Equilibrium 1Lecture 4
Architectural Structures IENDS 231
S2008abn
ARCHITECTURAL STRUCTURES I:STATICS AND STRENGTH OF MATERIALS ENDS 231
DR. ANNE NICHOLS
SPRING 2008
four
equilibriumof a particle
lecture
Equilibrium 2Lecture 4
Architectural Structures IENDS 231
S2008abn
Equilibrium• balanced• steady• resultant of forces on a particle is 0
X
X
=
Equilibrium 3Lecture 4
Architectural Structures IENDS 231
S2008abn
Equilibrium• analytically
• Newton convinces us it will stay at rest and won’t rotate
0==∑ xx FR
0==∑ yy FR
( )0==∑MM
Equilibrium 4Lecture 4
Architectural Structures IENDS 231
S2008abn
Equilibrium• collinear force system
0==∑ xx FR 0==∑ yy FR
0=∑ −lineinF
( )0==∑MM
Equilibrium 5Lecture 4
Architectural Structures IENDS 231
S2008abn
Equilibrium• concurrent force system
0==∑ xx FR
0==∑ yy FR
( )0==∑MM
Equilibrium 6Lecture 4
Architectural Structures IENDS 231
S2008abn
Free Body Diagram• FBD (sketch)• tool to see all forces on a body or a
point including– external forces– weights– force reactions– external moments– moment reactions– internal forces
Equilibrium 7Lecture 4
Architectural Structures IENDS 231
S2008abn
Free Body Diagram• sketch FBD• resolve each force into components
– known & unknown angles– known & unknown forces
• are any forces related to other forces?• write only as many equilibrium
equations as needed
Equilibrium 8Lecture 4
Architectural Structures IENDS 231
S2008abn
Free Body Diagram• solve equations
– most times 1 unknown easily solved– plug into other equation(s)
• common to have unknowns of– force magnitudes– force angles
Equilibrium 9Lecture 4
Architectural Structures IENDS 231
S2008abn
Cables• simple• uses
– suspension bridges– roof structures– transmission lines– guy wires, etc.
• have same tension all along• can’t stand compression
Equilibrium 10Lecture 4
Architectural Structures IENDS 231
S2008abn
Cables Structures• use high-strength steel• need
– towers– anchors
• don’t want movement
Equilibrium 11Lecture 4
Architectural Structures IENDS 231
S2008abn
Cable Structures
Equilibrium 12Lecture 4
Architectural Structures IENDS 231
S2008abn
Cable Loads• straight line
between forces• with one force
– concurrent– symmetric
Equilibrium 13Lecture 4
Architectural Structures IENDS 231
S2008abn
Cable Loads• shape directly
related to thedistributed load
Equilibrium 14Lecture 4
Architectural Structures IENDS 231
S2008abn
Cable Loads• trig:
• parabolic (catenary)– distributed uniform load
θcosTTx =
θ
22 /)(4 LxLxhy −=
θsinTTy =
h
y
x L)Lh
Lh(LLtotal 4
45
322
23
81 −+=
