What is Young's modulus for this alloy of titanium?

AI Thread Summary
The discussion focuses on calculating Young's modulus for a titanium alloy wire subjected to a load. Initial calculations for stress and strain were incorrectly applied, leading to erroneous results. After clarifications, the correct approach involves using the weight of the 109 kg mass to determine stress and the original length and elongation for strain. The recalculated Young's modulus yielded a value of approximately 1.2016E11, which aligns with expected values for titanium. The importance of using the correct mass and understanding the wire's geometry in calculations was emphasized.
badwallpaper0
Messages
4
Reaction score
0

Homework Statement


A hanging wire made of an alloy of titanium with diameter 0.15 cm is initially 2.8 m long. When a 109 kg mass is hung from it, the wire stretches an amount 1.41 cm. A mole of titanium has a mass of 48 grams, and its density is 4.54 g/cm^3.

Based on these experimental measurements, what is Young's modulus for this alloy of titanium?
From the mass of one mole and the density you can find the length of the interatomic bond (diameter of one atom). This is 2.60E-10 m for titanium. What is the k{s,i}?

Homework Equations


k{s,i}: stiffness of an interatomic bond in a solid.
Y= stress/strain = (tension force/cross section area)/(change of Length/Length)
Y = ((k{s,i}*s)/diameter^2)/(s/diameter) = k{s,i}/diameter


The Attempt at a Solution


stress = changeL/L = .0141m/2.8m = 5.035714E-3
strain = F{T}/A = 9.81*mass / pi*(7.5E-4)^2 = 5.532E11
4.5g/cm^3 = 4.5E3kg/m^3
(mass = Density*length^3 = 4.5E3kg/m^3*(2.8 m)^3 = 99662.08)

stress/strain = Y = 9.10195E-15


I'm pretty sure that's the wrong answer.
 
Physics news on Phys.org
Hi badwallpaper0,

The stress is F/A, and the strain is (change in L/L); you seem to have these reversed.

Also, what number did you use for the mass in calculating F/A? I don't see how you got the result 5.532e11.
 
alphysicist said:
Hi badwallpaper0,

The stress is F/A, and the strain is (change in L/L); you seem to have these reversed.

Also, what number did you use for the mass in calculating F/A? I don't see how you got the result 5.532e11.

Hey alphysicist,
First of all, thank you for the help.
Ok, I redid the problem making
mass = density*length^3 = 4.54E3 kg/m^3 * 2.8^3 = 9.966208E4
stress=F{t}/A = 9.81*mass / pi*(7.5E-4)^2 = 5.5325E11
strain=.0141m/2.8m = 5.0357E-3

stress/strain = 5.53256E11/5.0357E-3 = 1.098665E14

I changed the stress and strain as you suggested, and I redid the calculation for mass.
Does that seem like a reasonable/correct answer?
 
I think there are a couple of issues. First, to calculate the mass of the wire, you would multiply the density times the volume of the wire. The mass you found using length^3 would be the mass of a cube of titanium 2.8 m on each side. But this wire is a skinny cylinder, and so it's volume is its length times its cross-sectional area.

But the mass of the wire is not what goes into the Young's modulus calculation. The idea is this: first the wire had a length of 2.8 m, then because somebody hung a 109 kg mass on it, it stretched by 1.41 cm. So the force that we use in calculating the stress is the force that makes the wire stretch, so here the mass needs to be 109 kg (because the force causing the stretch is the weight of that mass).

(The reason we don't need to include the weight of the wire at all is because whatever effect it had on the length was already accounted for in the original 2.8 m length. Only the extra mass caused the extra length.)
 
That makes sense.

So instead of trying to figure out the wire's mass we would find the mass of the weight added to the end (though in this case it's given).

So it's simply (9.81m/s*109kg)/(pi*(7.5E-4)^2) = 6.0509E8 = stress

6.0509E8/5.0357E-3 = 1.2016E11 - this seems like a large number, though titanium is 105-120 GPa.

I think the units should be kg/(ms^2), though mine doesn't seem to work out that way.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top