Question

Sketch a tensile member with (a) a rectangular cross section, (b) a solid circular cross section, and (c) a circular tube cross section, and label the dimensions symbolically (e.g. label the radius for the solid circular case.) For each member, write out the definition of engineering stress in terms of the actual dimensions of the component. If the rectangular member has dimensions of width and thickness equal to 0.75 cm x 0.4 cm, what would be the radius of a solid circular member such that the stress is equal for an equal tensile load? If the tube has an outer radius equal to that of this same solid cylinder, what is the maximum inner radius such that the stress does not exceed 200% of the stress in the solid cylinder?

Answer 1

Answer:

See attachment for sketches:

1) $Stress_{1}=\frac{F}{ab}$

2) $Stress_{2} = \frac{F}{\pi   r^{2}}$

3) $Stress_{3} = \frac{F}{\pi (r_{o} ^{2}-r^{2} _{i})  }$

4) r = .309 [cm]

5) ri = .218 [cm]

Steps:

1) The tensile stress is given by the tensile Force over the object´s area

• $Tensile.Stress=\frac{F}{A}$

2) Sketch the figures and get their Area in terms of their dimensions to know their stresses

3) Draw the sketch of a rectangular cross section of sides "a" and "b". The cross sectional area of the rectangle is A1 = a*b

4) Draw the circle of radius "r", the cross sectional area of a circle is given by A2 = (PI)*r^2

5) Draw the tube, the cross sectional area of a tube is given by A3 = (PI)(ro^2-ri^2)

6) They give you the sides of a rectangle and ask you to get the radius of a solid circle such that their stresses will be the same when the same load is applied to them. You know the load will be the same, so in order to get the same stress, their areas must be equal.

7) .75*.4 = .3, equate the area of the circle to .3 and solve for "r". This will give you the radius of the circle to match the stress of the rectangular piece.

8) Now you need to equate the radius you just calculated to the outer radius of the tube such that the stress of the tube under the same applied load does not exceed two times the stress of the circle.

9) Equate the stress of the tube to 2 times the stress of the circle and solve for ri