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Solutions to Problems in

Strength of Materials SI UNITS

Solutions to Problems in

Strength of Materials SI UNITS

Y. BRONDIAL R. GALLARDO A. SY

HI

Copyright 1984 by Y. Brondial, R. Gallardo and A. Sy A ll rights reserved.

PREFACE This volum e contains solutions to problem s in Strength o f Materials taken from the te x tb o o k “ Strength o f Materials” b y Singers and Pytel, 3rd Edition, as w ell as solu­ tions to selected Board Problem s given over the past years. This review er is designed especially to facilitate self-study fo r students and graduates o f engineering w h o w ill take the board exam ination. T h e unique feature o f this review er is that all problem s and solutions are w ritten co m p letely in SI units. T h e authors believe that once the fundamentals have been thorou ghly mastered in one system o f units, extensions to oth er systems should n ot be difficu lt. The authors wish to express their indebtedness to Mr.

Ram on

Po

of

the

Popular B ookstore and to their

colleagues at the De La Salle U niversity. F in ally the authors Wisjr to acknow ledge the patience and forbearance o f their fam ilies during the many hours required to prepare this re­ viewer. A lth ou gh great care was taken to elim inate errors, it is inevitable that som e w ill still be found. Th e authors appreciate being in form ed about these and w elcom e any suggest­ ions fo r the im provem ent o f this w ork.

Y ola n d a Brondial R on ald o Gallardo A rsen io Sy

Manila, Philippines A p ril 1984

V

CONTENTS Chapter

J*age

1

Sim ple S t r e s s ..................................................................■• •

1

2

Sim ple S t r a in ..................................................................• • •

23

3

Torsion

...................................... ................................... • • •

77

4

Shear and M om ent in Beams

.......................................• • •

103

5

Stresses in B e a m s ........................................................... • • •

131

6

Beams D e f le c t io n s .........................................................• • •

183

7

Restrained Beams

219

8

Continuous B e a m s ........................................................ •••

235

9

C om bined S tre s s e s ........................................................ •••

257

10

R ein forced Beams

281

11

Columns

12

R iveted and W elded C onnections

13

Special T opics

14

Inelastic A c tio n

.........................................................• • •

.................................................. .. .• • • i

........................................................................•••

295

........... •.......... .. .• • •

307

............................................................... •••

327

.............................................. .............. •••

341

A p p en d ix A — Board Problem s

................................357

A p p en d ix B — SI Units, Conversion Factors

V II

...........•••

379

1 Simple Stress PROBLEMS 103. Determine the largest weight W which can be supported by the two wires shown in Fig. P-103. The stresses in wires A B and A C are not to exceed 100 M Pa and 150 MPa, respectively. The cross-sectional areas o f the two wires are 400 mm2 for wire A B and 200 mm2 for wire AC. Arts. W = 33.5 kN

W Flgww P-103. Solution: Pa

w W

Applying +he law o f sines !o ihe force triangle-. W J in TS*

.

Pac J in

66“

-

Pab Sir, 4 5 °

'

2 pAC = o . s j s e , PM

*

tv

O. 7 3 2 W

= _ fk . ^ ab to o

X

/O c

0 .7 3 2 W

Pa b

s

-p o o *

/O-

/o -6

& W

= 200

X / O '*

200

X

/O'

3 3 , V 6 0 /V '• IV = 3 3 , V G O N

= 3 3 . -Pfe A/V

105. For the truss shown in Fig. P-105, determine the cross-sec­ tional areas o f bars B E , BF, and C F so that the stresses will not exceed 100 M N / m 2 in tension or 80 M N / m 2 in compression. A reduced stress in compression is specified to avoid the danger of buckling. Ans. A be = 625 mm2; A BF = 427 mm2; A CF = 656 mm2

Use

t h e m e t h o d o f s e c t i o n s in ci'iviclincj the.

tr u s s . T a ke t h e r ic jh t s e c t io n o f t h e ir u s s fro m

a

- -*

*

P = 3 0 ,0 0 0 N

For B r o n z e :

t o o K /0 6= ---- ±P_-----

p_

SOO X ZO -‘

P=

I2 ,S O O N

-

i2 g o o N

•

IZ .S

kN

112. Determine the weight of the heaviest cylinder which can be placed in the position shown in Fig. P - l 12 without exceeding a stress of 50 M N / m 2 in the cable BC. Neglect the weight of bar AB. The cross-sectional area of cable B C is 100 mm1.

n#u.-» p - t i t

Solution:

Taking the FBO of AG : ^AC — »

PBC £M

-

a

Sgc too

X /O '* ( s o X / o 6)

5 ,0 0 0

/V

= O

A/() - 4 0 k N

8 *eo S so =

Pbo -

400 X I O * (S O X IO‘ ) = 2 0 k N

40+20

- 9.8 - P = O P * SO. 2 KN

iM „ = O 10(3.) - 9 .6 ( l ) ~ P ( x ) ~ O X = 0. 6016 m from ft

114. As in Fig. l-10c, a hole is to be punched out of a plate having an ultimate shearing stress of 300 MPa. (a) If the compressive stress in the punch is limited to 400 MPa, determine the maximum thickness of plate from which a hole 100 mm in diameter can be punched, (b) If the plate is 10 mm thick, compute the smallest diameter hole which can be punched. Ans. (a) I — 33.3 mm; (b ) d * 30.0 mm

Solution: P

£ 400 MPa. St < 300 MPa.

punched hole ha* tw o area* at shown., -the area. X to P U ~q-dx while ihe a rea parallel At P is

•trot. a.)

P,I

=

3

-Jj ^ d 2Sc - 'iTJiSs Ac

0.1 (400 1 I O* ) = t (30 0 X 1 0 *)* 4 t - 0. 0333 nrt * 35.3 m m

b)

^£d*Sc = TTdiS, d( ( O . 0 l 9 ')(l ‘/0 x io L)

= fe .o * =5 = 7 rivets

n ■

X

7 r ive ts for B C

use

For B E :

(S h e a rin g)

60,000 =

n (B e a rin g )

n x JL (0 .P I9 )Z(70 x 10*)

— 4.03

8 0 ,0 0 0 = n. •

=

rivets =£= S rivets

HX

O.

013 (o . 0 l l )(l 4 0 X IOL)

2.3/

rivets ={J= j

rivets

u se 5 rivets f o r B E

To get the Largest eu/erage ietvsile or compressive stress In BC and 0 E / P

The area- o f the angle section is taken from table B - S , p. 6 4 *

For BC.

^aurtq\* - g6V x 10'* ntl " 7

_5l£dt 96 X 10*

'(* 6 4 X /O-6)-(o .O I9 )(0 OO6) ST =

For

BE :

A.

IZ 8 M P a

u = 1,750 x IO~*mx 90

5C = =

X IO *

i,iso x io -* -(t > .o i* )(t> o n ) 52. Z MPm.

17 I \1. A cylindrical pressure vessel is fabricated from steel plates »lin It have a thickness o f 20 mm. The diameter of the pressure vessel is Mm nun mid its length is 3 m. Determine the maximum internal pressure Wln< li i nn he applied if the stress in the steel is limited to 140 MPa. If »h* Internal pressure were increased until the vessel burst, sketch the •vi,r " f fracture which would occur. Arts. 11.2 M P A

S tilu llo n :

Ui'mtj th e g i r t h j o i n t :

\

-

F° ‘t t

m o x to* p

p ^ °'s^ if ( O.Ol) 2 2 .1 - M P a.

=

t A in j the lo n g itu d in a l j o i n t :

l'/ o x /Ob =

p ^ ° 'f ^

1 (0.02.) p m ax U

=

11.2 MPa.

p -

I I .2

MPa.

ihie in te rn a l pressure u/ere. in crea sed u n til

I hr vessel b u r s t, t h e longitudinal, j o i n t will j i v e uitij tin d

/

th erefore

th e fr a c tu r e is a l o n j th e

luidincbl j o i n t .

135. The strength per meter o f the longitudinal joint in Fig. 1-16 u 4HO kN, whereas for the girth joint it is 200 kN. Determine the ni» hi mu m diameter o f the cylindrical tank if the internal pressure.is 1.5 M N / m 2.

18 Solution: L o n g itu d in a l

——

=

jo in t

480

:

kN

G irth, j o i n t : *

-f o

200 k N

F or the Longitudinal joint: : 2

P

pD

=

L

: - z ( { ) - p D Z x 4 8 0 x 10* =

1.5 X lO^D

-

D

O . b l m

F or the girth j o i n t : P =

p

x JL p *-

200 x 10i = /) Z/j£

=

o L = 3m

(J L L ) = (J L L )

_______S' tv._

I A t. Is -Sl‘ r

-r

I A t L

P i (3 ) = _A M Aa.X 7OXI0* A ix 2 0 0 x 10* -

W^Mg

+ )Z M

&

U a tta c h e d ,

, St - Sa ro d s

A t X 2 0 0 X 10*

X Ps X 6 A » X 7 O a IO*

= O

c.-3-

P ,C 5 )-

=»0

p r* = - -2l 6

* + )£ M = O Ps (3 ) P « -(Z ) = 0

Am.

bo

Ai

7

A* A*

0.5 7

A n *.

29 214. The rigid bars AB and C D shown in Fig. P-214 are supiwrted by pins at A and C and the two rods. Determine the maximum force P which can be applied as shown if its vertical movement is limited to 5 mm. Neglect the weights o f all members. A/is. P = 76.3 kN //^//Aluminum L - 2m A - 500 mm2 E = 70 GPa Steel L - 2m A = 300 mm2 E = 200 GPa

c

3m

3m

D

IP Flgur* P-214.

Solution: ////(///// ALum -0 Ps

- & fj) = O

P* = 2PS Sk

V

£\>'+

^ )r M c = 0 P ( 3 )~ P t ( 6 ) - O Ps = i P

By rrttio A n d proportion-,

P*. * P

i a - Sb 3

6

N o te ■■ Due to Ri, ihe aluminun, rod n/ill d eform by Sti . Correspondingly , B wttl move down by Sy due However, due to /2 , D will not m ove down by SyonUjt b u t by St T Ss.

Sy =

St „

Sj

3

Sb

6

St s 2-Se. + Ss = Ss ~ 1 (0 . OOS)

z fIL )

Z lAEU

+

= 2 (^ P )

l£ L )

IA E/s

= O.O!

30

P

X

Z

500 X10-*x 70 K JO9

3 0 0 X /0 -*K 2 0 0 X /O *

P=

7 b .3 b 3 .b V m

P=

7 6 .3 6 k N

Ams.

215. A round bar o f length L tapers uniformly from a diamet D at one end to a smaller diameter d at the other. Determine the elongation caused by an axial tensile load P . Ans. 6 - A P L / irE D d

Solution:

— H k— x —*

and proportion:

+

-4 k

«

w

- JA u

X ijL - d L

%

=

Dx-^x

s -

w L i - i-i-1 TTE(D-d) L D d J

4PL TTE(D-d) ^PL

P t^ J /l«s.

TTB.DJ

217. As shown in Fig. P-217, two aluminum rods A B and BC, hinged to rigid supports, are pinned together at B to carry a vertical load P — 20 kN. I f each rod has a cross-sectional area o f 400 mm2 and E -> 70 X 103 M N / m 2, compute the elongation o f each rod and the horizontal and vertical displacements o f point B. Assume a -* 30° and 9 “ 30°. Arts. 8h - 0.412 mm; 6„ - 3.57 mm

F ro m the 'force 'triangle. ; P ab

_

S 'm 6 0 “

Pec

2 0 ,0 0 0

S in 6 0 *

S in

(> O c

PAa = 1 0 ,0 0 0 * ( t ) Poe

= 2 0 ,0 0 0 * ( C )

The d efo rm a tio n s are

r?-

Is

P L

J1

AB

2. 0 ,0 00 ( 3) 400 x iO~h %70x ID9

2.143 x 10 * *

AhS.

( le n g th e n in g ) t

-

20.000 ( z )

* B C ~ t o o X to -*X 7 0x1 0 *

= [1.421 X I 0 '3n>

A h S.

( Shortening ) To a n alyse the. effects o f t h e d e formant io n s o n B's m o v e m e n t , im agine. A B a.nd B C to b e d isc o n n e c te d ast Q to ajlou/ th e d efo rm a tio n s ('g r e a t l y e x a g g e ra te d ) shoHun below. To re fasten, t h e b a r s , rota te t h e m cdoout A a n d C , respectively , t o m e e t a t B “- S in c e th e arcs g e n e r a te d , are $0 S m o -ll, th e y m a-tj be. re-p Laced b y stra ja k t lin e s th a t a r e clr*.tun, pe.rpenA 1 0 u.Lajr to A & a n d B C , i n t e r s e c tim j a t 8 ' ; SB =. B B

S b p ’ =*

.

CaC>

75#l \?' W

Is*?

.W/

/

*+2 .9

men

33

The de formations Sab And s (G O + f i )

w h ere,

COS ( G O - f i ) ~

c o s GO cos f i

+ iin

GO s in

astd c o s (G O + /3 ) — c*>i 60 COSfi — Siw (j O si** fi

ta + tfi = £• =

£

(,.5 S 3 °

__

£ab

.2./V3 x /O'5

cos (6 0 ~ f i ) cos (6 0 ~ 6. 5&3) SB

■•

-

3. 5 9 6 X /O '* t n

= Js

^

JSJi

= J. 596 X /0-J f*V* 6.

f H#i

= O . V / T X 10-3m.

j g(i = o. mz mm

*

A h .s .

~ &B "=*/3 J8„ = 5- 596 * / £ '’ «> « 6 .5 B 3 ° i Bv

-

3.572- k tO * n

S s,

=

5 5 7 2 H im

4/»S-

219. A round bar o f length L , tapering uniformly from a diame­ ter Z> at one end to a smaller diameter £jdv — p j Adx

d W * p y ^ f f 2dx

J d W = Jo P9 ^ y 2dx W

-

J

p i j l f

t j2

( p - d

)d tf

a

P 9 ir . _ f i _ D-d

w

=

w

= _£L II

. -± _ D -d

p t , jh b . - J ’ i ]

A D - 5 .V X 10 * m

A D ~ 0.50- m m

A n*.

C o n s id e rin g -the e f f e c t o f end. ik r u s t,

A D = p [ j & t -v -S t ) ] A D =