2007F ENGI 1313 Tutorial 06 Solution

7/26/2019 2007F ENGI 1313 Tutorial 06 Solution

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ENGI 1313 Mechanics I

Faculty of Engineering and Applied Science

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

Tutorial Problem Set #6 Page 1 of 25

TUTORIAL PROBLEM SET #6(WEEK 44:PERIOD ENDINGNOVEMBER 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 becomeengaged in the learning process.


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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.


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FEF 1 kN=

Given

Joint B F1 FBC+ 0=

F2 FBA 0=

Joint C FCD FBC FACcos ( ) 0=

F3 FACsin ( ) FCF 0=

Joint E FEF 0=

Joint D FCD FDFcos ( ) 0=

F4 FDFsin ( ) FED 0=

Joint F FAF FEF+ FDFcos ( )+ 0=

FCF

FDF

sin ( )+ 0=

FBA

FAF

FDF

FBC

FCD

FED

FAC

FCF

FEF

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

Positive means tension,Negative means compression.

FBA

FAF

FDF

FBC

FCD

FED

FAC

FCF

FEF

8.004.17

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: atanb

a

=

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=


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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 atAmust be zero. Why?


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FBA

FBD

FCB

FCD

1133.33

666.67

400.00

692.82

lb=

FBA

FBD

FCB

FCD

FindFBA FBD, FCB, FCD,( )=Positive means TensionNegative means

Compression

FBA FBDa

a2

b2

+ F1 0=FCB FBD

b

a2

b2

++ 0=JointB

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

Given

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

Initial Guesses

Solution:

60 deg=

b 3 ft=

a 4 ft=

F2 800 lb=

F1 600 lb=

Given:

kip 103

lb=

Units Used:


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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.


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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

FindFBD FCD, FAB, FCA, FBC,( )=

FCD FCA+( )sin ( ) FBC+ 0=

FCDcos ( ) FCA cos ( ) 0=Joint C

FBD FAB P2 sin 2( ) 0=

P2 cos 2 ( ) FBC 0=JointB

P1

2FBD sin 2 ( ) FCDsin 3 ( ) 0=JointD

Given

FBC 1 kN=FCA 1 kN=

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

Initial Guesses:

Take advantage of the symmetry.

Solution:

15 deg=

a 2 m=

P2 4 kN=

P1 2 kN=Given:

kN 103

N=

Units Used:


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4: Problem 6-33 (page 289) The roof truss supports the vertical loading shown. Determine the force in

members BC, CK, and KJand state if these members are in tension or compression.


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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

FindAx Ay, FKJ, FCK, FBC,( )=

FCK Ay+ b

b2

9a2

+

FBC+ 0=

FKJ Ax+3a

b2

9a2

+

FBC+ 0=

FKJ2b

3

Ax2b

3

+ 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:


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

in tension or compression.


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Positive (T)

Negative (C)

Dy

FBC

FFC

FFE

6

8.49

08.49

kN=

Dy

FBC

FFC

FFE

FindDy FBC, FFC, FFE,( )=

F

2

Dy

+ FFE

F

BC

+

( )sin ( )+ 0=

FFC FBC FFE+( )cos ( ) 0=

Dyb FFEcos ( )a 0=

F1 b F2 2b( ) Dy 3b( )+ 0=

Given

FFE 1 kN=FFC 1 kN=

FBC 1 kN=Dy 1 kN=

Initial Guesses

atana

b

=Solution:

b 3 m=

a 3 m=

F2 6 kN=

F1 6 kN=

Given:

kN 103

N=

Units Used:


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6: Problem 6-51 (page 292) Determine the force developed in members BCand CHof the roof truss and

state if the members are in tension or compression.


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FBC

FCH

3.25

1.923

kN=Ey 1.15 kN=

Ey

FBC

FCH

FindEy FBC, FCH,( )=Positive (T)Negative (C)

FBC sin ( ) FCHsin ( ) F1 Ey+ 0=

FBCsin ( ) c( ) FCHsin ( ) c b( )+ Eyc( )+ 0=

F2 d( ) F1c( ) Ey 2c( )+ 0=

Given

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

Initial Guesses:

atana

c b

= atana

c

=

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:


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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 atAneeded to do so, and also find his reaction on the seat.


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N 134lb=N P W2=

P N W2 0=Fy= 0;

Seat:

P 144lb=P 9T=

T 16lb=TW1 W2+

10=

10T W1 W2+=

T P+ W1 W2 0=Fy= 0;

Man and seat:

P 9T=Thus,

3R P 0=Fy= 0;

PulleyB:

3T R 0=Fy= 0;

Pulley C:

Solution:

W2 10 lb=

W1 150 lb=

Given:


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8: Problem 6-76 (page 315) The compound beam is fixed supported atAand supported by rockers at B

and C. If there are hinges (pins) at Dand E, determine the reactions at the supportsA, B, and C.


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By 1 N=

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

Given

Ay w2 a Dy 0= Ax Dx 0=

MA w2 aa

2

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

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

Ay

0

19

kN=

MA 26kNm=

By 51kN=

Cy 26kN=

Units Used:

kN 103

N=

Given:a 2 m= M 48 kN m=

b 4 m=w1 8

kN

m=

c 2 m=

w2 6kN

m=

d 6 m=

e 3 m=

Solution:

Guesses

Ax 1 N= Ay 1 N= MA 1 Nm=

Dx 1 N= Dy 1 N=


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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 pinsAand D. Also, what is the force in the cable at the winch W?


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MA= 0;

F b c+( ) TBDsin ( ) Tsin ( )( )b+ 0=

TBD

Fb c+

b

Tsin ( )+

sin ( )

=

TBD 2408.56 lb=

+ Fy= 0;

Ay TBDsin ( )+ Tsin ( ) F 0=

Ay TBDsin ( ) Tsin ( ) F= Ay 700.00 lb=

+ Fx= 0;

Ax TBDcos ( ) Tcos ( ) 0=

Ax TBDcos ( ) Tcos ( )+= Ax 1.878 kip=

AtD:

Dx TBDcos ( )= Dx 1.703 kip=

Dy TBDsin ( )= Dy 1.703 kip=

Units Used: kip 103

lb=

Given:

F 700 lb=

a 4 ft=

b 4 ft=

c 4 ft=

60 deg=

Solution:

PulleyE:

+ Fy= 0; 2T F 0=

T1

2F= T 350 lb=

This is the force in the cable at the winch W

MemberABC: atana

b

=


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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 Cwhen the boom is tied in the position shown.


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FAB

Cx

Cy

9.7

11.53

8.65

kN=Cx

Cy

FCB

a2

b2

+

b

a

=FAB

FCB

FindFAB FCB,( )=

M

2g sin 1( ) FAB sin 2( ) FCB

a

a2

b2

+

+ M g 0=

M

2g cos 1( ) FAB cos 2( ) FCB

b

a2

b2

+

+ 0=

Given

FAB 10 kN=FCB 10 kN=initial guesses:

Solution:

g 9.81m

s2

=

2 20 deg=

1 10 deg=

b 2.4 m=

a 1.8 m=

M 500 kg=

Given:

kN 103

N=

Units Used:


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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 shankAat the jaws.


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NA 36.0 lb=

NA Exd e+

sin ( )d cos ( ) a+

=

NA sin ( )d NA cos ( )a+ Ex d e+( ) 0=MB= 0;

From FBD(b)

Ex 34.286 lb=

Ex FCDc

c2

d e+( )2

+

=Fx= 0;

FCD 39.693 lb=FCD F b c+( )c2

d e+( )2

+

b d e+( )

=

F b c+( ) FCD

d e+

c2

d e+( )2

+

b 0=ME= 0;

From FBD(a)

Solution:

20 deg=

e 1 in=d 0.75 in=

c 3 in=a 1.5 in=

b 1 in=F 5 lb=

Given:


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12: Problem 6-125 (page 327) The four-member A frame is supported atAand Eby smooth collars and

at Gby a pin. All the other joints are ball-and-sockets. If the pin at Gwill fail when the resultant force there

is 800 N, determine the largest vertical force Pthat can be supported by the frame. Also, what are the x,

y, zforce components which member BDexerts on members EDCandABC? The collars atAand Eandthe pin at Gonly exert force components on the frame.


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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 2cb

b

2

c

2

+

Fmax c+ 0=

Mx= 0;

Solution:

c 600 mm=

b 600 mm=

a 300 mm=

Fmax 800 N=

Given:

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2007F ENGI 1313 Tutorial 06 Solution