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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
LECTURE
18 134
EQUILIBRIUM OF RIGHD BODY
It is necessary and sufficient for the equilibrium of rigid body that the resultant
of all forces acting on the body is zero and that the resultant of all moments
taken about any point inside or outside the body is zero.
0~
F~ 0
~M~
0
00
z
y
x
M
MM
Ox
FR
MR
MC
MF
O
MC
&
0
00
z
y
x
F
FF ...
...
...
...
...
...
1F~
iF~
nF~
iF~
nF~
1F~
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CE 201 - STATICS
Support for Rigid Bodies Subjected to Two-Dimensional Forces System.
Table 5 - 1
Type of Connection Reaction Numbers of Unknowns
One unknown. The reaction is the tension force which
acts away from the member in the direction of the cable
One unknown. The reaction is a force which acts
along the axis of the link.
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
One unknown. The reaction is a force which acts
perpendicular to the slot.
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.Rocker
Roller or pin in confined smooth slot
Roller
Weightless Link
Cable
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CE 201 - STATICS
Member fixed
connected tocollar on smooth rod
Support for Rigid Bodies Subjected to Two-Dimensional Forces System.
Table 5 - 1
Type of Connection Reaction Numbers of Unknowns
One unknown. The reaction is a force which acts
perpendicular to the rod.
Two unknown. The reaction are two components offorce, or the magnitude and its direction of the resultantforce. Note that and are not necessarily equal.(usually not, unless the rod shown is a link as in (2)).
Two unknowns. The reactions are the couple moment
and the force which acts perpendicular to the rod.
Three unknowns. The reaction are the couple and the
two forces components, or the couple and the magnitude
and direction of the resultant force.
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
Fixed support
Smooth pin or hinge
Member pin connected to collar on smooth rod
Smooth contacting surface
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
LECTURE
18 336
FREE BODY DIAGRAM
EQUILIBRIUM IN 2-DIMENIONS
FByFBxFB B
AD
CFc
OMo
Fy
Fx
FOR EQUILIBRIUM
OM
OF
OF
z
y
x
..(1)
..(2)
..(3)
RAx
RAyFcy RDy
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
LECTURE
19 137
EXAMPLE
Determine the support reaction in the beam (ABCD).Consider beam weight equals 50N.
A
C4
3
5100N
B
3m 2m 5m
DX
A
B
C D
RAx
RAy
60N80N
50N RDy
SOLUTION
First we draw FBD for beam (ABCD)
+
(1)
(2)
Solve (1) & (2)
RDy = 49N ( ) RAy = 81N( )
+
1007R3R-
07R2503R
0M
130RR
0R5080R
0F
)(06R060R
0F
DyAy
DyAy
B
DyAy
DyAy
y
N
Ax
Ax
x
a
+
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
LECTURE
19 238
A
B
C D
RAx
RAy
60N 80N
50N RDy
ALTERNATE SOLUTION I
SOLUTION
N
Ay
Ay
D
N
Dy
Dy
A
NAxAx
x
8110
560250R
010R780550
0M
)(4910
240250R
038055010R
0M
)(60R060R
0F +
+
+
CONCLUSION
0M0
M
0F
D
A
x
One force equationand two moment
equations on a linenot to the forcedirection.
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
LECTURE
19 339
A B C D
RAx
RAy
100N
50N RDy
+4m
E
43
5
3m 2m 5m
ALTERNATE SOLUTION II
CONCLUSION
0M0
M
0M
E
D
A
Three momentequations taken
at three points noton a straight line.
)(60
4
240
4
250490R
04R5501049
0M
)(81R
0M
)(49R
0M
N
Ax
Ax
E
N
Ay
D
N
Dy
A
+
+
+
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CE 201 - STATICS
Given:
kA = kB = 15kN/m
Required:
c = ?
40 9.8 = 392N
A B
C
40kg
392N
1m 3m
RA=KAA RB=KBB
C
0.1045m
0.0784m
m
C
C
m
A
AA
AAB
y
m
B
BB
A
0.6532
3
0.1045-
1
0.10450.0784
0.078415000
1176
k
R
1176R0392RR
0F
15680.104515000
4392
043921k
0M
N
N
BR
+
+
A
B
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CE 201 - STATICS LECTURE20 2
41
TWO FORCE MEMBERS:
OR
THREE FORCE MEMBERS:
AF~
AF~
AF~
AF~
AF~
BF~
BF~
BF~
BA
F~
F~
A
BB
A
B
A
BF~
AF~
BF~
AF~
CF~
CF~
0F
0F
y
x
O
BA
C
CF~
AF~
BF~
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
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CE 201 - STATICS Dr. Mustafa Y. Al-MandilDepartment ofCivil Engineering
DISTRIBUTED LOADING
A) Uniform Distributed Loading:
B) Linear Distributed Loading:
C) Trapezoidal Distributed Loading:
6m 3m
6m 4m
5N/m 30N
45N
15N/m
10N/m
60N
30N
20N/m
6m 4m
3m
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CE 201 - STATICS
Given: Beam ABCRequired: Find Reactions
)(408040R
0401005
4R
0F
)(306030R
0301005
3R
0F
100R
4004R
010405R5
4
0M
kN
Ay
Ay
y
kN
Ax
Ax
x
kN
B
B
B
A
a3
2d
2
ad
2
d
2
a
0)2
d
2
a
(W)2
a
(dW
0Mo
+
+
(
+
50kN
34
5m
5m
3m
4m
5mA B
D
C
B CA 5m
34
RBRAy
RAx
40kN
30kN
RB
RD3m
4mTwo Force Member
RB= RD d
D/2 D/2
Wa
o
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CE 201 - STATICS
Q.
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
Determine the support reaction of beam ABCD ?3m4m
B
D
CA
2m
w = 15N/m3
4
50N
3m4m2m
RAx
RByRAy
A B C60N 40N
30N
60Nm
Free body Diagram
A. Draw F.B.D.Transfer the (50N) force from point (D) to point (C)
)(85
4
340R
340602801204R
06070404R260
0M
)(30R
030R0F
N
By
By
By
A
N
Ax
Ax
x
)(15
408560R
0408560R
0F
N
Ay
Ay
y
+
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CE 201 - STATICS
Support for Rigid Bodies Subjected to Three-Dimensional Forces System.
Table 5 - 2
Type of Connection Reaction Numbers of Unknowns
One unknown. The reaction is a force which acts away
from the member in the direction of the cable.
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
Three unknowns. The reactions are three rectangular
force components.
Four unknowns. The reactions are two force and two
couple components which act perpendicular to the shaft.
Cable
smooth surface support
Roller on a smooth surface
Ball and socket
Single journal bearing
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CE 201 - STATICS
Support for Rigid Bodies Subjected to Three-Dimensional Forces System.
Table 5 - 2
Type of Connection Reaction Numbers of Unknowns
Five unknowns. The reactions are three force and
two couple components.
Five unknowns. The reactions are three force andtwo couple components.
Five unknowns. The reactions are three force and
two couple components.
Six unknowns. The reactions are three force and
three couple components.
Single thrust bearing
Single smooth pin
Single hinge
Fixed support
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
1m
1.5m
4m
D
A
C
B
y
x
z
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
A
0.3m
0.2m
0.6m
0.5m
1m
0.5m
100 kg
C
B E
D
x
y
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineeringz
x
y
A
C
B
3ft3ft
4ft
6ft
D
E
RAz
RAx
8ft
4ft
RAyT
CD
TCD
TCE
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CE 201 - STATICSDr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
TRUSSES
Chep. 6
Atruss is an assembly of prismaticmembers connected at their ends
by smooth pin.Forces can only be
applied at such connections (joints)
Simple truss:
Assumptions for Analysis & Design:
1- Loading and reactions are applied
at joints (nodes) only. Member weight is
distributed equally to the adjacent nodes.
2- Members are loaded at their ends only
(i.e. two-force members). So, they undergo
axial forces only (tension or compression). RxyRAy
RAx
B
A C E
D G
H
K
J
LECTURE
23 151
LECTURE
23 151
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CE 201 - STATICS LECTURE 232
52
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
Analyze the truss shown below:
)(40R0F
)(45R0F
)(45R
06203204R
0M
kN
Ayy
kNAxx
kN
Bx
Bx
A
F.B.D. For Whole Truss :
Take Joint (E) :
(C)15F0255
3F
0F
(T)25F020F54
0F
kN
EDED
x
kNECEC
y
+
+
4m
RAy
RAx
RBx
A C
B DE
3m 3m
20kN 20kN
FEC
FEDE
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CE 201 - STATICS LECTURE 23
Analyze the truss shown below:
252
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
Take Joint (C) :
(C)20F0F
(T)15F0F
kN
CDy
kN
CAx
Take Joint (A) :
0
0505
440
0
)(50
05
34515
0
AB
AD
y
kN
AD
AD
x
F
F
F
TF
F
F
+
FCA
FCD25kN
C
40kN
45kN 15kNA
FAD
FAB
CE 201 STATICSD M t f Y Al M dil
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CE 201 - STATICS LECTURE24 1
53
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering ZERO-FORCE MEMBERS
Members which carry no forces in a truss either because of particular
loading pattern or because they enhance geometric stability:.
EQUAL-FORCE MEMBERSF3
F1
F2
F2 F1
F1= F2
F3
= 0
F3
F1
F2F4
F1= F3
F2= F4
30kN
35kN25kN
30kN40kN
A
B
D I LP
T
G K
C E H J M N Q
ZERO-FORCE MEMBERS
BC, BEKM, KN
PQ, PN, LN
(C)40F
(T)25F
kN
IJ
kN
HG
CE 201 STATICSD M t f Y Al M dil
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CE 201 - STATICS
By sectioning the truss into two or more parts (each part containingmore then one joint), then we can find forces in certain numbers directly.
LECTURE
24 254
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering METHOD OF SECTIONS
Use Section A-A
Find FGH ?
(C)155F
04F480650
0M
kN
GH
GH
J
80kN
50kN
JG
FGH
+
4m
50kN80kN
3m
3m
3m
3m
3
m
A A
F3
F2
F1
0
0Mo
F1 =
CE 201 STATICSD M t f Y Al M dil
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CE 201 - STATICS LECTURE25 2
55
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering THEORY OF PLANE TRUSSES
M = Number of members
R = Number of Unknown ReactionsJ = Number of joints
if M + R > 2J Statically Indeterminate
if M + R = 2J Statically Determinate
if M + R < 2J Statically Unstable
M = 9
R = 3
J = 6
M + R = 2J
Determinate
M + R > 2J
Indeterminate
M = 10
R = 3
J = 6
M + R < 2J
UNSTABLE
M = 8R = 3
J = 6
B
C E
D
A G
CE 201 STATICSDr Mustafa Y Al Mandil
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CE 201 - STATICS
Frames are structural systemscomposed of members connected
and loaded at unrestricted
locations along their axis.
LECTURE
27 158
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
ANALYSIS & DESIGN
1- SOLVE FOR EXTERNAL REACTIONS
( partially or totally ).
2- DIS-ASSEMBLE & TREAT EACH COMPONENTAS RIGID BODY UNDER EQUILIBRIUM.
3- FORMULATE SETS OF SIMULTANEOUS
EQUATIONS & SOLVE FOR UNKNOWNS.
RAx
RAy RDy
A D
B
C
P2
P1
M
RAx RGx
RAy RGy
P1
P3P2
C
G
E
A
B
D
CE 201 STATICSDr Mustafa Y Al Mandil
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CE 201 - STATICS
Analyze Frame (ABCDE)
LECTURE
27 258
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering
Take whole Frame Take Part (ABC) As F.B.D.
kN
Ay
Ay
y
kN
Ey
Ey
A
25R
0R4015
0F
1512
180
R
03406012R
0M
)(3.75
8
120150R
06R8R340
0M)(15R
04025R
0F
kN
Ax
AyAx
C
kN
Cy
Ey
y
Take Whole Frame Again
)(3.75R03.75R
0F
kN
Exex
x
RAy
A E
C
40kN
60kNm
REy
4m
4m
DB
3m3
m
3
m
3
m
RAx
CRCy
RAy
RCx
A
B
45kN
40kN
+
+
CE 201 STATICSDr Mustafa Y Al-Mandil
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CE 201 - STATICS LECTURE25 3
56
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering GENERAL EXAMPLE
(a) (b)
(c) (d)
(e)
(g) (h)
(f)
Howe Truss Pratt Truss
Warren Truss Warren Truss with verticals
K Truss Sub-divided Warren Truss
Sub-divided Pratt Trussor Baltimore Truss
Baltimore Truss with inclinedchord or Petit Truss
(a)
(d)
(c)
(b)
Howe Truss
Pratt Truss
Fan Truss
Fink Truss
Conventional Roof Trusses
Conventional Bridge Trusses
CE 201 STATICSDr Mustafa Y Al-Mandil
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CE 201 - STATICS LECTURE25 4
57
Dr. Mustafa Y. Al-MandilDepartment of
Civil Engineering GENERAL EXAMPLE
CE 201 STATICS LECTUREDr Mustafa Y Al-Mandil
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CE 201 - STATICS LECTURE28 1
59a
Dr. Mustafa Y. Al MandilDepartment of
Civil EngineeringEXAMPLE
AnalyzetheFra
me
Take whole Frame :
)(35R
0R80115
0F
)(1158
920R
015404808R
0M
kN
Ay
Ay
y
kN
Hy
Hy
A
+
)(40R0F
)(40R
0M
kN
eyy
kN
Cy
E
Take ( C.D.E. ) :
+
RAx RHx
RAy RHy
80kN
C
H
G
A
B
E40kN D
12m
3m
6m
9m
4m 4m
RCx C
RCy
ED80kN
REy
REx
CE 201 - STATICS LECTUREDr. Mustafa Y. Al-Mandil
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CE 201 - STATICS LECTURE28 1
59b
Dr. Mustafa Y. Al MandilDepartment of
Civil EngineeringEXAMPLE
Take whole Frame :
kN
Exx
kNGB
GB
y
kNHx
Hx
J
120R0F
213.6F
040F8.54
3115
0F
80R
015R1640-16115
0M +
Analyze the Frame
RHx
H
G
6m
9m
115kN
FGB
JREx
40kN
16m E
3
8
CE 201 - STATICS LECTUREDr. Mustafa Y. Al-Mandil
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CE 201 - STATICS LECTURE28 2
60a
usta a a dDepartment of
Civil Engineering
Analyze the Frame
Take whole Frame :
N
AyAy
y
N
Ax
Ax
x
N
Ex
Ex
A
100R0100R
0F
5461R
05461R
0F
61.54R0810013R
0M +
Take Pulley as F.B.D. :
100N
100N
100N
100N
100N
2m6m
H
E
D
C
B
2m
4m
5m
2m
REx
RAx
RAy
3m
G
A
CE 201 - STATICS LECTUREDr. Mustafa Y. Al-Mandil
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CE 201 STATICS LECTURE28 2
60b
Department ofCivil Engineering
Take GCH as FBD :
Analyze the Frame
G C H 100N
100NTGB
RCy
RCx
N
Cy
Cy
y
N
Cx
Cx
x
N
GB
GB
c
300R
0100R2505
40F
50R
0100R2505
3
0FT250T
061003T5
4
0M +