I still can't visualize it... Can you help to draw out diagram? The thickness is 20?billy_joule said:
i still couldn't imagine , isn't the circle part represent the area ? in the diagram just above my circled part , i didnt see where is the (40-25mm)billy_joule said:
the total of 2 red part i add now represent 40 - 25 mm , but where is the 20mm? i didnt see itbilly_joule said:
the diagram is confusing ... From the top view , i found that the rod is quite thick (green line represent the thickness) , However , when we view it from front , it is a flat tod ( very thin) . Or you can draw me a 3d diagram please?billy_joule said:
Can you label in my diagram that i sketched , where is the 20mm ?billy_joule said:
It's the depth,or thickness of the rod end, into (or out of) the page.chetzread said:
Since it can't be shown on my diagram , then is my diagram wrong ? in the notes , the area is (20)(40-25) .. did I sketch the diagram wrongly?billy_joule said:
why we need to consider the stress for the area of side view ? Why not from front view ?billy_joule said:Your first sketch is right, you just need a top (or bottom, or end) view to show depth.
Alternatively, you can do an isometric sketch to show all the information.
You're looking for the area of the two cross hatched sections:
Note that we've taken the authors word that this is the smallest cross sectional area of the rod.
It is, but only by a very small margin, if the hole size was reduced to 24mm it wouldn't be the area we need to consider.
In future, you'll need to prove that you're using the correct cross section.
chetzread said:
stress perpendicular to areabilly_joule said:
I'm not sure what you're calling side or front view.chetzread said:
Well, new question here . how to get the circular cross section = 314x10^-6 (m^2) ?billy_joule said:
It's the cross sectional area of the circular part of the rod.chetzread said:
I have no idea what you're doing trying to do there.
diameter = 25mm ,billy_joule said:
chetzread said:
Then , what should it be ?billy_joule said:
Come on, It's shown two times on your first attachment! it's even labeled with a 'd' so you know its a diameter of a circular section!chetzread said:
To calculate the area of normal stress, you will need to use the formula A = σBC, where A represents the area, σ represents the normal stress, B represents the width of the rod, and C represents the thickness of the rod. Plug in the values for B and C and solve for σ to find the area of normal stress.
The units for the area of normal stress will depend on the units used for the width and thickness of the rod. For example, if the width and thickness are measured in meters, the area of normal stress will be in square meters (m^2).
Sure, let's say we have a thin rectangular rod with a width of 0.5 meters and a thickness of 0.2 meters. If we apply a normal stress of 500 Newtons per square meter (N/m^2), the area of normal stress would be calculated as follows: A = (500 N/m^2)(0.5 m)(0.2 m) = 50 N. So the area of normal stress in this example would be 50 square meters.
The area of normal stress represents the total amount of force applied to a specific area of the rod. In other words, it measures the intensity of the stress being applied to the rod.
Yes, the formula A = σBC can be applied to any shape with a defined area, as long as the normal stress is applied evenly across the surface. However, for irregular shapes, the calculation of the area may be more complex.
