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Unformatted text preview: CEE 4/5 Guam Ecﬁﬁmwnoc swam 8.82.3 an.qu H» Sim 33555 mam 35355
3% BE 5% ES w _
3 a N + 3 w 3
Sam 5ng “SE ﬂ Q can? SEE 333 N + an S Edam SEE .523 + an NEW someom £80 HU<
ma nommﬂm moﬁﬁﬁﬁoo mo £23 o>uomﬁm E maoswqmom Examples 3.3 and 3.4 _ T—sections in Positive and Negative Bending (a) Determine bl for a beam Tnsection that is part of a continuous ﬂoor system Consider the portion of the continuous floor system shown below and the central floor beam
spanning in the horizontal direction. Sketches of the beam sections corresponding to section
lines A—A and BB are given in parts (b) and (c) of this exampie, respectively. The ACI Code
rules for determining the effective width of the compression ﬂange for the beam section
corresponding to section line A-A give the following: E _ 24ﬁ.(12m./ ft.) be 5—.._-———mm—m72in. 4 4
be 5 bw + 2(8hf) z 12in.+16(5in.) = 92in.
be s bw + 2[1—0~ a 10 ft. = 120m. The first limit governs for this section, so in the following part of this exampfe it will be
assumed that be = 72 in. 10’ Figure for part (a) CEE415 Example 3.3: Assume the T~section shown below is subjected to positive bending. Calculate Mn, Aswan), and show that this is a tension-controlled section. Use f; m 4500 psi ([31 = 0.825) and fy = 60 ksi. 2.5” 24” Ignore the compression reinforcement Tension steel area, As 3 6 (0.60 in?) = 3.60 in.2
Effective ﬂexural depth, d E 11 —— 3.5 in. = 20.5 in. Calculation of Mn; assume that a S hf and as 2 8y = 0.00207
From T = CC,
Asf}. _ (3.60in.2)(60ksi) a 2 ﬁle 2 ~——- — Mm~—-—~—7—
0.85 ﬂbe 0.85(4.5ksi)(72m.) = 0.78111. < hf Clearly, as > ay and also at > 0005 (Note: dt 2 h w 2.5 in. m 21.5 in.)
So, M" :2 Asfy[d — z (3.60m?)(60ksi)[20.5in. m Mn = 4340 k —in. z 362 k-ft 0.78231.
2 For Ami“, 3J3? : 201 psi, so use 201 psi 2 201psi(b d) 2 20 1
As , min w
f}. 60, 000 (12")(20.5") : 0.82m.2 < AS (0.1.x) ...
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This note was uploaded on 03/29/2011 for the course CEE 415 taught by Professor Wight during the Spring '11 term at University of Michigan.
- Spring '11