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The load carried by each cable of the suspension is 2500 Ib per foot of horizon- bridge shown in Fig. tal length. The span of the bridge is 1500 ft, and the sag at midspan is 150 ft. Determine The tension in the cable at the middle of the span. 150 ft -750 ft- -1500 ft -750 ft-

The load carried by each cable of the suspension is 2500 Ib per foot of horizon- bridge shown in Fig. tal length. The span of the bridge is 1500 ft, and the sag at midspan is 150 ft. Determine The tension in the cable at the middle of the span. 150 ft -750 ft- -1500 ft -750 ft-

Structural Analysis
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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The load carried by each cable of the suspension is 2500 Ib per foot of horizon- bridge shown in Fig. tal length. The span of the bridge is 1500 ft, and the sag at midspan is 150 ft. Determine The tension in the cable at the middle of the span. 150 ft -750 ft- -1500 ft -750 ft-
Transcribed Image Text:The load carried by each cable of the suspension is 2500 Ib per foot of horizon- bridge shown in Fig. tal length. The span of the bridge is 1500 ft, and the sag at midspan is 150 ft. Determine The tension in the cable at the middle of the span. 150 ft -750 ft- -1500 ft -750 ft-
Expert Solution
Step 1

The tension at mid portion will be equal to the horizontal reaction H.

For uniformly distributed load the horizontal reaction is given by:

H=wl28h

Where, 

w = Load on the structure

l = span

h = sag

Step 2

Taking equilibrium condition,

∑V=0

VA+VB=2500×3000VA+VB=7500000 lb

Since it is a symmetrical structure,

Therefore,

VA=VB=3750000 lb

The tension at mid portion of the cable = 

H=wl28h=2500×150028×150=4687500 lb

This Horizontal reaction is equal to Tension at mid span

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