Answer this problem using the given guide:PROBLEM: An open tank in the shape of a frustum of a cone is bound by two hoops, one at the top and the other at the bottom. The diameters at the top and bottom are 2m and 3m, respectively. If the total bursting force is to be taken care of by the hoops, find tension in each hoop if water stands 2 m in the tank. Height of tank is 3.0 m.Use this guide to answer the problem and do not copy the given values on this given guide. Use this as a basis. A vertical cylindrical container, 1.60 m in diameter and 4m high, is held togetherby means of hoops, one at the top and the other at the bottom. A liquid of specific gravity1.40 stands 3m in the container. Calculate the tension in each hoop.1.60 m2T,A1.0 m3.0 m4.0 mhF 1.50Y, = 2.0s.g.= 1.403.0 me1.0 m2T,BSolution:F = Awh = (1.6)(3)(9.81)(1.40)(1.50) = 98.885 kN1/12(1.6)(3)Yp = ỹ +!Ig= 2.0 m= 1.5 +A(1.6) (3)(1.5)Location of F from A = 1+ 2 = 3.0.m{FH = 0EMĄ = 02T, + 2T, = F2T, = F - 2T,= 98.885 – 2(37.082)2T, (4) – F (3) =08T, = 98.885 (3)T, = 12.360 kNT2 = 37.082 kN

OVT ×DR ×HUSO = 1×122×H Take D = 12m H = 3.4724 m design of top dome w to W×R cos y×8.7 if cos…

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Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Answer this problem using the given guide:

PROBLEM: An open tank in the shape of a frustum of a cone is bound by two hoops, one at the top and the other at the bottom. The diameters at the top and bottom are 2m and 3m, respectively. If the total bursting force is to be taken care of by the hoops, find tension in each hoop if water stands 2 m in the tank. Height of tank is 3.0 m.

Use this guide to answer the problem and do not copy the given values on this given guide. Use this as a basis.

A vertical cylindrical container, 1.60 m in diameter and 4m high, is held together
by means of hoops, one at the top and the other at the bottom. A liquid of specific gravity
1.40 stands 3m in the container. Calculate the tension in each hoop.
1.60 m
2T,
A
1.0 m
3.0 m
4.0 m
hF 1.50
Y, = 2.0
s.g.= 1.40
3.0 m
e
1.0 m
2T,
B
Solution:
F = Awh = (1.6)(3)(9.81)(1.40)(1.50) = 98.885 kN
1/12(1.6)(3)
Yp = ỹ +!
Ig
= 2.0 m
= 1.5 +
A
(1.6) (3)(1.5)
Location of F from A = 1+ 2 = 3.0.m
{FH = 0
EMĄ = 0
2T, + 2T, = F
2T, = F - 2T,
= 98.885 – 2(37.082)
2T, (4) – F (3) =0
8T, = 98.885 (3)
T, = 12.360 kN
T2 = 37.082 kN

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