Design of Reinforced Concrete Structures

Dhanpat Rai & Sons
	CONTENTS
	Introduction
	2	Singly Reinforced Beams	49—85
	3.	Doubly Reinforced Beams	86—99
	4.	Shear in Beams	100—115
	7.	T-Beams
	8.	Axially Loaded Columns	154 -174
	9.	Combined Bending and Direct Stresses	175—183
	10,	Slabs	jm	^
	i I Beams	267-342
	12 Foundations	343 -512
	14.	Stair Cases	627—654
	15.	Water Tanks	655-917
	16.	Bonkers and Silos	918-959
	18.	Frames for Buildings	972— 1076
	19.	Reinforced Concrete Chimneys	1077—1084
	20.	Bridges	1085-1157
	2L Roads and Pavings	1158—1167
	22.	Prestressed Concrete	1168—132*
	DESIGN OF REINFORCED CONCRTE STRUCTURES
	Introduction
	§8	. White Cement
	§9	. Coloured Cement
	§10	. Sulphate Resisting Cement
	\"?’\"^ X 100%-
	§22	. Nominal Mix Concrete
	§27	. Test Strength of Sample
	§29	. Permissible Stresses in Concrete .	. „„„^..
	§33	. Loads
	Live Load
	Live Loads on Floors
	Analysis of Composite Sections by Elastic Theory
	”^w4, 1713 69 3951 ^cm-
	2
	Singly Reinforced Beams
	§39. Assumptions
	-(»«)
	.-H 7
	1 —~	...<«
	$42* Moment of Resistance
	§43 Balanced or Economic or Critical Section
	§47	. Types of Problems
	40- 3 )Ncm-
	m^Sm+ri-o
	54400X12. .	~
	«’] £ ■[’\" - i ] 4 -15 \'’-4
	280 3c
	9	20
	9	20
	3
	Doubly Reinforced Beams
	ÄÄÄiÄ.S\'g- — * \"** **
	&
	ä™äs \" mi“,bi
	L n J
	-10412\'8 £+5736*03 c -16148 83 cN cm.
	EXERCISE 3
	Shear in Beams
	I di
	r*n
	5
	Bond
	§66	. Local Bond
	s
	4.	Anchoring Reinforcing Bars
	6
	Reinforcement
	7
	T-Beams
	t)+—2—
	{80. Shear Stress in T-beams
	16
	18
	22
	23
	n S 12000	2
	±^„(2 * 0’5)1400 x 51’5 r. ~	6747
	10548000
	23440 x 5’7
	Hg. 63
	\"c“\"i8 • T^TFor-30\'^^
	i.•\"), 4 <> 552-0’25)* mare* 4	4
	mWWTnTftlHfflfHlflfll
	^\'^?J	 			 *	^J™
	if
	>	7 BARS
	I I
	J t r.t r~ . .
	Fig. 227
	Alternatively, V=-^~ .\'. M=j Vdx 0
	AF= 4=? .
	(fl)
	(fl)
	b Tm-
	4°‘ <(5+°’°926x)t M=™ ^O 429 *
	Number of bars of 22 mm. ^
	Minimum steel requirement
	—iZooV^iso““1213 cmt
	T
	i5^
	= \"\"t- h Atif at	•••(0
	Let the sizes of the footings under the columns A and B be /1XIf and kxh\' (See Fig 262).
	4-4)
	¿»■+V • 4]
	lb-0
	Max. Hogging B.M.
	[ W.-pan J x+ ) ~/[y + —/~ * “J- J
	Actual areas adopted
	§
	H \'5 2 mm. ^ bars be used as transverse steel,
	I derive depth «•52\'8 —1’2—0’6 = 51 cm.
	Spacing of 12 mm. ^ bars
	E( .^-^ 1±‘-fI>W
	bd 100x52«
	4]
	Fig. 287
	=0T5% of gross area
	Reinforcement in a pile
	2
	\' ■	\" v\'2	2 \'
	L	t
	Fig 294
	<1
	0*2
	/.	p —28’48 cm. say 28 rm,
	25 n2
	13
	Retaining Walls
	Hral
	or
	$148. Stability of Retaining Wilh
	5m
	Spacing of 16 mm. ÿ bars
	.*. Spacing of 20 mm. 4 bars
	.’.	i/\"35\'5O cm.
	ProvioMg an vnective cover of 4 cm. total thickness required
	‘5
	Check for slicing
	Equating the moment of resistance to the bending moment 85X100 ¿>-\'448725
	■	Fig. 337
	0’5 x 100
	“	7’5
	Design of the toe slab
	Design of heel slab
	teem 		jj
	KBTAIN1NO WALLS
	■1«.
	.-. r^nmiNi^
	.lx«“!\'«,),.
	—iwo-V111 ^
	0\'5x100 . ° 6\'2
	Solation.
	r-c. *
	(137266+^4986) xJxl„
	-393378 JV
	ai
	Total lateral pressure on the wall per metre run
	e-c,^
	(22
	.*. Tension transferred to the counterfort in one metre width j
	-50X5 (3-0 45) A#.
	-12967 kg.
	If the tie rods be provided at a spacing of x metres the tension in each-tie rod.
	Sometimes it is found economical to design the piles having
	Stair Cases
	The minimum width of a stair shall be 85 cm. In residential buildings «he minimum width is preferably 105 cm.
	Heat room over a stair The head room over a stair shall be at least 2‘10 metres,
	§158	. Design of Stairs
	Maximum bending moment
	~ .*. Maximum betiding moment
	56 SI
	\'At\' *1400x0-8 7 X 10 56
	§160	. Stairs Spanning Longitudinally
	gninnniinnni
	637
	Mobxo-ñxm em\'
	Vo
	Spacing of 12 mm. 4 bars
	113X100
	Fig. 402
	Fig. 403
	15
	Design of Water Tanks
	Hence let us provide an overall thickness of 15 cm.
	= 4375 kg
	Vertical reinforcement
	t=Th:ckness of the wall.
	^íó^oó”1412™\'
	1171. The I.S. Code Method
	0\'569 X 9810 X 4X ^N
	= 133965 N
	Hoop steel per metre height -|n96S-iTd 2
	“j^1^100“5’1 cm.2
	=0\'015 wH*
	- K4W
	The design of such a tank consists of :
	If 16 mm. 4 bars be used ata clear cover of 25 mm., the
	-20 -2 5-0 8-16 7 cm.
	The b.nv at the various conspicuous points are given by •
	n =
	P^H— ^s+Al+ yjj 2/
	¿»-(>.+4)]_ 2/
	Mfa+A*> 4	/
	~ r^t+W
	2
	“l\'“ior ]x 0 3 x 25000
	yr^oiT •x=1,Ow-
	V ——-9450N
	1.591
	- 9450N
	Foundation
		MM.
	“Ti lx A i
	Bnt	r«i sin 0
	0‘3
	<1’5X100
	In thia cat
	.R
	.113X100 103\'4 “
	“\"\'-s^r \' ,-Jf4r
	“ woo 5 24 em\'
	=0 0122 X I(W0 X 3a X100 k?
	s.-s+ve^
	<7 St- q\"~bd
	6000 x 4 23 “	015
	(ft) Ring beam at A supporting the dome
	l ig. 469
	Total hoop steel
	„	= 4A IpT^
	„3\'0565 ,11100 vi 00	.
	“ îiôô^-æïïôô xi5A/im‘
	Let the direct load transmitted to the column most distant from me centre of column be R Newton. Taking moments about the upper points of contraflexure.
	4£Sixiw“°^
	i«_i -«’J™0-	.	182575 I»
	^Thia horizontal load If» will produce a hoop tension in the- ring beam A
	.*. Stress on the seciion
	• Meridional stress
	At fo^oo\' 19 6 cm\'
	- »X -18\'*1
	Loads
	-1361820 2V
	- 5^t—33,863 N
	249647
	\" 0 9950
	3rRL
	22844	2
	tmk is full-7148 73 x 8-5718984 N
	(10% of column loads)	^71898 N
	Equivalent S.F.
	For M 15 concrete,
	w here.
	1509800
	17
	--255334O N cm.
	20885700 4-2553340 = 23439040 N cm.
	,	2ÍW90408
	’ 2’5 Jji«
	L\'2xo3^i4000	2 5 X 63 X 14íH?0j ¿
	; UW x 0\'87 x ¿V ~! 2 40	\"
	Design of the bottom Mab
	1 50 -0 40
	wR 0 5x510x8 61 , , 2 ”05 <	—THskglm‘
	=y	-
	(til) Cylindrical wall
	Tra^vh^ remfcrcedient
	3 14 = 25 12 cmr
	(x)	Design of foundation
	628725*?.
	Design of the circular girder
	Design at sapport section
	“\'n u
	Transverse reinforcement
	.. Extreme stresses in concrete are
	Net upward presure
	The cylindrical wall may be designed for hoop torsion. Ik circular slab may be supported by four beams of similar spaas
	=42919 N
	05X100 * 429 cm\'
	^7\'7 cm. say 7’5 cm.
	X 34X100 cm.*
	_0 79x100
	Maximum soil pressure
	16000 x 3 x ¡T MD S NI metre*
	1 I- sin 30
	= 9810 x 3 - 29430 N/metre*
	^ = 29430-16000
	£!?
	«800400 N cm.
	Maximum soil pressure
	-16000 N/metre*
	Cracking stress due to bending moment of 800400 N cm.
	Let the tensile stress in concrete be Ci
	“483’3 Ct
	Fig. 519
	X 12\'7 C‘ kg\"
	Base slab	-0*5XxX25000 #-12500 x#
	*»• k
	ii ÏX ICO if8®-76990
	.\' \'jvid ig v\'Tectiv-; cover of 4 u.7i. the effective depth available ■• ? ~4*=H ■•th.
	4 63OO(x-- ’nWOOO x
	w«j
	-J 548 ¿^M’, -6?20^,^.
	Reinforcement
	Case 1. Maximum bending moment at the bottom of the stem due to water pressure alone
	0*79X100 =	c*.
	Maximum available friction
	ReinfoKemeBt
	Pie. 56#
	-7’15	1+5 73
	-7\'15
	-15*09 rm.
	+ 7#
	c
	free B.M. fr . B
	• 15 09—51^9\'W tm
	Max. free B.M. for vertical wall
	=2’08 tm. at 1\'66 m. from A
	16
	Bunkers and Silos
	/
	hr/
	Design of the hopper bottom
	Sloping bottom as a slab spanning between Comers
	due to
	lining.
	Normal component of coal pressure
	0\'72
	A-3+i+V “3-86 m.
	Total normal pressure
	= im±^xmi.
	JI.- 10635x^9^222)	_47„ kf
	1463kg/m
	Fig. 318 sbdws the duuswn forces acting on the bunker.
	fig. $88
	n >nB,
	^31-5 kg/m*
	Fie- 595
	\' —	— \'8915	. s
	»distance between centres of steel =42 cm.
	\"-1Æ^ \"\"+Cl
	2
	’- ^ L1-\'
	t X^’^^r
	P* = —
	BUNKERS ANO SILOS	955
	0*2
	Vertical steel= 1nn x 15 x 100 cm.’ per metre of circumference
	Stress due to sell-weight ol silo wad
	1	x 20 > \'Uim A* (y im •
	= 48000 kg/m 2
	Stress due to self-weight and wheat load
	Design of the sloping bottom
	4
	5 67 Cm‘	Fig. 606
	$189. W. Airy’s Theory
	^=0
	t*+l*
	17
	Bus Stop Shelter Frames
	Load from the slab
	^710014375
	31s75
	9v<9e
	JmL 1 ■\' ‘ J k^me,re~
	Frames for Buildings
	§190. Frames for single storey buildings—Portal frames
	m	-12m-
	Let
	7^-7
	7*c-r 957
	Distribution factors. These have been computed in the following table :
	Fixed end moment for beam BC
	=2818 cm.2
	Section of the column just above the hinge
	Spacing of 6 mm. diameter bars
	Design of Beam at mid span section. Maximum B.M 13\'62 tm. (sagging moment).
	«70-4-66 cm.
	^+
	~--iTxiobo~{onnemetre
	Let C be vertically displaced to Ci.
	— 3 cosec 1-/ 2\"
	-4’12 3
	GEZ^
	77 4 : 10
	Fi*. 634
	Fig. 636
	cm.’=20\'19 cm.2
	MOD« .87x76 -16““
	B.M.
	40x25\'2*y( 749-”-)+ 17x22 81 X (-¿^j«
	(74-9-51)
	55110 c= 18\'69 X1000 x 1P0
	às-Km- l)?l+mÀ
	= 	12\'93x1000
	4o x 25\'2+17 X 22 81+18 x 22\'81
	and	Mrd=/+2G#
	2 . .jOTiror
	c ¡7) I 		4m.	H
	—40x2y.1 52 =-14-06 kNm.
	, 4ux?. 52xr5
	60>\'4 _ 8
	o
	an y
	Correcting factors
	Let/Uof j2/	• . he —21 and Nd—I
	The c »rrecting faciors are determined in the tabie beiow :
	Restraint moment at B~m - x 2 3’43-3u 0 — 6*5 7 kNm.
	Restraint moment at C^m +30\'C0—6’67 —+23’33 kNm.
	Cic=^ ^ilPIA^zzI’ShLPfo^HEliiZ”
	Restrain: moment at B-0-32*00=- -32 00 tm.
	Restraint moment at C«+ 32’00+0®+32’00 tm.
	y(-32)~y_ (+32)
	C.»= 	18
	36~1
	= + 12\'80 rm.
	0.0=2 G»=2X 12\'80-25\'6 tm.
	Cde— j	tm.=—12\'80 tm.
	44M
	“89
	B c D
	z£25+L?7 J5 11
	6t
	3m.
	198 f
	Fig. 659
	tribution his been carried out in the following table, in the usual nanner.
	IT
	m»a = +2‘67 tm.
	mrb\'=+4’00 tm.
	Fig. t68
	Free B.M. at the centre of span AB wl*
	7	.d + .M. at the te itrc of span AB
	5832+7180 .	, O1iri
	• 4625	2	*&m- \"\'+8115 ¿it m
	For instance,
	Mj] ¡2“ Af«2ai — - (numerically)
	BM in column just below 1;,- ‘ \'	in*
	1\'76 im.
	5x?‘52x PS
	42	^
	Afr
	= + 2’SH tn.
	r
	-\\W -/ \' ’
	Fig. $91
	3	I
	Fig. 692 Solution. Fixed moments
	Fig. 697
	Resultant fixed end moment at B
	=43 00--ri7=+W tm.
	Span BC. Resultant fixed end moment at B
	1æS2 I 6 x 2000 x 3500 x 0’25
	~	8 +	22 ÏUU/ tm‘
	= -2’504-2\'63 =+0’13 tm.
	Resultant fixed end moment at C
	& + 2 50+2 63 tm.~5 ;3 tm.
	Span Cl>. Fixed end moment at C
	Fixed end moment at D
	“Aide— +3*75 tm.
	Rotation factors. These have been computed in the following table :
	L^2L
	L U
	IH+-4
	Now the rotation contributions are computed. This computation is shown below
	5®iM un
	6 m.
	> IB
	¿ 3
	47	27
	6*3
	4f_2L
	6 \" 3
	3
	J
	3
	Fie. 702
	Fig. 703
	I
	Fig. 705
	B
	X
	The final moments can now shown in Figs. 709 and 710.
	be easily determined. These are
	Fig. 709
	i ig. 7il
	M&f^ — 36 tm.
	2 \' b 3
	fpEffim
	a)
	¿J
	--16’67 tm.
	.Af«d =>+16’67 tm.
	Now the rotair ; contributions can be easily computed. These computations are siu^n in Fig. 724.
	of all the columns of the rtn stoiey*
	Tne quantity Sr may be calkd the stwy >Lcar for the rili storey.
	and	Af^r~ \'U^ \' W » I If\'; Ar\\.
	For all the^ columns.
	.’. Displacement contribution for a column of a sto^y is proportional to the moment of inertia ol the senr n of the cdumn
	But relative stiffness A\'- J
	Al\" ? x K Hnce I has the same value for all the columns of the storey
	Bm. —
	r-
	The fixed end moments aie entered as shown in the commencement scheme (Fig. 730)
	Total fixed end moments at B and C are entered within the square boxes The rotation factors a»e entered at their usual places. The displacement factors are written by the side of each column.
	Now the iteration process starts. The various compulations may follow the following order :
	Joint B, Joint Ct member AB and member CD.
	O
	2X1
	Rotation factors
	These have been computed in ihc following table.
	The rotation contributions and displacement contributions can now be determined. See Fig /3j and exp’ natory tables.
	19
	Reinforced Concrete Chimneys
	From the geometry of the stress diagram, we have
	Af —	2[(«-a) COS2 « + ? sin A cos «
	1 xc^ a
	25
	20
	Bridges
	\"i^io^\"12 M™-2-012 Nimm*
	Fig. 747
	Fifc. 748
	1 — »1
	13X70- ni 1400	1-m
	13
	-0 87 d
	=~-267400 kg .cm.
	Cantilever bending moment
	Increase in bending moment due to 8 cm. increase in thickness
	Deck slab
	Cantilever
	Loads on the end beam
	Maximum shear force
	Fig. 751
	(v) Temperature and shrinkage effects
	BEAMS AND SLABS
	(i)	Effective span
	(#) Effective depth
	BRIDGES
	Fig. 761
	Dispersion of load along the span
	m
	—1524 kg.
	42i ^x r-
	= 75o9O’ Ky A c. cm.
	[ 756\' Wit’ 20x40\' cm\'
	=^97 cm.
	= WO cm
	= 87’75 cm.
	as cm. F7	?
	T~ ;
	Shear force at 1 metre from the support
	Shear force at 2 metres from the support
	»’OGB»	1157
	Check for bead. Perimeter of bars required at the bottom near the supports
	2-—A_	25230
	a»\"82\'23xlOcm\'
	“28\'6 cm.
	Perimeter provided by 6 bars of 25 mm.
	Roads and Pavings
	Concrde slab
	Bose
	Building paper -*	Subgrade soil
	Flexible filler1 and sealer
	la) Expansion Joint
	(6) Complete c on traction joint
	It) Partial contraction joints fig. 7*8.
	Expansion joint showing dowels as load transfer device.
	22
	Prestressed Concrete
	$200. Introduction
	•This chi pt^r deals with the elementary principles of prestressed concrete. For mo’e details and desiens the readers may refer to the text-book Prestressed Concrete by the author.
	203.	Tendon Placed at an Eccentricity
	r
	ifiW
	Direct stress due to prestressing force —
	Maximum B.M. due to external loading
	24000
	-± W
	3G0
	^okg/cm*
	33.3kg/Oft*
	»3.3kg/cm*
	sokg/cm*
	sis kg/cm*
	t3kg/cm*
	z
	«?*^®?^ ^f*9/^ fto-Bkg/cm^
	Fig. 805
	gocm
	Analysis of the end section
	4t.7kg/C^2
	Fig. 809 (i) shows a restressed beam carrying an external
	W PER UNIT RUN
	W
	40cm
	~± J -inMl o? ^”8 24 ^^
	Dead load of the beam— 0\'08 x 0’12 x 240= 23’04 kgim.
	Z
	..(0
	TST-259”
	...(a)
	96	192
	-13-SO
	% =246 29
	— 13 50
	Hi. 823
	From equations (i) and 00. we get
	I 5606-72 X 7 69 ^m ^ i^t
	Say î50750
	Z\'-^\'-^S2—’35”\'\"-\'
	Fip. 837. I reyasioet system.
	ate#l
	*«¿9*
	Fig 839. Magne! Blaton System
	\'-!-( *»+y)+( k*+** ^
	r—-i
	L >4 J
	At	x«»y-»7*5 m.
	P/ KTRHN O CÍNCMTE
	=-4Tx4r“89^-/cm-
	=	• x iooy
	The losses of prestress due to various causes are calculated below.
	.(5)
	140
	_ 43350 1110
	J217. LS. *	u.*i—
	Fig. 85 i. Computation of compressive stress due to initial prestress.
	0 i o
	Fig. «55
	|/2r^
	Fig. «56
	Fig. £59
	Fig 86)
	4 o
	Fip. 867
	where	g—Shear stress intensity at the level
	¿J ■»Static moment of the cross-sectional area\'~ of the beam above or below that levfl\'ata the centroidal axis.
	Ft«. MB	Pi#. S69
	The above results may also be obtained by Mohr\'s drcls.
	-^2i‘752+t “T“kglem\'
	Equivalent concrete area“Xi\".) — 175 kg.jcn2
	5166 <1AA
	Too“5166 cm-
	23
	Composite Constructor
	Note. Flange shall be oriented against the direction of horizontal shear, that is towards the centre in case of simply supported beams.
	Lei 20 mm. diameter studs be provided.
	Sr ^ ^ I01** °f ’^ ”® be resisted by the composite section
	Ultimate Load Design
	’ ‘	B1 0 85 /ci// b d
	Solution
	Span—/—6 m.
	Seiutioa.
	,	3 J i Fa
	2 y s
	At*=At—Atf =80—»5 60-34 40 cm*
	.’. The section is under-reinforced.
	25
	Limit State Method
	|250. Raml Member*
	as shown in Fig. 959.
	diagram Fig. 96?t stri|n and stress diagrams
	26
	Form Work
	Note. Ibc number of props, their sizes and disposition shall be such as to be able to safely carry the full dead of the slab, beam or arch as the case maybe.
	LS. recommendations {IS 456)
	Stan value of bent up bars
	14,000	1400
	^4
	^¿®n®i®!®
	TABLE 15 Fixed end moments M°ab for hinged members
	1425
	4?K
	Xr __ rut
	^iailllteih.
	^M
	-~r
	9*73 oxi Í 0x8 OX I
	IJHI
	table is (Contd.)
	r
	’^l «\'OSIN
	LOGARITHMS OF SINES
	»■ । sr
	4* । «
	H
	A
	4
	§
	TABLE 25
	POWERS. ROOTS AND RECIPROCALS
	TABLE 26 (Contd.)
	SQUARE ROOTS. From , to ..
	11
	12
	13
	I n 0
	TABLE 28
	RECIPROCALS OF NUMBERS. From i to io
	Index
	1453
	E