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Chapter 3. Equilibrium of Force Systems Equilibrium in Two Dimensions 3-1. Determine the reactions at the supports. Q By Ay Ax a) Resultant force of distributed load Q = qL = 75*8 = 600 N located at the midle of AB. constrain at A: smooth pin (or hinge) two unknwons constrain at B: roller / one unkn...

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Chapter 3. Equilibrium of Force Systems Equilibrium in Two Dimensions 3-1. Determine the reactions at the supports. Q By Ay

Ax

a) Resultant force of distributed load Q = qL = 75*8 = 600 N located at the midle of AB. constrain at A: smooth pin (or hinge) two unknwons constrain at B: roller / one unknown we have (Ax , Ay , Q, M , By )  0 Equations of Equilibrium  Fx  Ax  0  Fy  Ay  Q  B y  0  mA ()   4Q  8By  M  0

Ans. Ax  0, Ay  200, By  400 N

b)

c)

3-2. Thee slowly walking, a man having a total mass of 80 kg places all his weight on one foot. Assuming that the normal force N C of the ground acts on his foot at C, determine the resultant vertical compressive force FB which the tibia T exerts on the astragalus B, and the vertical tension FA in the achilles tendon A at the instant shown.

Ch 03

1

we have ( NC, FB, FA)  0 N C =mg = 80*g Equations of Equilibrium  Fy  FA  FB  NC  0  mA ()  (20  100)NC  20FB  0

Ans. (20  100) N C  6N C 20 FA  FB  N C  6N C  N C  5N C FB 

Prob. 3-2 3-3. Determine the reactions at the supports A and B. By M

Q

Ay  Ax

Prob. 3-3 Q = 500 N, M = 800 Nm constrain at A: smooth pin (or hinge) two unknwons constrain at B: rocker / one unknown calculate cos  

8 4 2

2

(8  4)  5

,

5

sin  

(8  4)2  52

AB  8 2  (5 * 8 /12)2  8.67 m

we have ( Ax , Ay , Q, By , M )  0 Equations of Equilibrium Ch 03

2

,

 Fx  Ax  Q sin  0  Fy  Ay  Q cos  B y  0  mA ()  ABQ  8By  M  0

Ans. Ax  192.31, Ay  180.13, By  641.67 N 3-4. Determine the horizontal and vertical components of force at the pin A and the reaction at the rocker B of the curved beam. Equations of Equilibrium  Fx  Ax  500 sin 10o  200 sin 15o  0

By

 Fy  Ay  500 cos10o  200 cos15o  By  0  mC ()  2 By  2 Ay  0

Ay

Ans. Ax  35.06, Ay  342.79, By  342.79 N

Ax

C

Prob. 3-4 3-5. The mass of 700 kg is suspended from a trolley which moves along the crane rail from d = 1.7 m to d = 3.5 m. Determine the force along the pin-connected knee strut BC (short link) and the magnitude of force at pin A as a function of position d. Plot these results of FBC and FA (ordinate) versus d (abscissa). Weight of load

Ay

S

P = mg = 700g Sx  35 S,

Ax

Sy  45 S

Equations of Equilibrium  Fx  Ax  S x  0  Fy  Ay  S y  P  0  mA()  dP  1.5Sy  0

P

Prob. 3-5 4

Ans.

3

Ax  3433.50 * d, Ay  6867.0  4578 * d, S  5722.50 * d

5

A  Ax 2  Ay 2  ( 3433.50 * d)2  (6867.0  4578 * d)2 S  5722.50 * d Ch 03

3

Plot these results of FBC and FA (ordinate) versus d (abscissa). > plot(S,d=1.7..3.5);

> plot(A,d=1.7..3.5);

Ch 03

4

3-6. When no force is applied to the brake pedal of the lightweight truc k, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is 3 N, determine the unstretched length of the spring if the stiffness of the spring is k = 80 N/m. relation Force on the spring and deformation F k 

F

  F /k

the force on the switch is 3 N (P = 3N)

P

Px  P cos 30o  3  3 / 2 Py  P sin 30o  3  1/ 2

Equations of Equilibrium  mD ()  50 F  40 Px  10Py  0

so

Prob. 3-6

F  (4 Px  Py )/ 5  2.378 N

deformation of the spring   F /k 

Ch 03

5

2.378  0.0297m  30 mm 80

3-7. Outriggers A and B are used to stabilize the crane from overturning when lifting large loads. If the load to be lifted is 3 Mg, determine the maximum boom angle  so that the crane does not overturn. The crane has a mass of 5 Mg and center of mass at GC, whereas the boom has a mass of 0.6 Mg and center of mass at GB .

Ay P=3 Mg By GB=0.6 Mg

GC=5 Mg

Prob. 3-7 Equations of Equilibrium

 mA ()  2  2.3By  GB (5 sin  0.7)  P (9.5 sin  0.7)  2.3GC  0

So By 

1 [2.3GC  GB(5 sin   0.7)  P(9.5 sin   0.7)]  0 2  2.3

  0.461 rad = 26.43o

ans.

Ch 03

 max  26.43o

6

3-8. The mobile crane is symmetrically supported by two outriggers at A and two at B in order to relieve the suspension of the truck upon which it rests and to provide greater stability. If the crane and truck have a mass of 18 Mg and center of mass at G 1, and the boom has a mass of 1.8 Mg and a center of mass at G 2 determine the vertical reactions at each of the four outriggers as a function of the boom angle  when the boom is supporting a load having a mass of 1.2 Mg. Plot the results treasured from   0o to the critical angle where tipping starts to occur.

By GL=1.2 Mg

Ay GB=1.8 Mg

1Mg = 106 g = 1 000 kg = 1 ton GC=18 Mg

Prob. 3-8 Equations of Equilibrium  Fy  Ay  B y  (GC  G B  G L )g  mA ()  4By  3gGC  gGB (6 sin  2)  gGL (12.25 sin  2)  0

So By  14 g[3 GC  GB (6 sin   2)  GL (12.25 sin   2)] Ay  (GC  G B  G L )g  B y

solution Ay  58.860  62.539 sin , By  147.150  62.539 sin 

the critical angle where tipping starts to occur : solve Ay  58.860  62.539 sin   0 sin max  1.226 rad = 70.25o

> plot(Ay,goc=0..theta_max);

Ch 03

7

> plot(By,goc=0..theta_max);

3-9. A cantilever beam, having an extended length of 3 m, is subjected to a vertical force of 500 N. Assuming that the wall resists this load with linearly varying distributed loads over the 0.15 m length of the beam portion inside the wall, determine the intensities w 1 and w 2 for equilibrium. W1  12  w1  0.15,

W2

W2  12  w2  0.15

Eq. of Equilibrium fy  W1  500 W2  0 mO  23  0.15W2  3  500  31  0.15W1  0

w1 = 413333.33 N/m W1

w2 = 406666.67 N/m

Prob. 3-9

Ch 03

8

3-10. The wooden plank resting between the buildings deflects slightly when it supports the 50-kg boy. This deflection causes a triangular distribution of load at its ends, having maximum intensities of wA and wB . Determine wA and wB , each measured in N/m, when the boy is standing 3 m from one end as shown. Neglect the mass of the plank.

W1

W2

Prob. 3-10 Solution W1  12  wA  0.45,

W2  12  wB  0.3

Eq. of Equilibrium f y  W1  50g  W2  0 mO   13  0.45W1  3  50g  (3  6  13  0.3)W2  0

wA = 1437.62 N/m, wB = 1113.57 N/m 3-11. The dimensions of a jib crane, which is manufactured by the Basick Co., are given in the figure. If the crane has a mass of 800 kg and a center of mass at G, and the maximum rated force at its end is F = 15 kN, determine the reactions at its bearings. The bearing at A is a journal bearing and supports only a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components. Eq. of Equilibrium fx  B x  Ax  0

Ax

fy  B y  G  F  0

By

mB  2  Ax  0.75  G  3  F  0 Ax  25443,

Bx

Bx  25443, By  22848 N

G

Prob. 3-11/12

Ch 03

9

3-12. The dimensions of a jib crane, which is manufactured by the Basick Co., are given in the figure. The crane has a mass of 800 kg, and a center of mass at G. The bearing at A is a journal bearing and can support a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components. Determine the maximum load F that can be suspended from its end if the selected bearings at A and B can sustain a maximum resultant load of 24 kN and 34 kN, respectively. 3-13. The winch cable on a tow truck is subjected to a for ce of T = 6 kN when the cable is directed at   60o . Determine the magnitudes of the total brake frictional force F for the rear set of wheels B and the total normal forces at both front wheels A and both rear wheels B for equilibrium. The truck has a total mass of 4 Mg (4*10^6 g) and mass center at G. > g:=9.81; theta:=Pi/3; q := g := 9.81

1 p 3

> T:=6*1000; G:=4*1000*g; T := 6000

G := 39240.00

> fx:=F-T*sin(theta); fx := F - 3000

3

> fy:=Ay+By-G-T*cos(theta); fy := Ay + By - 42240.00

> mB:=2.5*G-4.5*Ay-1.5*T*cos(theta)-3*T*sin(theta); mB := 93600.000 - 4.5 Ay - 9000

3

> res:=solve({fx,fy,mB},{Ay,By,F}); res :={ F = 5196.152423, Ay = 17335.89838, By = 24904.10162}

3-14. Determine the minimum cable force T and critical angle  which will cause the tow truck to start tipping, i.e., for the normal reaction at A to be zero. Assume that the truck is braked and will not slip at B. Ay

By

Probs. 3-13/14 G

Solution Ch 03

10

Eq. of Equilibrium f x  F  T sin  f y  Ay  B y  G  T cos mB  2.5G  4.5Ay  1.5T cos  3T sin

Solving F  T sin , Ay  21800  0.333333333T cos   0.6666666667T sin , By  17440  1.333333333T cos   0.6666666667T sin 

Consider Ay  21800  0.333333333T cos  0.6666666667T sin  0

21800 0.333333333 cos   0.6666666667 sin  1 0.333333333 cos   0.6666666667 sin  Y ( )   T ( ) 21800 T() 

d * Y ()  0    1.107148718  T min  29247.77 N . d

3-15. The operation of the fuel pump for an automobile depends on the reciprocating action of the rocker arm A B C , which is pinned at B and is spring loaded at A and D. When the smooth cam C is in the position shown, determine the horizontal and vertical components of force at the pin and the force along the spring DF for equilibrium. The vertical force acting on the rocker arm at A i s F A=60 N, and at Prob. 3-15 C it is FC = 125 N.

FD

Bx

By

Solution > > > >

FA:=60; FC:=125; # N fx:=Bx+FD*sin(Pi/6); fy:=By-FD*cos(Pi/6)+FA+FC; mB:=(20+10)*FC-10*FD*cos(Pi/6)-50*FA; fx := Bx +

1 FD 2

fy := By -

1 FD 2

3 + 185 mB := 750 - 5 FD

> res:=solve({fx,fy,mB},{Bx,By,FD}); res := { By = -110, Bx = -25

Ch 03

11

3 , FD = 50

3}

3

3-16. The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k = 5 N/m, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C. F FA

Prob. 3-16

FB

Solution we have (F, FA, FB )  0

so

FA  F,

FB  2F

deformation of springs A  FA / k  F / k, B  FB /k  2F /k C  A  2B  F / k  4 F / k  5F / k C  10 mm  0.01 m

determine the smallest force needed to close the contact gap at C F  k C / 5  5  0.01/ 5  0.01N

3-17. The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is 0.5 N. Originally the strip is horizontal as shown. F

Prob. 3-17 from result of 3-16, C  5 F / k  k  5 F / C  5  0.5/ 0.01  250 N/m

Ch 03

12

3-18. Determine the force P needed to pull the 50-kg roller over the smooth step. Take   60o . y

P x

60o NB O

B NA

Prob. 3-18

R A 20o

Determine angle AOB cos   (R  h)/ R  (0.6  0.1)/ 0.6  5/6

Equilibrium P cos   N B sin   W sin 20o  0 P sin   NB cos   W cos 20o  NA  0

Given P, W, , , we can find NA, NB Condition to roll over step NA 1.5Az = 0.5*300 == > Az = 100 N (3) == > T1z = T2z = 100 N == > T1 = T2 = 113.0388331 N == > T1x = T2x = 16.66666667 == > T1y = T2y = 50.00000000 Ax = 2*T1x = 33.3333334

Ch 03

17

3-23. The shaft assembly is supported by two smooth journal bearings A and B and a short link DC. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the bearings and the force in the link. The link lies in a plane parallel to the y-z plane and the bearings are properly aligned on the shaft.

S

S 20

Bz

C 30

Az By

Ay

Prob. 3-23 Equation of Equilibrium Fy  Ay  By  S cos 20o  0

(2)

o

Fz  Az  Bz  S sin 20  0

(3) o

mx ()  250  0.250 S cos10  0

(4) o

my ()  0.300 Bz  0.400 Az  0.120 S sin 20  0 o

mz ()  0.300 By  0.400 Ay  0.120 S cos 20  0

Ans. S = 1015.43 N, Ay = 572.51 N, Az = -208.38, By = 381.68 N, Bz = -138.92 N

Ch 03

18

(5) (6)

3-24. The member is supported by a pin at A and a cable BC. If the load at D is 300 N, determine the x, y, z components of reaction at these supports. (The length unit is m). Position vectors rB  0.4 i  0.6 j  0.0k

Az

rC  0.1i  0.0j  0.2k

Mz Ay

BC  rC  rB  0.3 i  0.6 j  0.2k

T

Direction unit vectors of T My

Ax

W=300N u1  

0.3 i  0.6 j  0.2k 0.32  0.62  0.22 0.3 i  0.6j  0.2k BC

Tx  (0.3/ BC)T

Prob. 3-24

Ty  (0.6/ BC )T Tz  (0.2/ BC )T

Equation of Equilibrium Fx  Ax  Tx  0

(1)

Fy  Ay  Ty  0

(2)

Fz  Az  Tz  W  0

(3)

mx ()  0.6Tz  0.6W  0

(4)

my ()  My  0.2W  0.4Tz  0

(5)

mz ()  Mz  0.4Ty  0

(6)

Solution (4) == > Tz  W  300 N T  Tz. BC / 0.2  300  0.7 / 0.2  1050 N Tx  T0.3 / BC  450,

(6)

Ty  T 0.6/ BC  900

Mz  0.4Ty  360 Nm

My  0.2W  0.4Tz

T = 1050,

Ch 03

My = -60,

Mz = -360, Ax = -450, Ay = 890, Az = 0.

19

3-25. Determine the x, y, z components of reaction at the pin A and the tension in the cable BC necessary for equilibrium of the rod. (The length unit is m) Position vectors rB  0.4 i  1.2 j  0.0 k

Mz

BC  rC  rB  0.0 i  1.2 j  0.5k



0.0 i  1.2 j  0.5 k 2

2

My

2

0.0  1.2  0.5 0.0 i  1.2 j  0.5k BC

Tx  (0.0/ BC )T,

Ty  (1.2 / BC )T

Tz  (0.5/ BC )T

Prob. 3-25

Equation of Equilibrium Fx  Ax  Tx  0

(1)

Fy  Ay  Ty  0

(2)

Fz  Az  Tz  F  0

(3)

mx ()  1.2Tz  0.4W  0

(4)

my ()  My  0.2W  0.4Tz  0

(5)

mz ()  Mz  0.4Ty  0

(6)

Ans. Mz = -112.00, My = -23.33 Nm Ax = 0.0, Ay = 280.00, Az = 233.33 , T = 303.33 N

Ch 03

Ay

Ax

Direction unit vectors of T u1 

Ay

Az

rC  0.4 i  0.0 j  0.5 k

20

T...