bco 3.2 Advanced Steel Design

7/22/2019 bco 3.2 Advanced Steel Design

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CONTENTS

Topic Page

Syllabus

Structural ConnectionsBeamsBeam ColumnsIndustrial BuildingsMulti-Storeyed BuildingsTransmission Line Towers

Light Gauge Steel SectionsPlastic Analysis and Design


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SYLLABUS ADVANCED DESIGN OF STEEL STRUC1"URES - (BCO 3.2)

Structural Connections: Design of high strength function grip bolts - Design of riveted andbolted connections at the junctions of beams and columns in frames - Design of un stiffened andstiffened seat connections - Welded connections - eccentric connections - Beam endConnections-Direct web fillet welded connections - Direct web Butt welded connections - Doubleplate web connection - Double angle web connection - Unstiffened and stiffened seat connection- Moment resistant connection - Behaviourof welded connections - Problems.Beams: Design of beams to resist biaxial bending moments - Design of section to resistunsymmetrical bending - Beam splices - Lattice beams - Elastic lateral torsional buckling.Beam Columns: Differential Equa tion s-Moment Magnification Factor for end moments - Sideway -Nominal strength - Interaction Equations - Biaxial bending.Industrial Buildings: Industrial Building Frames - General - Framing Bracing - Crane girdersand columns -analysis of trussed bents - Design example - Design of rigid joints knee for gableframesStructure of Multistoried buildings- Bracing of multistory frames - Loads -Lateral load of frame.Strong Structures and Transmission Tower: Design of steel bunkers and silos - Janssen'stheory - Ary's Theory - Design parameters-Design criteria - Analysis of Bins - Hopper bottom Design of bins Design and detailing of guyed steel chimneys.Transmission line Towers - Introduction, types of towers - tower configuration, load analysis anddesign of members.light Gauge Steel Sections: Design of cold-formed sections - concepts - effective width stiffened sections -multip le stiffened sections - design of flexure - design of light gauge columns beam column-connections.Plastic Analysis and Design: Plastic design of tension and compression members - Theory ofplastic bending - plastic hinge- redistribution of moments - failure mechanisms - plastic analysisand design of fixed beams. continuous beam and portal frames by mechanism method.

CHAPTER 1: STRUCTURAL CONNECTIONSSECTION -A

Multiple Choice Type Questions:The most common types o f structural steel connections are:1.a. Riveted Connectionsb. Bolted Connectionsc. Welded Connectionsd. All of the aboveIn case of Pinned Connections:2.a. Bolts are usedb Pins are usedc. Welding is usedd. None of the aboveWhich of the following statement is wrong in relation to Bolted connections?3.

4.

a.b.c.d.

Require less man-powerOperation is far more quicker than rivetingIs a hot processNone of the above

In case of Black Bolts, the diameter of hole is kept mm more than the diabolt.a. 1.5b. 2.0c. 2.5d. 3.0Which of the following bolt is classified according to type of shank?5. a. Ordinary structural boltb. Square boltc. Turned boltsd. None of the aboveIn common steel structural work, the type of bolt used is6.a. Square boltb. Ordinary structural boltc. High strength boltd. Hexagonal bolt

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7. The bolts which are manufactured from black round bars of low carbon steel is:a. Turned boltsb. High strength boltsc. Black boltsd. None of the above

8. The bolts which are manufactured from black round bars of mild steel containing lowcarbon content are:a. Turned boltsb. High strength boltsc. Black boltsd. None of the above

9. Bolts which are used for ordinary field work and light loads are:a. Turned boltsb. High strength boltsc. Black boltsd. None of the above

10. The strength of turned and fitted bolts are__ that of black bolts.a. Less than .b. More thanc. Equal tod. None of the above

11. The joint produced with the help of High Strength friction grip bolts is:a. ,Rigidb. Non Rigidc. Semi Rigidd. None of the above

12. The factor of safety considered during the design of bolted shear connection fo r all loads~ ~ ~ n ~ ,a. 1.2b. 1.4c. 2.0d. 2.2

2

13. The factor of safety considered during the design of bolted shear connection, for except wind load is taken as:a. 1.2b. 1.4c. 2.0d. 2.2

14. The factor of safety allows for the stress relaxation in the bolts which may be ofof:a. 1 0 percentb. 20 percentc. 25 percentd. 30 percent

15. Which of the following is a type of shear connections?a. Lap Jointb. Butt Jointc. Flange Plate Connectiond. All of the above

16. In case where, the beam is supported on ____ , the beam reaction are borend connections of the beam.a. Masonry wallsb. Steel columnsc. RCC columns .d. None of the above

17. When the beam is connected to a girder of a stanchion by means of two angles pthe two sides of the web the beam is called:a. Framed Connectionb. Seated Connectionc. Clip Angle Connectiond. None of the above

18. The connections in which clip angles are provided both at top and bottom of thof the beam in addition to web angles are known as:a. Framed Connectionb. Seated Connectionc. Clip Angle Connectiond. None of the above

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19, A connection, which essentially consists of two sets of angles along with gusset plates iscqlled as:a. Bracket connectionb. Split beam connectionc. Modified split beam connectiond. None of the above

20. In order to resist large moments, which type of connections are used?a. Clip Angle connectionsb. Large moment resistant connectionsc. Split beam connectionsd. Modified split beam connections

Key:1. (d) 2. (b) 3. (c) 4. (a) 5. (c) 6. (c)7. (c) 8. (a) 9. (c) 10. 11.b) (a) 12. (c)13. (a) 14. (a) 15. (d) 16. (b) 17. (a) 18. (c)19. (a) 20. (b)

SECTION -8True I False Type Questions:1. A bracket connection is essentially composed of two sets of angles with gusset pla2. Clip angle connections are used for resisting large moments.3. In a framed connection, it is assumed that the connections are flexible a

proportioned for transferring beam shears applied at appropriate eccentricities.4. To provide the flexibility in the framed beam connections, the depth of the co

angles is limited to 0.5 times the depth of the beam connected.5. When a beam is connected to a girder by means of two angles placed on the two

the web of the beam, it is known as seated connections.6. Actual connections are neither completely rigid nor completely flexible.7. High strength bolts gives high static strength due to high frictional resistance.8. The slip factor is defined as the ratio between the forces causing the large displ

between the two interfaces of the plates connected together and the force norminterfaces due to the tension in the bolts.

9. Bolting operation is far more quicker than riveting.10. Unfinished bolts have more strength because of non-uniform diameter.

Key:1. T 2. F 3. T 4. F 5. F6. T 7. T 8. T 9. T 10. F


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SECTION -C Q7. What do you mean by seated beam connections? Ans. When a beam is connected to the flange (or the web) of a steel stanchion, the Short Answer Type Questions: the flange (or the depth of the web) may be insufficient to accommodate the con

Q1. What are the advantages of bolted connections?Ans. Following are the advantages of bolted connections:

1. Bolting is a cold process, and hence there is no risk of fire.2. Bolting operation is far quicker than riveting.3. Less man-power is required in making the connection.

Q2. Classify the different types of bolt according to material and strength.Ans. According to material and strength, bolts are classified as follows:1. Ordinary structural bolt2. High strength steel bolt

Q3. What are the advantages of high strength bolts?Ans. 1. It gives rigid joint as there is no slip between plates at working loads.

2. It gives high strength due to high frictional resistance.3. There are no shearing or bearing stresses in the bolts.4. It has high fatigue strength.

Q4. What are the various types of shear connection?Ans. 1. Lap joint

2. Butt joint with double cover plates3. Moment connection4. Flange plate connection5. Bracket connection

Q5. List the direct & indirect method of nstallation of HTFG bolts.Ans. Direct Method

1. Load indicating bolt2. Load indicating washer.Indirect Method4. Torque control method5. Part turn method

Q6. What are the Pinned Connections?Ans. Pinned connections are used to connect tne members which are required

angles. In that case, framed beam connections are not su itable and seatedconnections are preferred. In its simplest form, a seated connection is the one inhorizontal angle with its horizontal leg at its top is used to receive the beam on it.

Qa. Name few structures in which pins can be used? Ans. 1. Crane booms

2. Arch hinges3. Diagonal bracings4. Expansion jOints5. Rocker supports

Q9. What are the disadvantages of pinned connections?Ans. 1. They should not be used for short span.

2. Pins cannot resist longitudinal tensionQ10. What do you mean by ordinary unfinished bolts?Ans: These are manufactured from black round bars of low carbon steel and the surfa

shank is left unfinished, that is rough as rolled. The head is formed by forgdiameter under the thread is usually 1.5 to 3 mm less than shank. They remainthe holes which are usually made 1.5 mm larger in diameter than the nominal diathe bolt.

Q.11. What do you mean by turned and fitted bolts?Ans: These are specially made from black round bars or mild steel containing low

content, but are turned down to exact diameter. The diameter of the shank is finturning to a diameter which is larger than the ,nominal diameter of the bolt by 1.2bolts M8 to M16 and by 1.3 mm for larger sizes.

Q12. What do you mean by High Strength Friction Grip Bolts?Ans: High strength friction grip bolts are comparatively a recent development. They a

of high strength steel and their surface is kept unfinished, i.e. as rolled and roughthis, they remain loose fit in the holes, similar to the unturned block bolts.

to rotaterelative to each other. Pins for structure purposes are cylinderical in shape, and are madeof structural carbon steel, forged and machined to accurate dimensions.

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SECTION-DLong AQswer Type Questions:Qi . What are the advantages of bolted connections?Ans. Advantages ofbolted connections:

1. The bolting operation is very silent, in contrast to the hammering noise in riveting.2. Bolting is a cold process, and hence there is no risk of fire.3. Bolting operation is far more quicker than riveting.4. There is no risk involved in the bolting, in contrast to the risk of flying rivets in

riveting work.5. Less man-power is required in making the connections.Disadvantages of bolted connections:1. The bolted connections, if subjected to vibratory loads, result in reduction in

strength if they get loosened.2. Bolted connections for a given diameter of bolt have lesser strength in axial

tension since the net area at the root of the threads is less.3. Unfinished bolts have lesser strength because of non-uniform diameter.4. In the case of black bolts, the diameter of hole is kept 1.5 mm more than the

diameter of the bolt, and this extra clearance does not get filled up, in contrast tothe riveted joints.

Q2. What do you mean by bolt? What are the different types of bolts used in steelstructural work?

Ans. A bolt is a metal in which a head is formed at one end and the shank threaded at theother end in order to receive a nut. Structural bolts are classifies as under:a) ~ c c o r d i n g to type of shank

i) Unfinished or black boltsii) Turned bolts

A BOLT ASSEMBl,.'(8

b) According to material and strength i) Ordinary structural bolts ii) High strength steel bolts

c) A cc or di ng t o s ha pe of head and nut i) Square bolts ii) Hexagonal bolts

d) According to pitch and fi t of thread i) Standard pitch bolts ii) Coarse pitch bolts iii) Fine pitch bolts

In common steel structural work, however, the following three bolt types are recognized :1. Ordinary unfinished or black bolts2, Turned and fitted bolts3. High strength boltsQ3. What do you mean by high strength friction grip bolts? Discuss its prope

detail.Ans. High strength friction grip bolts are comparatively a recent development. They are

of high strength steel and their surface is kept unfinished, Le. as rolled and roughthis, they remain loose fit in the holes, similar to the unturned black bolts, Howevinitial tension is developed in such bolts in the initial stage of tightening, and thisclamps the joining plates between the bolt head and the nut. The tightening of tha very high tension reaching their proof load, is done through calibrated torque wrThis high pre-compression causes clamping action due to which the load is trafrom one plate to the other by friction with negligible slip. The bearing of the bohole surface does not come to play at all. The joint so produced is a rigid oneremain fully tight even under dynamic load, free from fatigue.Fig. shows the load transmission by a friction grip bolt. In an ordinary bolted jforce from one side is transferred to the other side through the interlocking and bethe bolts. In a friction grip joint, however, the force is transmitted by virtue obetween the interfaces, To develop this friction a normal load is applied to the using high strength bolts tightened to proof load. By usual law of friction

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What are simple beam end connections? What are their types?QS.Ans. Simple beam end connections :

Steel beams are supported at their ends byi} masonry walls or piers,ii) steel columns, or(iii) heavier beams and girders running in transverse directions.

In the case where the beam is supported on steel supports, the beam are borne by the end connections of the beam with the supporting methe case of simple framing , the original angle between the members ma

Disadvantages of high strength bolts1. The material cost of these bolts is much higher It is about 50% greater

of ordinary bots and about 3 times that of rivets.2. Special attention is required for workmanship in installing and tightenin

bolts, specially in regard to giving them right amount of tension.

P=pT (1) ....... . upto 80% of the amount it would theoretically change if frictionlesWhere connections could be used. For beams, such a connection provides oT =clamping force induced by the bolt transfer at the ends. Simple beam end connections.jJ = coefficient of friction between the interfaces and Framed beam connections (Fig.)P=load carrying capacity of the Joint in shearHence if the actual applied load is equal to P or less, the Joint will withstand it, and transfer itwithout any slip. When the actual load exceeds this value, there cocurs a major slip, and as loadis further increased, gradual slipping brings the bolt in contact with the edges of the plate. Thecoefficient of friction is termed as slip factor. The slip factor is defined as the ratio between theforce causing the large displacement between the two interfaces of the plates connected togetherand the force normal to the interfaces due to the tension in the bolts. A slip factor of 0.45 isstipulated by IS : 4000-1967 for surfaces which are free of paint, dirt, loose rust and mill scale.The high tensile friction grip bolts are commonly abbreviated as HTFG bolts.Q4. What are the advantages & disadvantages of High Strength Bolts? Ans. Advantages of high strength bolts ---- ...... Fig. Framed Beam Connections1. It gives rigid Joint as there is no slip between plates at working loads.

2. It gives high static strength due to high friction resistance3. Smaller load is transmitted at net section of plates.4. There are no shearing or bearing stresses in the bolts.5. It has high fatigue strength.6. As the bolts are in tension upto proof load they do not permit loosing of the nut

and the washer. ... ....Fig. Beam Framing at the same level

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Q6. What are seated beam connection? Ans. When a beam is connected to the flange (or the web) of a steel stanchion, the width ofthe flange (or the depth of the web) may be insufficient to accommodate the connectingangles. In that case, framed beam connections are not suitable and seated beamconnections are preferred. In its simplest form, a seated connection is the one in which ahorizontal angle with its horizontal leg at its top is used to receive the beam on it, asshown in Fig. (A) in such a case it is called unstiffened seat connection. In addition to theseat angle, a web cleat is provided when the beam is connected to a beam (Fig. A) whilea flange cleat is used when the beam is connected to a stanchion. The angle cleats (Le.web cleat or flange cleat) are essential parts of seated connections because they keepthe beam stable in a vertical position and prevent it from lateral buckling.

HAM

SEAT ANGLE(8 ) BEAM CONNECTED TO STANe ...

S T ~ . O H '-_....

Fig. (A) Unstiffened Seated Connection

FLANGE CLEAT

LBUMSEAT ANGLESTIFFENERANGLES

'.....- - f ACK'"G

Fig. (8) Stiffened Seated Connection12

When the reaction to be transferred by the beam is so large that the seat angle support it, then the horizontal leg of the set angle is stiffened by means of one ostiffener angles, as shown in fig. (8) The stiffener angles should be tightly fitted undseating angle and suitable packing should be provided, as shown in fig. (8).Seated connections require more space in the vertical direction, and due to this. thnot commonly used for connecting the beam to a beam. Seated connections aresuitable for connecting the beam to either the flange or to the web of a steel stanSimilarly. framed connection is not suitable for connecting a beam to the web of a cbecause of the space limitation on either side of the beam.

Q7. Discuss Small Moment Resistant Connections in detail with diagram.Ans. Such a connection is also known as clip angle connection because clip angl

provided both at the top and bottom of the flange to the beam in addition to thangles. as shown in fig.

WEB AN ES

(0 ) (b'~ . ~ ~ .

(c )

Thus, four angles (Le.,Fig. Clip Angle Connectiontwo clip angles and two web angles) are used in

connection. The web angles, similar to those in framed connections, resist onlythey do not resist any moment. Similarly. the two clip angles resist only the momethey do not resist any shear. The vertical legs clip angles are connected to the flthe stanchion by two rivets provided in one gauge line, so that distribution of tinduced by moment, is uniform. If two gauge lines are used instead of odistribution of tension is not uniform (Fig. a). Similar argument applies if the flange is wide enough to accommodate four rivets in one gauge line on the lip, thmost rivets may take disproportionate share of load (Fig. b). For this reasonengineers contend that the number of rivets (equal to two) limits the moment r


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in, the rivets. Since eq. (iv) was obtained on the basis of detailed analysis, the design ofclip angle is based on this 8.M.

capacity of the connection; that's why such connection is used for resisting only smallmoments.

Ca) lb lF ~ g . shows modes of bending of clip angles. If initial tension in rivets is neglected thechp angle will under go simple cantilever flexure (Fig. a). If P is the tensile force in rivet Athe 8.M. to b e resisted by the angle is '

p.Q

1..1l p ~ .?

l!=:::::=:::t!:o _c pFig. Bending of Clip Angle M = P (a-.5t) " .(i)where t is the thickness of the angle.

If, however, the clamping action of the rivets is sufficient to prevent rotation at A and 8the angle will bend in double flexure (Fig. b), with the point of contraflexure midwa;between A and C. In that case the 8.M. is given byM = P x O.5(a O.5t) = O.5P(a - 0.5!) ..(ii) .The 8.M. g i v ~ n by (ii) is half of that obtained by (i). However, for safety, paint ofcontraflexure IS assumed at a distance 0.6 (a-0.5t), in which case, the 8,M. is given byM=0.6P (a-0.5t) ("'), .. . III Also, from Eq the maximum 8.M. in the angle is given by M = MA =O.6P.a .. (iv)Riveted 8eam conn.ectionThe different between Eqs. (iii), and (iv) is very small, both being based on initial tension

Ma = 0.6PaIf I is the length of the clip angle between the two rivets (equal to gauge g of thstanchion) its moment of resistance is equal to 1:..1t2 .G'bt'6

1 ,- It2.G'bl.=Ma=O.6Pa6t=./6Ma ~ 3 . 6 p a .... (v)

VI (J'bl I G'blIn the above expression O"bt may be taken as 185 N/mm 2

QB. What do you mean by split beam connection? How they are analysed?Ans. A split beam connection consists of

( i) two spl it beams consisting of either T-section or I-section cut into tprovided at the top and bottom flanges of the beam, and

(ii) pair of web angles, connecting the web of the beam to the flanstanchion. Such a connection is shown in fig.

A split beam connection is quite similar to the small moment connection, excthe place of clip angles of either T-seCtions, or else beam section splitted intoHere also, it is assumed that the split beams (flange clips), resist mom ent onlweb angles resists shear only. Let M be the moment at the connection. Dmoment, a tensile force P will be induced at each of the two rivests of the uppa compressive force P at each of the two rivets of the lower tee. The distorted lower tee. The distorted view of the upper T is shown in Fig.(a). thus the externM, will be resisted by a couple provided tensile and .compressive forces of2P (Fig. a), acting at a lever arm of (h + tws) where tws is the thickness of wbeam:2 P (h + lws) =M

. . (i)Also, the T-section will be subjected to a root moment Ms given by (Fig. b)

1Ms =P x2(g t ) O.5P(g-t J ... (iiws wLet t ft be the thickness of the flange of split beam. The moment Ms induce

1 2section is resisted by its length I between the centres of rivets. Hence 61tft .G.'. (ii


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P

.'- 2P-" - --0P

p- 'l

( 0 ) (t dFig. Split Beam Connection

The thickness tws of the split beam should be such th " . provides bearing strength to the rivets great th that It transmits the pull2P safely, and (ii) it the flange of the beam. er an e single shear strength of rivets joining it to Q9.

Fig. shows the lap joint. There is only one interface to transmit shear. Since there 1bolt, so load per bolt = -x2400 pN =1200 pN

Design a doubly bolted lap jOint for plates 16 mm thick t .. 0 carry Its full load. Takepermissible axial tension in plate 0 6 I" wh I" 2, . J x ere J x =250 N/mmSolution: .Load carried by the plate per pitch length==aat xpxtxN= (0.6 x 250)p x 16 == 2400pN

..~ C = I : l ==:::::ii=:JtSOLT LINE.. 16mm L I = ; = = 4 ~ 1 ~ ~ t c = = ~ ~I .IGmm

HTFGBOLTSM2411OtC)J601Ml..!.

2Hence from Eq., taking N =1,1200p='u xNxTFT = proof loadT = (1200p)F == 1 2 ~ ~ p ( 1 . 4 1

.uN 0.45xl =3733.33 p Providing bolts at 60 mm pitch. Proof load, T == 3733 x 60 == 224000 N = 224kN From table we get T = 232.5 kN for 24 mm dia. 10K bolt. Hence provide HTFG b(10K) with a pitch of 60 mm and edge distance of 40 mm.

Q10. A top column of section ISHB 400 @ 77.4 kg/m transmits a moment of 5 the bottom column of section ISHB 450 @ 87;2 kg/m. Design the mconnection.

Solution :. It is assumed that the direct load is transmitted from the upper column to thcolumn by web connection (shear connection). The moment of 5 kN-m will be transmittedflange connection, providing bolts in the flanges.

~ ! 5 I 1 N - mXING

.. I 'T'SHe 40 0 PACI l.... Ii- i-.400I - 1-

15MB 450I4::S0mm-.A , ....w I

Figure17
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Shear on the bolts at each flange = M == 5 X 103 == 12.5kN h 400

The bolts are in single shear, i.e. there will be one interface only at each flange,f. lxT=PF

Or 0.45 x Proof load::: 12.51.4Proof load ::: 12.5 x1.4 38.89kN

0.45Select HTFG bolts M12 (8G) 4 Nos. on each flange as shown in fig.Q11. A beam ISLB 400 @ 56.9 kg/m carrying a total U.D.L. of 280 kN over a.span of 7m,

is to be connected to the flange of a steel stanchion ISHB 250 @51.0 kg/m. Designthe connection taking 'ruf=100 MPa and fpy =300MPa fo r the rivets.Solution:For ISLB 400 @ 56.9 kg/m, we have: tw = 8 mm. h2 ::: 31.90 mmFor ISHB 250 @ 51.0 kg/m, t 9.7mm.fMaximum end reaction V ==!x280 140kN

2Using 16mm dia. rivets,Strength in single shear::: '::'(17.5)2 X 100 X 10-3 == 24.05kN

4Strength in double shear = 48.10 kN Strength in bearing on 8 mm thick web of beam ::: 17.5x8x300xlO-3 =42kN Strength in bearing on 9.7 mm thick flange :::17.5x9.7x300xlO-3 ::::50.93kN

The rivets connecting web of beam with the connection angles are in double shear.Rivet value::: 42 kNNo. of rivets ::: 140 = 3.33' say 4 rivets.

42 The rivets connecting the angles with the flange of stanchion are in single shear.Rivet value::: 24.05 NSince there are two arrgles transferring th'e end reaction to the stanchion, reactiontransfer by each angle = !x 140kN

2 No. of rivets ..!. x 140 = 2.91' say 3 rivets.

2 24.05

12mmt - 16mm RIVE'TS

00

ISL8400lSH82.50'

(b) SECnONAL PLANFigureHence provide 3 rivets on each angle. Using a pitch of 60 mm and an edge 30mm, minimum depth of angle required = 3 x60 +30+30=240mm, w.h'Ch 'I0.7 times the depth (=400) of the beam. Since there is only one row of rivets locations. use two ISA 90 x 90 x 8 mm angles. each 240 mm long. These I ted as near to the compression as possible to restrain it against latera; ~ : i d e a clearance of 12 mm between the flange of the stanchion and end ofFor ISA 90 x90 gauge distance is 50 mmCheck for shear in connection anglesFrom eqn, Max. shear

3- V 140xl0 54.7Nlmm2- 1.5-=1.5x 8 2402th 2x xPermissible shear stress=O.4fy 0.4x250=100Nlmm2Hence safe. The details of the connection are shown in fig. (a) - (b)
http:///reader/full/1.5-=1.5xhttp:///reader/full/1.5-=1.5xhttp:///reader/full/1.5-=1.5x


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Q12. Design a framed connecti on between a main beam (ISWB 600 @ 133.7) and twosecondary beams ISMB 400 @ 61.6 to kg/m, connected on either side of the mainbeam, keeping the top flanges of all the three beams at the same level. Eachsecond beam transmits an end reaction of 200 kN. The rivets on the web of ISMB400 are power shop rivets while those on the web of ISWB 600 are hand dr iven file drivets.

Solution:For ISWB 600 @ 133.7 kg/m, tw=11.2 mm; h2=42.9 mmFor ISMB 400 @ 61.6 kg/m, tw = 8.9 mm; h2=32.80 mm

----"'\ I - - ~ . I!1"1II

-_oJ.

I

r II'----'

Ib )

Figurea) Connection between angles and secondary beam

Let us use 20 mm dia. power driven shop rivets,2For which Tv! WON /mm2 and CJpI =300N /mm The rivets are in double shear.

Strength of rivets in double shear = 2x (21.5)2 X 100 X 10-3 =72.61kNStrength of rivets in bearing on 8.9 mm web == 21.5 x 8.9x300 x 10-3 57.405Rivet value = 57.405 kNNo. of rivets = 200 = 3.48. Provide 4 rivets.

57.405b) Connection between angles and the main beam

Let us provide 20 mm dia. hand driven field rivets,For which Tv! 80N / mm2 and Cpf =250N/mm2 The rivets are in double shear.Strength of rivets in double shear = 2 x!E.(21.5i x80 x10-3 =58.09kN 4 Strength of rivets in bearing on 11.2 mm web = 21.5 x 11.2 x 250 x 10-3 60.2 kN

20

(0 )

Rivet value = 58.09 kN. Total reaction to be transferred by the rivets = 2 x 200 = 400 kN No. of rivets = 400 =6.958.09Rivets on each side = .!.. x 6.9 =3.45, say 4 rivets.2Since only one row of rivets is required, provide connecting angles of size 90x90xUsing a pitch of 60 mm and an edge distance of 35 mm, length of each

(60 x 3) + (2 x 35) = 250 mm, which is less than 0.7 times the height (=400 secondary beams.For ISWB 600, h2=42.60 mm. For ISMB 400, h2=32.80 mm, Space available for connecting angles = 400 .42.90 32.80 = 324.3 mm, which is much more than the height of angles. The details of the joint are shown in fig. Check for shear in angles Max. Shear

V 1.5x200xl031.5-=-- - - 2th 2x8x = 75 N/mm2


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CHAPTER 2: BEAMSSECTION -A

Multiple Choice Type Questions:1. In rolled steel beams, shear force is mostly resisted by

a. Web onlyb. Flanges onlyc. Web and flanges togetherd. None of the above

2. For steel members exposed to weather and not accessible for repainting, the thickness ofsteel should not be less thana. 4.5mmb. 6mmc. 8mmd. 10mm

3. A beam is defined as a structural member subjected toa. Axial loadingb. Transverse loadingc. Axial and transverse loadingd. None of the above

4. A major beam in a building structure, is known asa. A girderb. A floor beamc. Amain beamd. All the above

5. The Indian standard code which deals with steel structures, isa. IS : 875b. IS : 800c. IS : 456d. IS : 1893

6. Stiffeners are used in plate girdersa. To reduce the compressive s t r ~ s sb. To reduce the shear stressc. To take the bearing stressd. To avoid buckling of web plates

The maximum shear stress of small beam shall not exceed7. a. 0.4 fyb. 0.66fyc. 0.5fyd. 0.45fyThe bearing stress is calculated on the8. a. Cross-sectional area of contactb. Mean of the cross-sectional area and the net projected area of contacc. Net projected area of contactd. None of the aboveSlendemess ratio of a compression member is the ratio of effective length of9.the Critical load.b. Radius of gyrationc. Area of cross-sectiond. Weight of sectionIn the plate girder, the vertical stiffeners are provided when ratio of clear dep10. thickness of web exceeds a. 50b. 85c. 75d. 65The maximum deflection, for a simply supported beam should not exceed11. a. 1/325 of spanb. 1/525 of spanc. 1/225 of span None of the above d.Section modulus is expressed in12.

mm. 2b. mm3mm.4d. mm


14/83

13. The permissible tensile bending stress in structural beam isa. 150 N/mm2b. 0.66 fy N/mm2c. 189 N/mm2d. 100 N/mm2

14. The permissible compressive bending stress of steel beams which are laterally supportedisa. 150 N/mm2b. 0.66 fy N/mm2c. 189 N/mm2d. 100 N/mm2

15. The moment carrying capacity of steel section is governed by the following stressa. Axial tensile stressb. Tensile bending stressc. Compressive bending stressd. Compressive or tensile bending stress

16. The rolled steel I-sections are most commonly used as beams because they providea. Greater lateral stabilityb. Larger moment of inertia with less sectional areac. Larger moment of resistance when compared with other sectionsd. All the above

17. Pick up the correct statement from the followinga. Vertical stiffeners may be placed in pairsb. Single vertical stiffeners may be placed oppositely on opposite sides of the webc. Horizontal stiffeners may be placed in pairs one on each side of webd. All of above

18. Web crippling in beams generally occurs at the point wherea. Deflection is maximumb. Bending moment is maximumc. Shear force is maximumd. Concentrated load is acting

19. Intermediate vertical stiffeners are provided in plate girder in order toa. Eliminate flange bucklingb. Eliminate web bucklingc. Transfer concentrated loads

. . 'ng of vertical stiffeners in plate girder is given byThe minimum spaci20.0.33da.0.5d.0.7d0.6d

c.d.

Key:5. (b)4. (d)(b)(c) 3.(a) 2. 11. (a)1. 10. (b)(b)(c) 9.7. (d) 8. 17. (e)15. (d) 16. (d)14. (b)13. (b)(b) 20. (a)19.

6.12.18.


15/83

SECTION -8 True I False Type Questions: 1. A beam is a structural member, the primary function of which is to support loads parallel

to its axis.2. The loads produce bending moment and shear force in the beam.3. Closely spaced beams supporting the floors and roofs of the buildings are known as

Purlins.4. The term beam - column is used for that structural element that supports both transverse

and axial loads.5. If one end of the beam is fixed and other end is free, then it is termed as overhanging

beam.6. The optimum section for flexural resistance is the one in which the material is located as

far as possible from the neutral axis. in the form o f flanges.7. Shear failure is due to crushing of compression flange or fracture of the tension flange of

the beam.8. Large beam deflections can also represent failure when the intended use of the beam

places limits on deflection.9. Section modulus represents the strength of the section.10. When a beam loaded transversely. it is subjected to both bending moment and shearing

force.

Key:1. F 2. T 3. F 4. T 5. F6. T 7. F 8. T 9. T 10. T

26

SECTION-CShort Answer Type Questions:Q1. What do you mean by beam? .Ans. Flexural members or bending members are commonly called beams. A b e ~ m ISstructural member, the primary function of which is to support loads normal to ItS aX

The loads produce bending moment and shear force in the beam. .What are the different types of sections used in beam? Show through sketches.Q2. Some of the commonly used beam sections: Ans.

L[I

Ct J eAR JOIST c"PLATt GIROIR ,a,DOUILE wa. . lOX.....

l i J CASTELLATED 8EAMQ3. Wha t are t he different mode of failure of beam?Ans. Modes of Failure of beam :i) Bending failure

ii) Shear failureiii) Deflection

27


16/83

Q4. On what factots, design process of beam is based?Ans. In the beam design process there are three factors of importance for determining the size

of he necessary structural steel beam for a given set of conditions. In order of prioritythey are:1. Design based on stress due to bending.2. Design based on deflection.3. Design based on shear.

QS. . What is the Bending Moment Equation?M cr Ens. -=-=I y RWhere M =bending moment at the section I = moment of inertia of the section. R = radius of curvature of the beam. E = modulus of elasti.city of the beam. a= bending stress or flexural stress, at any layer distant y from the neutral axis.

Q6. What is Section Modulus? Ans. At the extreme top fibre, the compressive stress is given by M M

C1'bc, cal ==T Ymax =2.Where Z = section modulus. The section modulus represents the strength of the section. Greater the value of Z, stronger will be t!1e section. The strength of the section does not therefore, depend on the section area but depends on the disposition of this area with respect to the centroidal axis.

Q7. What is the permissible Bending Stress fo r Laterally Supported and LaterallyUnsupported Beams?

Ans. For laterally supported beams, the permissible bending stress in tension ( abt ) or incompression (C1'bt) are given byabt or abc = 0.66/yFor laterally unsupported beams, a bt is taken equal to 0.66 /y but abc is given byabc =0.66 feb'!' lin [(feb)" + (fy)" ]

28

QB. Give Shear Stress Distribution for some typical sections.Ans. g- - t -8 tO.'a, c

,., ",Hatched parts show shear stress distribution.

Q9. What is bearing stress?Ans. Beams may either be supported directly on other structural members (such as st

stanchions etc.) or else they may rest on concrete or masonry supports such as wallspillars. In the latter case, the support is of a weaker material than steel, and it becomnecessary to spread the load (support reaction) over a larger area, so that the bearstress does not exceed a certain permissible value. The bearing stress in any part obeam when calculated on the net area of contact shall not exceed the value ofdetermined form the following formula:

a p : 0 .75 /yWhere a p = maximum permissible bearing stress, and

/y =yield stress of steelQ10. What is the Maximum deflection caused at the mid-span of the beam carry

U.D.L.? What is the Limiting Deflection?Ans. Deflection at the mid-span of the beam carrying U.D.L. is given by:

Ymax=5w14/384EIWhere, w =u.d.1. in N/m

I =span of the beam, in m E = young's modulus of elasticity, N/mm2 I = Moment of Inertia in mm4.

Limiting Deflection: Generally, the maximum deflection should not exceed 1/325 ospan.29


17/83

Q11. What is web crippling? How will you calculate web crippling stress?Ans. Web crippling is the localized failure of a beam web due to introduction of an excessive

load over a small length of the beam. It occurs at point of application of concentrated loadand at point of support of abeam. A load applied over a short length of beam can causefailure due to crushing due to high compressive stress in the web of the beam below theload or above the reaction. This phenomenon is also known as web crippling or webcrushing.

Q12. What is web buckling?Ans. The web of flanged beam is very thin and when placed in direct compression, then at the

points of the concentrated load and at supports, unstiffened webs of universal beams andcompound beams are likely to fail by buckling.Q13. What are Encased Beams?Ans. In the case of steel framed structures, where steel stanchions and beams frame into each

other, it is considered desirable to encase the stanchion as well as the beam intoconcreter to make these fire resistant and also to improve the appearance

If\ ...... '1. . ' #' , '11. . . . ., ... T,..0-..... :- 9. ,. ' . . . . ' . I' . .:.t...." . .. '." .:II &...,: .

..

. . . . , . ~ - 5 ..... . .." .: . . . . ..: .""T"\ " .... ,. . .. "#,a :-'-: . . . . . . '9 ...... 5" to '* .. . .. r

- w ....:. . ':.... .. ,STUtAUPS@"Ommc/

. . . .

Q15. Give function of Girders, Joists, Purlins, Rafters, Lintels, Girts&Spandrel beams.Ans. Beams structure may also be referred to by typical names that suggest their function in

the structure, as given below:i) Girders: Usually indicate a major beal'l) frequently at wide spacing that supports

small beams.ii) Joists: Closely spaced beams supporting the floors and roofs of buildings.iii) Purlins : Roof beams usually supported by trusses.iv) Rafters: Roof beams usually supported by purlins.v) Lintels: Beams over window or door openings that support the wall above.

30

vi) Girts : Horizontal wall beams used to support wall coverings on the sidindustrial building.

vii) Spandrel beam : Beam around the outside perimeter of a floor that supexterior walls and outside edge of the floor.

31


18/83

SECTION DLong AnswerType Questions:Q1. What do you mean by beam? Also give the history of beam.Ans. Flexural members or bending members are commonly called beams. A beam is a

structural member, the primary function of which is to support loads normal to its axis.The loads produce bending moment and shear force in the beam. The idea of beam .action is of great age but despite the long history of use, the systematic design of beamshad to await the development of theory of bending. Intellectual giants such as Leonardoda Vinci and Galilco concerned themselves with the strength of beams but it was not untilnearly 200 years after Galilco's death that Navier derived the correct flexural stressformula. Torsional stresses and lateral buckling were investigated by late nineteenth andtwentieth century workers.Beams structure may also be referred to by typical names that suggest their function ,nthe structure, as given belo w:i) Girder: Usually indicate a major beam frequently at wide spacing that supports

small beams.ii) Joist s: Closely spaced beams supporting the floors and roofs of buildings.iii) Purlins : Roof beams usually supported by trusses.iv) Rafters: Roof beams usually supported by purlins.v) Lintels: Beams over window or door openings that support the wall above.vi) Girts : Horizontal wall beams used to support wall coverings on the side of an

industrial building.vii) Spandrel beam : Beam a r o u ~ d the outside perimeter of a floor that support the

exterior walls and outside edge of the floor.The term beam-column is used for that structural element that supports both transverseand axial loads.

Q2. What are the different type of sections used in beam? Discuss.Ans. Fig. shows some of the commonly used beam sections. The efficient utilization of

material in a beam is determined by the geometrical layout of web and flanges. Theoptimum section for flexural resistance is the one in which the material is located as faras possible from the neutral axis, in the form of flanges. In practice, there will be need forsome web material to keep the flanges apart to resist shear. As a measure of beamefficiency, it is possible to relate the allocation of a given amount o f material to flange andweb to satisfy three different and generally mutually contradictory criteria of elasticbending strength, plastic bending strength and beam stiffness.

32

The angle section (fig. a) is not an efficient beam shape, though it may be good folightly loaded spans where the flat leg may be used to support some other elemenstructure, such as floor or roof deck. The channel section (fig. b) is also usedloads, such as purlins or girts. The I-section (Fig. c) known as universal beam, commonly used for wall supported structure. Fig .. (d) shows a composite section, I-section with thin web and with flat plates attached to flanges.

-l f BAR JOIST tt PLATE GIRDER lid DOUBLE WE.8C)XGlRDER

C< ) c85jL l i) CASTELLATED BEAMThis gives higher percentage of material concentrated in the flange, resultingelastic section modulus for the same mass per unit length. Fig. (e) shows

. beam section commonly used as gantry girders for cranes. The bar jois t shownis a light trussed beam, very widely used for floor and roof framing in lighbuildings. The flanges or chords of such a section may be made of angles,special hot-rolled or cold formed sections, while the webs are most often bent, rwelded to chords. Such a section can be commonly seen on railway platformshows double-web box girders particularly useful for heavy, flexural members also to torsion or direct stress. Fig. (i) shows a castellated beam giving andepth of the rolled beam by castellating. To obtain such asection, a zig-zagalong the beam web by an automatic flame cutting machine. The two hproduced are rearranged so that the teeth match up and the teeth are thtogether.

33


19/83

Q3. What are the different mode of failure of beam? On what factors, design process ofbeam is based?

Ans. Failure of a beam can occur in several ways, such as(i) bending failure (ii) shear failure and (iii) deflection failure.i) Bending failure: Bending failure may be due to crushing of compression flange

or fracture of the tension flange of the beam. Instead of failure due to crushing,the compression flange may fail by a column like action with sideway or lateralbuckling. Collapse would probably follow the lateral bucking. '

ii) Shear failur e: Bending failure would most likely be observed as buckling of webof the beam near locations of high shear forces. Near reactions of concentratedloads, the beam can fail locally due to curshing or buckling of web.iii) Deflection failure : Large beam deflections can also represent failure when theintended use of the beam places limits on deflection.

In the beam design process there are three factors of importance for determining the sizeof the necessary structural steel beam for a given set of conditions. In order of prioritythey are:1. Design based on stress due to bending.2. Design based on deflection.3. Design based on shear.

Q4. How will you analyse a member subject to bending stresses?Ans. A section resists the bending moment by setting up bending stress. Fig. shows the

distribution of bending stress across the section. which an be expressed by the followingwell known flexure formu la:

M - = = ~ ; " E ... (i)Iy R

r .J llie_ '0.4 .,t-e........-., T.A.1-

T

(a)Figure34

(b)

Where M=bending moment at the section I = moment of inertia of the section. R=radius of curvature of the beam. E =modulus of elasticity of the beam. a = bending stress or flexural stress, at any layer distant y from the neutral axis. If the applied B.M. is such that it causes concavity at top, the top fibres wcompression while the fibres situated below the N.A. will be in tension. At the extreme top fibre, the compressive stress is given by

M .... (ii)0" --xyce'cal- IM .... ii) (a)

O"be.cal Zc

Where Z =. section modulus of the section; for compression.c Yc

Similarly ,at the extreme bottom fibre, the tensile stress is given byM .... (iii)O"be'cal I x Yt

. .. iii) (a)

Where Z I = section modulus of the section, for tension.t

YtFor a section which is symmetrical about the N.A., = Y = Y Hence for such a section. Yc t max..

M M ... (iv)O"bc, cal I Ymax Z. Where Z = section modulus

The section modulus represents the strength of the section. Greater the vstronger will be the section. The strength of the section does not therefore, depesection area but depends on the disposition of this area with respect to the axis. It should be noted that the total compressive forae C, above the N.A., is equal tensile forae T below the N.A. for the beam to be in equilibrium. These two forin opposite directions, as marked in fig. (b) and form a couple =C (or T) xThis couple, also known as the moment of resistance (Mr), resists the externmoment. Maximum permissible bending stress

35


20/83

........ _ _ ~ t ......._...

For laterally supported beams, the permissible bending stress in tension (fbI) or in.compression (fbI) are given byU bf or Ubc =0.66 fy .. (v)For laterally unsupported beams, U bf is taken equal to 0.66 fy but Ubc is given by

=066 feb.fy O'be [ Jlln(febY+Uy )" ... (vi)Where n is assumed as 1.4, in the above expression, feb is the elastic critical stress.The values of Ubc as derived from eq. (vi). Some Indian standard available in structuralsteels are given in Steel Tables.

Q5. Discuss the variation of shear stress in a beam. Also give the relation to calculateMaximum permissible shear stress andaverage shear stress.

Ans. Shear Stress:- When a beam is loaded transversely, it is subjected to both bendingmoment as well as shearing force. For a simply-supported beam, with uniformlydistributed load, maximum B.M. usually occurs at mid-span, while maximum force isinduced at the supports. In general, every section of the beam is subjected to both B.M.(M) as well as shear force (V). The shear force causes shearing stress at the section, themagnitude of which varies across the depth of the beam, at that section.On any layer, at height y from N.A., the intensity of transverse shear stress (Tv) is given

V -by : Tv == lz (AY) .. (i)Where V = Transverse S.F. at section. I = Moment of inertia of the section, about the bending axis. Z = Width of the section at which Tv is being computed.AY=First moment of the outer area, above the point where Tv is being computed, aboutthe NA

36

Cd' ,. , II .It is to be noted that T does not vary uniformly across the depth of the secvshows the shear stress distribution for some typical sections. For a rectangulathe maximum shearing stress occurs at the N.A., and its magnitude is given by :

Vd 2 3 V ... (ii), =--max 81 2 bdb ---I . . . . - 8 ~

1 " d/2. t -_", \ lrIo,-...... --'''''''.

eb )

FigureWhere b = width of beam and d is the depth of the beam.The ratio of the maximum shear stress and mean shear stress is 1.5.For an I-beam (fig. b), the maximum shearing stress also occurs at the Nmagnitude is given by ;

... (iii)T = ~ [ B (D2 d2 )+d2]vmax 81 tMaximum permissible shear stress (IS: 800-1984)

37


21/83

The maximum shear stress in a member having regard to the distribution of stress inconformity with elastic behaviour of the member in flexure, shall not exceed the value tvagiven by

tva = 0.4 IyFor unstiffened web. The cross-section of the web shall be taken as follows: For rolled I-beams and channels: The depth of the beam multiplied by the we b thickness. For plate girders: The depth of the web plate multiplied by its thickness.

Q6. What is Bearing stress? How will you calculate the thickness of bearing plate?Ans. Bearing Stress: Beams may either be supported directly on other structural members

(such as steel stanchions etc.) or else they may rest on concrete or masonry supportssuch as walls or pillasters. In the later case, the support is of a weaker material thansteel, and it becomes necessary to spread the load (support reaction) over a larger area,so that the bearing stress does not exceed a certain permissible value. This is achievedby the provision of a bearing plate, as shown in fig.

':CRITICALr p l . . ~

: }+- k=hZ!11.="

", .1.T ' '; , " tp.9EARING [tt t, t , t f ,t ", t qTPLATE, tp ~ n . . ,. 1 . ~ - 1

(b )Fig. Bearing Stress

Permissible bearing stress (IS: 800-1984)The bearing stress in any part of a beam when calculated on the net area of contact shallnot exceed the value of a p determined form the following formu la:

Where a p =maximum permissible bearing s t r e s ~ , -andIy = yield stress of steel

Thickness of bearing plate (Fig.) Let the size of bearing plate be B x N, and its thickness be tp. Let Ip be the actual pressure. The critical section for bending will occur at a distance n form the edge, as

38

marked in fig.(b), where n= B _k.As per AISC practice, k is taken equal to h2 equ2Bto the distance from bottom of beam to web toe of the fillet. Hence n = _2

2Considering u nit length of the plate. we have M = + ~Jp 2

Hence the bending stress h is given by Iv = ( )

Or

It = 3/pn2o t 2pLimiting the value of h to abe' we have t = /3Jpn

p V To.

... (i)

Q.7.Ans.

What do you mean by laterally supported beams? Discuss in detail?Laterally Supported BeamsFor most of the rolled shapes, the permissible stress in bending (abc) is eq0.661y The important conditions associated with the use of this value is (i) the mmust have an axis of symmetry in the plane of the web. (ii) the member must be loathe plane of the web (iii) the compression flange must have lateral support, and section is compact. Loads on a beam cause bending, due to which tension is indone flange and compression' is induced in the other flange. The compressionbehaves somewhat like a column and will tend to buckle to the side, or laterally,stress increases, if it is not restrained in some say. Most beams have some laterMQst beams have some lateral support - speCially those which support the flooHowever, the degree of lateral support is often a matter of engineering judgmeshows lateral support conditions of beams. In some cases, it is conservative and assume no lateral support.

39


22/83

(0 J FULL LAT[RAL SUPPORT

'c ) NO LAT''''', SUfIPOIITC" INTIAIlITTINT LAT'''ALSUPPOMTFig. (a) shows two cases of full lateral support for the top flange, a s s u m ~ d to incompression. In fig. (c), no lateral support exists for top flange. Fig. (b) shows intermittentlateral support. . .Another important condition (condition no. iv mentioned above) for the section to qualifyfor the relation ()bc(or ()bJ=0.66 i y is that the section should be a compact section - acondition that deals with the response of the beam in an overload situation. To qualify ascompact, the section must be proportioned so that no local buckling can occur. In otherwords, the section must be proportioned so that no local buckling of the flange. or weboccurs before the full plastiC moment capacity is achieved. A cross-section that meetsthis criterion is said to be compact, and for that section, (Tbe (or (Tbt) =0.66 i y 'Give the step by step procedure fo r the design of laterally supported beams.Q8. Design of laterally supported beam:- The design of laterally supported beam is carried outAns.in the following steps:

Step 1 :Determine the effective span of the beam. Also, estimate self weight of the b e ~ m , andadd it to the super-imposed load to get the' total load on the beam. The self weight may

40

be assumed to be equal to total load on the beam. The self weight may be assumequal to total load W/300 to W/350 kN/m.

Step 2 : Compute the maximum B.M. (M) and shear force (V) in the beam.Step 3 : Take (Jbt =()bc =O.66fyStep 4: Find section modulus (Z) of the beam:

Z= Mabc

Step 5: From the steel tables, choose a suitable rolled beam section which has Z value to the one found above.

Step 6: Check for shear.Step 7: Check for deflection.

In addition to the above checks, the beam is also checked for web crippling bucking.

Q9. Give the IS recommendation for computing the length of compression fcase of simply supported beams and girder.

Ans. Effective length of compression flange:- The recommendations of IS : 800computing the effective length of compression flanges are as follows:1. For simply supported beams and girders, where no lateral restrai

compression flanges is provided, but where each end of the beam is against rotation, the effective length I of the compression flanges, to bshall be taken as follows (Steel Tables)

a) With ends of compression flanges unrestrained against lateral bendinfree to rotate in plane a t the bearings)I = span

b) With ends of compression flanges partially restrained against latera(that is, free to rotate in plane at the bearings)I=O.85 x span

c) With ends of compression flanges fully restrained against lateral bendinnot free to rotate in plan at the bearings)J =O.7x span.

Restraint against torsion can be provided by :i) Web or flange cleats, orii) Bearing stiffeners acting in conjunction with the bearing of the beam, oriii) Lateral end frames or other external supports to the ends of the co

flanges.Or

iv) Their being built into walls.41


23/83

Where the ends of the beams are not restrained against torsion, or where the load isapplied to the compression flange and both the load and flange are free to move laterally,the above values of the effective length shall be increased by 20 per cent.

Q10. What is web crippling? How will you calculate web crippling stress?Ans. Web Crippling : Web crippling is the localized failure of a beam web due to introduction of

an excessive load over a small length of the beam. It occurs at point of application ofconcentrated load and at point of support of a beam. A load applied over a short length ofbeam can cause failure due to crushing due to high compressive stress in the web of thebeam below the load (fig. a) or above the reaction (fig. b). This phenomenon is alsoknown as web crippling or web crushing.

J

- ......................_---_ ... - ..--.....

bd,..Fig. Web Crippling

Practical and commonly used bearing lengths b are usually large enough to prevent webcrippling from occurring. An assumption is m ade that the load spreads out along 30lines(fig. e), so that critical area of stress, which occurs at the toe of the fillet has a length of(b+2k) under the load and (b + k) at the end reaction, and with a width r w. Thecompressive stress at the toe of the fillet should not exceed the permissible bearingstress 0'p , which is taken equal to 0.75 iFor 30spread, and with hz as the depth of toe of the web, we have k = ~ . J 3

42

Hence under the concentrated load,R =. b +2k =. b+ 2h;.13Web crippling stress - p "50.75J, ..(i)( b + 2 ~ . J 3 ) t w yWhere P is the concentrated load, Similarly, at the supports, if b' is the width of the bearing plate, R' == b'+ k == b'+ h;.13 Web crippling stress R ..(ii)= ( b + 2 ~ . . J j ) t w 5.0.751;Where R is the end reaction.The value of h2 can be found form section tables, should the web crippling sexceSSive: the p ~ o b l e m may be corrected by (i) increasing the bearing length (ii)a beam With a thicker web, or (iii) providing bearing stiffeners.Web Crippling should invariably be checked at (i) all concentrated loads, ansupports where the beam is supported by walls or pedestal or at columns wconnection is a seated type.

Q11. What is depended beam? How it will be designed?Ans. A depended beam consil3t of two rolled steel sections, placed one above the oshown in fig., with the sale aim of increasing the depth of the beam. The moment of inertia of the built -u p beam is (h)2 (.). 1==21, +2 4 - ... I\ 2Where Ii and Ai are moment of inertia and area of cross-section of each primary b

. . . . b - ~ - I T...... ..- t1.Dividing both the sides of Eq. (i) by h, we get

43


24/83

... (ii)Z =Zi+!...Aih2It is found that for most of the I-sections, the term ! Ai.h ies between 1.5 Zi to 1.6 Zi.2Adopting a lower value of 1.5 Zj for safety, we get .. , (iii) Z =Zj +1.5Z;=2.5Zj

Z .. , (iv)Hence 2.=-I 2.5

Where

What are Encased Beams? Explain in detail.Q12. In the case of steel framed structures, where steel stanchions and beams frame into each Ans. other, it is considered desirable to encase the stanchion as well as the beam intoconcreter to make these fire resistant and also to improve the appearance. Fig. showssuch an encased beam. As per IS : 800-1984, beams and girders with equal flanges maybe designed as encased beams when t he following conditions are fulfilled.

---

... .."... .. .. ...'" .'I' - ........~ - ... : : - - ~ T' i \ . ~ f I I ... .. \ ' .. ". .. .... :.e.... f. '..., .. . . _. ..... ..

............-5"'m4STIRRUPS@1'Ommc/

A. ; . ' I I/' _ .& .. '''",-_ .... -. '.- . "," ""1.'" .....- ... .... ......-.(sor-

The section is of single web and I-form or of double open channel form with thea) webs not less than 40 mm apart.

44

b) The beam is unpainted and is solidly encased in ordinary dense concremm aggregate (unless solidity can be obtained with larger aggregategrade designation M15.

c) The minimum width of solid casing = (bo+ 100) mm, where bo is thesteel flange in mm;

d) The surface and edges of the flanges of the beam have a concrete coless than 50 mm, and

e) The casing is effectively reinforced with steel wire of at least 5 mm diamthe reinforcement shall be in the form of stirrups or binding at not m150mm pitch, and so arranged as to pass through the centre of coveredges and soffit of the lower flange.

Design of member: The steel section shall be considered as carrying the entireallowance may be made for the effect of the concreter on the lateral stabi

, compression flange. This allowance should be made by assuming for the pdetermining the permissible stress in compression that the equivalent momen(Iy) about y-y axis is equal to Ar/ where A is the area of steel section and. taken as 0.2 (bo+100) mm. Other properties required for referring to clause code may be taken as for un cased section. The permissible bending

d e ~ e r m i n e d shall not exceed 1.5 times that permitted for uncased section.Q13. A simply supported beam has an effective span of 7 m and carries a

distributed load of 50 kN/m. Taking i = 250N/mm2 and E= 2x 105 N/mmythe beam, if it is laterally supported.

Solution:Step-1: Effective span L =7 m.

Assume self weight = W = 50 x7 lkN Im 350 350 Total U.D.L. =w =50 +1 =5.1 kN/m.

Step 2: M WL2 51(7)2 =312.375 kN -m =312.375xl0 6 N-mm 8 8 V wL 51(7) = 178.5 2 ' 2

Step 3: ()bt = ()bc = 0.66 fy 0.66 x 250 = 165N Imm 2 6Step4: Required Z=M = 312.375x10 1893.2xIQ3 mm 3

()bc 165 45


25/83

Step 5: From steel tables, try ISWB 500 @ 95.2 kg/m, having the following properties :Zxx =2091.6cm3 =2091.6 x 103mm 3 ;1xx == 52209.9 X 104 mm4h=500mm; t 14.7mm: tw==9.9mmf

Step 6: Check for shear 3178.5 x 10 36.lN I mm2Average shear stress = I Vrva,ca ='-hxt., 500x9.92Permissible value t == OAf, == 004 x 250 == lOON I mmva y

Hence safe. Step 7: Check for deflection

5 wL4 5 (51 X 103)(7)4 3Y =__= -X x (1000)max 384 EI 384 (2 x 105)(52209.9 x104)

Allowable deflection == Span == 7000 21.54mm. Hence safe.325 325Q14. A beam, consisting of ISMB 600 @ 122.6 kg/m is simply supported over span of

8.5m. Determine the safe load the beam can carry, assuming that the beam is2laterally supported. Take Iy =250Nlmm2 and E=2xlOSN/mm

Solution: For ISMB 600 @ 122.6 kg/m, we have:I ==91813.0cm4 ==91813xl04 mm4 .xxZ ==3060Acm 3 =306004 x 103mm3xxh == 600mm ; b =210mmtw 12.0mm;tf ==20.8mm 2( fbc == ( fbi == 0.66 y == 0.66 x 250 == 165N ImmMr =( f bc 'Z ==165 X 3060Ax103 ~ 5 0 5 . 6 x l 0 6 N -mm==505.6kN-m

Hence Mmax Mr =505.6kN -m .M == WL2 == w(8.5iBut max 8 8w(8.5i = 505.68

w== 5 0 5 . 6 ~ 8 ==55.98kN 1mFrom which (8.5)

46

The above value of w will be acceptable only if the beam is safe in shear and deflectionCheck for shear: V wI == 55.98x8.5 _2 2 - 237.915 kN.

V 237.95 X 103 . 2-h-xt- - 600x12 33.04Nlmmw

1'va =Oo4l 0.4x250 100Nlmm2 Hence OKyCheck for deflection

5 wL4 5 (55.98 X 103)(8.5)4 3 Ymax 384 EI 384 (2x 105)(91813 x 104) x (1000) 20.7mm

Mass of beam/m =122.6 kgSelf weight of beam = 122.6x 9.81 x10-3 r: 1.2kN1mPermissible, superimposed load w = 55.98-1.2 = 54.86 kN/m.

Q15. A laterally supported beam having an effective span of 8 m consist of ISM1 . 3 7 . kg/m and cover plate of 250mm x 16mm connected to each flange bydla. n v ~ t s . Determine the safe U.D.L. which the beam can carry in additioown weight.

Solution: ISWB 550 @ 103.7 kg/m has the following properties; Ixx 64893.6xl04mm4 Zxx =2359.8xI03 mm3 h 550mm b 190mm =19.3mm; tw = 11.2mmf a 13211mm2

1 ~ - ~ n e EE e- 0 C\I++ - .,-.,.,

47

Ixx of built-up ~ e c t i o n = [64893.6xO4 + 2 x 250x 16(8+ 275)2] CHAPTER 3: BEAM COLUMNS
http:///reader/full/64893.6xhttp:///reader/full/64893.6x


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= 128965 x 104mm4(neglecting the moment of inertia of the plate about their own axes)Gross area of flange = Afg (190x19.3)+ (250x 16)::: 7667mm2Net area of flange =Afn = 7667 - 21.5(16 + 19.3)=6908mm2(only one row of rivets on each flange)abccol =( ~ } m a xM::: (J'bc.cal x l

YmaxLimiting (J'bl I::: 0.66!, = 0.66 x 250 = 165N I mm2,ca y

A 6908 2(J'b 1=(J'b /x i =165x--=148.67NlmmI,ea I,ca A 7667fg. 4M = 148.67x128965xlO .=659.87x106 N-mm=659.87 KN-m

291W= 8x659.87= 82.359kNlmL2 (8)2

Mass of ISMB 550 = 103.7 kg/m.Mass of 250mm x 16mm plates = 2[7.85X25X1.6 ]=62.8kg lm

Total mass/m = 103.7+62.8 = 166.5 kg/m Weightlm =166.5x9.81x10-3 =1.633kN 1m Permissible w::: 82.359 -1.63 3 :::80.726kN Im

48

SECTION -AMultiple Choice Type Questions:1. A structural member that is subjected to varying amounts of both axial compres

bending moment is termed as:a. Beamb. Columnc. Beam columnd. All ofthe above

2. Bending moment in a member is induced due to:a. Eccentric loadb. Non-symmetrical floor loadsc. Transverse loadsd. All of the above

3. If the axial load approaches to ~ e r o as a limit. the member is theoretically subB.M. only. and it is commonly known as;a. Beamb. Columnc. Beam Columnd. All of the above

4. When the B.M. approaches zero. as a limit. the member is theoretically subjecteload only. and it is commonly known as:a. Beamb. Columnc. Beam Columnd. All of the above

5. Axial compression and bending about the strong axis results in the:a. Failure by lateral torsional bucklingb. Failure by instability in one of the principal directionc. Failure by combined twisting and bendingd. None of the above

49

6. The primary.bending moment and deflection are maximum at the mid span for a member
http:///reader/full/N-mm=659.87http:///reader/full/8x659.87http:///reader/full/N-mm=659.87http:///reader/full/8x659.87


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7.

8.

9.

10.

11.

with:a. No end restrainedb. One end restrainedc. Both end restrainedd. None of the aboveThe value of factor of safety, to give consideration to end conditions and side sway of thecolumn is:a. 1/0.5b. 1/0.6c. 1/0.7d. 1/0.8The value of co-efficient, em for a membe r in frame where side sway is not prevented istaken as;a. 0.6b. 0.85c. 1.00d. None of the aboveAxial compression and biaxial bending results in:a. Failure by lateral torsional bucklingb. Failure by instability in one of the principal directionc. Failur4e by combined twisting and bendingd. None of the aboveA column is said to be short, when the slenderness ratio is:a. Less than 12b. Greater than 12c. Either less than 12 or greater than 12d. None of the aboveSteel Section commonly used for column is:a. I-sectionb. Channel section .c. L-sectiond. T-section

50

12. Magnification factor for th e deflection is given by:a. 1

( I -a)b. 1

( l+a)1. (a)

d. None of the above13. Maximum deflection in case of simply supported beam subjected to U.D.L. is g

a. w12/48b. 5 wl2/8c. 5w l2/384d. None of the above

14. Effective length of a column is defined as a length:a. Which takes part in bucklingb. Which will take all the loadc. Which will be small in lengthd. None of the above

15. The load which is concentrated at one point is called:a. Uniformly distributed loadb. Point loadc. Uniformly varying loadd. None of the above

Key:1. (c) 2. (d) 3. (b) 4. (a) 5. (a) 6.7. (b) 8. (b) 9. (c) 10. (a) 11. (a) 12.13. (c) 14. (a) 15. (b)

51


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SECTION -CShort Answer Type Questions:1.Ans:

What are the various causes of bending moment occurrence in a member?It is gen.erally known that axially (or concentrically) loaded, compression membnon-existent in actual structures and that all compression members are subjesome amount of bending moment induced due to (i) eccentric load (ii) non-symfloor loads (iii) transverse loads due to, wind or earthquake (iv) joint momencontinuous frame action (or building frame action) and (v) end reactive moments.

2.Ans: Under what conditions, a member is known as column or beam?Since a column may be subjected to both the axial load and bending momeextreme cases may exist. When the B.M. approaches zero, as a limit, the mtheoretically subjected to axial load only, and it is commonly known as cohowever, the axial load approaches zero as a limit, the member is theoretically sto B.M. only, and it is commonly known as a beam.

3.Ans:

Define the term Beam Column.A structural member that is subjected to varying accounts of both axial compresbending moment is commonly termed as a beam column. It is generally knaxially (or concentrically) loaded compression members are non-existent istructures and that all compression members are subjected to some amount ofmoment.

4.Ans:

What are the different categories of design procedures generally adopted?The design procedures generally are in one of the following three categories:i) Limitation on combined stressii) Semi-empirical interaction formulae based on working stress procedureiii) Semi-empirical interaction formulae based on ultimate strength

5.Ans:

How you will calculate critical buckling stress fo r a column?The following equation gives the value of critical buckling stress for a column and y-axis:f_ - "'IIe .

"'II.. - .t:6.Ans:

What are the conditions, under which a beam column may fail?A beam column may fail due to the following reasons:1. Failure by instability in plane of bending without twisting.

53

SECTION BTrue I False Type Questions:1. Column is a vertical tension member.2. Column may be subjected to both the axial load and bending moment.3. All the compression members are subjected to some amount of bending moment induceddue to eccentric load.4. 'Axial compression and bending about one axis results in failure by instability in plane of

bending without twisting.5. A structural member that is subjected to varying amounts of both axial compression andbending moment is termed as Beam6. The primary bending moment and deflection are maximum at the mid span for a memberwith both end restrained.7. Deflection in case of simply supported beam is maximum at edges.8. Uniformly distributed load is acting throughout the length of a member.9. Bending moment in a member is induced due to transverse load due to wind orearthquake.10. When the B.M. approaches zero, as a limit, the member is theoretically subjected to axialload only, and it is commonly known as column.

Key:1. F 2. T 3. T 4. T6. F 7. F 8. T 9. T

52

5. F10. F


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---

2. Failure by lateral-torsional buckling.3. Failure by instability in one of the principal directions.4. Failure by combined twisting and bending on these torsionally weak sections.5. Failure by combined twisting and bending when plane of bending does not

contain the shear centre.7. How you will design a beam column?Ans: In order to design a beam column, it is usually advantageous to approximately convert

the resulting B.M. into an equivalent axial compressive load, and then determine thecross-section of the column, using the usual column tables to find the permissiblecompressive stress. Sometimes, the conversion of the axial load into the equivalentmoment will be more helpful.

8. How you will go for simplified approach fo r magnification factor?Ans: In the simplified approach applicable for members with single curvature without end

translation, let us assume that the transverse load w(x) causes a deflection Yo at themid-span, and that the secondary B.M. varies as a sine-curve. Also, let Ys be the midspan deflection due to secondary B.M. Evidently according to moment-area-method Yxdue to secondary moment will be equal to moment of Ms IEl diagram between supportand mid-span, taken about the support.

9. Give the values of Co-efficient Cm, fo r the different end condition of columns.Ans: The following table gives the values of coefficient of em, for the d ifferent end conditions:F C a ~ ....------- c o n d i t T o n - - - - - - - ~ - - - - - C m

10. Calculate the value of magnification factor fo r deflection and moment fovalues of a and give the difference between the two.

Ans: The following table give you the values of magnification factor for deflection anwith the value of a

Magnification factor fo rTBuni f i c a t iOn factor for %. deflection =1/(1-a) moment. Differe=(1-0.1775a)/(l-a) .

1.091 0.1 11.111 1.81.2061.250 3.6

i 0.3 I 1.429 /1.353 I5.6I 0 1 667 1.548 7.71 .1 .4 I 9.80.5 2.000 1 8222.500 1 ..6 11.9.2340.7 3.333 .2.919 14.20.8 5.000 4.290 14.6' - - - -" - - -11. What is the difference between a short and a long column? Ans: Short Column: A column whose slenderness ratio is less than 12 is terme

column. The load carrying capacity of short column is more. Also, the bucklincolumn is negligible.Long Column: A column whose slenderness ratio is more than 12 is termecolumn. The load carrying capacity is less than the short column. Long columnbuckle and its failure is also due to buckling.

12. What do you mean by term "8eam"?Ans: A beam is a structural member, the primary function of which is to support loa

to its axis. The loads produce bending moment and shear force in the beams. T.beam action is of great age but despite of long history of use, the systematicbeams had to await the development of theory of bending.

13. What are the diffe rent types of beams according to end conditions?Ans : Depending upon the end conditions at the supports, beams may be classified a

a) Simply Supported beams b) Fixed Beams CanWeverBeams d) Propped Cantilever Beams e) Continuous Beams f) Over Hanging Beams

55

(a f - MemberSifllrames --where -side swayTs"rlotprevented

r - - - " ( b ~ ) - ~ M e m b e r s T n frames where side sway is preventedand not subject to transverse loading between theirsupports, in the plane of bendingMembers in frames wh-ere side sway is preventedin the plane of loading and subjected to transverseloading between their supports:i) For members whose e n ~ s are restrained

against rotationii) For members whose ends are unrestrainedagainst rotation

(c)

- - ~ - --- - - - - ~ - - - - - - - - ~ - ~ . - - - ~ - - ~ ~ - - - -54

_0.85 -

0.6 - 0.4

0.851.00

SECTION -0 The design procedures generally are in one of the following three categories.


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Long Answer Type Questions:Q1. What do you mean by Beam Column? What are the reasons of bending moment in

a member?Ans. A structural member that is subjected to varying amounts of both axial compression and

bending moment is commonly termed as a beam column.It is generally known that axially (or concentrically) loaded compression members arenon-existent in actual structures and that all compression members are subjected tosome amount of bending moment induced due to (i) eccentric load (ii) non-symmetricalfloor loads (iii) transverse loads due to wind or earthquake (iv) joint moment due tocontinuous frame action (or building frame action) and (v) end reactive moments.For example, the interior column will not receive concentric load if the live loads are notsymmetrical. Similarly bending mo ment will be induced in the column due to a load on abracket attached to it. Even in an ordinary truss, the purlins may not be at panel pointsgiving rise to moment in the main rafter.Since a column may be subjected to both the axial load and bending moment, twoextreme cases may exist. When the S.M. approaches zero, as a limit, the member istheoretically subjected to axial load only, and it is commonly known as column if,however, the axial load approaches zero as a limit, the member is theoretically subjectedto S.M. only, and it is commonly known as a beam. In a beam-column, the originalmaximum bending moment due to transverse loading is magnified due to the presence ofaxial load.

Q2. What are the different modes of failure of Beam Column?Ans. The following are categories of combined bending and axial load, along with the likely

mode offailure :1. Axial compression and bending about one axis: Failure by instability in plane of

bending without twisting.2. Axial compression and bending about the strong axis: Failure by l a t e r a l ~ t o r s i o n a l

buckling.3. Axial compression and biaxial bending (torsionally stiff sections): Failure by

instability in one of the principal directions.4. Axial compression and biaxial bending (thin walled sections): Failure by

combined twisting and bending on these torsionally weak sections.5. Axial compression, biaxial bending, and torsion : Failure by combined twisting

and bending when plane of bending does not contain the shear centre.

56

i) Limitation on combined stress ii) Semi-empirical interaction formula based on working stress procedure.iii) Semi-empirical interaction formula based on ultimate strength.

Q3. Discuss the differential equation approach formoment magnification.Ans. Let us consider a general case of bending where the lateral load w(x) in combend moments M1 and lor M2 constitute the primary bending moment (M1) function of x. This primary bending moment causes the member to deflecsection distant x from support, giving raise to a secondary moment P.y, wheaxial load.

d2yMx=El dx 2 ... (1)=Mj-P.yThe negative sign has been, used because the beam column has been bent its original centre line.d2y P M '"dx2 +EI Y =Ei (2) Where Mi is function of x. Differentiating Eq. (2) twice, we get d 4dy P d 2y I d 2Mi- -+- -=-- dx 4 ... (3)EI dx 2 EI dx1Sut from eq. (1), we have2d y _ Mx and d 4y 1 d2Mx dx2 - EI dx4 EI--;;;ZHence, substituting in (3), we have1 d2Mx P (M ld 2M EI dx2 + EI E;) dx1 '

k 2Substituting = P / El and simplifyingd 2Mx +k2M =: d 2Mjdx 2 x dx 2 . . (4)The moment differential equation (4) is of the same form as the deflection dequation (2). The homogenous solution for eq. 4 isMx =C1sinh + c2 cosh + .h,(x) ... (5)Where J;(x) = value of Mx satisfying eq. (4). When Mx is a continuous funmaximum value of Mx may be found by differentiation:

57

dM ' . kx dl"l(X) (6) Where a =P/PE. Hence


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_ _ =O=c kcoskx-czksln +_!l_Idx I dx ...For most ordinary cases of loading (such as U.D.L., point loads, end moments etc). it canbe shown that dJ;(x)/ dx = 0Hence for such situations, we get from eq. 6 :c1kcoskx:::: c2k sin gkx

Or tanh= C1c2At maximum Mx we get

.. , (7)

sinkx = c1~ c ~ + c ~

and coskx ... (8)

Substituting in eq. 5. we getc2 c2M I + 1 + I"(x}X.max ~ ~ JIVCI + C2 VCI + C2 .. , (9)

Mx.max :::: +ci +It(x)Q4.Ans.

Discuss the simplified approach fo r moment magnification.In the simplified approach applicable for members with single curvature without endtranslation, let us assu me that the transverse load w(x) causes a deflection Yo at the mid-span, and that the secondary B.M. varies as a sine-curve. Also, let y. be the mid-spandeflection due to secondary B.M. Evidently according to moment-area-method Yx due tosecondary moment will be equal to moment of Ms / El diagram between support andmid-span taken about the support.Ys== ~ ( y o + Y s { ~ ) ! ( ~ )

PLz== (Yo +Ys) :rzEl ... (1)Or PYs = (Yo + Ys) PE ... (2)

Where PE= ,,2EI =Euler's Load.L2Solving Eq. 2 for Ys' we getYs::::YOL p ~ ~ ~ J Y O C ~ a ) ... (3)

58

Ymax == Yo +Ys ::::Yo +Yo (1 a)Yo 1=--=y -- ... (4)I -a l - a

Thus _1 _ is the magnification factor f or the deflection.I -a

Hence, for a beam-column subjected to axial load P,Mx.max =Mo +Pymax ... (5)Substituting the value of Ymax and noting that p =aPE a:r 2ElWe get M El)(-2L)M +(a" 2. X.max 0 . L2 1-a

Or 2M =M [1+ a:r ElyO x_I_]x.max 0 M L2 1 - acOr ... (6)

2Where Fm = magnification factor =1+ _a:r Elyo x _1_ MoL2 I -a

... (7)

... (8)

Q5. Compare the moment magnification factors, obtained from differential approach and that from the simplified approach, for the case of beamcarrying U.D.L. w along with axial load P.

. Ans. From the differential equation, we have... (1)

Where kL ::::!:...JP : : : : ~ - J a2 2 E1 2

Hence F == _8-(sec!!.f;; -1) ... (2)m :r2a 2From the approximate solution F = em ... (3)

m I -a59

2


32/83

Where Cm=i+[n EIYo -lJaMoL2

5 wL 4 WL2Yo=--.and Mo=- (numerically)384 EI 8Y 5 L2_0 =_._Mo 48 El

nlEI 5 Ll )Cm 1+(T X 48E1-1 a=1+0.0281aF :=1+O.028Ia

m I - a... (4)

The values of the magni ficent factors, given by eq. 2 (exact method) and Eq. 4 (simplifiedmethod) are tabulated below for comparison. It is also to be noted that the magnificationfactor for deflection is 1/(1-a) , which is the same for all types of loading. The values ofmagnification factor for deflection have also been shown in the above table forcomparison.

Magnification factor-for deflections kL n ~ 1Va1uesoj Fm byec -=sec - a2 2 .~ ; ... - -Eq. (a) or Eq. (b) I -a2 or 4 - . --. 1.114 1.111.1 1.114.1371.250.2 1.310 1.257.257

1.533 1.441 1.441 1.429.3. ~ ~ - - -1.832 11.686 1.667.4 1.685

---.0.5 2.252 2.028 2.000 ~2.0302.884 2.546 2.542 2.500.6

0.7 3.941 3.405 3.399 3.333- - ~ - ~ ~ ~ .6.058 5.125 5.112 5.000.8

12.419 10.284 , 10.253 10.000.9From the table above, we find that there is no significant difference between values ofmagnification factors computed by the two methods. Hence the approximate (or simplifiedmethod can be safely u s ~ d . Also, these values are very near to the magnification factorfor deflection.

60

_ YoYmax --1-aHence magnification factor for deflection =

p

Q.6.

Ans.

.,..."""...... 'rtlGl

" " - - - L I 28

A beam column is subjected to axial load P and a transverse load W at ts ~ a n .. .Compute the magnification factors for deflection, and moment

s l m p h f l ~ d approach, and compare the two values of the magnification factoThe maximum deflection under any type of transverse loading is given by :

l - attl

Again, the magnification factor for moment is given by EquationF ~m l - awhere C 1 (, rElyo 'I= +l- - - l ja ML2oHereYo 1 L2-= - .Mo 12EI

(n2E1 1 'ICm 1+ l - - x -1)a=1-0 1775a4 12E1 .F =1 O.

m 1 a Thus we observe that the magnification factor for deflection (= is the sametypes of transverse loading while the magnification factor (Fm) for the moment deupon the type of loading. The values of magnification factors for deflection and moments, for various valuesare tabulated below:

61

Magnification factor fo r IMagnification factor for moment 0/0a It is to be noted that while the allowable stress ratio ((j / (j ) reduce


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=(1 O.l775a)/(1 a) Differenceeflection = 1/(1-a)1--.--+-1-:.1-1--1 - - - - - - - - - ~ - - + - - c 1 ~ . 0 : - : 9 c - : - 1 -------------0 1

0.20.30.4I - - - ~ - : : - - - - - ~ - - - _ t _ - - , - - : : - = : - - - - - - - ~ - -0.50.6

0.9

11.8------______-1-1_3._6_ _ _ ----1

15.617.7~ - - - l - - -9.8

- ~ - - - ~ - - - + I 11.9

-------I --+-c-.___--- \19.0Since the values of two factors are very near, the magnification factor for deflection canbe taken as the magnification factor for moment, as an approximation.

Q.7. Discuss the method of designing a beam column.Ans. In order to design a beam column, it is usually advantageous to approximately convert

the resulting B.M. into an equivalent axial compressive load, and then determine thecross-section of the column, using the usual column tables to find the permissiblecompressive stress. Sometimes, the conversion of the axial load into the equivalentmoment will be more helpful.For this purpose, Equation can be rewritten in the following form:

A:ac +O':zl(l-;"'"'J =1.0O.6/ c

where A =area of cross-section and Z is the section modulus about the axis of bending.Multiplying both the sides by A (joe we getPM(i)(O'oeV Cm 'A =P.+ Z lO'bJlI-O'ac,eo/10 .6jj 0'0" EQwhere PEQ = equivalent axial load.Hence ( 0 15. However when the ratio isac)cal ac 0.15, PEQ may be computed from the following expression corresponding to eqPEQ =P+M.Bf (::)To start with, the ratio (J'ac,cal / (J'ac is not known. Hence, equation may bdetermine PEQ , and based on this equivalent load, the section may be selehaving known the section, value (J'ac,cal / (J'ac can be determined, and if thfound to be greater than 0.15, Equation may be applied to find revised value ofExtending the above logic to biaxial bending, equation for PEQ take the form.BEAM COLUMNSPEQ (0 ' V C '] [ ( 0' )( C ']= P + Mbll ; " ) l-l-----"'m'---I + Mbll---E l --_m--)O'oe,coJo.6fce/ x O'be 1-O'ac,coI /0 .6/cc y

Iand p. = p + M B aa c + M B (jaeEQ xft. yfYa bcx (jbeyQ.S. A column of effective height 6m is subjected to an axial force of 56

bending moment of 25 kN-m. The section of the column consists of IS122.6 kg/m. Check the adequacy of the e c t i o n . Take C =0.85.mAns. For the ISMB 600 @ 122.6 kg/m section, we haveA = 15621 mm2;D = h = 600mm;b = 21Omm;Z =3060.4 x 103mm3 ,lxx = 91813 x 104mm4;rxx =242.4 mm;rY.)' = 41.2;T =if =20.8 mm;tw == I = 12.0 mm.d l =h - 2i f = 600-2 x 20.8 == 558.4 mm.

Hence T =tJ =20.8=1.73


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25 30 Extrapolation for 8 . 8 ~Df f - + IIIry-+ . ~. 97=1 98.38 140 103 I93.388L - - ~ ~ _ I . ~ . I I

O"bc ::::9S.6Nlmm2.Again for A 145.6,O"ac 'i:: N 1mm2

1(2E_1(2(2.0xI05)_ 2Also, fcc 2 - ( )2 -93.1NlmmA. 145.63P = 560 xl 0 = 35.85 N Imm 2A 15621

6a - M 25xl0 =8.2Nlmm 2 ac,eal - Z 3060.4 X 103

em = 0.85 (given) Hence from equation a C a ~ + m be,eal :>;1

aa e [abc 1- 0.61ccor 35.85 0.85 x 8.2 1

- -+


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a : 450 X 10 43.00 N Immae,eal 10466 SOx 106 2 cal:::::: 3 : 34.62 N / mmbe, 1444.2 x 10

Step 7 Determination of magnification factor Cm1- a ac,cal

0.6/eFor the first case, em == 0.6 - 0.4/3 == 0.6 +0.4 x 1 == 1.0

em 1 43.00 -1.05l - - ~ -0.6 x 1512.7

For second case Cm =0.6 - O.4x 1 =0.2 but 1- 0.4 Cm =0.4

Cm 0.4 0443.00 ==. 21- aae,eal 1----0.6/cc 0.6 x 1512.7Step 8 Check for the section

Here, O'ac,cal + O'bc,cal :::::: 43.00 0 . 6 4 2 ~ O . l 5 0'DC a ac 66.96

Hence equation applicable will be(a) for the f irs t case 10, I t'I

0.6/ccor 43.00 + 34.62 x 1.05 1 1\6.96 112.1 or 0.642 + 0.324 1 or 0.966 < 1. Hence safe. ~b) For the second case (a ) U.JPRlMARY Cc) SECONDARY43.00 + 34.62 x 0.42:S; 1 MOMENT MOMENT66.96 112.1 .., . . . .p., or 0.642 + 0.130 1

O. 772 1. Hence the section is over safe and needs revision.

66

Taking Cm=1.0 for bending about major axis and 0.8 for bending about mfind suitable H-section for the column.

Ans. Equivalent axial load for biaxial bending, letPEQ = (p + O.75P +O.75P) == 2.5P == 2.5 x'300 == 750kN, to start with.From steel tables we find that average value of fy is around 50 mm for ISHBHence A, =11 r =6000/50 =120. Corresponding to this value of .It , (jac =64

3 Area required == 750 x 10 == 11719mm2 64

Hence select ISHB 450@ 92.5 kgl m having A = 11789 mm 2, fy =50.8 mm, Zxx =1793.3 x 103 and Zyy =242.1 x 103 B 11789 6.57xlO-3 ft Z"" 1793.3 X 103

Bfy = ~ = ~ 7 8 9 N =48.69xl0-l A 1 S O ' L e / l a a c ~ ) =0.6Zyy 242.1 x Iv abc x When bending takes place about minor axis, lateral buckling is not a problem ab c =0.6fy:::::: 0.66 x 250 == 165 N Imm 2 ( O' x '1 = ~ = 0 . 3 8 8laJ 165 ySubstituting the values in equationP == P + JJ:xB aa e + M B aa eEQ f t .a y J y a hex bey = 300+ (50 X 103 x 6.57 X 10-3 x 0:6)+(10 x 103 x 48.69 X 10-3 x 0.0388) = 300+ 197.1 + 188.9 =6861cN

Step2 Selection of Section3 Area required :::::: 586 x 10 10718mm2

64 Hence select ISHB 450 @ 87.2 kg/m having the following properties: A == 11114 mm 2;h =450mm;tf == 13.7mm;t == 9.8mm;rx =187.8 mm;ry == 51.8mm;Zx =1742.7 x 103 mm 3;Zy == 238.8 x 103

Also, d 1 == h - 21f == 450 - 2 x 13.7 == 422.6 mm.

67

Step 3 Determinati6n of (y ac CHAPTER 4: INDUSTRIAL BUILDINGS
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T= tf =1,3.7 =140


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b. Large moment of resistance as compared to other section.c. Greater lateral stabilityd. All the above

8. The moment of the couple set up in a section of a beam by the longitudinal compressiveand tensile force, is known as:a. Bending momentb. Moment of resistancec. Flexural stress momentd. None of the above

9. In a rolled steel beams, shear force is mostly resisted by:a. Web onlyb. Flanges onlyc. Web and flanges togetherd. None of the above

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bco 3.2 Advanced Steel Design