1. SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS
Objectives:
a) Define a simple truss.
b) Determine the forces in members of a simple truss.
c) Identify zero-force members.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

2. APPLICATIONS Trusses are commonly used to support roofs.
For a given truss geometry and load, how can you determine the forces in the truss members and thus be able to select their sizes?
A more challenging question is that for a given load, how can we design the trusses’ geometry to minimize cost?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

3. APPLICATIONS (continued)
Trusses are also used in a variety of structures like cranes and the frames of aircraft or space stations.
How can you design a light weight structure that will meet load, safety, cost specifications, be easy to manufacture, and allow easy inspectioin over its lifetime?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

4. SIMPLE TRUSSES (Section 6.1)
A truss is a structure composed of slender members joined together at their end points.
If a truss, along with the imposed load, lies in a single plane (as shown at the top right), then it is called a planar truss.
A simple truss is a planar truss which begins with a a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2 J – 3.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

5. ANALYSIS & DESIGN ASSUMPTIONS
When designing both the member and the joints of a truss, first it is necessary to determine the forces in each truss member. This is called the force analysis of a truss. When doing this, two assumptions are made:
1. All loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members.
2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases where the joints are formed by bolting the ends together.
With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

6. THE METHOD OF JOINTS (Section 6.2)
A free body diagram of Joint B
When using the method of joints to solve for the forces in truss members, the equilibrium of a joint (pin) is considered. All forces acting at the joint are shown in a FBD. This includes all external forces (including support reactions) as well as the forces acting in the members. Equations of equilibrium ( FX= 0 and  FY = 0) are used to solve for the unknown forces acting at the joints.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

7. STEPS FOR ANALYSIS
If the truss’s support reactions are not given, draw a FBD of the entire truss and determine the support reactions (typically using scalar equations of equilibrium).
2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by inspection that the forces are compression loads.
3. Apply the scalar equations of equilibrium,  FX = 0 and  FY = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression).
4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

8. ZERO-FORCE MEMBERS (Section 6.3)
If a joint has only two non-collinear members and there is no external load or support reaction at that joint, then those two members are zero-force members. In this example members DE, DC, AF, and AB are zero force members.
You can easily prove these results by applying the equations of equilibrium to joints D and A.
Zero-force members can be removed (as shown in the figure) when analyzing the truss.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

9. ZERO – FORCE MEMBERS (continued)
If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member.
Again, this can easily be proven. One can also remove the zero-force member, as shown, on the left, for analyzing the truss further.
Please note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various different loading conditions.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

10. EXAMPLE Given: Loads as shown on the truss
Find: The forces in each member of the truss.
Plan:
Check if there are any zero-force members.
First analyze pin D and then pin A
Note that member BD is zero-force member. FBD = 0
Why, for this problem, do you not have to find the external reactions before solving the problem?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

11. EXAMPLE (continued) FCD
45 º
FCD D
450 lb
FAD
FBD of pin D
+   FX = – FCD cos 45° – FAD cos 45° = 0
+   FY = – FCD sin 45° – FAD sin 45° = 0
FCD = lb (Tension) or (T)
and FAD = – 318 lb (Compression) or (C)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

12. EXAMPLE (continued) Analyzing pin A: FAD FAB AY
45 º
FAB A
FBD of pin A
FAD AY
+   FX = FAB + (– 318) cos 45° = 0; FAB = lb (T)
Could you have analyzed Joint C instead of A?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

13. GROUP PROBLEM SOLVING Given: Loads as shown on the truss
Find: Determine the force in all the truss members (do not forget to mention whether they are in T or C).
Plan:
a) Check if there are any zero-force members.
Draw FBDs of pins D and E, and then apply EE at those pins to solve for the unknowns.
Note that Member CE is zero-force member so FEC = 0. If you didn’t see this simplification, could you still solve the problem?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

14. GROUP PROBLEM SOLVING (continued)
FBD of pin D
FDE Y D
600N X
FCD
26.57
From geometry, tan-1(1/2)=26.57
Analyzing pin D: → + FX = 600 – FCD sin 26.57 = 0
FCD = N = 1.34 kN (C) (Note that FCD = FBC!)
+  FY = cos 26.57 – FDE = 0
FDE = N = 1.2 kN (T)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

15. GROUP PROBLEM SOLVING (continued)
FBD of pin E
FEA Y E
900 N X
FEB
45
FDE
Analyzing pin E:
→ + FX = 900 – FEB sin 45 = 0
FEB = 1273 N = 1.27 kN (C)
+  FY = cos 45 – FEA = 0
FEA = 2100 N = 2.1 kN (T)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

16. Determine forces in truss members using the method of sections.
Objective:
Determine forces in truss members using the method of sections.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

17. So we need another method to determine such forces.
APPLICATIONS
Long trusses are often used to construct large cranes and large electrical transmission towers.
The method of joints requires that many joints be analyzed before we can determine the forces in the middle part of a large truss.
So we need another method to determine such forces.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

18. THE METHOD OF SECTIONS
Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newton’s third law.
In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

19. STEPS FOR ANALYSIS
1. Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general).
2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find).
3. If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

20. STEPS FOR ANALYSIS (continued)
4. Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially we may assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

21. STEPS FOR ANALYSIS (continued)
5. Apply the scalar equations of equilibrium (E-of-E) to the selected cut section of the truss to solve for the unknown member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut!
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

22. Given: Loads as shown on the truss.
EXAMPLE
Given: Loads as shown on the truss.
Find: The force in members KJ, KD, and CD.
Plan:
a) Take a cut through the members KJ, KD, and CD.
b) Work with the left part of the cut section. Why?
c) Determine the support reactions at A. What are they?
d) Apply the EofE to find the forces in KJ, KD, and CD.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

23. ∑MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = 45.6 kN
EXAMPLE (continued)
Analyzing the entire truss for the reactions a A, we get  FX = AX = 0. Then do a moment equation about G to find AY.
∑MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = kN
Now take moments about point D. Why do this?
+ MD = – 45.6 (9) + 20 (6) (3) – FKJ (4) = 0
FKJ = − 50.1 kN or kN ( C )
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

24. Now use the two force equations of equilibrium.
EXAMPLE (continued)
Now use the two force equations of equilibrium.
↑ +  FY = – 20 – 30 – (4/5) FKD = 0;
FKD = − 5.5 kN , or 5.5 kN (C)
→ +  FX = (– 50.1) + (3/5) ( –5.5 ) + FCD = 0;
FCD = kN (T)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

25. Find: The force in members IH, BH, and BC.
GROUP PROBLEM SOLVING
Given: The internal drag truss for the wing of a airplane is subjected to the forces shown.
Find: The force in members IH, BH, and BC.
Plan:
a) Take a cut through the members IH, BH, and BC.
b) Analyze the right section (no support reactions!).
c) Draw the FBD of the right section.
d) Apply the equations of equilibrium (if possible try to do it so that every equation yields an answer to one unknown.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

26. + → FX = FIH – 130– 255 cos 45º = 0; FIH = 310 lb (T)
SOLUTION
+ ↑ FY = – FBH sin 45º = 0;
FBE = 255 lb (T)
+  MH = – FBC (2) + 60 (2) + 40 (3.5) = 0;
FBC = 130 lb (T)
+ → FX = FIH – 130– 255 cos 45º = 0; FIH = 310 lb (T)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.4

27. FRAMES AND MACHINES Objectives:
a) Draw the free body diagram of a frame or machine and its members.
b) Determine the forces acting at the joints and supports of a frame or machine.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

28. Frames are commonly used to support various external loads.
APPLICATIONS
Frames are commonly used to support various external loads.
How is a frame different than a truss?
To be able to design a frame, you need to determine the forces at the joints and supports.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

29. APPLICATIONS (continued)
“Machines,” like those above, are used in a variety of applications. How are they different from trusses and frames?
How can you determine the loads at the joints and supports? These forces and moments are required when designing the machine members.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

30. FRAMES AND MACHINES: DEFINITIONS
Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members).
Frame
Machine
Frames are generally stationary and support external loads.
Machines contain moving parts and are designed to alter the effect of forces.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

31. STEPS FOR ANALYZING A FRAME OR MACHINE
1. Draw a FBD of the frame or machine and its members, as necessary.
Hints: a) Identify any two-force members, b) forces on contacting surfaces (usually between a pin and a member) are equal and opposite, and, c) for a joint with more than two members or an external force, it is advisable to draw a FBD of the pin.
FAB
2. Develop a strategy to apply the equations of equilibrium to solve for the unknowns.
Problems are going to be challenging since there are usually several unknowns. A lot of practice is needed to develop good strategies.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

32. Given: The frame supports an external load and moment as shown.
EXAMPLE
Given: The frame supports an external load and moment as shown.
Find: The horizontal and vertical components of the pin reactions at C and the magnitude of reaction at B.
Plan:
a) Draw FBDs of the frame member BC. Why pick this part of the frame?
b) Apply the equations of equilibrium and solve for the unknowns at C and B.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

33. EXAMPLE FBD of member BCCX CY B
45°
FAB
400 N
1 m
2 m
800 N m
FBD of member BC
Please note that member AB is a two-force member.
Equations of Equilibrium:
+  MC = FAB sin45° (1) – FAB cos45° (3) N m (2) = 0
FAB = N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

34. EXAMPLE FBD of member BCCX CY B
45°
FAB
400 N
1 m
2 m
800 N m
FBD of member BC
+  FY = – CY cos 45° – 400 = 0
CY = N
+  FX = – CX sin 45° = 0
CX = N
Now use the x and y direction Equations of Equilibrium:
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

35. Given: A frame supports a 50-kg cylinder.
GROUP PROBLEM SOLVING
Given: A frame supports a 50-kg cylinder.
Find: The reactions that the pins exert on the frame at A and D.
Plan:
a) Draw a FBD of member ABC and another one for CD.
b) Apply the equations of equilibrium to each FBD to solve for the six unknowns. Think about a strategy to easily solve for the unknowns.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

36. GROUP PROBLEM SOLVING (continued)
FBDs of members ABC and CD
50(9.81) N CY CX DX DY
0.7 m
1.6 m
1.2 m
Applying E-of-E to member ABC:
+  MA = CY (1.6) – 50 (9.81) (0.7) – 50 (9.81) (1.7) = 0 ;
CY = N
+  FY = AY – – 50 (9.81) – 50 (9.81) = 0 ; AY = 245 N
+  FX = CX – AX = 0 ; CX = AX
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

37. GROUP PROBLEM SOLVING (continued)
FBDs of members ABC and CD
50(9.81) N CY CX DX DY
0.7 m
1.6 m
1.2 m
Applying E-of-E to member CD:
+  MD = CX (1.2) + 50 (9.81) (0.7) – 735.8(1.6) = 0 ; CX = 695 N
+  FY = DY – (9.81) = 0 ; DY = 245 N
+  FX = DX – = 0 ; DX = 695 N
AX = CX = 695 N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6