SINCE 1995 Engineering Academy SINCE.1995, Hyderabad | New Delhi | Bengaluru | Bhubaneswar | Vijayawada | Visakhapatnam | Tirupati Pune | Chennai KI. Gop? Keishnw Ts (res) civil condact: 9502 54-08 F Design of Concrete and Masonry Structures (Theory with worked out Examples, Conventional & Objective Practice Questions) IES - Civil Engineering ACE is the leading institute for coaching in IES, GATE & PSUs HO: 204, 11 Floor, Rahman Plaza, Opp. Methodist School, Abids, Hyderabad-500001, ‘Ph: 040-24750242, 24750263, 24750437, 24752469. CONSISTENTLY TOP RANKS IN IES” _ All INDIA 15° RANK 26 TIMES IN GATEChapter Name Reinforced cement concrete Materials, Workmanship, Inspection and Testing Limit State Method of Design ue Limit State ~ Singly Reinforced Beams 2 = Basic Concept - Working Stress Doubly Reinforced Beams ~ Limit State Method Doubly Reinforced Beams - Working Stress Method Limit State Design - Flanged Sections Limit State of Collapse - Shear Bond Limit State of Collapse - Torsion Slabs Limit State of Collapse - Compression Footings hepa fs Pepys 8/8 )g)R Limit State of Serviceability Prestressed Concrete Fundamentals of Prestressed Concrete Analysis of Prestressed Concrete Members Losses in Prestressed Concrete Concrete Technology Cement 96-103 Aggregates 104 - 106 Concrete 107-110 Cement Mortar 111-116 Brick Masonry 117-120Materials, Workmanship, Inspection and Testing | OL Materials: early strength. Therefore preferred for high | a) Cement strength concretes, prestressed concretes etc, ) 'b) Aggregates: As per IS 383-1970. - Generally Type of IS. No. Purpose Hard Blasted granite chips (HBG) | Cement (i) Coarse aggregate: | OPC | 18269-1976 |, General + Nominal maximum size of coarse | constriction aggregate for RCC is 20 mm. Tow heat | 19269-1976 |, MASS Sac) noc preter thin ne forth: of ement niin | n thickness of mem Rapid i Bo-196 a reinforced members $ mm. less hardening : tndyal Of inimuim clear distance between i cement A form works or 5 mm less than the | Pozzolans 115 :4golfoay py Chemical ‘ to the ‘reidorectient Cement resistance i | High strength Prestressed " ly medium sand, | ve hs gy ad egtrcans : erally | Fiydrophobie Wars pepfa | egforcoment: seer | 18 1991 ae Mil and dum eset bas fi) Grades of Ceme Sad ii) Hot adden, 18 1139 iii) Cold twisted bars =1S 1786 Strength ‘ofa cet i iv) fos ire fabric - IS. 1566 70.71 mm ( surface ka tote”. A ‘tested at 28 days. TI iffersin | - | be free from loose terms of fineness of cer i Pa P = * oe \ = tust, oil etc.Modulus of expressed a8 specific oe re “Se s Peal a ee Specific surface is the gu re of steel since the linear particles’ in. 1 gram of E Sie rain curve ofelmost all the gram). Chemically. all the. three ie same. ‘cement viz. grade 33, grade 43 and nee a the increased: strength of are almost similar. (1S 516-1959) deformed bars viz. tor steel compared to mild ‘Their characteristics are listed below steel is because of the twisting given to the | Gr33- specific surface area. is minimum plain bars resulting in more dense crystalline i 2,250.cm*/ gram (1S: 269) ‘structure, | Gr43.-. specific surface area is minimum, ‘+ The increase of carbon content increases the | 3400 cm*/ gram (IS : 8112-1989): strength of steel but ductility decreases Gr 53." specific surface area is much greater ‘than 3400 om’/ gram. (IS: 12269- | d) Admixtures: These are the ‘chemical SN SS Se es Ton i 1987). compounds used for improving the | Grade 53 cements have more shrinkage characteristics of concrete such as workability, ————compared-to-other-grades; but having “higher—]~—setting etc. with out affecting the strengti of | the conerete.Orch Grostear RCC ‘Types: Retarders delays “the “setting of cement particularly in hot climates for certain ‘minimum time, Gypsum “is one such compounds, ‘ Accelerators, accelerates the ‘setting process particularly in cold:climates. Ex:- Cacl» Plasticisers, these “até air entraining agents improve the workability of concrete in case of rich mixes and congested reinforcement. ¢) Water : Potable water. P¥ value: tess than 6 as per 18456-2000 4. Water tanks; domes; ‘Types of concrete As per IS: 456-2000 Ordinary concrete MIO to M20 Standard concrete ‘M25 to M55 High strength concrete: M60 to M80 ©) Properties of concrete : i) Increase in strength with age (Age factors). I month -1.0, 3 month —1.10 6 months — 1.15, 12 months 1.20 ii) Tensile strength of concrete (fa) : tests ‘conducts are cural (modulus of rupture) test & itsensile strength test formula given by IS 456-2000 e other tests are Compacting gf ZARClor test and Vee-bee consistometer test. vii)Durability : The property by which concrete possesses same strength through out its life time. Without much of shrinkage and cracking. ‘© Factors effecting durability are w/e and ‘maximum cement content Maximum cement content as per 1S (RE eon en eee | ge] aT folded plates, shell roofs Mag. 456-2000 is (without fly ash and slag) *5. RCC. insea water | M30 for RCC & = 450 ke/m? ‘(General construction): M20 for FOC. ‘+ Minimum cement content is based on §; Bost tensioned pre- 0 exposure conditions Stressed concrete. om | Pi 7. Pre-tensioned pré- M40, stresses concrete+: 8 ‘Materials, Workman Ship, Inspection and Testing ¥ a Plain Conerete Reinforced Concrete Maximum Maximum gro) gome_tina|Sce™ mamin nw|"e”]nnen | water | Grade of water | Grade of content). Cement | Concrete | SR | Cement | Concrete Kom’ | Ratio Kom’ | Ratio GQ) Q) QB): (4) (3) (6) @). (8) i) Mild 220 0.60 s 300 055 | M20 i)_| Moderate [= 240 0.60 | Mis 300 0.50 | M25 iti) [~ Severe [= 250 050 | M20 | 320 0.45 [| M308 iv) _| Very severe | 260 045 | M20 | 340 0.45 | M35 v) [Extreme ig d) Proportions for concrete. i) Nominal Mixes: MS M20 ii) Design mixes for hig fof concrete : | ibe at-least two | > ° decreases strength increases ea better homogeneity. For example, cube of 100mm size will have 5% more strength than 150mm cube. Shape factor. Standard cylinder of 15cm diameter and 30 cm height will have strength of 80% of that of a standard cube of 150mm. Slendemess ratio: As slendemess ratio of a specimen increases, strength decreases. For example if compressive strength of a der of 15-cm diameter and 30.cm_| height (slendemess ratio 2) is 0.8 fy, the 3 beyond 4, the strength is yisThis is one of the (GATE-95) bys a direct ‘tensile strength of the concrete @) The fest compressive suength of the Brie Tensile strength of the concrete in bending. (@ The characteristic strength of concrete 02. The yield stress of a corrugated bar as compared to an ordinary mild steel bar is ly ES-92) (@) 50% more (b) 25% more | —(c}-50% ess __(d).25% less strength with slendemess ratio 3 is about 0.7 to is ‘Bengahar | Bbabancrvar | Vierooda | Vikhapan | Tupat | Pune [CheabalA pe 03. The ratio of direct tensile strength to compressive strength of concrete is taken as “ (ES-92) 0.05 . fo) 0.15 @ 035 04. The split tensile.-strength of MIS. grade concrete when expressed as a percentage of its compressive strength is Ad) 10 to 15% (©) 15 020% (©) 20 to. 25% (25 to 30% (©) 025 05. Compared to mild steel plain bars, high yield strength deformed bars are Ae) Less ductile but more strong (b) More ductile but less strong 06, The ratio of 7 days is @0s (©) Inferior compaction @ None of the above. 09. Approximate “value of shrinkage strain "6 conerete is (@)3x10> 43 x:104 (@ 0.002 ©) 0.0035 10: High yield strength deformed bars have a (@) a definite yield value (b) Chemical ‘composition different from mild RCC 11, Tensile strength of concrete is measured by, (@) Diréct tension test in the universal testing machine 6) Applying compressive load along the diameter of the cylinder (©) Applying third point load on a prism (@ Applying tensile load along the diameter of the diameter of the cylinder 12. Minimum grade of concrete for water tanks is @Mis Qo ©) Mas @) Mao 15 om and a cylinder of 15 5 cm height are tested for igth the strength of cube 9) lower : (@) difficult to assess 15, The flexural strength of M 30 concrete as per IS: 456 - 2000 > 3.83 MPa (©) 21.23 MPa (b) 5.47 MPa (@)30.0 MPa 16. Consider the following strengths of concrete: steel (©) percentage elongation less than that of 1, Cube strength mild steel 2. Cylinder strength ‘(@ Percentage elongation tore than that of 3. Splittensile strength mild steel ‘Bergalars | Bhobanerwar | Visyansda | Viakhapatrand | Ticpat Pune [Ci 4, Modulus of rupture@ wit Fiosen : 5 ‘Materials, Workman Ship, Inspection and Testing "The correct sequence in increasing order of | 24.The environment exposure condition of these strengths is concrete buried under aggressive sub soil is 3,4,2,1 (b) 3,4,1,2 classified as ©4321 431,212. (@) moderate () severe _Aoy very severe (@) extreme 17. Concrete strength determined from 150 @ x 150 mm height cylinders as compared | 25. The compressive strength of 33 grade cement to that of 150° x 300 mm height cylinder is at 7 days should not be less than | more — (b) less (c) equal (A) varies (2) 110 kg/m? (©) 175 kg/em | 40) 220 kg/om? (d) 275 kg/em* | 18. Flexural strength of M25 grade concrete is | (@) 1.5 MPa (b) 18 MPa 26. Consider the following statements: | (©)2.8 MPa _Say3.5 MPa ‘hiligeThe compressive strength of concrete | a ges with increase in water cement 19. According to IS 456-2000 | elasticity of concrete expressed as (@) E.= 5700 Fy. (c)E,= 5700 fx concrete mix dd, to. the. concrete mix for ment and workability 20. In M 7.5 nominal of water used-per then that used workability will be’ (@) less than 45 kg () equal to 45 kg 21. The maximum distance 2000 is ath of concrete increases with “increase in compressive strength of concrete 22. The minimum quantity of cement per meter ‘The true statements are; cube of reinforced conerete for mild exposure @ Wand 1 () I, Wand HL is Tand II (d) Land 1 (@)150 kg (0) 250 kg sch . kg. fay300kg, 28. A reinforced concrete structure has to be constructed along a sea coast. The minimum 23. The mean strength of the group of four non grade of concrete to be used as per IS 456- overlapping consecutive cubes, for M25 grade 3000 is conerete should not be less than MIS (b)M20 (¢) M25 30 (9) 28.3 MPa $0725 MPa be a i (©) 33.24 MPa (@ 29 MPa 29. Consider a reinforcing bar embedded in ~conerete-—In-a-marine-environment-this bar | is: [ New Delhi Bengaara | Bhubaneswar | Vinyoweda | Visahapatamm | Tirwja | Rane [Chem]“undergoes uniform corrosion. : Which leads to the deposition of corrosion products. on its surface and an increase in the apparent volume of ‘the bar. This subjects the surrounding: conerete to expansive pressure. As a result the corrosion induced cracks appear at the surface of concrete: Which of the following statements is TRUE | (@) corrosion. causes circumferential tensile stresses'in concrete and the cracks will be parallel to the corroded reinforcing bar (b) corrosion: causes radial-tensile stresses in. Which of the following statements regarding Ec are womg; (@) 1 and 3 only (b) | and 4 only (©) 2 and 3 only (@2 and 4 only 31. Which of the following is the effective modulus of conciete? ) Be / (148) (b) E,/ (1426) (©) Be / (1430) @E./ (1450) 03.6) 04.(a) 05.2) 08.(6) 09.0) 106) z 14.00) 15.(@) @ corrosion causes nsile stres i . . 19.6) 20.(b) my 2440) 25.(¢) 29.(a) 30.b) © corrosion causes ci stresses in concrete perpetidicular to :Limit State Method of Design 1) Phoilosphpy : Limit state design is a method | 2. For the limit states of serviceability, the values of designing structures based on a statistical of “yf. given in this table are applicable for concept of safety and the associated statistical short term effects. While assessing the long probability of failure. The method of design term effects due to creep the dead load and fora structure must ensure an acceptable that part of the live load likely to be permanent probability that the structure during its life may only be considered. will not become unfit for its intended use. + This. value is to be considered when stability against overtuming or "stress i :Thed ti 2) Principle limit states: The important ay eit states are 8) The limit state of collapse i i) Flexure (bending) iii) Shear b) The limit state of » Characteristic load: he Limit State method of 95% probability of not ft ‘ain xref) life time of structure 4 sé normal to the axis of the { . Zadar remain plane ater bending | Values of partial safety factor ‘y, for load ii) The tensile strength of concrete is ignored | ey iii) The maximum strain in concrete at the | Sa Toad Limit state of | Limit state of ‘outermost compression fiber is 0.0035 ‘combination collapse servis iv) The compressive stress strain curve may be 1 DL | LL | WL | DL | LL | WL assumed: to be rectangular, trapezoidal, buyers {is} = fio fig parabola or any other shape which results in the prediction of strength in substantial pi +wi | oe) | 1s ]10] ~ | 10 oe eth che vesles Gr a DLTEL+wo| 12 [12 [12 [io Tos [os + An acceptable stress strain curve (rectangular- parabolic) is shown aside. Ne + Compressive strength of concrete in the 1, While considering earthquake effects, | ~~ Strueture is-assumed to be" 0:67 times the substitute EL for WL. characteristic strength of conerete.REC strength of cohereté in addition toit. «Therefore, the desiga’ strength of concrete Fee 15 ie, 0.446 fog ~ 0.45 fag v)'‘The- ‘design * stress: ‘in reinforcement "is derived from the stress strain’ curves given below for mild steel and cold: worked deformed bars respectively. “The partial factors of safety ‘Yq’ equal to:1.15 is applied to the strength of reinforcement. ‘Therefore the design strength of reinforcement is fj / 15.ie. 087 9 STRESS 200000 Nini? (vi)The maximum strain in the tension reinforcement in the section at failure 087," should not be less than 0.002 +- serviceability ~ requirements before failure occurs is called (2) braking pa (b)failure point (@) ductility strain in the tension the section at failure shall not steel in limit state b) £0.87 4) 0876, beam carries a working and dead load is 3.5 load for limit state of @712 @i2 fy = 500 N / mm? the value a (a) 0.48 7046 (053 (@)0.43 06. The ‘partial. safety factor to be used in limit staté of deflection for strength of concrete is @12- YS O10 OS 07. The partial safety factor for stee! in limit state for serviceability is @12 O15 rio Mos 08. In limit state design of concrete structures, the a recommended partial safety factor ‘ym’ for Suan op steel according to 1S:456-2000 is ‘Steel Bar with Definite Yield Point @1s a 15 (10 (4)0.87 RSMO Nc Tovar | Vga | Waekdapaten [Tipe | Pane | enna@ iA Fictm 09, For avoiding the limit state of collapse, the safety of RC structures s checkéd for appriate | 14. In LSM of columns, the partial safety factor combination of dead load(DL) , imposed applied to steel and concrete are “sseGadQL), ‘wind load “(WL), and ‘earth’ quake} (a) 1.15 for concrete and 1.5 for steel load(BL), which of the following load (b) 1.15 for both concrete and steel, combinations are not considered AGS for concrete and 1.15 for steel (@)0.9 DL+ 1.5 WL (@) none @)LS DL +15 WL (AO NSDL 1S WLELS BL 15, Un-factored maximum bending moments at a @ L2DL+ 121412 WE section of a reinforced concrete beam resulting from a frame analysis are 50, 80, 120 and 10.In a random sampling procedure for cube 180kN-m under dead, live, wind and strength of concrete, one sample consists number of specimens. . These’ s tested at 28 days:and average, X specimens is Considered sample, provided the ii im) as per 1S:456-2000 for limit ge (in flexure) is, (GATE-08) 50 (c) 345 2 5, a Dome the strength of specim i/more than +¥ "| 16.‘The stress oe of the concrete as per IS percent of thé average The values of |, 456is S: (HES-03/11) Xand ¥ as per IS: 456 | (a) aperfect 1e up to failure (@)4 and 10 ©4and.15;:- 11. If the characteristic defined as the stren (b) straight line then parabolic\iip tb failure ge 1002 strain value and then higher ihiverage cube strength >» ob) Tower thanithe average cube strength 1°" (0)-the same-as the average cube strength «(ay higher:, 12. In LSM the characteristic mean load plus ‘k’ times characteristic load is not to excesd=only 5% times :the:expected. life -of :the: struc value of *k; is (a) zero." 1(b) 072 + JF1.64. (4) none 13. An RCC beam is subjected to the following moments 3!" Deal load moment = 20 kN.m; Live load moment + 30KN.th; 01. ©) 02.(@) 03.(¢) 04.(a) 05. (b) | Seismic load moment = 10 kN.m; | Wind load:SIN-m the’ design’ moment as per Seed eee eee | limit state of collapse is IL. ©, 12.@) 13.0) 14.) 15. @) | @)60KNm = QS 75Nm | 6 17.0 he Nf) BIN ef (ERGARRER ESR ess ey Date | Benet] Dhabas Vingovadh | Vininpendon | Tnapaly me (Cheam |Limit State — Singly Reinforced Beams 1, Definition: A’beam reinforced on tension side in the direction of beniding. 2. ‘Type of Sections: 2.1. Balanced section: ‘The 'tresses in ‘concrete and steel reach to * maximum values at thé same time Concrete and steel theoretically fail at the 087 fy Ax n compressive and Tequired for a balanced section "SS © The failure is in concrete, causes brittle/sudden failure. without any prior warning + As petLimit state method the design of over reinforced sections should be avoided +> The failure is by’ primary compression 6 u > Xooar Mo tinit = 0.36 fix b X vans [4~0.42 X y sed (or) Matinit = 0.87 fy Ag (4-042 x u madd Salient values are given below : Grade of steel | Xs max/d | Mu tinit 3, | Stress and Strain distribution for balanced Section : Fe 250 0.53 _| 0.148 fy bd? Fe 415 0.48 | 0.138 fax ba? Hom he iw tcl, e500 | 046 | 0.133 fix bd?Oi 2Gincer Limit State Singly Reinforced Beams Exanpesh Balanced percentage of steel with M20 concrete Sol: Balanced % steel gia 0.36x20 . — 0 0.87 0 = 0.96% Codal Specificatins: 1. Effective Span : Pot ‘Smaller of the following two a) a a5 ») Cher di distance bel 2. Reinforcement Details : —— =~ 8) Minimum tension reinforcement (Ay / bd) = (0.85 / fy) For mild steel , f, = 250 Nimm? = 0.34% eeanpem HYSD. grade Fe-415 what is the minimum percentage of steel? Sol: for HYSD grade Fe-415 {= 415 N/mm? Minimum tension reinforcement : P min= (Ay/bd)x 100 = (0.85/415) 100 =0.205% b) Maximum tension reinforcement = 4% of gross area, Moment ofResistance | Bal%0fAaPrim forSR]| 3. Cover Requirements : a Mpalte 7 Ov 5 The cover is based on durability 4 (Nima? act ir ifit is provided oven) [350 | 415 | 500 | Cm [350 | 415 | 500. ne i3_[224 [207] 200[ 13 [132 [0.72] 057 i. to have proper bond between concrete 20__| 2.98 | 2.76 | 2.66 {20 [1.76 [0.96 [0.76 and steel _ fi, to protect reinforcement against corrosion Nominal Cover or clear cover: is ‘the design thickness of concrete cover to all steel reinforcements including shear stirrups pes cee ties. jer is the distance from centroid stechto the extreme fibre of the For | different’ nominal cover the ‘minimum nominal cover diameter of main bar igher-cover to avoid buckling of #7 ae Botings minimum cover should be 50mm 4, Spacing of Reinforcements : a) Minimum Horizontal distance between ‘two parallel main bars shall be the greater of the following. @_ Dia of bar if bars are of equal dia. (ii) Dia of largest bar. ii) Smm more than nominal maximum size of coarse aggregates, ol eedie—vibrator—is—used~to~ compact concrete, */s of nominal max size of coarse aggregate jHiyderabad | New Delhi | Bergalarw | Bhubaneswar | Viayawada | Veakhapatnam | iOA Finin 1B: RCC b) Minimum Vertical Distance : 15mm or two- thirds of maximum size of coarse aggregate or maximum size of bar which ever is greater. Maximum distance between bars in tension depending on cracking of concrete Maximum’ crack width’ in mild’ & aggressive environments are 0.3 and 0.1 mm. d) Side Face Reinforcement: If wo) along, the two faces. side face reinforcem It should be equal ‘extreme compression fibre lies ata distance of (a) 0.367 xy (0) 0.446 xy @o: 02. In limit state design of concrete for flexure, the area of stress block is taken as. 9036 fara (b) 0.41 fxs (©)0.446 fax (4) 0.53 fa Xa 03, In limit state design, under-reinforced section is one in which tensile strain in steel reaches yield value while maximum compressive. strain in concrete is less than its ultimate crushing “|005, The lever arm compressive stress in concrete is less than its permissible value (©) maximum compressive strain in concrete reaches the ultimate crushing value while tensile stress in steel is less than its yield value. (@ maximum compressive stress in concrete reaches its permissible value while tensile stress in steel is less than its permissible value. limit state design, the limiting value of h of neutral axis for Mys and Fe 250 is 4 (0) 0.48 d x (4) 0.43 d it state design is (0) d-0.87 x A= 0.416%. \6.For a beam reinforced with steel of fy = 250 “Niinm®. The limiting: percentage of steel is (a) d- 0.446 (0) d-036 x" given by @ 27 faulty (bo) 19.82 fx! fy (©) 18.87 fa! fy (@) None 07.x.,/ d is greater than the limiting value of ‘séotion is designs ~-(a) under reinfc (b) over reinfe ‘section section value of Xy / d is less than limiting value then MR. is given by (@ MR = 0.87 (Au/ fa) [40.466 X4] (b) MR = 0.87 (fx /&) [4 - 0.466 xa} BOIMR = 0.87 fy Aa [d= 0.416%] (d) MR = 0.36 fy Ag [d- 0.416 x4] 09. Pick up to the correct pair of the following : Grade of Steel P,, limit percentage a) Fe 250 21.93 fax! fy b) Fe4is 19.86 fax! fy ~-¢}-Fe 500 19.03 fax fy ¢) All are correct ‘strain (b) maximum tensile stress in steel reaches its permissible value. while. maximum Fipdcaind | Now Debi | Beagians | Bhubanewvar | Viayarada | Visakbapamat | —a Bie 10. In coastal region minimum grade of concrete for R.C.C. is (As per IS 456-2000) > (@) Mis (b) Mao (© M25 407M30- 11. The purpose of providing certain minimum cover to reinforcements is to (@) have: better. bond between steel concrete (©) to prevent corrosion, fire hazard to steel and other adverse environmental effects. (AS Both a. and b (@) None and 12.In, LSD. of concrete structutes. distribution is assumed to be’ ‘Linear %) non linear (©) parabolic serviceability (@) Deflection AS Torsion 14, If fea and, fy are the, ive strength of coneree ang viel stt9s5, of steel respectively and ‘Es’ isthe modulus of clesticity of steel, | for all grades of concrete, strain in conerete can be & GES 95) (IES. 96) (0) £71000... (d) (0.87f/Es) =; 002. (a) 0.002 PF0.0035 .1S.1S : 456 permits maximum redistribution of ___ of moments @IS% —6)25% 30% (45% 16.In a RCC beam of breadth “b’ arid overall depth ‘D’ exceeding 750mm, side face reinforcement required and the allowable area ‘of maximum tension reinforcement shall be respectively (a) 0.2% and'0.026D _(b) 0.3%and 0.03bD (6) 0.1%and 0.046D (a) 0.4%and 0.016D Hiydetabad | New Delhi | Bengaluns | Dhobancevar | Vjsavade | Viathapamamy | Tia | 20. With the usu indeterminate beans — Limit State Singly Reinforced Beams, 17. The maximum strains in an extreme fiber in concrete and in the tension reinforcement (Fe 415 grade and B, = 200 kN/mm’) in a balanced section at limit state of flexure are respectively (a) 0.0035 & 0.0038 . (b) 0.002 & 0.0018 (6)0.0035 & 0.0041 (@) 0.002 & 0.0031 18. Por a singly reinforced over ~ reinforced section L. the lever arm will be less than for a balanced section Il. the maximum stress developed in concrete equal to allowable stress imum stress developed in steel illd be equal to allowable stress statements the correct statements are (b) Land It (I, Mand Or (@) Land I (©) Hand . In WSM and LSM of RCC "Yo, Plane sections remain plane after bending UI The tensile strength of concrete is ignored Of these'statements the correct ones are (@Talone (b) Malone (©) both T and I @none notations of symbols, the fitniting strain ip concrete is taken as @),.G E15 E,)40.002 (6) 0.002 (y0.0085 (4) 0.0003 21. The limiting siiiment of resistance of a singly forced’ rectangular beam with M20 “concrete and HYSD steel of grade Fe 415 is @)L. 498bd (b) 2.676bd? (©) 2.996bd @2.776ba 22, The maximum area of tension reinforcement in a rectangular beam of size b and D(total depth) shall not exceed @0.4bD (b) 0.3 bD AS004 D. (d) 0.02 BD 23. Maximum limit imposed by IS: 456-2000, on the redistribution of moment in a statically (10% (0) 20% fey 30% 40%£ x ea" & % ® sb Giostar ACE rds RCC 24. According to IS 456 — 2000 recommendations, the maximum depth of stress block for balanced section. of a beam of effective depth ‘d’ for mild steel reinforcement is (2) 0.43.4 P0534 (©) 068d (046d . The minimum tension reinforcement of Fe 500 steel in a 250 mm wide beam of 400 mm ctive depth is (a) 150 rat 10 mm? (©) 180 mm* @ 120mm, 26, Balanced ‘section is” that, maximum stresses’ in’ 3 simultaneously attain th ‘allowable stress (6) yield stress concrete at outermost (@) 0.002 $60.0035 29. The shape of idealized stress concrete is prescribed by IS 456 is () rectangular (b)parabolic \/@) rectangular-parabolic (d) none 30..1n a singly reinforced concrete beam section, ‘maximum compressive stress in concrete and tensile stress in steel reach their permissible stresses simultaneously. What is such a section called? (@ under reinforced section (b) economic section alanced-seetion (d) over reinforced section fiyderabad | New Delhi | Benga | Bhubaneswar | 31. Why is the design of a RC section as over reinforced section is undesirable? ((ES2008) (@ it consumes more concrete (b) it undergoes high strains (6) it fails suddenly @ its appearance is not good 32. The cover of longitudinal reinforcing bar in a beam subjected to sea spray should not be less than which one of the following? (IES 2007) (@) 30mm (b) 70 mm ATS mm (@ 80mm strain in the tension fa) >( fy MISES), {(b) $ (f /1.15 )40.002 (©) exactly equal to (fy /1.15 B,) +0.002 0.00 “T. actiakdepth . neutral axis is less than the critical depth of neutral axis _~ th balanced section following ‘statements are correct 35. Minimum tension steel in RC beam needs to be provided to $e) prevent sudden faiture (b) arrest crack width (©) control excessive deflection (@ prevent surface hair cracks 36. consider the following statements and select correct answers: {the limit state of collapse is. defined as the acceptable limit for the stresses in the materials javada | Vieapatamn | Tapa | Pune [CheamQian 37. An under reinforced sections are 01. The limiting moment of resistance of a singly 02, The limited steel for a singly reinforced beam _ With Mis and Fe 415 in mm’ is 2. limit» state method is one that ensures adequate safety of structure against collapse 3. in the limit state design method, actual stresses developed at collapse differ considerably from the theoretical values (IES-2004) (b) Land 3 @) none, @ Land 2 (©2and3 1. are deeper 2. stiffer 3.can undergo larger deflection a Which ofthe following are (@) 1,2.and3 © land3 reinforced beam of size 200 x 500 (effective) reinforced with Fe 415 is (M15 grade concrete) 6103.5 KN (©) 69 Nem (©) 90 kN-m (@)235.4 KN-m 215s Limit State Singly Reinforced Beams 03. A singly reinforced beam is of width 200 mm. If the factored moment of resistance is 138 kN-m Mbp and Fe 415, the effective depth of beam is (@) 400mm (67300 mm (©)657 mm (@) 456mm 04. A singly reinforced beam is of effective depth 500 mm. If, M20 and Fe 500 is used and factored moment of resistance is 149 KN-m, the width of the beam is (@) 200 mm (b) 300 mm. ae (@) 456 mm. ed beam is reinforced with 4 dimensions are 250 x 500 and Fe500. It is to be gnderreinforced and _over-reinforced "beams, in limit state design? 03. A singly reinforced rectangular section of 200 mm wide and 400 mm effective depth, reinforced with 4 bars of 20 mm diameter mild steel bars. The concrete used is M 15 Determine i) The ultimate moment of resistance of the beam fi) The ultimate moment of resistance if the reinforcement is 3 bars of 20 mm diameter Cn ag Serre (Gast 72KN=in and 22KN-m) pan @ 367 65-6 KIO sb Fisemr 216: RCC 04. A reinforced concrete beam, size 250° mm wide and 400 mm: deep overall is. simply supported over a span of 4.5m. It is subjected to two point loads P of equal magnitude placed at middle: third: points. The two: loads: are gradually incréased’simultaneously.’ Beam is reinforced with 2 HYSD bars of Fe 415 grade of 16 mm diameter placed at an effective cover of 40-mm’ on bottom face and nominal shear reinforcement. ‘The characteristic compressive strength and the bending tensile strength of the conerete are 20 MPa and 2 MPa respectively. i) Tgnoring the presence of reinforcement, fine the value; KN when the first develop on the beam ig ii) The theoretical failure attainment of limit: ‘A. rectangular under-reinforced concrete section of 250 mm width and 450 mm effective depth is reinforced with 3 bars of grade Fe 415, each of: 20 mm. diameter. Concrete mix is M 20. Determine i) The depth of the neutral axis from the compression fibre ? ii) What is the depth of the neutral axis obtained as per IS: 456-2000 differs from obtained in the a ‘sathe above problem (Ans: 142.3 mm and 46.7 mm) a 06. In the design of beams for the limit state of collapse in flexure as per IS: 456-2000, let the maximum strain in concrete be limited to 0.0027 (in place of 0.0035). for this situation, consider a rectangular beam section with breadth as 300 mm, effective depth as 450 mm, area of tension steel as 2000 mm’, and characteristic strengths of concrete and steel as 30 MPa and 250 MPa respectively. i) What is the depth of neutral axis for the section _,ii), At failure what is the force acting on the Borggresson zone ofthe section (Ans: 142.57mm and 435KN) S107. A sing o7 »péinforced rectangular concrete beam y d re" has a width: F:150-mm and_an effective depth acteristic: compressive 20 MPa, Adopt the s sSnerete as given in IS 456- 2000. Use Fe415 grade steel. ‘Detexmine: g ii) What is the limiting area of tension steel ? Hi / (Ans: 8,51 KN-m and 36503 mm’) Va Ae SOS Mn 08.A singly reinfotced section of 300 mm x 450 effective is, reinforced with 3 bars of 16 __- — ‘Area of stress block = 0.36 fol %y ACE ‘Engineering Academy rT: ‘Limit State Singly Reinforced Beams 20MPa yield stress of steel is 250MPa. Calculate its ultimate moment of resistance 10. Design’ a beam “of effective span 6m to support a. total design load of 12 kN/m including the self weight of the beam using limit state method of design. . Width of the beam is to be limited to 250 mm. Load factor for live load and dead load is 15. M20 concrete and Fe 415 steel are to be used. 11. A. rectangular RC. beam of concrete grade M_15 is 200 mm wide and 350 (effective depth)... It is. provi numbers of 12 mm diameter as. tension reinforcement moment of resistance state method. _ The i ‘by. limite | bldck parameters from IS 456-2000 are; ‘ig. 94) (20M) ] T : 42 X, a F Xy d & 1% Prove that the limiting sjement 0 “My’of singly (under-) reinforced concrete section, using parameters of code IS: 456, is given by 2( Pe Fy( .87f, bd?| —* | 1-1.005=4/ > 100, £4, {100 1 Where and dare width and effective depth of the beam and fy and fyy. are characteristic A,x100 strengths of steel and concrete Py Ag = area of steel. State the assumptions made. 48 Depth of compressive force from the extreme fibre in compression ~ 0.416 xy, 01. An RC beam of overall rectangular dimensions 300 x 600 mm is reinforced with 4 bars of 25 mm dia with an effective cover to the centre of reinforcements being 50 mm. Effective span of the beam is 6 m. Calculate ultimate moment capacity. Find the safe SEA Gs Design a concrete beam with balanced sectigh:forhflexure by Limit stress method for the dat below: Effective span (sittiply supported) 8 m [otal load 3 12kNim on of the beam 300 mm Concrete grade ; MIS Reinforcement steel grade Fe 4ls mm effective ig ram of fe 415, grade steel bars is made of M35, “grade concrete, jum strain in concrete is iiiited to 0.003 and partial safely factor for 1 7 Determine i) What is the critical neutral axis depth in mm for balanced faiture? ii) What would be the moment of resistance of the section ?ROC 01. (©). -2.(@)..03.(@). 04.(@) 06. (a) 07, ©) 08. (©): 09.@) 1.@.12.@) 3.©, 14.0 16-(@) 17. (@) “18: (b) 19.) 21.@- 22.0) 23. 24.0) 25:(a) 26: (@) °27.(@)” 28. (¢) 30.@. 31.0). 32.0. 3-@) 35. (a) 36.(¢) [ | O1.(@) 02. (c) 03.(b) 04. (a) 05. (@) | | iBasic Concept — Working Stress J.a) Working Stress. Method (or) Elastic Method’ (or) Modular. Ratio ‘Method (or) factor of safety method: This has been the traditional method. used... for reinforced concrete design where it is assumed that concrete’ is. elastic, steel and-=conerete act together elastically and the ‘relationship between loads and stresses is linear. basis of the method is’ that the stresses for concrete and exceed any where in the st b) Assumptions : 4) A section which is plane before bending remain plane after bending (based on theory of bending) (Bernoulli's assumption). ii) Bond between steel and concrete is perfect within the elastic limit of steel. All tensile stresses are taken by ement and none by concrete, pas otherwise specifically permitted. strain relationship of steel and working loads, is a straight of elasticity is constant. H0,‘m’ has the value (280/3 is the permissible due to bending in = C(d — kd/3) = Cd(1-k/3) = (U/2)opeb.kd.d(1-K/3) Ifj=1-1/,*7’ is called lever arm constant M= (1/2) 6a j.b.d? = Qbd? Where Q= % Otek} 2 d= (M/Qb, “Ag =Miog jdACE RCC ‘The value of design coefficients are compiled in table.1. (b) Neutral” Axis Depth, Moment © of Resistance, Under Reinforced, Balanced. and. over. Reinforced Sections: Ina given Reinforced Concrete Section, analysis for the depth of the actual neutral axis-and the moment of resistance of the section can be determined as follows: Let b= width of the section, = effective depth, m= modular ratio x= depth of actug et Sebo peri ‘Then by cot first areas of abor clow t The critical neutral axis depth’ depends Se a the permissible streis fede Ke {V/[L+ Gre! (mpl } dS If x <.xq then the section is considered as ‘Under Reinforced and the moment of resistance of the section is computed from the tension side with the steel reaching the ‘maximum permissible stress ‘oy’. Hence the ‘moment of resistance is given by the relation. Mr =64- Ag (@-W/3) «0... (0) Under reinforced beams will be subjected to primary tension failure. then the section is considered as forced’ and the moment of ‘the section is computed from the the steel and concrete sum permissible stresses simultgneously and the moment of resistance Of the section’ ean be estimated by either of the equations (1) aid.o ‘or (2) which yield the same Table 1: Constants for Critical or Balanced Section Grade of ‘Permissible in Mildstelo,-140 | ‘Tor Steel concrete concrete Nimm* «= 230 Nimm? (Nimm?) m ik. | Q P% \n 7 Qo |p% Mis 3 [is [04 [087 [087 j0714 [029 [09 065 [o3i5 20 7 (13 [oa [oa7 029 [09 [ost [044 Nhs LOA 087 029 {0.9 [0.535 PNeW Delis Bengalanr] Dhubancower | Vimwada | Visakhapatnam [Tirupati | Poi [Chena “p= balanced % reinforcement FenMl: ° Basic Concept - Working Stress 03. If the modular ratio is ‘m’ and ratio of allowable stress in steel to concrete (steel ratio) is ‘r’ the depth of critical neutral axis, Specifications: Tab 2: Permissible stresses in concrete ‘Giiic of ‘compression of the beam is (4! is the effective depth) " Bending direct (a) (elm) (b) (tri) .d concrete, Bie (©) (mtt).d @ none : 280 18ce MIS 3 al 04; The stresses developed in concrete:and steel that of (a).A section which is plane bet ‘> vbalaieed$eétion | remains plane after bending _BegSieess in steel. intension will reach its ‘maximum permissible value. (All. tensile stresses are. taken by # 3. The M.R. will be < that of balanced reinforcement and none by concrete, except an otherwise” specifically section. ae tea : 4. ‘The concrete on tension side is also to be (©) The modulated" “has the value of considered for calculating the MR. of 280/3 f. Where ‘f,’ is the allowable eee compressive ‘stress due to. direct ‘Which of these statements are correct? compression in conétete in N/mm”. @)1&2 1&4 (None 3&4 2&3 08. Flexural collapse in over reinforced beams is 02, Factor of safety for reinforcing steel in (in clastic: method) ehaitaciets due to _____(UES-98) Ls 3 . 7 (@) primary compression failure @ &)3 ©167 @LB Gye conn uate (Ea rai |New ent | Benga | Bhibaneswar | Visgawada | Visakhapat [ira ne neaROC (©) primary tension faiture (@ bond failure. 09. Side face reinforcement is provided in a beam when the depth of web exceed (IES-97& 98) (@) 300 mm (b) 450 mm (©) 500 mm (@ 750 mm 10. A beam section is having dimensions of b = 220mm, d = 400mm and is singly reinforced with: mild steel bars (1255.6: mm’). the section is 15. Modular ratio ‘m’ for M20 grade concrete is APPSC 2006 (BL) 9 @13 @7 16. When earthquake or wind load is taken into account, the permissible stresses in working @ 18 stress method are increased by (PSC-2006) (a) 33 13% (b) 25% (0) 0.4 times the yield stress. (4) 10 MPa 17. The stress developed in concrete and steel in reinforced concrete beam of 25 cm width and 4 fective depth are 625 kg/em” and (a) over reinforced“ (b) under ht 50 2 respectively. If modular ratio is 15 (c) balanced @n pj meg fh ofthe neutral axis is -(PSC-2008) Ly) (@) 256 (b) 30 cm 11,As: compared: to wor (©) 55.26 cme, @ none “design, limit state methé the value of mod 18. In a concrete structure the ratio of permissible stress in bending” compression and direct ion i (PSC-2007) 19. What is’ the momé 7 Sepacity of an under reinforced rectangular RCC beam? (IES 2008) where ote is the'al (b) An oajd compression of 0 @ Aeid above value of ‘m’ mi creep of concrete {6820 concrete is @no compensation — 19.33 ©) 1333 @IS (b) full compensation" oc (c) partial compensation. 21 __ ermissibie Bending stress for M15 concrete (d) the two are un-related. fneecicatr MPa is @S oT io (is 13, The effective depth of singly over reinforced section is 300 mm with neural axis at 120 mm below the top If the maximum stréss induced in the concrete is 3 MPa and modular ratio as 18 the stresses developed in steel would be OL.) 02.) 03. (c) 04. (b) 05. (¢) (a) 130 MPa (b) 135 MPa (©) 150 MPa (@ 180 MPa 06.(a) 07.(d) 08. (a) 09. (d) 10. (a) IL.@ 12.@ 13.0) 14. (a) 15.) 14, The permissible shear stress in a flexural sremiber of mild steel is APPSC 2006 (BL) 16.@)_17.@) 18.0) 19.@ (a) 140 MPa (b) 230 MPa 21. (a) (©) 190 MPa (@ 115 MPa PR a se [te [red ee i —considerations consideration. iii) Members subjected to Impact iv) Supports of continuous T-beams’ ii). Members subjected to stress reversal Definition: Reinforced on both tension and ji) To limit the depth from architectural and from head room a KE Aas N77 Basin ste Ags 3, MOMENT OF RESISTA DOUBLY REINFORCED SEC tension and compression ie, Mur = Muza +Muz a) MR ofa DRB-= balanced moi resistance of SRB. + additional moment of resistance due to additional steel in b) Mu,une =0.148 fx bd? for Fe 250 138 fx bd? for Fe 415 133 fx bd? for Fe 500 ©)Mur. = 0.87 f Aso (-d") (01) atten fa) (a ) cif ge is = neglected. fs |p Acrectany Doubly Reinforced Beams - Limit State Method Note: fj, = stress in compression steel which depends up on strain at its level for Tor. Steel (for both the grades of Fe 415 and Fe 500), use the following table depending up on strain level. 4'/4| 0.05 | 0.10 | 0.15 | 0.20 HYSD 355 | 353 | 342 | 329 fora S.R beam for My, lim 2. = additional tension steel for idular section 300. mm x 700mm overall “dimensions. 30mm effective cover to compression and tension steel is reinforced with 4 bars of 20 mm diameter in tension and 2 bars of 20mm diameter in compression respectively. Grade of steel used is Fe 415 and grade of concrete M25. ‘The actual depth of neutral axis is assuming compression steel also yielded (A) 83 mm B) 74mm (©) 93.mm___) 105mm 1) RRR ea New an Dearne Btabncwwar | Viewaved | Vika | Tirupati | Pune [Chena]i 2 @) Sol: 0:36f.4bXu + 0:87 fyAse = 0.87 fyAse 0.36x20x300xxq + 0.87 x415 x 2 x. | x20? O87 415 x 4x5 27 = 105 mm * doubly reinforced beam is (2) 0.002 (1 = 4" /xy (©)0.002 ae 02. The maximum stress mild steel is used'is, (seen in concrete ratio, (© Modular ratio and strain (@None of the above. 04, As the value of (4d) incteases:he stress in compression steel for Fe 415 bats: (@) Increase 6) Decrease (©) Weccan’t say 05, Doubly reinforced beams have 1, More ductility 2. Resistance against stress reversal 3, Shock resistance Of above statements the correct ones are: (a) Land 2 (b) 1 and 3 (©) 2and3 9 All of the above / ROC (a) more oles (©) equal (a) None 07. In limit state method stress in compression steel is based on (2) Modiilar ratio x. stress in concrete at its level ‘based on strain in concrete at its. level (©) either of the above (@) None of the above 08,.Which of the following statements about over forced beam is incorrect ection is sare compared to balanced section (@) Being subjected’ to brittle and sudden failure, not ‘recommended by IS 456 - (@) tensile steel required is more than that of a balanced section’ (b) compressive:stee! is under stressed Q ear reinforcement is more ‘concreteis not stressed to its full value What is the assumption in the stecl beam © theory of doubly reinforced beams? IES 2008 ).onlysteel bars will resist tension ¢ ly concrete resist tension "(6 stress in tension steel equal to, stress in compressive steel (@ both concrete and steel will resist compression . A doubly reinforced beam is recommended when (1ES2004) (a) the depth of the beam is restricted (b) the breadth of the beam is restricted (©) both depth and breadth is restricted (@) shear is very high z 06. Deflections of doubly reinforced beams are -vssem Compared to singly reinforced beams of same depth ‘Ses exalt] Benglors| Bhubaneswar | Viggvada | Veskoopatai | Trupal | Pune Chem] iDoubly Reinforce Beams - LSM 01. A reinforced beam of size 200mm width and 350nim’) overall “depth is. subjected to a working moment of 30 KN.m. If M15 and Fe 4153 is used, it is to be designed as (use eff. cover = 50mm) @ singly reinforced section _ Psy reinforced section (0) over reinforced section @ balanced section depth of 500 mm. i 2200 mm? of steel jon and 628 mit” of | steel in- compression: The effective céver for compression steel is 50.mm. Assumié that both | tension and comptession, steel. yie ‘grades of concrete aide] used for M20 and Fe250 _ respectively. stress “block parameters (rounded Off to first two decimal oins) for concrete and see shall be 25 per, | 18:456-2000 i) determine the depth ofneutral-axis. ii) Also determine monhétitzof. resistance of! the section is at ultimate co (Ans:160.91mm and 209.20 RNS) je: 02. A simply supported beam of 8 m effective span, RCC rectangular beam of 300 mm and 500mm (overall) section is reinforced with 4 number of 25mm bars at an effective cover of 375mm from tension face, M25 grade concrete and Fe415 grade steel are used. 2 bars of 16 mm diameter bars are used in compression side with an effective cover of 50mni from compression fibre. i) Determine ultimate moment of resistance, |_ ii) What is the superimposed live load the beam can cary with yas 1.5, for dead and live load combination (Use density of ‘concrete as 25kN/m’) (Ans: 270 KN-m and 18.82 KN/m) 03. A concrete beam has 350 mm width and 700 mm effective depth. It is subjected to a superimposed bending moment of 300 KN- mat service conditions. Material used are ‘MIS and Fe 415 with fj. = 353.7 MPa. Use ‘fective cover to compression and tensile 50 mm. Design suitable reinforcement stive cover to compression felis 40 mm. Compression iforced beam is made of M 20 conerete and teel...Find the moment of resistance of this beam’ by limit state method. Design codes will not be supplied; jf Reinforcement in MPa d'/d values 0.10 [0.15 353 [342 0.05 0.20 329 “= 600mm, A,=2060 mn, A, = 804 mm and effective cover of 50 mm for both tension and compression steels. The materials used are: M 20 concrete and HYSD steel of grade Fe 415. The salient points on stress strain curve are: ‘Strain 0.00144 0.00163 0.00192 ‘Stress, MPa 288 in kN-m, (Considering f,.= 0.85668,) 00241 0.00276 0.00380 351RCO OL. The “size of @ RC beam-is restricted: to 250:mm x 500.mm.. If carries a superimposed Toad of 25 kNim over a span of 6 m. Determine the reinforcements for the beam by LS.D. method. Mj, concrete and Fe 415 steel sate used: Effective cover to steel = 40 mm. Salient points‘on design stress-stress curve for steel : Percentage of steel P,, = (41.4 £,/6) x (X, mv), ‘f= stress in compression steel may be taken from the table: in N/mm’, aid fa 075 0.1 0.125 0.150 54353 347.5 342 |Doubly Reinforced Beams — Working Stress Method <_— bs Ie aly Aww As Consider a rectangular sect tension as well as compressi figure. d= effective cover to co x: = Stress in compressive Since strains are proporti NA. Strain in top force of concrete x/ iie., (Gae/E)-! (G/B) = x/x-d () Ge=Moue , x-d/x= Stress in concrete at the level of compression steel FM. Gite =. Sie The IS : 456 - 2000, the permissible compressive stress in bars, or in a beam or slab when compressive resistance of the conerete is taken in to account, can be taken ‘1.5m ‘times the compressive stress in surrounding concrete (1.5m. uSSibIS Stress in Steel in Compression (Gx.) which ever is less. MOMENT OF RESISTANCE OF DOUBLY REINFORCED SECTIONS: Ete MSlb8e672) (—x/3)+ (1.5m—1) Aw Pe ae (dx) (ded) = obd? + (1.5 m= 1) Age. Gate (xd /A(d—d) Mi Me veesscee Gv) Where My =MR of balanced section = Msat Me = MLR of compression steel Area of tensile steel Ag = Mi / ow. jd Area of tensile steel equivalent to compressive steel Are = Melon (@-d) Thus the total tensile steel Ag shall beRCC Aa= Asi + Aw ‘The area of steel required in compression face can be obtained from (iv). Altematively this steel Ase can also be obtained by taking moments of extra equivalent concrete area given by. compression steel and equivalent area of concrete given by extra tensile steel about N.A: (1.5m=1) Ave (x-d) = mAs (d-n) Age= Aw m/(1.5m~1) (G—1/ 2-4) srvssseen (W) NOTE: In WSM moment of resistance, compression sides, M. Thus doubly reinforc than singly reinforced b 01. Doubly. reinforced t (@) when the depth, room etc. (©) both @) ‘and o (@) none of the above 02. By over reinforcing the moment of resistance of -rginiforced |) ‘beam can be increased by not more tha (2) 30% (b)25% ~ (©) 20% @)3 re 03. The allowable compressive stress in high yield strength deformed bars used in doubly reinforced beams in ‘Nimm” is (IES-92) (a) 140 (b) 230 © moe (15 m oe 04.In a doubly reinforced beam, the allowable stress in compression steel is 05, Modified modulus ratio for compression zone is (@ equal to ‘m’ (© 2 times ‘m’ (b) 15 times ‘m’ (@) 5 times ‘m’ 06, Equivalent area of steel in compression is (@) (1.5m-1) Asc (b) m, Age © mAs @ (s/n 07. Transformed are of tension steel is equal to @mAw (b) (1.5m-1)Ase @ (x)/m area of compression steel is equal (b) (.5m-1) Ase @ (oxy/m 09. Depth of N.A fora doubly reinforced section is 20 om, The compression steel is being 4 ‘the compression edge. ssive stress in concrete the stress in the “at the level of compressit compression steel (@)5 MPa 9 3.MPa shal oe designed: b) Lebeam {ae2©) Singly reinforced (@) doubly reinforced section 02.) 03.(4) 04.(€) 05.(b) 07.(@) 08.(c) 09.(e) 10.(4) 01. ©) 06.(@) @v = ox (0) > o% (©) @) it Peso2 (or) Be>0.43 a Xe replace Ds by ‘yr in the above equation. (2) (AT ~ beam of, lange width 740mm, effective ‘depth 400mm, =0.879250x6 5 x 20? (400 - 0.42 x $1.29) ear STON ~ SRE a | Wastin | re | Pm 1:8L: * Limit State ~ Flanged Sections Ol. In the effective width of flange formule. Lo defined as distance between: points of zero . moments is considered due to LY Beyond Lo the: beam is in. hogging, and hhence flange: will be. in. tension and sineffective. <- (©) With in To stresses are uniform (©) Strain compatibility width of web = 250 of flange is % @im owt 25m reasofi for this asd (@ Stress is more either side comy (b) Nearer section “a farther one’s, © Slab ‘will be’ in beam (@ None of above 04, If neutral axis lies in can be treated as a rectan; @by xd b)dexd (© b, PES 05. If neuttal axis falls in the web” and flange thickness less than 0.2 d, and the section is balanced the stress in the flange will be (@ uniform and equal to 0.36 fx (b) uniform and equal to 0:45'fax (©) non-uniform (@uniform and equal to 0.416 fx 06: If neutral axis falls in the web and flange thickness larger than 0.2 d, and the section is balanced ,the stress in the lange will be @ 0-446 fa uniform A (@) None of above. a (b) 0.446 fax partly and more than 0.446 fx partly (©) 0.36 fek partly and less than 0.36 fix, partly (@) partly 0.446 fix partly less than 0.446 fy 07 If flange thickness larger than 0.2 d and u balanced and neutral axis falls in the web, the effective thickness. of flange Yris : (a) (0.15 xy max + 0.65 Dr) > Dy (&) (0.416 xu max +0.65 Deh 4 Dr (©) (0.15 xa + 0.65 Dr) 4 Dr @ @A16 xu + 0.65 Ds) m4 Dr (b) Df < 0.43 @ on 20.43 ‘and flange are more (a) Bending stress \ear stress respectively (0) Shear stress ani ling stress receptively (©) Both bending:and’shear (ayNone of abo |) 10. Doubly reinforced rectangular beams are to be SP) :edéSigned .in‘‘cerlain instances among. the _following»- °, a ipports-of continuous T-beams arn Subjected to reversal of stresses Ieest*Giy members ‘subjected to impact Of these statements: (a) ( only correct (b) (and Gi) are correct (©) Gil) also correct @ (ii) and (iii) only correct |. The effective flange width of T beams spaced at 3.25 m with web depth of 1 m, web width of 0.4 m spanning 12 m with a flange slab of 100 mm thickness is @in 325m ©)2.5m @2.0mOsh ivers 182: RCC 12, A'T- beam behaves-as a rectangular beam of width equal to its flange if its “NA? (IES 2006) (@) coincides with centroid of reinforcement (b) coincides with centroid of T-section: (c) remains with in the flange (@) remains in:the wed 13. A T-beam roof section has’ the’ following particulars: Thickness “of slab 100 mm; width of ib 300mm; depth of beam 500° min; ‘centre: to ‘centre distance of beams 3 tn; effective span of, beams 6m; distance between contraflexure 3.6m; The’ ef width is (2) 3000 mm (©) 1600 mm 01..A.T- beam with ef mum; effective dep 240. mm; depth i) What is the ultin of the above secti fi) If the the steel used of HYSD, Fe 415 ultimate moment of (Ans:12! 02. A hall of 10m x 15m consists of a beams 3m centre to centre parallel to the shorter span of the hall. Width of web is 250 ‘mm. Thickness of slab 100 mm, the beams are cast monolithic with the columns at their ends. What is The effective width’ of flange of an intermediate beam (Ans: 2.01m) 03, In the above problem the’beams are simply Vat ends: What is the effective width of flange of an end'beam ? ~ (Ans: 1.38 m) the following dimensions actual flange width is 2m; flange thickness 150mm; overall depth 750mm with effective cover of 40mm to compression and tension steel; rib width is 300mm; the beam is provided with 6 bars of 20 mm’ diameter in tension and 2 bars of 20 mm diameter in compression zone. M25 grade concrete and Fe415 grade steel are used. Both tension and compression steels are yielded. ‘What is the ultimate moment of resistance of ‘the beam. (Ans: 614 kN- 01. A Tee beam havi sam is required flange width of 2500 an ultimate moment of mine the area of ment. The following reinforcement requi formulae may be used: i ()) Rectangular, Seetions without compression i a reinforcement depth of neutral axis. i 190.36 fb less than the limiting value Aafy 187 £ Ast d [1-2 § ( a *e Gii)Moment of resistance for limiting value of depth of neutral axis Flanged section - if x, RCC | inclined stirrups. at 45°, where d_ is\ the effective depth of the section under | 02. Diagonal compression failure occurs when consideration. In no case shall the spacing @w>te (0) 40> Te, max exceed 300mm. © w36mm (a)2d (bo) 4d (0) Bd (A) 164 13. The lap length for bars in bending tension 07. The “bond strength between. steel shall not be less than reinforcement and concrete is affected by (@) Ly or 306 (b) 2Lg or 30 (@) Type of reinforcement (©) Ly or 249 (@) 2L4 or 246 (b) Grade of concrete (© Shrinkage of concrete 14, The lap length in compression is not less than @all the above (@) Ly or 309 (b) 2Lg or 306 (©) La or 24 (@) 2a or 246 08. ‘The main reason for providing certain minimum number of reinforcing bars at a | 15, The development length of each bar of bonded ‘support in'a simply supported beam is to resist |~—bars-shall” be that_for the individual’ bar, in that zone. increased by % for two bars in contact, i RN ea eta | We [Viator [Ta | Pe [che beeenBond Limit State and Working Stress nosed % for three bars in contact and... (@15MPa —(b) 3.75 MPa for four bars in contact, ()43MPa, — (d)2.4MPa (@)10, 20,30. -(b) 10,33,20 23. The splicing in reinforcement in flexural (©) 20,3010 (@) 10,20,33 members will be governed by which one of the following considerations 16. For full anchorage, the stuns should extend. (@) where BM is 10 % of the MR bya length of.......tithes diameter of bar (b) where SF is zero when bent at 135° (©) where BM is less than 50% of MR 492 ©) 86 © 126 -@ 6 (@ where BM is zero 17. The length of bar beyond theoretical cut off | 24. A bar is bent into an angle of 90 digress the point shall be anchorage value is (a) anchorage length (b) 4 times its dia (b) development length. itsdia _(d) 8 times its dia (c) bond length A (@) dowel length f.a reinforced concrete beam > Se bond is not getting 18. A bar of, 10mm. dia ied i isfied. ical option to satisfy the concrete for a dis (GATE 08) the maximum load i the bond stress is no (b) providing smallér diameter bars more in (a) 1.5 EN ©2kN eter bars “in less in 19. Bond’ between steel adic su “diameter bars more in ‘@) stress coms b 1}! 26. what is the borid'stress acting parallel to the @None feinforcementon,4he interface between bar correct? Pa : : Ina reinforced concrete riembeét : to ensure adequate bond is, See | 2. i (@) to provide’ minimum number “Of Wage. s}anii2When HYSD bars are used in place of mild diameter bars steel bars in a beam, the bond strength (b) to. provide large number of smaller (IES 2008) diametet bars : (@) does not change —_(b) increases (0)-to iticrease the cover for reinforcement (© decreases (a) becomes zero (@) to provide additional stirrups eee ae aL agence eer esee te tener ‘the following is correct a : expression to estimate the development length ina value equivalent to a straight length of deformed reinforced ber as per IS 456 in ae Oi O16 Bde limit state of design? (GES 2007) (a) $05/4.5t 0a (0) 655/5t 0a oihor 6dr @) 04! 8t08 sign 5 for deformed” bars in| compression is 3.0 MPa. The corresponding ‘bond stress for same bars in tension is Fipaerabad | Now Dah | Beapaor | Bhubanewar | Voerarads | Vikhapabiana | Tina)Drs Fses RCC 29. The distance between theoretical cut off point \ and actual cut off point in respect of the | curtailment of reinforcement of a'RC beam should not be less than (LES 2006) (@) development length | (b) 12 x dia of bar or effective length whichever is greater (©) 48 * dia of bar or effective length whichever is greater - (@) 30 x dia of bar or effective length ‘whichever is greater 30. Consider the following statements tf: correct statements: + 1. Reinforcement that is nogiongel ; for flexure, beyond’a however be" exten that section (@) Land 2 (©) 2and3 figure,” the sagging moment zone. beyondsthe- jnflection, The factored shear force doris ! i (150. RN If fx=15 N/mm, f= 250N/imm” i and Te = 1.0 MPa then the beam is Section of point inflection t fe 1 i 7 Cleat cover 25mm250mm 250mm () just safe in bond (b) over safe is bond (©) unsafe in bond (@ unsafe in bond even with additional anchorage. 02::Pick up the incorrect relation : (a) Long term modulus of clasticity = (short term modulus) / (1 + Creep coefficient ) (b) As IS 456 - 2000 Short term modulus of elasticity = 5000 ff, (c) Development length La = $ ( 0.87 )/ 4 ta imum shear reinforcement is governed anchorage length of 12 mm diameter bars/Of Fe 415 grade steel in M30 grade’ concrete with’ design bond strength of 2.4 MPa and a'90"dedree standard bend at the “ends” e + pong) 355.3 mm : 1mm (IES 2004) no longer required for Alexure bey: certain section, shall _/ nowever be extended by d or 120 js greater, before being urtailed. ifthe bars should be bent up at off point capacity at cut-off point should fitnes the shear force at that section. os pgs &2 () 1&3 ©283 @1,2&3 05. A bar of 20 mm diameter in a axially loaded short column is to be spliced with 16 mm diameter bar , the: lap length required for proper transfer for the load in mm is, (use Fe 415 grade steel and y= 1.2 MPa) (@) 752.18 mm (b) 553.6 mm (©) 576mm @ 601.75 iim, | |Bond Limit State and Working Stress: Oat Consider a bar of diameter D embedded in a Jarge concrete block as shown in the figure, ‘with a pull out force P being applied. Let oy aiid 6, be the bond strength (between the bar and concrete) and the tensile strength of the respectively. . If the block is held in ‘position and it is assumed that the material of “The block doesnot fail, which ofthe following se represents the maximum value of P (GATE 2011) and plain bars in te = 1.2 MPa. Fur this-design. bond. stres: by. 60% for HYSD HYSD reinforcing stee! 0, = 360 MPa, Find the diameter, 08.'A “rectangular: simply: supported ‘beam has factored shear force-at the centre of support is 220 KN. The beam has 2 bars of 16mm diameter Fe 415 steel at the centre of support, (on tension Zone. width of beam is 250 nim and effective “depth is'425° mm, M20- concrete grade With ‘tj = 1.2 MPa is used in beam. ‘What is the minimum length of extension of main steel’ bars with 90° bend to satisfy anchorage bond? eee 05.(b) 01.4) 026) 03.@)—04.@) 06.44) 07.44) 0B.) 09.4) 10.06) 1L@ 2@ 13@) 14@ 15.44) 16d) 17a) 18.(b) —-19.(b)_ 20.6) DUA) 22d) 23.66) 24.) 25.06) 30.) 01. © 02.@ © 06. (4). 0746.8 de, 7 05. (d)I t ‘ Limit State of Collapse - Torsion 1. DEFINITION: If the longitudinal axis of a structural member and loading. axis are perpendicular to each other, the structural ‘member will be subjected to twisting, called Torsion. Ex: a) Canopy beams (Isolated L-beain) si 3% — Be b) Curved bear tal plane (cing beam tank) <= ©) Grids 3 3 3 3 3 3 ) End beam of kj} End Bea} 2. CLASSIFICATION OF TORSION : a) Statically determinate or equilibrium or primary Torsion : A torsion that develops to ‘maintain static equilibrium in the structural assemblage. It arises as a result of primary action that the extemal load has no alternative to being resisted but by Torsion. Ex +-Torsion in-a Canopy-beam and Curved eam in horizontal plane are statically determinate or equilibrium Torsion, Sin: Ff New Dali Bailar PBhabascorar| Viwereday| Venkhapsnans | Trwpatl| Pune [Chennai | b) Statically Indeterminate or Compatibility ‘or secondary Torsion: A Torsion that arises as a result of secondary action ftom the requirement of continuity. The magnitude of this torsion in a member depends upon the magnitude of the torsional stiffness of the ‘member it self in relation to the stiffness of S'FWO,TORSIONS WE HAVE TO Re aprsc FOR ? WHY? Tt must be designed requirement or load ing resisted but by Design for this is of membetS is not considered in the analysis, the structure need no be ed for torsion. iffiness of members is Be analysis, the members for torsion. IS method is based on simplified skew bending theory proposed by ‘Hsu’ a) The Design Approach : It does not require determination of torsional _ reinforcement separately from that required for bending and shear. Instead, the B.M and S.F are modified to account for torsional moment. The reinforcement details are as follows. (@ Total longitudinal reinforcement is determined _for __ equivalent _moment ‘obtained from flexural moment and torsion.Gi) Transverse or web reinforcement is determined for equivalent shear obtained from shear and Torsion. It shall be in the form of closed stirrups _ placed perpendicular to the axis of the member. Step. 1: Determine Equivalent _ shear V = V, 4167 where Vy= Shear force, Ty = Torsional moment , b= width of beam for flanged (Lacor} beams) beams use b= by= width stre eS =F The value of “ye? other wise cross sect be revised. * I te < reinforcement, minimum ‘shear per bs, If Te > transverse provided, a) Tensile reinforcement: reinforcement shall be equivalent B.M (M,,) given = Ma = My + Mr, Where My = factored Bending moment M, B+) Where M, = B.M contributed by factored torsion moment (T,) D_ = Overall depth of beam b) Compression reinforcement : If My > Mu, longitudinal compression reinforcement. shall be provided on flexural compression face’ to st an equivalent moment M,2_given by. caustione fo | State of Collapse = Torsion Step 4: Transverse reinforcement : The cross ‘séctional area of two legged closed hhoops enclosing the comer longitudinal bars is given by, Sy = Spacing of stirrups, ‘ie = Equivalent shear stress , csbjested to torsion. Side face reinforcement of 0.1% of gross area (for both faces together) shall be Provided with a spacing not more than 300 mm center to center. Reason: Beams with greater depth are likely to buckle laterally. To take care of possible tensile forces on sides, side face reinforcement shall be provided. Ma = M.—RCC <(D/b)] 17 Ty @ 165 08. The transverse torsional reinforcement in 01. Shear stress for a rectangular beam subjected = : RCC beams shall” be provided as to torsion as shown in figure below is maximum at @A Ad (£6) Vertical sticrups _(b) inclined stirrups .B (©) bent up bars (@all the above @©c ce : 09. The minimum no. of torsional longitudinal i reinforcement in R.C.C beams is @D : Hoy (@) any of the above beam is subjected to torsion. moment of ¢ R.C.C Side “ees i igrooment cement shall be efficiently, 6 @An unsymmetrical 5) A box section (A solid rectangul (@ A symmetrical 03, The erack pattern at (a) vertical porora 04. Compatibility Torsio (@) Mes My p> Mu Sm (@) Edge beams: (© Canopy beams 05, Equivalent shear ‘V_ is ge 41.6 (Ty/8) Cm isto} 2 : ‘Which of the above statements are true ae j . 2 and 3 (b) Land 2 06. The spacing of stirrups in beams subjected to @1. fotsion sbi be (©) 2and 3 (@) Land 3 » @ aK >) #batyo/4 13. The value of tw < te i jected _ 83300: = (dail the above — lue of tye < te in a section subjected to ? < (a) torsional desi t 07. Equivalent B.M due to torsional moment ‘Ty’ yaaa design isnot « must isgivenby . (ES 2006) (c) the beam fail by diagonal compression failure (@) shear reinforcement isnot required - 7, (t+(@/b)) ahd 7 (EGR ne ese a oer [Vintfpm [ae | i et]if nominal shiear stress t, > témx_ section has tobe revised due to” 4 (@) diagonal tension failure Ao) diagonal compression failure (©) flexural shear failure @) all the above 15.A. Canopy. beam will be subjected to . type. of torsion, primary torsion. (b) secondary torsion (© only bending moment | @ only bending moment and shear 16. L — beams are subjected. to, (@ all the above 18, Beams of grid floor torsion. 4a) primary torsion () secondary torsion- (©) only bending moment @ only bending moment and si 19. Match the following with reference to R.C.C List -1 A. Torsional Analysis B, Shear stress analysis C. Limit state D. Working stress List - 11 1, Truss analogy 2. Skew Bending Theory AB CD (()Sesee bate eeeet eer 261 ce @® 21,4 3 @ 12 4 3 20, Primary torsion occurs in the case of (@) a simply supported, but laterally restrained ‘beam subjected to eccentric loading along ‘the span (b) the edge beam of a building frame shells elastically restrained by edge beams torsion i) (@ primary torsion Sar, Jin deep beam of 300 mm Seysecondary torsion wide, subjected'to ashear force of 150 KN and (©) only bending mom torsion 30 KNangeqhivalent shear is @ only bending m« shear,’ force (a) 180 kN 0Y310 kN (6) 246 kN ‘of size 300 x 1000 mm. It = 1SOKN, ‘and T = 30 KN-m. at limiting tign;-qiivalent bending moment and wialent’ shear force at ultimate loads are Sespectively THU G@sieNm,s10KN Me > Mut TOCA) (b) 226.5kN-m, 206 KN VF (6) 100.5kN-m, 310 kN (4) 226.5kN-m, 310kN- V2 = Vat hey 02. An RCC beam of size 300 mm x 600 mm is subjected to factored values of shear force 100 KN, B.M. of 100 KN — m and twisting moment of 34 KN ~ m. Longitudinal tension reinforcement shall be designed for a moment 3. Semi-probabilistic Approach. 4, Deterministic (a) 100 KN-m (©) 60 kN-m (6)120 KN-m ASFT6O KN250: RCC 03. In the above beam, if ultimate torsional moment is 68 KN ~m, longitudinal compression steel( shall be designed for a moment of ....kKN-m. 01. A reinforced concrete beam of rectangular gn (b) 30 40 OB section is 550mm wide and has an overall depth of 750 mm. It is subjected to an 04. A rectangular beam of 500 mm x 700 mm ultimate bending moment of 150KNm and an with effective cover of 35 mm is subjected to a lutimate twisting moment of SOkNan M 15 factored values. of shear force. 15 KN; a grade concrete and Fe415 grade steel are used sbending’mioment™ 100 KN- m and @ torsional Determine the necessary longitudinal ' ‘moment 10 KN-m the design bending moment reinforcement. A ting beam of a water tank is subjected to ete o values of bending moment of 100 (©) 128 44 15.@) 20.(a) f, 16.(b), -“17.(@) i 21.(b).- 22.) 05, The equivalent shear forcgiis (220 KN. () 34 * SERN J) 8 — 96.! THe SGhivatent Mexural. moment (Ma). for “OE oe designing the longitudinal tension steel is 06.() 07.) (@) 187.KN-th 46) 200 kN-m ° ©) 209 EN-m-5°y (4) 213. KNem 07. A reinforced.concrete. beam of 10 m effective span. and 1 m, effective depth is simply -supported:. If the total udl on the beam is 10 ‘MNN/m, the design shear force for the beam is ee) @ 30MIN ‘(Oy47SMN ———} a (©)325MN > (@)40 MN ii i [NSDA [Baan [Ear [Vipera [Venki Toop] Rene Tohewml |3¢ Dy ‘Types of Slabs : 4) Slabs spannig in one ditection : Supported at two opposite ends ») Slabs supporting on all four sides : These are further classified into two types based fon aspect ratio (1, / le) @ One way slabs 1f(J)/:) > 2 i) Two way slabs: 1674) > 2. 2) General notes on design of slabs" a)” Contrél of deflection i) Basic vatues of span _=— reinforcement shall IS- 456 q Borg any direction : Sie ~0.12 % of ross area WitnHEYSD bas (Tor steel) = 0.15 % of gross area with Mila: steel 3 d) Spacing - Main Reinforcement : 3 xd or 300 mm which ever less Minimum Steel: 5d or 450mm. which ever less 3) Slabs carrying concentrated loads : If a solid slab supported on two opposite edges, carries concentrated loads, the maximum, BM developed by the concentrated loads shall. be assumed to be resisted by an effective width of slab-(measured parallel to.the supporting edges) aw Slabs c)__ Minimum percentage of reinforcement it sf: as follows [ New Dati | Bengalara | Bhabanewar | Viayawada | Viakhapaain, ‘Tropa Poe [Chia | a) For single concentrated load : = effective width of slab k= constant depending on ratio of width of slab (1) to effective span (ler) distance of the centroid of the icentrated load from the nearer support. thvof contact area of the fenttated load measured parallel to ofie way slabs: In two fis transferred to all the “therefore, the bending fection ‘are considerably Can be used for any type "-ofilab (restediped or unrestrained ) ibs: (Comers are prevented * Maximum bending moments per unit width in a slab are My = aw Py, Myr ayw Py Where 0, and ay are coefficients based on ‘boundary conditions. ‘w= design load per unit area Slabs are considered as divided in each direction into Middle strips and-edge strips as shown below@ wb Grose 152: RCC = Maximum mon part: of the slab ‘continuous edge or = Over the contimious ¢ “the upper part. f isi} ~~~from support and at least. ee a distance of 0.3 /.. ‘Torsion reinforcement : Shall be provided where the slab is simply: supported on both ‘edges mecting'at that: comer. It consists of top ‘and bottom reinforcements ‘at comers extending from edges @ minimum distance of one-fifth of shorter span, The area of reinforcements in ‘each’. of these: four layers “shall be three ‘quarters of the area required for the maximum. mid span moments in the slab: = Torsion reinforcement need not be provided at a comer continued by edges over both of ‘which the slab is continuous. - Half the torsion reinforcement is required at a cover with one edge discontinuous and the ther continuous. i) Simply supported slabs -: When simply supported slabs do not have adequate provision to resist torsion at comers and to prevent the ‘comers from lifting, the maximum moment per unit width are given by . flowing, ‘equation. oe Wy say wl’ 5.50% of qeosion” Tenforcement id span should extend to the ‘ing 50% should extend 0.1 fy of the support, as maximum size of be used is (@)12 mm dia (b) 10 mm dia ‘(© 8mm dia (@) 6mm dia oa ges torsion Tejnforcement required is a). 0.7: area of steel provided at ie same direction imes the area of steel provided at ‘mid-span in the same direction (©) 0. 75 times the area of steel provided in the shorter span @Nil 03. A square slab 4 m x 4m is a simply supported slab. If it is subjected to a design load of 12 kPa (Including self weight) the moment capacity required as per IS:456-200; use ox = ‘ay = 0.062 @1.9 kN-m (b)15 kN-m Oe ENQ wt Slabs 04. The bending -moment - coefficients for continuous RC slabs in IS: 456 is based on IES 1998) (@) Pigeand’s method (b) Marcus's method (©) Yield- line theory (@ Westergaard’s mathematical analysis 05. As per IS: 456 — 2000 the vertical deflection limit for beams’ may gerierally be assumed to be ‘satisfied provided that ratio of span to effective depth of a continuous. beam of span 12 mis not to be greater than @35)26 21.6 (b) Primary function (©) Independent of (@ Most long span 07. The moment coeffi (©) Yield line theory (d) Westergaards math 08. The minimum thickness of according to IS; 456-2098“ ® a. a 11, Drops are provided in flat slabs to resist GES 97, 05) () shear @ torsion () bending moment (©) thrust 12. For maximum sagging bending moment in a given span of a multiple span beam (IES 1998) (@) every span as well as alternate spans are loaded (b) adjacent spans ate loaded (©) spans adjoining this span are loaded (dy adjacent spans are unloaded and next spans are loaded. Pe BB : ion reinforcement required ZZP for « two ‘way siniply supported slab is Where Astiy is the'iiain reinforcement in the shorter direction (a) % Ast, (0) 0.75 Aste (0) 4 Asti L (@) none Peay, 14. “TRiggminal cover fequired for a slab having mild “exposure, “with the diameter of reinforcing “bars usediequal to 10 mm should |) be not less than _ @30mm ) 25 mr, | O20mm @) none U a Ee 415.'A rectangulaf slab im x_6m supported on two ‘ofiposite shorteriedges should be designed as, (a)-2 one-way slab spanning along longer (a) 125. mm. wy edges: e ~~-(c)400-mm. (d),200 oe & actiio-way slab Nadie % i shorte 09;The ‘continuous slab designed secon ssn ws slab spanning along shorter moment coefficients given by IS :456 should (@none be @ One way slab (6) two way slab (©) any type (@ two way restrained slab 10, The isotropic slab can be (A) rectangular slab” (B) circular'slab © square slab : Which of the above statements is correct (a) AandB @BandC (A, Bandc (@ none 16. The negative reinforcement in a two .way restrained slab should be provided over {@) discontinuous edge (b) continuous edge (©)both a and b @aone 17. The thickness of slab depends on (= eff. Length; d= eff. depth) (©) spacing of reinforcement @noneRCC 18, The-minimum. reinforcement in a slab takes Seaie of (@) Temperature & shrinkage stresses, (b) homogeneity of slab (¢) support to main reinforcement “(@all the above 19..The moment coefficients given in IS: 456- 2000 for simply supported two-way slabs are ‘based on (a) Rankine- Grashoff’s method (b) westergaard’s method (©) Johansen’s yield line theory @ Bemnoullie’s theory 20. The main reinforcement ingR¢ beam is placed at Ae (a) top face along the fy () bottom face along the-width (o) top face perpendict @ bottom face perpendi 24, (©) Johansen’s yield @ plate theory 22, The: plaster thickness of th (a) can be included in the reinforcement (b) should not be inclu (©) provides nominal cover (@) none of the above 23. The minimum percentage of high yield strength deformed bars in RCC slabs is: (@)04 (@) 0.15 (©) 0:12 (60.23, 24, The torsional, reinforcement in a two way | restrained slab required for a corner with two \, continuous edges will be (20.75 times the area of steel provided at midspan in the same direction —-(0}-0:375-times-the-area-of-steel-provided at midspan in the same direction (c) 0.75 times the area of steel provided in the shorter span @Nil 25. 1n a two way restrained slab torsion steel is provided at (@) Top (b) bottom (©) ‘a" and ‘b’ (@) none 26. The critical section for shear in a flat slab is at a distance of , (d=effective depth) (a) effective depth from the face of a -eglumn or column drop |, 27.A simply supported slab of 10 m effective veelspan, the minimum effective depth to satisfy the Vertical deflectiom limits should be (a) 50m, (©) 60 cm” F; 28. In atwo way sla | (a) resultant i 4b) torsional (©) resultane’stress at the ends "_-«(@)unbalancedmoment on the slab ig of comers occur due to at the ends issiblpwWalue of deflection for a two- vay-simply supported slab with shorter span jneg@1ess than 3.5 m and the live load is less than 3 KNim’ using deformed bars is 35 (0)28—(©)40_— (4) 32 30, Which one of the following statements is correct? ES 2004) Temperature and shrinkage steel is provided in reinforced concrete slabs because (@) it occupies larger area (b) its thickness is less (©) itis a main structural element _(@itis.a flexural member. (pica [ New Dal] Bengalara | Bhubaneswar | Vinjagoda | Viniapatnan | Timpad | Pune |Chennat I a | ae es | [a | Vint [Toe Reco]