Project 1

DESIGN OF BOX CULVERT FOR 12.00x5.50
	1.0 Input Data
	INPUT DATA :
	No. of cells					=	1
	Clear Span					=	12.00	m
	Clear Height					=	5.50	m
	Width of Haunch					=	0.20	m
	Depth of Haunch					=	0.20	m
	No. of division of Bottom Slab					=	13
	Depth of Top Slab					=	0.400
	Depth of Bottom Slab					=	0.450
	Depth of External Vertical Wall					=	0.400
	Thickness of Wearing Coat					=	0.05	m
	Thickness of fill (on top slab)					=	4.404	m
	Safe bearing capacity					=	23	t/m2
	Allowable Settlement					=	75	mm
	Spring Constant					=	3066.67	KN/m
	Density of concrete					=	25	KN/m3
	Bulk Density of soil					=	20	KN/m3
	Density of submerged soil					=	10	KN/m3
	Density of Wearing Coat					=	22	KN/m3
	Density of fill					=	20	KN/m3
	Coefficient of earth pressure at rest					=	0.5
	Carriageway Width					=	13	m
	Crashbarrier					=	0.5	m
	Footpath					=	2	m
	Camber					=	2.5	%
	Maximum air shade temparature					=	46.5	oC
	Minimum air shade temparature					=	8.6	oC
	2.0 Design Parameter
	Grade of Concrete					=	M30
	Grade of steel ("m" for Mild Steel & "t" for HYSD)					=	t
	Type of exposure						Moderate
	Minimum grade of concrete						M 30
	Clear Cover for earth face structural component					=	75	mm
	Clear Cover for inside face structural component					=	40	mm
	Clear Cover for foundation					=	75	mm
	Impact For Live Load					=	1.25
	Thickness of top slab				=	400	mm
	Thickness of bottom slab				=	450	mm
	Thickness of external vertical wall				=	400	mm
	Thickness of internal vertical wall				=	0	mm

2.1 Dead Load
	Self weight of the structure has been calculated directly in STAAD file by the comment "SELFWEIGHT -1".
2.2 Super Imposed Dead Load
	(a) Wearing Coat						10.35
	Thickness of wearing coat					=	0.05	m
	Load ( UDL ) due to WC on top slab = 0.05 x 22					=	1.1	kN/m width
	(b) Earth fill on top slab
	Thickness of earth fill					=	4.40	m
	Load (UDL) due to earth fill =			4.404 * 20		=	88.08	KN/m
	(c) Parapet loading
	Load ( UDL ) due to Parapet					=	8.2875	KN/m
	Total SID load on top slab					=	96.3675	KN/m
2.3 Earth Pressure
	Thickness of top slab					=	0.400	m
	Height of top haunch					=	0.150	m
	Clear height between top & bottom slab					=	5.500	m
	Height of bottom haunch					=	0.150	m
	Thickness of bottom slab					=	0.450	m
	Height from top		Intensity of Earth pressure
	(m)		(KN/m2)
			Bulk soil				Submerged soil	Water
	4.654	4.654	0.5 * 20 * 4.654		=	46.54	0.00	0.00
	0.200	4.854	0.5 * 20 * 4.854		=	48.54	0.00	0.00
	0.150	5.004	0.5 * 20 * 5.004		=	50.04	0.00	0.00
	0.155	5.159	0.5 * 20 * 5.159		=	51.59	0.00	0.00
	2.445	7.604	0.5 * 20 * 7.604		=	76.04	12.23	24.45
	2.445	10.049	0.5 * 20 * 10.049		=	100.49	12.23	24.45
	0.155	10.204	0.5 * 20 * 10.204		=	102.04	0.77	1.55
	0.150	10.354	0.5 * 20 * 10.354		=	103.54	0.75	1.5
	0.225	10.579	0.5 * 20 * 10.579		=	105.79	1.13	2.25
2.4 Live Load Surcharge
	Equivalent height					=	1.2	m
	Uniform Intensity of loading =			1.2 * 20 * 0.5		=	12.00	KN/m2
2.5 Braking Load
Carriageway Live Load						=	700	KN
Dispersion width						=	8.554	m
Braking Load =		0.2 * 700/8.554				=	0.00	KN
FILL ON STRUCTURE IS MORE THAN 1M HENCE BRACING FORCE DIRECTLY TRANSFERED FROM SOIL HENCE BRAKING FORCE IS IGNORED

3)	LIVE LOADS DUE TO 70R TRACK LOADING
		Max. Wheel Load =	35.00 t
		Tyre Dimensions :
			a) Along Traffic Direction :	4.57 m		Distributed width =	6.670 m
			b) Perpendicular to Traffic :	0.84 m		Distributed width =	5.000 m
			c) Spacing of Wheels :	2.90 m
	Distribution along the span is contact area +2X(Fill Height)
	Hence,
			Width of Tyre after Distribution =	4.570 + 2 x ( 0.05 + 1 ) =		6.670m
						6.670m
	Distribution Perpendicular to span = beff =b1
	where,
	lo =	Effective Span =		12.40 m
	b1 =	Breadth of concentrated area = tyre width + 2 x Fill height.
	=	2.940m
	Hence, beff =	2.940m	For one wheel		Impact =	1.11125
					Hence the the dispersion in the perpendicular to the traffic =	5.000m
	Thus for the axle, the total Distribution area is as shown in the shaded portion is
	=	5.000 m	x	6.670m	for a load of
		Total wheel Load =	70.000 t
				Hence, equivalent unit area load =	2.332 t	/m2
		Hence, udl =	2.332 t	/m2
		Hence, Base udl =	2.332 t	/m2

4)	LIVE LOADS DUE TO 40T BOGIE LOADING - 20T AXLE LOAD
		Max. Wheel Load =	5.00 t
		Tyre Dimensions :				Tyre width =	36 cm
			a) Along Traffic Direction :	0.2634 m		Tyre Pressure =	5.273	kg/cm2
			b) Perpendicular to Traffic :	0.86 m
			c) Spacing of Wheels perpendicular drn. :	2.79 m		Crash Barrier (Typ)		CL of Deck
			c) Spacing of Wheels longitudinal drn. :	1.22 m
			Width of the Slab Panel :	13.00 m

		Dispersion along Traffic direction =	0.263+ 2 X 1 =		3.583 m
where,
lo =	Effective Span =		12.40 m
b1 =	Breadth of concentrated area = tyre width + 2 x Fill ht.
=	2.960m
Hence, beff =	2.960m		For one wheel	Impact =	0
				Hence the the dispersion in the perpendicular to the traffic =	4.890m
Thus for the axle, the total Distribution area is as shown in the shaded portion is
=	4.890 m	x	3.583m	for a load of
	Total wheel Load =	40.000 t
			Hence, equivalent unit area load =	0.000 t	/m2
			0.000 t

LIVE LOADS DUE TO CLASS A LOADING
	Max. Wheel Load =	5.70 t
	Tyre Dimensions :
		a) Along Traffic Direction :	0.25 m
		b) Perpendicular to Traffic :	0.50 m
		c) Spacing of Wheels perpendicular drn. :	2.30 m
		c) Spacing of Wheels longitudinal drn. :	1.20 m
		Width of the Slab Panel :	13.00 m

		Dispersion along Traffic direction =	0.250+ 2 X 1=		3.550 m
Distribution Perpendicular to span = beff = + b1
where,
lo =	Effective Span =		12.40 m
b1 =	Breadth of concentrated area = tyre width + 2 xFill ht.
=	2.600m
Hence, beff =	2.600m	For one wheel			Impact =	1.12
				Hence the the dispersion in the perpendicular to the traffic =	4.000m
But the pannel width is restricted =			13.00 m
Thus for the axle, the total Distribution area is as shown in the shaded portion is
=	4.000 m	x	3.550m	for a load of
	Total wheel Load =	22.800 t
			Hence, equivalent unit area load =	1.802 t	/m2
			1.802 t

1.802t Loading on top

Temperature Rise Case

Parameters:
Grade of Concrete			M =	30
Elasticity for Concrete			Ei =	3.100E+07	KN / m2
Modified Elasticity for Concrete			Ec =Ei	3.100E+07	KN / m3
Coefficient of Thermal Expansion			a =	0	/ oC

Member Description	h	h1	h2	h3	T1	T2	T3
	( m )	( m )	( m )	( m )	oC	oC	oC
Top slab	0.4	0.120	0.120	0.120	17.80	4.00	2.10

Member Description	CG of Section from TOP	CG of Section from BOTTOM	CG of TOP BLOCK from TOP	CG of MID BLOCK from TOP	CG of BOTTOM BLOCK from BOTTOM	x1	x2	x3	M
						m	m	m	T - m
Top slab	0.2000	0.2000	0.0473	0.1600	0.0400	0.1527	0.0400	0.1600	0.00

TEMPERATURE FALL CASE:-

Member Description	h	h1	h2	h3	h4	T1	T2	T3	T4
	( m )	( m )	( m )	( m )	( m )	oC	oC	oC	oC
Top slab	0.400	0.080	0.100	0.100	0.080	10.60	0.70	0.80	6.60

Member Description	CG of Section from TOP	CG of Section from BOTTOM	CG of TOP BLOCK from TOP	CG of MID BLOCK from TOP	CG of MID BLOCK from BOTTOM	CG of BOTTOM BLOCK from BOTTOM	x1	x2	x3	x4	M
							m	m	m	m	KN - m
Top slab	0.200	0.200	0.028	0.113	0.113	0.030	0.172	0.087	0.087	0.170	0.00

6.0 Design of Box Structure(ULS-Basic Combonation)
	Grade of Concrete				=	M 30
	Grade of Steel				=	Fe 500
	Stress in Concrete (Rare combination)				=	14.4
	Stress in Steel (Rare & QPC combination)				=	400
	Modular ratio				=	6.45
	Modular Ratio for Long term				=	12.90
	Stress in Concrete (QPC combination)				=	10.8
	Crack width (As per Table-12.1)				=	0.3	mm
	Min shear reinf. ratio (ρmin)				=	0.00091
	Design yield strength of shear reinforcement fywd = 0.8*fyk/γs				=	348
	modulus of elasticity of rienforcing of steel (Es)				=	200000	MPa
	modulus of elasticity of concrete (Ecm)				=	31000	MPa

6.1Check for ULS as per IRC:112-2011: Main Reinforcement

Section	Face	Mdes (KN.m)	Depth of NA (mm)	MOR (KN.m)	Doverall provd. (mm)	Deff provd. (mm)	Ast Req. (min) (mm2/m)	Bar Mark	Diameter	Spacing	Bar Mark	Diameter	Spacing	Ast provd. (mm2/m)	r (% steel)	Status (check for MOR)	Steel Min & Max Check
1	Top	54	48.0	173.4	400	354	460	w1	12	190	ts1	12	190	1190	0.34	OK	OK
	Bottom	31	22.8	86.3		360	468	-	0	200	ts2	12	200	565	0.16	OK	OK
2	Top	0	24.0	89.2	400	354	460	ts1	12	190	-	0	200	595	0.17	OK	OK
	Bottom	101	40.6	148.2		354	460	ts2	12	190	ts8	10	190	1009	0.28	OK	OK
3	Top	54	48.0	173.4	400	354	460	ts1	12	190	w1	12	190	1190	0.34	OK	OK
	Bottom	31	22.8	84.8		354	460	ts2	12	200	-	0	200	565	0.16	OK	OK
4	Outside	83	48.0	155.3	400	319	415	w1	12	190	ts1	12	190	1190	0.37	OK	OK
	Inside	0	24.0	80.1		319	415	w2	12	190	w2'	0	190	595	0.19	OK	OK
5	Outside	58	24.0	80.1	400	319	415	w1	12	190	-	0	200	595	0.19	OK	OK
	Inside	0	24.0	80.1		319	415	w2	12	190	w2'	0	190	595	0.19	OK	OK
6	Outside	80	46.8	151.6	400	319	415	w1	12	190	bs1	12	200	1161	0.36	OK	OK
	Inside	0	24.0	80.1		319	415	w2	12	190	w2'	0	190	595	0.19	OK	OK
7	Top	61	40.5	153.4	450	367	477	bs3	16	200	-	0	200	1005	0.27	OK	OK
	Bottom	54	46.8	176.9		369	480	w1	12	190	bs1	12	200	1161	0.32	OK	OK
8	Top	127	48.0	181.2	450	369	480	bs3	12	190	bs6	12	190	1190	0.32	OK	OK
	Bottom	0	24.0	93.1		369	480	bs1	12	190	-	0	190	595	0.16	OK	OK
9	Top	61	40.5	153.4	450	367	477	bs3	16	200	-	0	200	1005	0.27	OK	OK
	Bottom	54	46.8	176.9		369	480	bs1	12	200	w1	12	190	1161	0.32	OK	OK

Distribution Steel
Section	Face	Doverall provd. (m)	Deff provd. (m)	Ast reqd. (mm2/m)	Bar Mark	Diameter	Spacing	Bar Mark	Diameter	Spacing	Ast provd. (mm2/m)	r (% steel)	Status
1,2 & 3	Top	400	343	238	ts4	10	200	-	0	200	393	0.11	OK
	Bottom			202	ts5	10	200	-	0	200	393	0.11	OK
4,5 & 6	Outside	400	308	238	w4	10	200	-	0	200	393	0.13	OK
	Inside			238	w5	10	200	-	0	200	393	0.13	OK
7,8 & 9	Top	450	354	238	bs4	10	200	-	0	200	393	0.11	OK
	Bottom			232	bs5	10	200	-	0	200	393	0.11	OK

Check for Shear as per IRC:112-2011: Main Reinforcement
Section	Vdes (KN)	Axial Comp. Force (KN)	Shear resistance without shear reinforcement (VRd.c)	Minimum shear resistance (νRd.c)	Shear Resistance of concrete (KN)	Status	Dia of bar	No of legs	Spacing of bar	Spacing Required	Check for Spacing	Max Long. Spacing	ρmin provided	Check (ρmin)< (ρmin)pro	SF by shear reinf. (V Rd.s)	VNS - VRd.c	Status
1,2 & 3	166.4	93.4	160.6237	151.745	160.62	Provide Shear Rein.	12	4	200	851	OK	266	0.00226	OK	708.4	5.82	OK
4,5 & 6	65.8	255.8	170.8254	160.507	170.83	OK	0	2	200	0	Not OK	239	0.00000	Not OK	0.0	-105.00	OK
7,8 & 9	198.4	123.1	164.9478	158.479	164.95	Provide Shear Rein.	12	4	200	804	OK	277	0.00226	OK	796.997	33.40	OK

SLS-RARE COMBINATION:-

Tensile strength of concrete (fctm)			2.5	MPa
Clear Cover for earth face			75	mm
Clear Cover for inside face			40	mm
Clear Cover for foundation			75	mm

7.1 Check for SLS as per IRC:112-2011: Stress checking of Concrete and Steel
Main Reinforcement
Section	Face	Mdes (KN.m)	Depth of NA (m)	Doverall provd. (m)	Deff provd. (m)	Ast Prov. (mm2/m)	Max. comp. stress in concrete	Status	Max. Tensile stress in steel	Status
1	Top	38	65	400	354	1190	3.40	OK	97.49	OK
	Bottom	22	48		360	565	2.68	OK	113.30	OK
2	Top	0	47	400	354	595	0.00	OK	0.00	OK
	Bottom	72	60		354	1009	6.81	OK	214.23	OK
3	Top	38	65	400	354	1190	3.39	OK	97.24	OK
	Bottom	22	47		354	565	2.76	OK	115.27	OK
4	Outside	60	61	400	319	1190	6.29	OK	170.43	OK
	Inside	0	45		319	595	0.00	OK	0.00	OK
5	Outside	38	45	400	319	595	5.28	OK	208.76	OK
	Inside	0	45		319	595	0.00	OK	0.00	OK
6	Outside	58	61	400	319	1161	6.24	OK	169.04	OK
	Inside	0	45		319	595	0.00	OK	0.00	OK
7	Top	42	63	450	367	1005	3.89	OK	121.45	OK
	Bottom	38	66		369	1161	3.21	OK	94.35	OK
8	Top	90	73	450	369	1190	8.20	OK	215.10	OK
	Bottom	0	48		369	595	0.00	OK	0.00	OK
9	Top	42	63	450	367	1005	3.89	OK	121.45	OK
	Bottom	38	66		369	1161	3.20	OK	94.05	OK

8.1 Check for SLS as per IRC:112-2011: Crack width checking
											400
Main Reinforcement
Section	Face	Mdes (KN.m)	Depth of NA (m)	Doverall provd. (m)	Deff provd. (m)	Ast Prov. (mm2/m)	Max. comp. stress in concrete	Status	Max. Tensile stress in steel	Status	Equivalent Diameter	rr.eff	(esm-ecm)	Maximum crack spacing Sr,max	Crack width	Status
1	Top	32	65	400	354	1190	2.83	OK	81.17	OK	12	0.01066	0.00024	361.4	0.09	OK
	Bottom	22	48		360	565	2.68	OK	113.30	OK	12	0.00565	0.00034	496.8	0.17	OK
2	Top	0	47	400	354	595	0.00	OK	0.00	OK	12	0.00518	0.00000	564.1	0.00	OK
	Bottom	72	60		354	1009	6.81	OK	214.23	OK	11	0.00891	0.00064	347.7	0.22	OK
3	Top	32	65	400	354	1190	2.83	OK	81.17	OK	12	0.01066	0.00024	361.4	0.09	OK
	Bottom	18	47		354	565	2.29	OK	95.68	OK	12	0.00492	0.00029	550.9	0.16	OK
4	Outside	52	61	400	319	1190	5.45	OK	147.75	OK	12	0.01055	0.00044	363.4	0.16	OK
	Inside	0	45		319	595	0.00	OK	0.00	OK	12	0.00503	0.00000	541.8	0.00	OK
5	Outside	31	45	400	319	595	4.35	OK	172.12	OK	12	0.00503	0.00052	575.8	0.30	OK
	Inside	0	45		319	595	0.00	OK	0.00	OK	12	0.00503	0.00000	541.8	0.00	OK
6	Outside	50	61	400	319	1161	5.35	OK	145.06	OK	12	0.01028	0.00044	368.4	0.16	OK
	Inside	0	45		319	595	0.00	OK	0.00	OK	12	0.00503	0.00000	541.8	0.00	OK
7	Top	33	63	450	367	1005	3.08	OK	96.21	OK	16	0.00779	0.00029	485.2	0.14	OK
	Bottom	28	66		369	1161	2.39	OK	70.12	OK	12	0.00908	0.00021	394.7	0.08	OK
8	Top	84	73	450	369	1190	7.65	OK	200.70	OK	12	0.00947	0.00060	351.4	0.21	OK
	Bottom	0	48		369	595	0.00	OK	0.00	OK	12	0.00445	0.00000	628.8	0.00	OK
9	Top	33	63	450	367	1005	3.08	OK	96.21	OK	16	0.00779	0.00029	485.2	0.14	OK
	Bottom	28	66		369	1161	2.39	OK	70.12	OK	12	0.00908	0.00021	394.7	0.08	OK

Check for Safe Bearing Capacity of Soil

Support No.	DL+SIDL reaction (KN)	Live load Reaction (KN)
			1	13.053	0.735
2	22.822	1.284
3	23.132	1.301
4	26.583	1.495
5	39.785	2.236
6	53.05	2.981
7	53.025	2.979
8	53.05	2.981
9	39.785	2.236
10	26.583	1.495
11	23.132	1.301
12	22.822	1.284
13	13.053	0.735
SUM	409.9	23.0

MODEL CALCULATIONS:-

Model Calculation :
Maximum BM under Basic combination	=	5.41	t-m	(From analysis)
Maximum SF (VNS)	=	16.64	t
Total Depth D (mm)	=	400	mm
Effective Depth d (mm)		354	mm		Deck slab Thk =	0	mm
Web Width bw (mm)		1000	mm
Width bf (mm)		1000	mm
Grade of Concrete Used	=	M30
Grade of steel		Fe 500
Charactristic strength of concrete (fck)	=	30	MPa	Table no:6.5 (IRC:112-2011)
Charactristic strength of steel (fy)	=	500	MPa	Table no:18.1 (IRC:112-2011)
Tensile strength of concrete (fctm)	=	2.5	MPa	Table no:6.5 (IRC:112-2011)
Design yield strength of shear reinforcement fywd = 0.8*fyk/γs	=	348	MPa	Page 86 IRC-112-2011
Partial material safety factor for concrete (gm)	=	1.5	Basic	Page 49: (IRC:112-2011)
Partial material safety factor for Steel (gs)	=	1.15	Basic	Page 30: (IRC:112-2011)
Ultimate compressive strain in the concrete (єcu3)	=	0.0035	Up to fck ≤ 60Mpa	Table no:6.5 (IRC:112-2011)
modulus of elasticity of rienforcing of steel (Es)	=	200000	MPa	Clause 6.2.2 (IRC:112-2011)
modulus of elasticity of concrete (Ecm)	=	31000	MPa	Table no:6.5 (IRC:112-2011)
modular ratio αe (Es/Ecm)	=	6.45		page-126 IRC-112-2011
Ultimate tensile strain in the steel (єs) = [{fy/(gs xEs)}+0.002]	=	0.00417
Coefficient to consider the influence of the concret strength (a)	=	0.67	Cube	A2.10 Page : 244 (IRC:112-2011)		Cube	or
Factor (l )	=	0.8	0.8 Up to fck ≤ 60Mpa,Eq.A2-33 (IRC:112-2011); 0.8-((fck-60)/500) for 60<fck<110Mpa
Factor (h)	=	1	1.0 Up to fck ≤ 60Mpa,Eq.A2-35 (IRC:112-2011); 1.0-((fck-60)/250) for 60<fck<110Mpa
fcd = (a*fck/gm)	=	13.400		page-49 IRC-112-2011
Factor b = (l/2)	=	0.400
Cover	=	75	mm
Effective depth of member (d)	=	354	mm
Limiting Neutral axis depth (x lim) = (d*ecu3)/(es+ecu3)	=	161.46	mm
Calculation for check of Moment Of Resistance (MOR) Ultimate Limit State (ULS)
Force in compression = Force in Tension from Fig. A2-4 - rectangular Tensile Strength				and	IS 456: 2000 , Annexure-G
Fig- A2-4, page-243 IRC-112-2011
fy*As	=	hfcd * lxu* b	=	Fav * xu*b
gs
fcd	=	a*fck/gm	=	0.4467 fck
Fav	=	hfcd * l	=	0.36 fck
fy/gs	=		=	0.87 fy
As	=	(0.36 fck/0.87 fy ) Xu * b
X	=	0.87 fy * As / (0.36 fck * b )
Assume Provided reinforcement Ast	=	1190	mm2
Ast min	=	460	mm2	>	1190		OK
Ast max	=	10000	mm2	<	1190		OK
					Cl. 16.5.1.1 of IRC:112-2011
Neutral Axis Depth X (mm)	=			=	47.95	mm
Lie in flange	=	0.87*fy*Ast/0.36*fck*bf
Neutral Axis Xlim (mm)	=	(d*ecu3)/(es+ecu3)		=	161.46	mm
Neutral Axis Depth Xu (mm)	=	47.95<161.46		=	47.95	mm
MOR :	=	0.36 fck * Xu * bf * ( d -0.4 * Xu )			OR
		0.87 fy * As * ( d -0.4 * Xu )
	=	17.339	t.m
							Check for MOR
MOR Mu lim (t.m)	=	17.339	t.m	>	5.41		OK
Calculation of Stresses in the reinforcement and concrete (SLS- Rare combination)
Maximum BM under Rare combination	=	3.83	t.m	(From analysis)
Neutral Axis Depth X (mm)	=	47.95	mm
MOI of cracked section (mm4)	=	b*x^3/12+m*Ast*(d-x)^2
	=	7.56E+08	mm4
Stress in concrete (Mpa)	=	M*x/Icr	=	2.43	<	14.4	OK
Stress in Steel (Mpa)	=	m*M*(d-x)/Icr	=	100.06	<	400	OK
Calculation of Stresses & Crack width (SLS- Quasi permanent combination)
Maximum BM under Rare combination	=	3.19	t.m	(From analysis)
Neutral Axis Depth X (mm)	=	47.95	mm
MOI of cracked section (mm4)	=	b*x^3/12+m*Ast*(d-x)^2
	=	7.56E+08	mm4
Stress in concrete (Mpa)	=	M*x/Icr	=	2.02	<	10.8	OK
Stress in Steel (Mpa)	=	m*M*(d-x)/Icr	=	83.31	<	400	OK
Equi.diameter фeq (mm)	=	(n1ф12+n2ф22)/(n1ф1+n2ф2)			Eq. 12.9 of IRC:112-2011
	=	20	mm
Spacing of Bonded Reinf. in tension zone (mm)	<=	5*(C+ф/2)			Cl. 12.3.4 (3) of IRC:112-2011
	=	425	mm
rr.eff	=	As/Ac.eff	=	0.010	Eq. 12.7 of IRC:112-2011
As	=	1190	mm2
Ac eff	=	115000	mm2
Max. crack spacing Srmax (mm)	=	3.4c+0.17ф/ρp.eff			Eq. 12.11 of IRC:112-2011
	=	583.4	mm
(esm-ecm)	=	(ssc - kt fct.eff/rr.eff (1+aerr.eff))/Es			>=0.6 ssc / Es
ssc	=	83.31	Mpa		Eq. 12.6 of IRC:112-2011
ae	=	6.45
(esm-ecm)	=	0.000250
Crack width Wk (mm)	=	Sr,max (esm-ecm)			Eq. 12.5 of IRC:112-2011
	=	0.146	mm	<	0.30		OK
					Table-12.1 of IRC:112-2011
Check for Shear- Ultimate Limit State (ULS)
Shear Force VEd (t)	=	16.64	t
Max Shear Force Vmax (t)	=	0.5bwdνfcd			Eq. 10.5 of IRC:112-2011
	=	3856.38	t	>	16.64		OK
Where ν is a strength reduction factor	=	0.6(1-fck/310)fck			Eq. 10.6 of IRC:112-2011
	=	16.26
Longitudinal force (NEd) (t)	=	9.34	t
Shear resistance of section without shear reinf. (VRd.c) (t)	=	[0.12K(80ρ1.fck)0.33+0.15σcp]bw.d			Vide 10.3.2 of IRC:112-2011
Where K	=	1+	√200	<= 2	Where d is depth in mm
			√d		Eq. 10.2 of IRC:112-2011
νmin	=	0.031K3/2fck1/2			Eq. 10.3 of IRC:112-2011
Comp. stress as CG due to Axial load (σcp) Mpa	=	NEd/Ac	<=	0.2fcd
	=	0.23	Mpa
ρ1	=	Asl/bw.d	<=	0.02
	=	0.0034
VRd.c (t)	=	16.06	t
Minimum shear resistance (VRd.c) (t)	=	(νmin+0.15σcp)bw.d			Eq. 10.1 of IRC:112-2011
	=	15.17	t
Shear resistance (VRd.c) (t)	=	max (16.06,15.17)
	=	16.06
Check Shear Reinforcement Requirement	=	16.06	t	<	16.64	t
	=	Provide Shear Reinf.
If Shear reinforcement required then,
Net Design Shear force VNS (t)	=	16.64	t
Comp. stress as CG due to Axial load (σcp) Mpa	=	0.23	Mpa	(NEd/Ac)
Shear Force by shear reinforcement (V Rd.s)	=	Aswzfywdcotθ/s			Eq. 10.7 of IRC:112-2011
VNS	=	V Rd.max
	=	αcwbwzν1fcd/(cotθ+tanθ)			Eq. 10.8 of IRC:112-2011
	=	αcwbwzν1fcd/(2/sin2θ)
σcp	=	0.234			0.25 fcd	=	3.35
αcw	=	1.02			0.50 fcd	=	6.70
					1.0 Fcd	=	13.401
					Eq. 10.9 of IRC:112-2011
sin2θ	=	VNS/αcwbwzν1fcd/2
	=	16.64435*10000/((1.01742463249011*1000*0.9*400*0.6*13.4)/"2
	=	0.113
θ	=	3.25
Cotθ	=	17.64	<=	2.5	>=	1
	=	2.50		vide cl.10.2.2.2 of IRC-112-2011
Dia of bar used as shear reinforcement	=	8	mm
Nos of legs	=	2	Nos
Spacing of bars (s) mm	=	100	mm
Min shear reinf. ratio (ρmin)	=	0.072√fck/fyk		Eq. 10.20 OR 16.5 of IRC:112-2011
ρmin	=	0.00091
ρmin provided	=	0.00101	>	0.00091			OK
Shear Force by shear reinf.(V Rd.s) (t)	=	31.486	t
VNS - VRd.c	=	0.582	t
Check for Shear	=	31.486	t	>	0.582	t	OK
Asw max (mm2)	=	0.5*αcw bw s ν1 fcd/fywd		Eq. 10.10 of IRC:112-2011
	=	940.31	mm2
Check For Asw max > Asw pro	=	940.31	mm2	<	100.53	mm2	OK
Max Long. Spacing (mm)	=	266	mm	Eq. 16.6 & 16.8 of IRC:112-2011
Check for Spacing	=	266	mm	>	100	mm	OK
Surface Reinforcement Reqd (mm2)	=	4862.00	mm2	Cl.16.5.4 of IRC:112-2011
Dia. of bars for surface reinf. (mm)	=	12.00	mm
Spacing of bars ( mm)	=	90.00	mm
Surface Reinforcement Provided (mm2)	=	4940.15	mm2
Check for Surface Reinf.	=	4940.15	mm2	<	4862.00	mm2	OK

RESULTS:-

				I		II		III		IV		Axial Comp. Force(LL)	Axial Comp. Force(ULS)
			Node No	LL (KN-m)		ULS-Basic Comb.		SLS-Rare Comb.		SLS-Quasi Perm. Comb.
				BM 'KN-M'	SF 'KN'	BM 'KN-M'	SF 'KN'	BM 'KN-M'	SF 'KN'	BM 'KN-M'	SF 'KN'	KN	KN
At Face of wall outer		sec 1	51	1.29		51.583		36.637		31.906		0.253	93.15
At End of haunch outer			51	1.29
D from support outer		SF Sec 1	53		7.011		154.703	9.153		6.905		0.253	65.142
At center of top slab		sec 2	54	5.758		93.071		66.48		66.48		0.253	57.382
At Face of wall inner		sec 3	57	1.29		51.434		36.538		31.906		0.253	93.002
At End of haunch inner
D from support inner		sec 3	55		7.011		154.703	9.054		6.905		0.253	65.29
At Face of wall		sec 4	106	3.428		77.21		56.394		52.154		10.414	229.227
			206	3.428		77.09		56.315		52.154		10.414	229.227
At End of haunch
D from support			104		0.253		58.396	42.35		39.717		10.414	233.471
			204		0.253		58.396	42.3		39.717		10.414	233.471
At center of wall		sec 5	103	3.08		47.15		28.848		28.848		10.414	237.282
			203	3.08		47.15		28.848		28.848		10.414	237.282
At Face of wall		sec 6	100	2.732		75.779		55.276		50.092		10.414	245.338
			200	2.732		75.631		55.178		50.092		10.414	245.338
At End of haunch
D from support			102		0.253		65.271	38.803		37.45		10.414	241.094
			202		0.253		65.123	38.734		37.45		10.414	241.094
At Face of wall		sec10	2	0.476		53.174		37.003		28.138		0.253	90.978
At End of haunch
D from support			4		6.08		186.049	28.496		22.441		0.253	89.403
At center of bottom slab		sec 11	5	4.01		72.363		49.744		44.612		0.253	89.403
			6	5.917		108.438		76.751		74.39		0.253	89.403
			7	6.553		118.272		84.289		84.289		0.253	65.271
			8	5.917		108.258		76.63		74.39		0.253	57.511
			9	4.01		72.182		49.624		44.612		0.253	57.511
At Face of wall		sec 12	12	0.476		52.993		36.882		28.138		0.253	122.869
At End of haunch
D from support			10		7.451		186.048	28.375		22.441		0.253	57.511

STAAD PRO:-

LL due to 70R wheeled

LL due to Class A loading:-