Chapter 2 - Stiffnes, Stress & Strain

Question 1

What is the most important mechanical property of a golf ball and why?

Answer 1

Resilience, when there is an impact, the energy of the club will be transformed to the ball and stored to then fly as far as possible. If it was stiff, it wouldn’t absorb as much energy.

Question 2

What is resilience?

Answer 2

The possibility to store a lot of energy from a force and to be able to turn this into kinetic energy. Like a spring that will vibrate when the force is gone. The ability to turn back to its original form (so elasticity)

Question 3

What would be the best property for a spring: a strain of 0,005 or 0,002 and why?

Answer 3

Depends on the surface under the elastic part of the graph, but it is most likely to be the one with 0.005 strain

Question 4

F stretches a rod by 1 mm. How much will it be stretched if we double the load?

Answer 4

We don’t have enough information. If we stay in the linear elastic part (in that case it would be 2 mm). But that’s not mentioned in the question

Question 5

What is Hooke’s law

Answer 5

sigma_z = E * epsilon_z (Stress in z direction = Young’s modulus E x strain epsilon in z direction)

Question 6

What does a Poisson coefficient of 0 mean?

Answer 6

U = -(epsilon_x / epsilon_z) so no lateral contraction, the rod for example will not become thinner when it’s stretched.

Question 7

If we restrict lateral expansion or contraction: and I put 20 Mpa (compression) on it, which material will put the most pressure on the walls of the restriction (rubber, steel, aluminium or glass)?

Answer 7

Rubber: 
Uses extended version of Hooke's law. Except epsilon_x = epsilon_y = 0. And sigma_x = sigma_y
This gives (U*sigma_z)/1-U
Filling in values gives us rubber. Or theoretically, it wants to deform the most

Question 8

The difference between engineering (technical) and true stress/strain is negligible when?

Answer 8

When we are dealing with elastic deformation of metals that don’t deform a substantial amount.

Question 9

If you have the engineering strain, can the true strain be calculated? if yes, how?

Answer 9

Yes, ln (1 + epsilon_eng).

Question 10

Can we just add 2 consecutive TRUE strains epsilon_1 and epsilon_2? And why (not)? Will this work for engineering strains?

Answer 10

eps_1_true = Ln(L1/L0) and
eps_2_true = Ln(L2/L1)
eps_t_true = Ln(L2/L0) = Ln(L2/L0 * L1/L1)
= Ln(L2/L1 * L1/L0)
= Ln(L2/L1) + Ln(L1/L0)
= eps_1_true + eps_2_true.
So yes we can add them. Doing the same for the engineering strains will not work like this.

Question 11

Is natural rubber of 10m, 1cm diameter enough for a 50 kg lady to bungee jump from a bridge of 72m high [a graph shows a maximum of 600% strain which starts going up quicker at 500% and from 0 to 500% it goes up by 5 Mpa] (Its more about the approach in calculating not so much about the values)

Answer 11

No, the rope will only have a strain of 70m, so that’s not the problem. But if we look at the energy that it can absorb (the area under the curve) we see that it never exceeds the energy of the lady so it never fully stops her and she will smash on the ground.

Question 12

If we have a 10m rope that has a max strain of 600%, and we calculate that the elastic energy (the area under the graph) for a strain of 10 is smaller than the potential energy of the person bungee jumping (10mg). Does that mean that the rope will break since not all the energy is absorbed by the rope?

Answer 12

As long as the rope can still expand without breaking, it will not break. So for a 10m rope that breaks at 60%, this will be 70m