Question

This boat is normally used to tow medium-sized ships in short trips from the bay to the port, but a change of propeller to increase the load it can drag. Since the tug was purchased second hand, some important blueprints and design details were lost, but some data could be collected from an old analysis done to the axis. It is known by the material that the maximum shear stress must be less than 500 [MPa] and that the stress Maximum normal compression should not exceed 700 [MPa]. Also, for an image that it was found that in some plane the normal stress is 157.8 [MPa] and the stress shear of 136.75 [MPa] we need to calculate the angle of difference between the plane stresses of the different reference systems so it can be assumed that the normal stress aty is

the stresses change according to the reference system in place, the normal stress along the y axis is 0 if certain criteria are met, what are they? and the xy axes of the image how would they be used?

Answer 1

The solution is provided in the images below. Revert back in case of any query.

solution: case 1: T = 1857.8 MPa 136.75 mpa lay 2. case 2: 30 2 - ? Conditions: Ty max-compressive < 700 MPa ين Zmax s SoomPa. 7) boundary on the 5-2 plot, the wonditions (1) & (ii) would aut like conditions inside which the Moho write will be plotted. Emax = max = maximum radius of the circle.

Seo MPa (6,354.57) MPa. 5-678.2,13 136.79 20 = 136.3 -700 A( 157.8, -136.75) (-520.4,-354.57) MPa .soo mpa + we can plot pt. Al 187.8, -136-76) MPa from case li). Diametrically opposite pt. of A would lie on the line 136.'75 mpa Let us assume that that B1-700,0) lies on the urde 7-

be, we have we have two pts urde. A A B lying on the The equation the Wrle can be solved as to-157.8) ² (136.75) = (+700)² 2 G + 24 900.84 - 315.6 at 18 700.5625 o²+ 490000 + 14000 -44698.5975 → 1715.6 T :-260.199 mpa - 260.2 mpa We can the centre of Mohr wille = [- 260.2, o) mpa. = 0 now complete ürde to find the missing pts, which are = -678.2 MPa Ist 520.4 MPa. 354.57 MPa Zay And 20 = 107° =) 0 ~ 53.5°