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Hyperbolic Cosine help

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#1 3point5



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Posted 14 November 2011 - 07:33 PM

Hello smart people!

I ran into a issue in my study - I am using the CERM (11th Ed.) and I am having a serious issue determining the Hyperbolic Cosine, cosh and the Constant, c. Ref Chapter 41-19 in determining tension in catenary cables.

My question roots from my practice problems CERM Practice Problems (also the 11th Ed.)

How do you determine the Cosh - as in Practice Problem 1

"Two towers are located on level ground 100ft apart. They support a transmission line with a mass of 2 lbs/ft. The midpoint Sag is 10 ft."

a. What is the midpoint tension?
b. What is the max tension in the transmission line?

I understand the formula they use but am having issues with find the Constant c and Hyperbolic Cosine, cosh

thank you in advance!

Edited by 3point5, 14 November 2011 - 07:34 PM.

#2 Jayman_PE


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Posted 19 November 2011 - 09:33 PM

cosh(x) = .5[e^(x) + e^(-x)]

#3 palvarez83


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Posted 15 December 2011 - 04:39 PM

I have CERM 12. Look at equation 41.68: S =YD - c .... therefore c = 100ft - 10 ft = 90 ft. Read you calculator manual or just google it for your calculator as to how to enter hyperbolic cosine. I have both the casio and hp that are allowed for the test and both are capable of doing it, but as jayman mentioned : cosh(x) = .5[e^(x) + e^(-x)]

#4 DM79



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Posted 15 January 2012 - 08:30 PM


can you please give me the full name and specifications of the calculator you are using?

i am just getting started to prepare for my Civil Structural exam.


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