Statics meriam 5th edition

Statics meriam 5th edition

USE OF THE INSTRUCTOR’S MANUAL

The problem solution portion of this manual has been prepared for the instructor who wishes to occasionally refer to the authors’ method of solution

or who wishes to check the answer of his (her) solution with the result

obtained by the authors In the interest of space and the associated cost of educational materials, the solutions are very concise Because the problem solution material is not intended for posting of solutions or classroom presentation, the authors request that it not be used for these purposes

In the transparency master section there are approximately 40 solved problems selected to illustrate typical applications These problems are different from and in addition to those in the textbook Instructors who have adopted the textbook are granted permission to reproduce these masters for classroom use.

5/|_| From Table D/3, “the horitantal

Coordinate ot the centroid is

+ Œt(HÌ - sự 3

The vertical coordinate is

woo

[+ “37 = 3.67

5/2 | From Sample Problem 5,

Rl

nl cl 1

IN c “I (80 mm

_ 2023 _ ~f]6.4 mm

(Disk - shaped element viewed, edge-on.)

#8 ]zA:x4, A fer Bigs aly Spe [36

Sle ⁄ dm= fax = (P= mass ger unit length)

Ki m= fam = [f4x> [fL(c%)4x

Tre A= Sytn= “b sin Se dx

w= ———- ee Sect _ Ath fet 2+

A= Sydx = (Ta AE Jon `

= [ax- =Œœx` + asin”) | -a*(-4)

ƒ%c4A = >e* ~ {Tax a Aare] ax

i

NIH

Ser "5 + Notes Shaded, element

2, t ty =R p2]i8 a circular slice

529] ame f4V= f4A+ = Ef (b-a)ay

sbi | z = Seat x

dY* #(s7-4°)áz

= Fl(as Vara?) 0 Jaz

=E[ar2*+ zaya2* Jae

ABR z ake so Áz4/43 g ze Axe

Joo

he Sak= a (le fay) ay = Ệ huy £ ay

* soe dy = Gor [Hares oh ge

5/34

tự, Bela

sae

TP

Sense

Ve [mG Mr \de = «fl [a*- (a~ 2 ]á=

= af? (2az-2)de = Wfoe- ST

74-8 '% £rom Sample Problem 5/1, for eheme 9-4027) “my Gg ĐIDPY By

2E)

»5/37| Let f= mass per unit

& Shi |Arom Sample Problem 272 centreidal Coordinate

for elemental ring 7s igs arlir

a= ar l2ajdr = 20rVa*-(r-R} dr

Rea Sed = 4 VaR CRF tr

` ‘Ng Ve San - *x Ất = Henk’) amps #x(%

Jxcad = J (rsingssin 0)(r* cin gag a0 dr)

Compace to X= FR for wo hele.)

Note: A hemispherical shell of “radius

rand thickness dr would be @ better element.

SAT (Din im mm) Components :

5 (O 175% 210 rectangle

@® Semicicde 100( @ Tuo triangles

alo) + 5(99)-+ 1(180)

=10

2415+] Bea

Am kG mk mye đợt | xy | mm | em |@.mmMg.mm lo.àz|s.7? | 3ø [=?Zz|zose|=ø2#

mm Com A ok 3 Z AR AG AB

“¡1 > fa? 2610 (10°) "Sum?

foe = 78° ("*) ky Jom” (Note STZ must not be

Semicirewlor rod ive straight rods 3s semicirewlar pla

Length = 35 mm

“ " V„*- 5Œ3)*(2§) = - 15 10 mm

Zs &(Is+ 35) = 22,5 mm Length = 25mm

5/65

26-781 72.0 | 24.0 | 7720 | 4080 |

se-20| 7.0 | 18.0 | 190 | 3420 2-22) 52 | 704 | 210 | 204

ZV

+

lẽ R (Fairly close + =!)

So #(h*-4ah4 20%) =O) h= a(z‡{x)

The plus sign is rejected becawe h must

.2/63|@= Semé circular shell

5[14| vs OFA = zr(#*‡⁄2)#⁄/2122) = 3620 mmŠ

Tetol body surface aver is

Ap ALT Ags MPa t ]

3⁄80 | A=2u// + ath

= 2n/E.2)34 f77(#X/8) = 2204 ae

Neo of gal fer Zecats = 2224, 2+ 6102 25/

Z2 =

SE 2700.8) = 1.68 m From Sample Prob 5/1,

Syed = {fe Hsin E] [usin Tư]

-ấ ‘ab Ban + C= sin™ ay

2⁄1 Sguare ! A= 80*26400 m*

áez@:.A=4 (ao 722m 72m Jom 22: 4:47/27

5h2| From the solution + Trt 5/7 5

GE Ma =o: Rg (0.0) 1-8(9.18)#0, Re= O45 kN

+2ZF=o: O.45- 18+ Raz0, fÑa= l35 kN

R= Zk(w)('h) = kul CX=E

GXLMa= 0: 8s(0)— =wol (È)ze, R= 4 vạt MEE = 0: 0l =2wal+a =o, Ña =2 vạt

⁄2ZMạ=O: 24o()— 8oo(2.5)~ 329 (6.33)

+ 8u (S)=9

By = TẾ7 l6 ZFy =o + Ay +7157- 240~ 800-320 = 0

Ay = 603 ib

ZF, =o > Ax =o

Raz 241 KN

@ EMgso: ~ Sar (6.5) +P(4)- 20 =o

P=6,%6 KN

a

bes Re

The load reduces to 4 Couple )

Ss The reactions form a Couple,

Re= 7 kN

+ XF=o: 74 Ra #66, Ra = TKN

[Rs ye Wy ekint Kye hye?

(Alternatively , GEM =! Ma Car =0 Maz er)

5|II3 P GEMaro: Ret -P Ao

SIS] 4 kn akw ĐEMa9:4)v2l)7 48/75

5 [lle Ra =Re (load is a couple)

Mz 3!@e, Enea 3M 1)

At x= bes

sie 4

ee)

2M,20; 1.6 + LOR, 2(43) sở Rye Lt AN

From FBD et entire be, Ra=Z.#kN, Mạ=-W.4 kNm

At middie of beam + V= 2.4 kN

M= 4.3 KNem

5/123] (See beam elemert jouer left)

5/Iz4 z IMy=0; Fh-06 72

⁄5-2j 2,=8*44

S[l2T| AR MAK" From We Wy tka l95EkỦ:

At x= 6m:

= -b600 N M= 8400 ~ Go0(6)

5/130) R= oo (3) = 2490 N

M9: Re(4)~ 269 (4.6)

' 1

i Bool ÿ | 4499Nm ~ 4200 = 0, Re = 1400 N

“Ug XF=o > V=- I400Nn

Woon ZM=O => M= 12600 - loon

8/81 | (Ra tRo from Pab 5/3)

5/135] From areas under sheae diegram +

“XK Muy‡ IMỊ

= zu (oz)

= 0.024uLt

Do K

C= Wy cos8 + 2M sina, C= 201 , T (0 cos @+sind) d

Y= Wo si ao - 2m cos 9) V= 22 @sind - cos “ar @sind - exs8)

2

e TÑy

Tey = loll I5(5) = 236 |b

Ea, Si20: š= lO/74-sinh nang = ILS) ft

L=2s = 23.0 ft

Eliminate T, +o

obtain

AtA: 1” %g` ~ 350%, + [1 S0a=o

Quadratic selutibn: +g= 249 m (reject) , 60.4 m

From Eq, S/H @Bi z= Boy To= 2170 N

For sectin 08: (mmg)* = TH +(te)*

45/14 Please refer to The dingm in The

Solution ty a SÍ4I Eq 5ỈM :

#4 A: 1“ Ze Leosh Fe (ino ve) -1]

Solve These “tuto equations numencally ty obtain

Tos Tây soe = 21.0(10%) N

Maxim tension 18 TH, =VTo xÙ,

= NÍ[eLe(t93){” + [so (4.81) (S8.0)]* = 35.6(10*)N

OF "Tuy Ƒ 356 kN

los* 28 —(cash ose) _\)

Solve numerically t obtain Ty = 408 Ib

Poveiic Ry S/H: y= MEE AR

SAI sar bx? when x=0, wre,

*5|I45

Ey

*

At BGs & (ooh i") ()

Ae A lhe (coh =) &)

Solve Eqs (1) Š$ (3) numerically to obtain

Numerical or graphical solution: To = Silo N

Ey 5/2) 1 T= Te cosh = 3llo cosh ee

W= O.51(.8M100) = 500.3 N T= Mir = 400/25» /600N 460; P-16005in 30%: 0

2) 26,20; F- $00.3 ~ 1600 cos 30°20

v.v 892m

Hz jy-y, © 8% 7m

x5jI5| @) Use w= “Ar L281) = NLT Nha

= s[le FEY Bt] Sen

So the required length is tee 254 = l32 m

&) Eq 51: y= eesh “|

Ae A a tie [ewsh bạo, “ty

Numeriool solution : Ty = 65.5 N

Se yo = 5.57[cosh coment i (see pists)

Eq 5]Z2»: Sa * es sink wer = 5.10 m

The required legth is he=2Sa > [140 m

Even with an expanded vertical scake,

Yr and Yo are Nearly indistinguisheble from

each other Note that Yr is above Yo

except ot The end points @and ®) and the center, So the fact th Le>hp mokes Sense!

Bi Yatbo = + 2 (Cosh 7 (age 3)-| )

Eq, SÍA» : Ss TT” Sinh oe

Eq SIN © 8+ yqe5= FL cen AGarA) -i]

Fq 5/22@A! zoo=T, TAYA

Numerical solution of abore ‘three equations:

Te A(Kaw2) Wika

Sg-Sq = 13.02 = P| sinh TC ‘sinh Te,

Numerical solution of above three equations :

45/61 | Architects plan + (Ta)arh= @(109) = 600 N

Ey 5J22: T= Totpy ) so Ta = 65.5 +laa (I681%€)Z TA Ñ

Percent Increase n= TES (00) = 210% (I)

From FBD of junction ring at Ay

44EFzo: Z2TA singa- W=0

oe ETox nya | sin [tan ( sinh F* I} -X =o()

Ey SIV@ A: Yat Fe Leo oh XP, ma] @

Eq, SÏI\(@ ® : gạ+h = Flees oy Ay -l] 6)

Eq lao: Sp-Sa7 $2 [sion A627) - sin inn] = 5

Solution of ()-@) with Weo: h= 5.570

With Wto: h=6.30m

So S= 6.30-5,5] = 0.724 m

SỈk#] Force on bottom = weight of water

= £4 V= (ooo 53» (ta B) 3m) (sm)(s#m)

= S24N_ (doun, et center of bottom)

Force an Front f back = Py Ae = fgh_ Ae

Ay or C= 95.5 kN

2 tari 2/z 2 33.7

SUGT| 2 pas pghA =pghaet

- S.92 CŸ lê) = 0, P= 12.57 KN

Vorrzontal slice of block

Az We tạ (sa) fe, | water

Weignt of lok We T#A(6)= 667A CH)

Buayancy of oil Ba> ca (S6)= 8A (i)

Busyaney of weter By= ‘ar A(bA) = 5,33 (In-n) A

Linh \ hin frehes)

XF*: W[~6s~Bu*o: 66.1A~ZwA =5.33 (Irh)Are

h= 2.15 1n

Let y.set density of concrete

5/173 r

t = 150 16/644

Ty et 225/6 2/284 water = 624 bis?

—_( iw

4 Lelength of ylinder = 6H

3 r= radius of slider = 2 ft Fer equih, T2W-8 2 2

= RL - Ty we

: zr1⁄ (Te-z Fw)

= 7/2%)(6)\(259 - $27) = 8960 16

= $2809) 9 Yp0) = 25,270 6

te “6, Mie) = 4520

MF 11,520 (10) -25,270/4) 20 M1* //.82(/0%) 46-⁄

Horie Jensth of

(yoke 13 10 ft

5/180] Per fost of length

Moment arm of & abst G

w (isto) cos@ ~ (6-4) sin 0 =

S1184\| Submerged arca A of end + orca of

wt 120° sector minus area of

& FoOmm priangie OBC

S186] Take a vertical section of wate of unit

J} zim — horizontal length het (be

THe ator deqsity in tm

skit: o=k 23), ke # mw! +

Resa lian} 2t mj 4 Ry passes Through B, So

Ego, “us CAPPS (18) = 642 (6- la)

b= ahi m

wag ¥ IBS = os (util) Tay = 0 884m

F= 08 (8,4 Ba4 By) = O15 (19.28 +617 + 9 88m)

= I2.15+ 1333 mH

(m+m 3 57118) = 55.1 kN

TZF =0: FrBi + B+ 83 — Cimam)g- my =? I2N5 +O 1333m 4 19.28 + 61272: 9.84 m~SS.4~ 8Ì m=0

m= 4.24 My

51188 | The gage presse IZ m below the surface

tet A=cress-sectiona/ area of plank

2o For equilibrium M@+/= Ö,+8,

o/s

steel 5S pgs tonaity of Steel = 283 adam

4 8 gait wearers L238 naglen*

¢

at 1s Pi " fead = 1.37 Mg/mF

peleom, 2 £20,035

kee vit: as 8 I= macs of (ead = PV Ys volume of (20d, m4

so ag varie + Aas pus Ê(rrŸ| 9V

SLI} The pressure at the bsttom of the Som

Wall is pe gh = 2400(4.81)(s) = 10 coo Nin®

Each tie contmls an area Ñ given by

part, A= P= FoGoo ~ 00920 w This square area has a side d given by

r= AQ ) A= 0,303 m

Using the pressure at The very bsttom of the,

Wall gives us a Conservotve desi; + Good

Figo for A wtovih ke AS _9 $00 m

si] Metind Z: Direct inteyratin

Na P= water density | p= P9(b+2,)

Ze htkee= F(x) (i) Cateulte dR ie elemental

he 99% zảx+ { TỦ Z

(2) Integrate ¢ get R=fdR

(s) RB= fear

a Method IL: Geometry of pressrerarce snes

© het Ag@ F234 | ALE S-4-5

+ T5” - 2 (4)(4)

= 3.66 in

¬ « k es

ae

Bợlczar| (mm) 20

All corner radii = 10 rm

ẤM = 2w 8 = 2x (ae+ %)j4e(s)~ 4(Ie")

= +1((o°) ] = 280 000 mm$

A = 2w?L= (ao + ®)|z2):2(3)+ 2ml9 |

= 30 700 mm~