CE 102 Statics

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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CE 102 Statics

Chapter 6

Analysis of Structures



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Contents

Introduction

Definition of a Truss

Simple Trusses

Analysis of Trusses by the Method of Joints

Joints Under Special Loading Conditions

Space Trusses

Sample Problem 6.1

Analysis of Trusses by the Method of Sections

Trusses Made of Several Simple Trusses

Sample Problem 6.2

Analysis of Frames

Frames Which Cease to be Rigid When Detached From Their Supports

Sample Problem 6.3

Machines



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Introduction

• For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered.

• In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

• Three categories of engineering structures are considered:

a) Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces.

b) Trusses: formed from two-force members, i.e., straight members with end point connections

c) Machines: structures containing moving parts designed to transmit and modify forces.



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• A truss consists of straight members connected at joints. No member is continuous through a joint.

• Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only two-force members are considered.

• Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure.

• When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression.



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Members of a truss are slender and not capable of supporting large lateral loads. Loads must be applied at the joints.



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Simple Trusses

• A rigid truss will not collapse under the application of a load.

• A simple truss is constructed by successively adding two members and one connection to the basic triangular truss.

• In a simple truss, m = 2n - 3 where m is the total number of members and n is the number of joints.



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Analysis of Trusses by the Method of Joints• Dismember the truss and create a freebody

diagram for each member and pin.

• The two forces exerted on each member are equal, have the same line of action, and opposite sense.

• Forces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite.

• Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m + 3. May solve for m member forces and 3 reaction forces at the supports.

• Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations.



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Joints Under Special Loading Conditions• Forces in opposite members intersecting in

two straight lines at a joint are equal.

• The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load).

• The forces in two members connected at a joint are equal if the members are aligned and zero otherwise.

• Recognition of joints under special loading conditions simplifies a truss analysis.



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Space Trusses

• An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron.

• A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time.

• Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations.

• In a simple space truss, m = 3n - 6 where m is the number of members and n is the number of joints.

• Conditions of equilibrium for the joints provide 3n equations. For a simple truss, 3n = m + 6 and the equations can be solved for m member forces and 6 support reactions.



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Sample Problem 6.1

Using the method of joints, determine the force in each member of the truss.

SOLUTION:

• Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.

• Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements.

• In succession, determine unknown member forces at joints D, B, and E from joint equilibrium requirements.

• All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.



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Sample Problem 6.1SOLUTION:

• Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.

ft 6ft 12lb 1000ft 24lb 2000

0

E

MC

lb 000,10E

xx CF 0 0xC

yy CF lb 10,000 lb 1000 - lb 20000

lb 7000yC



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Sample Problem 6.1

• Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements.

534

lb 2000 ADAB FF

CF

TF

AD

AB

lb 2500

lb 1500

• There are now only two unknown member forces at joint D.

DADE

DADB

FF

FF

532

CF

TF

DE

DB

lb 3000

lb 2500



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Sample Problem 6.1

• There are now only two unknown member forces at joint B. Assume both are in tension.

lb 3750

25001000054

54

BE

BEy

F

FF

CFBE lb 3750

lb 5250

375025001500053

53

BC

BCx

F

FF

TFBC lb 5250

• There is one unknown member force at joint E. Assume the member is in tension.

lb 8750

37503000053

53

EC

ECx

F

FF

CFEC lb 8750



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Sample Problem 6.1

• All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

checks 087507000

checks 087505250

54

53

y

x

F

F



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Analysis of Trusses by the Method of Sections

• When the force in only one member or the forces in a very few members are desired, the method of sections works well.

• To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side.

• With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD.



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Trusses Made of Several Simple Trusses• Compound trusses are statically

determinant, rigid, and completely constrained.

32 nm

• Truss contains a redundant member and is statically indeterminate.

32 nm

• Necessary but insufficient condition for a compound truss to be statically determinant, rigid, and completely constrained,

nrm 2

non-rigid rigid32 nm

• Additional reaction forces may be necessary for a rigid truss.

42 nm



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Sample Problem 6.2

Determine the force in members FH, GH, and GI.

SOLUTION:

• Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L.

• Pass a section through members FH, GH, and GI and take the right-hand section as a free body.

• Apply the conditions for static equilibrium to determine the desired member forces.



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Sample Problem 6.2

SOLUTION:

• Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L.

kN 5.12

kN 200

kN 5.7

m 30kN 1m 25kN 1m 20

kN 6m 15kN 6m 10kN 6m 50

A

ALF

L

L

M

y

A



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Sample Problem 6.2

• Pass a section through members FH, GH, and GI and take the right-hand section as a free body.

kN 13.13

0m 33.5m 5kN 1m 10kN 7.50

0

GI

GI

H

F

F

M

• Apply the conditions for static equilibrium to determine the desired member forces.

TFGI kN 13.13



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Sample Problem 6.2

kN 82.13

0m 8cos

m 5kN 1m 10kN 1m 15kN 7.5

0

07.285333.0m 15

m 8tan

FH

FH

G

F

F

MGL

FG

CFFH kN 82.13

kN 371.1

0m 15cosm 5kN 1m 10kN 1

0

15.439375.0m 8

32

m 5tan

GHF

GHF

LM

HIGI

CFGH kN 371.1



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Analysis of Frames• Frames and machines are structures with at least one

multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit and modify forces.

• A free body diagram of the complete frame is used to determine the external forces acting on the frame.

• Internal forces are determined by dismembering the frame and creating free-body diagrams for each component.

• Forces between connected components are equal, have the same line of action, and opposite sense.

• Forces on two force members have known lines of action but unknown magnitude and sense.

• Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components.



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Frames Which Cease To Be Rigid When Detached From Their Supports

• Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies.

• A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions.

• The frame must be considered as two distinct, but related, rigid bodies.

• With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components.

• Equilibrium requirements for the two rigid bodies yield 6 independent equations.



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Sample Problem 6.3

Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD.

SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

• Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C.

• With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C.

• With member ACE as a free-body, check the solution by summing moments about A.



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Sample Problem 6.3SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

N 4800 yy AF N 480yA

mm 160mm 100N 4800 BM A N 300B

xx ABF 0 N 300xA

07.28tan150801

Note:



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Sample Problem 6.3• Define a free-body diagram for member

BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C.

N 561

mm 100N 480mm 06N 300mm 250sin0

DE

DEC

F

FM CFDE N 561

• Sum of forces in the x and y directions may be used to find the force components at C.

N 300cosN 561 0

N 300cos0

x

DExx

C

FCF

N 795xC

N 480sinN 5610

N 480sin0

y

DEyy

C

FCF

N 216yC



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Sample Problem 6.3

• With member ACE as a free-body, check the solution by summing moments about A.

0mm 220795mm 100sin561mm 300cos561

mm 220mm 100sinmm 300cos

xDEDEA CFFM

(checks)



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Machines• Machines are structures designed to transmit

and modify forces. Their main purpose is to transform input forces into output forces.

• Given the magnitude of P, determine the magnitude of Q.

• Create a free-body diagram of the complete machine, including the reaction that the wire exerts.

• The machine is a nonrigid structure. Use one of the components as a free-body.

• Taking moments about A,

Pb

aQbQaPM A 0

29

Problem 6.4

The pin at B is attached to member ABCD and can slide along a slot cut in member BE. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium.

A B C D

E

15 in15 in15 in

80 lb40 lb

36 in

M

30

Solving Problems on Your Own

The pin at B is attached to member ABCD and can slide along a slot cut in member BE. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium.

A B C D

E

15 in15 in15 in

80 lb40 lb

36 in

MFor this problem we note that there are no two-force members. In solving this problem, we

1. Dismember the frame, and draw a free-body diagram for each member.

Problem 6.4

2. To simplify the solution, seek a way to write an equation involving a single unknown.

31

+

Dismember the frame, and draw a free-body diagram for each member.

B

E

15 in

36 in

M

5

12

13

Ex

Ey

Free-Body: Member BE

BE = (152 + 362)1/2 = 39 in

15

3639

5

1213

To simplify the solution, seek a way to write an equation involving a single unknown.

ME = 0: M - B (39 in) = 0 M = B (39 in) (1)

Problem 6.4 Solution

32


Dismember the frame, and draw a free-body diagram for each member.

B5

12

13

Free-Body: Member BE

A CD

15 in15 in15 in

80 lb40 lb

Dx

Dy

+


MD = 0: (40 lb)(45 in) + (80 lb)(15 in) - B(30 in) = 0513

B = 260 lb

33


B5

12

13A C

D

15 in15 in15 in

80 lb40 lb

Dx

Dy

B = 260 lb

B

E

15 in

36 in

M

5

12

13

Ex

Ey

M = B (39 in)

From EQ (1)

M = B (39 in) = (260 lb) (39 in) = 10,140 lb-in

M = 10.14 kip-in

34

Problem 6.5

For the frame shown and neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE.

12 in

A B

C

D

E

6 in

4 in

50 lb

6 in

2 in

6 in

35


For this problem we note that there are no two-force members. In solving this problem, we

1. Dismember the frame, and draw a free-body diagram for each member.

For the frame shown and neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE.

12 in

A B

C

D

E

6 in

4 in

50 lb

6 in

2 in

6 in

Problem 6.5

2. To simplify the solution, seek a way to write an equation involving a single unknown.

36

Problem 6.5 SolutionDismember the frame, and draw a free-body diagram for each member.

A

C

6 in

50 lb

6 in

2 in

18 inA

DD

Cx

CyH

Cx

Cy

Bx

By6 in

37


A

C

6 in

50 lb

6 in

2 in

18 inA

DD

Cx

CyH

Cx

Cy

Bx

By

6 inFree-Body: Member ACD


+ MH = 0: Cx(2 in) - Cy(18 in) = 0 Cx = 9Cy

38


+

A

C

6 in

50 lb

6 in

2 in

18 inA

DD

Cx

CyH

Cx

Cy

Bx

By

6 in


MB = 0: Cx(6 in) + Cy(6 in) - (50 lb)(12 in) = 0

Free-Body: Member BCE

Cx = 9Cy

Substitute Cx = 9Cy: 9Cy(6 in) + Cy(6 in) - 600 = 0

Cy = + 10 lb; Cx = 9Cy = 9(10) = 90 lb

39


C

6 in

50 lb

Cx

Cy

Bx

By

6 in

Free-Body: Member BCE

C

C

90 lb

10 lb C = 90.6 lb6.3o

Fx = 0: Bx - 90 lb = 0 Bx = 90 lb Fy = 0: By + 10 lb - 50 lb = 0 By = 40 lb

+

+

B

B

90 lb

40 lb B = 98.5 lb 24.0o

40

Problem 6.6

Using the method of joints, determine the force in each member of the truss shown.

AB C

D

E

F

G

2 m12.5 kN

2.5 m

2 m 2 m12.5 kN 12.5 kN 12.5 kN

41


1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports.


AB C

D

E

F

G

2 m12.5 kN

2.5 m

2 m 2 m12.5 kN 12.5 kN 12.5 kN

2. Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use this free-body diagram to determine the unknown forces in each of the two members. Assuming all members are in tension, if the answer obtained from Fx = 0 and Fy = 0 is positive, the member is in tension. A negative answer means the member is in compression.

Problem 6.6

42


3. Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it as indicated in step 2 to determine the two unknown forces.


AB C

D

E

F

G

2 m12.5 kN

2.5 m

2 m 2 m12.5 kN 12.5 kN 12.5 kN

4. Repeat this procedure until the forces in all the members of the truss have been determined.

Problem 6.6

43

+


Draw the free-body diagram of the entire truss, and use it todetermine reactions at the supports.

AB C

D

E

F

G

2 m12.5 kN

2 m 2 m12.5 kN 12.5 kN 12.5 kN

2.5 m

Ax

Ay

E

MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m) - (12.5 kN)(6 m) = 0 E = 60 kN

Fy = 0: Ay - (4)(12.5 kN) = 0 Ay = 50 kNFx = 0: Ax - E = 0 Ax= 60 kN

+

+

44


Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use the free-body diagram to determine the unknown forces in each of the two members.

12.5 kN

FCD

FGD

2.56

6.5

Fy = 0: FGD - 12.5 kN = 0

+

+

2.56.5

66.5

FGD = 32.5 kN C

Fx = 0: FGD - FCD = 0

FCD = 30 kN T

Joint D

A B C

D

F

G

2 m

12.5 kN

2 m 2 m

12.5 kN 12.5 kN 12.5 kN

2.5 m

E60 kN

60 kN

50 kN

45


A B C

D

F

G

2 m

12.5 kN

2 m 2 m

12.5 kN 12.5 kN 12.5 kN

2.5 m

E60 kN

60 kN

50 kN

FCG

FFG

F = 0: FFG - 32.5 kN = 0

FFG = 32.5 kN C

Joint G

Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it to determine the two unknown forces.

32.5 kN

F = 0: FCG = 0

46


FBC

FCF

Fy = 0: - 12.5 kN - FCF sin= 0 - 12.5 kN - FCF sin39.81o= 0 FCF = 19.53 kN C

Joint C

FCD = 30 kN

Repeat this procedure until the forces in all the members of the truss have been determined.

12.5 kN

A B C

D

F

G

2 m

12.5 kN

2 m 2 m

12.5 kN 12.5 kN 12.5 kN

2.5 m

E60 kN

60 kN

50 kN

= BCF = tan-1 = 39.81o

23

BF2

BF = (2.5 m) = 1.6667 m

Fx = 0: 30 kN - FCF cos -FBC = 0 30 kN - (-19.53) cos39.81o-FBC = 0 FBC = 45.0 kN T

+

+

47


FEF

Fy = 0: FBF - FEF - (32.5 kN) - (19.53) sin = 0

Joint F=39.81oA B C

D

F

G

2 m

12.5 kN

2 m 2 m

12.5 kN 12.5 kN 12.5 kN

2.5 m

E60 kN

60 kN

50 kN

Fx = 0: - FEF - (32.5 kN) - FCF cos = 0+

+

FBF

2.56

6.5 FFG = 32.5 kN

FCF = 19.53kN

66.5

66.5

FEF = -32.5 kN - ( ) (19.53) cos39.81o FEF = 48.8 kN C6.56

2.56.5

2.56.5

FBF - (-48.8 kN) - 12.5 kN - 12.5 kN = 0 2.56.5

FBF = 6.25 kN T

48


FBE

Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kN

Joint B

A B C

D

F

G

2 m

12.5 kN

2 m 2 m

12.5 kN 12.5 kN 12.5 kN

2.5 m

E60 kN

60 kN

50 kN

+

+

FAB FBC = 45.0 kN

FBF = 6.25kN

12.5 kN

tan = ; = 51.34o2.5 m2 m

Fx = 0: 45.0 kN - FAB + (24.0 kN) cos 51.34o = 0 FAB = 60.0 kN

FBE = 24.0 kN C

FAB = 60.0 kN T

49


Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0

Joint E

A B C

D

F

G

2 m

12.5 kN

2 m 2 m

12.5 kN 12.5 kN 12.5 kN

2.5 m

E60 kN

60 kN

50 kN

+

FAEFEF = 48.75 kN

FBE = 24 kN

2.56.5

FAE = 37.5 kN FAE = 37.5 kN T

2.56

6.560 kN

= 51.34o

50

Problem 6.7

A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI.

D

E

F

G

1.5 kN

3 m

A

B

C

3 m 3 m 3 m 3 m 3 m

H

J

I K

L

3 kN

3 kN

3 kN

6.75 m

3 kN

3 kN

1.5 kN

51


1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports.

2. Pass a section through three members of the truss, one of which is the desired member.


D

E

F

G

3 m

A

B

C

3 m 3 m 3 m 3 m 3 m

H

J

I K

L

3 kN

3 kN

3 kN

6.75 m

3 kN

3 kN

1.5 kN1.5 kN

Problem 6.7

52


3. Select one of the two portions of the truss you have obtained, and draw its free-body diagram. This diagram should include the external forces applied to the selected portion as well as the forces exerted on it by the intersected members before these were removed.

4. Now write three equilibrium equations which can be solved for the forces in the three intersected members.


D

E

F

G

3 m

A

B

C

3 m 3 m 3 m 3 m 3 m

H

J

I K

L

3 kN

3 kN

3 kN

6.75 m

3 kN

3 kN

1.5 kN1.5 kN

Problem 6.7

53


Draw the free-body diagram of the entire truss, and use it todetermine reactions at the supports.

Ay = L = (18kN) = 9 kN

+

D

E

F

G

A

B

C

H

J

I K

L

3 kN

3 kN

3 kN

3 kN

3 kN

1.5 kN1.5 kN

3 m 3 m 3 m

Ay

Ax

3 m 3 m 3 mL

Fx = 0: Ax = 0

Total load = 5(3 kN) + 2(1.5 kN) = 18 kN

By symmetry

12

54


D

E

F

G

A

B

C

H

J

I K

L

3 kN

3 kN

3 kN

3 kN

3 kN

1.5 kN1.5 kN

3 m 3 m 3 m 3 m 3 m 3 m

Pass a section through three members of the truss, one of which is the desired member.

9 kN 9 kN

55


H

J

I K

L

3 kN

3 kN

1.5 kN

3 m 3 m 3 m

9 kN

Select one of the two portions of the truss you have obtained, and draw its free-body diagram.

G

FFFH

FGI

FFI

a

6.75 m

3

4

5

Slope of FHJL

Free Body: Portion HIL

3

4

56.759.00

34

=

tan = = = 66.04oFGGI

6.753.00

Now write three equilibrium equations which can be solved for the forces in the three intersected members.

56


+

H

J

I K

L

3 kN

3 kN

1.5 kN

3 m 3 m 3 m

9 kN

G

FFFH

FGI

FFI6.75 m

3

4

5Free Body: Portion HIL

Write the three equilibrium equations.

66.04o

Force in FH:

MI = 0: FFH ( x 6.75 m) + (9 kN)(6 m)

- (1.5 kN)(6m) - (3 kN)(3m) = 0

45

23

FFH (4.5 m) + 36 kN-m = 045

FFH = -10.0 kN FFH = 10.0 kN C

57


+

H

J

I K

L

3 kN

3 kN

1.5 kN

3 m 3 m 3 m

9 kN

G

FFFH

FGI

FFI6.75 m

3

4



66.04o

Force in FI:

ML = 0: -FFI sin 66.04o(6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0

FFH = 10.0 kN C

FFI sin 66.04o(6 m) = 27 kN-m

FFI = + 4.92 kN FFI = 4.92 kN T

58


+

H

J

I K

L

3 kN

3 kN

1.5 kN

3 m 3 m 3 m

9 kN

G

FFFH

FGI

FFI6.75 m

3

4



66.04o

Force in GI:

MF = 0: -FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m) - (1.5 kN)(9 m) + (9 kN)(9 m) = 0

FFH = 10.0 kN CFFI = 4.92 kN T

FGI (6.75 m) = +40.5 kN-m

FGI = + 6.00 kN FGI = 6.00 kN T

59

Problem 6.8

A pipe of diameter 50 mm is gripped by the Stillson wrench shown. Portions AB and DE of the wrench are rigidly

attached to each other and portion CF is connected by a pin at D. Assuming that no slipping occurs between the pipe and the wrench, determine the components of the forces exerted on the pipe at A and at C.

50 mm 40 mm

20 mm

400 mm

500 N

A

B

C

D

EF

60


1. Dismember the machine, and draw a free-body diagram of each member.

2. First consider the two-force members. Apply equal and opposite forces to each two-force member where it is connected to another member.



50 mm 40 mm

20 mm

400 mm

500 N

A

B

C

D

EF

Problem 6.8

61


3. Next consider the multi force members.

4. Equilibrium equations can be written after completing each free-body.



50 mm 40 mm

20 mm

400 mm

500 N

A

B

C

D

EF

Problem 6.8

62


90 mm

20 mm

Dismember the machine, and draw a free-body diagram of each member.

First consider the two-force members. Ax

Ay

Dy

Dx

Free Body : Portion ABDE

A

B

D

E

Ay

20 mm= ;

Ax

90 mm

Dy= Ay : Dx= Ax= 4.5 Dy (1)

Ax= 4.5 Ay90 mm

20 mmA

D

63


+

20 mm

400 mm

500 NF

Next consider the multi force members.

Equilibrium equations can be written after completing each free-body.

Dy

Dx

Cy

Cx

Free Body : Portion CF

Dy= Ay : Dx= Ax= 4.5 Dy (1)

MC = 0: Dx(20 mm) - Dy(40 mm) - (500 N)(440 mm) = 0

Substitute from (1) 4.5Dy(20) - Dy(40) - 220 x103= 0

Dy = 4400 N = 4.4 kN Dx= 4.5 Dy = 19.8 kN

Fx = 0: Cx - 19.8 kN = 0 Cx = 19.8 kN Fy = 0: Cy - 4.4 kN -0.5 kN = 0 Cy = 4.9 kN

+

+

Using (1) Ax = Dx = 19.8 kN Ay = Dy = 4.4 kN

40 mm

64


20 mm

400 mm

500 NF

A

B

D

E

4.4 kN

19.8 kN

4.4 kN4.4 kN

19.8 kN

19.8 kN

19.8 kN

4.9 kN

All components act in the directions shown. Components on the pipe are equal and opposite to those on the wrench.

4.4 kN

19.8 kN

19.8 kN

4.9 kN Ax = 19.8 kNAy = 4.4 kNCx = 19.8 kNCy = 4.9 kN

40 mm

CE 102 Statics

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