PROCEDURE: t = (nominal wall thickness) - (grout thickness ...
Transcript of PROCEDURE: t = (nominal wall thickness) - (grout thickness ...
DATA SET:Height of wall, h 19 FTNominal thickness of wall 16 INgrouted cells o.c. spacing 16 INMasonry compressive strength, f'm 4000 PSIThe wall DL 82 KLFThe wall LL 103 KLF
PROCEDURE:
r = = = 4.9348 in
t = (nominal wall thickness) - (grout thickness) = 16 - 0.375 = 15.625 in1. Find the actual wall thickness, t (IN)
2. Find the net area per foot of wall, An (IN2)
An = 112.4 in2
This information can be found in TEK 14-1B (�le is on canvas). Make sure you’re using thecorrect table from this section! You’ll want to use the table for your wall thickness.
3. Find the net moment of inertia per foot of wall, In (IN4)
In = 2737.2 in4
This is also provided in the TEK 14-1B.
4. Find the radius of gyration per foot of wall, r (IN)In
An
2737.2112.4
Q3
Q2
DATA SET:Height of wall, h 19 FTNominal thickness of wall 16 INgrouted cells o.c. spacing 16 INMasonry compressive strength, f'm 4000 PSIThe wall DL 82 KLFThe wall LL 103 KLF
5. Find the Ratio of h/r
h/r = = 46.2024
This is just the given height of the wall, h, divided by the radius of gyration, r, which wefound in Q4.Note: Remember to convert! The two values you have are in di�erent units!
(19)(12)4.9348
6. Which TMS equation used? 11 or 12Compare your h/r from Q5 to 99. This will give you the axial strength equation to use.
If h/r < 99, use TMS 402 eq. 3- 11:
If h/r > 99, use TMS 402 eq. 3-12: 0.80(An)(F’m) 70( r )h
2Pn = 0.80
h140( r )
21 - 0.80(An)(F’m)Pn = 0.80
7. Find the nominal axial strength, Pn (KLF)Use the equation per Q6. Remember to convert where required!
h140( r )
21 - 0.80(An)(F’m)Pn = 0.80
19(12)140(4.9348)
21 - 0.80(112.4)(4000)Pn = 0.80 = 256,405.3137 plf = 256.4053 KLF
Q4Q2
DATA SET:Height of wall, h 19 FTNominal thickness of wall 16 INgrouted cells o.c. spacing 16 INMasonry compressive strength, f'm 4000 PSIThe wall DL 82 KLFThe wall LL 103 KLF
8. Find the factored nominal axial strength, ΦPn (KLF)Φ for axial force is equal to 0.9ΦPn = 0.9(256.4053) = 230.7648 KLF
Q7
9. Find the required axial strength, Pu (KLF)
10. Does the wall pass or fail? 1=pass 0=fail
Using our given DL and LL to determine this:
Compare the answers from Q8 and Q9. If Q8 > Q9, then it passes.
Pu = 1.2(DL) + 1.6(LL) = 1.2(82) + 1.6(103) = 263.2 KLF
Q8 ? Q9230.7648 ? 263.2230.7648 < 263.2The wall fails.