Midterm Tufree Mechanics 1/2558

7/23/2019 Midterm Tufree Mechanics 1/2558

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MIDTERM EXAMT U - F R E E

STATICS&DYNAMI


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!" !"#$ MECHANICS MIDTERM EXAM

1 STRUCTURAL ANALYSIS TRUSS

Truss !"#$%&'()*$+,*-.(/0( )(12(34"567*589:59"5 ;0%*(?*5@ 34"5+A54* +B%16* CDE*9"5F5 12(G( H581"*I/I)JK% 9"5L*-I( 89?,#.(/0(M&CN 1O%C!P(0Q?%R*.(/0(S()#1T-+*-+B%1!U*

Statically determinate

m + 3 = 2j :

METHOD OF JOINTP(0Q+* 9"5L*-I((internal force) )*$6V$*"6VW,X%5 joint 9?,# joint

! F x = 0

! F y = 0

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!"!"#$ MECHANICS FINAL EXAM

1. [midterm 1/2553] Determine the force in all member of thetruss. Is the structure statically determinate? Why?



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2. Using the method of joints, determinethe force in members EG , and ED , andstate if the members are in tension orcompression.


A EB C D

I J

30 kN20 kN 20 kN

40 kN

H G F

4 m

16 m, 4@4m


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METHOD OF SECTION

12(YZ$*"Y14"*#[9"5L*-I(3\-$*";\ section X%5 member 86(I)%%$12(6%5/0( )*$S(P(0Q+*9"5L*-I()*$6V$*"

3. Determine internal force of members EF and GI

! F x = 0

! F y = 0

! M = 0



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4. )5+*9"5I(.(/0( FH 9,# GH



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FRAMEFrameK%34"567*58 member L*-I(M&9"5V*$$R* 2 9"5 /0(I+]67*5V*1O%"%5M&9"5$"#^)*$

L*-(%$ (9? member L*-I( Truss )#M&9"5F5_\ 2 9"5)5. Determine the force supported by the roller

at E



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MACHINE MachineK%34"567*58 member 6*V*"`14a%(8Cb 9,#c$%%$9&&V*1O%/5d*-9"5

6. [midterm 1/2009] The levermechanism for a machine press servesas a toggle which develops a large forceat E when a small force is applied at thehandle H . To show that this is the case,determine the force at E if someoneapplies a vertical force of 80 N at H . Thesmooth head at D is able to slide freelydownward. All members are pinconnected.



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7. Determine the moment M which must be applied at A tokeep the frame in static equilibrium in the position shown.Also calculate the magnitude of the pin reaction at A.


150 kg

M

C D

E

B A

1.5 m

1 m 1 m

1.25 m

60 60


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2 FRICTIONFRICTION COEFFICIENT

!"#$%&'()*+,-. K%9"51T-\e*(XQ#8fgh+i\j5(k%($*"14a%(8)!"#$%&'()*/01 K%9"51T-\e*(XQ#8fghlA514a%(8 m5)#no*p$R*9"51T-\e*(6qgr8s(t01u-0v(

)*$ Free-body Diagrams X%5fgh )#1w(R*x*"0V 9"51T-\e*5 F 1y*v& 9"5 N )#Cb9"5A>z R m5^{V v& N

m59|5%%$12( 2 }~\g*V9"51T-\e*(K%

F = N

P

mg

Pm

tan ! =F

N

tan ! s = s

tan ! k = k



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WEDGE (!" #$)V12(14%5 (9"5s(*(8I/5d*-9"5%-@C!12(9"5V*$ 3\-I+A$$*"e*5g3$Q

Self-Locking

8. l+(\IV!"#e40*V1T-\e*56qgrX%5V1*v& 0.39,#V!"#e40*V1T\e*(6qgr"#+R*5 $U%5(500 kg) v&s(12( 0.6 )5+*9"5%-8_\8^I$U%5 500 kg 14a%(8

P

W

u!

!

P

W

u!

!


5

P

500 kg


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SCREW (&'())(%$)*$)#6$ 81"*I\.(5*(1y*b0-v(9 0 56*V*"`6$ V*!"#i$II($*"/5d*-9"5 +B%/5d*-

$*"14a%(8Cb$b0-

Self-Locking

! = tan -1l

2 " r # $ %

& ' (



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9. The square threaded screw of the clamp has a mean diameter of 14mm and a lead of 6 mm. If ! s = 0.2 for the threads, and the torqueapplied to the handle is 1.5 Nm, determine the compressive force F onthe block.

10. 6$ A 9,# B nMn1 - 6 VV. "#-#"#+R*51$-0 2VV. 6$ A 12(6$ 1$-0X0* /0( B 12(1$-0*- V!"#e40*V1T-\e*(X%56$ v&!,%$1+$12( 0.12

)5+*!1,3V1V(8^v&!,%$1+$1O%+{(I6$ 56%51y*+*v(.


1.5 N m

F

F

Prob. 877

2 kN 2 kN A B

Fig. P8.70


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FLEXIBLE BELT$"1%$+B%6*->*(n$*"C`,1 \'( ^I9"5I(1(1%$1u-0v(no*C1*v( m5+*Cb)*$

11. l+(\V!"#e9"51T-\e*("#+R*51%$v&t0"%$12( 0.3 )5+* ($) 9"5 P8I-$V+$ 100 $$. '( (X) 9"5 P 8I!U%- 100 $$. ,5b0-40*V10458 l+(\ ! =0

T 2

T 1

= e !

T 2 T 1

Impendingmotion

b


Or

P

100 kg


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12. [midterm 2557] 1%$\1B%I,%-1-&* ("%&g%+%5 ! +*$%%$9"555CN 200 N )5+*9"5)*$1B% T V*$8_\8 5 C^I1B%,%-%%$)*$* l+(\V!"#e40*V1T-\e*("#+R*5 1%$ v& g%+%12( 0.30

JOURNAL BEARINGJournal bearing 12( support +M&1>,*+{(8M&9"5g*V9(05 *$v&$*"+{(


200 NT


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13. 3V1V( M=1510 N.m $"#^81>,*1(*((r$,*5 50 VV. 1O%-$V+$ 500 $$. '(b0-40*V10458 l+(\I1>,*+$ 100 $$. )5+*V!"#e 40*V1T-\e*(X%59&58"%5M&1>,*S(

THRUST BEARINGThrust bearing 12( support 8M&9"5g*V9(0$*"+{(X%51>,* 1O%)#P(0Q+* moment 86*V*"`^I

1>,*+{(Cb )#G%5I$X%59"51T-\e*(9,#6V$*"6VW,5

+B%

P P

M

M

M =2

3 Pr

2 R i 2 R o P

M

M =2

3 P

r o

3 ! r i 3r

o

2 ! r i 2" # $

% & '


500 kg

50mm

300mm

M


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14. 9(1-nMn 100 VV. 3\-89"5$\ 5+V\1*v& 70 N x*V!"#e9"51T-\e*(),"#+R*59(1-v&3#12( 0.6 )5+*3V1V( M 8V%1g%G%5I

15. [midterm 1/2552] 9&5059+0( 3 05"%5M&9"59(09$(X(*\ 800 lb +*$V!"#e40*V1T-\e*(6qgr12( 0.35 )5+*3V1V(X%59"5\8II($*"^I1>,*+{(



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3 AREA MOMENT OF INERTIA

Radius of GyrationK%"#-#)*$\+{(C! 5\"0V9"5$"#)*-1+U*S(

I x = y 2 dA

A!

I y = x2 dA

A

!

J O = I x + I y

Figure Area Moment of Inertia

r

y

C x

x

x0

y0

h C

b

k = I

A



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Parallel-Axis Theorem$"9$(+{(C%$,*5X%5s(8+*;\ 1"*6*V*"`IeX%5 Parallel axis 1y*V*P(0Q+* moment of

inertia X%59$(S(Cb

Ex.

Composite Area1%s(8+*;\C6VV*g" 1"*6*V*"`9-$+* moment of inertia !


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16. [midterm 1/49] )5+* I x 9,#Mn-1"( k x X%5s(8+*;\b*(U*5 (+0-12(,1Vg" )

17. )5+*o* I x X%5s(8+*;\b*(U*5 (X(*\12(,1Vg" )



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PRODUCT OF INERTIA

Ex . Determine the product of inertia of the area shown with respect to the x-y axes.

I xy = I xy + Ad xd y

2 m

3 m

x

y y

x

r = 1.5 m

2 m

1.5 m

x

y y

x

2 m

1 m


2 m

3 m

y

x

r = 1.5 m

3 m


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ROTATING AXES&*545I($*"I5*()5 9$(+{(%*)C12(9(05+B%9(0(%( 9?12(9$(1-5 m56*V*"`P(0QCb)*$

6V$*"

MOHRS CIRCLE)*$6V$*" rotating axis 1%-$lA56%5X%56V$*"9"$9,#6V$*"86*V9 0V*&0$v()#Cb5

)#1w(R*o* I x , I y ,I xy 12(o*458 /0( I x 9,# I xy S('(%v&{V81-5C! ! m5)#n{V1-58I moment of inertia 8V*$8_\ 1"*)#1-${V1-5S(R* Principal Axes 9,#1-$ I max , I min "0V@R* PrincipalMoment of Inertia

'*(+*, Mohrs Circle1) 0*\ 9$( x ! I, 9$( y ! I xy

2) >%g \ A( I x ,I xy ) 9,# \ B( I y , -I xy ) 3) +*\(r$,*505$,V O

4) +*Mn05$,V R 5) +* 2 ! p6) I max = O + R 9,# I min = O - R

I x =I x ' + I y '

2

+I x ' ! I y '

2

cos2 " ! I x ' y ' sin2 "

I y =I x ' + I y '

2! I x '

! I y '2

cos2 " + I x ' y ' sin2 "

I xy =I x ' ! I y '

2sin2 " + I x ' y ' cos2 "

I !x , I !y

I x , I y

"

I !x "

I x + I y 2

# $ %

& ' (

2

+ I !x !y 2

=I x " I y

2

# $ %

& ' (

2

+ I xy 2


"

x

x

yy


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18. )5+*o* Principal Moment of Inertia X%5s(8+*;\ (X(*\12(,1Vg" ) (10 4#9(()



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19. )5P(0Q+* Principal Moment of Inertia 9,# Principal Axes X%5s(8+*;\X%54*(I( !


x

y

5 mm

30 mm

5 mm

50 mm7.5 mm

C

17.5 mm

Prob. 1072


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!""#$"%&'(&)*)+,-./ *012$3-4(05(1&26$'78#9:-"; +7?3................................................

?4 @-AB [ ] @A 1 [ ] @A 2 [ ] @A 3 [ ] @A 4 [ ] @. 92-84*

C$-D+ E 1 BC.&*++,*"+8+;#:B&8)D EF78@G::H*)I*D"81#8+;#:B&8)D$+ +7?3.....................................

(E4?B+713F23*2K":2; =-678(6"P$LEL4,O61+H?)

Midterm Tufree Mechanics 1/2558

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