Buckling of web plates (www.theengineeringcommunity.org).xls

	B	C	D	E	F	G	H	I	K	T	U	W	AB	AC	AD	AE	AF	AG	AH	AI	AJ	AK	AL	AM	AN	AO	AP	AQ	AR	AS	AT	AU	AW	AX	AY	BA	BB	BC	BE	BF
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2	BUCKLING OF WEB PLATES
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4	ELEMENT ID :				Girder at Mezzanine (1000 kg/m2)								Logic Chart																	Special Cases to get K for Case 2
5													Case		3	Case		1	a	>	1	K	=	84,00		K	=	84,00		K'	=	7,64	K'	=	7,64	K	=	23,90
6	SECTION DIMENSIONS															Case		1				K	=	109,96		K	=	84,00		K'	=	10,00	K'	=	7,64
7	Upper Flange Width						=	320	mm							Case		2	----			K	=	23,90		K	=	23,90											3	23,90
8	Upper Flange Thickness						=	22	mm							Case		2	----			K	=	23,90		K	=	23,90											3	23,90
9	Web Depth (b)						=	1000	mm							Case		3	a	>	2/3	K	=	23,90		K	=	23,90		K"	=	23,90	K"	=	23,90				3	23,90
10	Web Thickness (t)						=	8	mm										a	>	2/3	K	=	23,90						K"	=	23,90							3	23,90
11	Lower Flange Width						=	320	mm													K	=	41,37						K"	=	41,37							N.A
12	Lower Flange Thickness						=	22	mm													K	=	41,37						K"	=	41,37							N.A
13	Distance between Vertical Stiffeners (If not found, to be taken as Span Length), (a)									=	1,70	m	SHEAR						a	>	1	K	=	6,72		K	=	6,72
14													SHEAR									K	=	5,85		K	=	6,72
15	ACTING FORCES
16	Compression (P)				=	0,00		t
17	Moment (M)				=	196,00		mt					fcr,i,red.		=	2.4 - 0.0000445			l2	=	2,08		t/cm2
18	Shear (Q)				=	0,00		t					2,08
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20	SECTION PROPERTIES
21	Y		=	52,20		cm							CHECK SAFETY AGAINST WEB BUCKLING
22	Area (A)		=	220,80		cm2							y	=	f2 / f1		=	-1,00
23	Ix		=	434381,824		cm4							a	=	a / b		=	1,70
24													fE	=	1898 (t/b)2			=	0,12		t/cm2
25	CALCULATION OF STRESSES												For Normal Stresses:					K	=	23,90
26	f1	=	-2,36		t/cm2								For Shear Stresses:					K	=	6,72
27	f2	=	2,36		t/cm2								fcr	=	K . fE		=	2,90		t/cm2
28	q	=	0,00		t/cm2								qcr	=	K . fE		=	0,82		t/cm2
29													fcr,I	=	2,90		t/cm2
30													fp	=	0.8 fy		=	1,92		t/cm2		(St. 37)
31													fcr,i	>	fp			Case of Inelastic Buckling - Reduction in the equivalent stress must be calculated
32													l	=	( p2 . E / fcr,I )1/2				=	84,49
33													fcr,i,red.		=	2.4 - 0.0000445				l2	=	2,08		t/cm2
34													Factor of safety against buckling (F.S)									=	fcr,i,red.		/ (f12 + 3q2)1/2			=	0,88
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