 In this problem we have a long uniform aluminum alloy bar, which is subjected to a force F equal to 3-gluonutons, and we need to find the change in dimensions of the bar due to this force. Then of course we need to consider the change in dimensions in the direction of L, in the longitudinal direction, but also in the direction of B, and the direction of H. We can start, for example, calculating this change in dimensions in the direction of L. We know that the internal force is equal to F, then the average internal stress is equal to F divided by A, then this is equal to 3-gluonutons divided by the area which is 5 times 4 millimeters squared, then this is equal to 150 megapascals, and now we know that the stress is equal to the modulus of elasticity times the strain, so from here we have that the strain is equal to sigma divided by E, then sigma is equal to 150 megapascals, and the modulus of elasticity is equal to 69 gigabascals, then 10 to the power of positive 9 pascals, then this is equal to 0.0022, and from here we know that the strain is equal to the increment of L divided by L, then the increment of L is equal to epsilon L, and this is equal to, then from here we have that delta L is equal to 1.1 millimeters, so this is positive, then we have elongation. And remember that instead of following this procedure, we know that in general, for an axi-loaded member, the increment of length is equal to directly F times L divided by AE, which is exactly the same formula that we have used before. Well now as I said, we can calculate the change in the direction of B, now we know that the length B contracts due to Poisson effect, because we don't have any force applied in this direction, so we know that the strain in this direction B is equal to, and here using Hooke's law, is equal to minus B times the strain in the longitudinal direction, then this is equal to minus 33, this is the Poisson ratio, times 0.0022, and this is equal to the increment of B divided by B, then from here we have that the increment of B is equal to, as I said before, this is a contraction and you can see here that the sign is negative, so this is what we were expecting, the change in dimensions in the direction of H. As we said before, the length H contracts due to the Poisson effect, then this strain in the H direction is equal to minus Poisson ratio, times this strain in the longitudinal direction, that it is the direction in which we are applying the forces, then this is equal to minus 0.33 times 0.0022, and this is equal to delta H divided by H, so from here we have that delta H is equal to 0.33 times 0.0022 times H, which is 5 millimeters, then this is equal to minus 0.0036 millimeters, so we have here a negative sign, then this is a contraction.