Chapter 5.ppt
-
Upload
aravind-bhashyam -
Category
Documents
-
view
217 -
download
1
Transcript of Chapter 5.ppt
![Page 1: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/1.jpg)
CIVL3310 STRUCTURAL ANALYSISProfessor CC Chang
Chapter 5: Chapter 5: Cables and ArchesCables and Arches
![Page 2: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/2.jpg)
Cables: Assumptions Cable is perfectly flexible & inextensible No resistance to shear/bending: same as
truss bar The force acting the cable is always
tangent to the cable at points along its length
Only axial force!
![Page 3: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/3.jpg)
Example 5.1 Under Concentrated Forces
Determine the tension in each segment of the cable. Also, what is the dimension h?4 unknown external reactions (Ax, Ay, Dx and Dy)3 unknown cable tensions1 geometrical unknown h8 unknowns8 equilibrium conditions
![Page 4: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/4.jpg)
Solution
mmhFinally
kNT
o
BAo
BA
7428532
906 and 853
Bat mequilibriuJoint Similarly,
..tan)(,
..
BCBA
kNTmkNmkNmTmT
M
CD
CDCD
A
796048235554253
0
.)()().)(/())(/(
kNT
TkNkNTkN
BCo
BC
BCBC
BCBC
824 and 332
0854796053796
Cat mequilibriuJoint
..
sin)/(.cos)/(.
![Page 5: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/5.jpg)
Cable subjected to a uniform distributed load
Consider this cable under distributed vertical load wo
The cable force is not a constant.
0sincos)2/)((0
ve as clockwise-antiWith
0)sin()()(sin
0
0)cos()(cos0
0
xTyTxxwM
TTxwT
F
TTTF
o
o
y
x
3eqn tandxdy
2eqn wdx
)sinT(d
1eqn 0dx
)cosT(d
o
FH
wo
cos)cos(coscos TTT
![Page 6: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/6.jpg)
Cable subjected to a uniform distributed load
From Eqn 1 and let T = FH at x = 0:
Integrating Eqn 2 realizing that Tsin = 0 at x = 0:
Eqn 5/Eqn 4:
4eqn HFttanconscosT
5eqn sin xwT o
3eqn tandxdy
2eqn wdx
)sinT(d
1eqn 0dx
)cosT(d
o
FH
6eqn tanH
o
Fxw
dxdy
FH
TTsin
![Page 7: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/7.jpg)
Cable subjected to a uniform distributed load
Performing an integration with y = 0 at x = 0 yields
7eqn 2
2xFwyH
o
6eqn tanH
o
Fxw
dxdy
FH
Cable profile:parabola
8eqn 2
2
hLwF o
H y = h at x = L
9eqn 22 xLhy
![Page 8: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/8.jpg)
Cable subjected to a uniform distributed load
Where and what is the max tension?
T is max when x=L10eqn 22 )Lw(FT oHmax
11eqn 21 2)h/L(LwT omax
FH
4eqn cos HFT
5eqn sin xwT o
8eqn 2
2
hLwF o
H
Tmax
max
22 )( xwFT oH
![Page 9: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/9.jpg)
Cable subjected to a uniform distributed load
FH
cos HFT
sin xwT o
tanH
o
Fxw
dxdy
T
2
2
hLwF o
H
22 xLhy
)2/(1 2max hLLwT o
![Page 10: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/10.jpg)
Cable subjected to a uniform distributed load
Neglect the cable weight which is uniform along the length
A cable subjected to its own weight will take the form of a catenary curve
This curve ~ parabolic for small sag-to-span ratio
axcoshay
Hangers are close and uniformly spaced
If forces in the hangers are known then the structure can be analyzed
1 degree of indeterminacy
Determinate structure
hinge
Wiki catenary
![Page 11: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/11.jpg)
Example 5.2The cable supports a girder which weighs 12kN/m. Determine the tension in the cable at points A, B & C.
12kN/m
![Page 12: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/12.jpg)
SolutionThe origin of the coordinate axes is established at point B, the lowest point on the cable where slope is zero,
Assuming point C is located x’ from B:
From B to A:
(1) 62
12kN/m2
222 xF
xF
xFw
yHHH
o
(2) '0.1'66 22 xFxF HH
FH
mxxx
xx
xFH
43.12'0900'60'
)]'30(['0.1
612
)]'30([612
2
22
2
FH
kN.4154
203890 x.
![Page 13: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/13.jpg)
Solution
FH=154.4kN
203890 x.y
x.dxdytan 07770
kN.cosFT
.
.dxdytan
A
HA
oA
.xA
4261
7953
36615717
12.43m17.57m
A
C
CT
AT
kN..cos.
cosFT
.
.dxdytan
oC
HC
oC
.xC
6214044
4154044
96604312
cos HFT
![Page 14: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/14.jpg)
Example 5.3 Determine the max tension in the cable
IHAssume the cable is parabolic(under uniformly distributed load)
75180 .AIM yyB
HyyC F.AIM 66700
kN.FH 1328
![Page 15: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/15.jpg)
Example 5.3
kN.FH 1328
2
2
hLwF o
H m/kN.wo 133
kN46.9 21 2 )h/L(LwT omax
![Page 16: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/16.jpg)
Cable and Arch
flip
What if the load direction reverses?
FH
FH
![Page 17: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/17.jpg)
Arches An arch acts as inverted cable so it receives
compression An arch must also resist bending and shear
depending upon how it is loaded & shaped
![Page 18: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/18.jpg)
Arches Types of arches
indeterminate
indeterminate
indeterminatedeterminate
![Page 19: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/19.jpg)
Three-Hinged Arch
Ax
AyBy
Bx
![Page 20: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/20.jpg)
Problem 5-30Determine reactions at A and C and the
cable force3 global Eqs1 hinge condition
Ay
Ax
Cy
Ay
Ax
By
Bx
T
![Page 21: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/21.jpg)
Example 5.4The three-hinged arch bridge has a parabolic shape and supports the uniform load. Assume the load is uniformly transmitted to the arch ribs.Show that the parabolic arch is subjected only to axial compression at an intermediate point such as point D.
![Page 22: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/22.jpg)
Solution
=160 kN=160 kN
=160 kN
=160 kN10 m
=160 kN
=160 kN
160 kN =
0 =
2
22010 xy
x
y
o10
2
626
5020
20
.
.x)(dx
dytan
D
mxD
00
9178
D
D
D
MV
kN.N
8 kN/m
![Page 23: Chapter 5.ppt](https://reader036.fdocuments.us/reader036/viewer/2022062520/55cf91ee550346f57b91e548/html5/thumbnails/23.jpg)
Reflection: What Have You Learnt?