tutorial problem set #6 (week 44: period …spkenny/courses/undergraduate/engi1313/tutorial... ·...

ENGI 1313 Mechanics I Faculty of Engineering and Applied Science

Shawn Kenny, Ph.D., P.Eng. Fall 2007 Revision 0

Tutorial Problem Set #6 of 25

TUTORIAL PROBLEM SET #6 (WEEK 44: PERIOD ENDING NOVEMBER 2, 2007) The problem set provides a representative sample of questions on relevant course material and concepts covered in the lectures. The tutorial problems sets are intended to develop good study habits and become engaged in the learning process.




1: Problem 6-7 (page 279) Determine the force in each member of the truss and state if the members are in tension or compression.




FEF 1 kN=

GivenJoint B F1 FBC+ 0=

F2− FBA− 0=

Joint C FCD FBC− FAC cos θ( )− 0=

F3− FAC sin θ( )− FCF− 0=

Joint E FEF− 0=

Joint D FCD− FDF cos θ( )− 0=

F4− FDF sin θ( )− FED− 0=

Joint F FAF− FEF+ FDF cos θ( )+ 0=

FCF FDF sin θ( )+ 0=

FBA

FAF

FDF

FBC

FCD

FED

FAC

FCF

FEF

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find FBA FAF, FDF, FBC, FCD, FED, FAC, FCF, FEF,( )=

Positive means tension,Negative means compression.

FBA

FAF

FDF

FBC

FCD

FED

FAC

FCF

FEF

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

8.00−

4.17

5.21

3.00−

4.17−

13.13−

1.46−

3.13−

0.00

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

kN=

Units Used:

kN 103 N=

Given:

F1 3 kN=

F2 8 kN=

F3 4 kN=

F4 10 kN=

a 2 m=

b 1.5 m=

Solution: θ atanba

⎛⎜⎝

⎞⎟⎠

=

Initial Guesses

FBA 1 kN= FBC 1 kN= FAC 1 kN=

FAF 1 kN= FCD 1 kN= FCF 1 kN=

FDF 1 kN= FED 1 kN=




2: Problem 6-18 (page 281) Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The horizontal force component at A must be zero. Why?




FBA

FBD

FCB

FCD

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

1133.33−

666.67

400.00−

692.82−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

lb=

FBA

FBD

FCB

FCD

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find FBA FBD, FCB, FCD,( )=Positive means TensionNegative means Compression

FBA− FBDa

a2 b2+− F1− 0=FCB FBD

b

a2 b2++ 0=Joint B

FCD− F2 sin θ( )− 0=FCB− F2 cos θ( )− 0=Joint C

Given

FCD 1 lb=FCB 1 lb=FBD 1 lb=FBA 1 lb=

Initial Guesses

Solution:

θ 60 deg=

b 3 ft=

a 4 ft=

F2 800 lb=

F1 600 lb=

Given:

kip 103 lb=

Units Used:




3: Problem 6-28 (page 282) Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 2 kN, P2 = 4 kN.




FFD

FED

FGF

FEG

FFE

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

11.46−

6.69

13.46−

6.69

3.46−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

kN=

FBD

FCD

FAB

FCA

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

11.46−

6.69

13.46−

6.69

3.46−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

kN=Positvive means Tension,Negative means Compression

FFD

FED

FGF

FEG

FFE

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

FBD

FCD

FAB

FCA

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

FBD

FCD

FAB

FCA

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find FBD FCD, FAB, FCA, FBC,( )=

FCD FCA+( )sin θ( ) FBC+ 0=

FCDcos θ( ) FCA cos θ( )− 0=Joint C

FBD FAB− P2 sin 2θ( )− 0=

P2− cos 2 θ( ) FBC− 0=Joint B

P1−

2FBD sin 2 θ( )− FCDsin 3 θ( )− 0=Joint D

Given

FBC 1 kN=FCA 1 kN=

FAB 1 kN=FCD 1 kN=FBD 1 kN=

Initial Guesses:

Take advantage of the symmetry.

Solution:

θ 15 deg=

a 2 m=

P2 4 kN=

P1 2 kN=

Given:kN 103 N=

Units Used:




4: Problem 6-33 (page 289) The roof truss supports the vertical loading shown. Determine the force in members BC, CK, and KJ and state if these members are in tension or compression.




Positive (T)Negative (C)

Ax

Ay

FKJ

FCK

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.00

6.67

13.33

0.00

14.91−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

kN=

Ax

Ay

FKJ

FCK

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find Ax Ay, FKJ, FCK, FBC,( )=

FCK Ay+b

b2 9a2+

⎛⎜⎝

⎞⎟⎠

FBC+ 0=

FKJ Ax+3a

b2 9a2+

⎛⎜⎝

⎞⎟⎠

FBC+ 0=

FKJ2b3

⎛⎜⎝

⎞⎟⎠

Ax2b3

⎛⎜⎝

⎞⎟⎠

+ Ay 2a( )− 0=

F2 3a( ) F1 4a( )+ Ay 6a( )− 0=

Ax 0=

Given

FKJ 1 kN=

FCK 1 kN=FBC 1 kN=

Ay 1 kN=Ax 1 kN=

Initial Guesses

Solution:

b 3 m=

a 2 m=

F2 8 kN=

F1 4 kN=

Given:

kN 103 N=

Units Used:




5: Problem 6-39 (page 290) Determine the force members BC, FC, and FE, and state if the members are in tension or compression.




Positive (T)Negative (C)

Dy

FBC

FFC

FFE

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

6

8.49−

0

8.49

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

kN=

Dy

FBC

FFC

FFE

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find Dy FBC, FFC, FFE,( )=

F2− Dy+ FFE FBC+( )sin θ( )+ 0=

FFC− FBC FFE+( )cos θ( )− 0=

Dy b FFE cos θ( )a− 0=

F1− b F2 2b( )− Dy 3b( )+ 0=

Given

FFE 1 kN=FFC 1 kN=

FBC 1 kN=Dy 1 kN=

Initial Guesses

θ atanab

⎛⎜⎝

⎞⎟⎠

=Solution:

b 3 m=

a 3 m=

F2 6 kN=

F1 6 kN=

Given:kN 103 N=

Units Used:




6: Problem 6-51 (page 292) Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression.




FBC

FCH

⎛⎜⎜⎝

⎞⎟⎟⎠

3.25−

1.923⎛⎜⎝

⎞⎟⎠

kN=Ey 1.15 kN=

Ey

FBC

FCH

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find Ey FBC, FCH,( )=Positive (T)Negative (C)

FBC− sin θ( ) FCH sin φ( )− F1− Ey+ 0=

FBC sin θ( ) c( ) FCH sin φ( ) c b−( )+ Ey c( )+ 0=

F2− d( ) F1 c( )− Ey 2c( )+ 0=

Given

FCH 1 kN=FBC 1 kN=Ey 1 kN=

Initial Guesses:

φ atana

c b−⎛⎜⎝

⎞⎟⎠

=θ atanac

⎛⎜⎝

⎞⎟⎠

=

Solution:d 0.8 m=

c 2 m=

b 1 m=

a 1.5 m=

F2 2 kN=

F1 1.5 kN=

Given:kN 103 N=

Units Used:




7: Problem 6-70 (page 314) The 150-lb man attempts to lift himself and the 10-lb seat using the rope and pulley system shown. Determine the force at A needed to do so, and also find his reaction on the seat.




N 134 lb=N P W2−=

P N− W2− 0=ΣFy = 0;

Seat:P 144 lb=P 9T=

T 16 lb=TW1 W2+

10=

10T W1 W2+=

T P+ W1− W2− 0=ΣFy = 0;

Man and seat:

P 9T=Thus,

3R P− 0=ΣFy = 0;

Pulley B:

3T R− 0=ΣFy = 0;

Pulley C:

Solution:

W2 10 lb=

W1 150 lb=

Given:




8: Problem 6-76 (page 315) The compound beam is fixed supported at A and supported by rockers at B and C. If there are hinges (pins) at D and E, determine the reactions at the supports A, B, and C.




By 1 N=

Ey 1 N= Ex 1 N= Cy 1 N=

Given

Ay w2 a− Dy− 0= Ax− Dx− 0=

MA w2 aa2

− Dy a− 0= Dy w1 b c+( )− By+ Ey− 0=

w1−b c+( )2

2By b+ Ey b c+( )− 0= Dx Ex+ 0=

Ey w1d e+

2− Cy+ 0= Ex− 0=

w1−d e+

2⎛⎜⎝

⎞⎟⎠

d e+3

⎛⎜⎝

⎞⎟⎠

Cy d+ M− 0=

Ax

Ay

MA

Dx

Dy

By

Ey

Ex

Cy

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find Ax Ay, MA, Dx, Dy, By, Ey, Ex, Cy,( )=Ax

Ay

⎛⎜⎝

⎞⎟⎠

0

19⎛⎜⎝

⎞⎟⎠

kN=

MA 26 kN m=

By 51 kN=

Cy 26 kN=

Units Used:

kN 103 N=

Given:

a 2 m= M 48 kN m⋅=

b 4 m= w1 8kNm

=

c 2 m=

w2 6kNm

=d 6 m=

e 3 m=

Solution:

Guesses

Ax 1 N= Ay 1 N= MA 1 N m=

Dx 1 N= Dy 1 N=




9: Problem 6-83 (page 316) The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W?




Σ MA = 0;

F− b c+( ) TBD sin φ( ) T sin θ( )−( )b+ 0=

TBD

Fb c+

b⎛⎜⎝

⎞⎟⎠

T sin θ( )+

sin φ( )=

TBD 2408.56 lb=

+↑Σ Fy = 0;

Ay− TBD sin φ( )+ T sin θ( )− F− 0=

Ay TBD sin φ( ) T sin θ( )− F−= Ay 700.00 lb=

+→ Σ Fx = 0;

Ax TBDcos φ( )− T cos θ( )− 0=

Ax TBDcos φ( ) T cos θ( )+= Ax 1.878 kip=

At D:

Dx TBD cos φ( )= Dx 1.703 kip=

Dy TBD sin φ( )= Dy 1.703 kip=

Units Used: kip 103 lb=

Given:

F 700 lb=

a 4 ft=

b 4 ft=

c 4 ft=

θ 60 deg=

Solution:

Pulley E:

+↑Σ Fy = 0; 2T F− 0=

T12

F= T 350 lb=

This is the force in the cable at the winch W

Member ABC: φ atanab

⎛⎜⎝

⎞⎟⎠

=




10: Problem 6-102 (page 321) The pillar crane is subjected to the load having a mass of 500 kg. Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position shown.




FAB

Cx

Cy

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

9.7

11.53

8.65

⎛⎜⎜⎝

⎞⎟⎟⎠

kN=Cx

Cy

⎛⎜⎝

⎞⎟⎠

FCB

a2 b2+

b

a⎛⎜⎝

⎞⎟⎠

=FAB

FCB

⎛⎜⎝

⎞⎟⎠

Find FAB FCB,( )=

M−2

g sin θ1( ) FAB sin θ2( )− FCBa

a2 b2++ M g− 0=

M−2

g cos θ1( ) FAB cos θ2( )− FCBb

a2 b2++ 0=

Given

FAB 10 kN=FCB 10 kN=initial guesses:

Solution:

g 9.81m

s2=

θ2 20 deg=

θ1 10 deg=

b 2.4 m=

a 1.8 m=

M 500 kg=

Given:

kN 103 N=

Units Used:




11: Problem 6-107 (page 322) A 5-lb force is applied to the handles of the vise grip. Determine the compressive force developed on the smooth bolt shank A at the jaws.




NA 36.0 lb=

NA Exd e+

sin θ( ) d cos θ( ) a+⎛⎜⎝

⎞⎟⎠

=

NA sin θ( )d NA cos θ( )a+ Ex d e+( )− 0=ΣMB = 0;

From FBD (b)

Ex 34.286 lb=

Ex FCDc

c2 d e+( )2+

⎡⎢⎣

⎤⎥⎦

=ΣFx = 0;

FCD 39.693 lb=FCD F b c+( )c2 d e+( )2+

b d e+( )

⎡⎢⎣

⎤⎥⎦

=

F b c+( ) FCDd e+

c2 d e+( )2+

⎡⎢⎣

⎤⎥⎦

b− 0=ΣME = 0;

From FBD (a)

Solution:

θ 20 deg=

e 1 in=d 0.75 in=

c 3 in=a 1.5 in=

b 1 in=F 5 lb=

Given:




12: Problem 6-125 (page 327) The four-member “A” frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is 800 N, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pin at G only exert force components on the frame.




Bx Dx= 0=

Dy 283 N=

By 283 N=Dy By=ByFmax b

2 b2 c2+=

Dy By=By Dy+ Fmaxb

b2 c2+− 0=

Dz 283 N=

Bz 283 N=Dz Bz=BzFmax c

2 b2 c2+=

Dz Bz=Bz Dz+ Fmaxc

b2 c2+− 0=

P 282.84 N=PFmax b

2 b2 c2+=

P− 2 cb

b2 c2+Fmax c+ 0=

ΣMx = 0;

Solution:

c 600 mm=

b 600 mm=

a 300 mm=

Fmax 800 N=

Given:

tutorial problem set #6 (week 44: period …spkenny/courses/undergraduate/engi1313/tutorial... ·...

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