statically determinate plane trussesauthor.uthm.edu.my/uthm/www/content/lessons/566/chapter7.pdf ·...

7 STATICALLY DETERMINATE

PLANE TRUSSES

OBJECTIVES:

This chapter starts with the definition of a truss and briefly explains various types of plane truss. The determinancy and stability of a truss also will be discussed. The procedures of analyzing statically determinate trusses will be developed using the method of joints and the method of sections. For advance, the alternative computation using joint equilibrium method will be included in this topic.

6.1 Introduction

A truss is defined as a structure composed of slender elements joined together at their end points. The members commonly used in construction consist of wooden struts, metal bars, angles or channels.

Plane or planar truss composed of members that lie in the same plane and frequently used for bridge and roof support.

Loads that cause the entire truss to bend are converted into tensile and compressive forces in the members.

compression tension

Loading causes bending of truss, which is develops compression in top members, tension in bottom members.

Fig. 7-1

Mechanics of Materials

203

6.2 Types of trusses and its application

6.2.1 Roof trusses Often used as part of an industrial building frame as shown in Fig.

7-2.

Fig. 7-2

The roof load is transmitted to the truss at the joints by series of purlins.

Generally, the roof trusses are supported either by columns of wood, steel or reinforced concrete.

The roof truss along with its supporting columns called as a bent. To keep the bent rigid and capable of resisting horizontal wind

forces, knee braces are sometimes used at the supporting columns. The space between adjacent bents is called a bay. Bays are often

tied together using diagonal bracing to maintain rigidity of the structure.

The gusset plate is the connections which formed by bolting or welding at the ends of the members to a common plate.

Fig. 7-3 : Gusset plate

Trusses used to support roofs are selected on the basis of the span,

the slope and the roof material. Some of common types of trusses are displayed in the Fig. 7-4.


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Fig. 7-4: Common types of plane trusses

1.2.2 Bridge trusses

The main structural elements of a typical bridge truss are shown in Fig. 7-5.

Fig. 7-5 : The structure elements of typical bridge truss

The load on the deck is first transmitted to stringers then to floor beam and finally to the joints of the two supporting side trusses.

The top and bottom cords are connected by top and bottom lateral bracing which serves to resist the lateral forces caused by wind and the sidesway caused by moving vehicles.

Additional stability is provided by the portal and sway bracing.


205

The typical forms of bridge trusses currently used for single span are shown in Fig. 7-6.

Fig. 7-6 : The forms of bridge trusses

6.3 Assumptions in analysis

The assumptions are necessary to determine the force developed in each member when the truss is subjected to a given loading.

C A to C, C to B, B to A

the connection between them is called as truss member

Fig. 7-7 : Truss member

the assumptions are; i) All members are connected at both ends by smooth

frictionless pins.


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ii) All loads are applied at joints (member weight is negligible).

ASK STUDENTS: Why the analysis assume that the smooth pin is frictionless?

Why the analysis assume that the member weight is negligible?

Because of these two assumptions, each truss acts as an axial force member and the forces acting at the ends of the member must be directed along axis of the member.

If the force tends to elongate the member, it is tensile force (T) as shown in Fig. 7-8.

If the force tends to shorten the member, it is compressive force (C).

In analysis it is important to state whether the force is tensile or compressive.

The forces inside the member and outside should opposite

T T

C C

Smooth pin

Fig. 7-8 : The tensile and compressive force 6.4 Stability and Determinancy

6.4.1 General The structures mechanics involves determination of unknown forces

on the structures. Some of these structures can be completely analyzed by using the

equations of equilibrium.

ΣFx = 0 (F = F) ΣFy = 0 (F = F ) ΣMz =0 (M = M)

The structures are known as statically determinate.

On the other hand, if there exist extra redundant reaction

components, then the structure is said to be statically indeterminate.


207

6.4.2 Determinancy Criteria for Structures To be in a state of static equilibrium, a structure must meet the

requirements of stability. A statically indeterminate structure is a structure that had more

unknown forces.

Eg; ( to determine whether the truss is determinate or indeterminate)

For a pinned joint frame (trusses); i) m+r = 2j ……just stiff (statically determinate) ii) m+r < 2j ……under stiff (form a mechanism)…unstable (it will collapse since there will be an insufficient number of bars @ reactions to constraint all the joints). iii) m+r > 2j ……over stiff (statically indeterminate)

-A truss can be unstable if it is statically determinate or statically indeterminate. -A truss is externally unstable if all of its reaction is concurrent or parallel.


208

EXAMPLE 7.1 Determine number of redundant and state the determinancy criteria for the truss as shown below;

a) m=10 r =3 j =6 n= m + r – 2j = 10 + 3 – 2(6) m + r = 13 = 1…..has a redundancy of 1th degree 2j = 12 =m+r > 2j ….over stiff (statically indeterminate)

b)

EXAMPLE 7.2 Define whether the truss is statically determinate or indeterminate

m = 7 m+r = 2j r = 3 7 + 3 = 2(5) j = 5 10 = 10 …….statically determinate truss But this truss is unstable since the support reactions are parallel.


209

m = 7 m+r = 2j r = 3 7 + 3 = 2(5) j = 5 10 = 10 …….statically determinate truss But this truss unstable since the support reactions are concurrent.

6.5 Method of Joints

EXERCISE 7.1

EXAMPLE 7.3 Calculate all member forces by using method of joints.

(b)

(a) (c)

(d)

VCVA

HA


210

Determine whether the truss is determinate or indeterminate. m = 3 r = 3 j = 3 m+r =2j 3+3 = 2(3) 6 = 6 ……the truss is determinate and stable. Calculate the support reactions. F = F H = H

0 kN = VA + VC 500 N = HA + M = 0 MA = VC ( 2) + 500 (2)=0 VC = - 500N ( ) VA -500 = 0 VA = 500N( ) -ve sign indicated that the original direction is upward

VC = 500N ( )

change the direction;

Calculate all member forces Joint A

FAB ΣFx → = ← ΣFx


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↑ ΣFy = ↓ ΣFy FAB = 500N (T)

Joint C

NOTES: -ve sign indicated the force tends to shorten the member,

so the force is compression. How? All loads are assumed moved outward from the

joint. B The calculation shows the force is

compression, so;

C Change the direction;

To stabilize inside the member; To stabilize inside and outside the member;

The loads should be opposite at joint

FCB

500 500

↑ ΣFy = ↓ ΣFy 500 + FCB sin 45 = 0 = - 707 N (C)

CHECKING :

Joint B: ΣFx → = ← ΣFx 500 =707 sin 45 0 = 0 (BALANCE)

(707) 500

500


212

Procedure for Analysis: The following procedures can be used as the guidance to solve the problems using the method of joints.

Determine the truss is determinate or indeterminate by using the stability and determinancy equation.

Calculate the support reactions if necessary.

Draw the free-body diagram of a joint having at least one known force and

at most two unknown forces. If the joint is located at the supports, it may be necessary to find the external reactions at the support.

All loads are assumed moved outward from the joint.

The x and y axes can be resolved into their x and y components. Two force

equilibrium equations ΣFx =0 and ΣFy =0 are applied.

Continue to analyze the member forces at the other joints where again it is necessary to choose a joint having at most two unknowns and at least one known force.

Indicate whether the member in compression or tension.


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EXAMPLE 7.4

Determine the support reactions and member forces using method of joints. All members are connected at both ends by smooth frictionless pins.

A

C D

E B

F

8 kN

6 kN

VA VB

HA

4m 5m 4m

5m

Solution; Determine the truss is determinate. Calculate the support reactions.

[Ans: VA = 3.2 kN( ), VB = 4.8kN( ), HA = 6 kN( )]

Calculate the member forces Joint A

↑ ΣFy = ↓ ΣFy 3.2 + FAC sin 51.34 = 0 FAC = - 4.1 kN (C) ΣFx → = ← ΣFx FAF + FAC cos 51.34 = 6 FAF = 8.6 kN (T)

-4.1

FAF

FAC

6 θ

3.2 tan θ = 5/4 θ = 51.34°


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Joint F FFE = 8.6 kN (T) FFC = 8 kN (T) Joint C Joint B Joint D

8.6 8

FFC

FFE

θ1 θ2

FCD

FCE

8 -4.1

6

tan θ2 = 4/5 θ = 38.66°

tan θ1 = 5/5 θ = 45°

↑ ΣFy = ↓ ΣFy 0 = 8 + FCE sin 45 + (-4.1) cos 38.66 0.707 FCE = -4.8 FCE = -6.8 kN (C) ΣFx → = ← ΣFx FCD + FCE cos 45 + 6 = (-4.1) sin 38.66 FCD + 1.19 = -2.56 FCD = -3.75 kN (C)

-6.8

FBD tan θ= 5/4 θ= 51.34°

↑ ΣFy = ↓ ΣFy 4.8 + FBD sin 51.34 = 0 FBD = - 6.15 kN (C) ΣFx → = ← ΣFx 0 = FBE + FBD cos 51.34 FBE = 3.84 kN (T)

-6.15

θ FBE 4.8

CHECKING:

-3.75 θ θ = 180-90 -51.34 =38.66° -6.15

FDE

↑ ΣFy = ↓ ΣFy 0 = FDE + (-6.15) cos 38.66 FDE = 4.8 kN (T)


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EXERCISE 7.2

[VK = 0kN, VA = 7kN (↑), HK = 9.6kN (←), HA = 9.6kN (→)]

A

C D

E B

F

8 kN

6 kN

(4.1) (6.15)(6.8)

4.8 8

(3.75) Checking: Joint E ΣFx → = ← ΣFx 3.84 + 6.8 cos 45 = 8.6 8.6 = 8.6 (BALANCE)

8.6 3.848.6

HK

HA

VK

VA

Determine the support reactions and member forces of GF, EF, GE, HG, HD, GD and HI. The method of joints is selected to solve the problem. 15

5HD: ?

15 = 65 ? ? = 2m

6


216

EXERCISE 7.3 A pin-connected truss is loaded and supported as shown in figure; a) prove that the plane truss is a statically determinate structure. b) determine all member forces by using Method of Joint.

3m

[Ans: FAC = 33.33kN (T), FAB = 6.66kN (C), FDC =26.67kN (T), FBC = 20kN (T)] 6.6 Method of Sections EXAMPLE 7.6

Determine the forces of member BC, CG and FG by using method of section. Prove the truss is statically determinate. State whether the members are in tension or compression.

2m


217

Solution; Prove the truss is statically determinate. Calculate the support reaction if necessary. Cut the section of the truss through the members where forces are to be determined. Choose one section whether left or right sections. If choose right section, the external reactions at support have to calculate.

Calculate the member forces.

C

+ MG = 0 FBC (2) – 100 (2) = 0 FBC = 100 N (T) ↑ ΣFy = ↓ ΣFy FGC sin 45 = 100 FGC = 141.42 N (T) + MC = 0 -FGF (2) – 100 (4) = 0 FGF = - 200 N (C)

All member loads are assumed moved outward from the ‘cut’.

FBC

FGC

FGF

Checking: ↑ ΣFy = ↓ ΣFy FGC sin 45 = 100

141.42

100 = 100 (BALANCE)


218

Procedure for Analysis: The following procedure provides the guidance to determine the forces in the members of a truss by using the method of sections.

Determine the truss is determinate or indeterminate by using the stability and determinancy equation.

Calculate the support reactions if necessary.

Make a decision as to how to ‘cut’ the truss through the members where

forces are to be determined.

Draw the free-body diagram of the part of the section truss whether left or right section which has the least number of forces on it.

Moments should be summed about a point that lies at the intersection of

the lines of action of two unknown forces. The third unknown force is determined directly from the equation.


219

EXAMPLE 7.7

Determine the force in members CA, CF and FE of the bridge truss by using method of section. State whether the members are in tension of compression. The support reactions have been calculated.

4m 4m 4m Solution:

A

C D

EB

F

8 kN

6 kN

6kN 3.2kN 4.8kN

C D6

A BF E

8 Choose the left section

+ MC = 0 -FFE (5) + 3.2 (4)+ 6(5) = 0 FFE = 8.6kN (T) + MA = 0 -FFC (4) + 8 (4) = 0 FFC = 8 kN (T) ↑ ΣFy = ↓ ΣFy 3.2 + 8 + FAC sin 51.34 = 8 FAC = - 4.1 kN (C)

C FAC FFC6 A FFE 3.2 8

5m


220

EXAMPLE 7.8

Solution:

Determine the member forces of HI, ID and CD by using the method of sections.

Choose right section

tan θ = 2/3 θ =33.7°

I FHI

FDI

FDC

+ MI = 0 FDC sin 18.43(9) + 1(6) + 2(3)= 0 FDC = -4.22 kN (C) ↑ ΣFy = ↓ ΣFy FDI cos 56.3 = 1+2 + FDC cos 71.6

tan θ = 3/1 θ =71.6º

15 = 6 = 9 5 HD IC HD = 2m, IC = 3m

0.55 FDI = 1.668 FDI = 3kN(T)

+

MD = 0 -FHI (2) + 1(3) = 0 FHI = 1.5 kN (T)

-4.22

tan θ = 5/15 θ = 18.43°


221

EXAMPLE 7.9

Determine the force in members HG, BC and BG of the truss. State whether the members are in tension or compression.

6 kN 8 kN 2 kN 3m 3m 3m 3m Solution; Determine the truss is statically determinate. Calculate the support reactions. [Ans: VA = 9 kN (↑),VE = 7 kN (↑)] Cut the section of the truss through the members where forces are to be determined.

6 kN 8 kN 2 kN 3m 3m 3m 3m


222

Choose one section whether left or right sections. G A B 9kN 6kN

FBG

FBC

θ2

θ3= θ2

tan θ1 = 4.5/3 = 56.3º tan θ2 = 1.5/3

θ1

= 26.6º

FHG

Calculate the member forces. + MB = 0 (9) 3 + FHG cos 26.6(3) = 0 2.68 FHG = -27 FHG = - 10.1 kN (C) ↑ ΣFy = ↓ ΣFy 9 + FHG sin 26.6 + FBG sin 56.3 = 0

-10.1

9-4.5-6 = -0.832 FBG FBG = 1.8 kN(T)

+

MG = 0 9(6) – 6(3) –FBC (4.5) = 0 FBC = 8 kN (T)


223

EXERCISE 7.3 A pin-connected compound truss is loaded and supported shown in figure, calculate the member forces of HG, JC and BC by using the method of section.

[Ans: FJC = 1.15kN (T), FBC = 2.9kN (T), FHG = 3.47kN (C)] 6.7 Alternative Computation using Joint Equilibrium Method

An alternative method can be applied to determine the member forces. The purpose is to reduce the time calculation. If the problems stated need to solve by using the method of joint or method of section, please follow the instruction.

EXAMPLE 7.10 Calculate all member forces.

A

D

E B

8 kN

6 kN

VA VB

HA

4m 5m 4m

5m


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Solution; The calculation can be started at joint A or B. 6 3.2 4.8

1st

4.8x

EBx

DE=

8.4

458.4 x=

x = 3.84

FBD

FBD= 22 8.484.3 + = 6.15 kN

Opposite direction

C D

A

E B

F

8 kN

6 kN 2nd

FBD

4.8

3.84

A

C D

E B

F

8 kN

6 kN

4.8

3.84

4.8

3rd

4th

3.84

5th

x

558.4 x=

x = 4.8

4.8

FEx

DE=

8.4

FCE

22 8.48.4 +FCE= = 6.8 kN


225

6th

A

C D

E B

F

8 kN

6 kN 7th

6 8.6 6 3.2 = 8.6 so 8.6 - 6 = 2.6

= (to balance) FEF= 8.6kN

3.84

4.8FEF= FFA =8.6kN

4.88th

3.84

4.8

FCF =8 kN

10th 2.6 + 6 4.8 = (to balance) FCD =3.8 kN

A

C D

E B

8 kN

10thFCD

6 kN

9th

y

CFy

AF=

6.2

546.2 y=

y = 3.25

2.6

FAC= 22 6.225.3 + = 4.16 kN

9th

2.6 4.8


226

TUTORIAL 6 1) The pin-jointed truss shown in figure is subjected to a uniformly distributed load of 40kN/m along member CD and a vertical point load of 90kN at D.

a) Determine the reactions at A and F. b) Determine the forces in member BC, BE and EF by using the

method of joint. c) Calculate the forces in members CD and DE using the method of

section.

40kN/m

90kN

[Ans: VA = 98.89kN (↓),VF = 374.89kN (↑), FDE = 330 kN (C), FEF = 435 kN (C)] 2) The truss shown in figure is loaded with concentrated loads at B, D, E and G. The truss is pinned to the foundation at A and supported on rollers at F. Determine the internal forces in members LK, LC and BC using the method of joint and checks your answers using the method of section. State if the forces are tensile or compressive.

30kN 30kN 30kN 30kN

[Ans: FBC = 36 kN (T)]


227

3) Figure shows a pin-jointed plane truss pinned at A and supported on rollers at B. Determine the forces in each member of the truss by using the method of joint. Recalculate the force in member BC by using the method of section.

60kN

30kN

[Ans: FAC = 117.24kN (C), FAB = 38.14kN (T), FDE = 141.5 kN (T), FBD = 88.8kN (T)] 4) A pin-jointed is supported on roller support and pinned support at A and B respectively as shown in figure. Determine the forces in each member of the truss by using the method of joint. Recalculate the force in members AB and DE by using the method of section.

20kN

30kN

[Ans: FAB= 5.3kN (C), FED = 20kN (C), FEB = 26.51 kN (C), FDC = 20kN (C)]


228

5) (a) List two assumptions of analysis to determine the member’s force of the truss. (b) Identify each of the trusses in Figure Q5(a) to Q5(d) as a determinate or

statically indeterminate and stable or unstable. c) In Figure Q5(e), a statically determinate pin-jointed plane truss is pinned

at A and supported on rollers at B. It carries two loads of 5000kg at joints F and G. (i) Prove that the plane truss is a statically determinate structure.

(ii) Using the method of joint determine all member forces.

(iii) Using the method of section, determine the member forces of CD and FG.

L= 5m for each member.

Figure Q5(a) Figure Q5(b) Figure Q5(c) Figure Q5(d)


229

5000kg 5000kg

Figure Q5(e)

(FINAL EXAM SEM I 2007/2008 UTHM) 6) The pin-jointed plane frame shown in figure carries a vertical load 10 kN at D. The frame is supported by a hinge at support A and by a roller support at B. The roller support at B is designed to ensure that the raction at B is in the direction BC. (a) Determine the reaction at support A and B. (b) Determine the internal force in all member.

D

AB

C10kN

4m

3m 8m

2m

2m

(FINAL EXAM SEM II 2007/2008 UTHM)


230

REFERENCES

1) R.C.Hibbeler, Mechanics of Materials, Sixth Edition, 2005. 2) Ferdinand P.Beer, Johnston Jr, T.Dewolf, Mechanics of Materials, Fourth

Edition, 2006


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