Why do you say the angle doesn't go all the way to 90 degrees? The resultant moment with zero moment about the Y axis will be parallel to the Y axis, and zero moment about the X axis will be parallel to the X axis.
Yes, I should have read the notes, I assumed continuity at point A. If it is pinned at A, it is in static equilibrium in the same way that a free-standing column that is pinned at the base is in equilibrium under a pure axial load; i.e. it is highly unstable.
Yes, assuming we are ignoring non-linear effects (and obviously there isn't enough information to evaluate those), it's hard to see how there could be any dispute that the resultant force has to be 5/sin(30) = 10.
Because we are working from the point where the NA crosses the Y axis, and using the strain gradients along the Y axis ("vertical"), and perpendicular to the Y axis ("horizontal"), so we use the distance from the point of zero strain to the bar in the "horizontal" and "vertical" directions...
Why?
Perhaps I should expand on that: Why should it be transferred to the plastic centroid, rather than the centroid of the uncracked concrete section?
I think it does matter.
The calculation is being done to compare the moment capacity of the column with the maximum applied bending moment, so they both need to be taken about the same axis.
Bending moments in a frame analysis are about the centroid of the uncracked sections, so that is what...
The direction of the resultant is measured as the anti-clockwise rotation from the X-axis, but it doesn't really matter as long as your angles are always consistent.
In a column with the NA near the centroid of the section there will be significant cracks in the concrete at the ultimate moment...
No, look at the plot of the unrotated section. The direction of the resultant moment relative to the X axis is less than 90 degrees, and relative to the NA it is more than 90 degrees.
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...
