Strength of materials

In summary: Therefore, the ratio of internal diameter to external diameter of the hollow shaft for equal strength is 0.8. Additionally, using this ratio, you can calculate the percentage saving in weight by finding the difference between the volume of the solid shaft and the hollow shaft and dividing it by the volume of the solid shaft. This would give you the percentage saving in weight, which can be calculated as follows:Percentage saving in weight = ((pi/4)(R^2 - Ri^2)/((pi/4
  • #1
MMCS
151
0
A mild steel solid shaft is to be replaced by a stainless steel hollow shaft of the same
outside diameter. Calculate the ratio of internal diameter to external diameter of the
hollow shaft for equal strength. Also find the percentage saving in weight if density of
mild steel is the same as stainless steel.Take maximum shear stress for Mild Steel as
34 MPa and that for Stainless Steel as 57 MPa

I assume that for the strength to be the same the max torsion should be the same so i use the max allowed shear stress in this equationrearranged for torsion and equate them both

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi*(R-Ri)^4)/2)/R - Ri

RI = inside radius

Is this the correct step? how would i get a ratio from this?
 
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  • #2
MMCS said:
A mild steel solid shaft is to be replaced by a stainless steel hollow shaft of the same
outside diameter. Calculate the ratio of internal diameter to external diameter of the
hollow shaft for equal strength. Also find the percentage saving in weight if density of
mild steel is the same as stainless steel.Take maximum shear stress for Mild Steel as
34 MPa and that for Stainless Steel as 57 MPa

I assume that for the strength to be the same the max torsion should be the same so i use the max allowed shear stress in this equationrearranged for torsion and equate them both

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi*(R-Ri)^4)/2)/R - Ri

RI = inside radius

Is this the correct step? how would i get a ratio from this?
(R^4 - Ri^4) is not the same as (R - Ri)^4.
And in both cases, max shear stress occurs at the same distance fron the centroid of the circular cross section.
 
  • #3
Ok so

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi/2*R^4)-(pi/2*Ri^4))/R

to simplify and multiply out pi terms i get

(-23*pi/2*r^4)/r = (57*pi/2*ri^4)/r

-36.128*r^4 = 89.535ri^4

so as a ratio i get

36.128/89.535 = 0.404

but i have the answer to be 0.8
 
  • #4
You've got negative signs appearing and disappearing at random.

You should review basic algebra.
 
  • #5
SteamKing said:
You've got negative signs appearing and disappearing at random.

You should review basic algebra.

That is irrelevant. -36.128/89.535 produces the same ratio as 36.128/89.535.
 
  • #6
And yet you have not solved this problem.
 
  • #7
SteamKing said:
And yet you have not solved this problem.

Another irrelevant response. I do not know how to solve the problem otherwise i wouldn't have posted the question, however, Your input had nothing to do with getting closer to the solution.
 
  • #8
MMCS said:
Ok so

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi/2*R^4)-(pi/2*Ri^4))/R

to simplify and multiply out pi terms i get

(-23*pi/2*r^4)/r = (57*pi/2*ri^4)/r

-36.128*r^4 = 89.535ri^4

so as a ratio i get

36.128/89.535 = 0.404
the plus and minus signs heretofore not withstanding, you have calculated that (ri^4/r^4) = (ri/r)^4 = 0.404. So to calculate (ri/r), you need to take the 4th root of that number.
but i have the answer to be 0.8
which is correct.
 

1. What is the definition of strength of materials?

The strength of materials is a branch of engineering that deals with the behavior of solid objects subjected to stresses and strains. It is concerned with understanding how materials deform and fail under different types of loads, such as tension, compression, bending, and torsion.

2. What are the types of stresses that materials can experience?

Materials can experience different types of stresses, including tensile stress (pulling forces), compressive stress (pushing forces), shear stress (parallel forces acting in opposite directions), and bending stress (combination of tensile and compressive stresses due to bending).

3. How is the strength of a material determined?

The strength of a material is determined by its ability to withstand stresses and strains without breaking or deforming permanently. This is measured using mechanical testing methods, such as tensile testing, compression testing, and bending testing.

4. What factors affect the strength of materials?

The strength of materials can be affected by various factors, such as the type and composition of the material, its microstructure, the type and magnitude of the applied load, and the environmental conditions (e.g. temperature, humidity, etc.). Manufacturing processes and defects can also impact the strength of materials.

5. Why is understanding the strength of materials important?

Understanding the strength of materials is crucial in engineering and construction, as it allows engineers to design structures and machines with the appropriate materials and dimensions to withstand expected loads. It also helps in predicting and preventing failures, ensuring the safety and reliability of structures and products.

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