Truss Analysis

Ahmad Shufni bin Othman

Mei 2014

ANALYSISOFSTATICALLYDETERMINATE TRUSSES(Part 3)

Truss Analysis

Techniques for Truss Analysis

Truss Analysis

METHOD OF SECTIONS

Learning Objectives

Upon completion of this course, students should be able to;

• Explain the Method of Sections.• Determining the truss members forces using the Method

of Sections.

Truss Analysis

Assumptions for Design

Truss Analysis

Assumptions for Design• Axial Force Members

– Tension / Compression

– Compression Members Usually Thicker

Tension

Compression

Tensile (T) axial member force is indicated on the joint

by an arrow pulling away from the joint.

Compressive (C) axial member forceis indicated by an

arrow pushing toward the joint.

Truss Analysis

The Objectives:

Students will be able to determine forces in truss members using the method of sections.

In-Class Activities:

•Applications

•Method of sections

•Concept quiz

•Group Problem solving

• Attention quiz

Truss Analysis

APPLICATIONS

Long trusses are often used to construct bridges.

The method of joints requires that many joints be analyzed before we can determine the forces in the middle part of the truss.

Is there another method to determine these forces directly?

Truss Analysis

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 member will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newton’s third law.

Truss Analysis

Truss in Equilibrium Each PART in Equilibrium

THE METHOD OF SECTIONS

Efficient when forces of only a few members are to be found

Truss Analysis

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 the necessary support reactions by drawing the FBD of the entire truss and applying the EofE.

Truss Analysis

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 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.

PROCEDURE (continued)

Truss Analysis

PROCEDURE (continued)

5. Apply the equations of equilibrium (EofE) to the selected cut section of the truss to solve for the unknown member forces. Please note that in most cases it is possible to write one equation to solve for one unknown directly.

Truss Analysis

0

0

0

M

F

F

y

x

Equations of Equilibrium

Take moments about a

point that lies on the

intersection of the lines of

action of two unknown

forces

PROCEDURE (continued)

Truss Analysis

SOLUTIONS

NF

F

Mc

GF

GF

2000

0)4(1000)2(

0

(compression)

NF

F

M

BC

BC

G

1000

0)2(1000)2(

0

(tension)

NF

F

F

GCY

GCY

Y

1000

01000

0

NF

F

FFF

GCYGC

GCXGCYGC

21.141482

10008

2

228

(tension)

FGCY

FGCX

Truss Analysis

Example 1

For this truss :

1. Calculate the forces in

members AB, BE and DE.

2. Determine whether the

member is in compression or

tension.

20 kN

10 kN 10 kN

2 m 2 m

4 m

AB

CDE

Truss Analysis

Solution 1 :

For this truss :

1. Determine support reaction (if

needed)

2. Apply ‘a cut’ through the required

members (x – x)

3. Determine the forces in the members

(state : compression or tension)

Truss Analysis

Solution 1 : (Right Side)

Truss Analysis

Solution 1 : (Left Side) cont.

Truss Analysis

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.

Truss Analysis

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

Truss Analysis

Example 2

For this truss :

1. Calculate the forces in

members CD, CG and GH.

2. Determine whether the

member is in compression

or tension.

Then ?

Which side of the “cut” ? WHY ?

Calculate the support reaction at A

Truss Analysis

Taking the right side of the truss :

1. Calculate FCD

2. Calculate FHG

3. Calculate FCG

Solution 2 :

Truss Analysis

By taking a moment to the joint G

+MG = 0 ;

300(6) + FCD(3) = 0

FCD = 600 N (compression)

By taking a moment to the joint C

+MC = 0 ;

300(6) – FCG(3) = 0 FED = 600 N (tension)

Solution 2 : (cont.)

Truss Analysis

Define FCGy:

Fy = 0 : -FCGy + 300 = 0

FCGy = 300 kN

18

F

3

F

3

FCGCGCG XY

YCGCG F*)3

18(F

(tension) N424.26N)(300*)3

18(

Solution 2 : (cont.)

Truss Analysis

CONCEPT QUIZ

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

Truss Analysis

Given: Loads as shown on the roof truss.

Find: The force in members DE, DL, and ML.

Plan:

a) Take a cut through the members DE, DL, and ML.

b) Work with the left part of the cut section. Why?

c) Determine the support reaction at A. What are they?

d) Apply the EofE to find the forces in DE, DL, and ML.

Example 3

Truss Analysis

Analyzing the entire truss, we get FX = AX = 0. By symmetry, the vertical support reactions are

AY = IY = 36 kN

+ MD = – 36 (8) + 6 (8) + 12 (4) + FML (5) = 0

FML = 38.4 kN ( T )

Solution 3 :

Truss Analysis

+ ML = –36 (12) + 6 (12) + 12 (8) + 12 (4) – FDE ( 4/17)(6) = 0

FDE = –37.11 kN or 37.1 kN (C)

+ FX = 38.4 + (4/17) (–37.11) + (4/41) FDL = 0 FDL = –3.84 kN or 3.84 kN (C)

Solution 3 : (cont.)

Truss Analysis

Given: Loading on the truss as shown.

Find: The force in members BC, BE, and EF.

Plan:

a) Take a cut through the members BC, BE, and EF.

b) Analyze the top section (no support reactions!).

c) Draw the FBD of the top section.

d) Apply the equations of equilibrium such that every equation yields answer to one unknown.

GROUP PROBLEM SOLVING

Truss Analysis

+ FX = 5 + 10 – FBE cos 45º = 0 FBE = 21.2 kN (T)

+ ME = – 5(4) + FCB (4) = 0 FCB = 5 kN (T)

+ MB = – 5 (8) – 10 (4) – 5 (4) – FEF (4) = 0

FEF = – 25 kN or 25 kN (C)

SOLUTION

Truss Analysis

For the truss shown, find the internal fore in member BE

Class Problem 1

Truss Analysis

The structure shown, is pinned to the floor at A and H. Determine the magnitude of all the support forces acting on the structure and find the force in member BF.

Class Problem 2

Answer : FBF = 128.06 kN (Tension)

Truss Analysis

Class Problem 3

Using method of sections determine the forces in members BC, GC and GF of the pin-jointed truss shown below.

Answer : FGC = 7.07 kN (Comp.) FBC = 25 kN (Tension) FGF = 20 kN (Comp.)

Truss Analysis

The truss below is pinned to the wall at point F, and supported by a roller at point C. Calculate the force (tension or compression) in members BC, BE, and DE.

Class Problem 4

Answer : FBC = 120 kN (Comp.) FBE = 150.78 kN (Tension) FDE = 64 kN (Tension)

Truss Analysis

The roof truss shown below is pinned at point A, and supported by a roller at point H. Determine the force in member DG.

Class Problem 5

Answer : FDG = 18.14 kN (Tension)

Truss Analysis

Determine the force in member BE and BC of the truss shown. State whether each member is in tension or compression. Member DE and EG is equal in length.

Answer : FBE = 4.17 kN (Comp.) FBC = 6.67 kN (Comp.)

Class Assignment 1

Truss Analysis

Class Assignment 2

Determine the forces in members BC, BF, and EF of the truss shown in Fig. below by using the method of sections.

Answer:

TBF = 3.75 kN (T)

TBC = 2.25 kN (T)

TEF = 4.5 kN (C)

Truss Analysis

A symmetrical fink roof truss is subjected to vertical loadings as shown. Determine the force in members IM, IJ and CJ.

GROUP PROBLEM SOLVING

Truss Analysis

Excercise : Determine the forces in the members of the truss

Truss Analysis

Answer :

Truss Analysis