SRSD 2093: Engineering Mechanics

2SRRI SECTION 19ROOM 7, LEVEL 14, MENARA RAZAK

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

In-Class Activities:

• Applications• Simple Trusses• Zero-force Members• Group Problem Solving

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.

APPLICATIONS

For a given truss geometry and load, how canyou determine the forces in the trussmembers and thus be able to select theirsizes?

Trusses are commonly used to support roofs.

A more challenging question is, that for a givenload, how can we design the trusses’ geometry tominimize cost?

Trusses are also used in a variety of structureslike cranes and the frames of aircraft or thespace station.

How can you design a light weightstructure satisfying load, safety, costspecifications, is simple to manufacture,and allows easy inspection over itslifetime?

APPLICATIONS

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 theirend points.

A simple truss is a planar truss which begins with atriangular element and can be expanded by adding twomembers and a joint. For these trusses, the number ofmembers (M) and the number of joints (J) are relatedby the equation M = 2 J – 3.

SIMPLE TRUSSES

When designing the members and joints of a truss, first it is necessary todetermine the forces in each truss member. This is called the force analysisof 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 astwo-force members. They are loaded in eithertension or compression. Often compressivemembers are made thicker to prevent buckling.

ANALYSIS AND DESIGN ASSUMPTIONS

1. If the truss’s support reactions are not given, draw a FBD of the entiretruss and determine the support reactions (typically using scalarequations of equilibrium).

2. Draw the free-body diagram of a joint with one or two unknowns. Assumethat all unknown member forces act in tension (pulling on the pin) unlessyou can determine by inspection that the forces are compression loads.

3. Apply the scalar equations of equilibrium, FX = 0 and FY = 0, todetermine the unknown(s). If the answer is positive, then the assumeddirection (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 requiredforces are determined.

STEPS FOR ANALYSIS

If a joint has only two non-collinearmembers and there is no external load orsupport reaction at that joint, then thosetwo members are zero-force members. Inthis example members DE, DC, AF, and ABare zero force members.

Zero-force members can be removed (asshown in the figure) when analyzing thetruss.

ZERO-FORCE MEMBERS

Please note that zero-force members are used toincrease stability and rigidity of the truss, and toprovide support for various different loadingconditions.

If three members form a truss joint for whichtwo of the members are collinear and there isno external load or reaction at that joint, thenmember DA and CA are zero force members.

ZERO-FORCE MEMBERS

1. Check if there are any zero-force members. (Note that member BD iszero-force member. FBD = 0)

2. First analyze pin D and then pin A.

Given: Loads as shown on the truss

Find: The forces in each memberof the truss.

Plan:

EXAMPLE 1

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

+ FY = – FCD sin 45 – FAD sin 45 = 0

FCD = 318 kN (Tension) or (T)

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

45 º

FCD

D 450 kN

FAD

FBD of pin D

45 º

EXAMPLE 1: SOLUTION

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

Could you have analyzed Joint C instead of A?

45 º

FAB

A

FBD of pin A

FAD

AY

Analyzing pin A:

EXAMPLE 1: SOLUTION

a) Check if there are any zero-force members. Is member CE zero-force member?

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

Given: Loads as shown on thetruss

Find: Determine the force in allthe truss members (do notforget to mention whetherthey are in T or C).

Plan:

GROUP PROBLEM SOLVING

Analyzing pin D:+ FX = FDE (3/5) – 600 = 0

FDE = 1000 N = 1.00 kN (C)

+ FY = 1000 (4/5) – FCD = 0

FCD = 800 N = 0.8 kN (T)

FBD of pin D

FCD

Y

D 600N

X

FDE3

45

GROUP PROBLEM SOLVING

Analyzing pin C:

→ + FX = FCE – 900 = 0

FCE = 900 N = 0.90 kN (C)

+ FY = 800 – FBC = 0

FBC = 800 N = 0.80 kN (T)

FBD of pin C

FBC

Y

C 900 N

X

FCE

FCD = 800 N

GROUP PROBLEM SOLVING

Analyzing pin E:

→ + FX = FAE (3/5) + FBE (3/5) – 1000 (3/5) – 900 = 0

+ FY = FAE (4/5) – FBE (4/5) – 1000 (4/5) = 0

Solving these two equations, we getFAE = 1750 N = 1.75 kN (C)

FBE = 750 N = 0.75 kN (T)

FBD of pin E

FAE

Y

E FCE = 900 N

X

FDE = 1000 N

FBE

3

4 5

3

4 5

3

4 5

GROUP PROBLEM SOLVING

In-Class Activities:• Applications• Method of Sections• Group Problem Solving

Today’s Objectives:Students will be able to determine:1. Forces in truss members using the method of sections.

THE METHOD OF SECTIONS

Long trusses are often used to construct large cranes and largeelectrical transmission towers.

The method of joints requires that many joints be analyzed before wecan determine the forces in the middle of a large truss.

So another method to determine those forces is helpful.

APPLICATIONS

In the method of sections, a truss is divided into two parts by taking animaginary “cut” (shown here as a-a) through the truss.

Since truss members are subjected to only tensile or compressive forces alongtheir length, the internal forces at the cut members will also be either tensileor compressive with the same magnitude as the forces at the joint. This resultis based on the equilibrium principle and Newton’s third law.

THE METHODS OF SECTIONS

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 ofunknowns does not exceed three (in general).

2. Decide which side of the cut truss will be easier to work with (minimizethe number of external reactions).

3. If required, determine any necessary support reactions by drawing theFBD of the entire truss and applying the E-of-E.

STEPS FOR ANALYSIS

4. Draw the FBD of the selected part of the cut truss. You need to indicatethe unknown forces at the cut members. Initially, you may assume all themembers are in tension, as done when using the method of joints. Uponsolving, if the answer is positive, the member is in tension as per theassumption. If the answer is negative, the member must be incompression. (Please note that you can also assume forces to be eithertension or compression by inspection as was done in the figures above.)

STEPS FOR ANALYSIS

5. Apply the scalar equations of equilibrium (E-of-E) to the selected cutsection of the truss to solve for the unknown member forces. Pleasenote, in most cases it is possible to write one equation to solve for oneunknown directly. So look for it and take advantage of such a shortcut!

STEPS FOR ANALYSIS

a) Take a cut through 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 E-of-E to find the forces in KJ, KD and CD.

Given: Loads as shown on thetruss.

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

Plan:

EXAMPLE

Analyzing the entire truss for the reactions at A, we getFX = AX = 0. A moment equation about G to find AY results in:

∑MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = 56.7 kN

Now take moments about point D. Why do this?

+ MD = – 56.7 (9) + 20 (6) + 30 (3) – FKJ (4) = 0FKJ = − 75.1 kN or 75.1 kN ( C )

56.7 kN

EXAMPLE: SOLUTION

Now use the x and y-directions equations of equilibrium.

↑ + FY = 56.7 – 20 – 30 – (4/5) FKD = 0;

FKD = 8.38 kN (T)

→ + FX = (– 75.1) + (3/5) (8.38) + FCD = 0;

FCD = 70.1 kN (T)

56.7 kN

EXAMPLE: SOLUTION

a) Take the cut through members GF, GB, and AB.

b) Analyze the left section. Determine the support reactions at A. Why?

c) Draw the FBD of the left section.

d) Apply the equations of equilibrium (if possible, try to do it so that every equation yields an answer to one unknown.

Given: Loads as shown on the truss.

Find: The force in members GB and GF.

Plan:

GROUP PROBLEM SOLVING

1) Determine the support reactions at A by drawing the FBD of the entire truss.

+→ FX = AX = 0

+ MD = – AY (28) + 600 (18) + 800 (10) = 0;

AY = 671.4 kN

Why is Ax equal zero by inspection?

GROUP PROBLEM SOLVING: SOLUTION

2) Analyze the left section.

+ MB = – 671.4 (10) + FGF (10) = 0;

FGF = 671 kN (C)

↑ + FY = 671.4 – FGB = 0;

FGB = 671 kN (T)

GROUP PROBLEM SOLVING: SOLUTION

In-Class Activities:•Applications• Analysis of a Frame/Machine•Group Problem Solving

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.

FRAMES AND MACHINES

How is a frame different than atruss?

To be able to design a frame, youneed to determine the forces at thejoints and supports.

Frames are commonly used tosupport various external loads.

APPLICATIONS

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

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

APPLICATIONS

Frames are generally stationary and support external loads.

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

Frames and machines are two common types of structures that have at leastone multi-force member. (Recall that trusses have nothing but two-forcemembers).

Frame

Machine

DEFINITION: FRAMES AND MACHINES

1. Draw a FBD of the frame or machine and itsmembers, as necessary.

Hints:

a) Identify any two-force members,

b) Note that forces on contacting surfaces(usually between a pin and a member) areequal and opposite, and

c) For a joint with more than two members or anexternal force, it is advisable to draw a FBD ofthe pin.

FAB

STEPS FOR ANALYZING A FRAME OR MACHINE

2. Develop a strategy to apply the equations ofequilibrium to solve for the unknowns. Lookfor ways to form single equations and singleunknowns.

Problems are going to be challenging sincethere are usually several unknowns. A lot ofpractice is needed to develop good strategiesand ease of solving these problems.

FAB

STEPS FOR ANALYZING A FRAME OR MACHINE

a) Draw FBDs of the frame member BC. Why pick this part of theframe?

b) Apply the equations of equilibrium and solve for the unknowns atC and B.

Given: The frame supports an externalload and moment as shown.

Find: The horizontal and verticalcomponents of the pinreactions at C and themagnitude of reaction at B.

Plan:

EXAMPLE

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

FBD of member BC(Note AB is a 2-force member!)

CX

CY

B

45FAB

400 N

1 m 2 m1 m

800 N m

Equations of Equilibrium: Start with MC since it yields one unknown.

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

FAB = 1131 N

.

EXAMPLE: SOLUTION

FBD of member BC

CX

CY

B

45FAB

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

.EXAMPLE: SOLUTION

a) Draw a FBD of member BC and another one for AC.

b) Apply the equations of equilibrium to each FBD to solve for the four unknowns. Think about a strategy to easily solve for the unknowns.

Given: A frame supports a 50-kNload as shown.

Find: The reactions exerted by thepins on the frame members atB and C.

Plan:

GROUP PROBLEM SOLVING

FBDs of members BC and AC

Applying E-of-E to member AC:

+ MA = – CY (8) + CX (6) + 50 (3.5) = 0 (1)

+ FX = CX – AX = 0

+ FY = 50 – AY – CY = 0

CY

CX

AX

AY

3.5 m

8 m

6 m

50 kN

GROUP PROBLEM SOLVING: SOLUTION

FBDs of members BC and AC

Applying E-of-E to member BC:

+ MB = – 50 (2) – 50 (3.5) + CY (8) = 0 ; CY = 34.38 = 34.4 kN

From Eq (1), CX can be determined; CX = 16.67 = 16.7 kN

CY

CX

AX

AY

3.5 m

8 m

6 m

50 kN

+ FX = 16.67 + 50 – BX = 0 ; BX = 66.7 kN

+ FY = BY – 50 + 34.38 = 0 ; BY = 15.6 kN

GROUP PROBLEM SOLVING: SOLUTION

The End