- #1

wrw103

- 6

- 0

## Homework Statement

A Bimetallic torsion bar consists of a steel core (Gs = 75 [GPa]) of diamter di = 25 [mm] around which is bonded a titanium sleeve Gt = 45 [GPa] of inner diameter di = 25 [mm] and outer diameter do = 40 [mm].

a) If the maximum sher stress in the steel is 50 [Mpa], what is the total torque, T, applied to the bimetallic bar? Ans. T = 664.5 [N*m]

b) What is the total angle of twist of the composite bar if it is 2 [m] long? Ans. phi = 6.108 [degrees]

## Homework Equations

tau = T*r/Ip -> tau = shear stress, T = torque, r = radius, Ip = Polar Moment of Inertia

phi = T*L/G*Ip -> T = torque, L = Length of bar, G = shear modulus, Ip = Polar Moment of Inertia

## The Attempt at a Solution

To determine the Torque, I tried using the first equation, using the moment of inertia of the whole rod, and solving for T.

T = tau*Ip/r -> (50*10^6*pi*(0.040)^4/32)(0.25/2) = 1005 [Nm]

This is the wrong answer. I've tried other things as well, but I don't know why any of them do not work. I've tried using the moment of inertia of the the steel core and the moment of inertia of only the titanium, but they all produce incorrect results. Am I approaching the problem from the right angle?

For the second part, I want to use the second equation, but I don't know how to account for the different shear moduli.

Can i solve it for the torque and write out an expression for each part of the rod and then equate the two and solve for phi

T= (phi*Gs*Ips)/Ls=(phi*Gt*Ipt)/Lt -> phi = Gt*Ipt-Gs*Ips

Is that a correct approach?

Thanks for all the help!