JEC Prog - Design of Corbel (www.theengineeringcommunity.org).xlsx

	B	C	D	E	F	G	H	I	J	K	S	T
1	Design of Concrete Corbel
2	Design of Concrete Corbel
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4
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6
7	UNITS:	SI					1			Normal-weight	SI	English
8	SI	1	E =	200000	MPa		Agregate, λ =	Normal-weight		Sand-weight
9	English		fc' =	30	MPa		1,4	1		Lightweight	21	5000
10			fy =	450	MPa		Friction, μ =	1.4λ for Concrete cast monolithically.			480	60000
11	Dimensions:						1.4λ for Concrete cast monolithically.	1
12			a =	250	mm		1.0λ for Concrete placed against hardened rough concrete.				75	5
13			d =	500	mm		0.6λ for Concrete placed against unroughened hardened concrete.				150	14
14			h =	600	mm		0.7λ for Concrete anchored to structural steel.				190	18
15			b =	500	mm						150	10
16			Remarks: Safe!
17	Reinforcement sizes:
18			Primary tension:	500	mm	196349,5408					8967	0,875
19			Closed stirrups:	10	mm	157,0796327					9,375	0,375
20			Framing:	10	mm	78,53981634					9,375	0,375
21			Welded anchor:	22	mm	380,1327111					21,875	0,875
22			Concrete cover =	30	mm						40	2,75
23			Allowable clear spacing =	25	mm						25	1
24			Remarks: not OK!
25	Loads:
26			VDL =	50	kN	50000					15	1,823976
27			VLL =	300	kN	300000					23,12856	23,12856
28			[Optional] VU =	0	kN	0					140	80
29			[Optional] NU =	0	kN
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31
32			Vu =	1.4VDL + 1.7VLL
33			=	60	kN	or	60000	N
34			Nu =	0	kN	or	0	N
35
36			Vn ≥	Vu	=	60000
37			Vn ≥	Ø	=	0,75
38			Vn =	712,000	N
39
40			0.2f`cbwd =	1500000	MPa	> Vu = 60000 N OK!
41		5,5158068	bwd =	1378951,7	MPa	> Vu = 60000 N OK!
42
43			Avf =	Vu
44			Avf =	Øfyμ
45			=	712,000
46			=	450(1,4)
47			=	1,130	mm²
48			Af =	Mn	=	Vua+Nu(h-d)
49			Af =	Øfyjd	=	Øfyjd
50				where: jd ≈		0.85d
51				where: jd ≈		0.85d
52					jd =	425	mm
53				Act Nu	=	0	N
54				Min Nu	=	0.20Vu
55					=	12000	N
56				Use: Nu	=	12000	N
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58
59
60			Af =	60000(250)+12000(600-500)
61			Af =	0.90(450)(425)
62			=	16200000
63			=	172125
64			Af =	94,118	mm²
65			An =	Nu	=	12000
66			An =	Øfy	=	0,75(450)
67			=	35,556	mm²
68
69		Primary steel:
70		(a)	As =	2/3Avf + An	=	2/3(1,13) + 35,556
71			=	36,309
72		(b)	As =	Af + An	=	(94,118) + 35,556
73			=	129,673	mm²
74			min As = 0.04		f`c	bd
75			min As = 0.04		fy	bd
76			= 0.04		30		(500)(500)
77			= 0.04		450		(500)(500)
78				Min As	=	666,667	mm²
79			Therefore: since case (a) governs
80			As =	129,673	mm²
81			n =	0,001	say	2
82			Soc =	-60,000	mm
83			Scl =	-560,000	mm	Not OK!
84
85		Horizontal closed stirrups:
86			since case (a) governs
87					Ah =	1/2Af
88					=	47,059
89					=	47,059	mm²
90					Min .Ah =	0.5(As -An)
91					=	0.5(36,309 - 35,556)
92					=	0,377	mm²
93
94			Use: Ah =	47,059	mm²
95			n =	0,300	say	2
96			Soc =	333,333	mm		spread above 2/3d =	333,333	mm
97			Scl =	323,333	mm	OK!
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