# Homework Help: Cable tension from UDL

1. Nov 29, 2008

### Els27

1. The problem statement, all variables and given/known data
Ok guys my first ever post so be nice!

Im doing an experiment using a wind tunnel looking at the effects of vortices on cables. Basically the experiment willconsist of 2 vertical cantilever beams supporting a cable between them (So if you like it looks like a portal frame only the top beam is a cable). The cable's longitudinal axis is perpendicular to the flow. The flow creates a lifting force in the form of a UDL with respect to time. The cable is treated as pinned at each end. I am measuring vibration of the cable by relating the strain in the 2 supports to the amplitude of vibration.

Thats the setup, now the actual problem is determining the model dimensions so that i get a givenstrain value. I am aiming for 0.0001 strain (thats what ive been told to go for given the strain guage accuracy)

2. Relevant equations

I have 3 equations for the setup one for the UDL force of the flow on the cable, one for the tension force that UDL puts on the cable (which is transmitted to the supports) and finally one that governs the strain in the pole given a ceratin point load at the top(from tension in cable).

Force of flow per meter

0.5*p*d*u^2*Cl*sin(ws*t)
where p=air density =1.2
d=cable diameter =say 1mm
u=free stream velocity= up to 28m/s
Cl=coefficient of lift= call 1.0
ws = shedding frequency= not needed for now
and t= time in s

Strain in pole(cylindrical support) from stress = moment *y/ moment of Inertia

e=32*P*(L-z)/E*pi*D^3
where e=strain=0.0001
P=point load at top of beam
L= length of pole
L-z= lever arm point load has to where strain guage is
E=youngs modulus
pi=3.14....
D=diameter of support

finally the equation i currently have (but have beef with) for tension at supports from cable with UDL

w*a^2/2h
where w=UDL force=equation above from flow
a= distance max amplitude of cable = half cable length =0.15m
h = amplitude of displacment

3. The attempt at a solution

Ok this is a lengthy thread i know sorry but need help big time!!
So given my equation for strain, i know what strain i need and ive assumed a support diameter of 5mm and steel modulus 200e^6. This gives a point load of 9.82e^-4 N. Given other similar shapes of supports gives point load to the order of e^-4.

So i know what force at the top of the cantilever i need (~9.82e^-4) to give a strain of 0.0001. Now i need the tensile force in the cable that will give me the point load at my supports. The flow force equation is dynamic and all i am interested in is the amplitude (i.e. not the sin(ws*t) bit right?) This gives a max UDL on the cable at some point in time doesnt matter what time thats the force im interested in =0.5*p*d*u^2*Cl. Now as Cl and p are constant (1.0 and 1.2 respectively) and given the max flow speed 28 m/s the only variable is 'd' and the equation is now 470.4*d. Lets say for now a 1mm cable is used (its gonna be around 0.5mm to 1.5mm i think) that gives a UDL of 0.47 N/m.

So now i have a simply supported cable with UDL across entire length. So now what is the tension in the cable and hence the point load at supports? This is where my third equation and major problems come in. My thoughts at this point are given the applied load (the UDL) the cable will deflect i think parabolically is the word but it will remain almost stationary at supports and deflect most in centre. This will be resisted by the stiffness of the cable. the force at one of the pinned end is

P= w*a^2/2*h (this is taken from course notes so i assume they are right)

But one would assume as you increase the deflection 'h' the material would be stretching more and increasing in tension. However the equation says otherwise. (Panic time! and no i dont have a panic button on my keyboard...) I know what P i want to achieve through my strain equation (9.82e^-4)All i need is h the deflection at the position of max amplitude (the centre). I have tried simplifying the string into a 'single mass on a spring' system which gives a max deflection of 0.0018m or 1.8mm .but when plugging the numbers in i get 5.875N and i dont believe thats the case at all! i cant believe that wind flow is causing that much tension in the wire

Does anyone know if my tension form UDL equation is correct????
Can someone make sense of this and tell me where ive gone wrong / give me some pointers please????

All help much appreciated
cheers

Elliot