# Math Help - Determine Apsl

1. ## Determine Apsl

Want to calculate Apsl?

Mu1 = fpb Apsl (d-dn1)

Where
Mu1 = 56.6*106
fpb = 0.95*1750
d = 156.7
dn = fpb Apsl / 0.9 fcu b
fcu = 50
b = 1168

Mu1 = 56.6*106 = ( 0.95*1750* Apsl*156.7 ) – (0.95*1750* Apsl)^2 / 0.9*50*1168

The values I have worked out are not appropriate to the solution I am trying to solve. I the areas of the equation where I don’t understand is moving 156.7 & ^2

If anybody could help I would be grateful

2. I have no idea what any of your variables mean. Please post the original problem word-for-word.

3. the variables are the material properties for engineering design.

the answer to this particular solution Apsl = 227mm^2

4. You are being highly incoherent, and you're not doing what I asked. See here and read rule # 11.

5. ## Q in full

6. Please post a picture or a pdf. Word documents are frowned on in this forum. It's an unwritten rule around here.

Thanks!

7. ## Q in full .bit file

Example 4.10

8. Sorry to be so picky, but could you please blow up the picture a bit? I need more resolution for my 32-year-old eyes to read that, I'm afraid.

9. ## Question & Solution

Sorry about delay the file was too big.

Thanks for taking the time to look at this with me and trying to help me understand

10. Thank you for posting the better pictures.

This looks like a straight-forward algebra problem. What I would do is work with the equations 4.24, 4.25, and 4.26, and combine them in order to eliminate the variables you don't know. Your goal is, apparently, to find $A_{\text{psl}}.$ So solve for that variable in terms of variables you know. Do this algebraically, without plugging in any numbers. What do you get?

11. why is (fpb Apsl )^2 & the d=156.7 moved.

12. Moved from where to where? Your Mu1 equation in the OP appears correct to me.

13. Mu1 = fpb Apsl (d-dn1)

Mu1 = 56.6*106 = ( 0.95*1750* Apsl*156.7 ) – (0.95*1750* Apsl) ^2 / 0.9*50*1168

Queries..........................................d =156.7...................&........^2

$\displaystyle{d_{n1}=\frac{f_{pb}A_{psl}}{0.9\,f_{ cu}\,b}},$ and

$\displaystyle{M_{u1}=f_{pb}A_{psl}(d-d_{n1}).}$

Now you simply plug the first equation into the second and distribute:

$\displaystyle{M_{u1}=
f_{pb}A_{psl}\left(d-\frac{f_{pb}A_{psl}}{0.9\,f_{cu}\,b}\right)=
f_{pb}A_{psl}d-f_{pb}A_{psl}\,\frac{f_{pb}A_{psl}}{0.9\,f_{cu}\,b }=
f_{pb}A_{psl}d-\frac{(f_{pb}A_{psl})^{2}}{0.9\,f_{cu}\,b}
}.$

Therefore,

$\displaystyle{M_{u1}=
f_{pb}A_{psl}d-\frac{(f_{pb}A_{psl})^{2}}{0.9\,f_{cu}\,b}
}.$

This last expression is what you have plugged your parameters into. You will, incidentally, need to use the quadratic formula in order to solve for the target variable.

Does it make sense now?