Subject
Year
: S0695 / STATICS
: 2008
Cables & Arches
Session 11 - 12
4.1. Cables
high strength
steel wires are completely flexible
Cables constructed of
with high
Bina Nusantara
tensile strength
Sources :
Daniel L Schodek. (2004). Structures. 5th Ed. Pearson Prentice Hall. New Jersey.
4.1. Cables
Engineer use cables to construct long span
structures, suspension bridges and roof ( large
area e.g arena, convention hall etc )
Bina Nusantara
4.1. Cables
Bina Nusantara
Sources :
Daniel L Schodek. (2004). Structures. 5th Ed. Pearson Prentice Hall. New Jersey.
4.1. Cables
Variation in cable force
:
If cable support vertical load only…..
Horizontal forces H of cable tension T is constant at all
section along the axis of the cable.
Va
Ta
Σ Fx = 0 Æ to a segment of cable
Hb
Tb
Bina Nusantara
αb
a
αa
w
b
Vb
Ha
4.1. Cables
Variation in cable force
:
If cable tension is expressed in Horizontal forces H and the
cable slope α Æ
H = T cos α
Since : cos α = 1 Æ
H=T
The
maximum value of T -Æ where the cable
slope is largest
Bina Nusantara
4.2. Arches
Arch construction uses material
effectively
becauce….. Applied load create mostly
axial
compression on all cross section
Bina Nusantara
4.2. Arches
Designer uses arches as the main structural
elements
in long span bridge & large coulumnfree areas ( airplane Hangers, convention hall,
etc)
Bina Nusantara
4.2. Arches
Types of Arches :
1. Three Hinged Arch :
Hinged 3rd
S
A
Hinged 1st
( Supported )
Bina Nusantara
B
Hinged 2nd
( Supported )
4.2. Arches
Types of Arches :
2. Two Hinged Arch :
indeterminate structure 1st degree
A
Hinged 1st
( Supported )
Bina Nusantara
B
Hinged 2nd
( Supported )
4.2. Arches
Types of Arches :
3. Fixed-end Arch :
indeterminate structure 3rd degree
Fixed-end 1st
( Supported )
Bina Nusantara
Fixed-end 2nd
( Supported )
4.3. Structural Analysis for Cables & Arches
4.3.1. Structural Analysis for Cables
Problem :
l1 = l2 = 3.00 m
l3 = 4.00 m
h1 = 3.00 m
A
B
α1
h2
h1
α3
C
α2
D
P
P
l1
Bina Nusantara
l2
l3
4.3. Structural Analysis for Cables & Arches
4.3.1. Structural Analysis for Cables
TA
H
A
TB
B
α1
h2
h1
α3
C
P1
VA
l1
Bina Nusantara
α2
P1
l2
D
VB
l3
H
4.3. Structural Analysis for Cables & Arches
4.3.1. Structural Analysis for Cables
Step 1 :
Compute force Vertical Reactions Æ VA , VB
TA
H
A
TB
B
α1
H
h2
h1
α3
C
α2
VA
P
l1
Bina Nusantara
P
l2
D
VB
l3
4.3. Structural Analysis for Cables & Arches
4.3.1. Structural Analysis for Cables
TA
Step 1 :
Compute
Vertical Reactions
Æ VA , VB
H
A
TB
3.0
C
α3
α2
VA
ΣMB = 0
-P.3m –P.7m + VA.10m = 0
10 P = 10 VA
Bina Nusantara
H
h2
P
3.0
VA = +P
B
α1
P
4.0
D
VB
3.0
ΣMA = 0
P.3m P.7m - VB.10m = 0
10 P = 10 VB
VB = +P
4.3. Structural Analysis for Cables & Arches
4.3.1. Structural Analysis for Cables
TA
Step 2 :
Compute
Horizontal Reaction
ÆH
H
ΣMC = 0
VA.3m + H. 3m= 0
3P=3H
H = +3 P
Bina Nusantara
A
TB
B
α1
H
h2
3.0
C
α3
α2
VA
P
3.0
P
4.0
D
VB
3.0
ΣMD = 0
- VB.3m –H.h2 = 0
3 m = 3 h2
h2 = 1.00 m
4.3. Structural Analysis for Cables & Arches
4.3.1. Structural Analysis for Cables
Step 3 :
Compute The cable’s forces Æ TAC & TDC
& TBD
TBD
TAC
B
A
H
h2 = 1m
α1
h2 = 1m
3.0
3.0
3.0
tan α1 = 3/3 = 1
H =TAC Cos α1
TAC = +3 P
TDC
C
VA
Bina Nusantara
D
C
TDC
D
α3
3.0
α2
4.0
TAC
tan α2 = (3-1)/4 = ½
H =TDC Cos α2
TDC = +1 ½ 5 P
TBD
VB
tan α3 = 1/3
H =TBD Cos α3
TBD = + 10 P
H
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
q[t/m]
S
½L
A
B
L
Bina Nusantara
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
1 [t/m]
S
½L
HA
A
B
10
VA
Bina Nusantara
VB
HB
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
Step 1 :
1 [t/m]
Equilibrium Eq. at A
Æ Eq.(1)
ΣMA = 0
(1 t/m.10).5m –VB.10m+ HB.0 = 0
50 t = 10 VB
HA
S
½L
A
VB = 5 t
10
VA
Bina Nusantara
B
VB
HB
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
Step 2:
1 [t/m]
Equilibrium Eq. at A
Æ Eq.(2)
ΣMB = 0
-(1 t/m.10).5m –VA.10m +HA.0 = 0
50 t = 10 VA
HA
S
½L
A
VA = 5 t
10
VA
Bina Nusantara
B
VB
HB
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
Step 3 :
1 [t/m]
Equilibrium Eq. at SR
Æ Eq.(3)
ΣMSR= 0
-(1 t/m.10).5m –VA.10m = 0
50 t = 10 VA
HA
S
½L
A
VA = 5 t
10
VA
Bina Nusantara
B
VB
HB
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
Step 4 :
1 [t/m]
Equilibrium Eq. at SR
Æ Eq.(3)
ΣMSR= 0
(1 t/m.5).2.5m –VA.5m + HA.5 = 0
-12.5 t + HA.5 = 0
5 t = 12.5 HA
SR
½L
HA
A
5
HA = 2.5 t
Bina Nusantara
VA
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
1 [t/m]
Step 5:
Equilibrium Eq. at SL
Æ Eq.(4)
ΣMSL= 0
(1 t/m.5).2.5m –VB.5m + HB.5 = 0
-12.5 t + HB.5 = 0
5 t = 12.5 HB
SL
½L
B
5
VB
HB = 5 t
Bina Nusantara
HB
4.3. Structural Analysis for Cables & Arches
4.3.2. Structural Analysis for Arches
Step 6:
1 [t/m]
Check all the
Equilibrium Eq.
ΣV = 0
VA + VB = 1 t/m . 10 m
5 t + 5 t = 10 t
10 t = 10 t
ΣH= 0
HA = HB
2.5 t = 2.5 t
Bina Nusantara
S
½L
HA =2.5 t
A
B
HB =2.5 t
10
OK !!!
VA = 5 t
VB = 5 t
SR
1 [t/m]
1 [t/m]
SR
SL
½L
HA
½L
A
B
5
VA
Bina Nusantara
5
VB
HB