The lambda factor (0.8) is to account for the effect of replacing a varying, parabolic-linear, stress distribution with a uniform stress. It is based on a rectangular section but it is applicable to any shape.
I was talking about the additional 10% reduction in your 3rd paragraph. This...
Yes, reducing the steel stress and reducing the concrete area are effectively the same thing for force, and will give exactly the same answer for a rectangular stress block.
The difference is in the bending moment, and reducing the steel stress for bars in compression will give a more...
Celt83's diagram provides the answer, but in words:
We calculate the concrete force based on the full area of the concrete, but there is no concrete at the bar locations, so the additional compression force provided by the reinforcement is: (Steel stress - concrete stress) x bar area.
The...
Hi Pretty Girl, in reply to your post above:
Another way to do the calculation, which might be more intuitive, is to take moments about the NA, then add the moment due to the net axial force x distance NA to centroid. This will give exactly the same results. It follows that for a section with...
Further to my previous post, I have reviewed the calculation method in my spreadsheet, which is:
For a section with a defined NA angle:
1: Rotate the section about the uncracked centroid, so the NA is horizontal and find the coordinates of all the reinforcement in the new orientation.
2: Find...
The problem is, the reinforcement and the concrete stress block are in general not symmetrical about the perpendicular to the NA, so in general the direction of the resultant moment about the NA is not perpendicular to the NA.
The simplest way to handle that is to take moments about the...
I now agree with your calculated bar forces.
For the bending moments you should be taking moments about the X and Y axes passing through the centroid of the uncracked section (i.e. the mid-point of the rectangle).
Do the same for the concrete and you can then find the resultant moment magnitude...
It is wrong. The strain at the corner with maximum tensile strain is not -.0035 unless the NA is exactly mid-way between the two corners.
Use the maximum compressive strain as the starting point for both bars, then the strains are:
1) 0.0035 * (1 - (110-95)/110) = 0.00302273
2) 0.0035 * (1 -...
To my mind, by far the nicest GUI for engineering/maths calculations is Excel combined with pyxll or xlwings. Yes, managing external library updates can be a pain, but it's well worth the effort.
I had a look at the spreadsheet. Not everything is clear to me, in particular I don't know what the dis dimension is, and I don't know why you have two different dimensions from the max compression fibre to the NA. There is only one NA.
I get exactly the same compression strains as you, but...
@rb1957 - but if there is a distributed loading between the plates, they will have a different shear force distribution, so they will have different curvatures, so they can't be in contact other than at points, so there can't be a distributed loading.
A simple thought experiment that convinced...
Things do get complicated when some of the beams do not extend full length, as in Greg's example and the discussion at:
https://www.eng-tips.com/threads/beam-on-top-of-beam-not-connected-and-of-different-length.379229/
But for two or more stacked beams or plates spanning between the supports...
On reflection my previous post was a little over-conservative.
The bottom plate will deflect downwards until the upper plates have the same curvature and are in contact over the full length, so the load will be distributed between all the plates in the stack.
Assuming weightless plates of zero depth and a true point load, the top two plates will curve upwards at the point of contact and lift off the plate below. All the load will therefore transfer to the lower plate at the load position, which will deflect exactly the same as if the top two plates...
Not much I can do about the spreadsheet not working on Apple Excel I'm afraid. If you can find specific code lines it doesn't like I may be able to do something, but even then I don't have access to Apple computers, and I'm not familiar with their version of VBA.
I had a quick look at results...
I don't have time for a detailed answer now, so I'll just say:
1) If your results don't follow the form of the chart from the book, there is definitely something wrong.
2) My blog has several spreadsheets for biaxial bending, including VBA versions. You can search for biaxial bending, or just...
I don't know what you mean by "relative distance". All the moments are taken about the centroid of the rectangle, and I have taken the origin as the bottom left corner of the rectangle, so the rectangle centroid is at 4, 5.5. The triangle is 5 wide and 7 high, so the triangle centroid is at...
You are using the wrong shapes; you need to think about what they represent.
If the rectangle had a uniform load creating unit stress over the whole area the force would be 88 and moment about the centre 0.
Now imagine the bottom left corner is cut, and that section of load lifted off. The...
It doesn't assume the entire section is in compression. The triangular area is treated as having a compressive force and an equal and opposite tensile force, so the net force on the triangle is zero, and the centroid of the combined forces is the centroid of the pentagonal stress block.
You...
