Axially Loaded Member and Tensile Stress oh so lost

In summary, the homework statement says that the axial load for a given test sample carries 1590 N. The sample thickness is 5mm, so the tensile stress at sections 1 and 2 is calculated assuming the sample thickness is 5mm.
  • #1
noboost4you
61
0

Homework Statement


"The axial load for a given test sample carries 1590 N. Calculate tensile stress at sections (1) and (2) assuming the sample thickness is 5mm. (Rectangular cross section).

http://img208.imageshack.us/img208/121/tensilestressby1.th.jpg [Broken]


Homework Equations


normal stress = P/A
Pallow = (normal stress)allow * A


The Attempt at a Solution


This is a new one for me. All the previous homework problems involved a prismatic bar (a bar of uniform cross-section) and not this 'necked' bar.

Since P are pulling forces, the bar is in tension.

I know my first step would be to convert 5mm into meters = 0.005m

But then I'm sort of lost.

Do I just use the normal stress equation above for each section; 1 and 2? Keeping P = 1590N and then changing A for section 1 to 30mm = 0.03m, and for section 2 to 10mm = 0.01m?

But then where does 5mm come into play?

Thanks for all the help.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You will use the cross sectional area at both locations to calculate the normal stresses. The 5mm is required to calculate the area. The 30 mm and 10 mm dimensions are not areas. How do you calculate the area of a rectangle?
 
  • #3
Area of a rectangle is length * width

So at section 1, the area A would be 30mm*5mm = 150mm = 0.15m ?
and
Section 2, the area A would be 10mm*5mm = 50mm = 0.05m ?

Then I would just plug those areas into the normal stress = P/A equation for both sections?

Thanks for the quick reply
 
  • #4
Those above areas should read m^2 and not just meters.
 
  • #5
normal stress of section 1 = 1590N / 0.15m^2 = 10600 N/m^2
normal stress of section 2 = 1590N / 0.05m^2 = 31800 N/m^2

Could it be that easy? Feel like I'm missing something still
 
  • #6
It is in this case. You will never see an easier theoretical state of stress than that of a uniaxial tension test. There is no shear until you start dealing with transformations and Mohr's Circle stuff.
 
  • #7
Wow, well I expected that to be worse.

Thank you very much
 

1. What is an axially loaded member?

An axially loaded member is a structural element that experiences forces only along its longitudinal axis, also known as the axial direction. This means that the forces acting on the member are either pulling or pushing in one direction, with no bending or twisting forces involved.

2. What is tensile stress?

Tensile stress is a type of mechanical stress that occurs when a material is pulled or stretched. It is equal to the force applied divided by the cross-sectional area of the material. Tensile stress is a measure of the internal resistance of a material to external forces.

3. How is tensile stress calculated?

Tensile stress is calculated by dividing the force applied to a material by its cross-sectional area. The resulting unit of measurement is typically in units of force per unit area, such as pounds per square inch (psi) or newtons per square meter (Pa).

4. What factors can affect the tensile stress of an axially loaded member?

The tensile stress of an axially loaded member can be affected by various factors, including the material properties of the member (such as its yield strength and modulus of elasticity), the geometry of the member (such as its cross-sectional area and length), and the magnitude and direction of the applied force.

5. How is the safety of an axially loaded member determined in relation to tensile stress?

The safety of an axially loaded member can be determined by comparing the maximum tensile stress experienced by the member to its ultimate tensile strength. If the maximum stress is lower than the ultimate tensile strength, the member is considered safe. However, if the maximum stress exceeds the ultimate tensile strength, the member may experience permanent deformation or failure. It is important to design and choose materials for axially loaded members that can withstand the anticipated tensile stress without compromising safety.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
660
  • Engineering and Comp Sci Homework Help
Replies
1
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
10K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
19K
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
5K
Back
Top