RCC & STEEL STRUCTURES · RCC: Working stress, ... transfer and service loads. Steel Structures: Working stress and Limit state design concepts; Design of tension and compression

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RCC & STEEL STRUCTURES

SYLLABUS RCC: Working stress, Limit state and Ultimate load design concepts; Design of beams, slabs, columns; Bond and development length; Prestressed concrete; Analysis of beam sections at transfer and service loads.

Steel Structures: Working stress and Limit state design concepts; Design of tension and compression members, beams and beam- columns, column bases; Connections - simple and eccentric, beam- column connections, plate girders and trusses; Plastic analysis of beams and frames.

ANALYSIS OF GATE PAPERS

Exam Year 1 Mark Ques. 2 Mark Ques. Total 2003 3 5 13

2004 2 5 12

2005 2 6 14

2006 1 5 11

2007 - 6 12

2008 2 5 12

2009 2 2 6

2010 1 2 5

2011 4 2 8

2012 2 3 8

2013 4 1 6

2014 Set-1 2 3 8

2014 Set-2 5 - 5

2015 Set-1 4 2 8

2015 Set-2 1 3 7

2016 Set-1 1 2 5

2016 Set-2 1 1 3

2017 1 2 5

RCC & STEEL STRUCTURES

Topics Page No

1. BASICS

1.1 Plain and Reinforced Concrete 01 1.2 Compressive Strength of Concrete in Structure 02 1.3 Flexural Strength of Concrete (Modules of Rupture) 02 1.4 Tensile Strength of Concrete 02 1.5 Stress-Strain Curve of Concrete 02 1.6 Concrete Strain at Ultimate Strength 03 1.7 Design Stress-Strain Curve 04 1.8 Shrinkage and Creep in Concrete 04 1.9 Design Stress Strain Curve for Steel 05

2. DESIGN OF RCC STRUCTERES

2.1 Introduction 07 2.2 What is Limit State Method 07 2.3 Characteristic Strength of Method 07 2.4 Characteristic Load 08 2.5 Partial Safety of Factor 08 2.6 Assumption Limit State of Collapse: Flexure 09 2.7 Working Stress Method 12 2.8 Deficiency in Working Stress Method 12 2.9 Permissible Stresses 12 2.10 Permissible Stresses in Concrete 13 2.11 Permissible Stress in Steel Reinforcement 13

3. CONCRETE TECHNOLOGY

3.1 Introduction 15 3.2 Chemical Composition of Raw Materials 15 3.3 Type of Cements 16 3.4 Admixture 18 3.5 Different Type of Admixture 18 3.6 Aggregates 20 3.7 Permissible Limits for Impurities in Water 21 3.8 Concrete 21 3.9 Methods of Proportional Concrete 21 3.10 Water Cement Ratio 21

4. REINFORCED SECTION

4.1 Balanced, Under-Reinforced And Over Reinforced Section 23

CONTENTS (RCC)

4.2 Moment of Resistance Calculation 24 4.3 Doubly Reinforced Section 25 4.4 Analysis Doubly of Reinforced Section 26 4.5 Moment of Resistance Doubly of Reinforced Section 27 4.6 Design of Doubly of Reinforced Section 27 4.7 Flanged Beam 28 4.8 Effective Flange Width 29 4.9 IS Code Specification 29 4.10 Design of Flanged Section 32 4.11 Limit State of Collapse in Shear 33 4.12 Diagonal Tension and Diagonal Compression 34 4.13 Mechanism of Shear Resistance 36 4.14 Shear Stress 36 4.15 Design Shear Strength of Concrete in Beams 37 4.16 Shear Reinforcement in Beam 39 4.17 Minimum Shear Reinforcement 41 4.18 Design Steps 42 4.19 Bond and Anchorage 43 4.20 Torsion 45 4.21 Effect of Torsion Moment 46

5. BEAM COLUMN

5.1 Design of Beam and Slab 48 5.2 I.S 456 Provisions 48 5.3 Control of Deflection 49 5.4 One-way Slab 50 5.5 Introduction 50 5.6 One-way and Two-way Slabs 51 5.7 Code Requirements on Reinforcement and Detailing 51 5.8 Column Subjected to Axial Compression and Uniaxial bending

(clause 39.5 ) 55 5.9 Column Subjected to Axial Compression and Biaxial bending

(clause 39.6) 56 5.10 Code Procedure for Design of biaxilly Loaded Columns 56 5.11 Depth of Foundation 58 5.12 Design Consideration 58

6. BASIC ELEMENT OF PRESTRESS CONCRETE

6.1 Losses in Priestess 62 6.2 Loss of Prestress Due to Friction 62 6.3 Loss of Prestress Due Anchorage Slip 63 6.4 Loss of Prestress due to Creep of Concrete 64 6.5 Loss due to Shrinkage of Concrete 64 6.6 Loss of Prestress due to Relaxation of Steel 65 6.7 Deflection of Restressed Beam 66

6.8 Prestressed Concrete 68 6.9 Requirement of High Strength Steel & Concrete in Prestressing 69

8. GATE QUESTIONS 71

Topics Page No

1. GENERAL DESIGN SPECIFICATION

1.1 Design Objective 90 1.2 Methods of Design 90 1.3 Loads and Forces 90 1.4 Load Combinations 90 1.5 Geometrical Properties 90 1.6 Classification of Cross Sections 90 1.7 Limit State Designs of Steel Structures 91 1.8 Connections: Riveted Connection 93 1.9 Terminology in Bolted Connection 93 1.10 IS 800-2007 Specification for Spacing and Edge Distances

of Bolt Holes 94 1.11 Types of Joints 95 1.12 Design Strength of Plates in a Joint 96 1.13 Block Shear Strength 96 1.14 Design Strength of Bearing Blots 96 1.15 Bearing Capacity or Bolts Vdpb 97 1.16 Tensile Capacity of Bolts 97 1.17 Welded Connection 97 1.18 Types of Welded Joints 98 1.19 Specification for Welding 98 1.20 Fillet Weld 98 1.21 Reduction in Design Stresses for Long Joints 100 1.22 Eccentric Connection - Plane of Moment and The Plane of

Welds is the same 100

2. DESIGN OF TENSION MEMBERS

2.1 Tension Member 101

CONTENTS (STEEL STRUCTURES)

2.2 Design Strength Due to Yielding of Gross Section 101 2.3 Design Strength Due to Rapture of Critical Section 101 2.4 Design Strength Due to Block Shear 103 2.5 Lug Angles 103

3. DESIGN OF COMPRESSION MEMBERS

3.1 Introduction 105 3.2 Slenderness Ration 105 3.3 Design Compressive Stress and Strength 106 3.4 How to Select the Shapes of Compression Members 106 3.5 Steps for Design of Compression Members 107 3.6 Battens 107 3.7 Design of Battened Columns 107 3.8 Design of Slab Base 108

4. DESIGN OF BEAMS

4.1 Introduction 110 4.2 Bending Strength of Laterally Supported Beam 110 4.3 Design Procedure 111 4.4 Shear Strength of Laterally Supported Beam 111 4.5 Web Buckling Strength 112 4.6 Web Crippling 112

5. GATE QUESTIONS 113

6. ASSIGNMENT QUESTIONS (RCC & STEEL STRUCTURES) 132

RCC

1.1 PLAIN & REINFORCED CONCRETE

Plain Concrete: It is a mixture of sand, gravel, cement, and water which results in a solid mass. Concrete is strong in compression but weak in tension. Its tensile strength is approx, one tenth of compressive strength. Plain concrete is mostly used in mass concrete work. (As in dams)

1.1.1 REINFORCED CONCRETE

1) It is a concrete with reinforcementembedded in it. The embeddedreinforcement makes it capable ofresisting tension also.

2) Steel bars embedded in the tensionzone of concrete, relieves concrete ofany tension and takes all tensionwithout separating from concrete.

3) The bond between steel andsurrounding concrete ensures straincompatibility i.e., the strain at any pointin the steel is equal to that in theadjoining concrete.

4) Reinforcing steel imparts ductility toconcrete which is otherwise brittlematerial.

(5) Here ductility means large deflection owing to yielding of steel, thereby giving ample warning of impending collapse.

6) Tensile stress in concrete arises onaccount of direct tension, flexuraltension, diagonal tension (due to shear),temperature and shrinkage effect,restraint to deformation.

7) Under these conditions, reinforcementsmust be provided across potentialtensile crack.

1.1.2 GRADE OF CONCRETE

Compressive strength of concrete isthe most important property ofconcrete. Because other properties like

tensile strength, shear strength, bond strength, density, impermeability, durability etc. can be inferred from the compressive strength using established correlations.

Compressive strength can bemeasured by standard test on concretecube. (or cylinder) specimen.

Strength of concrete in uniaxialcompression is determined by loadingstandard test cube (150 mm size) tofailure in compression testing machine.

The test specimen in generally tested28 days after casting (and continuouscuring)

Cube is always tested on sides i.e., facein touch with mould.

Strength of cube is expressed to benearest of 0.5N/mm2.

As per IS 456: 2000, three specimen of asample is taken.

Additional samples may be required forvarious purposes such as to determinethe strength of concrete at 7 days or atthe time of striking from work, or todetermine the duration of curing, or tocheck the testing error. Additionalspecimen may also be required fortesting samples cured by acceleratedmethods.

To report, strength of cube, we takeaverage of three specimen of a sample.

1 BASICS

© Copyright Reserved by Inspiring Creativity & Endeavour Gate Institute. No part of this material should be copied or reproduced without permission. 1

Individual variation should not be more than ±15% of average if variation is more, test results of the sample are invalid.

1.2 COMPRESSIVE STRENGTH OF CONCRETE IN STRUCTURE Strength of concrete is found to decrease with increase in the size of specimen. However, beyond 450mm size, there is no decrease in the compressive strength of concrete. Thus, compressive strength of concrete in structure is taken as 0.67 fck. 1.3 FLEXURAL STRENGTH OF CONCRETE (MODULES OF RUPTURE) Tensile strength of concrete in flexure is called flexural strength.

3

2

M W×100= ×10

Z 6×100× 100

6

N/mm2 [if W is in

kN] [Assuming linear stress strain curve and contribution of steel area to be negligible] fcr = 0.4W N/mm2 [W is in kN for onset of cracking.] However, stress strain variation is not linear hence as per IS code

fcr = 0.7 ckf N/mm2 N/mm2

Flexural strength is used to determine the onset of cracking or the loading at which cracking starts in a structure. 1.4 TENSILE STRENGTH OF CONCRETE

Tensile strength of plain concrete is obtained y the splitting test.

Splitting tensile strength ct

2Pf

πdL

Fct = splitting tensile strength = 0.66 fcr. Modulus of rupture

Direct tensile strength = [0.5 – 0.625] fcr.

1.5 STRESS-STRAIN CURVE OF CONCRETE

It is found by testing cylinder under

compression. The max. Strength obtained is the cylinder strength. Cylinder strength = 0.8 × cube strength (150 mm , 300 long)

Cylinder is tested to obtain stress strain curve, because curve, we have to obtain condition for uniaxial stress condition. In case of cubes, due to friction between the concrete surface and the steel plate of testing machine, lateral restraints occurs.

The effect of this lateral restraint is to compressive strength in longitudinal direction. This effect dies down with increasing distance from the friction

© Copyright Reserved by Inspiring Creativity & Endeavour Gate Institute. No part of this material should be copied or reproduced without permission.

Basics

2

surface [called platen restraint]. Thus as the distance from the friction surface increases (i.e., as height/width ratio increases), compressive strength decreases.

From the above stress-strain curve, following points must be noted.

1) Max compressive stress occurs at astrain value of 0.002. i.e., 0.2%. Thevalue of stress at 0.002 strain iscalled compressive strength ofconcrete.

2) Lower strength concrete has greatdeformability i.e., ductility than highstrength concrete.

3) Descending part of high strengthconcrete is sleeper.

4) High strength concrete gets crushedat smaller strain.

5) The point where curve ends is calledcrushing strain.

6) High strength concrete is morebrittle as compared to low strengthconcrete.(Crushing strain is 0.3%-0.5%)

7) Curves are generally linear upto astress of 0.6 times the peak stress.

8) Modulus of elasticity of concrete forall practical purpose is taken assecant modulus at a stress of around0.33 ckf . This Ec is generally found

acceptable in representing an average value of Ec under service load condition (static loading).

Modulus of elasticity is primarilyinfluenced by the elastic properties ofaggregate and to a lesser extent by theconditions of curving, age of concrete,mix proportion and type of cement.

As per IS code:

c ckE 5000 f ( cE and ckf are in

N/mm2)Short term modulus of elasticity ofconcrete

Long term modulus of elasticity

including creep cce

EE

1+θ

Where, θ = creep coefficient

cE = short term modulus of elasticity

θ = creep coefficient ultimate creep strain

elastic strain at the age of loading

Age at loading Creep coefficient 7 days 2.2 28 days 1.6 1 year 1.1

1.6 CONCRETE STRAIN AT ULTIMATE STRENGTH

If a concrete cylinder is axially loadedthe ultimate strength is at 0.2% strain.

For flexure, crushing is assumed tooccur at 0.35% strain.

Note:- Actually it is seen that if the stress sit the distribution is


Basics

3

a) rectangular, strain = 0.2% b) Triangular, strain = 0.35% c) Trapezoidal, strain = 0.2 – 0.35% 1.7 DESIGN STRESS-STRAIN CURVE Ascending part is taken as 2nd degree

parabola. 0.67 fck = Strength of concrete in

structure ym = partial safety factor for material

strength. ym = 1.5 for limit state of collapse ym = 1.0 for limit state of serviceability

1.8 SHRINKAGE & CREEP IN CONCRETE 1.8.1 Creep When concrete is subjected to sustained

loading, its deformation keeps on I increasing with time, even though the stress level is not altered.

Time dependent component of total strain is called creep.

Creep is thought to occur due to:

1) Internal movement of adsorbed

water

2) Viscous flow or sliding between gel

particles

3) Moisture loss

4) Growth in micro cracks

Effect of creep are 1) Increase in deflection of beams and

slabs 2) Increased deflection of slender

column that may lead to buckling 3) Gradual transfer of load from

concrete to reinforcing steel in comp. members

4) Loss of prestress

Beneficial effects of creep are: 1) Reduction in stress induced by

2) In indeterminate structure, stress

induced due to settlement of

support is reduced due to creep.

Factors influencing creep Cree increase when:- (a) Cement conte4nt is high

(b) W/c ratio is high

(c) Aggregate content is low

(d) Air entertainment is high (e) Relative humidity is low (f) Temperature (causing moisture loss)

is high (g) Size/thickness of member is small (h) Loading occurs at early age (i) Loading sustained over a long period

Creep coefficient for design

As long as stress in concrete does not exceed one-third of its characteristic strength, creep may be assured to be proportional to stress. Thus under service load condition, creep will be proportional to stress. This concept can be used to compute total deflection (initial + creep) by usual linear elastic analysis with reduced modulus of elasticity. The reduced modulus of elasticity

cc

EE

1+θ

Where cE = reduced modulus of

elasticity taking into account long term effect of creep


Basics

4

cE short term modulus of elasticity

θ creep coefficient

Age at loading θ

7 days 28 days 1 Year

22 16 1.1

Intermediate value of creep coefficient

may be interpolated by assuming that the creep coefficient decreases linearly with the long of time in days. Thus, creep coefficient for age of loading at 15 days in

10 10

15

10 10

0.6 log 28-log 15θ = 1.6

log 28-log 7

i,e., 0θ = C - θ log t

Effect of creep can be reduced by 1) Using high strength concrete 2) Delaying the application of finishes,

partitions walls etc. 3) Adding reinforcement 4) Steam curing under pressure.

Note: Steam curing under pressure reduces drying shrinkage and moisture movement

1.8.2 SHRINKAGE The shortening in length of a member or contraction of the concrete per unit length due to drying when concrete sets are known as shrinkage. Shrinkage can be classified as a) Plastic shrinkage b) Drying shrinkage 1.8.2.1Plastic shrinkage This type of shrinkage manifests itself soon after the concrete is placed in the form while the concrete is still in plastic state.

Loss of water by evaporation from the surface of concrete or by absorption by aggregate or sub grade is believed to be the reason for plastic-shrinkage. As aggregate & steel restrain this effect, cracks appear at the surface or internally around aggregate. Thus shrinkage is prevented by using aluminum powder and expanding cement. 1.8.2.2Drying shrinkage It is an everlasting process and occurs

mainly due to loss of water held in the gel pores when concrete is kept in drying condition.

The finer the gel, the more is the shrinkage.

Harder aggregate leads to lower shrinkage but higher shrink but higher shrinkage stress. Reverse is true for softer aggregate.

Shrinkage decreases with increase in size of member.

Concrete made with smaller size aggregate shrinkage more than concrete made with larger size aggregate.

Shrinkage produces tensile cracks in any member which is restrained.

For a given environment, the total shrinkage depends more on the total amount of water present in concrete at the time of mixing and to a lesser extent on the cement content.

In the absence of test data, an apex value of total shrinkage strain for design may be taken as 0.0003 shrinkage decreases with increase in relative humidity.

½ of total shrinkage assumed in 1st month and 3/4th in 1st 6 moths.

1.9 DESIGN STRESS STRAIN CURVE FOR STELL

m Factor of safety

m 1 for limit state of serviceability


Basics

5

= 1.5 for limit state of collapse Increase of stress in strain hardening

region is neglected In RCC steel never reaches its ultimate

strength because strain at which ultimate strength is reaches will never be reached.

Concrete will get crushed before steel reaches its ultimate strength.

Note:-Failure of RCC beam always occurs due to crushing of concrete. It never occurs due to failure of steel. (i.e., shaping of steel) We may use same curve for tension and

compression for both grades of steel (Hot rolled or cold worked.) This may lead to insignificant error. (in case of compression)


Basics

6

Topics Page No

1. WORKING SRESS & LIMIT STATE METHOD 72

2. SHEAR, TORSION, BOND, ANCHORAGE & DEVELOPMENT LENGHTH 79

3. FOOTING, COLUMNS, BEAMS AND SLABS 82

4. PRESTRESSED CONCRETE 86

GATE QUESTIONS

71

Q.1 The following two statements are made with reference to a simply supported under reinforced RCC beam: I. Failure takes place by crushing

of concrete before the steel has yielded.

II. The neutral axis moves up as theload is increased.

With reference to the above statements, which of the following applies? a) Both the statements are falseb) I is true but II is falsec) Both the statements are trued) I is false but II is true

[GATE-2001]

Q.2 Maximum strains in an extreme fibre in concrete and in the tension reinforcement (Fe-415 grade and Es-200kN/mm2) in a balanced section at limit state of flexure are respectively a) 0.0033 and 0.0038b) 0.002 and 0.0018c) 0.0035 and 0.0041d) 0.002 and 0.0031

[GATE-2003]

Q.3 List-I contains some properties of concrete cement and List-II contains list of some tests on concrete/cement Match the property with the corresponding test. List-I A. Workability of commie B.Direct tensile strength of coma-etc C. Bond between concrete and steel D. Fineness of cement List-II 1. Cylinder splitting lest2. Tiee-Bee test3. Surface area test4. Fineness modulus test5. Pull out. Lest

Codes: A B C D

a) 2 l 5 3 b) 4 5 1 3 c) 2 1 5 4 d) 2 5 1 4

[GATE-2003]

Q.4 For avoiding the limit state of collapse, the safety of RC structures is checked for appropriate combinations of Dead Load (DL), imposed Load or Live Load (M), Wind Load (WL,) and Earthquake Load (EL). Which of the following load combinations is NOT considered? a) 0.9 DL + 1.5 WLb) 1.5 DL + 1.5 WLc) 1.5 DL + 1.5 WL + 1.5 ELd) 1.2 DL + 1.2 IL + 1.2 WL

[GATE-2004]

Q.5 In the limit state design method of concrete structure, the recommended partial material safely factor (yn) for steel according to IS: 436-2000 is a)1.5 b)1.13c)1.00 d)0.87

[GATE-2004]

Q.6 The partial factor of safety for concrete as per IS : 456-2000 is a) 1.50 b) 1.15c) 0.87 d) 0.446

[GATE-2005]

Q.7 The flexural strength of M30 concrete as per IS: 456-2000 is a) 3.83 MPa b) 5.47 MPac) 21.23 MPa d) 30.0 MPa

[GATE-2005]

1 WORKING SRESS & LIMIT STATE METHOD


Q.8 In a random sampling procedure for cube strength of concrete, one sample consists of X number of specimens. These specimens are tested at 28 days and average Strength of these specimens is considered as test result of the sample, provided the individual variation in the strength of specimens is not more than ∓ Y 5 percent of the average strength. The values of X and Y as per IS : 456 – 2000 are

a) 4 and 10 respectively b) 3 and 10 respectively c) 4 and 15 respectively d) 3 and 15 respectively

[GATE-2005]

Q.9 As per IS : 456 – 2000, consider the following statements: 1. The modular ratio considered in

the working stress method depends on the type of steel used.

2. There is an upper limit on the nominal shear stress in beams (even with shear reinforcement) due to the possibility of crushing of concrete in diagonal compression.

3. A rectangular slab whose length is equal to its width may not be a two way slab for some support conditions.

The TRUE statements are a) 1 and 2 b) 2 and 3 c) 1 and 3 d) 1, 2 and 3

[GATE-2006]

Q.10 Consider the following statements: 1. The compressive strength of

concrete decreases with increase in water-cement ratio of the mix.

2. Water is added to the for hydration of workability.

3. Creep and shrinkage are independent of the water –cement ratio in the concrete mix.

The TRUE statements are a) 1 and 2 b) 1, 2 and 3 c) 2 and 3 d) only 2

[GATE-2007]

Q.11 A reinforced concrete structure has to be constructed along a sea coast. The minimum grade of concrete to be used as per IS : 456-2000 is

a) M 15 b) M 20 c) M 25 d) M 30

[GATE-2008] Q.12 Un-factored maximum bending

moments at a section of a reinforced concrete beam resulting from a flame analysis are 50, 80, 120 and 180 kN-m under dead, live, wind and earthquake loads respectively. The design moment (kNm) as per IS: 456-2000 for the limit state of collapse (flexure) is

a) 195 b) 250 c) 345 d) 372

[GATE-2008]

Q.13 The modulus of rapture of concrete terms of its characteristic cube compressive strength (fck) in MPa according to IS 45 6:2000 is

a) 50000 fck b) 0.7 fck

c) 50000ckF d) 0.7 ckF

[GATE-2009]

Q.14 For limit state of collapse, the partial safety- factors recommended By IS 456:2000 for estimating the design strength of concrete and reinforcing steel are respectively

a) 1.15 and 1.5 b) 1.0 and 1.0 c) 1.5 and 1.15 d) 1.5 and 1.0

[GATE-2009]

Q.15 Match List-I (List of test methods for evaluating properties of concrete) with

List-II (List of properties) and select the correct answer using the codes given below the lists:


Working Sress & Limit State Method

73

List-I A. Resonant frequency test B. Rebound hammer test C. Split cylinder teat D. Compacting factor lest List - II 1. Tensile strength2. Dynamic modulus of elasticity3. Workability4. Compressive strengthCodes:

A B C D a) 2 4 1 3 b) 2 1 4 3 c) 2 4 3 1 d) 4 3 1 2

[GATE-2009]

Q.16 As per IS 456:2000, in the Limit Suite Design of a flexural member, the swain in reinforcing bars under tension at ultimate state should not be haw than a) f/E b) f/E + 0.002c) f/1.15E d) f/1.15E + 0.002

[GATE-2012]

Q.17 Maximum possible value of Compacting Factor for fresh (green) concrete is: a) 0.5 b) 1.0c) 1.5 d) 2.0

[GATE-2013]

Q18. Match the information given in Group – I with those in Group II.

Group - I Group - II (p) Factor to decrease

ultimate strength to design strength

(1) Upper bound on ultimate load

(q) Factor to increase working load to ultimate load for design

(2) Lower bound on ultimate load

(r) Statical method of ultimate load analysis

(3) Material partial safety factor

(s) Kinematical mechanism method of ultimate load analysis

(4)Load factor

a) P – 1; Q – 2; R – 3; S – 4b) P – 2; Q – 1; R – 4; S – 3c) P – 3; Q – 4; R – 2; S – 1

d) P – 4; Q – 3; R – 2; S – 1[GATE-2014]

Q19. While designing, for a steel column of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in N/mm2) in limit state method of design as per IS:456-2000 is _________

[GATE-2014]

Q.20. The first moment of area about the axis of bending for a beam cross-section is a) moment of inertiab) section modulusc) shape factord) polar moment of inertia

[GATE-2014]

Q.21. Group I contains representative stress-strain curves as shown in the figure, while Group II gives the list of materials. Match the stress-strain curves with the corresponding materials.

Group I Group II (p) Curve J (1) Cement paste (q) Curve K (2) Coarse aggregate (r) Curve L (3) Concrete a) P-1; Q-3; R–2 b) P-2; Q-3; R-1c) P-3; Q-1; R-2 d) P-3; Q-2; R-1

[GATE-2014]

Q.22. The target mean strength fcm for concrete mix design obtained from the characteristic strength fck and standard deviation σ, as defined in IS:456-2000, is

a) ckF +1.35σ b) ckF +1.45σ

c) ckF +1.55σ d) ckF +1.65σ

[GATE-2014]


Working Sress & Limit State Method

74

Q.1 (b)

Q.2 (a)

c 0.0035

steelE = 5

fy

1.15E + 0.002

3

4150.002

1.15 200 10

= 0.0038

Q.3 (a)

Q.4 (c)

Q.5 (c)

Q.6 (b)

Q.7 (d)

Q.8 (d) Refer : IS : 456 – 2000 clause 15.3

Q.9 (d)

Q.10 (a)

Q.11 (d)

Q.12 (d) Load combinations (1) 1.5 (DL + LL)

= 1.5 (50 + 80) = 195 kN-m (2) 1.5 (DL + EQ) = 1.5 (50 + 180) = 345 kN-m (3) 1.2 (DL + LL + EQ)

= 1.2 (50 + 80 + 180) = 372 kN-m Use maximum of above combination.

Q.13 (a)

= 0.7 ckf

Q.14 (a)

Q.15 (c)

Q.16 (b)

Q.17 (b) Compaction factor = weight of partially compacted concrete/ weight of fully compacted concrete For full workability weight of partially compacted concrete will be equal to weight of fully compacted concrete, therefore maximum possible value of compaction factor is equal to 1

Q.18. (c) Static method → Upper bound method of ultimate load analysis Kinematic method → Lower bound on ultimate load QDesign = Qw × load factor M0 = Mu × Ym' Ym= material partial safety factor

Q.19 (9 to 9) Permissible bearing strength = 0.45

fck 20.45 20 9N / mm

Q.20 (b)

Section modulus, z =2I A.r

= =A.rr r

,

i.e. Moment of Area.

Q.21 (b) So, P = 2,

Q = 3 R = 3

EXPLANATIONS


STEEL STRUCTURE

89

S

1.1 DESIGN OBJECTIVE

The objective of design is the achievement of an acceptable probability that structures will perform satisfactorily for the intended purpose during the design life. With an appropriate degree of safety, they should sustain all the loads and deformation during construction and use and have adequate resistance to accidental loads and fire.

1.2 METHODS OF DESIGN

Structure and its elements shall normally be designed by limit state method as per IS 800-2007. Where the limit state method cannot be conveniently adopted; working stress method shall be used.

1.3 LOADS AND FORCES

For the purpose of designing any element, member or structure, the following loads and their effects shall be taken into account, where applicable, with partial safety factors and combinations : a) Dead loads;b) Imposed loads; (Live load, crane load,

snow load etc.)c) Wind loadsd) Earthquake loadse) Erection loadsf) Accidental loads such as those due to blastg) Secondary effects due to contraction or

expansion resulting from temperaturechanges, differential settlements of thestructure as a whole or of itscomponents, eccentric connections.

1.4 LOAD COMBINATIONS

The following load combinations with appropriate load factors may be considered in designing.

a) Dead load + Imposed loadb) Dead load + Imposed load + Wind or

Earthquake loadc) Dead load + Wind or Earthquake loadd) Dead load + Erection load

1.5 GEOMETRICAL PROPERTIES

IS 800-2007 gives the concept of the gross and effective cross -sections of a member. The properties of the gross cross-

section shall be calculated from thespecified size of the member or readfrom appropriate table.

The effective cross -section of a memberis that portion of the gross cross –section that is effective in resisting thestresses.Highlighting this concept of effectivecross -section, IS 800-2007 hasclassified the members cross-section asfollows.

1.6 CLASSIFICATION OF CROSS SECTIONS

Basis of classification The plate elements of a cross -section

may buckle locally due to compressivestresses.

When plastic analysis is used, themembers shall be capable of formingplastic hinges with sufficient rotationcapacity (ductility) without localbuckling to enable the redistribution ofbending moment required beforeformation of failure mechanism.

When elastic analysis is used, themember shall be capable of developingthe yield stress under compressionwithout local buckling.

On the above basis, four classes of section are defined as follows :

a) Plastic : Cross-sections, which candevelop plastic hinges and have the

1 GENERAL DESIGN SPECIFICATION


rotation capacity required for failure of the structure by formation of a plastic mechanism.

b) Compact : Cross -sections, which can develop plastic moment of resistance, but have inadequate plastic hinge rotation capacity for formation of a plastic mechanism.

c) Semi-compact : Cross -Sections, in which the extreme fiber in compression can reach, yield stress, but cannot, develop the plastic moment of resistance, due to local buckling.

d) Slender : Cross -sections in which the elements buckle locally even before reaching yield stress.

1.7LIMIT STATE DESIGNS OF STEEL STRUCTURES Basis for Limit State Design In the limit state design method, the structure shall be designed to withstand safely all loads likely to act on it throughout its life. It shall also satisfy the serviceability requirements, such as limitations of deflection and vibrations and shall not collapse under accidental loads such as from explosions or impact or due to consequences of human error to an extent not originally expected to occur. The acceptable limit for the safety and serviceability requirements before failure

occurs is called a limit state. The objective of design is to achieve a structure that will not become unfit for use with acceptable target reliability. In other words, the probability of a limit state being reached during its lifetime should be very low. In general, the structure shall be designed on the basis of the most critical limit state and shall be checked for other limit states. In limit state design, structures are designed on the basis of safety against failure and are checked for serviceability requirements. 1.8 CONNECTIONS:RIVETED CONNECTION

Riveting is a method of joining together pieces of metal by inserting ductile metal pins called rivets into holes of pieces to be connected and forming a head at the end of the rivet to prevent ach metal piece from coming out.

Rivet holes are made in the structural members to be connected by punching or by drilling. The size of rivet hole is kept slightly more (1.5 to 2.0 mm) than the size of rivet.

After the rivet holes in the members are matched, a red hot rivet is inserted which has a shop made head on one side and the length of which is slightly more than the combined thicknesses of the members to be connected.

Then holding red hot rivet at shop head end, hammering is made.

It results in to expansion of the rivet to completely fill up the rivet hole and also into formation of driven head.

Desired shapes can be given to the driven head.

The riveting is done may be in the workshops or in the field.

Riveting has the following disadvantages. a) It is associated with high level of

noise pollution. b) It needs heating the rivet to red hot.


General Design Specification

91

c) Inspection of connection is a skilled work.

d) Removing poorly installed rivets is costly.

e) Labour cost in high. Production of wieldable quality steel

and introduction of high strength frication grip bolts have replaced use of rivets.

Design procedure for riveted connections is same as that for bolted connection except that the effective diameter of rivets may be taken as rivet hole diameter instead of nominal diameter of rivet.

IS 800-2007 do not discuss riveted connection in it hence it is not discussed further here.

1.8.1 BOLTED CONNECTIONS

A bolt is a metal pin with a had formed at one end and shank threaded at the other in order to receive a nut bolts are used for joining together pieces of metals by inserting them through holes in the metal and tightening the nut at the thread ends.

Bolts are classified as : a) Unfinished (black) bolts b) Finished (turned) bolts c) High strength friction grip (HSFG) bolts

a) Unfinished/Black Bolts

These bolts are made from MILD steel rods with square or hexagonal head. The shank is left unfinished i.e. rough as rolled.

In structural elements to be connected holes are made larger than nominal diameter of bolts.

As shank of black bolts is unfinished, the bolts may not establish contact with structural member at entire zone of contact surface.

Joints remain quite loose resulting into large deflections.

These bolts are used for light structures under static loads such as trusses, bracings and also for temporary connections required during erections.

It is not recommended for connection subjected to impact, fatigue or dynamic loading.

Bolt of property class 4.6 means, ultimate strength is 400 N/mm2 and yield strength is 400 × 0.6 = 240 N/mm2

If a bolt is designated as M16, M20, M24, M30, it means shank diameter of 16 mm, 20 mm, 24 mm, and 30 mm respectively.

b) Finished/Turned Bolts

These bolts are also made from mild steel, but they are formed from hexagonal rods, which are finished by turning to circular shape.

Tolerance available for fitting is quite small (0.15 mm to 0.5 mm)j.

It needs special methods to align bolt holes before bolting.

As connection is tighter, it results in to much better bearing contact between the bolts and holes. These bolts are used in special jobs like connecting machine parts subjected to dynamic loadings.

c) High strength Friction Grip (HSFG) Bolts High Strength bolts

a) Made from bars of medium carbon steel.

b) Normally class 8.8 and 10.9 are commonly used

c) Less ductile than black bolts d) Material of the bolt does not have well-

defined yield point. Instead of yield stress, proof load in used



92

e) As per IS 8000: 2007 proof load is taken as 0.7 × ultimate tensile stress of bolt

f) M16, M20, M24, M30, are generally used

g) Designated like 8.8S, 10.9S denotes high strength bolt.

h) Percentage elongation of these bolts at failure is approx 12%

i) Special techniques are used for tightening the nuts to induce specified initial tension in the bolts, which caused sufficient friction between the flaying forces.

i) These bolts with induced initial tension as called high strength friction grip (HSFG) bolts.

k) Due to friction, the sleep in the joint is eliminated hence; connection in this case is called nonslip connection or friction type connections.

1.9 TERMINOLOGY IN BOLTED CONNECTION 1. Pitch of the bolts (P)

It is the centre to centre spacing of bolts in a row, measured in the direction of load.

2. Gauge (g) It is the distance between the two consecutive bolts of adjacent rows and is measured at right angle to the direction of load.

3. Staggered Pitch (Ps) It is the centre to centre distance of staggered bolts measured in the direction of load.

4. Diameter of Bolt Hole Diameter of bolt hole is larger than the nominal diameter (shank diameter) of the bolt to facilitate erection and to allow for in assurances in fabrication. Holes are

a) Standard clearance hole normal

b) Oversized holes (i.e. holes of size larger than standard clearance hole)

used in slip resistant connection.

c) Short and long slot used slip

resistant connection Following table gives the diameter of holes for bolts.

Clearances for fastener Holes

Sl No.

Nominal size of Fastener, d

mm

Size of the Hole = Nominal Diameter of the Fastener + Clearances

Mm Standard

Clearance in Diameter and Width of Slot

Over size Clerance in Diameter

Clearance in the Length of

the slot Short Slot

Long Slot

(1) (2) (3) (4) (5) (6) i) 12 - 14 1.0 3.0 4.0 2.5 d ii) 16 - 22 2.0 4.0 6.0 2.5 d iii) 24 2.0 6.0 8.0 2.5 d iv) Larger than 24 3.0 8.0 10.0 2.5 d

From the above table:

Diameter of Normal Bolt Holes is: Nominal size of Bolts in mm

12 14 16 20 22 24 30 36

Diameter of Bolt hole in mm

13 15 18 22 24 26 33 39

5. Area of Bolt at Root (Anb)

Area of bolt at root of the thread is less than that at shank of the Bolt. It is taken approximately equal to 0.78 times the shank area i.e. Anb = 0.78 × Asb Where Asb = Area of bolt at shank d=Nominal diameter of Bolt(shank diameter) Anb = Area of bolt at root 1.10 IS 800-2007 SPECIFICATION FOR SPACING AND EDGE DISTANCES OF BOLT HOLES

1. Pitch 'P' shall not be less than 2.5 d, where d is the nominal diameter of bolt.

2. Pitch 'P' shall not be more than a) 16 t or 200 mm, whichever is less, in

case of tension members. b) 12 t or 200 mm, whichever is less, in

case of compression members where t is the thickness of thinner plate.



93

c) In case of staggered pitch, pitch may beincreased by 50 percent of valuesspecified above provided gaugedistance is less than 75 mm.

3. In case of compression member whereforces are transferred through buttingfaces, i.e., (butt joints), maximum pitchis to be restricted to 4.5 d for a distanceof 1.5 times the width of plate from thebutting surface. (Refer Figure Below).

4. The gauge length 'g' should not be morethan 100 + 4 t or 200 mm whichever isless in compression and tensionmember where t is the thickness ofthinner outside plate.

5. Minimum edge and end distance shallnot bea) Less than 1.7 × hole diameter in case

of sheared or hand flame out edges.b) Less than 1.5 × hole diameter in case

of rolled, machine flame cut, sawnand planed edges.

6. Maximum edge distance (e) should notexceed

a) 12 tε, WHERE y

250ε

f and t is the

thickness of thinner outer plate. This recommendation does not apply to fasteners interconnecting the components of back to back tension members.

b) Where the members are exposed tocorrosive environment max edge qdistance 40 mm + 4 t, where t is

the thickness of thinner connectedplate.

7. Apart from the required bolt from theconsideration of design forces,

additional bolts called tacking fasteners should be provided as specified below. a) Tacking rivets should be providing.

i) At 32 t or 300 mm, whichever isless, it plates are not exposed toweather.

ii) At 16 t or 200 mm, whichever isless, if plates are exposed toweather?

8. In case of a tension member made up oftwo flats or angles or tees or channels,tacking rivets are to be provided alongthe length to connect its components asspecified below:a) Not exceeding 1000 mm, if it is

tension member.b) Not exceeding 600 mm, if it is

compression member.Countersunk heads

• is neglected in calculating length offastener in bearing

• For Fastener in tension havingcountersunk heads, tensile strength isreduced by 33.3% and no reduction inshear strength calculation.

1.11 TYPES OF JOINTS

Types of joints may be grouped into the following two: a) Lap jointb) Butt jointa) Lap Joint:

It is the simplest type of joint. In this theplates to be connected overlap one another.



94

b) Butt Joint In this type of connections, the two main plates butt against each other and the connection is made by providing a single cover plate connected to main plate or by double cover plates, one on either side connected to the main plates.

1.12DESIGN STRENGTH OF PLATES IN A JOINT

Plates in a joint made with bearing bolts may fail due to any one of the following 1. Shearing or bursting of the edge. 2. Crushing of plates. 3. Rupture of Plates. 4. Block shears failure of plates in tension.

The brushing or shearing and crushing failures are avoided if the minimum edge/end distance as per IS 800-2007 recommendations are provided. As per IS 800-2007: The minimum edge distance and end distances from the centre of any hole to the nearest edge of a plate shall not be less than. 1.7 times the hole diameter in case of sheared or hand – flame cut edges; and 1.5 times the hole diameter in case of rolled, machine-flame cut, sawn and planed edges. The maximum edge distance to the nearest line of fasteners from and edge of any un-stiffened part should not exceed

y

15012tε,wherε=f and t is the

thickness of the thinner outer plate.

This would not apply to fasteners interconnecting the components of back to back tension members. Where the members are exposed to corrosive influences, the maximum edge distance shall not exceed 40 mm plus 4t, where t is the thickness of thinner connected plate. S

It the maximum distances are ensured in a joint, the design tensile strength of plate in the joint is the strength of the thinnest member against rupture. This strength is

given by. 1

n udn

m

0.9A fT

γ

Where γm1

= Partial safety factor for failure at

ultimate stress = 1.25

fu = Ultimate stress of the material An = Net effective are of the plate at critical section, which is given by b = Width of plate t = Thickness of thinner plate in joint dn = Diameter of the bolt hole (2 mm in addition to the diameter of the hole, in case of directly punched holes) g = Gauge length between the bolt holes Ps = Staggered pitch length between lines of bolt holes n = Number of bolt holes in the critical section It may be noted that, if there is no staggering Psi = 0 and hence An = (b – ndh)t, which is the net area at critical section.



95

1.13 BLOCK SHEAR STRENGTH

Block Shear failure of palate occurs as shown below. It's a combination of yielding and rupture. Block shear failure of a plate occurs along a path involving tension on one plane and shear on a perpendicular plane along the fasteners.

Block Shear strength at an end connection is calculated as given below. It should be taken as the smaller of

0

1

u tndb vg y m

m

0.9f AT = A f 3 γ +

γ

Shear yielding + tensile rupture OR

1

y bgu tndb

m m0

f A0.9f AT =

γ γ

Shear rupture + tensile yielding Where Avg, Avn = Minimum gross and net area in shear along a line of transmitted force respectively (1-2 and 4-3 as shown in figure above.) Atg, Atn = Minimum gross and net area in tension from the hole to the toe of the angle or next last row of bolt in plates, perpendicular to the line of force respectively (2-3). fu, fy = Ultimate and yield stress of the material respectively.

0 1m mγ , γ = Partial factor of safety in

yielding and rupture respectively

0 1m m(γ 1.1,γ 1.25).

1.14 DESIGN STRENGTH OF BEARING BLOTS

The design strength of bearing bolts under shear is the least of the following : a) Shear capacity (strength). b) Bearing capacity (strength) a) Shear Capacity (Strength) of Bearing Bolts in a joint

Design strength of the bolt, Vdsb is given by

nsb

dsb

mb

VV =

γ

Where Vnsb, nominal shear capacity of bolt and

mbγ partial safety factor of material of

bolt= 1.25. In the above expression Vnsb is

given by ub

dsb n nb s sb

fV = η A +η A

3

Where, fub=Ultimate tensile strength of a bolt

nη = Number of shear planes with threads

intercepting the shear plane.

sη = Number of shear planes without

threads intercepting the shear plane Asb = Nominal shank are of bolt

2d=

4

Anb = Net shear area of the bolt at threads, may be taken as the area corresponding to root diameter of the thread.

2 2

nb nb

π πA =0.78 d or A = (d –0.382p)

4 4

Where p is the pitch of thread, d = nominal dia. of bolt ie shank dia.

1.15 BEARING CAPACITY OR BOLTS(Vdpb)

IS 800-2007 Suggest the following procedure to find bearing strength of bolts.

npb

dpb

mb

VV =

γ

Where, Vdpb = Design bearing strength Vnpb = Nominal bearing strength

mbγ = Partial safety factor of material = 1.25

Nominal bearing strength may be found from the following relation.

npb b uV = 2.5 K dt f

Where Kb is smaller of



96

ub

h h u

fe P, – 0.25, , 1.0

3d 3d f

In which e, p are end and pitch distances respectively dh = diameter of hole fub, fu = ultimate tensile strength of bolt and plate respectively d = nominal diameter of bolt t = summation of the thickness of the connected plates experiencing bearing stress in the same direction. If bolts are countersunk, it is to be reduced by the half depth of countersinking. In plates the bearing strength is a linear function of and distances both bolts and plates are subteated to significant triaxial containment. Due to this, bearing behavior of plates in influenced by the proximity of neighboring holes or boundary (edge distance)

1.16 TENSILE CAPACITY OF BOLTS

A bolt subjected to a factored tensile force Tb shall satisfy:

b dbT T

Where, Tdb = nb mbT γ

Tnb = Nominal tensile capacity of the bolt

mb

nb ub n yb sb

m0

γT = 0.9 f A < f A

γ

Where, fub = Ultimate tensile stress of the bolt fyb = Yield stress of the bolt An = Net tensile stress area, shall be taken as the area at the bottom of the threads Asb = Shank area of the bolt.

mb m0γ , γ = Partial safety factors = 1.25, 1.1

respectively It means that tensile capacity of bolts is minimum of

yb sbtb ndb

mb m0

f A0.9f AT Min ,

γ γ

tb n

mb

0.9f A

γ= tensile rupture

yb sb

m0

f A

γ= tensile yielding

1.17 WELDED CONNECTIONS

Welding consists of joining two pieces of metal by establishing a metallurgical bond between them. The elements to be connected are brought closer and the metal is melted by means of electric arc or oxyacetylene flame along with weld rod which adds metal to the joint. After cooling the bond is established between the two elements.

1.18 TYPES OF WELDED JOINTS

There are three types of welded joints: 1. Butt weld 2. Fillet weld 3. Slot weld and Plug weld

1. Butt Weld Butt weld is also known as groove weld. Depending upon the shape of groove made for welding butt welds. Types of Butt Welds

2. Fillet Weld



97

Fillet weld is a weld of approximately triangular cross-section joining two surfaces approximately at right angles to each other in lap joint, tee joint or corner joint.

When the cross-section of fillet weld is isosceles triangle with face at 45º. It is known as standard fillet weld. In special circumstances 60º and 30º angel are also used. A fillet weld is known as concave fillet weld, convex fillet weld or as miter fillet weld depending upon the shape of weld face.

3. Slot Weld and Plug Weld Figure below shows a typical slot weld in which a plate with circular hole is kept with another plate to be joined and then fillet welding is made along the periphery of the hole.

Figure below shows typical plug welds in which small holes are made in one plate and is kept over another plate to be connected and then the entire hole is filled with filler material.

1.19 SPECIFICATION FOR WELDING

Important specification regarding butt weld, fillet weld and plug and slot weld as per IS 800 – 2007 is:

Butt Weld

1. The size of butt weld shall be specified by the effective throat thickness. In case of a complete penetration butt weld it shall be taken as thickness of the thinner part joined. Double U, double V, double J and double level butt welds may be generally regarded as complete penetration butt welds. The effective throat thickness in case of incomplete penetration butt weld shall be taken as the minimum thickness of the weld metal common to the parts joined excluding reinforcement. In the absence of actual data it may be taken as 5/8th of thickness of thinner material.

2. The effective length of butt weld shall be taken as the length of full size weld.

3. The minimum length of butt weld shall be four times the size of the weld.

4. If intermittent butt welding is used, it shall have an effective length of not less than four times the weld size and space between the two welds shall not be more than 16 times the thickness of the thinner part joined.

1.20 FILLET WELD

1. Size of Fillet Weld a) The size of normal fillet weld shall be

taken as the minimum weld leg size. b) For deep penetration welds with

penetration not less than 2.4 mm, size of weld is minimum leg size + 2.4 mm.

c) For fillet welds made by semi-automaticor automatic processes with deep penetration more than 2.4 mm, if purchaser and contractor agree.

S = minimum leg size + Actual penetration

2. Minimum size of fillet weld specified in 3 mm.



98

To avoid the risk of cracking in the absence of preheating the minimum size specified are: For less than 10 mm thickness plate – 3 mm For 10 to 20 mm thickness plate – 5 mm For 20 to 32 mm thickness plate – 6 mm For 32 to 50 mm thickness plate – 8 mm

3. Effective throat thickness:It shall not be less than 3 mm and shall not generally exceed 0.7 t (or t under special circumstances) where t is the thickness of the thinner plate at the elements being welded. If the face of plates being welded is inclined to each other, the effective throat thickness shall be taken as K times the fillet size where K is as given in table below:

4. Effective Length:The effective length of the weld is the length of the weld for which specified size and throat thickness exist. In drawings only effective length is shown. While welding length made is equal to effective length plus twice the size of the weld. Effective length should not be less than 4 times the size of the weld.

5. Lap joint:The minimum lap should be four times the thickness of thinner part joined or 40 mm whichever is more. The length of weld along either edge should not be less than the transverse spacing of welds.

6. Intermittent welds:Length shall not be less than 4 times the weld size or 40 mm whichever is more. The minimum clear spacing of intermittent weld shall be 12 t for compression joints and 16 t for tensile joints, where t is the thickness of thinner plate joined. The intermittent welds shall not be used in positions subject to dynamic repetitive and alternating stresses.

1.20.2 DESIGN STRESSES IN WELDS

Butt Welds Butt welds shall be treated as parent metal with a thickness equal to the throat thickness, and the stresses shall not exceed those permitted in the parent metal. It means that the strength of Butt weld is equal to the strength of the parent metal.

Fillet Weld, Slot or Plug Welds Design strength shall be based on its throat area and shall be given by

wn

wd

mw

ff =

γ

Where, u

wn

ff =

3fu = Smaller of the ultimate stress of the weld or of the parent metal

mwγ = 1.25 for shop welds= 1.5 for field

welds IS 800-2007 gives the following provisions for the fillet welds:

1. If a fillet welds is applied to the squareedge of apart, the specified size of theweld should generally be at least 1.5mm less than the edge thickness.

2. If fillet weld is applied to the roundedtoe of a rolled section, the specified sizeof the weld should generally not exceed3

4th of the thickness of the section at the

toe.

3. In members subjected to dynamicloading, the fillet weld shall be of fullsize with its leg length equal tothickness of plate.



99

4. End fillet weld, normal to the direction of force shall be of unequal size with throat thickness not less than 0.5 t. The difference in the thickness of weld shall be negotiated at a uniform slope.

1.21 REDUCTION IN DESIGN STRESSES FOR LONG JOINTS

If the length of the welded joint lj is greater than 150 tt , where tt is throat thickness, the design capacity of weld fwd shall be reduced by the factor.

j

lw

t

0.2l=1.2 1.0

150t

lj = length of the joint in the direction of the force transfer and tt = throat thickness of the weld

The reduction is design strength of long joint is to cater for non uniform mobilization of stress in a long joint causing ineffectiveness of certain length of joint. 1.22ECCENTRIC CONNECTION – PLANE OF MOMENT AND THE PLANE OF WELDS IS THE SAME

Figure above shows a typical case. The eccentric load P is equivalent to i) a direct load P at the centre of gravity of

the group of weld and ii) a twisting moment P × e

Let a weld of uniform size be applied throughout and 't' be the effective throat thickness. It’d’ is the depth of weld and 'b' is the width as shown in the figure, the direct shear stress in the weld is

1

Pq =

2b+d t

The stress in the weld due to twisting moment is the maximum in the weld at the extreme distance from the centre of gravity of the group of weld and acts in the direction perpendicular to the radius vector. The maximum stress due to the moment.

max

2

zz

P×e×rq =

I

where rmax is the distance of the extreme weld from the c.g. of the group. Izz = Ixx + Iyy, the polar moment of inertia. The vector sum of the stress is

2 2

1 2 1 2q = q +q +2q q cosθ.

For the safe design in should be less than the resistance per unit area.



100

Topics Page No

1. STRUCTURAL FASTENERS 114

2. TENSION MEMBER 121

3. COMPRESSION MEMBER 123

4. BEAMS 125

5. PLATE GIRDERS & INDUSTRIAL ROOTS 128

6. PLASTIC ANALYSIS 129

GATE QUESTIONS

113

Q.1 Identify the most efficient but joint (with double cover plates) for a plate in tension from the patterns (plan views) shown below, each comprising 6 identical bolts with the same pitch and gauge.

a)

b)

c)

d)

[GATE –2001]

Q.2 ISA 100 x 100 x 10mm (Cross sectional area = 1908mm2) is welded along A and 13 (Refer to figure in the below question) such that the lengths of the weld along A and 13 are 4 and 12 respectively. Which of the following is a possibly acceptable combination of I1 and I2?

a) I1 = 60 mm and I2 = 150 mmb) I1 = 150mm and I2 = 60mmc) I1= 150mm and I2 = 150 mmd) Any of the above, depending on

the size of the weld[GATE – 2002]

Q.3 ISA 100 x100 x 10mm (Cross sectional area = 1908mm2) serves as

tensile member. This angle is welded to a gusset plate along A and B appropriately as shown. Assuming the yield strength of the steel to be 260 N/mm2 the tensile strength of the member can be taken to be approximately

a) 500kN b) 300kNc) 225kN d) 375kN

[GATE – 2002]

Q.4 When designing steel structures, one must ensure that local buckling in webs does not take place. This check may not be very critical when using rolled steel sections because a) Quality control at the time of

manufacture of rolled sections isvery good

b) Web depths available are smallc) Web stiffeners are in-built in

rolled sectionsd) Depth to thickness ratios (of the

web) are appropriately adjusted[GATE – 2002]

Q.5 An ISMB 500 is used as a beam in a multi-storey construction. From the view point of structural design, it can be considered to be 'laterally restrained' when, a) The tension flange is' laterally

restrained'b) The compression flange is

`laterally restrained'c) The web is adequately stiffenedd) The conditions in (a) and (c) are

met[GATE – 2002]

1 STRUCTURAL FASTENERS


Q.6 Rivet value is defined as a) Lesser of the bearing strength of

rivet and the shearing strength of the rivet

b) Lesser of the bearing strength of rivet and the tearing strength of thinner plate

c) Greater of the bearing strength of rivet and the shearing strength of the rivet

d) Lesser of the shearing strength of the rivet and the tearing strength of thinner plate

[GATE –2004]

Q.7 A moment M of magnitude 50 kN-m is transmitted to a column flange through a bracket by using four 20 mm diameter rivets as shown in the figure. The shear force induced in the rivet A is

a) 250 kN b) 176.8 kN c) 125 kN d) 88.4 kN

[GATE – 2004]

Q.8 Which one of the following is NOT correct for steel sections as per IS: 800-1984? a) The maximum bending stress in

tension or in compression in extreme in fibre calculated on the effective section of a beam shall not exceed 0.66fy.

b) The bearing stress in any part of a beam when calculated on the area shall not exceed 0.75fy.

c) The direct stress in compression on the gross sectional area of axial loaded compression member shall not exceed 0.6fy.

d) None of the above. [GATE -2005]

Q.9 A fillet-welded joint of 6 mm size is shown in the figure. The welded surfaces meet at 60-90 degree and permissible stress in the fillet weld is 108 MPa. The safe load that can be transmitted by the joint is

a)162.7 kN b)151.6 kN c)113.4 kN d)109.5 kN

[GATE – 2005]

Q.10 In the design of welded tension members, consider the following statements: I. The entire cross-sectional area

of the connected leg is assumed to contribute to the effective area in case of angles.

II. Two angles back-to-back and tack-welded as per code requirements may be assumed to behave as a tee section

III. A check on slenderness ratio may be necessary in some cases. The TRUE statements are

a) only I and II b)onlyII and III c) only I and III d) I, II and III

[GATE – 2006] Q.11 A bracket connection is made with

four bolts of 10mm diameter and supports a load of 10 kN a an eccentricity of 100mm. the maximum force to be resisted by any bolt will be

a)5 kN b)6.5 kN c)6.8 kN d)7.16 kN

[GATE –2007]


Structural Fasteners

115

Q.1 (a) For main plate critical most section is outermost section for that a outer most section net area should be more to resist more load in all four case we will get more netA in case (a)

Q.2 (a) The C.G of the angle is close to B compared to A. Therefore, the length of the weld, 2l shall be more than 1l

so that the algebraic sum of moments of the two welds 1l & 2l

about C.G of the angle will be zero and there is no eccentricity.

Q.3 No correct answer.

Q.4 (b)

Q.5 (d)

Q.6 (a)

Q.7 (b)

Force due to moment effect,

m 2

M.rF

r

2 2r 50 50 70.71mm 0.0707m

m 2

50 0.0707F

4 0.0707

176.8kN

Q.8 (b)

Q.9 (c)

Total length of weld, l = 100+50+100 = 250 mm Throat of fillet, t = 0.7 * size

= 0.7 * 6 = 4.2 mm

vfStrength, P = l t τ

= 250 * 4.2 * 108

= 113.4 kN

Q.10 (d)

Q.11 (d) Force due to axial load effect,

a

w 10F = = 2.5 kN

N 4

1 2 3 4 5 6 7 8 9 10 11 12 13 14

(a) (a) * (b) (d) (a) (b) (b) (c) (d) (d) (c) (b) (b)

15 16 17 18 19 20 21 22 23

(b) (a) (b) (a) (d) 156 (b) (b) (c)

ANSWER KEY:

EXPLANATIONS



118

Force due to moment effect,

m 2

M.rF = ;

r

2 2r = 30 +40 = 50mm = 0.05m M = W e = 10 0.1 = 1 kNm

m 2

1 0.05F =

4 0.05

mF = 5 kN

Resultant force, 2 2

r a m a mP F +F +2F F cos θ

40Cos θ 0.8

50

2 2

r P = 2.5 5 (2 2.5 5 0.8)

Pr 7.16 kN Q.12 (c) Q.13 (b) Gross diameter of rivet, D = 16 + 1.5 = 17.5 mm Strength of rivet in s hearing,

2

s

πP = 2× ×d ×τvf

4 (Double shear

for but joints)

2π

=2 17.5 904

sP 43.29 kN

Strength of rivet in bearing,

b pfP = d.t.σ = 17.5 12 270

bP 56.7 kN

rivet value vR is smaller of s bP & P

vR 43.29 kN

Q14 (b)

Max. size of weld as per code

maxS = T - 1.5 = 10 - 1.5 = 8.5 mm

Size, S = 8.5 mm

Throat thickness ‘t’ = 0.7 * 8.5 = 5.95 mm Strength of the plate ‘P’ is

stP = (b.T) σ

= (100 10) 150 3P = 150 ×10 N:

Let l = total length of the weld required. Strength of the weld,

s vfP = lt × τ = l × 5.95 × 110

Equating sP = P 3150 × 10 = l 5.95 110

l = 229.2 mm Q.15 (b) Throat, t = k.S

Angle between fusion faces

k

60° - 90° 0.70 91° - 100° 0.65 101° - 106° 0.6 107° - 113° 0.55 114° - 120° 0.50

Q.16 (a) Q.17 (b) Design strength of fillet weld,

d w.t

mw

fuP =L

3.γ

3

W270 10 = 2L (0.7410

3 2

05

1 )1.

wL 101.84mm

Rounded off to 105 mm Q.18 (a)

As per clause 9.2.1 (IS: 800-2007) for combined shear and bending: Factored value of applied shear force is greater than or equal to shear strength for high shear. i.e., V > 0.6 Vs

Q.19 (d)



119

Q.1 A singly reinforced balanced section designed by limit state method will have 1. Lesser depth compared to designed

by working stress method2. Higher value of steel compared to

designed by working stress method3. Lower cost compared to designed

by working stress methodWhich of these statements is/are correct? a) 1 and 2 b) 2 and 3c) 3 and 1 d) Only 3

Q.2 Which one represents correct deflected profile corresponding to section 1-1for a two way simply supported RCC slab subjected to UDL as shown in the figure below?

a) b)

c) d)

Q.3 Which expression represents effective flange width of isolated reinforced concrete T- beam? Symbols have their usual meaning.

a) 0w f

l+b +6d

6Or `b’ whichever is lesser

b) 0w f

l+b +3D

12Or `b’ whichever is lesser

c) 0w

0

0.5l+b

l+4

b

Or `b’ whichever is lesser

d) 0w

0

l+b

l+4

b

Or `b’ whichever is

slesser

Q.4 A deep continuous beam has effective depth of 500 mm and effective cover of 50 mm. Its maximum effective length is a) 1375 mmb) 1100 mmc) 1275 mmd) cannot be determined from the

given data

Q.5 If a beam of cross-section 350 mm × 600 mm is subjected to torsional moment of v 50 KN-m then equivalent shear will be a) 129 kN b) 229 kNc) 329 kN d) Zero

Q.6 For a continuous RC beam, match list-1(condition) with list-2 (Placement of live load) and select he correct answer using the codes given below the lists: List-1 A. For maximum sagging moment

in the span B. For maximum hogging moment

at a support C. For maximum hogging moment

in a span List- 2 1. The spans adjoining the span as

well as alternate span2. The same span as well as

alternate span3. The adjacent span on both side

of this support as well as spansalternate to these

4. Span next to the adjacent spans ofthe support plus alternate spans

A B C a) 2 3 1 b) 1 4 2 c) 2 4 1 d) 1 3 2

ASSIGNMENT QUESTIONS


Q.7 For a continuous slab of 3 m × 3.5 m size the minimum overall depth of slab to satisfy vertical deflection limits is a)5 cm b) 7.5 cm c) 10 cm d) 12 cm

Q.8 Which one of the following

statements about the percentage of tensile steel required to produce a balanced reinforced concrete section is correct? The required percentage of steel a) Reduces as the yield strength of

steel increases b) Remains unchanged irrespective

of the yield strength of steel c) Is the same for a given quality of

steel irrespective of whether working stress method is followed or ultimate load method used

d) Is only function of the modulus of elasticity of steel

Q.9 The effective depth of the singly

reinforced rectangular beam is 30 cm. The section is over reinforced and the neutral axis is 12 cm below the top. If the maximum stress

attained by concrete is 50 kg/cm 2 and modular ratio is 18, then stress developed in the steel would be

a) 1800 kg/cm 2 b) 1600 kg/cm 2

c) 1350 kg/cm 2 d) 1300 kg/cm 2

Q.10 If permissible stress in steel in

tension is 140 N/mm 2 , then the depth of neutral balance section using working stress method is a) 0.35d b) 0.40d c) 0.45 d d) Dependent on grade of concrete

also Q.11 As per the provision of IS: 456-2000,

in the limit state method for design of beams, the limiting value of the

depth of neutral axis in a reinforced concrete beam of effective depth `d’ is given as a) 0.53 d b) 0.48 d c) 0.46 d d) Any of the above depending on

the different grade of steel

Q.12 The spam o depth ratio limit is specified in IS: 456-2000 for the reinforced concrete beam, in order to ensure that the a) Tensile crack width is below a

limit b) Shear failure is avoided c) Stress in the tension

reinforcement is less than the allowable value

d) Deflection of the beam is below a limiting value

Q.13 A floor slab of thickness t is cast monolithically transverse to a rectangular continuous beam of span L, and width, B. If the distance between two consecutive points of

contra flexure is, L 0 , the effective

width of compression flange at a continuous support is

a)B b)3

L

c)B+ 12t d) B+ 6t+ 0

6l

Q.14 The total compressive force at the

time of failure of a concrete beam section of a width `b’ without considering the partial safety factor of the material is

a) 0.36 ckf b uX b) 0.54 ckf b uX

c) 0.66 ckf b uX d) 0.8 ckf b uX 4

Q.15 What is the modular ratio to be used

in the analysis of RC beam using working stress method if the grade of concrete is M 20? a)18.6 b)13.3 c)9.9 d)6.5


Assignment Questions

133

Q. 1 (b)

Q. 2 (a)

Q. 3 (d) See clause 23.1.2 of IS 456 : 2000

Q. 4 (a) L

D<2.5 for continuous beam

So L < 2.5×D So Maximum effective length = 2.5× (500+50)=1375 mm

Q. 5 (b)

ue u

TV V 1.6

b

Here uV 0 3

e

50 10V 1.6 229kN

350

Q. 6 (a)

Q. 7 (c) L

D≤28 simply supported slab

L

D ≤ 32 continuous slab

L is the short span of slab

∴ 3000

32d

d ≥ 9.37 cm = 10 cm

Q. 8 (a) For a balanced section

ck y st0.39f *b*X 0.87f A

ckst

y

0.36f *b*XA

0.87f

As yf increases stA decreases.

Q. 9 (c) The limiting value l/D for simply supported and continuous span deep beams are as follows:

For simple span deep beam, l/D ≤2.0 For Continuous span deep beam, l/D ≤.5 Where, l = Effective span taken as centre to centre distance between the support of 1.15 time the clear span, whichever is smaller. D = Overall depth D = 30 cm xu= 12 cm

c = 50 kg/ cm 2

U

U

XC

t d Xm

t = 18 50

12

× (30-12)

18 50t (30 12)

12

= 1350kg/ cm 2

Q. 10 (b) Depth of the balanced neutral section using working stress method is given by

Umin

SR

280X d

280 3

280d 0.4d

280 3 140

Q. 11 (d) The limiting value of depth of neutral axis depending on the different grades of steel is as follows:

yf UminX / d

250 0.53 415 0.48 500 0.46

Q. 12 (d) Basic value of span to effective depth ratios for spans up to 10 m to satisfy vertical deflection limits are for

EXPLANATIONS


Cantilever beam 7 Simply supported 20 Continuous 26

Q. 13 (d)

For T-beams effective width of compression flange

0f

Lb B 6t

6

For L- beam

0f w f

Lb b 3D

6

Q. 14 (b)

Total compressive force without considering the partial safety factor of material 0.36 ck Uf b X ×1.5

= 0.54 ck Uf b X

Q. 15 (b)

As per IS: 456-2000 modular ratio is given by

M = 280

3 cbc

For M20 concrete

cbc =7N/mm 2

∴ m=280

3 7 = 13.3

Q. 16 (a)

As per the clause 22.2 of IS: 456-2000, for simply supported beam and slab, the effective span of a member that is not built integrally with its supports shall be taken as clear span plus the effective depth of slab or beam or centre of supports, whichever is less. ∴ Effective span = 5 × 1000 + 40

= 5400 mm And effective span

= 5 × 1000 + 300

2 +

300

2= 5300 mm

Q. 17 (d)

As per clause 26.5.1.5 of IS: 456-2000, the maximum spacing of shear reinforcement measured along the axis of member shall not exceed 0.75 d for vertical stirrups and

1d’ for inclined stirrups at 45 , where 1d’ is the effective depth of the section under consideration. In n case shall the spacing exceed 300 mm. ∴ Maximum spacing = 0.75 d = 0.75 × 300 = 225 mm

Q. 18 (b)

Moment of resistance for a balanced section is given by

M ,limU = 0.36 f ck ×,limUX × (1- 0.42 ×

,limUX

d) × bd 2

But for FE 415

,limUX = 0.48 d

M ,limU = 0.36 f ck × 0.48 × (1- 0.42 ×

0.48) × bd2

M ,limU = 0.13796 f ck × bd 2

On comparison, we get

K = 0.13796 f ck

= 0.13796 × 20 = 2.76

Q. 19 (a)

To ensure ductility the maximum strain in tension reinforcement in the section at failure shall be less

than 1.15

y

S

f

E + 0.002

Q. 20 (b)

For under- reinforced beam

i) UX < ,limUX or CX

ii) sc Steel reaches ultimate

stress before concrete reaching the ultimate stress

iii) uM < ,limuM


Assignment Questions

146

Documents

RCC & STEEL STRUCTURES · RCC: Working stress, ... transfer and service loads. Steel Structures: Working stress and Limit state design concepts; Design of tension and compression