..,e ... be.rs ABC ....... J 8Ft) Q.S -t .. ee ~oc:Iy,

~ LMF = "

(O. 'SoX -1 F~) - (0.<00)( * 1='1») - (1.IJ';'o - O.SSo)(SOo; ': 0

-, A = -1f el 'l. = -l(::/o )l/O~)2- :: 3/~. Il, xie> • ""

S. - FQ6. :: - 1$00 = 1» - A 3.1'llbxlo-C: " -4-.77)(10 Po.

(6) 0DG- = - 4. 77 MP~ ..

n o o o n o c o o o o o o

ID ID

o o

let

'A

~

PROBLEM 1.15

2-1 kN

24 kN

1',15 The wooden members A and B are to be joined by plywood spliee plates which will be fully glued on the surtàces in Contact As part of the design of the joint and knowing that the clearance between the ends of the members is to be 8 mm, determine the smallest a1lowable length L if the average shearing stress in the glue is not to exceed 800 kPa,

SOLUTION

The".., ~ .foIN' sep<>.",,Je Q.. .. ee..s ~ 1J\Je. ~ o.re~ ""\)1).+ t~",~ ... it V!<lif! "f +la. ~".IJ Jo.....J. T~e .... b"" F -= I~)IJ -= !7J. 10

3 N

F A~ T -1

': /t»(lo

J -= ;"""a-tI. of vvO: a.~eQ. a.,v.J 'II" W/";1-/11 = 100 ... "" .., 0.1 ....

'" Jw " J= os 'J..e +- ~~

.,.

PROBLEM 1.16

SOLUTION

A ': TIdt

r:., v .JI' ~Ô A '1

S...flli,,~ .for J:

A 15\(/0-S /':'-0 'I./o-!. ~ ISO ... .., = ':

'" W o., (:1 )(/SOo ) ... 8 - 80& ..,.., ..... -

1.16 Detennine the diameter of the largest circular hole which cao be punched inla a sheet of polystyrene 6-mm thick, knowing that the force exerted by the punch is 4S kN and that a 55-rvfPa average shearing stress is required to cause the material to rail.

il , --J

l '; r' JI (

ID 0 0 01

0 0 0 0 D 0 0

iD

Cl DI D,

0

0.4 in.

PROBLEM 1.17

)

1:::: p A

• ..

PROBLEM 1.18

1

tm .

Clue

A=

~ 0.25 in.

t

'U

1.17 Two wooden plank" each 'le - in. thick and 6 in. wide, are joined by the glued mortise joint shown. Knowing that the joint wiU fait when the average shearing stress in the glue reaches 120 psi, determine the smallest aHowable length d of the cuts if the joint is to withstand an axialload of magnitude P = 1200 lb.

)

SOLUTION

Seve,., suV'f",-c:e~ c..o.rf'y t~e -ho+ .. P ).,....,/ :?., 1 ~ ()O .fI..

d

1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 0.6-in.-diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the largest lüad P which may be applied to the rod.

SOLUTION

FOY' s+eei

• .. A, -= "TT Jt = "TT (o. ")(0. '1)

= 0.;5",+0 j,,' 'P -= A, 1:'1 = (o. 7S'\lO)( I~')

-:: /3.57 k;ps

1relt .. lT(I.(:)(().~S')::- , 1 ,.,

w • zz- 1 -

iJ

[J

[J

lJ

[J

o [J

o o

ID C

ID ID 1

C o o

PROBLEM 1.19

p

PROBLEM 1.20

"

1.19 The axial force in the column supporting the timber beam shown is P = 75 kN Determine the smallest allowable length L of the bearing plate if the bearing stress in the timber is not to exceed 3 0 MPa

SOLUTION

C"' -.E. ~ J:.. u .. - I~ Lw

So 1 v""1 .fo.... L: L~ c;.w -<"3.0 >(fO"'XO.1 fo)

'" 178.6 x10-3 ",

L -= /78.6 ......

1.20 An axialload P is supported by a short W250 x.67 column ofcrosSrSectionai area A = 8580 mm' and is distributed to a concrete foundation by a square plate as shown. Knowing that the average normal stress in the column must not exceed 150 MPaand that the bearing stress on the concrete foundation must not exceed 12,5 MPa, determine the side a of the plate which will provide the most econooùcal and safe design.

SOLUTION

A",ec.. .,f c"jl,)",,-"'" A :" 8SS0,..,"" .... = Mao >II oc. No",."",) 'st-I"e .. s t'" eoJuJ-\t'I ~ ô'::- 15"0 'rl/o" Pl:\. ~-E. \;;) - A • p: A 6'": (gS80'fC/O·)(JS'O "'Ot.)

: 1.:u17 Il 10= N

Sb" k

ru

I:ZS7 'II JO· I~.s 'K 10'

MS " rz , oS

( "

1

1

[J

[J

lJ

0\ [J

1 [J

[J

o

[J

o []

D o

ID 1

o DI

1

\ D, 1

PROBLEM 1.21

, ~ ~lIi" '~ q ", I:~' .• ( ' . .. ' .. ~. '. . ~. ~~~ .. '" .J. , . ~ ;; ~~ . .Lin • ...... - .,,~" t1

, .. " 1 ! , '1 '

, 1 :

, '

1.21 Three wooden planks are fastened together by a series of bolts to forrn a column. The diameter of each boit Îs Y2 in. and the inne!" diameter of each wa'iher is SIe in., which is slightly largerthan the diameter of the holes in the planks. Determine the smallest a110wable outer diarneter d of the washers, knowing that the average nonnal stress in the bolts is 5 ksi and that the beariog stress between the washers and the planks mUst not exceed 1.2 ksi.

SOLUTION

A _JI _1 Z. _ 1r. ( .L)a. - () 1 Q , 3'- ' • bJt - 'if" orb - "t I,:l - • ". .;) li.

6b=f Te".,sde fOl'Ge il'\. boIT 'P =,6b t. = (S)(O.'/"I(;3SJ = 6.9817S-k,ps

IV c:o..5her: o<.T/-side J,'a. ... e:fc'" = do

Be",(';"'~ <:t ... e4 Aw = t ( do 2. - d.è.. 2. )

E'1 u d-;ns ~(do2. - J;..l) = ~ J,l.. 4 P " (JJ:.)l + (1f~~O.=t8 :7$) =

.. 11 (5'10 S 7T /::/. S)

L-Olo '

1.'+3:<3 , t ,,,

PROBLEM 1.22 1.22 Link AB, ofwidth b:::: 2 in. and thickness 1::;; 1/4 in., is used to support the end ofa horizontal beam. Knowing that the average normal stress in the link is - 20 ksi and that the average shearing stress in each of the two pins is 12 ksi, de termine (t1) the diameter d of the pins, (b) the average bearing stress in the link

SOLUTION

Rao! AB is ,1'\ Co ... fl"€SS ,"'<1.

A = bt: w~e"'( b,. ;(j", «<.,J t=* j.,

P = -C5A " - (-;( 0 )( 2. )( t ) :- /0 1< il' ~

Pin : 'LI' ::' E. ONId Ap = f dt. Ar'

( o.) d=~i} :: f;{ If 1:', (4)(/0) --rr(I~)

1.030 'OI -4

-

•• 7 ,- l' t $

n [J

[J

o o c o o o o o o o o o o o

ID 1

iD 1

1

20

PROBLEl\f 1.1.3 1.8 Each of the four vertical links bas an 8 x 36- mm unifomt" rectangular cross section and each of the four pins has a 16- mm diameter

0.4

O~~ 1" ' , 1

1 B'

c

1.23 For the assembly and loading of Prob. 1 8, determine Ca) the average shearing stress in the pin at B, (b) the average bearing stress at B in link BD, (c) the average bearing stress at B in member ABC, knowing that this member has a 10 x 50-mm unifonn rectangular cross section.

SOlUT/Oc\'

Ose bal' ABC o..s Q" +..-e~ bodt

(), 0 2S -i~*'1 "Ol--- 0 _ O'fo --71

A 8 c

(O,O'/O)F'D -(O. OZS"+O.Ollo)(;l.O >1103 ) ::. 0

Feo: 3~.Sl( 10 3 N

(bl 8e~ ... in'l :.Pi .. /t BD A = cJt • (0.0/6 }(O.OOc!.)" 1:28 ~/D" ",'

~ ':. *F.~ _ (OS)( 3~.S)l 1(:1) :: l:<b.CfS >'ID' 11.7.0 M?Q..4 b A - J~g )( ID"

(e) 8e.a.l"i ~ '1 j 11\ ABc 0.+ B

A = dt -= (o.O/€.)(O.OIO)::- l'O'tf/O·'- ",,'

r )' '1- $-

1

1 c,

( ,

l 1 1

l 1 r

1

1

~,-

D ,

[J

0 lJ

D' 10 l ' , ,

~

n n 0 0 0 DI

1 .

Ir, [ ; ,-'-'

1

0 D,

DI 1 (

iD 1

ID '\,

PROBLEM 1.24

20

1.8 Each of the four vertical links has an 8 x 36~ mm uniform rectangular cross section and each of the four pins has a 16~ mm diameter

1.24 For the assembly and loading of Prob. 1.8, determÎne (a) the average shearing stress in the pin at C, (b) the average bearing stress al C in link CE. (c) the average bearing stress at C in member ABC. knowing that this member has a 10 x 50-mm uniform rectangular cross section.

SOLUTION

O. o~.5'· ---..+-- D.Mo ---l 6 c

(a) she~"" ;0, pi"l a..t c

le)

A = olt = 6"",,, 1: F.,

A

Be~"",' T\~ 11<'1

A ': dt

6i.= Fee A

1;(.5:.<103 :'

(2)(:lO/.D" x/O")

·t 31.1></0

3/.1 t-1P~

(0.01& )(O.OOg) = ln 'lf10·'- "",'a.

: (Q.S")(I~.S')(/o3) = Y8.8 li' JO" \~2 ~ '0' ,----

ABC 4-t C (O.OI{' )(0.010)

... l. - - 16o )(/D M - -':'

I~.S IIIC? IGo )(iO-C;

:- 78./ )t 10'

F t t

78./ M?Cl ~

.n o o o o [J

D ID o o o o

ID

1

o o o

!O

o 1

ID

-- ---~ ---~-~ -'. ""'-"'..,., .. , _._~"..,-".~.,

PROBLEM 1.25

LaI\! of S ; .. es

FAB_. = F8G si", ifS' s;.., Go·

\.9 Two horizontal5-kip forces are applied to pin B of the assembly shown. Knowing that a pin of 0 8~in, diameter is used at each connection, deterrnine the maximum value of the average normal stress (a) in link AB, (b) in link Be

\.25 For the assembly and loading ofProb. 19, determine (a) the average shearing stress in the pin at A, (h) the average beating stress at A in member AB.

SOLl'T10N

U "e ,j 01'" + B 0. ~ {'>rH b 0"\1'

(a..) SheCi("'.,~ stress j", pi", 0.+ ,r-I . . ~'''' l'''t= t d z. :: ~ (0_ 8 ) .. ::

A "t =

(bl

7_:Z~

Be",":,,~ sfr~~, a,:+ A '" "'e""he .... AS

~:: t cl : (0-5)( o. 8):: 0.4

7,3.'(oS ::

0.,+ 18.'30

f • t

t , .. 1~.30 Ws; ~

(

l 1 1

1

r (

1

,:'.,....

70 o o o 0/

o o o o o

,0

o o O• 1

l '

o o 01

(

01 Of

PROBLEM 1.26

0.5 in.

1.9 Two horizontal 5-kip forces are applied to pin B of the assembly shown Knowing that a pin ofO.8-in. diameter is llsed at each cOMection, determine the maximum value of the average normal stress (a) in link AB, (h) in link BC

1.26 For the assembly and loading ofProb. 1.9, determine (a) the average shearing stress in the pin at C, (h) the average bearing stress at C in member BC, (c) the average bearing stress at B in member Be.

SOLUTION

10 kifS

t../t.IV ot Sines

1='48 = 10

FQf'ce ~\,,, .. ~j~

Flle. = 8. ~bS'8

(Cl,.)

si" 'iS"

She"lf'\',,~ S+l"'e~ i.., 1'; III "1-+ C

A _JI: .J'l. - lC. (,~ 8)'l. -f' - 't a - 't u. -

BeCOf""'n~ st!'eSS J C. tOI "'e'" bef' sc

À:: t 01 = (0.5)(0.81 ': o. '+ j" ..

0."1

.s+~ss ~+ 8

 ': 2t J :- 210.5)(0. &) =

Il.2./

li 'fi

1

.dt »

0.8 , 1-, ..

\l."<1 ks,'

D o o

o n o [J

D o o

.0 D '0

D o

10

(Q)

PROBLEM 1.27

p

p

AJ)olV<o..b1e pin

't = F ... t3 : 2. Ar

d '2. =- ~ I='A8 '" 'ITt

},]7 Knowing that8 = 40° and P = 9 kN, deterrnine (a) the smallest allowahle diameter of the pin at B if the average shearing stress in the pin is not to exceed 120 MPa, (b) the corresponding average bearing stress in member AB at B, (c) the correspondiog average beariog stress in each of the support brackets at B.

SOLUTION

Fol"ce nioV\,,\~e ,

GeOMe.+v.y: T'/"io,,,,~.Pe ABC is CV'l i sose)es +V'iCc ... ~J~ wi+l.,

o."':lles ~ho"m ne'M!..

c

La.w of' Si l'les. Q..{l{ll je cl +0 -t'''I''c.~ +1"1·&<.Vl~Je.

p =...EaL_ lSi ... ~O# 5i'l1/0"

l' Si" 110 0

.sih~Ù· .. = (9)si~ 110 =

si", ~ùo 2'/, T3 kl\l

..,I,e,'" FM?,:: ~'t. i$ >,/0.3 }I

1 -c: ~ 13 ./8 >< /0 M

j -3 (.1 = Il. 'IS "1 () ".., 11.45' ... .., -.

(bJ 8e"''''''TlCj S+ress ,,, A8 r4 A.

Ab'" t ol ~ (Cl. 01'- )Cil.'!$" ~(o'\) :'

El. - FAB _ 'Z'f.73 "10'3 :=

1. - -;;::;: - I??,.'<''''to·&

-, 183. U )(/0

/3'1.9 "'10'

c) Bea"'" n'( s-h-e.ss '" sr.JPfJo,..,f j,t'<A."'~ e-ls Q,,+ '8

A:: toi = (O.OI~ )(11.45'>1/0.3) = 1~7.4 )(/0-"''''''''

0,= ~~B - (O.S)(:l,'f,7~ .. ~Jo.!1 :::- qo.O)(LO~ C A - 137. 't >tUF"

1 3't. '1I1Pc!l.. ...

90.D MPGl.. ~

(

1

r 1 l 1 r

(

1 1

1 1 _________ W_"_.='m_. __ ~-'m----------------------

l' r/ n .. . r n DI

0 0 0 Di

10 J

0 0 0

ID 0 0 0 D, DI

(

D\ r ,

D 1

ci ~._-. --~~-- ~'~"-.'-- .~'~-"-,,........-- .. _~ . - - c - ~--- ••

r-----------~------------------------------------------~

PROBLE:\<I 1.28

p

-

1.18 Determine the largest laad P which may be applied at A when () ~ 60 ,knowing that the average shearing stress in the IO-mm-diameter pin at B must not exceed 120 MPa and that che average bearing stress in member AB and in the bracket at B must not e<ceed 90 MPa"

SOLUTION

A GI!O\N\et"'l: Tv"'''''''''l;~ À'S(. i.

~ i soSe..;e$ tV'I'q >'l'J...Ie \',11 +~

M~Je~ SholoJ\'l he~

':---:::-0----"" C 750

Use joi~+ A a.ç ff'-e .. body ~ClIV "f sjnes CA.fP /; e../ f..

-toHe tV','.:, 11\ ,Je p

I:lO· FloS

FOV'l:I!

r1>6 +V-I'''''~ Rt

If sheQ"' .... ~ st~ss i", pi" J g is Cr .. h'e",)

~ :: ld1,:: tlO,OID)t :: 78.S'/ x/O·' ",,1-

F,.13 ~ ?À't. :: (~)(7B.S'/"/O·'I(I';i.O'>t'IO~i '"

= _F.,..,A:::C,=---_

n: beo.",; .. ~ S.+~S i", VI'IQ"" be .. AS A:f- bM"itd 0.+ A Î$ c. .. ,'f,·e..../

Ab': tel ~ (0.01(,)(0.010).- 1'0></0·' .....

FAIS ~ At,6' .. :: (/'O>f/O·' 'X. '10 >tI06) = 1'1. IfO ></Q" N

Ir bh"""~ ~,s 'YI +~e b~ke+ A+ 13 ;$ C:t-'+ic-J

A~" 2 t J :('2. )(0.01'4)(0.010) =: ::«0)</0.6 ~:z.

FM3 =- Ab6h =(2'/o"ID'~)("f,O)(/O') " ~'.r.,)(/O" IV

Alio..; JoJe FAe. 'S tl.e s ..... Jjest,; ,.e. 1"/.40 ~ IO! IV

n~"I)rt",c.o-, S+J.'C$ 'P..JI..,. ~ (O.S77S5' )(l'f'.'l-ox 103)

~.g) 1N

o D o D o o

D D D o o o o o

PROBLEM 1.29

PROBLEM 1.30

125 mm

75 II\~ /",, ___ /

7rJ'

p

et

~.- .-., ________ r __ ,_._" ~ ____ ~ ____ • _____ • _,_~.,.. ,_.~ , ___ .'.

1.29 The 6- kN lo.d P is supported by IWo wooden members of 7S x 125- mm unifonn rectangular cross section which are joined by the simple glued scarf splice shown. Determine the normal and shearing stresses in the glued splice.

SOLUTION

AD ~ (O. 07S )(0. I~sî = 1. '675 "'IO-~ "","

p Z ~(Ô)(/O'!>c.;s~ ~O" _ o -= AD CoS e -= 9. 37SY lcr:s -

G,: S6.s- k PQ.

(b .... ,O'!) s,~ Ljo~ r -=

(~l("I.'!.7S lt 10-3 )

-r -= :< O~ /.I:PQ..

1.30 Two waoden members of75 x 125· mm uniform rectangular cross section are joined by the simple glued scarf spliee shown. Knowing that the maximum a/lowable tensile stress in the glued splice is 500 kPa, determine (a) the largest load P which can he safely supported, (h) the corresponding shearing stress in the spliee.

SOLUTION

Ao ;:

e =

6' -=

p=

(0.075"')(0.1:25'): "1."375" "'/0-'" VV\'t,

Cfo" - 70" ; :<'0° 6" -:: .5"00 x /ù3 fb..

-t~s"e AoO' =-

t ..... GO$ .....

('0 )

('1.375">< 10-$ )(~oo)C' Id) = s. '308S">< l'C''' CO~· !to

?= 5.?./ kN ~

(S:30~S,..10~) s,'., 40· ~ 181."1"'1 ~I09 (:l)(Cf.31S" 1<'/0·\)

rc;: J 2,".{.O k flo.. ~

if" i 11

nf o 1][

01 DI

101 '1 ,- ,

DI

'0 o o

PROBLEM 1.31

bl 0= p e~a9

A..

PROBLEM 1.32

P

P'

::

P'

!

--, ~--,-~'" -- --1 1.31 Two wooden members of3 x 6- in. uniform rectangular cross section are joined

by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the giued splice is 90 psi, determine (0) the larges! load P which can be safely applied, (b) the corresponding tensile stress in the spliee

SOLUTION

e = orO· - '/00 =- $0·

Ao -= (3)(b) = 18

?:: :: lA S ,''1 :l..8

, ( ,,,

-= ..2.1:..:C (;{ )(/11) (90) = 3Z Cf 0 f si", 1e - si~ 100°

p::-. 3'<'''10 J/" --3~"1o coS 1. SO~

/8 = 7$.5

1.32 Two wooden members of3 x 6- in. uniform rectangular cross section arejoined by the simple glued searf spliee shown, Knowing that P ~ 2400 lb, determine the normal and shearing stresses in the glued splice.

SOLtTlON

ê ::: 900 - tlC° '" $0°

Ao: (3)(') '" I~ ;,,1 p.: ::<,-'100 Ah,

1: eo.s"'S ~'too) (j;)$'LSQo := -SS".I 6" = - p.., 1'&

cs= SS.I psi

7: -::: .f. $,'" 29 =- ( ~ '100 .J:p, "',-'-00 0 "- GS.7 ~A ~)(;s)

...

,. .H o o o

o []

o o o

o o o o o

PROBLEM 1.33

in.

PROBLEM 1.34

p

! 1

, ,- . ,,- -~---~ ______ ~--~.1

1.33 A centric load P is applied to the granite block shown. Knowing that the resulting maximum value of the shearing stress ln the black is 2 5 ksi, deterrnine (a) the magnitude ofP, (b) the orientation of the surface on which the maximum shearing stress occun, (c) the normal stress exerted on that surface, (cl) the maximum value of the normal stress in the block.

SOLUTION

A,,:- (b)(t;)

El"" IfSo fo"" l') .... ~ e J "[ .......

(!:t.) y ...... -= ~o .: Ipl=- ~Ao i~...,. = :: /80 k,'p s

(b) si .... Z9 ,. 1

le)

-/80 :;

36 - ~- J<sj

...

1.34 A 240- kip load P is applied to the granite block shown. Detennine the resu!ting maximum value of(a) the normal stress, (b) the shearing stress. Specify the orientation of the plane on which each of these maximum values aCCUTS.

SOLUTION

A" ,. ('-)( (,),. 36 i .. l.

P' • e . - 1 'to >.""'" 6 = - c..,~ - 3& ~!> <J' '" - E;.67 A(J (Q..) Mo-,(.. +ev. sde S+re.~$, ;:' 0 ",t 8 = ~(.)o

~. co--p~:!osj,,~ ~fv..e~ = ~.67 k~l'

Ihf e:o O·

,

o c C D

0 !

0 :0 :0 ie 1

o l o [J

(

o '1

10 1

1

PROBLEM 1.35

p

1.35 A steel pipe of300- mm outer diameter is f.bric.ted !Tom 6- mm- thick plate by welding along a helix which forms an angle of25 0 with a plane perpendicular to the axis of the pipe. Knowing that a 250- kN axial force P is applied to the pipe, determine the normal and shearing stresses in directions respectively normal and tangential to the weld.

SOLUTION 6mm

do -= o. '300"" ro ;, 1 Jo ~ D. IS-o /Vl -11--

"'I-W.Id~

PROBLEM 1.36

p

6mm

'(';. .... 1"0 - t = O. 150 - O.OOb = 0.14 Lf .1'))

Ao"" 1T(rc>~- na.) = TI (O./So'2. - O.I'I'i J,)

e :: :l5 0

:::: S~S'l x/o-s "" ~

- :?S-o )< lo~ C 0$' ?S0 S.S'f )( 10 -)

D :, '- 37./ HP(\. ....

- ~SO )( 103 si ... SO" ('7.. )( 5'. S'f xl 0-3 )

=-/7.:?S x/O ~ 1:: = 17. ~8 Mfb.. 4

1.36 A steel pipe of 300- mm outer diameter is fabric.ted from 6- mm- thick pl.te by welding along a helix wruch farms an angle of25 0 with a plane perpendicularto the axis of the pipe. Knowing that the maximum allowable nonnal and shearing stresses in directions respectively normal and tangential to the weld are a = 50 ~1Pa and t = 30 MPa, detemllne the magnitude P of the largest axial force that can be applied to the pipe.

--H---- SOLUTION

~weld~ d .. : 0.300..,

r.: = ro- t '" fo:: *"/0:: c. 15"0 ""

O.ISO - O.()O.(, = 0.1'11./..,

Ao." 1f (t .. ~- r',.:"):: rr( (J. ISO"- o.''1C1'') = S- S'I ~/O·5 J,ot"

BaseJ 0.. \ rs \ '" 50 Mf",: (5 '" ~ cos"e

_ (S.~ ItIO'''J(So)( 19') po: !:/!fJ

00'1 l't \ = .30

;( A. 1:: p= ~t''I ~6

- eos1. UO -

MA::.. ?:. '= -.E St'" :le ~Ao

::: (1,)(5:5'/)(/ o-!.)( 1,O)C Id" '" 5;" 50·

Jjlo .... .J,'e yo..l.x. ,,1' "P /. 'P = 331 kN

, [1

o o o o

ID

o o o o o [J

o [J

[J

c o o

ID 1

1

1

'1.:~~l~k BC is 6 mm truck, has a Mdth w = 25 mm,:dis~ad: :'f~ steel Mth-a~~ '------1 PROBLEM 1.37 480~ MPa ultimate strength in tension What was the safety factor used if the structure 1

shown was designed to support a l6,kN load P ? ,

i 600mm i

D

p

SOLUTION

Use 60.\1' AC D c..s Q --\'t'e~ body' <1",01 note +-1,,,,,+ 1"'\ e ... loer ra D ,., <t. tiVO­fOol'c:.e "'e~b ... oI'

- (bOO)P = 0 = (boo)(J6)/IO~) ::' F ~p tic =

"110 '180

O.P+i ... .1e }otJ.J {'<.l(' Me...;be.t- 'Be F .. '" 6"t.I A F", -= (4<gOlC/O')(O.ô06)(o.02S)= 7'J..'l.ldN

r-s= F .... : r.. I;~

p

PROBLEM 1.38 1.38 Link BC is 6 mm truck and is made ofa steel with a 450- MFa ultimate strength in tension. What should be ilS width w if the structure shown is being designed to support a 20,kN lo.d P with a factor of safety of J ?

D

p

SOLUTION

Use b<t..- ACD o..s. co. ty.ee. boJy (1)\ cl ~ 0+ e +1, .. 1 "" e,.,. I.e'" ta ~ i, 0. TWO - ~., ... c:.C. "" .. -10 .. "..

'1iO Fa" - '00 f : 0

..

F"." ~,,('oQH:lO!!l9~2. = ,"S>tlt,)'" N "'" '180 lof 80

p

Fo'f' c.. +~.:+,,~ of' s",A eT'j F.S.:: 3 tl the I))+i,., ... +« i_.J .,f' .... e_ bet­

FIJ ~ (F. S,.)(F ... .î' = (3 )(:lS)( 10$) :: 75)( IO/J N

8(.1+ FI> -= 6 u A ' A '" ~ """4;~"!~;& :: 'bf..b7'" 10-' ""," ., A = wt o~ !loi = A = {".'1)(IO

t. O.iJO"

Be

, ,

lJ

[J \

o [J

D o D o o o o DI DI DI

ID 1

DI (

DI DI

PROBLEM 1.39 1.39 Member ABC, which is supported by a pin and bracket at (' and a cable BD. was designed to support the 4-kip load P as shown. Knowing that the ultimate load for cable BD is 25 kips. determine the factor of safety with respect to cable failure.

«1' p

A

T~~B IS in. l_. __ _

18in.--+~12in.~

SOLUTION

Ose "'em ber ABC G.$ ~ -t-V'ee 'body O-i'ld ~ ate -rh ... t 1"'1 ~"" be;- 13 t) i s Q.

-\"'0 - -toV'c.e .... e"" Jo'!".

p

('P e~ 'io·) (30 ', .. ) "l'P .si", LIo° )(15 i ... ) - lFeto co'; 30°)( IS ; .. )

-(Fel> si", 30·)(12 i .. ) ': 0

h.- _ 3:t.':ta '? " (3.(.'~3)(1f) ': b fms- j(' w \t."I'Io \'6.'1'10 • 'P'

e

Fu.+w- of sa:f:e.+y t,,,.. c .... tJe '80 F.~ ': ~ :' ,.;;'5':' 3.''+ ...

PROBLEM 1.40

A~~~:"f-L T B

15 in.

L---l 18 in.-_~12 in.-l

1.40 Knowing that the ultimate load for cable BD is 25 kips and that a factor ofsafety of3.2 with respect to cable failure is required, determine the magnitude of the largest force P which can be safely applied as shown to member ABC.

SOLUTION

0!;~ yr)e,..,~ ABC o..s a. ft"ee. body a..J \'lOfE. '"'~+ ~e""bu Be i.s Go tvlQ - Wc.4!. l'VI e ... b e ....

p

A If:::::.==::::.'( •

(PcoS '10°)(30 ; .. ) +(P si .. '10°)( 15' i .. ) - (FllO GO$ 300 )(I5'i,,)

- (FSD St'", 30°)( 1:1 ,"') :' 0

P :' ,go q'to F. 3~. ,,:è BD -

A2IQwa.Me Poc:ul -t"()I" me .... bëi'

O.Sg oZ,' FalO

aD "il Fil) = J~~~ = F.~.

:lS = 7. 81~5"' /t.ps 3.:<

'P = (O.SU" )(7. 8/;(S ) :' If. SS /,(,'p~ ~

f 'D ' 1 ! -1

[l , J

o o o o

o o o '0

--.- - --."",-, "-.• ' -~ .-. --. --_. '- - -._" --------,.-,,~--<

PROBLEM 1.41

t--==-:L r 'J~m 1.4 m

L 2m

1.41 Members AB and AC of the truss shown consist of bars of square cross section made of the same alloy. It is known that a 20· mm- square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded [fbar AB has a 15. mm· square cross section, detennine (a) the factor of safety for bar AB, (h) the dimensions ofthe cross section ofbar AC ifit is to have the same factor ofsafety as bar AB.

SOLUTION

Le"'j+h Q1 me ... ber Ag

; AB -:: -/0.75 t + o. 'f~ = 0 _ &5' ,.,

~I"Mc: = 0

.. f'tFY = 0

\.1./ AM - (O,7S)(:?8) = 0

Ay-l8::0

A;I.:- 15 I<N

A'I :: n JIN

At 4. 2' F" -= 0 ~ F. A D.85 1.8 - )(" 0 c c

F;.s F. - (0..a.sJ(IS) = /7 kN Ali - O~7S

F~ Ay- FA<. - ~.~ F-'S -= Q

F -= 28 - (".~)(l7) -:: 20 k N M. 0, as

Fol" the +es+ b ... 1'

Foil" tJ, e ~J e",'",P 5'" ~ l'. -... A-

(a.) FOIr bc..r AB F. ..EL. . S.'" Fi. " (aOOIf,o'l(O.OIS)l , 7 Il 10

(b) ro," b4(, AC F.. S. -:: .& = <5"A = . F.ic. F.,

0..:1.. ~ (J::.s.) ~4" = (3. q7)(~O )(/Oa) 6"v 300)( 10'

-3 Cl..:: 1 G.~? )1/0

0,,0..1

!=Ac

'" 3."17

= l'If.7 '1110" MI.

,c; .2? ""... ....

D ID 1 ___________ --,-_______________ -'

!O 1

. 1 o o [l 1

: -~

o DI D,

[J

01 DI Of

ID 1

o DI

(

ID 1

ID 1 !

PROBLEM 1.42

r:-~.75m

0.4 m

--*-B r 1.4 m

1.42 Members AB and AC of the truss shown consist afbars of square cross section made of the same alloy_ It is kno'NTl that a 20-mm-square bar of the same alloy was tested to failure and that an ultimate [oad 01'120 kN was recorded. If a factor of safety of3.2 is to be achieved for bath bars, deterrnine the required dimensions of the cross section of(a) bar AB, (h) bar AC.

SOLUTION

L e,.~+)., o-J: IV> e .... bel' A 13

)". :~O.1S'l.+O.'1& = l the ent.·V'& tf'ÙSS o.!. 0.. +~e /ocdf

Âx-;' 15")< IV '9 :!Mc :> 0

+t~Fy = 0

Use. joi .... t A

I.L\A" -(O.15")(~B) =0

At - U =0 At '28 \(N

Ait t Ay

~s -fret body

~·Z:F,,=o O.7Sr:- A - 0 0.-8-S nI! - )( - e

~~ Floc. t

For- tl,e t-es+ ha. ..

(CI.) h>f' "" ~M b,... AS

0.. ... '" tl=.s.) FM> 6'"

Cl- = (b) Folf' l''le ... be.- AC

b'" :: (F.S.) FA<. 6'"

h =

~ - (O.85")(IS) :: 17 ItN r~8 ~ Q.7S"

Av - FAc. - 0.1i ~a = 0 1 O.IS'

F• :: 2g - ~()(u1. =: 30 kN ,.0:. O.&î

~ -~ A :: (o.ozo) ::- Lloo ></0 M' "Pu ::- 120)( IO~ N

'Pu J;(O '>1/03 =

6'"u : A = ,+00'tC 10·' 300><10' 'PG\.

F: S =.flt.. " ~: '1.

<S"a.. ., FAt. F ... F_.

:: ~3.~)( 17 )( 10') 300 )( (0'

~ ·4 '1.

1 li 1.33 ilia ""

13. '17 'tC/o· 3 PI

F S :: 'P ... ": 6uA =- ~b? ., FA<.. Fk.. FA<. (3. '1..)(20 )( lOS) .4 ...

= =- ~I~ .'33 )1(0 M 2,00 '/CIO"

14. b 1 )/ IO-~ M

•

o o o o o o o o o o o o o D

iD

10

PROBLEM 1.43

:!u kN 120 mm

1.43 The two wooden members shown, which support a 20-kN load, are joined by plywood splices fully glued on the surfaces in contact. The ultimate shearing stress in the glue is 2 8 MPa and the clearance between the members is S mm. Detennine the factor of safety, knowing tbat the length of each splice is L ~ 200 mm.

~L

> SOLUTION

The.,.4. "',-c 1.} So.p ...... "h QV'<o.s cf ~Jùe.. Ecv:J.

Le"'jth of' s"J,ee de~",c.è.. .e A reo... 01' 'j 11,,<

U/+;""de }olo.J

PROBLEM 1.44

20~:'\ 120 mm

'3JVft .,."..ec:>. ....... ~+ -rr"", S "" .. + 10 I<~ of' 5~e ......

Joo.J. :!Cl kN p:: 10" 10& N

l", -:< ~ + C W \'~M. ; ': )e,,~+h ot3J,;e ... "'.) c ':

= i ( L - c) ":: -4:(0.'-00 - O. 00&)::' 0_ 0% "".

A -=)w = (0.0%)(0./:<0) = /I.Sl ></O-S />'IL

Pli -= 'tù k ~ (~.! ltl06 )( Il.5":2. '>iIO~ ) = 32. ~S' ., /c:? N

t'"s -= 'Pu r.. 'P = 3.?3

20kN

1.43 The two wooden membeTS shown, whieh support a 20-kN load, are joined by plywood spliees fully glued on the surfaces in contact. The ultimate shearing stress in the glue is 2.8 MPa and the clearance between the members is 8 mm.

1.44 For the joint and loading ofProb. 1.43, determine the required length L of each spliee if a factor of safety of). 5 is ta be .chieved

SOLUTION

Thell'e O"..t If sepq",o,,+e Q.V'e~ of 'j.Pve ~ éAc..~

~Ive cu'ee.. ~ud +rAl'IS .... ;-f 10 kk) of' oShe" .. .voa.d.

P "" 10 l< 103

N

r?e'fvireJ IJ.H;tn.J., io,..l Pv" (F.S.)(p) -= (3.S)(IO)l/0') = 3S")l/u·· tJ

Re"lvireJ Je"""H ; o~· ec..J,' :iv~ <:\. .... a. fi O· R _ 3S)lld~ ='

Pli: 't'u A" ~ A 'N ..{. :0 " -u 1:"" W - (2.8X/Of»(O./20)

(~)(lo~.IÎ ></0-1

) ... 0.008

·3 /0<1.11)<10 '"

= Z Ib_ '3 )< /o-s '" 2./G "'""

(

i ______________________ ~------------------------~------------

.. .." , ... ~--"' .•.. - ... ,- ...

PROBLEM 1.45

24 kipli

PROBLEM 1.46

1.45 Three t -in. -diameter steel bolts are to be used to attach the steel plate shown

to a wood en beam. Knowing that the plate will support a 24-kip load and that the ultimate shearing stress for the steel used is 52 ksi, determine the factor of safety for this design.

SOLUTION

Fo... eCAGk boJt

Pu = A 't'., =

Per boJt Pol _

F. S. = P -

A =!fd l = -f(~t ,. O,4~18 (6.4LfJ~)(S,n:- :n. ct7 w.'pt

, 1 1"

1.46 Three steel bolts are to be used to attach the steel plate shown to a wooden bearn. Knowing that the plate will support a 24-kip load, that the ultimate shearing stress for the steel used is 52 ksi, and that a factor of safety of 3.37 is desired, determine the required diarneter of the bolts,

SOLUTION

For eCtc~

Re, ... it'cd

boH 'P =.2f = 8 kirs

Pu = (FoS'>? • (3.37)(8):-

/'V. _.& . ,'-LI - A •.

A=~cI~:.

A : ft = _~~1'- = 0.,5 18 ~(;

.J _ J 'f A _Il" le o·S.'1 'ft.) !;J- 11' -T 'T(

:: O.81~S' j""

-

, '\ , ..

~n

[J

o o o o o [J

o o

.0 o o o o

o

PROBLEM 1.47

wood

1.47 A load P is supported as shown by a steel pin which has been inserted in a short . wooden member hanging from the ceiling The ultirnate strength of the wood used is 60 ~a in tension and 7,5 MFa in shear, while the ultimate strength of the steel Îs 150 MPa in shear. Knowing that the diarneter of the pin is d = 16 mm and that the magnitude of the load isP= 20 kN, detennine (al the factor ofsafety for the pin, (bl the required values of band c if the factor ofsafety for the woodenmemberis to be the same as that found in part a for the pin.

SOLUTION

fi: ~O kN 0, • rll"l· A=

Cb) T eVls "0"' ; ..

60

= Pu A = . ...:.. ~~( ':'"'b --d~) W ':' 40 "' ..... .., O.D40 .....

b : r:J + ?u 'N6o

Shectll' lOI wooJ

Do\.! bJe sÀeev'; N _ Pl) L.u - 2 A

= O.Olb + C)o.SI<ï>lIO.3 ... '11./ x/Os "'" (O.O/fO)('-O'CfD') -

b";: y /, 1 .... "" 41

Pù ~ '0.3/ q "1 d N +00' SC'tIOl€ F. S.

e (.o.c.~ a.,>"ec... .$ A = 'fol C

Pu ~wc

- ,,0.3/1)( Ip3 ":' 1 0 O.s 1- JD·~ M - ("),'t..0.OYO)(7.$"'/05.)

C : /(JD.~ ""'111

r

1

L

(

,

1

I]T 1 .

Jo

!

d o o o o o o o o o

ie 1

1

iD! , ~

1

o ID

[J

o o

• -o,~'--------O"-:!o>""",,'~" -~~----o------- -- ~",~-~~-------,-=-"_~co--._,-.:--"".,.._, .... ----"--_____ --- -" -- :-- CO;", -_;~ --,.,..--"-_~'__-_-""'""".ë_ .. -. ____ ~_< _____ •. _

PROBLEM 1.48 1.47 A load Pis supported as shown by a steel pin which has been inserted in a short

wooden mernber hanging from the ceiling. The ultimate strength of the wood used is 60 MPa in tension and 7,5 MPa in shear, while the ultimate strength of the steel is 150 MP. in shear. 1.48 For the support ofProb. 1.47, knowingthat b ~40 mm, c~ 55 mmandd~ 12

mm, delermine the aUowable load P if an overan factor ofsafety 00.2 is desired.

SOLUTION

SQSed 0... JovUe she",,,. i ... pi...,

'Pu = ~A t'v -:: 2lf cJlrv

= :[f(:J..)(O.OlZY(J..$'O>l/O') '= 33.Q3></O' N

?",:, A !Su == w lb - d )<S"u

:: (0.0'10)(0.0'10- 0.01-<)(60)<10&)

.. G7.'- ></0 3 N

8a.,se d Ol'l ~()vbJe sheQtf- i '" i-~e WOOe:!

?u= 2A"l"v:' ~wctf) '= (::l,)(O.O'lO)(D,OSS){7S>t/o6 )

= 33.0 >1/0'3 N

Use oS f'fl-.JJ esf 'Pu = 33.0 )(/ùJ N

AJ)Q",~.bje p.,. p" = 33.0 lt/O3 ~ /0.31></0

1 N F. S. 3,'t

Jo.!.1 kN

U"f]" 7 ..1

lo :

o o o o o o o o o o o o o o o o o o

PROBLEM 1.49

Top<iew

1.49 In the structure shown, an 8- mm- diameter pin is used at A, and 12- mm­diameter pins are used at Band D. Knowing that the ultimate shearing stress is 100 MPa at ail connections and that theultimate normal stress is 250 MPa ineach of the tWQ

links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired.

t-2OO mm-t-1BO mm 112 mm

Bmm<l= te ~ )~ " A B cl

A B c

~. (:;L;]---l ° 1-----1'

- -20mm P

Front view

B

Bmm_ -Bmm

f- D

12 mm ---1 1-Sicle view

SOLUTION

stcd-\c.s: IJse A8C CL!. he .. body.

AS' C

FO,20+0.'B =t p

F. F." Z Ma = C O_~o FA - O./i P :0 0

P=~F"

0.10 Fa" - 0.38 P ... Q

8(t,sed CM dovble shec ..... 'III pi ... A ~ :~ F-A ~ 14,P = :jf(o.OO!)1. :: SO.2"xIO-'~"

~" = ~ToA :: (::I~ 100 )(/0')( S'o.~" )1/0-<:') ... 3.'35/)J Jo 3 N F.S, ~.O

P =: -If F;. =- 3.7::1 )(/O~ N

B ... ~ed ov) Jooble s"'e«1I' i", p''''S ,...1 B o,,,,J D A -= t dl. ': ~ (0.01:1)' ;" 113.10 )(/0';; ",1.

Flio

= :ttoA :: (:l.)('ooltIO' )(113./Ol</o·6) ~ 7.S~ )i/0~ N F: s. 3.0

P = :~ Fiio '" 3. <)7 )( 103 N

So.secl 0'" c.o""pr-esSiolfl , .... J,,,,!<s BD FoY' on~ .Pi .. k A = (0.0.1.0)(0.008) ::

-, 2-160 )1/0 ...,

Fso

= 2.S"A:: (:l)(ltSo lC/o')('ColC/O-CO ) = F.S. 3.0

~,.7 )JIIO~ N

(') - 10 F-r - fëï 80 =

1'= 3.7~ ""0" N

3.12 k~

. $ - .

( '.

i J

o

1 0

1 0

o

DO

o

o

o

o o o

l, 0 1 C

1 )

'.

1]

"J 1

PROBLEM 1.50 1.49 In the structure shown, an 8. mm· diameter pin is used at A, and 12· mm· diameter pins are used at Band D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimatenormal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall mctor of safety of 3.0 is desired

Topview

1.50 In an alternative design for the structure ofProb.l.49, a pin of 10-mm-diameter is to be used at A. Assurning that ail other specifications remain unchanged, determine the allowable load P if an overall factor ofsafety of3.0 is desired.

r 200 mm-t-180 mm--j12 mm

Bmm<Î= .tç : )..1 ,; A B cl

J A B c fi;} 0

if ~ 1--- 20mm P

D 0

Frontview

B

Bmm Bmm

D

12mm....j 1 __ Side view

SOLUTION

stt>.h·c.s: IJse ABC o.s f .. ee body.

A S Co

FO.lO 4= 0./8 =t P

FA Fe" ZMs = C o.~o F" - 0./'& ? :: 0

'P:: 1;f F~ 0.20 Fao - 0.38 p: 0

l=A = 'l'tuA :: (:lX 100 )(/0')( TS..s*t )I/!?-c.) ... F.S. '3.0

p:: if f; :- $.8:1)( IO~ N

BG.sed 0111 Jouble she«1I' i", p;~S ~i B Q. . ..,J D . .. "-

PI '" ~ d1. ': ~ (O.OI,)l ::' 118. 10 )(/0 /'I?

F. - ::tt.A _ (:I.)(/00 lt /O')(113./0)</0-6) ~ lit) - -F. s. 3_0

7.5,+ )C 10'" N

p :: ~ Fat) ~ 3.C!7 ></0 3 N BaSee! 0'"' c:.o"'pr-essiolll in i'oIIks BD

FoY' one ..Pi,,1c A = (o.o~o)(o.008) :: .,.z L.

160 )(/0 ~

Fso

=- 2.SIJ A: (~)(~So 1</0')("°1</0-10) :-

F_S. 3.0 ~'.7 ')1/05

N

'P = /~ Fel) - l'i.04)< lOS N

AJJo",Q,bJe VQ,/oJe of' P ès .s ........ JJest :. 1'-=- 3.'7 ~/O$ N

3.'1 ktJ

L

PROBLEM 1.51

lin.

LH "'f.'

T=\~"D p

1.51 Each of the steel links AB and CD is connected ta a support and ta member BCE

by t -in. - diameter steel pins acting in single shear. Knowing that the ultirnate shearing

stress is 24 ksi for the steel used in the pins and that the ultimate normal stress is 60 ksi for the steel used in the links, determine the allowable load P if an overall factor of safety of3.2 is desired. (Note that the links are not reinforced around the pin hales.)

410 .. ' .~ ~_ 0

~=-:-'

~LJ 1-8 in. 12 in.

SOLUTION

U~e ,~C E A.!) +~e 'tloc1y S c. e

<t=========:::3 ,.--8· .Ie \~--J

'! Ms:: Q

a Fe.;, - ~o 'P " Q

? = fi Fe .. F~1!o

8 F,.,. - I~ P = 0

p

SQ+t, Ri"ks hA..,e +I.e So. ... e we.... ... ... .:1 s~...,e f'''I aI,' ...... e.+eJr j h\Mc.~..) be.iVli) '*+"e S.,. ... e l'Vlo..te.r'· ... .I, T~e'l wiJl hQl/e +he sCt."'e I./l+iWl.hJI)....t.

Bo..seJ 01'\ pi", i", 5 i"'j je sI. e<>J'

A-=-1fol"", .jf(~)2. = O.l%~i ... " FI> = 1:'1.1 A = (2. '1)( o. l't'as) ::-

Ba.seJ 0" t~",~ .. o", '''' Pi h k

A '" (b-d) t = (f - -t )(t) - 0./:<5" ;","

FoJ = fiùA ". (Go )(O.I~.s) = 7.Sb kipS

UJri ;"4te iodJ .f., .. A" k Îs s"'lttlJest; Fu -= 1+. 71~'" k:ps.

APJo"",J.Pe j~ ~ jJ. k F;:: .Et. = 1!t·7L~'1 - 10117'" 1.,'0° ,1'" 'F;S.'3.:t-' "''''''''',''

A~)O""6.,Ufl Poo."'/ fo". stNc:,.+1JI"e -p = ,. f F'::" o. S'89 1(;"$

F .. Sil:( lb.

[

(J [' o o o o o o o

00 [J

o o o o o

1

DI Op)

1.1 ~I

I-------------------------------------r----~

] 1

1

iD 1

1 rJ

o o ~D

o DO

,0 D

iD

D

iUel , 1 1 _~ __ _

J 1

l

PROBLEM 1.52

lin.

1.51 Each of the steel links AB and CD is connected to a support and to rnernber BCE

by t -in.- diarneter steel pins acting in single shear. Knowing that the ultimate shearing

stress is 24 ksi for the steel used in the pins and that the ultimate normal stress is 60 ksi H '4 "i"

f9\~ D

p for the steel used in the links, deterrnine the allowable load P if an overall factor of

,m. ~. of-"."""_,, "",:-.. ,-"'-

'lJ o A -'J."

~Bin. 12 in.

safety of3.2 is desired. (Note that the links are not reinforced around the pin holes, 1.52 An alternative design is being considered to support member BCE ofProb. 1.51

in which Iink CD will be replaced by two links, each of t x l-in. cross section, causiPl~ the pins at C and D to be in double shear. Assuming that ail other specifications rernai n unchanged, determine the allowable load P if an overall factor of safety of 3.2 is desired.

SOLUTION

~~o- ~O 'P = 0

2 F~B - I~ 'P :' 0

B ... ud 0" p"", Â ; ... S'I'l!jill. ~helll'" A ': ~J~ ~ 1f(t)"'::" O./9fôSS ;",t.

Ft,I = ""(liA .. (~'+)(O,I"I~35)= 4. 711"l1<;~

l3~ec:l 0" +e"'s'·o .... , ... .9;",k. AB A:, ('b-cl)t = CI -~ )(~) =- O.I~S' ;","

FU" OuA r (GO)(O.I~S) = 7.5'ol6ps

UJ+i ... .tc. ~u.J *i)'" 'j;", li: Â8!;ts .s~Jlesti..e. fli; =J4~ 11~4 k,'ps ,..., ., . ,

t.oi'~esponJ"1'I1 uP+;I"I<t.fe Joa..l ~f stNJ-V'('-(; ~,=- i Fu = 2Yfl' kips

l3A.SeJ 0111 ?, .. .s a;/- Ca ... "" D 'III dotJbJe shea.r A :: tol'l.. = t(~t ~ 0."'3.5'" ;",a.

Fu" :2 '(vA = ('<)(~'f)(O.I"gs) - 'r.'+~48 j<ifs

(3<Uie~ 0 .... te.ns ;0"" , .... li Il ks '8 C.

A = (b - ~t) t " (1 - 1:)( t),. o. o,~S" IV. .. (one iioll k)

~:: '2 0.. A " (:{.)(IÔO)(O.OG.:lS) = 7.50 kips (tohe" ... bo+!" 'p;t\ks) UJ+i"".te JQ-.J fo~ i:"ks Bt ,s :,W),Jjes.+.; I.e. Fil = 7.S0 U'F$

CO ...... es~""l'Id v.PhIlflA+e Jo~ .fOr !.tr~~ 'p,) = i- Fù= 3.00 ).I.'·rs. AC.+ll.J u}'''''' ... +e ;~.1 tS S~~ies1-.. i.e. PI) = 3.00 l<lF$ AiJowr;..bJe loo..J t-c:>t" strlJc,,-tvre "P .. ft -:. i:~9 : O.'~'a kif

'P -::. '138 A. .04

PROBLEM 1.53 1.53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a 10 x 40- mm uniform rectangular cross section and is made of a steel with h an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20·

250 mm ~ mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa. 1·_________ Determine the overall factor of safety for the links CF and the pins connecting thern to

IIIII\~- 400 mm the horizontal members.

~250mm SOLUTION

"1

F

1-- 0.40

1)~ MIO :: 0

0.110 Fe .. -(O.6S)(a't\fIOS) .. 0

ga..sed 0", talrts;Ojll i" .P;" ks C.F A = (/:,- d) t := (0.0'10 - o.o:<)(o.Oto)::: '2.00 ~/O-' ""'\. (aroeii"k)

!=/J :26"uA= (-2(/.lOO)t/O~)C:lOO""O-') ~ I(;().O)( 103 N

BllI.seJ Oh dOl) !cj~ shec,y';", p,'ns

A :JI: t 2. 7T')Z v 10-' ~ .. '= If Of = "if (0.020 = 314.lb ~ ",

Fu '" 2. tu A = ('2.)(1501</0')(3/'1, Il. )t/O-") = "t't. 211 8 l</Os N

F,) is sfrlaJJeV' VAlve.) i.e. Fu:' qlf.2'18 "'/03

N

"1f.~'18 )1/03

3c:t ~Ioa : 2.4-2

1

0 0 0 0 0 0 0 0

00 n 0 0 0 0 0

Ol] "'- 17)

~I L-

I

1

cr: D.

(

Û 0 0 0 0 [J

0 0 (

0 D 0

;0 l , ,

[J

II n

( 1

\ 1

iD ,

!ID 1

- - ---- " .. -.. _-- .-- -,- --~- -- ... """, ... _._.- --- - ------- - - '-., ---- -.C'- - -

PROBLEM 1.54

8 CI.'5eJ 01/1 teh s,' ....

A :. (1:. - J)t

F'loI : 2 S'vA ::

Bo.sej 0111 double

A ":. JL~t. _ '+ -

rYl

...

1..53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a 10 x 40- mm uniform rectangular cross section and is made ofa steel with an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20~ mm diameter and is made of a steel with an u1timate strength in shear of 150 MPa. Determine the overall factor ofsafety for the links CF and the pins connecting them to the horizontal members.

1.504 Solve Prob. 1.53, assuming that the pins at C and Fhave been replaced by pins

);" h. CF

Use member

F ••

~ (J. 'tO ~---.~I·-D. 2S ...... 11

2.L/ kN

o.qo I='cr - (O.6S)(~'f )fld) - 0

Fcp :. 3<=r 'Il /02 N

(O.()tfO - 0.0$0)(0.010')" 100 )I/o·GWot. (0"" },'",k)

(2. )( I.J 0 0 11/0' ) ( , 00 Il 10- C ) .s

:: 80.0){ 10 N

sheo.o" 1\"1 p'~S

t(O.030)t. 706. gt,; ·G &

= 't-/o m

Ac.+-1J41 F ... ;$ 5Wt.A JJu' V A.JoJ e.) I.e.

FA.et~r of' s~.te+y FoS.:: ~F := 2.0S

; . treXCWfn

l 1 1

D

D

o o o o o o o o o D D o o o D

PROBLEM 1.55

p

tin.

1.55 A steel plate & -in. thick is embedded in a horizontal concrete slab and is used

to anehor a high-strength vertical cable as sho\.Vl1. The diameter of the hale in the plate

is t - in., the ultimate strength of the steel used is 36 ksi, and the ultimate bonding

stress between plate and concrete is 300 psi. Knowing that a factor of safety of 3.60 is desired when P = 2.5 kips. determine (a) the required width a of the plate. (b) the minimum depth b to which a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the lower end of the plate)

SOLUTION

SO.PII,·n~ -tv",

Q.=- cl +

'B c..S e~ 01'1 +e ... s ,0"" 1'1 p.PA-+e

b

A ... (o..-d ') t

Pu ':: fiv A F. s ,. Pu ... .• f'

(FoS.) P ,.. ô.Jt

a. - I.SS'O ;'"

Conc.~+e sj ... ~

ü.,{a - d) t 'P

A:: pe" ... ·;,..,"'feV''' J€-f+/' =

Pu= 't'uA ::2Z'u(a.+tH F .. ::. ? ... -~. P

O. '300 KSI'

b h = (ES.) f :;>(0.+ t) '4J

_ (g.') (2.S) - ('2. )(/.55'0+4)(0.300)

b:, 8. os ;'"

l

l 1

1

,

] 1-

iD o

o o o o o

iD 1

:0 DO o o o o

PROBLEM 1.56

p

; 11 tin. " " " " " "

1.55 A steel plate ~- in. thick is embedded in a horizontal concrete slab and is used

ta anchor a high-strength vertical cable as shawn. The di'meter of the hale in the plate

is i -in .• the ultimate strength of the steel used is 36 ksi, and the ultimate bonding

stress between plate and concrete is 300 psi. 1.56 Determine the factor ofsafety for the cable anchor ofprob. 1.55 when P ~ 3

kips, knowing that a ~ 2 in. and b ~ 7.5 in.

SOLUTION

A = (0.- eI)t , . ,''' =- (2 -~ )(k):- O.3"t 0 ~

Po": ~oA =- (3' )(0.3906): \4.0' I<;rt

t:'S ... &- 11·0' - " ""0 r.. f> - 3 - T. "'"1

Bo..seal 011 shea.r betwe.eVI pPate al1 cl cov>cl"ete s'pQ,.b

~(~+t)b = 2(:).-I-f)(7.S)

"t oJ .,. o. 300 ks i

A :: pe-r-"WI~+e'(' )1 clep+h ::

A" 3'+.(;Q ;",2.

Pu = -ruA = (o. goo) (31.f.i;/1) - 10.lfl kips

F.S. -:: .Et. " 10. 'f 1 .. 3.'+7

P .3

A c+uQ,} .1:~cA-"1I' of sCl.+~+y ioS +A~ $ rv, rJJe". 1/ J..J'l. F.S.=.3.'17 ~

PROBLEM 1.57

!---2.4m_

Fol" dQcwf ..Pouli ... ~

*1.57 A40-kg platform is attached to the endB ofa 50-kg wooden beamAB, which is supported as shown by a pin at A and by a slender st~el rod BC with a 12-kN ultimate load. (a) Using the Load and Resistance Factor Design method with a resistance factor t/>= 0.90 and load factors YD = 1.25 and YL = 1.6, determine the largest load which can be safely placed on the platform. (b) What is the corresponding conventional factor of safety for rod BC ?

SOLUTION

2./f "li; - 1.:1. Wz .­Wl = (Ij.O) ( ,. 8 1)'" 3 cp :(. 4 N

"lV1,:' ($0)( flt. ~I)::' 1./.'0. oS N

"Pc :' (~X'3".t,I) +(f)(I4'O.S).

For .Pive. j)oQ,,J .. I'\~

PL: 3' Mg

-w; = m5 -F,.~"" \.i"'h l'''~

p es l ~ 1'1 c.r \ tell"\' 01'\

Y"?D ... Y'" 'P... = rp 'Pli q> PL! - 1'9 'B. =

y ...

op ..

F. s.

(P.'lO)(I~')(lol) - (J.~S")O.O'~8)1(/O-S) I.G

'= 1.7/8

(JO o o o o o o o

00 n [J

o o lJ o

!

D' OUi

l\

l . J

II o o o o o o o o JO o o o o o o o

,--..\ ( 1

c / ---J -,-

.~!

] 1

1

PROBLEM 1.58

p p

l'

*1.58 TheLoad and Resistance Factor Design method is to be used to select the two cables which will raise and lower a platform supporting two window washers. The platform weighs 160 lb and each of the ",indow washers is assumed to weigh 195 lb with his equipment. Since these workers are free to move on the platform, 75% oftheir total weight and ofthe weight oftheir equipment will be used as the design live load of each cable .. (a) Assuming a resistance factor </>~ 0.85 and load factors rD ~ 1.2 and YL ~ 1.5, determine the required minimum ultimate load of one cable. (b) What is the conventional factor of safety for the selected cables?

SOLUTION

1"1) :Po + -r: Fi = q-J H,

?u ': "fo 'Po + Y'L 'PL cp

'" (1.'4)( ~ li Ibo) + (,.5)( i li ~)I IQS-)

O.8S

= b~'f lb. Co.,,,e ... +,'oll1ai -ttA.,+oV' of s«.fdy

? -::. 'Po + 'PI. '" ~)I 80 ... 0.7.5')(::U 1 qS = 3ns 11:.

~s. Pli --p

PROBLEM 1.59

A B lBOkN 1

·2.25m

~DJ 180kN 1

2.25m

J t=~~t':

By ,si"" .. JGt" +"'·GtVl~..oes ho::- 180

3.7.)' .3

~SOO

1.59 For the truss and loading shown, detennine the average normal stress in member DF, knowing that the cross-sectional area ofthat member is 2500 mm2.

SOLUTION

Usi\'\~ Vhe.+hoJ of joints +0 f;fld Me""be .. -f"""es

JOint 13: A'8 Q .. ."I8D Q\f"e zeof'O .çà'i'G.e me"",be",~.

1". "''';3''+ 2.25"" "" 3.7S ....

Illo

FAr>:- ~~S JI:.. (e ..... p.-ess.'o'" )

137 s;"';}&/.o" +rù""''1,pes

;ns" =

~F = 3.7S

13S lN (co,,",p)

135'></03 N

oc. .. :- ;( SOO )< 10 l'I1

13S><I03

:' - :?SOO ""0'" - S'/.O HP&;.. ~

()

o o o o o o

D D

rJ

lJ [J

o fJ

LJ \~j

1

J~ 1

o o o o o o OC)

o o o o o o o

1 Ci \Il \~

]

PROBLEM 1.60

1201h

Y~M1i!. = 0

1.60 Link AC has a uniform rectangular cross section tin. thick and 1 in. wide. Determine the normal stress in the central portion of that link.

SOLUTION

Use the. FRette. to~e.He'" wiH, +"'0 FIJ.J)e'lS c..s ..... f'"ee \'ooly. Note. H.:\ the c.t<,biq +e...~I'O'" c"-,,ses ",t 1;(.00 iJ..i., cR .. t:.k ",,'se c.ovrJe +0 <Ut ()\;\ ~e hodj.

- (I~ + <.J)( I='~ c.os 300) + (10)( FlIC sin 30°) - 1 ZOO = 0

r __ ~I~Zo==o ____ ~~ rA<.:= IG c.>S30 o - 105; .. 30° = -135. So .Pb.

Av-t! 01. of Pi", k AC: A = , in \t k;., " O. 1 ~S . "-/1'1

PROBLEM 1.61

~ . r P' ,~ ,. -'. -~' p '-'~~

Steel ~ Wood

PROBLEM 1.62

SOLUTION

=- 13S.52 O.12~

1.61 When the force P reached 8 kN, the waoden specimen shawn faited in shear along the surface indicated by the dashed tioe. Determine the average shearing stress along that surface at the time of failure.

A~ec.. bein~ shectl"ed A. '10"'", " 15...... = 1'35"0 ......... = l '3~o ,,/ o' Go l'YI"

'P .. 8,. Id' N

,. J...!!..LO_=-- ." 5. 93 "'lo~ Pa ",5.93 M1h. ~ 13So ... /o·"

1.62 Two wooden planks, each 12 mm thick and 225 mm wide, are joined by the dry mortise joint shawn. Knowing that the wood used shears off along its grain when the average shearing stress reaches 8 1v1Pa, determine the magnitude P of the axial load which will cause the joint to fail.

5i)( CU'eu.s, e~ 16 ... '" " I~ "'... o.o-e be'''~ s"eo. ... e~.

A":"A: A" CS')(I')(I~): \lS~ ... ",,'" :- IIS~ )t10·" .....

i = t.-op ~ '( A l' (8)( 10 t. ) (1I.s";I. ~ 10·') = "1. ~ IdJ

p 225mm--'"

PROBLEM 1.63 1.63 The hydraulic cylinder CF, which partially controls the position of rod DE, has

been locked in the position shown. Member BD is tin. thick and is connected to the

vertical rod by a t· in.' diameter boit. Determine (a) the average shearing stress in the boit, (b) the bearing stress at C in member BD.

I-l l.Bin.

SOLUTION

Use Me'" ber BCD as G\. +t-ee body" ",-",d YI oie th "'+ A ra ; s. &<. -+ .... 0 .fo"'~Q Me .... b e .....

.Q ~e, = 1/ 8" ~ 1. '8'"

= 8.::( ;"'.

8.~ F.~ (4cl>s ~o·)( g~~ F;e) - (4-s;~ ~OO)( ~:; F .... )

_ (7 Co>$Zoo)(ifOO s;~ 7.!>-o) - (7 $ .... 20°)(1100 C-O!; 7S") = c 3.3'0& Fi.r. - 47~9' .. 3S:- 0 :. FIIe ::- 8~8 .If,. Jh.

::"ZF" = 0 - ~ J;B ~ ex + 4-00 Cos 7[;" '" 0

C" :0

.dZ~y :0 0

Cf ::

C "

('oi' g:Z8.4!r..2 _ ,+00 CAS 75":" 78.3'1 il,. 8. ;Z

- g~Â ~g ('8)(822. 'l'll +

il. ;Z 1/ 9'/-. (.5 lb.

l' ct 7. '2.. O.IIO'tS'"

'P :: 1/97.2 it.

0.110"15' ;" ..

10. <B"I /(s,'

(6) 8 e","',' ~~ sfr e ss ,J- Ci",,,,, e"" b &- '8 C D, : 'P = Il'''17. ':l .eh.

Ab =: dt:: (~)(f) :: 6.:Z3 '1-37.5" ;" '-

~ ':' 'P : 11'(7. ~ _ ::' s: Il )( {03 FS,' = S. JI ks,' A o.n'f37S

1

0 0: o o o

1

1

DI DI fll L, ,1

'1

DI (JOI

o Dil

DI 1

0\ DI

DI IJ 1

(J~J !

LI [

________________________________ ~--------I

o o o o o DO o o o o o o

PROBLEM 1.64

7: :::

6 --E­= :?t't

1.64 A + -in.- diameter steel rod AB is fitted to a round hole near end C of the waoden member CD. For the laading shown, determine (a) the maximum average nonnal stress in the wood, (b) the distance b for which the average shearing stress is 90 psi on the surfaces indicated by the dashed lines. (c) the average bearing stress on the wood.

SOLUTION

(QI M <MI i "" ... '" Vlo",,,,,,,i s+r-ess ln +he wood

A .... t : ~ (3- ~ ') :: 1.87S" . \. ,'"

(S"-.E. ,~OO .. - A .. t /.8ÎS = S3S ,

p!.'

-:: 1000 :: (~)(\)(GfO)

7.41 ,,, .... "" L: dt tCI "L A/,

/000

(t )(ir ')

PROBLEM 1.65

P'

r 5mm

'Pu :: :< kt/J

F. s ." ~ = .' p

PROBLEM 1.66

'P =

P

;1.100 15'00

1.65 Two plates, each 3-mm thick, are used to splice a plastic strip as shown. Knowing that the ultimate shearing stress of the bonding between the surfaces is 900 kPa, determine the factor ofsafety with respect to shear when P = 1500 N.

SOLl'TION

B" .. J ~~"': (See -f'-~c) ... eJ

 ': ~ ('0)(:/") +(lS)(6. î "- , ~

-:' \ SOO ..... ., ~ 1 SOc> "'ID ...

1.2>00

,,-----/ 2J>

L Gf~-I--+-,---,,~

.1 VV\,1\;"""e.~ -

1.66 Two wooden members of3.5 x 5.S-in. uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 75 psi, determine the largest axialload P which can be safely applied.

SOLUTION

1'\0 = (3.5 )(5.5) = 1 q. 1.5' ,,,'"

e = '100_ 20· : 70·

& ~I'", ~B (~)( \ "1.:1 S" )(7S2

5,,,, 1400 -= 44'1~ .Rb 4. 4'1 !('(>s

1.67 A steelloop ABCD oflength 1.2 m and of 10-mm diameter is placed as shown around a 24-mm-diameter aluminum rod AC. Cables BE and DF, each of 12-mm diameter, are used to apply the load Q. Knowing that the ultimate strength of the aluminum used for the rod is 260 MPa and that the ultimate strength of the steel used for the loop and the cables is 3.$0 MP,a, determine the largest load Q which can be applied ifan overall factor ofsafety of])s desired.

SOLUTION

U.5i"~ jOi",t 'B o.stl~"'U_ bo.t~ ~~01 c.OIl\"'''f';''''~ $~"'IWI..,fvq

2· t 'F;.. - Q -= 0

Go -:: -! fi"

Os.'" ~ j o.'", + A as CL -t,.ee boJy OlVl"/ co ... si'" e.t';o/I~ S'I ~ .... e,.-fy-1

2. • .! h., - 1=" .. <.. = 0

j.-fQ

8o..seJ 0 ... sfV'e..,~+~ oP c~ le 8P'

. . , Q. = t F:c.

Qu= Ôu A::- Ô;; 1f d '2. = (ttaolf/64) f (D_ 01.1)2. = S".7.q~/03 N

Ba.se J 0'" s-tv-I!"'l,-I-l, o-l' sfeeJ Joof'

Q - G C _ • .s r-:- A _ G r.:- J[ .2. iJ - s' ~ ~ S Wu - S Vu If r:A

-:: 1=(Lf80><IO')fCo_OIO)2.. =

BQ.SeJ 01'\ .s+V-evt,+-~ QP '('oJ AC

Q - a F. _..i c-' A - ..3. ~ JI.j'" ~ - Cf k.,V • 't \Jo - Cf 1;)" .,. (X

= ~ (uo)<'o40)f(o.o~'f)'2..::-

Ac.+c)6.2 vi+: ... ..Je Je...) O'>'~ +I.e s"",Jest ' Q,,::- ~S. 2.'1 )(' Io'~ N

<> IS". 08 Y-/O tJ

" 1S'.08 k tJ

,1 1

Llo

o 1 Il

I-D ) il

PROBLEM 1.68

r 8 in.

t c D

p

1.68 Link AC has a uniform t x t- in. uniform rectangular cross section and is made of a steel with a 60-ksi ultimate normal stress. 1t is connected to a support at A and to member RCD at C by t -in. -diameter pins, while member RCD is connected to

a support at B by a fc; -in.-diameter pin. AH ofthe pins are in single shear and are made of a steel with a 25-ksi ultimate shearing stress. Knowing that an overall factor of safety of3.25 is desired, determine the largest load P which can be safely appliedat D. Note that link AC is not reinforced around the pin holes.

SOLUTION

'DL Me .., 0 (G'fJ F", ) - 10 'P " 0

. F~e. = <. 0833'? p., 0.11-80 1=;. ..

.:!:..Z- F,. .. 0 '9". -! F,. ... ~ 0

'8~ :: f FA<. ." (·i'x.~-()a33 'P) ~ ,. ~S" P

"'''2: f) .. 0 BI' + : Ft.<. - 'P " 0

St' p- f(2.08$f» " - 0.""7 P

a = ~ e~ "'+ Btl. ~ 1. if/ "7 P .)

Bt>SeJ 0", .s+~.,H of ,i,,,,/( Ac: 6" = 60 /1$,'

A ... 't-= (";)(t-I) .. 0.03/25' .... : F~IJ =~A..;t .. C&ol(o.OS/l.r):: 1. a 7S"" k·p.

pc) '" (0.480)(1. 87S') ':" O. "foo /!;p.

\2ASe...t On sf,.e..,fI.. "F pi", ",t c .' Ap;~ " -lf J1. .. tf(ïY'·:: O. 110<f5";","

t:J ': :<.5' lài F~. u " r:, Afi~ ~ (.<5)( O. Il O'fS) ~ ;(. 76/ kir.

'Po) '" <O.410)(2.7'1)"!" /.825" 1<:1',

6o.se~ 0 .... .s+ ... e."\jH rJr p,' ........ t la: Ar;'" = ~ d Z ."

1:3" -: -r:, Af';~ ~ (:lS) (O. 07670 ) ~ /. '7175" k:r.

?" <:: (0_70 588)(I."1/7S) .. 1.353.5" k:f

Ae..fù ... ) 'PIJ ,'s +le SIIvto.)JeJ1·: -PIJ" O. 'tOO k;f'

A)jowoJ,je ".JlJe fo ... -p; -p ~ P ... ~ O.~D ': 0.1..77 b p , FS .3.

' .... '"

/

PROBLEM 1.69 1.69 The two portions of member AB are glued together a10ng a plane forming an angle 8with the horizontal. Knowing that the ultimate stress for the glued joint is 17 MFa in tension and 9 MFa in shear, determine the range of values of 8 for which the factor ofsafety of the member is at least 3.0.

SOLUTION

A. -= (O.030)(O.OSO) -= J.SO)t/O·S .... 1-

P -= lox/oS N Pu ': (F..5)P =' go lCld' /II

8 ~seJ 0'" te.v.d Je 51v-e." ~ 'Pu 2.e \,)(1 ~ A. c.os

cos/è = 5'p~P:: (17 1110")(1.50;</O''!.) = Q. gS 3D ></0'"

cos e = D. "1 ~ 1"1 S' e :- ;(';1. 7"f. e ~ :1..:2.7"t"

B",se>J ..... .ske .. "':t'\~ .sfr.4!~ '2:"tJ :" 11.. si,> e c..>S e :: Pu si",:le A~ :lA

PROBLEM 1.70

III kN

= (;('(I.$'O ><10''- 'tqx 10') __ O. "f00 30" 10 3

e ::- 3:2.0:3-

1. 70 The two portions of member AB are glued together a10ng a plane forming an angle 8with the horizontal. Knowing that the ultimate stress for the glued joint is 17 MPa in tension and 9 IvIPa in shear, determine (a) the value of Oror which the factor of safety of the member is maximum, (b) the corresponding value of the factor of safety. (Hint: Equate the expressions obtained for the factors ofsafety with respect to normal stress and shear.)

SOLUTION

fl o

()[]

o o o o o

[

o o

1

o 1

ID 1

ID

1

OC)

o o o o

ID 1

:0

o

PROBLEM 1.Cl

n

Element 1

A rr:. Ci. = A· :::-1..-

i?R 0 6'RA'f'vl

1.C1 A solid steel rod consisting of n cylindrical elements welded to­gether is subjected to the loading shown. The diameter of element i is denoted by di and the load applied to its lower end by Pi' with the magnitude Pi of this load being assumed positive if Pi is directed downward as shown and negative otherwise. (a) Write a computer program which can be used with either SI or V.S. customary units to determine the average stress in each element of the rod. (h) Use this program to solve Probs. 1.1 and 1.3.

SOLUTION

r OR CEl Iv ~ L 1::' M 'E ri TL:

ft is rhé' 5um 0J the forL.es appJ;ed 10 Hwt eJement and ail I()vlj("ro(j~~:

t.-

F[ ~ ~ Fk

~ fld~ ~ L

!<. = 1

{N EL E M Cf" Ti' -_.~--- ---

Ave stress = fi Ai

OUTPUT~

Problem 1.1 Element Stress (MPa)

problem 1.3 Element Stress (ksi)

1 2

84.883 -96.766

1 2

22.635 17.927

PROBLEM I.C2

20 'N 1/,>., '.' ,

11'/ LINKS p= 20 kN

f.

1.C2 A 20-kN force is applied as shown to the horizontal member A8C. Member A8C has a 10 X 50-mmuniform rectangular cross section and is sup­ported by four vertical links, each of 8 X 36-mm uniform rectangular cross section, Bach of the four pins at A, 8, C, and D has the same diarneter d and is in double shear, (a) Write a computer program to calculate for values of d from 10 to 30 mm, using I-mm increments, (1) the maximum value of the av­erage normal stress in the links connecting pins 8 and D, (2) the average nor­mal stress in the links connecting pins C and E, (3) the average shearing stress in pin 8, (4) the average shearing stress in pin C, (5) the average bearing stress at B in member ABC, (6) the average bearing stress at C in member ABC. (h) Check your program by comparing the values obtained for d = 16 mm with the answers given for Probs, 1.8, 1.23, and 1.24, (c) Use this program tofind the permissible values of the diarneter d of the pins, knowing that the allow­able values of the normal, shearing, and bearing stresses for the steel used are, respectively, 150 MPa, 90 MPa, and 230 MPa, (d) Solve part c, assuming that the thickness of member ABC has been reduced from 10 to 8 mm,

SOLUTION

FB,]){{::\Gkfilv\ OF' ABr: +) 2.f\;1 c.'= 0: 2 Fflp (.B C)-- P( PI c}= 0

(ri~---o-:-:13------- C PB]) =- P (A Ô/2 (B C/ (TENSION) A~----~~------~~

I~ Q,Z5\'Y1 0,4- rn ~ +, 2: f'1B = 0; 2 ~l (BC)- P(AB) =0

• 2 F3J) Fe[;::;- P (AB)/Z (B C) (c0(v1 p.) (i) LINk BD

~B.l; 7'A ic.k ne'3tS = t I

~~.. f1f3])'~ t L (-WL -d) "':1 cL.

!-w['"f C'BIJ =- -+- ~D/ABD (3) P (f! B

'lB '::: fb]) / ('ffd/l;- )

Thicklle'ss = tL

Ac~ -= tL ~L Ci'è, E - - Fè r.iAce

(5) BEARIAjG- ST1,\[SS AT B T.0/'c7::neSs 6/ ,(1)(:'Il)hfî f\C:::- tAC

S'seear B~- FBD/ ( d [l'oc) (6) .l3l:.A'i<./NG STREsS fH C

. S L~ Bea. r C ::: Fet:! Cd t;..c)

(JO o J ]

D

[1

o (JI]

[J

Il o Di

1

U1 [J)

o (JO

III - - 1

----------------------------------------~----------\

J

o o

o o o \JO

1

o o o o

ifJ i

PROBLEM l.C2 CONTINUED PROGRAM OUTPUTS

INPUT DATA FOR PARTS (a), (b), (ch P = 20 kN, AB = 0.25 m, BC = 0.40 m, AC = 0.65 m, TL= 8 mm, WL= 36 mm, TAC = 10 mm, WAC= 50 mm

d Sigma BD Sigma CE Tau B Tau C SigBear B SigBear C

1 1 1 1 1 1 1

0.00 1. 00 2.00 3.00 4.00 5.00 6.00

II. uu 18.00 19.00 20.00 21.00 22.00 23.00 24.00 25.00 26.00 27.00 28.00 29.00 30.00

78.13 81.25 84.64 88.32 92.33 96.73

101.56 lUb.~.l

112.85 119.49 126.95 135.42 145.09

-21.70

~ -21.70 -21.70 -21.70 ~ -21.70 -21.70 -21.70 tlu.tl2 21. 10 JI.59

-21.70 63.86 -21.70 57.31 -21.70 51.73 -21.70 46.92 -21. 70 42.75 -21.70 39.11 -21.70 35.92 -21. 70 33.10 -21.70 30.61 -21.70 28.3.8 -21.70 26.39 -21.70 24.60 -21.70 22.99

79 .. 58

m 125.00

65.77 113.64 55.26 104.17 47.09 96 .15 40.60 89.29 35.37 83.33 31. 08 203.12 78.13 2) . 54 191.18 13.53

~(b) 24.56 180.56 69.44 22.04 1 7l. 05 65.79 19.89 162.50 62.50 18.04 154.76 59.52 16.44 147.73 56.82 15.04 141.30 54.35 13.82 135.42 52.08 12.73 130.00 50.00 11.77 125.00 48.08 10.92 120.37 46.30 10.15' 116.07 44.64

9.46 112.07 43.10 8.84 108.33 41.67

(c) ANSWER: 16 mm ~ d ~ 22 mm .. cc.) CHECK: For d = 22 mm, Tau AC = 65 MPa < 90 MPa OK

INPUT DATA FOR PART (d): P = 20 kN, AB = 0.25 m, BC = 0.40 m, AC =0.65 m, TL= 8 mm, WL=36 mm, TAC = 8 mm, WAC = 50 mm

d Sigma BD Sigma CE Tau B Tau C SigBear B SigBear C

10.00 11.00 12.00 13.00 14.00 15.00 16.00 17.00 18.00 19.00 20.00 21.00 22.00 23.00 24.00 25.00 26.00 27.00 28.00 29.00 30.00

78.13 8l.25 84.64 88.32 92.33 96.73

10l.56 106.91 112.85 119.49 126.95 135.42 145.09

-21. 70 -21.70 -21.70 -21.70 -21.70 -21.70 -21.70 -21. 70 -21.70 -21. 70 -21.70 -21.70 -21.70 -21.70 -21.70 -21.70 -21.70 -21.70 -21.70 -21.70 -21. 70

~ i4 80.82 71.59 63.86 57.31 51.73 46.92 42.75 39.11 35.92 33.10 30.61 28.38 26.39 24.60 22.99

79.58 65.77 55.26 47.09 40.60 35.37 31.08 27.54 24.56 22.04 19.89 18.04 16.44 15.04 13.82 12.73 11.77 10.92 10.15

9.46 8.84

156.25 142.05 130.21 120.19 111.61 104.17

97.66 91.91 86.81 82.24 78.13 74.40 71.02 67.93 65.10 62.50 60.10 57.87 55.80 53.88 52.08

CHE (d) ANSWER: 18 mm ~ d ~ 22 mm CK: For d = 22 mm, Tau AC = 81.25 MPa < 90 MPa OK

PROBLEM l.e3

0.5 in.

1.C3 Two horizontal 5-kip forces are applied to pin B of the assembly shown. Each of the three pins al A. B, and Chas the same diameter d and is in double shear. (a) Write a computer program to calculate for values of d from 0.50 to 1.50 in., using 0.05-in. increments, (1) the maximum value of the av­erage normal stress in member AB, (2) the average normal stress in member BC, (3) the average shearing stress in pin A, (4) the average shearing stress in pin C, (5) the average bearing stress at A in member AB, (6) the average bear­ing stress al C in member BC, (7) the average bearing stress at B in member BC. (h) Check your program by comparing the values obtained for d = 0.8 in. with the answers given for Probs. 1.9, 1.25, and 1.26. (c) Use !his program to find the permissible values of Ihe diameter d of the pins, knowing that the al­lowable values of the normal, shearing, and bearing stresses for the steel used are, respectively, 22 ksi, 13 ksi, and 36 ksi. (tf) Solve part c, assunting Ihal a new design is being investigated, in which the thickness and width of the two members are changed, respectively, from 0.5 to 0.3 in. and from 1.8 in. to 2.4 in.

SOLUTION

fORCES {III /'v1EN1BER5 AB AN.D Be FRE.E BOj)y~ PINB FRD/'1 FOR ce. Tf<.1 A (\/ G' L.. ë. :

13

/~~ ~; ZP :s:(.~BC 1 f".!; F" r

0-- 000 ,'i5" 2P ~.

fA8 _ Fee _ 2? Sin If,' - sin 60'- $iffi7Y

~ =- ZP(Sin45,/sin 75")

f8c = 2. p(s in 60/.:5 in 75'

(j)fi1AX,flVE,5TRE.SS JN?lB (z) AVE, S TR =: 5S IN Be rWB lA! /e( fh =- W­

Thickne..~s =t AA8 = (W--d) t

f;,.e 0';1 B:: Ffr B / A M:3-

(3) PIN A f:A= UAB/2 )/(rrdj4)

('.JJBfA'P..ING srR~<;5 ArA 5ZJ3 Bear fl :::- (-AB / ci t (7) B ËC.A 1<.11-1(.. S Tk:Hc) AT 6

IN (vi E/v/5E. R Be 5113 &Ar B:: FB C / 2ci 'C

F&~ n3~

(4-) PUll C

ABC. = 1).) t ~G= FEe/ABC

'èc .:::- ('F5 ': /2)/( rr cl ïlf ) (Ç,) BEt-i(\JNG 5mrS5 Il T' C

SiS Bear C:::: Fic/dt

(CONTINUED)

[

[J

o

o

.00 1 /

o

o o

o ID '1

iD

PROBLEM I.C3 CONTINUED PROGRAM OUTPUTS

INPUT DATA FOR PARTS (a), (b), (c): P = 5 kips, w = 1.8 in., t = 0.5 in.

D in.

SIGAB SIGBC ksi. ksi

TAUA TAUC SIGBRGA SIGBRGC SIGBRGB ksi ksi ksi ksi ksi

0.500 11.262 -9.962 29.282 35.863 17.932 ~.550 11.713 -9.962 26.620 32.603 16.301 D.600 12.201 -9.962 5 24.402 29.886 14.943 0.650 12.731 -9.962 11.030 '~~~~ 22.525 27.587 13.793 0.700 13.310 -9.962 9.511 - 20.916 25.616 12.808 0.750 13.944 -9.962 8.285 10.147 19.521 23 .. 909 11.954 ----(h' 0.800 14.641 -9.962 7.282 8.918 18.301 22.414 11.207 -r-' ~ ~0~.~8~5~0--~1~5~.~4~1~2--~9~.~9~6~2--~6~.~4~5~0--~7~.i9~0~O--T1~7~.~2~2~5--~2Ti~.~O~9~6--Ti~o~.~5~4~8--0.900 16.268 -9.962 5.754 7.047 16.268 19.924 9.962 0.950 17.225 -9.962 5.164 6.324 15.412 18.875 9.438 1.000 18.301 -9.962 4.660 5.708 14.641 17.932 8.966 1.050 19.521 -9.962 4.227 5.177 13.944 17.078 8.539 1.100 20.916 -9.962 3.852 4.717 13.310 16.301 8.151 1.150 -9.962 3.524 4.316 12.731 15.593 7.796 1.200 -9.962 3.236 3.964 12.201 14.943 7.471 1.250 -9.962 2.983 3.653 11.713 14.345 7.173 1.300 -9.962 2.758 3.377 11.262 13.793 6.897 1.350 -9.962 2.557 3.132 10.845 13.283 6.641 1.400 ~. 0 -9.962 2.378 2.912 10.458 12.808 6.404 1.450 . 1 -9.962 2.217 2.715 10.097 12.367 6.183 1.500 ij .871 -9.962 2.071 2.537 9.761 11.954 5.977

(ci ANSWER: 0.70 in. ~ d ~ 1.10 in. .04

INPUT DATA FOR PART (d): P = 5 kips, w = 2.4 in., t = 0.3 in.

D SIGAB SIGBC in. ksi ksi

0.500 0·550 C.600 0.650 0.700 0.750 0.800 0.850 0.900 0.950 1.000 1. 050 1.100 1.150 1.200

12.843 -12.452 13.190 -12.452 13.556 -12.452 13.944 -12.452 14.354 -12.452 14.789 -12.452 15.251 -12.452 15.743 -12.452 16.268 -12.452 16.829 -12.452 17.430 -12.452 18.075 -12.452 18.771 -12.452 19.521 -12.452 20.335 -12.452

i:;;~ ~21:'219 =i~::;~ 1.350 ~'-12.452 1.400 -12.452 1.450 -12.452 1.500. -12.452

TAUA ksi

4 11.030

9.511 8.285 7.282 6.450 5.754 5.164 4.660 4.227 3.852 3.524 3.236 2.983 2.758 2.557 2.378 2.217 2.071

TAUC ksi

10.147 8.918 7.900 7.047 6.324 5.708 5.177 4.717 4.316 3.964 3.653 3.377 3.132 2.912 2.715 2.537

SIGBRGA SIGBRGC SIGBRGB ksi ksi ksi

29.886 27.169 24.905 22.989 21.347 19.924 18.679 17.580 16.603 15.729 14.943 14.231 13.584 12.994 12.452 11.954 11. 4 95 11.069 10.674 10.305

9.962

(d) ANSWER: 0.85 in. ~ d ~ 1.25 in. ....

(c)

(ci)

PROBLEM l.C4

a p

Ac!B""~ T B

15 in.

L __ -_18 in,---+~12 in.-\

~ - ( B t n. *",,..---,4 C JZin.

1.C4 A 4-kip force P fonning an angle" with the vertical is applied as shown to member ABC. which is supported by a pin and bracket at C and by a cable BD fonning an angle (3 with the horizontal. (a) Knowing that the ulti­mate load of the cable is 25 kips. write a computer program to construct a table of the values of the factor of safety of the cable for values of " and (3 from 0 to 45°. using increments in lX and {3 corresponding to 0.1 increments in tan" and tan (3. (h) Check that for any given value of" the maximum value of the factor of safety is obtained for (3 = 38.66° and explain why. (c) Deter­mine the smallest possible value of the factor of safety for (3 = 38.66°. as weil as the corresponding value of lX, and explain the result obtained.

SOLUTION

F +YL MC-=: 0: (P:;i17c0(;"SlnJt(Pc'oJ00(30iy

- (F cO.:,~)(61n.) -(Fsin (3)(IZ in) =0

F= P

F. 5, =

/,Sil1(X + 30 C050<..

1 S CoS ~ + r 2 5 i il (J Fuit / F

OUTPUT FOR P- Ij kifs AND 6dt =-g.QJ:0f-~ VALUES OF FS

BETA 0 5.71 11.31 16.70 21. 80 26.56 30.96 34.99 p-8:'66 . 41. 99 45.00

ALPHA 0.000 3.125 3.358 3.555 3.712 3.830 3.913 3.966 3.994 ~. 002 3.995 3.977 5.711 2.991 3 :214 3.402 3.552 3.666 3.745 3.796 3.823 ~. 830 3.824 3.807

11.310 2.897 3.113 3.295 3.441 3.551 3.628 3.677 3.703 .710 3.704 3.687 16.699 2.837 3.049 3.227 3.370 3.477 3.553 3.600 3.626 .633 3.627 3.611 21.801 2.805 3.014 3.190 3.331 3.438 3.512 3.560 3.585 .592 3.586 3.570

65 2.795 3.004 3.179 3.320 3.426 3.500 3.547 3.572 .579 3.573 3. 5581 30.964 2.803 3.013 3.189 3.330 3.436 3.510 3.558 3.583 .590 ,~:~~; ~:~~~ 34.992 2.826 3.036 3.214 3.356 3.463 3.538 3.586 3.611 .619 38.660 2.859 3.072 3.252 3.395 3.503 3.579 3.628 3.653 .661 3.655 3.638 41.987 2.899 3.116 3.298 3.444 3.554 3.631 3.680 3.706 .713 3.707 3.690 45.000 2.946 3.166 3.351 3.499 3.611 3.689 3.739 3.765 .773 3.767 3.750

tCb) (6) Wh en (3 =- 3 8,66)" tûn(3 -= Q,8 and cable BD is fer ff'n t!:I,'c u lar

le: fhe lever Q.r/Y] Be,

Cc) r. s. :- 3.579 for CV = Z 6.6 0,; P /5 ferpend (cvlcu- te. fhé le ver Qrm Ac NOTE:

The va.lve P.S. =-3.579 ;s il4e Sf'YJ 4 //f5t Of theVtJ./ue5 of r.S. é_O(re5p0Vld"I)~ to (3 =- .3 8.66 0 cuje! the la'(ged- Df -rhDS-e Cc rrH pc /Ici ;'\5' tù eX: 26,6

0• The point 6(--1, G. b ~ f3 = 380 bb 6

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PROBLEM I.CS

p

She6...-; nj

0) Po~: ~fùr rSN

1.C5 A load P is supported as shown by two wooden members of uni­form rectangular cross section which are joined by a simple glued scarf splice. (a) Denoting by Uu and TU, respectively, the ultimate strength of the joint in tension and in shear, write a computer program which, for given values of a, b, P, uu.and 'Tu. expressed in either SI or V.S. customary nnits, and for val­ues of" from 5 to 85' at 5' intervals, can be used to calculate (1) the normal stress in the joint, (2) the shearing stress in the joint, (3) the factor of safety relative to failure in tension, (4) the factor of safety relative to failure in shear, (5) the overall factor of safety for the glued joint. (h) Apply this prograrn, us­ing the dimensions and loading of the members of Probs. 1.29 and 1.32, know­ing that Uu = 1.26 MPa and TU = 1.50 MPa for the glue used in Prob. 1.29, and that Uu = 150 psi and Tu = 214 psi for the glue used in Prob. 1.32. (c) Verify in each of these two cases that the shearing stress is maximum for Cl:' = 45°,

SOLUTION

.Dra.w fhe FB. d t'Off rQtl1 of /OVler mem he r:

~ Z~.:::: o~.' - V + PcosD\.= 0 V=- Pco-.J 0(

-t/'Z.~ =; 0: r - P6iI1CX::: 0 f == PsinO(

Ar~a.-::: Q. bl sin lV

N tJrrrJa. / .s rreS5 :

CI =.L - (P/a-b) Sffl?:O( Arec<-

S7r e.55: cc-- ~ ~ = CP/ab) SihO< CDSC{' Arec<.-

fensie>T/ (nvrlYlC{.! ,5fn=.5Ses)

- Œu/O'

U) r. 5. for 5 hea..r ; F 55:::- 'Cu / {;

ej OVERALl f.S,: FS ::: The smo.//er ()f r 51'1 and F.55,

(CONTINUED)

PROBLEMl.CS CONTINUED f.pOG>Rf\1V\ OùTPùTS

For Pn'. 1,20: p:: 01<N O\..~ 12.5m1lÎ) b=75mY'rÎ, 0<=70°) Ç:-!.26 MPc:1-> '2'u-= 1. 50 MPQ..,

ALPHA

5.0000 10.0000 15.0000 .. 0000

:<5.0000 30.0000 35.0000 40.0000 45.0000 50.0000 55.0000 60.0000 65.0000 :ZO POOO 75.0000 80.0000 85.0000

SIG(MPa)

0.0049 0.0193 0.0429 0.0749 0.1143 0.1600 0.2106 0.2644 0.3200 0.3756 0.4294 0.4800 0.5257 0.5651 0.5971 0.6207 0.6351

TAU (MPa)

0.0556 0.1094 0.1600 0.2057 0.2451 0.2771 0.3007 0.3151 0.3200 0.3151 0.3007 0.2771 0.2451 0.2057 0.1600 0.1094 0.0556

FSN

259.1782 65.2905 29.3899 16.8301 11.0229

7.8750 5.9842 4.7649 3.9375 3.3549 2.9340 2.6250 2.3968 2.2296 2.1101 2.0300 1. 9838

for Prob, 1.:'2: P= 2~l)of.b

FSS

26.9942 13.7053

9.3750 7.2925 6.1191 5.4127 4.9883 4.7598 4.6875 4.7598 4.9883 5.4127 6.1191 7.2925 9.3750

13.7053 26.9942

FS

26.9942 13.7053

9.3750 7.2925 6.1191 5.4127 4.9883 4 .. 7598 3.9375 3.3549 2.934()· 2.6250 2.3968 2.2296 2.1101 2.0300 1.9838

~l (c)

~r (b)

Ct. =- 6 in.) b = 3 in.) 0( = 't-O~ ~ =- /50 p6G." 'Zu -= 2ft;- p3~,

ALPHA

5.0000 10·0000

.. 0000 ~.j.0000

25.0000 30.0000 35.0000 40.0000 45.0000 50.0000 55.0000 60.0000 65.0000 70.0000 75.0000 80.0000 85.0000

SIG (psi) TAU (psi)

1.0128 4.0205 8.9316

15.5970 23.8142 33.3333 43.8653 55.0901 66.6667 78.2432 89.4680

100.0000 109.5192 117.7363 124.4017 129.3128 132.3205

11.5765 22.8013 33.3333 42.8525 51.0696 57.7350 62.6462 65.653§ 66.6667 65.6538 62.6462 57.7350 51.0696 42.8525 33.3333 22.8013 11.5765

FSN

148.1018 37.3089 16.7942

9.6172 6.2988 4.5000 3.4196 ,? 7228 2.2500 1.9171 1.6766 1.5000 1.3696 1.2740 1.2058 1.1600 1.1336

FSS

18.4857 9.3854 6.4200 4.9939 4.1904 3.7066 3.4160 3.2595 3. 2100 3.2595 3.4160 3.7066 4.1904 4.9939 6.4200 9.3854

18.4857

FS

18.4857 9.3854 6.4200 4.9939 4.1904 3.7066 3.4160 2.7228 2.2500 1.9171 1.6766 1.5000 1. 3696 1.2740 1. 2058 1.1600 1.1336

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PROBLEM l.C6

Top view r Bmm

1-200 mm-j-lBO mm112 mm

l L~ -1 <F Cp . j'.; , %1;M~t

l.C6 Member ABC is supported by a pin and bracket at A and by two links which are pin-connected to the member atB and to a fixed support atD. (a) Write a computer program to calculate the allowable load P", for any gi .-en values of (1) the diarneter d, of the pin atA. (2) the common diarneter li, of the pins atB and D, (3) the ultimate normal stress (Ju in each ofthe two links, (4) the ultimate shearing stress ~u in each ofthethree pins, (5) the desired overall factorofsafety F.S. Your prograrn should also indicate which of the following three stresses is critical: the normal stress in the links, the shearing stress in the pin at A, or the shearing stress in the pins at B and D. (b and c) Check your prograrn by using the data ofProbs. 1.49 and 1.50, respectively, and comparing the answers obtained for p .. with those given in the tex!. (d) Use your prograrn to determine the allowable 10adP"" as well as which of the stresses is critical, when d, ~ d, ~ 15 mm, (Ju ~ 110 MPa for aluminum links, ~u ~ 100 MPa for steel pins, and F.S. ~ 3.2.

Bmm

Frontview 12mm

SOLUTION B

Bmm i'

, Side view

la) f-.]3. ]) II-\G RAM

,A,._---:~-o._ t ZOD mm' t';/-a=o=rn=;+f ~ F.ED P

'P - 200 F. 1 - /BO ft

(2) Ëùr 5'IV€11 clpof r.:;,if)$ jj andJ).· 'E])=- 2. (1:u /F.5){fïdt;lf),) Pz = 5~~ &.1J (3) FOr ()/titl1c..te..5f~':'$ in lin1;$ B.D: lé!)= 2(~/F5)(O.02)(D,OD8)) p=2:00 f$ (4) Dr vi t, .s/l~trr 'fil( .::;iYe5! in f'. in!;: D • f'/ fi f P :3 }90

cr ............... '-- rf(. 1$ 1e $ma er 0 CIno P-U) rD( de5;rf'd 6vecaJ/ [S,,: ?5" if, fhe srna/ler- of F; o.~j Pif 2

If Fl < Pif, 5 fre!,s if; cr/fi cal ; (') li () ks 1 f p.'i < P,'i cU.ld P, < P. l 5f(('$$ tS crihca l ln pin A If P" < ~ rU) el R.: < ,?, ) S h .... ~% ls c ri Nud in p inr..B CtYl ci D ?RO&rRAM OUTPUTS

(6) ['Db, /,49, UlrA:d,-=amll1)c\z=12nlmJC:)=2>OMPa,ft;.:; 1 OOYl Pû , r:S,=3,D PalL=-3,72/:...N. stress in pil1 A ;, Cr/tiGa/ ~

(c) .?ïob. l, 51XM TA: dr~~ fDf/ln), dL ==12. mW), (10-= ~5'Oft1Pa., 'l'v=- IDoMPiJ-J:S, = 3.0

F:rl/: 5:~7kN.: Hrf'.5.S in pins Band]) ;$ cr/Nc..aL ~

(cl) PfJ Tft: ~ = d2: 15 mm j Vu = J/O l'1Pa.. 7:;, =/QJtlJ fVlPa, P.S .. :: 3. L

Pa,1{:::-5,79kN. Stress in fil1ks ls critiLaL ....

I~------------------------------------------~------------