COLUMNA CIRCULAR.xls

	A	B	C	D	E	F	G	H	I	J	K	L	M	N	P	R	T	U	V
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2						PROJECT :									PAGE :
3						CLIENT :									DESIGN BY :
4						JOB NO. :				DATE :					REVIEW BY :
5	Circular Column Design Based on ACI 318-05
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7	INPUT DATA & DESIGN SUMMARY
8	CONCRETE STRENGTH				fc'	=	5	ksi
9	REBAR YIELD STRESS				fy	=	60	ksi
10	COLUMN DIAMETER				D	=	20	in
11	FACTORED AXIAL LOAD				Pu	=	480	k
12	FACTORED MAGNIFIED MOMENT				Mu	=	200	ft-k
13	FACTORED SHEAR LOAD				Vu	=	20	k									2	1	3
14	COLUMN VERT. REINFORCEMENT				8	#	7											5
15	LATERAL REINF. OPTION (0=Spirals, 1=Ties)						1	Ties										5
16	LATERAL REINFORCEMENT			#	4	@	12	in o.c.										6
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18		THE COLUMN DESIGN IS ADEQUATE.
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22	ANALYSIS
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29	f Pn (k)
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31															f Pn (kips)	f Mn (ft-kips)
32												AT AXIAL LOAD ONLY			954	0
33												AT MAXIMUM LOAD			954	80
34												AT 0 % TENSION			763	161
35												AT 25 % TENSION			633	197
36												AT 50 % TENSION			523	215
37												AT e t = 0.002			361	225
38					f Mn (ft-k)							AT BALANCED CONDITION			356	228
39												AT e t = 0.005			146	229
40												AT FLEXURE ONLY			0	164
41	CHECK FLEXURAL & AXIAL CAPACITY
42		f Pmax =0.85 f [ 0.85 fc' (Ag - Ast) + fy Ast] =							953.6522423	kips., (at max axial load, ACI 318-05, Sec. 10.3.6.1)
43		where	f =	0.70	(ACI 318-05, Sec.9.3.2.2)								>	Pu	[Satisfactory]
44			Ag =	314	in2.					Ast =	4.80	in2.
45		a = Cbb1 =	8	in (at balanced strain condition, ACI 10.3.2)
46		f = 0.57 + 67 et =		0.709		(ACI 318-05, Fig. R9.3.2)
47		where	Cb = d ec / (ec + es) =				10	in		et =	0.002069		ec =	0.003
48			d =	17.56	in, (ACI 7.7.1)					b1 =	0.8	( ACI 318-05, Sec. 10.2.7.3 )
49		f Mn = 0.9 Mn =		164	ft-kips @ Pn = 0, (ACI 318-05, Sec. 9.3.2) ,& et,min = 0.004, (ACI 318-05, Sec. 10.3.5)
50		f Mn =	218	ft-kips @ Pu =			480	kips			>	Mu		[Satisfactory]
51		rmax	=	0.08	(ACI 318-05, Section 10.9)					rprovd	=	0.015
52		rmin	=	0.01	(ACI 318-05, Section 10.9)									[Satisfactory]
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54	CHECK SHEAR CAPACITY
55		f Vn = f (Vs + Vc) =		52	kips, (ACI 318-05 Sec. 11.1.1)
56				>	Vu		[Satisfactory]
57		where	f =	0.75	(ACI 318-05 Sec. 9.3.2.3)
58			A0 =	242	in2.			Av =	0.40	in2.		fy =	60	ksi
59			Vc =	2 (fc')0.5A0 =			34.3	kips, (ACI 318-05 Sec. 11.3.1)
60			Vs =	MIN (d fy Av / s , 4Vc) =					35.1	kips, (ACI 318-05 Sec. 11.5.6.2)
61		smax	=	14	(ACI 318-05, Section 7.10.5.2)						sprovd	=	12	in
62		smin	=	1										[Satisfactory]
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