mechanics:-sample problems chapter 8

8/18/2019 Mechanics:-Sample Problems Chapter 8

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Two forces PI and Pz, of magnitude PI::::: 15kN and

P2 ::::: 8 eN, are applied as shown to the end A of bar AB,

which is welded to a cylindrical member BD of radius

c

:::::

20 mm (Fig. 8.21). Knowing that the distance rom A to

the axis of memberBD is a

:::::

50 mm and assuming hat all

slressesemainbelow the propOrtional imit of the material,

determine a) the normal and shearingstresses t point K of

the transverse ection of memberBD located at a distance

b

:::::

60 mm from end B, (b) the principal axesand principal

stresses t K, (c) the maximumshearing tressat K.

rt'I jf()~ ~

F J'( J.- 6

([)

~t

fo P,

.

,s K N

15 t O. 0

.5

~ 'IS () (VrO

~ cljf"tct/ct')

Du~

PJ..

..

,-

".

.,.

-

,

G

0

(t ,,~

-"vt

v

,

...

.".

.,.

,.

0

12,

(-,<-It

ffy-IA

....

,.

-'"

I ft 1<N

".

:. 7 , () IV ""

-

I

- : -

- - - _: - - - - - - ::-: - - -- -- - -: --

-.';- :- -".', -::-::-.' /"':.'.-_:->:' . :', :.;:-;--:':-',"':\:';~~-:'..fU;;,:,:,;~;{~,:",/~;-':i:'~) 1'-;~:::.I'~;:;.. :~'.~~{ :.~'Jd<)"~}:::.'i;:;::;':~;j/i.~i1.i:_:~u.S'YH ~:~\~~.~)~"i"<i.'U~.)Y'i~ '~Ik»'~\t:~,~;?~~~~ :,t~_;:~;i~:~~(Nt{,

i

Fig. 8.21 P;2 '" H) kN

My

~T

a

0

F

-

M.

v

#./ftO.D6

1081) N'rn

(5

d.iff dlUt')

,r- KN

",gxO.O.5

Fig. 8.22

,...

...

y/

-

-

::< 7.50 N.II)

My

2.-

.,.

-

N~

,so

".

...

)l

v = IS kN

Fig. 8.23

N

. 75. 0

~k

)

(vA

t &'jY)( e.-

N ft- do~

J\Jr()

)

f.. IJ. t<

(' D J- p(JJA



~

Ne}-

~x ::

() Wdij

(J,;::. + 107

,-..

""

- l, .

y ~ -- r()rA((f(\

+ Q i J

.4 f'fPd..

f "6 heCJ..I'

-

52.5

~ '[ "'y

&p ~

~

f~ - fr:J

(6'7C-Q'~

o'(>z e s :.

2.-

75.1 MP",--

(2~. t 1'1PrA.

-2./.1i. MP'1

g

.,.

-

Of

0

-

..

2-

-<910u eA

r."'J=

-52.5 MPa

Fig. 8.24

I~ PIA

t07

MPa..

8p

==

22.2°

15kN

(

;11in= -21.4 MPa.

Fig. 8.26

)

= 75.1 MP~l

'T,n,,~

as=22.8°



A

~

r

Ii

.Sin. ~

5001b

n. -.. .. .. ... ..

4.5 n.

4.5 in.

A

A == 2501b

B = 2.50b

= 62.51bill.

M f= 625

Ii

~/V:250lb

..-T =9001b.in.

~O.9-in. diameter

G

T:: 6290 psi

--LH

T = 6290 psi

,L

(0)

T =6290psi

K

= 62-90 si

H

T = 524 psi

"l

(b)

T=524psi

:,.

- (T = 8730 psi

.,.:::'0

K

(1';:0

(c)

u = 8730 psi

-u = 8730 psi

L

..,. =

6810 psi

.u =" 8730psI

, "" ~'"~ ""~;";, ',;~i;;:.;{;:.1~.i; ;f,<",'i:kv:, ,; J';':' ~>:,;i;\:i.:;:' ;,./i;'~~iD~iifi~t~;ii.fiii~Xl~At#wkMi.{~;~.&.~;~;.,{~;

SAMPLE PROBLEM 8.4

A horizontal 500-lb force acts at point D of crankshaft AB which is held in

static equilibrium by a twisting couple T and by reactions at A and B. Know-

ing that the bearings are self-aligning and exer~no couples on the shaft, de-

termine the normal and shearing stressesat points H, J, K, and L located at the

ends of the vertical and horizontal diameters of a transverse section located

2.5 in. to the left of bearing B.

SAMPLE PROBLEM 8.4

SOLUTION

Free Body. Entire Crankshaft. A =

B

= 250 Ib

+ ~2:M~= 0: -(500 Ib)(1.8 n.) + T = 0 T = 9001b

.

n.

Internal Forces n TransverseSection. We replace he reactionB and

the twisting couple T by an equivalent orce-couple ystemat the centerC of

the transverse ectioncontainjngH. J, K, and L.

%

v = B = 250 Ib T = 900 Ib . n.

My =

(250 b)(2.5 n.) = 6251b in.

The geometric properties of the O.9-in.-diameter section are

A = 7T(O.45n.)2

=

0.636 n2 1= 7T(O.45n.)4 = 32.2 X 10-3 n4

J = 7T(O.45n.)4= 64.4X 10-3 n4

Stresses Produced by Twisting Couple T. Using Eq. (3.8), we

determine he shearingstresses t points H, J, K, and L and show them in

Fig. (a).

Tc (900 b . n.)(0.45 n.)

=

6290psi

'T = J = 64.4 X 10 3 n4

Stresses Produced by Shearing Force V. The shearing force V pro-

duces no shearing stressesat points J and L. At points Hand K we first com-

pute Q for a semicircle about a vertical diameter and then determine the shear-

ing stress produced by the shear force V = 250 lb. These stressesare shown

in Fig. (b).

Q =

(

.1Tc2

)( 4C)= ~c3 = ~(O45 n)3 =60 7 X 10-3 in3

,2 31T 33' . .

VQ (250 Ib)(60.7 X 1O-3in3)

.

T

= - ::::: = 524pSi

It (32.2 X 10-3 in4)(O.9 n.)

,

StressesProduced by the Bending Couple M,.. Since the bending cou-

ple My acts in a horizontal plane, it produces no stressesat Hand K. Using Eq.

(4.15), we detennine the normal stressesat points J and L and show them in

Fig. (c).

IMylc (625lb . in.)(O.45 in.) .

(J'

= - = =

8730

PS

I 32.2 X 10-3 in4

Summary. We add the stressesshown and obtain the total normal and

shearingtressest pointsH. J, K, andL



4O~mm

y

~~

MJ :; t).,'5 kN. III

G

x

T'"m

I ~;:~8°

O~

I~

O"'"iL1

SAMPLE PROBLEM 8.5.

Three orces are appliedas shownat pointsA, B, and Jj of a short steelpost.

Knowing that the horizontal crosssectionof the post s a 40 X 140-mm ec-

tangle, determine he principal stresses, rincipal planesand maximumshear-

ing stressat point H. .

SOLUTION

Internal Forces n Section EFG. We replace he three applied orces

by an equivalent orce-couple ystemat the centerC of the rectangular ection

EFG. We have. .

Vx

=

-30 kN P

=

50kN V~

=

-75 kN

Mx =

(50 kN)(0.130m) - (75 kN)(0.200m)

=

-8.5 kN

.m

My =

0

M~ =

(30kN)(0.1O0 ) = 3 kN . m

We note that there s no twisting couple about he y axis. The geometric

propertiesof the rectangular ectionare

A= (0.040m)(0.140m) = 5.6X 10-3 m2

Ix= i2(0.040m)(0.140m? = 9.15

X 10-6 m4

lz = i2(0.140m)(0.040m? = 0.747 X 10-6 m4

Normal Stress at H. We note that normal stresses J'yare produced by

the centric force P and by the bending couples Mx and M<;.We determine the

sign of each stress by carefully examining the sketch of the force-couple sys-

tem at C.

P IMtla IMx1b

(J' = +-+---

Y A I I

,t x

50 kN (3 kN

.

m)(0.020 m) (8.5 kN

.

m)(O.O25m)

= + -

5.6X 10-3 m2 0.747 X 1O-6m4 9.15 X 10-6 m4

Uy = 8.93MPa + 80.3MPa - 23.2MPa cry=

66.0 MPa. <1

Shearing Stress at H. Considering first the shearing force Vx, we note

that Q = 0 with respect to the z axis,. since H is on the edge of the cross sec-

tion. Thus V

x

produces no shearing stressat H. The shearing orce Vz does pro-

duce a shearing stress at H and we write

Q

=

A1YJ ~ [(0.040 m)(0.045m)](0.0475m)

=

85.5 X 10-6 m3

VzQ (75 kN)(85.5 X 10-6 m3) .

Tyz=

Ixl= (9.15X 1O-6m4)(0.040m) Tyz= 17.52MPa <J

Principal Stresses~ rincipal Planes, and Maximum Shearing Stress

at H. We draw Mohr's circle for the stresses t point H

mechanics:-sample problems chapter 8

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