Post on 18-Mar-2020
1
3. Structures
2142111 S i 2011/22142111 Statics, 2011/2
© Department of Mechanical Engineering Chulalongkorn UniversityEngineering, Chulalongkorn University
2
Objectives Students must be able to #1Objectives Students must be able to #1
Course ObjectiveAnalyze structures (trusses and frames/machines) in equilibrium Analyze structures (trusses and frames/machines) in equilibrium
Chapter Objectives Differentiate connecting bodies in equilibrium into
frames/machines and trusses For 2D/3D frames/machines State appropriate action/reactions between joints/members and State appropriate action/reactions between joints/members and
disassemble the structure by drawing FBDs of members and important jointsId tif t t ith t diti l i th ff t f Identify structures with symmetry condition, explain the effects of symmetry and identify the equal-values and zero-components due to symmetry
2
Analyze structures for unknown loads/reactions by appropriate FBDs
3
Objectives Students must be able to #2Objectives Students must be able to #2
Chapter ObjectivesF 2D/3D tFor 2D/3D trusses Describe characteristics and approximation justification for
members as 2-force members Analyze for forces in members by the method of joints and
method of sections with appropriate FBDs Explain the 2 causes of zero-force members and determine Explain the 2 causes of zero-force members and determine
zero-force members in trusses Identify trusses with the symmetry condition and explain the
ff t f t th bleffects of symmetry on the problem
3
4
ContentsContents Analyses of Frames and Machines
Analyses of Trusses Truss Method of Joints Method of Sections
4
5
Equilibrium ObjectsEquilibrium Objects
So far, you analyze simple rigid bodies Draw FBD Draw FBD
Apply equations of equilibrium
What’s you are going to do with complex systems Draw FBD of each members
I l t d b Isolated member Identify two and three-force members Add external and support loads Add external and support loads
Apply equations of equilibrium For two force members, also determine the internal
f
5
forces.
6
Equilibrium StructuresEquilibrium Structures
Members are interconnected parts in structures.
Structures are any connecting system of members that is built to support or transfer forces and safely withstand the applied loads.
We will study statically determinate We will study statically determinate Frames & Machines Trusses
6
7
Frames and MachinesF&M
Frames and Machines Consider structures of
interconnected members that do not satisfy the definition of a truss.
Frames remain stationary and support l dloads.
Machines are designedMachines are designed to move and apply loads.
7
8
Frames & Machines Procedures
F&M
Frames & Machines Procedures
Analyze the entire structure and find as many reaction forces at the supports as you canforces at the supports as you can.
Analyze individual members Disassemble structure and isolate members Identify two-force members
Identify important joints which connect three or more Identify important joints which connect three or more members/supports
Apply action/reaction forces on any two contacting members
Reassemble members to check for errors. All internal and action/reaction forces must cancel out.
8
9
Example F & M 1 #1
F&M
Example F & M 1 #1
Determine the magnitudes of horizontal and vertical components pof force which the pin at B exerts on member CB.
9
10
Example F & M 1 #2
F&M
p3 equations, 4 unknowns
Do you really unable to find reactions?
= − = 0 0x x xF A C = + − = = + °
0 2000 N 0
0 (2 m)(2000 N) (3sin60 m)
x x x
y y y
C
F A C
M A
10
+ − + ° =
0 (2 m)(2000 N) (3sin60 m)
(4 m 3cos60 m) 0C x
y
M A
A
11
Example F & M 1 #3
F&MApproach 1
Example F & M 1 #3
= Consider FBD of
0C
BC
M
− =
=
(2 m)(2000 N) 4 0
1000 N Ansy
y
B
B
= − + =
0
(2000 N) 0y
y y
F
B C
=
= 1000 N
0y
x
C
F
11
− = 0 (1)x xB C
12
Example F & M 1 #4
F&MApproach 1
Example F & M 1 #4
Consider FBD of AB = − =
=
Consider FBD of
0 0
1000 Ny y y
y
AB
F A B
A
= ° − ° =
=
000
0 (3sin60 m) (3sin60 m) 0
1000 / 3 N
y
A x yM B B
B ==
= − =
1000 / 3 N577 N Ans
0 0
x
x
BB
F A B
=
=
0 0
1000 / 3 N
From (1) 1000 / 3 N
x x x
x
F A B
A
C
12
=From (1), 1000 / 3 NxC
13
Example F & M 1 #5
F&MApproach 2
Example F & M 1 #5
is a two-force member.AB
= − ° = Consider FBD of
0 (2 m)(2000 N) (4 m) sin60 0C
BC
M AB
=
°
2000 / 3 N is the force that pin exerts on ,
Horizontal component cos60 577 N
ABAB B BC
AB
13
= ° == ° =
Horizontal component cos60 577 NVertical component sin60 1000 N Ans
ABAB
14
Example F & M 2 #1
F&M
Example F & M 2 #1
Disassembling the structure
14
15
Example F & M 2 #2
F&M
Example F & M 2 #2
Disassembling the structure
15
16
Example F & M 2 #3
F&M
Example F & M 2 #3
Isolate the the member ACED
16
17
Example F & M 2 #4
F&M
Example F & M 2 #4
Disassembling the structure
17
18
Example F & M 2 #5
F&M
Example F & M 2 #5
Disassembling the structure
18
19
Example Hibbeler Ex 6-10 #1
F&M
Example Hibbeler Ex 6 10 #1
A constant tension in theA constant tension in the conveyor belt is maintained by using the device shown.Draw the free-bodyDraw the free-body diagrams of the frame and the cylinder which supports the belt The suspendedthe belt. The suspended block has a weight of W.
19
20
Example Hibbeler Ex 6-10 #2
F&M
Example Hibbeler Ex 6 10 #2
20
21
Example F & M 4 #1
F&M
Example F & M 4 #1
The mechanism is used to weigh mail A package placed at A causesmail. A package placed at A causes the weighted pointer to rotate through an angle a. Neglect the weights of the members except for themembers except for the counterweight at B, which has a mass of 4 kg. If α = 20°, what is the mass of the package at A?of the package at A?
21
22
Example F & M 4 #2
F&M
Example F & M 4 #2
22
23
Example F & M 4 #3
F&M
Example F & M 4 #3
Given 20α = °Consider FBD of
0 0E EC x
AEC
M r C = = 0
0 cos20 0x
x x
C
F EF C
=
= − ° − = 0
0 0y y
EF
F W C
=
= − − =
23
(1)yC W= −
24
Example F & M 4 #4
F&M
Example F & M 4 #4
Consider FBD of BDC =
° + ° =
Consider FBD of
0
(0.1 m)sin20 (4 N) (0.1 m)cos10 0D
y
BDC
M
g C
° ⋅= −°
( ) ( ) ( )
0.4sin20 N m0.1cos10 m
y
y
g
gC
° ⋅= =°
Substitute (1)0.4sin20 N m
0 1cos10 mgW mg
°= =°
0.1cos10 m0.4sin20 1.3892 kg0.1cos10
m
24
Mass of is 1.39 kg Ansm A
25
Example Hibbeler Ex 6-15 #1
F&M
Example Hibbeler Ex 6 15 #1
Determine the horizontal and vertical components of force which the pin at C exerts on member ABCD of theforce which the pin at C exerts on member ABCD of the frame shown.
FBD
25
FBD
26
Example Hibbeler Ex 6-15 #2
F&M
p
Whole frame in equilibrium = +
− + = 0
( 981 N)(2 m) (2.8 m) 0A
x
M
D
=
= → + 700.71 N
0 x
x
D
F
− ==
↑
0700.71 N
x x
x
A DAFBD
= ↑ + − =
0
981 N 0y
y
F
A
26
= 981 NyA
27
Example Hibbeler Ex 6-15 #3
F&M
Example Hibbeler Ex 6 15 #3
Apply equilibrium equations for member CEF = +
−
for member
0
( 981 N)(2 m)C
CEF
M
− ° == −
( 981 N)(2 m)( sin45 )(1.6 m) 0
1734.2 NB
B
FF
= → + − − ° = 0
cos 45 0x
x B
F
C F= − = −
= ↑ + 1226.3 N 1.23 kN Ans
0 x
y
C
F
27
− ° − =
= − = −
sin45 981 N 0
245.26 N 245y B
y
C F
C N Ans
28
Example Hibbeler Ex 6-17 #1
F&M
Example Hibbeler Ex 6 17 #1
The smooth disk shown is pinned at D and has a weight of 20 lbThe smooth disk shown is pinned at D and has a weight of 20 lb. Neglecting the weights of the other member, determine the horizontal and vertical components of reaction of pins B and D.
28
FBD
29
Example Hibbeler Ex 6-17 #2
F&M
p
Equilibrium of entire frame =
− + =q
0
( 20 lb)(3 ft) (3.5 ft) 0A
x
M
C=
= 17.143 lb
0x
x
C
F
− ==
0 17.143 lb
x x
x
A CA
= − =
0
20 lb 0 y
y
F
A
29
= 20 lbyA
30
Example Hibbeler Ex 6-17 #3
F&M
p
30
31
Example Hibbeler Ex 6-17 #4
F&M
Example Hibbeler Ex 6 17 #4
Equilibrium of member AB =
− + =
Equilibrium of member
0
( )(6 ft) (3 ft) 0B
D
AB
M
A N+
=
=
( )(6 ft) (3 ft) 0
40 lb
0
y D
D
x
A N
N
F − == =
0
17.143 17.1 lb Ans
x
x x
x
A BB
= − + =
0
0y
y d y
F
A N B
31
= 20 lb Ansy y
yB
32
Example Hibbeler Ex 6-17 #5
F&M
p
E ilib i f di k D = → + Equilibrium of disk
0
0 Ax
D
F
D =
= ↑ + 0 Ans
0
20 lb 0
x
y
D
F
N D− − =
− − =
=
20 lb 0
40 lb 20 lb 0
20 lb Ans
D y
y
N D
D
D = 20 lb AnsyD
32
33
Example Hibbeler Ex 6-17 #6F&M
p
33
34
Example Hibbeler Ex 6-13 #1
F&M
Example Hibbeler Ex 6 13 #1
Draw the free-body diagrams of the bucket and the vertical boom f th b k h h i th h t Th b k t d it t tof the back hoe shown in the photo. The bucket and its contents
have a weight W. Neglect the weight of the members.
34
35
Example Hibbeler Ex 6-13 #2
F&M
Example Hibbeler Ex 6 13 #2
35
36
Example F & M 3 #1
F&M
Example F & M 3 #1
For P = 150 N squeeze on the handles of the pliers, determine the force F applied by each jawdetermine the force F applied by each jaw.
36
37
F & M Example3 #2
F&M
& p
Due to horizontal symmetry of foces acting on
37=
Due to horizontal symmetry of foces acting onon jaws and handles: 0xC Unit in mm
38
Example F & M 3 #3
F&M
Example F & M 3 #3
= FBD of the upper handle
0M = − =
=
0
(0.03 m) (0.18 m) 0
6 (1)
C
y
M
B P
B P=
=
6 (1)
FBD of the upper jaw
0
y
A
B P
M
Unit in mm
− =
= =
0
(0.06 m) (0.02 m) 0
3 18
A
y
y
M
B F
F B P
38
=
3 8
2.7 kN Ansy
F
39
Example Hibbeler SI Ex 6-13 #1
F&M
Example Hibbeler SI Ex 6 13 #1
Disassembling the structure.
39
40
Example Hibbeler SI Ex 6-13 #2
F&M
Example Hibbeler SI Ex 6 13 #2
40
41
Example Meriam Ex 47 #1
F&M
Example Meriam Ex 47 #1
Disassembling the structure.
41
42
Example Meriam Ex 47 #2
F&M
Example Meriam Ex 47 #2
42
43
Example Meriam Ex 47 #3
F&M
Example Meriam Ex 47 #3
43
44
Example Hibbeler Ex 6-20 #1
F&M
Example Hibbeler Ex 6 20 #1
Disassembling the structure.
44
45
Example Hibbeler Ex 6-20 #2
F&M
Example Hibbeler Ex 6 20 #2
45
46
Example Hibbeler Ex 6-20 #3
F&M
p
= + − = → = FBD of lever
0 (1 in) (8 lb)(4 in) 0 32 lbB EA EA
ABG
M F F
= ↑ + ° − ° = → = FBD of pin
0 sin60 sin60 0 y ED EG ED EG
E
F F F F F
46
= → + ° + ° − = → = 0 cos60 cos60 0 32 lbx ED EG EA EDF F F F F
47
Example Hibbeler Ex 6-20 #4
F&M
p
FBD of arm DC = +
− ° =
FBD of arm
0
(6 in) + cos30 (3 in) 0C
DC
M
F F
==(6 in) + cos30 (3 in) 013.9 lb
S ED
S
F FF
47
48
Example Bedford 6.114 #1
F&M
Example Bedford 6.114 #1
Disassembling the structure.
48
49
Truss Definition
Truss
Truss Definition
A framework composed of bars joined at their ends bybars joined at their ends by smooth pins to form a rigid structure.
Trusses are supported and loaded at their joints.
If we neglect the weights of the bars each bar is a two-force
River Kwai Bridgebars, each bar is a two-force member.
49
50
Truss Joints
Truss
Truss Joints
50
51
Truss Type #1
Truss
Truss Type #1
51
52
Truss Type #2
Truss
Truss Type #2
52
53
Truss Type #3
Truss
Truss Type #3
Howe Bridge Truss
K Bridge Truss
53Subdivided Warren Bridge Trusswww.nasa.gov
54
Truss Categories
Truss
Truss Categories
Plane Truss – 2D
Space Truss – 3D
54
55
Plane Truss Rigid Frames
Truss
Plane Truss Rigid Frames
Rigid frames do not collapse and have negligible deformationdeformation.
55
56
Plane Truss Simple Trusses
Truss
Plane Truss Simple Trusses
The basic element is a triangle.Rigid simple trusses are built by adding units of two end Rigid simple trusses are built by adding units of two end-connected bars.
56= −SD plane truss: 2 3m j
57
Plane Truss Stability
Truss
Plane Truss Stability
> −2 3m j SI truss with redundant member(s).
= −2 3m j SD plane truss
< −2 3m j Truss collapses under load.
SI with redundant supports
SI with improper supports
57
p p pp
58
Plane Truss Solving Procedures
Truss
Plane Truss Solving Procedures
Main AnalysesMethod of joints disassembling all joints and Method of joints – disassembling all joints and members
Methods of sections – sectioning truss as needed
Helping HandsZero force members identifying members that Zero-force members – identifying members that do not support loads
58
59
Method of Joints Procedure
Truss
For SD trussesIsolate the truss draw FBD and find the support Isolate the truss, draw FBD and find the support reactions
Isolate each joint, draw FBD and find forces of members on joints
59
60
Method of Joints Demo #1
Truss
Draw FBD of the entire truss
Method of Joints Demo #1
Determine the axial forces in each member Draw FBD of the entire truss
60
61
Method of Joints Demo #2
Truss
Method of Joints Demo #2
Determine the reactions at its supports.
= + = 0 500 N 0x xF A = −
= − + = 500 N
0 (2 m)(500 N) 2 0x
A y
A
M B =
= + = 500 N
0 0y
y y y
B
F A B = −
500 N
y y y
yA
61
62
Truss
FBDs of members and joints Select order of joint analyses
62
63
Method of Joints Demo #4
Truss
Method of Joints Demo #4
Draw the FBD of a joint having at least one known force and at most two unknown forcesforce and at most two unknown forces.
Orient x and y axes such that the forces on the FBD can be easily resolved into their x and y components. S l f thSolve for them.
= − = At joint
0 500 N 0x
A
F AB =
= − =
500 N
0 500 N 0
x
y
AB
F AC
63
=
500 N
y
AC
64
Method of Joints Demo #5
Truss
Method of Joints Demo #5
Continue with the rest of joints using the same procedureprocedure.
At joint B = ° + =
= −t jo t
0 sin45 500 N 0
707.1 NyF BC
BC
Axial forces in members are==
member : 500 N : 500 N : 707 N Ans
AB AB TAC AC TBC BC C
64
= : 707 N AnsBC BC C
65
Method of Joints Tips & Hints
Truss
p
Procedural tips to avoid confusionp Assign tension on joints and members in the original FBD
Check for their correct senses later + = tension, pulling on joints and members + tension, pulling on joints and members − = compression, pushing on the joints & members
65
66
Method of Joints WWW Interactive
Truss
Bridge Designer @ Johns Hopkins UniversitySmall package Small package
http://www.jhu.edu/~virtlab/bridge/truss.htm FRAME3DD @ Duke University @ y
http://frame3dd.sourceforge.net/
D t F Doctor Frame http://www.drsoftware-home.com/index.html
66
67
Example Method of Joints 1 #1
Truss
Example Method of Joints 1 #1
Determine maximum axial forces in each member when 0 ≤ θ ≤ 90°member when 0 ≤ θ ≤ 90 .
67
68
Example Method of Joints 1 #2
Truss
p
θ
= + =
0
(10 kN)sin 0xF
A θθ
+ == −
=
(10 kN)sin 010sin kN
0
x
x
A
AA
M
θθ
− − =
=
(4 m) (8 m)(10 kN)cos 0
20cos kN
A
y
y
C
C
θ
= + − =
0
(10 kN)cos 0
y
y
y y
F
A CWhich joints should be chosen?
And in what order?
68
θ= −
( )10cos kN
y y
yAAnd in what order?
69
Example Method of Joints 1 #3
Truss
p
α −= 1At joint , tan (3 / 4)A
α θ
= − =
j , ( )
0
sin (10 kN)cos (10 kN) 0yF
ABθ=
=
( ) ( )(50 cos ) / 3 kN
0x
AB
F
α θθ θ
+ − == −
cos (10 kN) sin 010 sin (40 cos ) / 3 kN
AC ABAC
69
70
Example Method of Joints 1 #4
Truss
p
At joint C
θθ
= + =
0
20cos kN 020 kN
yF
BCBC θ= −
= 20cos kN
0
0x
BC
F
CD ACθ θ
− == −
010 sin (40 cos ) / 3 kN
CD ACCD
70
71
Example Method of Joints 1 #5
Truss
Example Method of Joints 1 #5
−1At joint tan (3 / 4)D α
α θ
=
=
1At joint , tan (3 / 4)
0
sin (10 kN)cos 0y
D
F
BD α θθ
− ==
sin (10 kN)cos 0(50 cos ) / 3 kN
BDBD
When you get used to the procedure
71
When you get used to the procedure, you can draw only FBDs of joints.
72
Example Method of Joints 1 #6
Truss
Example Method of Joints 1 #6
θ= =Magnitude of forces in members are
50 cos 3 kNAB BD θθ θ
θ
= == = −= −
50 cos 3 kN10 sin 40 cos 3 kN
20cos kN
AB BDAC CDBC
72
73
Example Method of Joints 1 #7
Truss
Example Method of Joints 1 #7
θ =Maximum magnitude of when 0.F dF dθ≤ ≤ °
= == =
max max
Thus, in the range 0 90 ,16.7 kN 13 3 kN
AB BD TAC CD C
θ
= === °
max max
max
13.3 kN 20 kN
all at 0 Ans
AC CD CBC C
73
74
Zero-Force Member identification #1
Truss
Zero Force Member identification #1
They support no load.They are used to increase stability of the truss during They are used to increase stability of the truss during construction.
They provide support if the applied load changes.
= → = 30 0yF F
74
= → = 1 20xF F F
75
Zero-Force Member identification #2
Truss
Zero Force Member identification #2
= → =
10 0
0 0xF F
F F′ = → = 20 0xF F
= → = 3 40xF F F
′
→ = → =
3 4
1 2
0
0x
x
F F F
F F F
75
76
Zero-Force Member Removing
Truss
Zero Force Member Removing
= → = →
10 0
0 0yF F
F F
76
= → = 20 0xF F
77
Example Zero-Force Member 1 #1
Truss
p
77
78
Example Zero-Force Member 1 #2
Truss
p
78
79
Example Zero-Force Member 1 #3
Truss
p
79
80
Example Zero-Force Member 2 #1
Truss
Example Zero Force Member 2 #1
80
81
Example Zero-Force Member 2 #2
Truss
Example Zero Force Member 2 #2
Howe Bridge TrussZero-force members in standard trussesstandard trusses
K Bridge Truss
81Subdivided Warren Bridge Truss
82
Method of Sections Basics #1Imaginary Sections
Truss
Method of Sections Basics #1
Imaginary cut through the considered region If the whole body is in equilibrium, its parts must also be
in equilibrium.
Built-in Support
82
83
Method of Sections Basics #2
TrussImaginary Sections
Method of Sections Basics #2
Find resultant internal load by method of section Internal loads are normal force N, shear force V, torsion T
and bending moment M
83
84
Method of Sections Concept
Truss
Method of Sections Concept
If a body is in equilibrium, any partIf a body is in equilibrium, any part of the body is also in equilibrium.
We can draw an imaginary section th h th b dthrough the body.
If bodies are two-force members, there are only normal internal forces.
84
85
Method of Sections Internal Forces
Truss
Method of Sections Internal Forces
85
86
Method of Sections Procedure #1
Truss
Method of Sections Procedure #1
86
87
Method of Sections Procedure #2
Truss
Method of Sections Procedure #2
87
88
Method of Sections Demo #1
Truss
Method of Sections Demo #1
Determine the axial forces in members BD, CD and CE.
88
89
Method of Sections Demo #2
Truss
Method of Sections Demo #2
Draw FBD of entire structurestructure.
Choose an imaginary li t “ t” tiline to “cut” or section through the members where forces are to be d t i ddetermined.
89
90
Method of Sections Demo #3
Truss
Method of Sections Demo #3
Consider FBDs of sections
90
91
Method of Sections Demo #4
Truss
Method of Sections Demo #4
Draw FBD of that part of the sectioned truss with the least number of forces on itleast number of forces on it.
A l ilib i ti t th h FBD
91
Apply equilibrium equations to the chosen FBD.
92
Method of Sections Demo #5
Truss
Method of Sections Demo #5
= + ° − =
0
( )(1 m)sin45 (10 kN)(1 m) 0DM
CE
′
=
= 10 2 kN
0y
CE
F
° − ° ==
sin45 (10 kN)sin45 010 kN
CDCD
′ = − − ° − − ° = 0
cos 45 (10 kN)cos 45 0
20 2 kN
xF
BD CD CE
BD= 10 2 kN CE T
92
= −20 2 kNBD ==
20 2 kN 10 kN Ans
BD CCD T
93
Space TrussesTruss
Space Trusses
Members are connected by joints that do not resist moments.
The basic element is a tetrahedron. Rigid simple space trusses can be built by Rigid simple space trusses can be built by
adding units of three end-connected bars. Analyze in the same way as 2D trusses.
93= −SD space truss: 3 6m j
94
Example Space Trusses 1 #1
Truss
Example Space Trusses 1 #1
A space truss is placed on a smooth floor. Joint A is supported ppby the corner where the smooth walls meet, and joint C rests against the smooth back wall.against the smooth back wall. Determine axial forces in member AC.
94
95
Example Space Trusses 1 #2
Truss
p p
Applied loadˆ ˆ ˆ2 6 kNF i j k= − − −
All smooth supportsJoint is nested in a corner
ˆ ˆ ˆ kN
A
A A i A j A k+ +
kNJoint is placed on a floor
ˆ kN
x y zA A i A j A k
B
B B j
= + +
=
kN
Joint rests against a wallˆ ˆ kN
yB B j
C
C C j C k
=
= +
95
kNy zC C j C k= +
96
Example Space Trusses 1 #3
Truss
Example Space Trusses 1 #3
= − + + + − + + − = ˆ ˆ ˆ0 ( 2) 6) ( 1) 0 (1)x y y y z zF A i A B C j A C k
= × + × + × =
0 ( ) ( ) ( ) 0 (2)A AB AC ADM r B r C r F
Solving (1) and (2)ˆ ˆ ˆ2 4 1 kNA i j k= + +
ˆ1 kN ˆ1 kN
B j
C j
=
=
96
97
Example Space Trusses 1 #4
Truss
Example Space Trusses 1 #4
97
98
Example Space Trusses 1 #5
Truss
Example Space Trusses 1 #5
At joint C
jˆˆ
ˆ ˆ ˆ ˆˆ ( 2 3 ) ( 0 55470 0 83205 )
CA CA CA CA
CB
F F n F iFF F n i k F i k
= = −
= = + = +
( 2 3 ) ( 0.55470 0.83205 )13
ˆ ˆ ˆˆ ( 2 3 1 )14
CB CB CB CB
CDCD CD CD
F F n i k F i k
FF F n i j k
= = − + = − +
= = − + +
98
14ˆ ˆ ˆ( 0.53452 0.80178 0.26726 )
CD CD CD
CD CDF F i j k= − + +
99
Example Space Trusses 1 #6
Truss
Example Space Trusses 1 #6
=
0F
+ + + =
− − − +
0
ˆ( 0.55470 0.53452 )CA CB CD y
CA CB CD
F F F C
F F F i
+ +
+ =
ˆ(0.80178 1)ˆ(0.83205 0.26726 ) 0
CD
CB CD
F j
F F k
= −Thus, 1.24722 kN 0 40062 kN
CDFF =
=0.40062 kN0.44444 kN
CB
CA
FF
99
= 444 N AnsCAF T
100
Trusses Tips & HintsTruss
p
Members All member are of two-force member types. Look for zero-force members to simplify problems
Internal forces Use tension as +ve and compression as −ve
Visualize the ‘flow’ of forces/loads through the trusses Visualize the ‘flow’ of forces/loads through the trusses Check the results
100
101
SummarySummary Frame: A structure which is designed to remain stationary
and support loadsand support loads. Machine: A structure which is designed to move and exert
loads. Truss: A structure which is comprised entirely of two-force
members.
Tension: axial forces at the ends are directed away from each othereach other.
Compression: axial forces at the ends are directed toward each other
101
102
ConceptsReview
Concepts Structures are systems of interconnecting rigid bodies and
fastening (pins joints etc ) Structured can be analyzes byfastening (pins, joints, etc.). Structured can be analyzes by disassembling the system into inter-related FBDs and considering the equilibrium of individual members.
Frames and machines are structures that are not trusses. Frames stay stationary while machines can move.
Trusses are structures which comprise of only Trusses are structures which comprise of only interconnecting 2-force members.
102