Transcript 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
I-Beam (I-Kiriş)
Channel Beam (U-Profil)
A
Gusset Plate (Bayrak)
Angled Beam
(Köşebent – L profil)
Bar (Çubuk))
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 Typical Roof Truss
F
Joint
(Düğüm)
External Force
G
E
A
B
Support Reactions
Member
(Çubuk)
D
C
Support Reaction
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
m=2j-3
j : number of joints
should be satisfied for rigidity
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 q 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.
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.
F1 and F2 , F3 and F4 collinear
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 (SFx=0, SFy=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 (SFx=0, SFy=0,
SM=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
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).
4m
4m
4m
E
D
4m
r=400 mm
H
G
C
F
4m
B
16 kN
4m
A
2. Determine the forces in members BC and FG.
FCJ
Cut
FBC
FFJ
FG
3. Determine the forces in members CD, CJ and DJ, state whether they work
in tension (T) or compression (C).
I. Cut
3m
T
FJI
FDJ
FCD
Ax
Ay
T
FKJ
FCJ
FCD
Ax
Ay
II. Cut
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.
I. Cut
Ax
II. Cut
Ay
By
5. Determine the force acting in member DK.
Ux
Uy
Vy
I. Cut
Uy=15 kN
III. CutII. Cut
Vy=20 kN
6. Determine the forces in members DE, EI, FI and HI.
II. Cut
I. Cut
Gx
Ay
Gy
7. Determine the forces in members ME, NE and QG.
I. Cut
II. Cut
III. Cut
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.
4 kN
10 kN
6 kN
F
E
H
G
2m
B
C
D
J
2m
N
K
L
2m
A
3m
20 kN
M
P
3m
4m
4m
4m
4m
Radii of pulleys H, F and K 400 mm
(I)
(II)
(IV)
By
4 kN
10 kN
6 kN
F
E
10 kN
10 kN
(III)
2m
B
C
G
D
10 kN
Bx
10 kN
2m
J
20 kN
N
K
L
2m
Ax
H
A
3m
10 kN
M
P
3m
4m
4m
4m
4m
Radii of pulleys H, F and K 400 mm
C
2 kN
2 kN
2 kN
D
E
F
5 kN
3
G
4
1 kN
4m
O
N
H
B
M
4m
A
I
L
2 kN
3m
J
K
2 kN
2 kN
3m
3m
3m
9. Determine the forces in members EF, NK and LK.
C
2 kN
2 kN
2 kN
3 kN
D
E
F
G
4 kN
From the
equilibrium of
whole truss
1 kN
4m
I. Cut
N
Top Part
O
M
B
FMN
FBN
FHO
FMO
FHI
A
I
L
Ay
2 kN
3m
J
K
2 kN
2 kN
3m
3m
are determined
H
FBA
Ax
Ax, Ay and Iy
3m
Iy
4m
I. Cut
SMH=0
FAB is determined
C
2 kN
2 kN
D
E
3 kN
2 kN
FEF
G
F
4 kN
1 kN
FMF
II. Cut
Top Part
4m
II. Cut
N
O
M
B
FMN
FBN
SMM=0
H
FMO
FBA
4m
A
I
L
2 kN
3m
J
K
2 kN
2 kN
3m
3m
3m
FEF and FMF are
determined
C
2 kN
2 kN
D
E
3 kN
2 kN
FEF
G
F
4 kN
1 kN
FMF
N
FMO
M
B
4m
O
H
III. Cut
SMN=0
4m
FNK
A
I
L
2 kN
3m
K
J
2 kN
2 kN
FLK
3m
III. Cut
Left Side
3m
3m
FLK and FNK are
determined
25 2 kN
G
1m
H
I
F
P
E
10 2 kN
10 2 kN
1m
N
M
O
1m
L
J
K
D
C
25 2 kN
20 2
2m
B
A
2m
kN
1m
1m
2m
10. Determine the forces in members KN, FC and CB.
25 2 kN
G
III. Cut
I. Cut
H
I
F
P
1m
E
10 2 kN
10 2 kN
1m
N
M
J
K
O
1m
D
C
L
II. Cut
25 2 kN
20 2
IV. Cut
Ax
kN
2m
B
A
2m
1m
Ay
2m
1m
By
Forces in KN, FC and CB.