Analysis of Truss

Fig.6

Fig.3Fig.4

Fig.5

Fig.1

Fig.2

3 .ANALYSIS OF TRUSS

3.1 .Introduction

Trusses are structures that are widely used in civil engineering applications, such as bridges, steel buildings and roof structures.

Figure 9 A truss with parallel chords

upper chord (top chord)

web members

vertical

Span lower chord diagonalpanel

Figure 10: triangular truss

Figure 11: Trusses according to the type of the web (a, b, c)

a

b c

What is a Truss?

A truss is a structure usually consisting of straight members that are connected to each other at the two ends of each member.  All members of a truss structure are connected together with pin joints, such that for the purpose of the design of these structures we assume the joints cannot carry or resist any moments (Fig.9).  All external loads acting on a truss are assumed to be acting only at the joints, and therefore, all members of a truss are two-force members (Fig.8).

3.2. Types of Trusses

The following five criterions may serve as a basis for the classification of trusses:

1) The shape of the upper and lower chords; (Fig.9, 10,image 1, 4, and 6).2) The type of the web; (Fig.11a, b, c and image 2).3) The conditions of the supports; (Fig.12)4) The purpose of the structure; (Images 1, 2, and 7).5) The level of the floor (lane, road) (Image 5)

Basic Truss Fig.13:

Fig.15: Statically Determinater=3, m=17, j=10

3.3. Determinacy and Stability

Basic Truss Element ≡three member triangular truss (Fig.13).

Simple Trusses–composed of basic truss elements

m = 3 + 2(j –3) = 2j –3

For a simple truss

m≡total number of members

j≡total number of joints (Fig.14).

Fig.14: Simple Truss

●Statically Determinate Truss ≡if all the forces in all its members as well as all the external reactions can be determined by using the equations of equilibrium.

Fig16:External Statically Indeterminater=4, m=17, j=10

Fig.17: Internal Statically Indeterminater=3, m=18, j=10

Fig. 18

Fig.19: Internal Instabilityr=3, m=17, j=10

●Statically Indeterminate Truss ≡if all the forces in all its members as well as all the external reactions cannot be determined by using the equations of equilibrium.

●External Indeterminacy ≡excess number of support reactions(Fig.16)

●Internal Indeterminacy ≡excess number of members(Fig.17)

●Redundants≡excess members or reactions

●Number of redundants defines the degree of static indeterminacy

●External Instability ≡instability due to insufficient number or arrangement of external supports (Fig.18).

●Internal Instability ≡number and arrangement of members is such that the truss does not change its shape when detached from the supports (Fig.19).

Fig.20: Members subjected to forces; Tension and Compression.

Tension (+) Tension

Compression Compression(–)

L

L

L

LP

HA

RC RA

A

B

C

DE

θ =600

3.4. Axial force determination in the members of simple trusses 1. Sign convention A tensile force in a member is considered positive, and a compressive force is considered negative. Forces are depicted as acting from the member on the joints as shown in figure 20.

3.4 .Analysis of Trusses

3.4.1. Method of Joints1. First find the reaction at the supports.

2. Then choose the joints where there are only two unknowns‟ forces. The forces in the unknowns truss are found from horizontal forces ΣH = 0 and vertical forces ΣV = 0.

3. An inclined member force will produce a vertical force and horizontal force.

4. Assume forces in the members as compressive or tensile force. Resolve the forces in the members.

5. Find forces in the unknown members as done in engineering mechanics using step 2.

6 .If the answer is negative then the assumption is incorrect; in such case simply change the sign.

Fig.22

FAE

FAB

Joint A

HA θ = 600

RA

FEDJoint E

θ = 600

FEBFEAJoint B

FBC

θ = 600

FBE

FBA

P

FBDFCD

FCBJoint C

RC

θ = 600

Figure 23: Zero force members

S1

S2h

S2S2v

S2

S3

S1

S3h

S3v

S2h

S2S2v

FS1

b)a)

c)

■A tower with many zero force members. ■Although they carry no load they prevent structural collapse.■A common use of zero-force members is to brace long bars that are in compression and which would otherwise buckle (pop out to one side).

P

Figure 21: A simple truss model supported by pinned and roller support at its end. Each triangle has the same length, L , where θ is 60° on every angle. The support reactions, RA and RC can be determined by taking a point of moment either at point A or point C.

Zero force members 1) Unloaded two members joint (Fig. 23a). It follows from the equilibrium equations. From V=0 follows that S2v=0 so S2=0, fromH=0 follows that S1=0.

2) Unloaded three members joint (Fig. 23b), when two of the three members are collinear (two members have the same direction).

The force in the third member is zero. From ΣV=0 follows that S3v=0 from where S3=0. 3) Two members joint when the external concentrated force is in the direction of one of the members, the force in the second member is zero – S1=0.

Fig.25

120kN

80kN

D B

C

A

3

44

RA =30 kN

RB =90 kN

HA =80 kN

120kN

80kN

D B

C

A

ACy

ACx

A80kN

30kN

BCx BCy bcy bcy

B

90kN bcy bcy

Application:

(1 )Use the method of joints to determine the value and type of force in all members of the truss given in Fig.25 .

Reactions

-∑Fx=0→ HA=80kN←

-∑MA=0: 80×3+120×4-RB×8=0 RB=(240+480)/8=90kN

-∑Fy=0:RA=120-90=30kN

Member forces:

-Joint A :

∑Fy =0: +30- FACy=0 → F ACy=30kN

-FACx=30×4/3=40kN, FAC=40×5/4=50kN(compression)

∑Fx=0: Facx+HA=FAD=120kN

-Joint B:

-∑Fy=0: FBCy=90kN, FBC=90×5/3=150kN

120

kN

120kN120kN

150kN50kN

30kN90kN

80kN

120kN

80kN

DB

C

A

Fig.26

50kN

A

D

B

C

6 6

3

3

Joint B56.28 kN

56.28 kN

31.28 kN56.28 kN

θ

α3

6

6

25kN

25 kN 25 kN

50 kN

50kN

A

D

C

B

-FBCx=150×4/5=120kN

-∑Fx=0:FBD=120kN

Joint C:

-∑Fy=0: FCD=30+90=120kN

(2)Use the method of joints to determine the value and type of force in all members of the truss given in Fig.26 .

-Reactions

-∑Fx=0: HA=50kN

-∑MA=0: RB=50×6/12=25kN( ↑ ), ∑Fy=0: Ra=25kN( ↓ )

Member forces:

Joint B:

-sin α=0.707=cos α

AD

15kN105kN30kNC

B

62.56 kN

62.81 kN

79.61 kN

62.81 kN

8.88 kN

25 kN 25 kN

50 kN

50kN

A

D

C

B

2

2

12kNC

A

D

B

F

E

L KM

-sin θ =0.49776, cos=0.8955

-∑Fx=0: 0.707FBC=0.8955FBD

-FBC= 0.8955FBD/0.707 (1 )

-∑Fy=0: +25+( 0.49776)FBD(0.707)FBC=0 (2), substitute (1) into (2):

-FBD=62.81kN, FBC=79.61kN

-Joint D:

-∑Fy=0: FDC=62.56 kN

Joint A:

∑-Fx=0: FAC(x) =6.28 kN; FAC =8.88 kN

The member force( its value and type) for all members of the truss are shown in the figure below:

P

HD

RD RA

A

B C

DE

3.4.2. Method of Sections

1. First find the reaction at the supports.

2. Pass a section through a maximum of 3 members of the truss, 1 of which is the desired member where it is dividing the truss into 2 completely separate parts,

3. At 1 part of the truss, take moments about the point (at a joint) where the 2 members intersect and solve for the member force, using ∑ M = 0,

4. Solve the other 2 unknowns by using the equilibrium equation for forces, using ∑ Fx = 0 and ∑ Fy = 0.

5. The advantage of this method is forces of any member can be found using this method.

Note: The 3 forces cannot be concurrent, or else it cannot be solve.

Virtual Cut

P

HD

RD RA

A

B C

DE

Fig.29

Fig.30

3

3

4 4200kN

80 kN

40 kN 160 kN

80kN

2

2

1

1

FA B

CD

E

G H

Figure 28: Using the same model of simple truss, the details would be the same as previous figure with 2 different supports profile. Unlike the joint method, here we only interested in finding the value of forces for member BC, EC, and ED.

Procedure:

Firstly, the support reactions of RA and RD should be determined.

A virtual cut is introduced through the only required members BC, EC, and ED (Fig.29) .

Taking moment at joint E (virtual pint) on clockwise for the whole RHS part

Then, either joint D or C can be considered as point of moment.

Taking Fy=0 For RH or LHS to find the force in member EC .

Note: Each value of the member’s condition should be indicate clearly as whether it is in tension (+ve) or in compression (-ve)

state.

Application:

(1)Determine the value and type of force in the member EH of the truss shown in Fig.30.

FEH sinα

160 kN

H

B

FEH

O1

FFH

80 kN

FEH

α

α

B

D

H

160 kN

80 kN

Fig.31

GC

E

D

L

B

M

NI

J

H

A

K

F

2kN2kN 2kN

Fig.31a

GC

E

D

L

B

M

NI

J

H

A

K

F

2kN2kN 2kN

3kN 3kN

Solution:

Reactions:∑Fx=0, HA=80kN←

∑MA=0, 200×4+80×6–8RB=0, ►RB=160kN↑

∑Fy=0, ►RA=40kN↑

Member force FEH:

Method No.1: section S1-S1, right part of the truss

∑Mo1=0: 160×4+FEHsinα×8=0

FEH= –133.3kN (Compression)

Method No.2:Joint H:

∑Fx=0: FEHcosα+F1cosα=0

FFH=–FEH

Section 2-2, right part, ∑Fy=0

0.6×(2FEH)=–160

FEH= –133.33kN

(2) The truss shown in Fig.31 consists of-4 bays of ‘K’ structure. Each bay has 2m long horizontal and vertical members. Find the force in members DE and DG.

Solution

The reactions are calculated as shown in Fig.31a

3kN

3kNC D

B

A

First, we need to find the support reactions at points A and N. We do this by using the equilibrium equations, and in this truss, because the load is acting symmetrical, RA and RN are equal and each reaction is 3kN.

Now, we proceed to find the forces in the members DE, and DG. For this purpose,

we make a cut in the truss such that it cuts members DG, DE, BF and AF.

Next, we draw the free-body diagram of the left portion of the truss and use the equilibrium equations to find the required forces.

ΣMF=0: RA(2) - FDG(2)=0; FDG=3kN(C)

The force in member AF is zero. The force in

Member BF is resolved into two components(FBFx and

FBFy). sinα==0.4464, cosα=0.89286 FBFx=FBFcosα, FBFy= FBFsinα,

ΣFx=0: FBFx - FDG=0; FBFx=3kN

FBF= FBFx ⁄ cosα = 3⁄ 0.89286=3.36kN(T);

FBFy=3.36(0.4464)=1.5kN

ΣFy=0: +3- 1.5- FDE=0, FDE=1.5kN( C )

3.4.3. Graphical Method (Maxwell’s Diagram)

The method of joints could be used as the basic for a graphical analysis of trusses. The graphical analysis was developed by force polygons drawn to scale for each joint, and then the forces in each member were measured from one of these force polygons.

The number of lines which have to be drawn can be greatly reduced, however, if the various force polygons are superimposed. The resulting diagram of truss analysis is known as the Maxwell’s Diagram.

F

Fig.32

1

23

4 5

76

RA =30 kN RB =90 kN

HA =80 kN

120kN

80kN

D B

C

A

4-6 =8cm AC=80kN ( C )6-2 =12cm AD=120kN ( T )5-7 =15cm BC=150kN ( C )7-6 =12cm DC=120kN ( T )1-7 =12cm BD=120kN ( T )

34

1

2

7

5

6

Fig.33

6

4.5

6

B

A

D

C

F

E

30kN30kN

B120kN 4D F

In order to draw the Maxwell diagram directly, here are the simple guidelines:

Solve the reactions at the supports by solving the equations of equilibrium for the entire truss,

Move clockwise around the outside of the truss; draw the force polygon to scale for the entire truss,

Take each joint in turn (one-by-one), then draw a force polygon by treating successively joints acted upon by only two unknown forces,

Measure the magnitude of the force in each member from the diagram,

Lastly, note that work proceeds from one end of the truss to another, as this use for checking of balance and connect to other end.

Application (1):(Fig.32):

(2)Find, by graphical analysis, the value and type of member forces on the truss given in the figure 33.

Fig.33b

8

4 9

5

6

2

7 3

1

C150kN

A B

D

EFig.3.5.1

4 8

2

4

120kN

αF(BD) y

F(BD) x

115kN

B

75kN

150kNC

The force polygon for the external load and reactions (1-2-3-4-5) is shown in Fig.33b.

To complete the Maxwell diagram, the force polygon for member forces is added from joint consideration, staring from joint A, B, C, D,F, and E. The member forces are listed below:

3.5.Examples:

3.5.1. Determine the member forces in the truss shown in Fig.3.5.1, by using the method of joints. State whether these forces are in tension or compression.

6-1 AC =12kN( C ) 8-5 CE= 40kN (C )

2-6 AB = 0 4-9 DF =0

4-7 BD =40kN (T) 9-8 DE=50kN (T)

7-6 BC =66.67kN (T) 9-4 EF =0

7-8 CD =30kN( C )

αA

75kN

150kN

5kN

D

150kN

75kN 115kN

230kN

80kN

80kN

Fig.3.5.1a 120kN257.15kN75kN

230kN

167.7kN

230kN

113.14kN

150kN

120kN

150kN

115kN5kN

BE

C

A

D

Solution

Reactions: ∑Fx =0: HA =150kN←

∑MA =0: 150×6+120×4–12RB =0, RB=115kN↑

∑Fy =0: RA =5kN↑

Member forces:

Joint B: ∑Fy=0, FBDy=115kN↓

FBD=115∕sinα=115∕.4474=257kN(compression)

FBE=FBDx=FBD×cosα=257×.895=230kN(tension)

Joint C: ∑Fx =0, FCDx=150kN

FCD=FCDx⁄cosα=150⁄0.895=335.3kN(C)

FCDy=FCD×sinα=335.3×0.4474=75kN

∑Fy =0, FCA=75kN (T)

Joint A: ∑Fy =0, FADy=80kN

FADx =FAD y(α=45°) =80kN►FAD=80√2kN( C )

∑Fx =0, FAE=230kN (T)

Joint D: ∑Fy=0↑, +80–75+115–FDE=0

FDE=120kN

The final result of the member forces is given in Fig .3.5.1a.

Fig.3.5.2

4.8 4.8

B

C

D

20kN

A

4

4

Fig.3.5.2a

20kN

15.63kN

15.63kN

15.63 15.63kN

20kN20kN

40kN

A

B

C

D

C

12kN

20kN

10kN

10kN

12kN

FABy

FACx

FABx

FACy

20kN

A

Fig.3.5.4

G

A

B F

E

C

D

H

Fig.3.5.3

E

H

C D

I JG

A B

F

3.5.2.Determine the value and type of force on the truss member shown in Fig.3.5.2.

Joint A: sinα=0.64, cosα=0.768

∑Fx =0, FAC x=FABx ►FAC=FAB

∑Fy =0↑, FaC×sinα+FAB×sinα – 20=0►FAC=FAB=20⁄(2×0.64)=15.63kN

In joint D( from symmetry), FDC=fAC, and FDB=FAB= 15.63kN

Joint C: ∑Fy=0, FCB=20kN ( T )

3.5.3. The trusses in the figures 3.5.3 and 3.5.4are subjected to an arbitrary load. Detect the zero force members, using the rules

explained in 3.4.1.

Fig.3.5.5 2

4×2=8

40kN 40kN40kN20kN 20kN

J

D

F

EA B

I

C

H G

Fig.3.5.5a

80kN80kN

40kN 40kN40kN20kN 20kN

S1

S1 S2

S2

G

C E

D

J HI F

BA

Solution:

Reaction forces at supports:

∑Fx =0, HA =0

∑M A=0, RB =(20×9.6) ⁄4.8

RB =40kN↑

∑Fy =0: R A=20kN↓

Member forces:

Fig.3.5.3: joint C, ∑F =0, FCA=0 Fig.3.5.4: joint C, FCE=0, fCG=0

Joint D, ∑F =0, FDH=0 joint G, ∑F =0, FGF=0

Joint B, ∑F =0, FBE=0

Joint G, ∑F =0, FGA=0 joint H, FHD=0

Joint I, ∑F =0, FIE=0

3.5.4.Calculate the member forces JI, CI, DG, and DE on the truss shown in Fig.3.5.5

Solution:

The vertical reactions at supports are equal and each one is 80kN.

Fig.3.5.5a:

FA: section 1-1, left part, ∑MO1=0, 80×2–20×2–2FA=0 ►FA=60kN (C )

FB: section 1-1, left part, ∑Fy=0, –20+80–FB =0►FB =60kN ( C )

FC- joint O2. ∑Fy =0►F =40kN ( C )

FD: section 2-2, right part, ∑MO2=0, 80×4–40×2–20×4–2FD=0► F =80kN(T)

First, we need to find the support reactions at points A and N. We do this by using the equilibrium equations, and in this truss, because the load is acting symmetrical, RA and RN are equal and each reaction is 3kN.

Now, we proceed to find the forces in the members DE, and DG. For this purpose,we make a cut in the truss such that it cuts members DG, DE, BF and AF.

Next, we draw the free-body diagram of the left portion of the truss and use the equilibrium equations to find the required forces.

ΣMF=0: RA(2) - FDG(2)=0; FDG=3kN(C)

The force in member AF is zero. The force in member BF is resolved into two components(FBFx and FBFy). sinα==0.4464, cosα=0.89286 FBFx=FBFcosα, FBFy= FBFsinα,

ΣFx=0: FBFx - FDG=0; FBFx=3kN

FBF= FBFx ⁄ cosα = 3⁄ 0.89286=3.36kN(T);

FBFy=3.36(0.4464)=1.5kN

ΣFy=0: ++3- 1.5- 2+FDE=0, FDE=0.5kN( C )

3.6.Problems:

Fig.3.6.1

4 2.7

4

100kN50kN

D

CB

A

Fig.3.6.3

180kN

D

C B

A

4 4

2

Fig.3.6.23

2kN4kN 4kN

DC

BA E

4 4

Fig.3.6.4

3

D

CB

A E

4 4

15kN15kN

Fig.3.6.5

D

CB

AE

80kN

80kN4

4

4

5

Fig.3.6.6

10kN

D C

B

A

F

G

E

3×5=15

3.6.1 Find analytically, using method of joints, all the member forces of the trusses shown in Figs.3.6.1 to 3.6.3.State whether the member forces are tension or compression.

3.6.2.Determine the force in all members of each truss shown in Figs.3.6.4 and 3.6 .5. Indicate if the members are in tension or compression.

3.6.3. Determine the force in each member of the truss in Fig.3. 6. 6, and state if the members are in tension or compression.

F

D

I J

E

H G

BA

C

30kN

60kN30kN40kN

3

6

3 3 3

Fig.3.6. 7

A

G

CED

L

B

M N

I

J

H

K

F

100kN

50kN

100kN100kN

4

3

4

3

4 4

Fig.3.6. 8

B

A

Fig.3.6..10

4

4

50kN

40kN

33

B

A

Fig.3.6.. 9

4 4

3

3

200kN200kN80kN

3.6.4.Determine the value and type of the force for the member HI in Fig.3.6.7.

3.6.5.Determine the force in the members GF, MG, and EB in Fig.3.6. 8, and state if the members are in tension or compression.

3.6.6 .Determine the value and type of the force for the members A and B marked in Figs.3.6. 9 and 3.6.10.