Finding Average Normal Stress in Rod

rugerts · Mar 1, 2019

Homework Statement

Here's a snapshot of the problem:

Engineering figure of a rigid cap G joining two copper rods AB and CD of 30 mm diameter to a steel rod EF of 40 mm diameter, with 300 mm and 450 mm dimensions, and 40 kN loads applied at the cap and at the end of the rod.

Homework Equations

Equation for axial deformation: delta equals the integral from 0 to L of N(x) dx divided by A(x) times E(x), with the integral limits 0 and L.

+ Newton's 2nd Law.

The Attempt at a Solution

Worked engineering solution showing equilibrium equation 2F + F_EF - 2[40(10^3)] = 0, a compatibility equation with superposition terms, and the result F_EF = 42 483.23 N.

My question is: why does the delta P term have only a single 40 kN force considered, whereas for delta EF there's an F/2?
Thanks for your time.

Chestermiller · Mar 2, 2019

I can't figure out what they did, but this is definitely not how I would have solved this problem. I would have written $$F=\frac{\frac{\pi}{4}(0.03^2)(101)(10^9)}{300}\delta_p$$and $$F_{EF}=\frac{\frac{\pi}{4}(0.04^2)(193)(10^9)}{450}\delta_p$$where ##\delta_p## in mm is the displacement that the plate shifts to the left. Then I would have substituted for F and ##F_{EF}## into Eqn. 1 to solve for ##\delta_p##.

This gives ##\delta_p=0.0874\ mm##, ##F_{EF}=38380 N##, and ##F=20810 N##. So the answer given in your book doesn't agree with this.

Finding Average Normal Stress in Rod

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

Similar threads