SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS

In-Class Activities:• Check Homework, if

any• Reading Quiz• Applications• Simple Trusses• Method of Joints• Zero-Force Members• Concept Quiz• Group Problem Solving• Attention Quiz

Today’s Objectives:

Students will be able to:

a) Define a simple truss.

b) Determine the forces in members of a simple truss.

c) Identify zero-force members.

READING QUIZ

1. One of the assumptions used when analyzing a simple truss is that the members are joined together by __________.

A) Welding B) Bolting C) Riveting

D) Smooth pins E) Super glue

2. When using the method of joints, typically _________ equations of equilibrium are applied at every joint.

A) Two B) Three

C) Four D) Six

APPLICATIONS

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?

Trusses are commonly used to support roofs.

A more challenging question is that for a given load, how can we design the trusses’ geometry to minimize cost?

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?

SIMPLE TRUSSES (Section 6.1)

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 truss is a structure composed of slender members joined together at their end points.

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 equationM = 2 J – 3.

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.

THE METHOD OF JOINTS (Section 6.2)

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.

A free body diagram of Joint B

STEPS FOR ANALYSIS

1. 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.

ZERO-FORCE MEMBERS (Section 6.3)

You can easily prove these results by applying the equations of equilibrium to joints D and A.

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.

Zero-force members can be removed (as shown in the figure) when analyzing the truss.

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.

EXAMPLE

1. Check if there are any zero-force members. 2. First analyze pin D and then pin A3. Note that member BD is zero-force member. FBD = 0

4. Why, for this problem, do you not have to find the external reactions before solving the problem?

Given: Loads as shown on the truss

Find: The forces in each member of the truss.

Plan:

EXAMPLE (continued)

+ FX = – 450 + FCD cos 45° – FAD cos 45° = 0

+ FY = – FCD sin 45° – FAD sin 45° = 0

FCD = 318 lb (Tension) or (T)

and FAD = – 318 lb (Compression) or (C)

45 ºFCD

D 450 lb

FAD

FBD of pin D

45 º

EXAMPLE (continued)

+ FX = FAB + (– 318) cos 45° = 0; FAB = 225 lb (T)

Could you have analyzed Joint C instead of A?

45 º

FAB

A

FBD of pin A

FAD

AY

Analyzing pin A:

CONCEPT QUIZ

1. Truss ABC is changed by decreasing its height from H to 0.9 H. Width W and load P are kept the same. Which one of the following statements is true for the revised truss as compared to the original truss?

A) Force in all its members have decreased.

B) Force in all its members have increased.

C) Force in all its members have remained the same.

D) None of the above.

H

P

A

B C

W

CONCEPT QUIZ (continued)

2. For this truss, determine the number of zero-force members.

A) 0 B) 1 C) 2

D) 3 E) 4

F F

F

GROUP PROBLEM SOLVING

a) Check if there are any zero-force members.

b) Draw FBDs of pins D and E, and then apply EE at those pins to solve for the unknowns.

c) Note that Member CE is zero-force member so FEC = 0. If you didn’t see this simplification, could you still solve the problem?

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:

GROUP PROBLEM SOLVING (continued)

Analyzing pin D:→ + FX = 600 – FCD sin 26.57 = 0

FCD = 1341 N = 1.34 kN (C) (Note that FCD = FBC!)

­ + FY = 1341 cos 26.57 – FDE = 0

FDE = 1200 N = 1.2 kN (T)

FBD of pin D

FDE

Y

D600NX

FCD26.57

From geometry, tan-1(1/2)=26.57

GROUP PROBLEM SOLVING (continued)

Analyzing pin E:

→ + FX = 900 – FEB sin 45 = 0

FEB = 1273 N = 1.27 kN (C)

­ + FY = 1200 + 1273 cos 45 – FEA = 0

FEA = 2100 N = 2.1 kN (T)

FBD of pin E

FEA

Y

E900 NX

FEB45

FDE

ATTENTION QUIZ1. Using this FBD, you find that FBC = – 500 N.

Member BC must be in __________.

A) Tension

B) Compression

C) Cannot be determined

2. For the same magnitude of force to be carried, truss members in compression are generally made _______ as compared to members in tension.

A) Thicker

B) Thinner

C) The same size

FBD

FBC

B

BY

THE METHOD OF SECTIONS

In-Class Activities:• Check Homework, if any• Reading Quiz• Applications• Method of Sections• Concept Quiz• Group Problem Solving• Attention Quiz

Today’s Objectives:

Students will be able to determine:

1. Forces in truss members using the method of sections.

READING QUIZ

1. In the method of sections, generally a “cut” passes through no more than _____ members in which the forces are unknown.

A) 1 B) 2

C) 3 D) 4

2. If a simple truss member carries a tensile force of T along its length, then the internal force in the member is ______ .

A) Tensile with magnitude of T/2 B) Compressive with magnitude of T/2 C) Compressive with magnitude of T D) Tensile with magnitude of T

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.

THE METHOD OF SECTIONS

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.

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.

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.

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.)

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!

STEPS FOR ANALYSIS (continued)

EXAMPLE

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.

Given: Loads as shown on the truss.

Find: The force in members KJ, KD, and CD.

Plan:

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 = 45.6 kN

Now take moments about point D. Why do this?

+ MD = – 45.6 (9) + 20 (6) + 30 (3) – FKJ (4) = 0 FKJ = − 50.1 kN or 50.1 kN ( C )

EXAMPLE (continued)

Now use the two force equations of equilibrium.

↑ + FY = 45.6 – 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 = 53.4 kN (T)

CONCEPT QUIZ

1. Can you determine the force in member ED by making the cut at section a-a? Explain your answer.

A) No, there are 4 unknowns.

B) Yes, using MD = 0 .

C) Yes, using ME = 0 .

D) Yes, using MB = 0 .

CONCEPT QUIZ

2. If you know FED, how will you determine FEB ?

A) By taking section b-b and using ME = 0

B) By taking section b-b, and using FX = 0 and FY = 0

C) By taking section a-a and using MB = 0

D) By taking section a-a and using MD = 0

GROUP PROBLEM SOLVING

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.

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:

SOLUTION

+ ↑ FY = 80 + 60 + 40 – 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)

ATTENTION QUIZ

1. As shown, a cut is made through members GH, BG and BC to determine the forces in them. Which section will you choose for analysis and why?

A) Right, fewer calculations.

B) Left, fewer calculations.

C) Either right or left, same amount of work.

D) None of the above, too many unknowns.

ATTENTION QUIZ

2. When determining the force in member HG in the previous question, which one equation of equilibrium is best to use?

A) MH = 0

B) MG = 0

C) MB = 0

D) MC = 0

FRAMES AND MACHINES

In-Class Activities:• Check Homework, if any• Reading Quiz• Applications• Analysis of a Frame/Machine• Concept Quiz• Group Problem Solving• Attention Quiz

Today’s Objectives:

Students will be able to:

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.

READING QUIZ

1. Frames and machines are different as compared to trusses since they have ___________.

A) Only two-force members B) Only multiforce members

C) At least one multiforce member D) At least one two-force member

2. Forces common to any two contacting members act with _______ on the other member.

A) Equal magnitudes but opposite sense

B) Equal magnitudes and the same sense

C) Different magnitudes but opposite sense

D) Different magnitudes but the same sense

APPLICATIONS

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.

Frames are commonly used to support various external loads.

APPLICATIONS (continued)

How can you determine the loads at the joints and supports? These forces and moments are required when designing the machine members.

“Machines,” like those above, are used in a variety of applications. How are they different from trusses and frames?

FRAMES AND MACHINES: DEFINITIONS

Frames are generally stationary and support external loads.

Machines contain moving parts and are designed to alter the effect of forces.

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

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.

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.

FAB

EXAMPLE

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.

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:

EXAMPLE

Please note that member AB is a two-force member.

FBD of member BC

CX

CY

B

45°FAB

400 N

1 m 2 m1 m

800 N m

Equations of Equilibrium:

+ MC = FAB sin45° (1) – FAB cos45° (3) + 800 N m + 400 (2) = 0

FAB = 1131 N

EXAMPLE

FBD of member BC

CX

CY

B

45°FAB

400 N

1 m 2 m1 m

800 N m

­ + FY = – CY + 1131 cos 45° – 400 = 0

CY = 400 N

® + FX = – CX + 1131 sin 45° = 0

CX = 800 N

Now use the x and y direction Equations of Equilibrium:

CONCEPT QUIZ

1. The figures show a frame and its FBDs. If an additional couple moment is applied at C, then how will you change the FBD of member BC at B?

A) No change, still just one force (FAB) at B.B) Will have two forces, BX and BY, at B.C) Will have two forces and a moment at B.D) Will add one moment at B.

2. The figures show a frame and its FBDs. If an additional force is applied at D, then how will you change the FBD of member BC at B?

A) No change, still just one force (FAB) at B.B) Will have two forces, BX and BY, at B.C) Will have two forces and a moment at B.D) Will add one moment at B.

CONCEPT QUIZ (continued)

D

GROUP PROBLEM SOLVING

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.

Given: A frame supports a 50-kg cylinder.

Find: The reactions that the pins exert on the frame at A and D.

Plan:

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 = 735.8 N

+ FY = AY – 735.8 – 50 (9.81) – 50 (9.81) = 0 ; AY = 245 N

®+ FX = CX – AX = 0 ; CX = AX

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 – 735.8 + 50 (9.81) = 0 ; DY = 245 N

®+ FX = DX – 695 = 0 ; DX = 695 N

AX = CX = 695 N

ATTENTION QUIZ

2. For the above problem, imagine that you have drawn a FBD of member AB. What will be the easiest way to write an equation involving unknowns at B?

A) MC = 0 B) MB = 0

C) MA = 0 D) FX = 0

1. When determining the reactions at joints A, B, and C, what is the minimum number of unknowns for solving this problem?A) 3 B) 4C) 5 D) 6