Transcript AR 231 Structures in Architecture I

IZMIR INSTITUTE OF TECHNOLOGY
4.0
16.07.2015
Department of Architecture
AR232 Spring2006
INTERNAL FORCES AND MOMENTS IN
MEMBERS
4.1
Introduction
4.2
Idealization of Structural Systems and Loads
4.3
Method of Analysis
4.4
Differential Equailibrium Relations
4.5
Examples
Dr. Engin Aktaş
1
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
4.1. Introduction
• Structural members are subjected to many different types of
external forces. In the design of these members, it is essential
to compute the stresses and deformations produced by the
external loads
The study of structural members
√ 1.
√ 2.
3.
4.
21.02.2005
Idealization of the structural system and the loads
Computation of the external reactions
Determination of the internal forces
Calculation of the stresses and deformations caused by these
internal forces
Dr. Engin Aktaş
2
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
4.2. Idealization of Structural Systems and
Loads
4.2.1 Types of Supports
(1) Roller (Link) Supports
B
A
A
(a) Link
(b) Rod
(c) Roller
21.02.2005
B
(d) Roller
Dr. Engin Aktaş
(a) and (b) can resist
the force only in the
direction of line AB
(c) and (d) can only
resist a force which ‘s
perpendicular to the
direction of movement
of the roller.
3
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
(2) Pin or Hinge Supports
(a) and (b) can resist
a force acting in any
direction in the plane.
Therefore, in general,
hinged or pinned
supports have two
unknowns
(a)
(b)
(3) Fixed, clamped, built-in or encastre supports
This type of support is capable of resisting a force in any direction and is also
capable of resisting a moment or a couple.
21.02.2005
Dr. Engin Aktaş
4
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
4.2.2 Types of Loads
A force is transmitted to a member through a post, hanger or another
member. In such cases the load is applied over a very limited portion of
the member. Such forces are idealized as concentrated forces and
shown with a single vector acting at a point.
In general, forces and loads are distributed over a finite length or area of
the member. These are idealized as continuously distributed loads.
dF
y
General Load Distribution on Member
x
dx
If the total force on length Dx is denoted as DF, then the intensity of loading,
q is defined as
DF dF

Dx 0 Dx
dx
q  lim
21.02.2005
Dr. Engin Aktaş
5
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
y
y
q
q(x)=constant
AR232 Spring2006
q0
x  xl 
l  xl
q0
x
xl
x
l
(b) Triangularly distributed
(a) Uniformly distributed
y
q q
q  q1  2 1 x
l
y
q  q0 sin
x
l
q0
q2
q1
x
x
l
(d) Sinusoidal
(c) Trapezoidal
21.02.2005
Dr. Engin Aktaş
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
4.3. Method of Analysis
Internal forces in members are determined by passing an imaginary section
across the member at the point concerned, and writing the equilibrium
equations for the free body either to left or right of the cut.
y
My
Fy
Fx
Mz
z
21.02.2005
Fz
Mx
x
Fx – Normal Force
My, Mz – Bending Moments
Fy, Fz – Shear Force
Mx – Twisting moment of torque
Dr. Engin Aktaş
7
IZMIR INSTITUTE OF TECHNOLOGY
2 kN/m
Example
Compute the internal
forces at point D for
the beam given.
Department of Architecture
10 kN
1m
3m
REy
RBy

226  5RBy  10 4  43 1  0
2
F
At D (use right portion)
x
N
1m
RBy  14 kN
43 4  5R  0







M

0

2
2
1

10
1

 B
Ey
2
M=6 kNm
REy  6 kN
D
 Fx  0  REx  5 kN
N=5 kN
+
M
D
C
A
2m
+
4 kN/m
E
REx
B
5 kN
ME  0
AR232 Spring2006
D
E
Q
6 kN
21.02.2005
E
5 kN
6 kN
 0   5  N  0  N  5 kN (comp)
43
6Q  0  Q  0
5 kN
2
 43 
M

0



2  6  3  M  0  M  6 kNm
 D
 2 
 Fy  0  
Dr. Engin Aktaş
8
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
4.4. Differential Equilibrium Relations
q(x)
q
y
Q+DQ
M
x
x
O
Q
M+DM
Dx
Dx
DQ
 Fy  0  Q  DQ   qDx   Q  0  Dx  q
Dx
 Dx 
 M O  0  Q  DQ 2  Q 2   M  M  DM   0
Q Dx  DQ  Dx   DM  0
 2 
 Dx  is very small compared to the other
D
M
DQ  

 Q
 2  two terms and will be neglected
Dx
dM
dQ
 Q


q
&
In the limit, when Dx approaches zero
dx
dx
These are basic differential equations relating the external load, q(x), the shear force, Q,
and the bending moment, M.
21.02.2005
Dr. Engin Aktaş
9
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR232 Spring2006
Let’s write this relations in integral form as
i 1
i 1
 dQ    qdx
i
i 1
or
Qi 1  Qi    qdx
i
i
i 1
 dM    Qdx
i
i 1
i 1
or
M i 1  M i    Qdx
i
i
i and i+1 are any two sections along the member
First equation shows that the change in shear between any two sections is
equal to minus the area of the load diagram between these two sections.
Second equation shows that the change in moment between any two sections
is equal to minus the area of the shear diagram between these sections.
21.02.2005
Dr. Engin Aktaş
10
IZMIR INSTITUTE OF TECHNOLOGY
Example
Department of Architecture
Sketch the shearing force and bending moment diagrams
4 kN
0.5 kN/m
for the beam given
 0 RA  RD  4  05 2  0
A
RA  RD  5 kN
+  M A  0 4RD  43  05 21  0
RD  3.25kN
RA  1.75kN
R
F
AR232 Spring2006
D
y

I
B
2m
C
1m
1m
RD
A
II
0.5 kN/m
MBA
A
x
QBA
1.75 kN
between A and B 0  x  2
QBA  1.75  0.5x 2
x
M BA  1.75x  0.5
2
0.5 kN/m
MCB
A
x
QCB
1.75 kN
between B and C 2  x  3
QCB  0.75kN
M CB  1.75x  0.52x 1
@ x = 2m
@ x = 3m
QB  0.75kN M B  2.5kNm
QC  0.75kN M C  3.25kNm
It is to be noted that because of the 4 kN load at
C, these values correct strictly just short of C.
21.02.2005
Dr. Engin Aktaş
11
IZMIR INSTITUTE OF TECHNOLOGY
III
Department of Architecture
between C and D
4 kN
MCD
0.5 kN/m
A
B
x
@ x = 4m
1.75 kN
3 x  4
QCD  3.25
M CD  1.75x  0.52x 1  4x  3
QCD
C
AR232 Spring2006
QD  3.25kN (just shortof D)
MD  0
QB  QA  Area of theload diagram between A and B
4 kN
0.5 kN/m
D
A
B
C
1.75 kN
QB  1.75   0.5  2 QB  0.75kN
3.25 kN
M B  M A  Area of theshear diagram between A and B
3.25 kN
Q
1.75 kN
+
___
0.75 kN
2.5 kNm
M
21.02.2005
+
  1.75   0.75 
M B  0  
 2  M B  2.5 kNm
2


3.25 kNm
M=0.75x+1
M=13-3.25x
Dr. Engin Aktaş
12
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
RA=4 kN
Example
y
4 kN/m
AR232 Spring2006
RB=8 kN
q can be written as
q
x
Therefore Q
6m
RA
RB
8 kN
4 kN
3.46 m
+
__
Q
x
Q  QA   qdx
0

4
x
6
x
4
x2
Q  4   xdx  Q  4 
6
3
0
x3
0  4 
 x  12  3.46 m where Q  0
3
@x 6m
Q  4  12  8 kN
x
M  M A    Qdx
9.24 kNm
0

x2 
x3
M      4   dx  4 x 
3
9
0
@x 6m
M 0
x
+
M
at Q  0 max.momentocurs@ x  12
M max  4 12 
21.02.2005
Dr. Engin Aktaş
12 12
 9.24 kNm
9
13
IZMIR INSTITUTE OF TECHNOLOGY
Example
Department of Architecture
3 kN
B
C
A
1.9 kN
6.1 kN
AR232 Spring2006
QB  QA  thearea of theload diagram A toB
QB   1.9   5  QB  3.1kN
1m
5m
The maximum moment occurs where the shearing
force is zero, @ x=1.9 m, and is given by
3.1 kN
1.9 kN
1.9 m
+
__
__
3.1 m
Q
  1.9 1.9 
M max  M A  

2


3 kN
M max  0 
1.81 kNm
3.61
 M max  1.81 kNm
2
+
3.8 m
__
and for MB
 3.1 3.1 
M B  M max  

2


3 kNm
M B 1.81  4.81  M B  3 kNm
21.02.2005
Dr. Engin Aktaş
14