ENGINEERING STRUCTURES

ENGINEERING STRUCTURES

In the previous chapter we have employed the equations of equilibrium in order to

determine the support / joint reactions acting on a single rigid body or a system of

connected members treated as a single rigid body.

Determination of the support / joint reactions constitutes only the first step of the analysis

in engineering structures. From now on we will focus on the determination of the forces

internal to a structure, that is, forces of action and reaction between the connected

members.

An engineering structure is any connected system of members built to support or transfer

forces and to safely withstand the loads applied to it. To determine the forces internal to

an engineering structure, we must dismember the structure and analyze separate free body

diagrams of individual members or combination of members which are mostly connected

each other using smooth pin connections. Determination of joint reactions is of great

importance in the selection of connection members that hold the structure as a whole.

Joint reactions always occur in pairs that are equal in magnitude and opposite

in direction. If not isolated from the rest of the structure or the environment by

means of a free body diagram, joint forces are not included in the diagram

since they will be internal forces.

In order to determine the joint reactions, the structures must be separated into

at least two or more parts. At these separation points the joint reactions become

external forces and thus, are included in the equations of equilibrium.

In this chapter trusses, frames and machines will be examines as engineering

structures.

TRUSSES (KAFES SİSTEMLER)

A framework composed of members joined at their ends to form a rigid structure

is called a “truss”. Bridges, roof supports, derricks, grid line supports, motorway

passages and other such structures are examples of trusses. Structural members

commonly used are I-beams, channels, angles, bars and special shapes which are

fastened together at their ends by welding, riveted connections, or large bolts or

pins using large plates named as “gusset plates”.

For bridges and similar structures, plane trusses are commonly utilized in pairs

with one truss assembly placed on each side of the structure. The combined

weight of the roadway and vehicles is transferred to either side.

http://www.youtube.com/watch?v=VZLc8odACUw

http://www.youtube.com/watch?v=VZLc8odACUw

I-Beam (I-Kiriş) Channel Beam (U-Profil)

Angled Beam

(Köşebent – L profil)

Bar (Çubuk))

A

Gusset Plate (Bayrak)

Simple Trusses

The basic element of a plane truss is the triangle. Three bars joined by pins at their

ends constitute a rigid frame. In planar trusses all bars and external forces acting

on the system lie in a single plane.

P A

B C

A truss can be extended by additing extra tirangles to the system. Such trusses

comprising only of trianges are called “simple trusses – basit kafes”. In a simple truss it

is possible to check the rigidity of the truss and whether the joint forces can be

determined or not by using the following equation:

m : number of members j : number of joints

m=2j3 should be satisfied for rigidity

A Typical Roof Truss

A

B CD

E

F

G

Support Reactions

Support Reaction

External Force

Member (Çubuk)

Joint (Düğüm)

Assumptions

1. In a truss system it is assumed that all bars are two forces members. The weights of

members are neglected compared to the forces they are supporting. Therefore members

work either in tension or compression.

(Çeki) (Bası)

2. When welded or riveted connections are used to join structural

members, we may usually assume that the connection is a pin joint

if the centerline of the members are concurrent at the joint. In this

case the joint does not support any moment since it allows for the

rotation of the members.

3. It is assumed in the analysis of simple trusses that all external

forces are applied at the pin connections.

4. Since bars used in trusses are long, slender elements they can

support very little transverse loads or bending moments.

Determinaton of Zero-Force Members

(Boş Çubukların Belirlenmesi)

Determination of the zero-force members beforehand will

generally facilitate the solution of the problem

1.Rule: When two collinear members are under compression, it

is necessary to add a third member to maintain alignment of the

two members and prevent buckling. We see from a force

summation in the y direction that the force F3 in the third member

must be zero and from the x direction that F1=F2. This conclusion

holds regardless of the angle and holds also if the collinear

members are in tension. If an external force in y direction were

applied to the joint, then F3 would no longer be zero.

2. Rule : When two noncollinear members are joined as shown, then in the absence of an externally load at this joint, the forces in both members must be zero, as we can see from the two force summations.

F1 and F2 , F3 and F4 collinear

Equal Force Members (Eşit Yük Taşıyan Elemanlar)

When two pairs of collinear members are joined as shown, the forces in each pair must be equal and opposite.

SOLUTION METHODS

1) METHOD OF JOINTS (Düğüm Yöntemi)

This method for finding the forces in the members of a truss consists of satisfying the

conditions of equilibrium for the forces acting on the connecting pin of each joint. The

method therefore deals with the equilibrium of concurrent forces, and only two

independent equilibrium equations are involved (Fx=0, Fy=0).

Sign convention (İşaret anlaşması): It is initially assumed that all the

members work in tension. Therefore, when the FBDs of pins are being

constructed, members are shown directed away from the joint. After

employing the equations of equilibrium, if the result yields a positive value

(+), it means that the member actually works in tension (T) (çeki), if the

result yields a negative value (-), it means that the member works in

compression (C) (bası).

1. Determine the force in each member of the loaded truss. Make use of the symmetry of

the truss and of the loading.

2) METHOD OF SECTIONS (Kesim Yöntemi)

When analyzing plane trusses by the method of joints, we need only two of the three

equilibrium equations because the procedures involve concurrent forces at each joint. We

can take advantage of the third or moment equation of equilibrium by selecting an entire

section of the truss for the free body in equilibrium under the action of nonconcurrent system

of forces.

The Method of Sections is often employed when forces in limited number of members are

asked for and is based on the two dimensional equilibrium of rigid bodies (Fx=0, Fy=0,

M=0). This method has the basic advantage that the force in almost any desired member

may be found directly from an analysis of a section which has cut that member. Thus, it is

not necessary to proceed with the calculation from joint to joint until the member in question

has been reached. In choosing a section of the truss, in general, not more than three members

whose forces are unknown should be cut, since there are only three available independent

equilibrium relations.

Once a truss is cut into two parts, one of the parts is taken into consideration and all the

internal forces now become external from where the cut was passed. The forces are

initially assumed as working in tension, so, they are shown directed away from the

FBD. After employing the equations of equilibrium, if the result yields a positive value

(+), it means that the member actually works in tension (T) (çeki), if the result yields a

negative value (), it means that the member works in compression (C) (bası).

Before starting to solve with method, if necessary the support reactions can be

determined from the FBD of the whole truss and also zero-force members can be

identified. It is very important to recognize that, only the forces acting on the part are

considered, the forces acting on the other part, which is not considered, should not be

included. The moment center can be any point on or out of the part in consideration.

TRUSSES SAMPLE QUESTIONS

4 m

4 m

4 m 4 m

4 m

4 m

r=400 mm

16 kN

A

C

D

B

E

FGH

1. The crane in the figure consists of a planar truss. Determine the forces in members DE,

DG and HG when the crane supports a 16 kN load, indicate whether the members work in

tension (T) or compression (C).

2. Determine the forces in members BC and FG.

CutFBC

FCJ FFJ

FG

3. Determine the forces in members CD, CJ and DJ, state whether they work

in tension (T) or compression (C).

Ax

Ay

T

I. Cut3 m

FCD

FDJ

FJI

Ax

Ay

T

II. Cut

FCD

FCJ

FKJ

4. The truss shown consists of 45° triangles. The cross members in the two

center panels that do not touch each other are slender bars which are incapable

of carrying compressive loads. Identify the two tension members in these panels

and determine the forces they support. Also determine the force in member MN.

Ax

AyBy

I. Cut

II. Cut

5. Determine the force acting in member DK.

Ux

Uy

Vy

Uy=15 kNVy=20 kN

I. Cut II. CutIII. Cut

6. Determine the forces in members DE, EI, FI and HI.

Gx

Ay

I. Cut

Gy

II. Cut

7. Determine the forces in members ME, NE and QG.

I. Cut III. CutII. CutFDE

FME

FMB

FLB

FEK

FNF

FAF

FEK

FFQFFG

2 m

2 m

2 m

4 m3 m 3 m 4 m4 m4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN6 kN

Radii of pulleys H, F and K 400 mm

4 kN

8. In the truss system shown determine the forces in members EK, LF, FK and CN,

state whether they work in tension (T) or compression (C). Crossed members do not

touch each other and are slender bars that can only support tensile loads.

20 kN

2 m

2 m

2 m

4 m3 m 3 m 4 m4 m4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN6 kN

20 kN

4 kN

10 kN10 kN

10 kN

10 kN

10 kN

Ax

By

Bx


2 m

2 m

2 m

4 m3 m 3 m 4 m4 m4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN6 kN

20 kN

4 kN

10 kN10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

1st cut


FEF

FFL

FEK

FKL

2 m

2 m

2 m

4 m3 m 3 m 4 m4 m4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN6 kN

20 kN

4 kN

10 kN10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

2nd cut


FEF

FFL

FFK

FJK

2 m

2 m

2 m

4 m3 m 3 m 4 m4 m4 m

A

BC D

E F

G

N

M

L K

J

H

P

10 kN6 kN

20 kN

4 kN

10 kN10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

3rd cut


FCD

FDN

FMN

FPM

2 m

2 m

2 m

4 m3 m 3 m 4 m4 m4 m

A

BC D

E F

G

N

M

L K

J

H

P

10 kN6 kN

20 kN

4 kN

10 kN10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

4th cut


FCD

FCN

FPN

FPM

9. Determine the forces in members ON, NL and DL.

Ax

Ay Iy

kNIIAF

kNAAMkNAF

yyyy

yyA

xx

60100

40)3(2)6(2)9(4)15(2)2(6)18(060

From equilibrium of whole truss;

FON

FOC

FBC

I.cut

I.cut

)(014.9

0)3(64

6264

4)3(2)2(6)6(02222

4

nCompressiokNF

FFAM

ON

ONON

kN

yC

Joint M 4 kN

FMLFMN

)(605.3064

4240

064

6

64

60

22

2222

CkNFFFF

FFFFF

MLMNMNy

MLMNMLMNx

II.cut

FMN

FNL

FDL

FDE

)(0054

64

420

)(5.4

0)4(464

6364

4)2(6)6(2)9(0

22

22605.3

22605.34

memberforceZeroFFFAF

CkNF

FFFAM

DLDLMNyy

NL

NLMNMN

kN

yD

II.cut

20 kN

10. Determine the forces in members HG and IG.

20 kN

I.cutII.cut

20 kN

20 kN20 kN

20 kN

20 kN

20 kN

20 kN

I.cutII.cut

FCD

20 kN

20 kN20 kN

20 kN

20 kN

20 kN

FHG

FGI FGJ

I.cut MG=0 FCD=54.14 kN (T)

II.cut MA=0 FHG=81.21 kN (C)

I.cut Fx=0 FGI=18.29 kN (T)

FCD

FHG

FHIFBA

11. Determine the forces in members EF, NK and LK.

C

B

A

D E F G

HO

L K JI

N

1 kN

2 kN 2 kN2 kN 5 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

M

34

From the equilibrium of whole truss

Ax, Ay and Iy are determined

I. CutMH=0FAB is

determined

C

B

A

D E F G

HO

L K JI

N

1 kN

2 kN 2 kN2 kN 3 kN

4 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

I. Cut Top

Part

Ay Iy

M

Ax

FHI

FHOFMOFMNFBN

FBA

II. CutMM=0

FEF and FMF are determined

C

B

A

D E F G

HO

L K JI

N

1 kN

2 kN 2 kN2 kN 3 kN

4 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

II. CutTop

PartM

FEF

FMF

FMOFMNFBN

FBA

III. CutMN=0

FLK and FNK are determined

C

B

A

D E F G

HO

L K JI

N

1 kN

2 kN 2 kN2 kN 3 kN

4 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

III. CutLeft

Side

MFMO

FLK

FNK

FMF

FEF

12. Determine the forces in members KN and FC.

kN

kN

kN

kN

kN

1 m

1 m

1 m

2 m

2 m1 m1 m2 m

A B

C D

O

E

G

P F

NM

I

JK

L

H

225

210

220

210 210

kN

kN

kN

kN

kN

1 m

1 m

1 m

2 m

2 m1 m1 m2 m

A B

C D

O

E

G

P F

NM

I

JK

L

H

225

210

220

210 210

I. Cut

II. Cut

III. Cut

ByAy

Ax