1. Analysis of Structures
CE 102 Statics
Chapter 6
Analysis of Structures

2. 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

3. 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:
Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces.
Trusses: formed from two-force members, i.e., straight members with end point connections
Machines: structures containing moving parts designed to transmit and modify forces.

4. Definition of a Truss
A truss consists of straight members connected at joints. No member is continuous through a joint.
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.
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.
When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression.

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

6. Definition of a Truss

7. 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.

8. 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 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.

9. 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.

10. 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.
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.
Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations.

11. Sample Problem 6.1 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.
Using the method of joints, determine the force in each member of the truss.
All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

12. Sample Problem 6.1 SOLUTION:
Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.

13. Sample Problem 6.1
Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements.
There are now only two unknown member forces at joint D.

14. Sample Problem 6.1
There are now only two unknown member forces at joint B. Assume both are in tension.
There is one unknown member force at joint E. Assume the member is in tension.

15. 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.

16. 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.

17. Trusses Made of Several Simple Trusses
Compound trusses are statically determinant, rigid, and completely constrained.
Truss contains a redundant member and is statically indeterminate.
non-rigid
rigid
Additional reaction forces may be necessary for a rigid truss.
Necessary but insufficient condition for a compound truss to be statically determinant, rigid, and completely constrained,

18. 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.
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.
Determine the force in members FH, GH, and GI.

19. 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.

20. Sample Problem 6.2
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.

21. Sample Problem 6.2

22. 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 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.
Forces between connected components are equal, have the same line of action, and opposite sense.

23. 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.

24. Sample Problem 6.3 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.
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.
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.

25. Sample Problem 6.3 SOLUTION:
Create a free-body diagram for the complete frame and solve for the support reactions.
Note:

26. 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.
Sum of forces in the x and y directions may be used to find the force components at C.

27. Sample Problem 6.3
With member ACE as a free-body, check the solution by summing moments about A.
(checks)

28. 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,

29. Problem 6.4
15 in
15 in
15 in
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.
80 lb
40 lb A B C D
36 in M E

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.
80 lb
40 lb A B C D
36 in
For this problem we note that there are no two-force members. In solving this problem, we M E
1. Dismember the frame, and draw a free-body
diagram for each member.
2. To simplify the solution, seek a way to write an equation
involving a single unknown.

31. BE = (152 + 362)1/2 = 39 in S ME = 0: M - B (39 in) = 0
Problem 6.4 Solution B 13
Dismember the frame, and draw a free-body diagram for each member. 5 12
Free-Body: Member BE
BE = ( )1/2 = 39 in
36 in M 15 36 39 5 12 13 Ex E Ey
15 in
To simplify the solution, seek a way to write
an equation involving a single unknown. +
S ME = 0: M - B (39 in) = 0
M = B (39 in) (1)

32. S MD = 0: (40 lb)(45 in) + (80 lb)(15 in) - B(30 in) = 0
Problem 6.4 Solution
Dismember the frame, and draw a free-body diagram for each member.
15 in
15 in
15 in
Free-Body: Member BE
40 lb
80 lb
To simplify the solution, seek a way to write an equation involving a single unknown. Dx A C D 13 5 B Dy 12 + 5 13
S MD = 0: (40 lb)(45 in) + (80 lb)(15 in) B(30 in) = 0
B = 260 lb

33. M = B (39 in) = (260 lb) (39 in) = 10,140 lb-in M = 10.14 kip-in B B M
Problem 6.4 Solution 13
15 in
15 in
15 in 5 12
40 lb
80 lb Dx A C D 13
36 in 5 B Dy 12 M
B = 260 lb
From EQ (1) Ex E
M = B (39 in)
= (260 lb) (39 in)
= 10,140 lb-in Ey
15 in
M = B (39 in)
M = 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
50 lb
6 in
6 in E D
4 in
2 in C
6 in A B

35. Solving Problems on Your Own
50 lb
6 in
6 in
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. E D
4 in
2 in C
6 in A B
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.
2. To simplify the solution, seek a way to write an equation
involving a single unknown.

36. Dismember the frame, and draw a free-body diagram for each member.
Problem 6.5 Solution
Dismember the frame, and draw a free-body diagram for each member.
50 lb Cy Cy D H D C
2 in Cx Cx
6 in A A
18 in Bx By
6 in
6 in

37. S MH = 0: Cx(2 in) - Cy(18 in) = 0 Cx = 9Cy
Problem 6.5 Solution
To simplify the solution, seek a way to write an equation involving a single unknown.
50 lb Cy Cy D H D C
2 in Cx Cx
6 in A A
18 in Bx
6 in
6 in By
Free-Body: Member ACD +
S MH = 0: Cx(2 in) - Cy(18 in) = Cx = 9Cy

38. S MB = 0: Cx(6 in) + Cy(6 in) - (50 lb)(12 in) = 0
To simplify the solution, seek a way to write an equation involving a single unknown.
Problem 6.5 Solution
50 lb Cy Cy D D H C
2 in
Cx = 9Cy Cx Cx
6 in A A
18 in Bx
6 in
6 in By
Free-Body: Member BCE +
S MB = 0: Cx(6 in) + Cy(6 in) - (50 lb)(12 in) = 0
Substitute Cx = 9Cy:
9Cy(6 in) + Cy(6 in) = 0
Cy = + 10 lb; Cx = 9Cy = 9(10) = 90 lb

39. C S Fx = 0: Bx - 90 lb = 0 Bx = 90 lb S Fy = 0: By + 10 lb - 50 lb = 0
Problem 6.5 Solution
Free-Body: Member BCE C
50 lb
10 lb
C = 90.6 lb Cy
6.3o C C Cx
90 lb +
S Fx = 0: Bx - 90 lb = 0
Bx = 90 lb
S Fy = 0: By + 10 lb - 50 lb = 0
By = 40 lb Bx +
6 in
6 in By B
40 lb
24.0o
B = lb B
90 lb

40. Problem 6.6
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Using the method of joints, determine the force in each member of the truss shown.
2 m
2 m
2 m A B C D
2.5 m G F E

41. Solving Problems on Your Own
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Using the method of joints, determine the force in each member of the truss shown.
2 m
2 m
2 m A B C D
1. Draw the free-body diagram of
the entire truss, and use this
diagram to determine the
reactions at the supports.
2.5 m G F E
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 SFx = 0 and SFy = 0 is positive, the member is in
tension. A negative answer means the member is in
compression.

42. Solving Problems on Your Own
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Using the method of joints, determine the force in each member of the truss shown.
2 m
2 m
2 m A B C D
2.5 m G F E
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.
4. Repeat this procedure until the forces in all the members of
the truss have been determined.

43. S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m)
Problem 6.6 Solution
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Draw the free-body diagram of the entire truss, and use it to
determine reactions at the supports.
2 m
2 m
2 m A Ax B C D Ay
2.5 m G F E E +
S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m)
- (12.5 kN)(6 m) = E = 60 kN +
S Fy = 0: Ay - (4)(12.5 kN) = Ay = 50 kN
S Fx = 0: Ax - E = Ax= 60 kN +

44. S Fy = 0: FGD - 12.5 kN = 0 FGD = 32.5 kN C S Fx = 0: FGD - FCD = 0
Problem 6.6 Solution A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
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.
Joint D
12.5 kN
2.5
6.5 +
S Fy = 0: FGD kN = 0
FCD
FGD = 32.5 kN C
6.5
2.5 6
6.5 6
S Fx = 0: FGD - FCD = 0 +
FGD
FCD = 30 kN T

45. S F = 0: FCG = 0 S F = 0: FFG - 32.5 kN = 0 FFG = 32.5 kN C
Problem 6.6 Solution A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
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.
Joint G
FCG
S F = 0: FCG = 0
32.5 kN
S F = 0: FFG kN = 0
FFG = 32.5 kN C
FFG

46. S Fx = 0: 30 kN - FCF cos b - FBC = 0 A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem 6.6 Solution
Repeat this procedure until the forces in all the members of the truss have been determined.
Joint C 2 3
BF = (2.5 m) = m
12.5 kN BF 2
b = BCF = tan = 39.81o
FBC
FCD = 30 kN b +
S Fy = 0: kN - FCF sin b = 0
kN - FCF sin 39.81o = 0
FCF = kN C
FCF
S Fx = 0: 30 kN - FCF cos b - FBC = 0
30 kN - (-19.53) cos 39.81o - FBC = 0
FBC = 45.0 kN T +

47. S Fx = 0: - FEF - (32.5 kN) - FCF cos b = 0A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem 6.6 Solution
FCF = 19.53kN
Joint F
FBF
b=39.81o
6.5
FEF
FFG = 32.5 kN
2.5 6 + 6
6.5 6
6.5
S Fx = 0: FEF (32.5 kN) - FCF cos b = 0
6.5 6
FEF = kN - ( ) (19.53) cos 39.81o FEF = 48.8 kN C
2.5
6.5
2.5
6.5 +
S Fy = 0: FBF FEF (32.5 kN) - (19.53) sin b = 0
2.5
6.5
FBF (-48.8 kN) kN kN = 0
FBF = 6.25 kN T

48. S Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kNA B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem 6.6 Solution
12.5 kN
Joint B
FAB
FBC = 45.0 kN g
FBF = 6.25kN
FBE
2.5 m
2 m
tan g = ; g = 51.34o
S Fy = 0: kN kN - FBE sin 51.34o = 0
FBE = kN +
FBE = 24.0 kN C +
S Fx = 0: kN - FAB + (24.0 kN) cos 51.34o = 0
FAB = 60.0 kN
FAB = 60.0 kN T

49. S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem 6.6 Solution
Joint E
FBE = 24 kN
FAE g
FEF = kN
6.5
2.5
60 kN 6
g = 51.34o
2.5
6.5 +
S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0
FAE = 37.5 kN FAE = 37.5 kN T

50. Problem 6.7
3 kN
A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI.
3 kN
3 kN F
3 kN
3 kN D H
6.75 m
1.5 kN
1.5 kN B J A L C E G I K
3 m
3 m
3 m
3 m
3 m
3 m

51. Solving Problems on Your Own
3 kN
Problem 6.7
Solving Problems on Your Own
3 kN
3 kN F
3 kN
A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI.
3 kN D H
6.75 m
1.5 kN
1.5 kN B J A L C E G I K
3 m
3 m
3 m
3 m
3 m
3 m
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.

52. Solving Problems on Your Own
3 kN
Solving Problems on Your Own
3 kN
3 kN F
A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI.
3 kN
3 kN D H
6.75 m
1.5 kN
1.5 kN B J A L C E G I K
3 m
3 m
3 m
3 m
3 m
3 m
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.

53. S Fx = 0: Ax = 0 Ay = L = (18kN) = 9 kN
Problem 6.7 Solution
3 kN
3 kN
Draw the free-body diagram of the entire truss, and use it to
determine reactions at the supports. F
3 kN
3 kN D H
1.5 kN
1.5 kN B J A L
S Fx = 0: Ax = 0 C E G I K + Ax Ay L
3 m
3 m
3 m
3 m
3 m
3 m
Total load = 5(3 kN) + 2(1.5 kN) = 18 kN
By symmetry 1 2
Ay = L = (18kN) = 9 kN

54. Pass a section through three members of the truss, one of
3 kN
Problem 6.7 Solution
3 kN
3 kN F
Pass a section through three members of the truss, one of
which is the desired member.
3 kN
3 kN D H
1.5 kN
1.5 kN B J A L C E G I K
9 kN
9 kN
3 m
3 m
3 m
3 m
3 m
3 m

55. tan a = = a = 66.04o Free Body: Portion HIL
Problem 6.7 Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Select one of the two portions of the truss you have obtained,
and draw its free-body diagram.
3 kN H
FFI
6.75 m
1.5 kN J a L
Slope of FHJL G I
FGI K
6.75
9.00 3 4 5
9 kN 3 = 4
3 m
3 m
3 m FG GI
6.75
3.00
tan a = = a = 66.04o
Now write three equilibrium equations which can be solved for
the forces in the three intersected members.

56. S MI = 0: FFH ( x 6.75 m) + (9 kN)(6 m)
Problem 6.7 Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Write the three equilibrium equations.
3 kN H
FFI
6.75 m
1.5 kN J
66.04o L G I
FGI K
9 kN
Force in FH:
3 m
3 m
3 m 4 5 2 3
S MI = 0: +
FFH ( x 6.75 m) + (9 kN)(6 m)
- (1.5 kN)(6m) - (3 kN)(3m) = 0 4 5
FFH (4.5 m) + 36 kN-m = 0
FFH = kN FFH = 10.0 kN C

57. FFH = 10.0 kN C S ML = 0: -FFI sin 66.04o(6 m) + (3 kN)(6 m)
Problem 6.7 Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Write the three equilibrium equations.
3 kN H
FFH = 10.0 kN C
FFI
6.75 m
1.5 kN J
66.04o L G I
FGI K
9 kN
Force in FI:
3 m
3 m
3 m
S ML = 0: +
-FFI sin 66.04o(6 m) + (3 kN)(6 m)
+ (3 kN)(3 m) = 0
FFI sin 66.04o(6 m) = 27 kN-m
FFI = kN FFI = 4.92 kN T

58. -FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m)
Problem 6.7 Solution 5 3
Free Body: Portion HIL 4
3 kN F
FFH
Write the three equilibrium equations.
3 kN H
FFH = 10.0 kN C
FFI = 4.92 kN T
FFI
6.75 m
1.5 kN J
66.04o L G I
FGI K
9 kN
Force in GI:
3 m
3 m
3 m
S MF = 0: +
-FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m)
- (1.5 kN)(9 m) + (9 kN)(9 m) = 0
FGI (6.75 m) = kN-m
FGI = 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 A F E
500 N C
20 mm
400 mm
50 mm
40 mm
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.

60. Solving Problems on Your OwnD
A pipe of diameter
50 mm is gripped by the Stillson wrench shown. Portions AB and DE of the wrench are rigidly A F E
500 N C
20 mm
400 mm
50 mm
40 mm
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.
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.

61. Solving Problems on Your OwnD
A pipe of diameter
50 mm is gripped by the Stillson wrench shown. Portions AB and DE of the wrench are rigidly A F E
500 N C
20 mm
400 mm
50 mm
40 mm
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.
3. Next consider the multi force members.
4. Equilibrium equations can be written after completing each
free-body.

62. Ay 20 mm Ax 90 mm = ; Ax= 4.5 Ay Dy= Ay : Dx= Ax= 4.5 Dy (1)
Problem 6.8 Solution Dy B
Dismember the machine, and draw a free-body diagram of each member. A Ay D Dx Ax
20 mm
First consider the two-force members. E
90 mm
Free Body : Portion ABDE D Ay
20 mm Ax
90 mm
= ;
20 mm A
90 mm
Ax= 4.5 Ay
Dy= Ay : Dx= Ax= 4.5 Dy (1)

63. S MC = 0: Dx(20 mm) - Dy(40 mm) - (500 N)(440 mm) = 0
Dy= Ay : Dx= Ax= 4.5 Dy (1)
Problem 6.8 Solution Dy Dx
20 mm
Next consider the multi force members. F Cx
500 N
Equilibrium equations can be written after completing each free-body. Cy
40 mm
400 mm
Free Body : Portion CF +
S MC = 0: Dx(20 mm) - Dy(40 mm) - (500 N)(440 mm) = 0
Substitute from (1)
4.5Dy(20) - Dy(40) x103= 0
Dy = 4400 N = 4.4 kN Dx= 4.5 Dy = 19.8 kN +
S Fx = 0: Cx kN = Cx = 19.8 kN
S Fy = 0: Cy kN -0.5 kN = Cy = 4.9 kN +
Using (1)
Ax = Dx = 19.8 kN Ay = Dy = 4.4 kN

64. 4.9 kN Ax = 19.8 kN 19.8 kN Ay = 4.4 kN Cx = 19.8 kN Cy = 4.9 kN
Problem 6.8 Solution
4.4 kN
4.4 kN B
20 mm
19.8 kN A
4.4 kN D
19.8 kN F
19.8 kN
500 N
19.8 kN E
4.9 kN
40 mm
400 mm
All components act in the directions shown. Components on the pipe are equal and opposite to those on the wrench.
4.9 kN
Ax = 19.8 kN
Ay = 4.4 kN
Cx = 19.8 kN
Cy = 4.9 kN
19.8 kN
19.8 kN
4.4 kN