1. 6.5 Space Trusses
A space truss consists of members joined together at their ends to form a stable 3D structure
The simplest space truss is a tetrahedron, formed by joined 6
members as shown
Any additional members
added would be redundant
in supporting force P

2. 6.5 Space Trusses Assumptions for Design
The members of a space truss may be treated as two force members provided the external loading is applied at the joints and the joints consist of ball and socket connections
If the weight of the member is to be considered, apply it as a vertical force, half of its magnitude applied at each end of the member

3. 6.5 Space Trusses Procedure for Analysis Method of Joints
To determine the forces in all the members of the truss
Solve the three scalar equilibrium ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 at each joint
The force analysis begins at a point having at least one unknown force and at most three unknown forces
Cartesian vector analysis used for 3D

4. 6.5 Space Trusses Procedure for Analysis Method of Sections
Used to determine a few member forces
When an imaginary section is passes through a truss and the truss is separated into two parts, the below equations of equilibrium must be satisfied
∑Fx = 0, ∑Fy = 0, ∑Fz = 0
∑Mx = 0, ∑My = 0, ∑Mz = 0
By proper selection, the unknown forces can be determined using a single equilibrium equation

5. 6.5 Space Trusses
Example 6.8
Determine the forces acting in the members
of the space truss. Indicate
whether the members are
in tension or compression.

6. 6.5 Space Trusses
View Free Body Diagram
Solution
Joint A

7. 6.5 Space Trusses
Solution
Joint A
To show

8. 6.6 Frames and Machines
Composed of pin-connected multi-force members (subjected to more than two forces)
Frames are stationary and are used to support the loads while machines contain moving parts, designated to transmit and alter the effects of forces
Apply equations of equilibrium to each member to determine the unknown forces

9. 6.6 Frames and Machines Free-Body Diagram
Isolate each part by drawing its outlined shape
- show all the forces and the couple moments that act on the part
- label or identify each known and unknown force and couple moment with reference to the established x, y and z coordinate system

10. 6.6 Frames and Machines Free-Body Diagram
- indicate any dimension used for taking moments
- equations of equilibrium are easier to apply when the forces are represented in their rectangular coordinates
- sense of any unknown force or moment can be assumed

11. 6.6 Frames and Machines Free-Body Diagram
Identify all the two force members in the structure and represent their FBD as having two equal but opposite collinear forces acting at their points of application
Forces common to any contracting member act with equal magnitudes but opposite sense on the respective members

12. 6.6 Frames and Machines Free-Body Diagram
- treat two members as a system of connected members
- these forces are internal and are not shown on the FBD
- if the FBD of each member is drawn, the forces are external and must be shown on the FBD

13. 6.6 Frames and Machines Example 6.9
For the frame, draw the free-body diagram of
each member,
the pin at B and
the two members
connected together.

14. 6.6 Frames and Machines Solution Part (a)
members BA and BC are not two-force members
BC is subjected to 3 forces, the resultant force
from pins B and C and
the external P
AB is subjected to the
resultant forces from the
pins at A and B and the
external moment M

15. 6.6 Frames and Machines Solution Part (b)
Pin at B is subjected to two forces, force of the member BC on the pin and the force of member AB on the pin
For equilibrium, these
forces and respective
components must be
equal but opposite

16. 6.6 Frames and Machines Solution Part (b)
But Bx and By shown equal and opposite
on members AB ad BC results from the
equilibrium analysis of
the pin rather from
Newton’s third law

17. 6.6 Frames and Machines Solution Part (c)
FBD of both connected members without the supporting pins at A and C
Bx and By are not shown since
they form equal but
opposite collinear pairs
of internal forces

18. 6.6 Frames and Machines Solution Part (c)
To be consistent when applying the equilibrium equations, the unknown force components at A and C must act in the same sense
Couple moment M can be
applied at any point on
the frame to determine
reactions at A and C

19. 6.6 Frames and Machines Example 6.10
A constant tension in the conveyor belt is
maintained by using the device. Draw the
FBD of the frame and
the cylinder which
supports the belt.
The suspended black
has a weight of W.

20. 6.6 Frames and Machines Solution Idealized model of the device
Angle θ assumed known
Tension in the belt is the same on each side of the cylinder since it is free to turn

21. 6.6 Frames and Machines Solution FBD of the cylinder and the frame
Bx and By provide equal but opposite couple moments on the cylinder
Half of the pin reactions at A act on each side of the frame since pin connections occur on each side

22. 6.6 Frames and Machines Example 6.11
Draw the free-body diagrams of each part of
the smooth piston and link mechanism used
to crush recycled cans.

23. 6.6 Frames and Machines Solution Member AB is a two force member
FBD of the parts

24. 6.6 Frames and Machines Solution
Since the pins at B and D connect only two parts together, the forces are equal but opposite on the separate FBD of their connected members
Four components of the force act on the piston: Dx and Dy represent the effects of the pin and Nw is the resultant force of the floor and P is the resultant compressive force caused by can C

25. 6.6 Frames and Machines Example 6.12
For the frame, draw the free-body diagrams of (a)
the entire frame including the pulleys and cords, (b)
the frame without the pulleys and cords, and (c)
each of the pulley.

26. 6.6 Frames and Machines Solution Part (a)
Consider the entire frame, interactions at
the points where the pulleys and cords
are connected to the frame
become pairs of internal
forces which cancel
each other and not
shown on the FBD

27. 6.6 Frames and Machines Solution Part (b) and (c)
When cords and pulleys are removed, their effect on the frame must be shown

28. 6.6 Frames and Machines Example 6.13 Draw the free-body
diagrams of the bucket and
the vertical boom of the back
hoe. The bucket and its
content has a weight W.
Neglect the weight of the
members.

29. 6.6 Frames and Machines Solution Idealized model of the assembly
Members AB, BC, BE and HI are two force members

30. 6.6 Frames and Machines Solution FBD of the bucket and boom
Pin C subjected to 2 forces, force of the link BC and force of the boom
Pin at B subjected to three forces, force by the hydraulic cylinder and the forces caused by the link
These forces are related by equation of force equilibrium

31. 6.6 Frames and Machines Equations of Equilibrium
Provided the structure is properly supported and contains no more supports and members than necessary to prevent collapse, the unknown forces at the supports and connections can be determined from the equations of equilibrium
The selection of the FBD for analysis are completely arbitrary and may represent each of the members of the structure, a portion or its entirety.

32. 6.6 Frames and Machines Equations of Equilibrium
Consider the frame in fig (a)
Dismembering the frame in fig (b), equations of equilibrium can be used
FBD of the entire frame in fig (c)

33. 6.6 Frames and Machines Procedures for Analysis FBD
Draw the FBD of the entire structure, a portion or each of its members
Choice is dependent on the most direct solution to the problem
When the FBD of a group of members of a structure is drawn, the forces at the connected parts are internal forces and are not shown
Forces common to two members which are in contact act with equal magnitude but opposite sense on their respective FBD

34. 6.6 Frames and Machines Procedures for Analysis FBD
Two force members, regardless of their shape, have equal but opposite collinear forces acting at the ends of the member
In many cases, the proper sense of the unknown force can be determined by inspection
Otherwise, assume the sense of the unknowns
A couple moment is a free vector and can act on any point of the FBD

35. 6.6 Frames and Machines Procedures for Analysis FBD
A force is a sliding vector and can act at any point along its line of action
Equations of Equilibrium
Count the number of unknowns and compare to the number of equilibrium equations available
In 2D, there are 3 equilibrium equations written for each member

36. 6.6 Frames and Machines Procedures for Analysis
Equations of Equilibrium
Sum moments about a point that lies at the intersection of the lines of action of as many unknown forces as possible
If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reserve of that shown on the FBD

37. 6.6 Frames and Machines Example 6.14
Determine the horizontal and vertical
components of the force which the pin C
exerts on member CB
of the frame.

38. 6.6 Frames and Machines Solution Method 1
Identify member AB as two force member
FBD of the members AB and BC

39. 6.6 Frames and Machines
Solution

40. 6.6 Frames and Machines Solution Method 2
Fail to identify member AB as two force member

41. 6.6 Frames and Machines
Solution
Member AB

42. 6.6 Frames and Machines
Solution
Member BC

43. 6.6 Frames and Machines Example 6.15
The compound beam is pin connected at B.
Determine the reactions at its support.
Neglect its weight and thickness.

44. 6.6 Frames and Machines Solution FBD of the entire frame
Dismember the beam into two segments since there are 4 unknowns but 3 equations of equilibrium

45. 6.6 Frames and Machines
Solution
Segment BC

46. 6.6 Frames and Machines
Solution
Member AB

47. 6.6 Frames and Machines Example 6.16
Determine the horizontal and vertical
components of the force which the pin at C
exerts on member ABCD of
the frame.

48. 6.6 Frames and Machines Solution Member BC is a two force member
FBD of the entire frame
FBD of each member

49. 6.6 Frames and Machines
Solution
Entire Frame

50. 6.6 Frames and Machines
Solution
Member CEF

51. 6.6 Frames and Machines Example 6.17
The smooth disk is pinned at D and has a weight of
20N. Neglect the weights of others member,
determine the horizontal and vertical components
of the reaction at pins B and D

52. 6.6 Frames and Machines Solution FBD of the entire frame
FBD of the members

53. 6.6 Frames and Machines
Solution
Entire Frame

54. 6.6 Frames and Machines
Solution
Member AB

55. 6.6 Frames and Machines
Solution
Disk

56. 6.6 Frames and Machines Example 6.18
Determine the tension in the cables
and also the force P required to
support the 600N force using the
frictionless pulley system.

57. 6.6 Frames and Machines Solution FBD of each pulley
Continuous cable and frictionless pulley = constant tension P
Link connection between pulleys B and C is a two force member

58. 6.6 Frames and Machines
Solution
Pulley A
Pulley B
Pulley C

59. 6.6 Frames and Machines Example 6.19
A man having a weight of 750N supports
himself by means of the cable and
pulley system. If the seat has a
weight of 75N, determine the force
he must exert on the cable at A and
the force he exerts on the seat.
Neglect the weight of the cables
and pulleys.

60. 6.6 Frames and Machines Solution Method 1
FBD of the man, seat and pulley C

61. 6.6 Frames and Machines
Solution
Man
Seat
Pulley C

62. 6.6 Frames and Machines Solution Method 2
FBD of the man, seat and pulley C as a single system

63. 6.6 Frames and Machines
Solution

64. 6.6 Frames and Machines Example 6.20
The hand exerts a force of 35N on the grip of the
spring compressor. Determine the force in the
spring needed to maintain equilibrium of the
mechanism.

65. 6.6 Frames and Machines
Solution
FBD for parts DC and ABG

66. 6.6 Frames and Machines
Solution
Lever ABG
Pin E

67. 6.6 Frames and Machines
Solution
Arm DC

68. 6.6 Frames and Machines Example 6.21
The 100kg block is held in equilibrium by
means of the pulley and the continuous
cable system. If the cable is
attached to the pin at B,
compute the forces which this
pin exerts on each of its
connecting members

69. 6.6 Frames and Machines Solution FBD of each member of the frame
View Free Body Diagram
Solution
FBD of each member of the frame
Ad and CB are two force members

70. 6.6 Frames and Machines
Solution
Pulley B

71. 6.6 Frames and Machines Solution Pin E
Two force member BC subjected to bending as caused by FBC
Better to make this member straight so that the force would only cause tension in the member

72. Chapter Summary Truss Analysis
A simple truss consists of triangular elements connected by pin joints
The force within determined by assuming all the members to be two force member, connected concurrently at each joint
Method of Joints
For the truss in equilibrium, each of its joint is also in equilibrium

73. Chapter Summary Method of Joints
For a coplanar truss, the concurrent force at each joint must satisfy force equilibrium
For numerical solution of the forces in the members, select a joint that has FBD with at most 2 unknown and 1 known forces
Once a member force is determined, use its value and apply it to an adjacent joint
Forces that pull on the joint are in tension

74. Chapter Summary Method of Joints
Forces that push on the joint are in compression
To avoid simultaneous solution of two equations, sum the force in a direction that is perpendicular to one of the unknown
To simplify problem-solving, first identify all the zero-force members

75. Chapter Summary Method of Sections
For the truss in equilibrium, each section is also in equilibrium
Pass a section through the member whose force is to be determined
Draw the FBD of the sectioned part having the least forces on it
Forces that pull on the section are in tension

76. Chapter Summary Method of Sections
Forces that push on the section are in compression
For a coplanar force system, use the three equations of equilibrium for solving
If possible, sum the force in a direction that is perpendicular to two of the three unknown forces
Sum the moment about a point that passes through the line of action of two of the three unknown forces

77. Chapter Summary Frames and Machines
The forces acting at the joints of a frame or machine can be determined by drawing the FBD of each of its members or parts
Principle of action-reaction should be observed when drawing the forces
For coplanar force system, there are three equilibrium equations for each member

78. Chapter Review

79. Chapter Review

80. Chapter Review

81. Chapter Review

82. Chapter Review