middotGLOBAL INSTITUE OF HIGHER EDUCATION

INSTITUTE MEMERSHIP OF STUDENT CHAPTER INSITUTE OF CIVIL ENGINEERS (INDIA)

TC 24

MECHANIC OF SOLIDS

Mechanics of Solids(TC-24) (Short)

Q2 ~-~14 Failure of Riveted Joint ~ f

bull A riveted joint may fail due to many ways Buube followingare more important from the subject ~middotpoint of view

I Failure of the rivets 1 Failure of the plates

We shall discuss bolh the above mentioned cases offrulure in more details one by one

~2915 Failure of the Rivets A rivet may fail due to either of the oUowin~ two reasollS

1 Shearing of the rivet 2 Crushing of the rivet Now we shall discuss both the cn~ one by one

~1 Sharin9 ofth~~ The plates which are connected by the rivets exert tensile stress on the rivets If the rivets are

unable to resist Ihe stresses they may be sheared off as shown in Fig 2910 It will be interesting to

know that

p

p

Fig 2910 Shearing of rivets 1 The rivets are in single shear in a lap joint and in a single cover butt joint

2 The rivets are in double shear in a doable coverbuUjoinL

2917 Crushing of the Rivets Sometimes the ri vets do not actually shear off under the tensile stress but are crushed alt shown

in Fig 2911 Such a failure of rivet is called crushin of the rivet

p

p

Fig 2911 Crushing of rivets

2918 Failure of the plates A plate may fail in many ways BuUhe following are impoItlUlt from the subject point of

1 Tearing off the plate across a row of rivets 2 Tearing off the plate at an edge

2919 Tearing off the Plate across a Row of Rivets Due to the tensile stresses in the main plates the main plate or cover plates may tear off

a row of rivets alt shown in Fig 29121n such cases we consider only one pitch length oflhe since every rivet is responsible for that much length of plate only

pp

Flg 2912 Tearing across a row of rivets

2920 Tearing off the Plate at an Edge A plate may also fail due to tearing at an edge as shown in Fig 2913 This can be avoidedmiddot

keeping the centre of the nearest rivet from theedge of the plale at least two times the diameter rivet

tpf J

Ag 2S13 Tearing off at an edge

2921 Strength of a Rivet We have seen in arts 2916 and 2917 that a rivet may fail either due to its shearing off or due lI)

its crushing Thus while calculating tJyen strength of a rivet we see as to how much resistance it can offer The resistance offered by a rivet is known as its strength or the value of the rivet Following Ivu values of a rivet are imponant from the subject point of view

1 Shearing value 2 Bearing value t Sheoring laUI

The resistance offered by a rivet to be sheared off is known as its shearing value MathematicaU) pull required to shear off the rivet

P = -xti2 Xt 4

where d = Diameter of the rivet and t = Safe pennissible shear stress for the rivet material

(f the rivet is in double shear (Lebull in the case of a double cover bUll joint) the corresponding

2

Mechanics of Solids(TC-24) (Short)

Q2 ~-~14 Failure of Riveted Joint ~ f

bull A riveted joint may fail due to many ways Buube followingare more important from the subject ~middotpoint of view

I Failure of the rivets 1 Failure of the plates

We shall discuss bolh the above mentioned cases offrulure in more details one by one

~2915 Failure of the Rivets A rivet may fail due to either of the oUowin~ two reasollS

1 Shearing of the rivet 2 Crushing of the rivet Now we shall discuss both the cn~ one by one

~1 Sharin9 ofth~~ The plates which are connected by the rivets exert tensile stress on the rivets If the rivets are

unable to resist Ihe stresses they may be sheared off as shown in Fig 2910 It will be interesting to

know that

p

p

Fig 2910 Shearing of rivets 1 The rivets are in single shear in a lap joint and in a single cover butt joint

2 The rivets are in double shear in a doable coverbuUjoinL

2917 Crushing of the Rivets Sometimes the ri vets do not actually shear off under the tensile stress but are crushed alt shown

in Fig 2911 Such a failure of rivet is called crushin of the rivet

p

p

Fig 2911 Crushing of rivets

2918 Failure of the plates A plate may fail in many ways BuUhe following are impoItlUlt from the subject point of

1 Tearing off the plate across a row of rivets 2 Tearing off the plate at an edge

2919 Tearing off the Plate across a Row of Rivets Due to the tensile stresses in the main plates the main plate or cover plates may tear off

a row of rivets alt shown in Fig 29121n such cases we consider only one pitch length oflhe since every rivet is responsible for that much length of plate only

pp

Flg 2912 Tearing across a row of rivets

2920 Tearing off the Plate at an Edge A plate may also fail due to tearing at an edge as shown in Fig 2913 This can be avoidedmiddot

keeping the centre of the nearest rivet from theedge of the plale at least two times the diameter rivet

tpf J

Ag 2S13 Tearing off at an edge

2921 Strength of a Rivet We have seen in arts 2916 and 2917 that a rivet may fail either due to its shearing off or due lI)

its crushing Thus while calculating tJyen strength of a rivet we see as to how much resistance it can offer The resistance offered by a rivet is known as its strength or the value of the rivet Following Ivu values of a rivet are imponant from the subject point of view

1 Shearing value 2 Bearing value t Sheoring laUI

The resistance offered by a rivet to be sheared off is known as its shearing value MathematicaU) pull required to shear off the rivet

P = -xti2 Xt 4

where d = Diameter of the rivet and t = Safe pennissible shear stress for the rivet material

(f the rivet is in double shear (Lebull in the case of a double cover bUll joint) the corresponding

2

2918 Failure of the plates A plate may fail in many ways BuUhe following are impoItlUlt from the subject point of

1 Tearing off the plate across a row of rivets 2 Tearing off the plate at an edge

2919 Tearing off the Plate across a Row of Rivets Due to the tensile stresses in the main plates the main plate or cover plates may tear off

a row of rivets alt shown in Fig 29121n such cases we consider only one pitch length oflhe since every rivet is responsible for that much length of plate only

pp

Flg 2912 Tearing across a row of rivets

2920 Tearing off the Plate at an Edge A plate may also fail due to tearing at an edge as shown in Fig 2913 This can be avoidedmiddot

keeping the centre of the nearest rivet from theedge of the plale at least two times the diameter rivet

tpf J

Ag 2S13 Tearing off at an edge

2921 Strength of a Rivet We have seen in arts 2916 and 2917 that a rivet may fail either due to its shearing off or due lI)

its crushing Thus while calculating tJyen strength of a rivet we see as to how much resistance it can offer The resistance offered by a rivet is known as its strength or the value of the rivet Following Ivu values of a rivet are imponant from the subject point of view

1 Shearing value 2 Bearing value t Sheoring laUI

The resistance offered by a rivet to be sheared off is known as its shearing value MathematicaU) pull required to shear off the rivet

P = -xti2 Xt 4

where d = Diameter of the rivet and t = Safe pennissible shear stress for the rivet material

(f the rivet is in double shear (Lebull in the case of a double cover bUll joint) the corresponding

2

Q3 2921 Strength of s Rivet

We baveseen in arts 2916 and 29I7 that a rivet may fail either due to its sbearing off or due to

its crushing Thus while cahuJatiag tlJp strength of a nvel we see as to how much resistance it cal4

offer The resistance offered by a rivedslrnown as itsstrength or the value of (be rivCI FOllowing twQ

values of a rivet are important from the subject point of view

1 Shearing value 2 Bearing value I Sliamprring _lEU

The resistance offend by a rive~ to be sheared off is known as its s11earing value Muthematicall) pull required to shear off the riv~

P = Exd1xt 4 where d Diameter of the rivet and

t = Safe permissible sbear stress for the rivet material If the rivel IS in double shear (iebullbull in the case of a double cover butl joint) the corresponding

equation becomes

2 Blaquorring VIIIIue

Ibe resistance offend by a rlvet to be crushed is known liISmiddot tbebearing value Mathematically pull required to crush tbe rivet

PIgt OJ xrgtltd where (11) = Safe permissible bearing stress for the rivet material

I = Thickness of the main plate and d = Diameter of the rivet

NOTE Somet the bearing Slnmgtb ofa rivet $ also termed as its crushing stnngtb And pennissible bearin~ stress is tenned aspennissiblc crushing streS (0)

~22__~trength of a Plate We have seen in Art 2919 that the plate may fail due to its tearing off across the row of rivets

Thus while calculating tbe strength of a plate we see as to how much reistance it can offer This 1eSu(a1lce offeredby the plate against tearing off is known as the strength of the plareor value of the plate Mathematically pull required to tear off the plate

P = 0t (p-d) I where a

f = Pennissible tensile stre s fot the plate material

p = Pitch of the rivet and I = ThiCkness of the plate

2923 Strength of a Riveted Joint The strength of n joint may be definild as themaxinium force which it can 1ransmit without

causing it to faiL We have seen in Arts 2921 IUld 2922lhat p~ PIgt andP are die pulls required to shear of the rivet crushingofiberlvet and tearing of theplllte A little conIidera1ioo will show dult if we go on increasing the pun od a rivetedjoint itwill fail when the least oflhesethreepulls is reached because a higher value of the otherpuUs will never reach since tbe jOinthao already failed either by shearing of the rivet or by crusJiUng of the rivet or by tearing of the plate

fthc joint is continuous (as in the caseofboilers) the strength is calculated perpitch length But if the length of lhe joint u small the strength is calculaIed for the whole length of the plate

2924 Efficiency of sRlveted Joint ~ The efficiency of a riveted joint is the dtiO of the strength of the joint to lhe strength of the

unriveted plate Mathematically

Least of P Po und PEfficiency Tl = P

where [ = CJXPXI

p = Pull required to tear off the unriveted plate Pennissibte tensile stress for the plate materialdeg1 = Pitch of the rivets andP =

t = Thickness of the plate

3

Q4 ExAMPtE291 A single riveted lap joim i~ l1WiJe in J2 mm thick plates with 22111ll~~wm

rivets Determine the strength of the rivet if the pit~h of the rillets is 60 mm TaIw stresses ttl shearing as 60 MPtl in bearing as 150MPa and iHearing as 80 MPa relive(~tiVl

~ Gtven Thickness of plates (t) = 12 mm Diameter of rivets (41 =22mm Pitch 60 mm AUowableshear stress (t) 60 MFa 6() Nmm2

Allowable stret in bearing(ab) = 150 Nmm2 and allowable stress in tearing (at) 80 MPa 80 Nmm

First of all let us calculate the pulls required for shearing and crushing of the rivets a well teanng of the maIO plates

heariltJf 01 TiICr We know that in a lap joint the rivets are in single shear Thus in a riveted lap ]OinL the strength of one rivet in single hear i taken Therefore shearing the rivet

P tx 14 X (d) =60x ~ X (22)2 22 810 N = 2281 kNbull 4 2 Bearing of riyenets We know that in a single riveted joinl the strength ofone rivet in be1llrinri

taken Therefore bearing strength of the riVet

Ph = atgtxtxJ=150x12x22=39600N=396kN

3 Tearing i1f tJlt plaM We atsoknow mal in a continuous joint the strength ofone pitch taken Therefore tearing strength of the main plate

PI = (il (p -4) t= 80 (60- 22) 12 = 36480N =3648 kN

Thus stre1llttth afthe rivet joint is the least of the above mentioned three values P~ P and Pr 2281 kli Ans

Q5 ~925 Design of a Riveted Joint

__~ bull _ _~_~bull bullbull _ bullbull _ bullbull ~ __~A__ ~______ ___ ~_

The design of riveted joints is an important job in one modern d~igJl offi~e these days A fauiry de~ign can Itad to lot many complication Whilc designing a riveted joint for structural use we usually tnilke die foU)wing assumptions

1 The load anjoint is equally shared by allthcrivets 2 Initial tensile or shearing stress in the rivets is neglected

3 Frictional forces berween the plates are neglected

t Piates are rigid

5 The shearing stress in aU the rivets is unifoml 6 The bearing stress is unifoml 7 Bending of rivels is neglected

4

Q6 2920 Eccentric Riveted Connections ---~--

mOle previous3rtUles wehave~the~ when1herivetltbaveoo1y toresilttbeljneardi$p~ of the plates But sometimes a riVet tgay havettgtn$ist rotary displacement also in addition to lin~t displacement

Consider abrackel eonnection middotwJtnJI~middotmiddotglrtJel~$gt shown in Fig 2917 it may be rioted UIllli$IJlUlIIlf~ have to resist the followfug two types of displacement

iLinear disp13cementand 1 Rotary displacement

1 UteM di$pltlc~l

All the rivetS have to offer resistatlee torbilinear displacement due to tbeload P~ 1l1i~resiitanceie fonre is assumed to be unifonn for all the rivets andiSequltl to

n where P = Toral JoadontheJoint and n = No ofriyetsonthejoint 2 RVltlT) tlipoundplacemeul

All the rivets of the joint have aha tooffer~Si$llhcentet01he rotary disPlacement due to the load P FoUo~ing two assumptions are made for findiII~ourtbereSi~ to the rotary displaceJIlcnt

1 The force on a rivet to beresistedisproportlonallQmiddotthe distance of the centre of the rivet tromthecentreofa)1 the rivets (iebull centroid ofthe rivet system)

2 The direction of the foree on a rivet to be resisted is perpendicular to the line joining theC$lttCoftbetivet and the centre Of all the rivets (iebullbull am~dof the rivet system)

Now consider an eccentric riveted connection a shown in Fig 2918

Let P = Eccentric load on the Fig 2918 joint

e = Ecccntricityofthe load ie thedisrance betwtentheJme of action of the load and the centroid of the rivetsystem iebull G

Consider a ri vet at a distance r from the cenwjdof the rNa system G The forceto be resisted by this rivet due to themomtmr of P x e ie load X distance) is directly proportional to the distance between it centremiddot and Gie

P

= kr (1) wherek is a constant

The momentofthismiddotresistanceabQul G =PRxrkrxr=krl

and total moments of resistance by all the rivets about G

=ni =kli (0

5

Q4 ExAMPtE291 A single riveted lap joim i~ l1WiJe in J2 mm thick plates with 22111ll~~wm

rivets Determine the strength of the rivet if the pit~h of the rillets is 60 mm TaIw stresses ttl shearing as 60 MPtl in bearing as 150MPa and iHearing as 80 MPa relive(~tiVl

~ Gtven Thickness of plates (t) = 12 mm Diameter of rivets (41 =22mm Pitch 60 mm AUowableshear stress (t) 60 MFa 6() Nmm2

Allowable stret in bearing(ab) = 150 Nmm2 and allowable stress in tearing (at) 80 MPa 80 Nmm

First of all let us calculate the pulls required for shearing and crushing of the rivets a well teanng of the maIO plates

heariltJf 01 TiICr We know that in a lap joint the rivets are in single shear Thus in a riveted lap ]OinL the strength of one rivet in single hear i taken Therefore shearing the rivet

P tx 14 X (d) =60x ~ X (22)2 22 810 N = 2281 kNbull 4 2 Bearing of riyenets We know that in a single riveted joinl the strength ofone rivet in be1llrinri

taken Therefore bearing strength of the riVet

Ph = atgtxtxJ=150x12x22=39600N=396kN

3 Tearing i1f tJlt plaM We atsoknow mal in a continuous joint the strength ofone pitch taken Therefore tearing strength of the main plate

PI = (il (p -4) t= 80 (60- 22) 12 = 36480N =3648 kN

Thus stre1llttth afthe rivet joint is the least of the above mentioned three values P~ P and Pr 2281 kli Ans

Q5 ~925 Design of a Riveted Joint

__~ bull _ _~_~bull bullbull _ bullbull _ bullbull ~ __~A__ ~______ ___ ~_

The design of riveted joints is an important job in one modern d~igJl offi~e these days A fauiry de~ign can Itad to lot many complication Whilc designing a riveted joint for structural use we usually tnilke die foU)wing assumptions

1 The load anjoint is equally shared by allthcrivets 2 Initial tensile or shearing stress in the rivets is neglected

3 Frictional forces berween the plates are neglected

t Piates are rigid

5 The shearing stress in aU the rivets is unifoml 6 The bearing stress is unifoml 7 Bending of rivels is neglected

4

Q6 2920 Eccentric Riveted Connections ---~--

mOle previous3rtUles wehave~the~ when1herivetltbaveoo1y toresilttbeljneardi$p~ of the plates But sometimes a riVet tgay havettgtn$ist rotary displacement also in addition to lin~t displacement

Consider abrackel eonnection middotwJtnJI~middotmiddotglrtJel~$gt shown in Fig 2917 it may be rioted UIllli$IJlUlIIlf~ have to resist the followfug two types of displacement

iLinear disp13cementand 1 Rotary displacement

1 UteM di$pltlc~l

All the rivetS have to offer resistatlee torbilinear displacement due to tbeload P~ 1l1i~resiitanceie fonre is assumed to be unifonn for all the rivets andiSequltl to

n where P = Toral JoadontheJoint and n = No ofriyetsonthejoint 2 RVltlT) tlipoundplacemeul

All the rivets of the joint have aha tooffer~Si$llhcentet01he rotary disPlacement due to the load P FoUo~ing two assumptions are made for findiII~ourtbereSi~ to the rotary displaceJIlcnt

1 The force on a rivet to beresistedisproportlonallQmiddotthe distance of the centre of the rivet tromthecentreofa)1 the rivets (iebull centroid ofthe rivet system)

2 The direction of the foree on a rivet to be resisted is perpendicular to the line joining theC$lttCoftbetivet and the centre Of all the rivets (iebullbull am~dof the rivet system)

Now consider an eccentric riveted connection a shown in Fig 2918

Let P = Eccentric load on the Fig 2918 joint

e = Ecccntricityofthe load ie thedisrance betwtentheJme of action of the load and the centroid of the rivetsystem iebull G

Consider a ri vet at a distance r from the cenwjdof the rNa system G The forceto be resisted by this rivet due to themomtmr of P x e ie load X distance) is directly proportional to the distance between it centremiddot and Gie

P

= kr (1) wherek is a constant

The momentofthismiddotresistanceabQul G =PRxrkrxr=krl

and total moments of resistance by all the rivets about G

=ni =kli (0

5

Q6 2920 Eccentric Riveted Connections ---~--

mOle previous3rtUles wehave~the~ when1herivetltbaveoo1y toresilttbeljneardi$p~ of the plates But sometimes a riVet tgay havettgtn$ist rotary displacement also in addition to lin~t displacement

Consider abrackel eonnection middotwJtnJI~middotmiddotglrtJel~$gt shown in Fig 2917 it may be rioted UIllli$IJlUlIIlf~ have to resist the followfug two types of displacement

iLinear disp13cementand 1 Rotary displacement

1 UteM di$pltlc~l

All the rivetS have to offer resistatlee torbilinear displacement due to tbeload P~ 1l1i~resiitanceie fonre is assumed to be unifonn for all the rivets andiSequltl to

n where P = Toral JoadontheJoint and n = No ofriyetsonthejoint 2 RVltlT) tlipoundplacemeul

All the rivets of the joint have aha tooffer~Si$llhcentet01he rotary disPlacement due to the load P FoUo~ing two assumptions are made for findiII~ourtbereSi~ to the rotary displaceJIlcnt

1 The force on a rivet to beresistedisproportlonallQmiddotthe distance of the centre of the rivet tromthecentreofa)1 the rivets (iebull centroid ofthe rivet system)

2 The direction of the foree on a rivet to be resisted is perpendicular to the line joining theC$lttCoftbetivet and the centre Of all the rivets (iebullbull am~dof the rivet system)

Now consider an eccentric riveted connection a shown in Fig 2918

Let P = Eccentric load on the Fig 2918 joint

e = Ecccntricityofthe load ie thedisrance betwtentheJme of action of the load and the centroid of the rivetsystem iebull G

Consider a ri vet at a distance r from the cenwjdof the rNa system G The forceto be resisted by this rivet due to themomtmr of P x e ie load X distance) is directly proportional to the distance between it centremiddot and Gie

P

= kr (1) wherek is a constant

The momentofthismiddotresistanceabQul G =PRxrkrxr=krl

and total moments of resistance by all the rivets about G

=ni =kli (0

5

TIlis total must be equal to the moment of the lOud ahout G

Pl wn Equating equations Iii) and (Uil

or k p t ) ~ r

Substituting the value of k in equation (il

bullbull(i-)

1f x and yare the co-ordinates of rivet (taking Gas originl then

c + y

Substituting the value of rl in e~uation (iv)

Per PR 2 (x- + y-)

The dirccion of this force will be at right angles to the line joining the centre of the rivet and (Cntroid of the rivet system Since this force is dire(~ly proportional to r (a is clear iTom the relation) therefore the rivet which is at the forrhtst dist1l1tc from the centroid of the rivet system will be subjected to the maximumfClrce It is mus obvious that while cakulating the resistance of a rivet or mfety of the connection a rivet which is at th~ farthest distance from G is studied because aU the other ri vets will be suhjected to a legter forte than the farthe one

The load PR is resolved horizomally as well as vertically The resultant load on a rivel will be

given by the relation

where W Horizontal component of Piland

LV PI j Vertical component of PRo

Norc Care should ulwnys be taken for the +Ie or -Ie sign In generu if the rivel under considerulioll is between G and the load +ve sign is takemiddotn But if the ooel i~away from G then-vI loign i~ taken

6

-r--J 341 introduction

A tnutural member subjected to an axial c(Ympregtltive force is called a strut As per dtfishynition a ~tnlt may be horizontal inclined Of even verticaL BUI a vertical stnn used in buildings or fnllnes is middotcalled a column

Q8 J~----~----~------~--~----------~ ~_~_~~fe~_~~lu~~Tbeo~

The flr1 lahonal attempL to studythestabilityoflollg coiulllrnS wa made by Mr Etder He derjved in equation for the huclding load ()fIong columns based 00 the bending stress While derivshying tllis equution he effectof dlrettstressrs~eeted This may be justified with the statement lOOt the direcl stress induced in a long colwnn isnegflgibleas compared to the bending stress It may be noted that the Etilets fomnllacrumot be used inthe case of sOOn coiunms bocause the direct stress is considerable and hence cannOt benegl~

~~~__A~~mp~i~ls n 111amp_ Euter~$ Colu~n ~~ The foHowing simplifying assumption~ aemade in the Etliers column theory

L Iuniall] lllc colulln i~ perfectly straight and the load ~pphcd i truly axial

Tilt (f(1gt ltcctfon it the column ii uniform t1r)UghltiU ib lengih

1)( c-)Iumn matria is perf~tly elastic homJgenc(ju and isotropic and fuUli obeys Hookes Jaw

a The knglh of column i very farge as compared t~ illgt cmslt-sltctional dimensions

- The~hortening ofcolumn duetQ direct compregtltion (being very small) is neglected

6 The failure of column occurs due to buckling alone

7

Q9

11 Eulermiddots Formula lind b(lentaength of a Column In the previous articles we rujVederived tOO relations (or the aiP9UinggtJ~ under vnous ~nd

conditions SOmetimes all by ageDeca1 eqUation C8lledEulcrs fonnula

whereL~ is the equivalent lengthofcolumn The is another way of representing the equatiODtforthe cripplin~loadt)y anequivalcnt Jength

of cffecti~ length ofa oolnmn ibeequivalent lcDgthofagiven column witllgiven end Conditions is the length of an equivalent coiumn of the same~iiatand cross~section wilhbotb ends hinged and having me value of the cripp1ingloed equal to that of thegiveo column

The equivalent lengths (L) fortbe given endcopditioos are given below

NcrT1 The vertka1 column will have two momeniS~mcma (~ilt In and Lw Sincethe colwnn will tend to buctde in the direction of leas moment ()f~ tletetoo he eati value of1be twO momenlsof ine11la i~ to be used in tIhc nlaIiOll

3412 Slenderness Ratto ~-------~

We have already discussed in Art 34nth$ltitlj~ee~formula forth~crippling load Jt2 El

PE = i 2 (i) t

We know that thebuckling~a~lumn~1he~pllIigload wi1l~~PlitceabpUfllaxis pf least resistance Now sUbStituting I =Akl (where A is the area and k is the least radius of gYJatiQnof the section) in the above equati0llbull

Jt2 E(Ak~t -p2EI1 (ii)

L2 (~ Ji where E isknowll as slendernessraUo ThU$gtlendeffiessratiois defined as ratio of equivalent(or

k unsuppoited) length of column tolbeleast radjus ofgyration of the section

Slenderness ratio does not h~veanY uriit~ Nmf It may be noted that the fOnnul(1or crlppiing Ipid in the pervious articles balcJxxu aerlveaorlthe

assumption the the sendeJIl~~r4io iSW~ that the failure of the CQ1Utin~~nly~~1O bending the effect of direct stress (ie ~lbclngnegligible

8

SNo E7ul conditi~ll$

B(th ends hinged

One end fixed and the ~

Both end~ fixed

free

4 One end fixed and the other hinged

--Relationmiddotbetwi eqllivaiimr Crippling load (P)

andaclIwllellglll (J

L~=J

IL

fL=-1(

I Lt= fi

H The ertlal column will hgtlVC two ll(gtlt1en~ of iooltia (VIbull J and Lyy) Since ti1e column VilI tend to buckle tI lhe directiQu of lea ummem t)f inenia hrefore the ealt value of the two moments of inertia is 10 ht ued in Ilhc relation

9

307 Technical Terms Before entering imo the details of the welded joints let us

discuss some of the tedmical tenTIS wbich are important poundrom the subjecf point of view

L 1~t1 ~j the lyenld ~The sides cOlluuning the right angle are caned legs of the weld In Pig 304 the sides AS and Be are called the legs of the weld Fig 304

l sitl oJtkl filkt wWl

The mininmm length of the leg of a weld is called size of the weld In Pig 304 the side is called size of the fillet weld

t ThIQt tlti~lt

It is the pupendicular distance between the comer and hypotenuse of the weld cross-section as shown in Fig 304 The thickness of reinforcement is not included in the d1roat thickness Effective throat thickness = k x minimum leg length The value of k for different angles between fusion faces is gilen below

Angle 601090deg 91 Co 100 101tolO6 107deg to U3~ 114deg to 120

k 07 065 000 055 050

un I The fillet weld shook not be used for connccring parts whose fusion faces make an angle less than 6(r or more thllll 120

2 1f no nngJebetv1en fusion faces is given it is taken as 90 and the value of k ill taken lIS 07

4 poundffo~ lrNi~ of the wdd

The actuallcngth of the weld whicll iso specified size and required thickness is called effecshytive length of the weld For the design purpose the effective length of a weld is taken as the actual length of the weld minus twice the si7e of the weld

gt Sj(l~ jilMl ~flftJ

The fiilet weld placed pandlel to the diTeCion of me force is caJIed 11 silkfiIkt Wfld

In Fig 305 the welds A and C are the side fillet welds

t Em 1l1~1 p]d

The fdlet weld placed at the end of the memshyber so that i[ is perpendicular to the force is called fndfllJel wtM In fig 305 the weld B is

Fig 305the end fiUe weld

10

Q13 I Ir-------------------------------------------------------------~

1--- 3010 Eccentric Welded Joints In tbeprevious articles webave discussedtbe cases where the weld bas to resist only the linear

displacement of the plate or metJ]ber ~ BUt simletimes a weld may have to offer resistance to bending or torsion inaddUion to the uneardispIaeementQt the plate Sucll welded connections are called eccentric welded connectiOns Though there are many types of welded connections yet the following two are important from the subject point of view

1 Eccentric welded joint subjected to moment and

2 Eccentric welded joint subjected to torSion

Q14amp15 1---___-____fea~S of Composite Section (FIftChed Beams)

A composite section may be defined as a sediooinadc up of two or lllOIC diffcrentmaterials joined together in such a manner thaI they bebave like a single piece and each material bends to the iiame radius of curvature Such beams are used when Ii beam ofone material if used alone requires quite a large ltross-sectional area which does t10tstlit thespaee available A material is1heDreioforced with some other marerial of bigher ~ in crdjilrJo~the cross-sectional area of the beam and to suit Ihe space available (as is done iMOO c8Se ofteinforcedcement conctefebeams)

In such cases thetotal moment ofremtanlewnI~~to ilie sum Gfthemoments ofindividual sections

Consider a beam of a composite section made up of~Qifferent materials as shown in Fig 1512 Let EI = Modu1nsof~cityofpart I

i I = Moment ofiooctia of the part ] M I = Moment of resistance for pan I cr I Stress in part 1 Zj = Modnlusof section for part I

E~ 11 M1bull 0Z ~ = Correspanding values for part 2 and R = Radius of the bend up be-un

Fig 1512 We know that the moment ofresisnmce for beam J

M j = 01 xZ1 (M=crxZ)

Simliarly M) = Oz X Z2

Total moment of resistance of the composite section M Ml + Ml = (crj x Zj) + (02 x Zz) W

We also know that at any distance from the neutral axis the strain in both the materials will be the

arne OJ

pound 02

~ or ~ a 1 = shyE x 0 III X 0- bull

where m= i ie Modulus ratio pound2

From the above two relations we can fmd out lhe total moment of rejStance of a composite beam or SIre5Ses in the two materials But if the sections of both the materials are not synunetrical then one area of the componenfs isc(tVerted into an equivalent area of the other

1 1

-r--J 341 introduction

A tnutural member subjected to an axial c(Ympregtltive force is called a strut As per dtfishynition a ~tnlt may be horizontal inclined Of even verticaL BUI a vertical stnn used in buildings or fnllnes is middotcalled a column

Q8 J~----~----~------~--~----------~ ~_~_~~fe~_~~lu~~Tbeo~

The flr1 lahonal attempL to studythestabilityoflollg coiulllrnS wa made by Mr Etder He derjved in equation for the huclding load ()fIong columns based 00 the bending stress While derivshying tllis equution he effectof dlrettstressrs~eeted This may be justified with the statement lOOt the direcl stress induced in a long colwnn isnegflgibleas compared to the bending stress It may be noted that the Etilets fomnllacrumot be used inthe case of sOOn coiunms bocause the direct stress is considerable and hence cannOt benegl~

~~~__A~~mp~i~ls n 111amp_ Euter~$ Colu~n ~~ The foHowing simplifying assumption~ aemade in the Etliers column theory

L Iuniall] lllc colulln i~ perfectly straight and the load ~pphcd i truly axial

Tilt (f(1gt ltcctfon it the column ii uniform t1r)UghltiU ib lengih

1)( c-)Iumn matria is perf~tly elastic homJgenc(ju and isotropic and fuUli obeys Hookes Jaw

a The knglh of column i very farge as compared t~ illgt cmslt-sltctional dimensions

- The~hortening ofcolumn duetQ direct compregtltion (being very small) is neglected

6 The failure of column occurs due to buckling alone

7

Q9

11 Eulermiddots Formula lind b(lentaength of a Column In the previous articles we rujVederived tOO relations (or the aiP9UinggtJ~ under vnous ~nd

conditions SOmetimes all by ageDeca1 eqUation C8lledEulcrs fonnula

whereL~ is the equivalent lengthofcolumn The is another way of representing the equatiODtforthe cripplin~loadt)y anequivalcnt Jength

of cffecti~ length ofa oolnmn ibeequivalent lcDgthofagiven column witllgiven end Conditions is the length of an equivalent coiumn of the same~iiatand cross~section wilhbotb ends hinged and having me value of the cripp1ingloed equal to that of thegiveo column

The equivalent lengths (L) fortbe given endcopditioos are given below

NcrT1 The vertka1 column will have two momeniS~mcma (~ilt In and Lw Sincethe colwnn will tend to buctde in the direction of leas moment ()f~ tletetoo he eati value of1be twO momenlsof ine11la i~ to be used in tIhc nlaIiOll

3412 Slenderness Ratto ~-------~

We have already discussed in Art 34nth$ltitlj~ee~formula forth~crippling load Jt2 El

PE = i 2 (i) t

We know that thebuckling~a~lumn~1he~pllIigload wi1l~~PlitceabpUfllaxis pf least resistance Now sUbStituting I =Akl (where A is the area and k is the least radius of gYJatiQnof the section) in the above equati0llbull

Jt2 E(Ak~t -p2EI1 (ii)

L2 (~ Ji where E isknowll as slendernessraUo ThU$gtlendeffiessratiois defined as ratio of equivalent(or

k unsuppoited) length of column tolbeleast radjus ofgyration of the section

Slenderness ratio does not h~veanY uriit~ Nmf It may be noted that the fOnnul(1or crlppiing Ipid in the pervious articles balcJxxu aerlveaorlthe

assumption the the sendeJIl~~r4io iSW~ that the failure of the CQ1Utin~~nly~~1O bending the effect of direct stress (ie ~lbclngnegligible

8

SNo E7ul conditi~ll$

B(th ends hinged

One end fixed and the ~

Both end~ fixed

free

4 One end fixed and the other hinged

--Relationmiddotbetwi eqllivaiimr Crippling load (P)

andaclIwllellglll (J

L~=J

IL

fL=-1(

I Lt= fi

H The ertlal column will hgtlVC two ll(gtlt1en~ of iooltia (VIbull J and Lyy) Since ti1e column VilI tend to buckle tI lhe directiQu of lea ummem t)f inenia hrefore the ealt value of the two moments of inertia is 10 ht ued in Ilhc relation

9

307 Technical Terms Before entering imo the details of the welded joints let us

discuss some of the tedmical tenTIS wbich are important poundrom the subjecf point of view

L 1~t1 ~j the lyenld ~The sides cOlluuning the right angle are caned legs of the weld In Pig 304 the sides AS and Be are called the legs of the weld Fig 304

l sitl oJtkl filkt wWl

The mininmm length of the leg of a weld is called size of the weld In Pig 304 the side is called size of the fillet weld

t ThIQt tlti~lt

It is the pupendicular distance between the comer and hypotenuse of the weld cross-section as shown in Fig 304 The thickness of reinforcement is not included in the d1roat thickness Effective throat thickness = k x minimum leg length The value of k for different angles between fusion faces is gilen below

Angle 601090deg 91 Co 100 101tolO6 107deg to U3~ 114deg to 120

k 07 065 000 055 050

un I The fillet weld shook not be used for connccring parts whose fusion faces make an angle less than 6(r or more thllll 120

2 1f no nngJebetv1en fusion faces is given it is taken as 90 and the value of k ill taken lIS 07

4 poundffo~ lrNi~ of the wdd

The actuallcngth of the weld whicll iso specified size and required thickness is called effecshytive length of the weld For the design purpose the effective length of a weld is taken as the actual length of the weld minus twice the si7e of the weld

gt Sj(l~ jilMl ~flftJ

The fiilet weld placed pandlel to the diTeCion of me force is caJIed 11 silkfiIkt Wfld

In Fig 305 the welds A and C are the side fillet welds

t Em 1l1~1 p]d

The fdlet weld placed at the end of the memshyber so that i[ is perpendicular to the force is called fndfllJel wtM In fig 305 the weld B is

Fig 305the end fiUe weld

10

Q13 I Ir-------------------------------------------------------------~

1--- 3010 Eccentric Welded Joints In tbeprevious articles webave discussedtbe cases where the weld bas to resist only the linear

displacement of the plate or metJ]ber ~ BUt simletimes a weld may have to offer resistance to bending or torsion inaddUion to the uneardispIaeementQt the plate Sucll welded connections are called eccentric welded connectiOns Though there are many types of welded connections yet the following two are important from the subject point of view

1 Eccentric welded joint subjected to moment and

2 Eccentric welded joint subjected to torSion

Q14amp15 1---___-____fea~S of Composite Section (FIftChed Beams)

A composite section may be defined as a sediooinadc up of two or lllOIC diffcrentmaterials joined together in such a manner thaI they bebave like a single piece and each material bends to the iiame radius of curvature Such beams are used when Ii beam ofone material if used alone requires quite a large ltross-sectional area which does t10tstlit thespaee available A material is1heDreioforced with some other marerial of bigher ~ in crdjilrJo~the cross-sectional area of the beam and to suit Ihe space available (as is done iMOO c8Se ofteinforcedcement conctefebeams)

In such cases thetotal moment ofremtanlewnI~~to ilie sum Gfthemoments ofindividual sections

Consider a beam of a composite section made up of~Qifferent materials as shown in Fig 1512 Let EI = Modu1nsof~cityofpart I

i I = Moment ofiooctia of the part ] M I = Moment of resistance for pan I cr I Stress in part 1 Zj = Modnlusof section for part I

E~ 11 M1bull 0Z ~ = Correspanding values for part 2 and R = Radius of the bend up be-un

Fig 1512 We know that the moment ofresisnmce for beam J

M j = 01 xZ1 (M=crxZ)

Simliarly M) = Oz X Z2

Total moment of resistance of the composite section M Ml + Ml = (crj x Zj) + (02 x Zz) W

We also know that at any distance from the neutral axis the strain in both the materials will be the

arne OJ

pound 02

~ or ~ a 1 = shyE x 0 III X 0- bull

where m= i ie Modulus ratio pound2

From the above two relations we can fmd out lhe total moment of rejStance of a composite beam or SIre5Ses in the two materials But if the sections of both the materials are not synunetrical then one area of the componenfs isc(tVerted into an equivalent area of the other

1 1

Q9

11 Eulermiddots Formula lind b(lentaength of a Column In the previous articles we rujVederived tOO relations (or the aiP9UinggtJ~ under vnous ~nd

conditions SOmetimes all by ageDeca1 eqUation C8lledEulcrs fonnula

whereL~ is the equivalent lengthofcolumn The is another way of representing the equatiODtforthe cripplin~loadt)y anequivalcnt Jength

of cffecti~ length ofa oolnmn ibeequivalent lcDgthofagiven column witllgiven end Conditions is the length of an equivalent coiumn of the same~iiatand cross~section wilhbotb ends hinged and having me value of the cripp1ingloed equal to that of thegiveo column

The equivalent lengths (L) fortbe given endcopditioos are given below

NcrT1 The vertka1 column will have two momeniS~mcma (~ilt In and Lw Sincethe colwnn will tend to buctde in the direction of leas moment ()f~ tletetoo he eati value of1be twO momenlsof ine11la i~ to be used in tIhc nlaIiOll

3412 Slenderness Ratto ~-------~

We have already discussed in Art 34nth$ltitlj~ee~formula forth~crippling load Jt2 El

PE = i 2 (i) t

We know that thebuckling~a~lumn~1he~pllIigload wi1l~~PlitceabpUfllaxis pf least resistance Now sUbStituting I =Akl (where A is the area and k is the least radius of gYJatiQnof the section) in the above equati0llbull

Jt2 E(Ak~t -p2EI1 (ii)

L2 (~ Ji where E isknowll as slendernessraUo ThU$gtlendeffiessratiois defined as ratio of equivalent(or

k unsuppoited) length of column tolbeleast radjus ofgyration of the section

Slenderness ratio does not h~veanY uriit~ Nmf It may be noted that the fOnnul(1or crlppiing Ipid in the pervious articles balcJxxu aerlveaorlthe

assumption the the sendeJIl~~r4io iSW~ that the failure of the CQ1Utin~~nly~~1O bending the effect of direct stress (ie ~lbclngnegligible

8

SNo E7ul conditi~ll$

B(th ends hinged

One end fixed and the ~

Both end~ fixed

free

4 One end fixed and the other hinged

--Relationmiddotbetwi eqllivaiimr Crippling load (P)

andaclIwllellglll (J

L~=J

IL

fL=-1(

I Lt= fi

H The ertlal column will hgtlVC two ll(gtlt1en~ of iooltia (VIbull J and Lyy) Since ti1e column VilI tend to buckle tI lhe directiQu of lea ummem t)f inenia hrefore the ealt value of the two moments of inertia is 10 ht ued in Ilhc relation

9

307 Technical Terms Before entering imo the details of the welded joints let us

discuss some of the tedmical tenTIS wbich are important poundrom the subjecf point of view

L 1~t1 ~j the lyenld ~The sides cOlluuning the right angle are caned legs of the weld In Pig 304 the sides AS and Be are called the legs of the weld Fig 304

l sitl oJtkl filkt wWl

The mininmm length of the leg of a weld is called size of the weld In Pig 304 the side is called size of the fillet weld

t ThIQt tlti~lt

It is the pupendicular distance between the comer and hypotenuse of the weld cross-section as shown in Fig 304 The thickness of reinforcement is not included in the d1roat thickness Effective throat thickness = k x minimum leg length The value of k for different angles between fusion faces is gilen below

Angle 601090deg 91 Co 100 101tolO6 107deg to U3~ 114deg to 120

k 07 065 000 055 050

un I The fillet weld shook not be used for connccring parts whose fusion faces make an angle less than 6(r or more thllll 120

2 1f no nngJebetv1en fusion faces is given it is taken as 90 and the value of k ill taken lIS 07

4 poundffo~ lrNi~ of the wdd

The actuallcngth of the weld whicll iso specified size and required thickness is called effecshytive length of the weld For the design purpose the effective length of a weld is taken as the actual length of the weld minus twice the si7e of the weld

gt Sj(l~ jilMl ~flftJ

The fiilet weld placed pandlel to the diTeCion of me force is caJIed 11 silkfiIkt Wfld

In Fig 305 the welds A and C are the side fillet welds

t Em 1l1~1 p]d

The fdlet weld placed at the end of the memshyber so that i[ is perpendicular to the force is called fndfllJel wtM In fig 305 the weld B is

Fig 305the end fiUe weld

10

Q13 I Ir-------------------------------------------------------------~

1--- 3010 Eccentric Welded Joints In tbeprevious articles webave discussedtbe cases where the weld bas to resist only the linear

displacement of the plate or metJ]ber ~ BUt simletimes a weld may have to offer resistance to bending or torsion inaddUion to the uneardispIaeementQt the plate Sucll welded connections are called eccentric welded connectiOns Though there are many types of welded connections yet the following two are important from the subject point of view

1 Eccentric welded joint subjected to moment and

2 Eccentric welded joint subjected to torSion

Q14amp15 1---___-____fea~S of Composite Section (FIftChed Beams)

A composite section may be defined as a sediooinadc up of two or lllOIC diffcrentmaterials joined together in such a manner thaI they bebave like a single piece and each material bends to the iiame radius of curvature Such beams are used when Ii beam ofone material if used alone requires quite a large ltross-sectional area which does t10tstlit thespaee available A material is1heDreioforced with some other marerial of bigher ~ in crdjilrJo~the cross-sectional area of the beam and to suit Ihe space available (as is done iMOO c8Se ofteinforcedcement conctefebeams)

In such cases thetotal moment ofremtanlewnI~~to ilie sum Gfthemoments ofindividual sections

Consider a beam of a composite section made up of~Qifferent materials as shown in Fig 1512 Let EI = Modu1nsof~cityofpart I

i I = Moment ofiooctia of the part ] M I = Moment of resistance for pan I cr I Stress in part 1 Zj = Modnlusof section for part I

E~ 11 M1bull 0Z ~ = Correspanding values for part 2 and R = Radius of the bend up be-un

Fig 1512 We know that the moment ofresisnmce for beam J

M j = 01 xZ1 (M=crxZ)

Simliarly M) = Oz X Z2

Total moment of resistance of the composite section M Ml + Ml = (crj x Zj) + (02 x Zz) W

We also know that at any distance from the neutral axis the strain in both the materials will be the

arne OJ

pound 02

~ or ~ a 1 = shyE x 0 III X 0- bull

where m= i ie Modulus ratio pound2

From the above two relations we can fmd out lhe total moment of rejStance of a composite beam or SIre5Ses in the two materials But if the sections of both the materials are not synunetrical then one area of the componenfs isc(tVerted into an equivalent area of the other

1 1

SNo E7ul conditi~ll$

B(th ends hinged

One end fixed and the ~

Both end~ fixed

free

4 One end fixed and the other hinged

--Relationmiddotbetwi eqllivaiimr Crippling load (P)

andaclIwllellglll (J

L~=J

IL

fL=-1(

I Lt= fi

H The ertlal column will hgtlVC two ll(gtlt1en~ of iooltia (VIbull J and Lyy) Since ti1e column VilI tend to buckle tI lhe directiQu of lea ummem t)f inenia hrefore the ealt value of the two moments of inertia is 10 ht ued in Ilhc relation

9

307 Technical Terms Before entering imo the details of the welded joints let us

discuss some of the tedmical tenTIS wbich are important poundrom the subjecf point of view

L 1~t1 ~j the lyenld ~The sides cOlluuning the right angle are caned legs of the weld In Pig 304 the sides AS and Be are called the legs of the weld Fig 304

l sitl oJtkl filkt wWl

The mininmm length of the leg of a weld is called size of the weld In Pig 304 the side is called size of the fillet weld

t ThIQt tlti~lt

It is the pupendicular distance between the comer and hypotenuse of the weld cross-section as shown in Fig 304 The thickness of reinforcement is not included in the d1roat thickness Effective throat thickness = k x minimum leg length The value of k for different angles between fusion faces is gilen below

Angle 601090deg 91 Co 100 101tolO6 107deg to U3~ 114deg to 120

k 07 065 000 055 050

un I The fillet weld shook not be used for connccring parts whose fusion faces make an angle less than 6(r or more thllll 120

2 1f no nngJebetv1en fusion faces is given it is taken as 90 and the value of k ill taken lIS 07

4 poundffo~ lrNi~ of the wdd

The actuallcngth of the weld whicll iso specified size and required thickness is called effecshytive length of the weld For the design purpose the effective length of a weld is taken as the actual length of the weld minus twice the si7e of the weld

gt Sj(l~ jilMl ~flftJ

The fiilet weld placed pandlel to the diTeCion of me force is caJIed 11 silkfiIkt Wfld

In Fig 305 the welds A and C are the side fillet welds

t Em 1l1~1 p]d

The fdlet weld placed at the end of the memshyber so that i[ is perpendicular to the force is called fndfllJel wtM In fig 305 the weld B is

Fig 305the end fiUe weld

10

Q13 I Ir-------------------------------------------------------------~

1--- 3010 Eccentric Welded Joints In tbeprevious articles webave discussedtbe cases where the weld bas to resist only the linear

displacement of the plate or metJ]ber ~ BUt simletimes a weld may have to offer resistance to bending or torsion inaddUion to the uneardispIaeementQt the plate Sucll welded connections are called eccentric welded connectiOns Though there are many types of welded connections yet the following two are important from the subject point of view

1 Eccentric welded joint subjected to moment and

2 Eccentric welded joint subjected to torSion

Q14amp15 1---___-____fea~S of Composite Section (FIftChed Beams)

A composite section may be defined as a sediooinadc up of two or lllOIC diffcrentmaterials joined together in such a manner thaI they bebave like a single piece and each material bends to the iiame radius of curvature Such beams are used when Ii beam ofone material if used alone requires quite a large ltross-sectional area which does t10tstlit thespaee available A material is1heDreioforced with some other marerial of bigher ~ in crdjilrJo~the cross-sectional area of the beam and to suit Ihe space available (as is done iMOO c8Se ofteinforcedcement conctefebeams)

In such cases thetotal moment ofremtanlewnI~~to ilie sum Gfthemoments ofindividual sections

Consider a beam of a composite section made up of~Qifferent materials as shown in Fig 1512 Let EI = Modu1nsof~cityofpart I

i I = Moment ofiooctia of the part ] M I = Moment of resistance for pan I cr I Stress in part 1 Zj = Modnlusof section for part I

E~ 11 M1bull 0Z ~ = Correspanding values for part 2 and R = Radius of the bend up be-un

Fig 1512 We know that the moment ofresisnmce for beam J

M j = 01 xZ1 (M=crxZ)

Simliarly M) = Oz X Z2

Total moment of resistance of the composite section M Ml + Ml = (crj x Zj) + (02 x Zz) W

We also know that at any distance from the neutral axis the strain in both the materials will be the

arne OJ

pound 02

~ or ~ a 1 = shyE x 0 III X 0- bull

where m= i ie Modulus ratio pound2

From the above two relations we can fmd out lhe total moment of rejStance of a composite beam or SIre5Ses in the two materials But if the sections of both the materials are not synunetrical then one area of the componenfs isc(tVerted into an equivalent area of the other

1 1

307 Technical Terms Before entering imo the details of the welded joints let us

discuss some of the tedmical tenTIS wbich are important poundrom the subjecf point of view

L 1~t1 ~j the lyenld ~The sides cOlluuning the right angle are caned legs of the weld In Pig 304 the sides AS and Be are called the legs of the weld Fig 304

l sitl oJtkl filkt wWl

The mininmm length of the leg of a weld is called size of the weld In Pig 304 the side is called size of the fillet weld

t ThIQt tlti~lt

It is the pupendicular distance between the comer and hypotenuse of the weld cross-section as shown in Fig 304 The thickness of reinforcement is not included in the d1roat thickness Effective throat thickness = k x minimum leg length The value of k for different angles between fusion faces is gilen below

Angle 601090deg 91 Co 100 101tolO6 107deg to U3~ 114deg to 120

k 07 065 000 055 050

un I The fillet weld shook not be used for connccring parts whose fusion faces make an angle less than 6(r or more thllll 120

2 1f no nngJebetv1en fusion faces is given it is taken as 90 and the value of k ill taken lIS 07

4 poundffo~ lrNi~ of the wdd

The actuallcngth of the weld whicll iso specified size and required thickness is called effecshytive length of the weld For the design purpose the effective length of a weld is taken as the actual length of the weld minus twice the si7e of the weld

gt Sj(l~ jilMl ~flftJ

The fiilet weld placed pandlel to the diTeCion of me force is caJIed 11 silkfiIkt Wfld

In Fig 305 the welds A and C are the side fillet welds

t Em 1l1~1 p]d

The fdlet weld placed at the end of the memshyber so that i[ is perpendicular to the force is called fndfllJel wtM In fig 305 the weld B is

Fig 305the end fiUe weld

10

Q13 I Ir-------------------------------------------------------------~

1--- 3010 Eccentric Welded Joints In tbeprevious articles webave discussedtbe cases where the weld bas to resist only the linear

displacement of the plate or metJ]ber ~ BUt simletimes a weld may have to offer resistance to bending or torsion inaddUion to the uneardispIaeementQt the plate Sucll welded connections are called eccentric welded connectiOns Though there are many types of welded connections yet the following two are important from the subject point of view

1 Eccentric welded joint subjected to moment and

2 Eccentric welded joint subjected to torSion

Q14amp15 1---___-____fea~S of Composite Section (FIftChed Beams)

A composite section may be defined as a sediooinadc up of two or lllOIC diffcrentmaterials joined together in such a manner thaI they bebave like a single piece and each material bends to the iiame radius of curvature Such beams are used when Ii beam ofone material if used alone requires quite a large ltross-sectional area which does t10tstlit thespaee available A material is1heDreioforced with some other marerial of bigher ~ in crdjilrJo~the cross-sectional area of the beam and to suit Ihe space available (as is done iMOO c8Se ofteinforcedcement conctefebeams)

In such cases thetotal moment ofremtanlewnI~~to ilie sum Gfthemoments ofindividual sections

Consider a beam of a composite section made up of~Qifferent materials as shown in Fig 1512 Let EI = Modu1nsof~cityofpart I

i I = Moment ofiooctia of the part ] M I = Moment of resistance for pan I cr I Stress in part 1 Zj = Modnlusof section for part I

E~ 11 M1bull 0Z ~ = Correspanding values for part 2 and R = Radius of the bend up be-un

Fig 1512 We know that the moment ofresisnmce for beam J

M j = 01 xZ1 (M=crxZ)

Simliarly M) = Oz X Z2

Total moment of resistance of the composite section M Ml + Ml = (crj x Zj) + (02 x Zz) W

We also know that at any distance from the neutral axis the strain in both the materials will be the

arne OJ

pound 02

~ or ~ a 1 = shyE x 0 III X 0- bull

where m= i ie Modulus ratio pound2

From the above two relations we can fmd out lhe total moment of rejStance of a composite beam or SIre5Ses in the two materials But if the sections of both the materials are not synunetrical then one area of the componenfs isc(tVerted into an equivalent area of the other

1 1

Q13 I Ir-------------------------------------------------------------~

1--- 3010 Eccentric Welded Joints In tbeprevious articles webave discussedtbe cases where the weld bas to resist only the linear

displacement of the plate or metJ]ber ~ BUt simletimes a weld may have to offer resistance to bending or torsion inaddUion to the uneardispIaeementQt the plate Sucll welded connections are called eccentric welded connectiOns Though there are many types of welded connections yet the following two are important from the subject point of view

1 Eccentric welded joint subjected to moment and

2 Eccentric welded joint subjected to torSion

Q14amp15 1---___-____fea~S of Composite Section (FIftChed Beams)

A composite section may be defined as a sediooinadc up of two or lllOIC diffcrentmaterials joined together in such a manner thaI they bebave like a single piece and each material bends to the iiame radius of curvature Such beams are used when Ii beam ofone material if used alone requires quite a large ltross-sectional area which does t10tstlit thespaee available A material is1heDreioforced with some other marerial of bigher ~ in crdjilrJo~the cross-sectional area of the beam and to suit Ihe space available (as is done iMOO c8Se ofteinforcedcement conctefebeams)

In such cases thetotal moment ofremtanlewnI~~to ilie sum Gfthemoments ofindividual sections

Consider a beam of a composite section made up of~Qifferent materials as shown in Fig 1512 Let EI = Modu1nsof~cityofpart I

i I = Moment ofiooctia of the part ] M I = Moment of resistance for pan I cr I Stress in part 1 Zj = Modnlusof section for part I

E~ 11 M1bull 0Z ~ = Correspanding values for part 2 and R = Radius of the bend up be-un

Fig 1512 We know that the moment ofresisnmce for beam J

M j = 01 xZ1 (M=crxZ)

Simliarly M) = Oz X Z2

Total moment of resistance of the composite section M Ml + Ml = (crj x Zj) + (02 x Zz) W

We also know that at any distance from the neutral axis the strain in both the materials will be the

arne OJ

pound 02

~ or ~ a 1 = shyE x 0 III X 0- bull

where m= i ie Modulus ratio pound2

From the above two relations we can fmd out lhe total moment of rejStance of a composite beam or SIre5Ses in the two materials But if the sections of both the materials are not synunetrical then one area of the componenfs isc(tVerted into an equivalent area of the other

1 1

152 Types of Composite -Th~~h-ili~~-~-~ ~ype$ of beams that we come across yet the followinQ important from the subject point of view

1 Beams of unsymmetrical sections

1 Beams of uniform strength

3 Fl itched beruns

153 Bfsms f~nsym_t~cal Sections We have already di~sed in the last c1Utp~ that ina symmetrical section the distance of

extreme fibre from the egofthe section y dI2~ Butthisis not the case in an lJ1lSymmetrical section (L 1 T etc) since the neutral axismiddot6fsucb a secentonJiooSnot pass through the geometrical centre of the section In such cases tllSt thetentre ofgravit)ot tlie section is obtaine-d as diSCUssed in Chapter 6 and then the values ofy in the tenSion and compression sides is srudied For obtaining the bending stress in a beam the bigger value of y (in tension or compression) is used in the equation This will be i11ustratedby the following examples

~54 ~ms of Un1fonn Strength We have already discussed tba~ ina simply suppmted beam carrying a mllif(mnly(listJrioo1te4t

load the maximum bending tnomenfWill occuratits ~tre It is thus obviousthitttbe tierlding $1re is also maximiHn atfheceillm-ofthe-fiealnAsweproceedfrom tbeamtre of tbe beam nuArrill

supports the bending moment decreases and MOO the maximtun -stress develOped kshypermissible linllit It results in theWastageof material This wastage is negligible in case of SparlS but considerable in -easeoflspan~ __

The~of lspammiddotari~t~instren a Way that their eros~onll8Ieajs iWTfil1j~ tow~ the snpports so that the _intmnbendiJtg~ developed is eqtml rolhe allQW8ble ($ is done at the centreQftbebeam~Smb a~inwhich bendingstreiS developed isconstanl is equal (0 the allowable stress at every section isca1led a beam of unifonn strength The section beam ofunifmm strength may be-Varied mtbefollowmg ways

J By keeping the widthunifontHmd varyingthe depth

2 By keeping the depth unifolnHUW varying the width

3 By varying both width and depth

The most common way of keeping the beaniQf unifonn strength is by keeping the width ltmd varying the depth

12

155 Beams of Composite sectiOn (FI1tched Beams) A composite section maybedefmedaS a seetioumade up of two or more differem materials

joined together in sucb a manner that tbcentybehave like a single piece ano each material bends to the same radius of curvature Such beams ate used when a beam ofone material if used alone requires quite a large cross--sectional area which doesnoistiitthe space available A material is then reinforced with some other mareriat of higher su-ength inorderlo reducemiddot the cross-sectional area of the beam and to suit the space available (as is done in the ca~ of reinfoned cement concrete beams)

]n such cases the total moment of te$istance will beequalto the sum of themoments of individual sections

Coasidef a beam ofa compositpoundsection made up of two differentmateriaJs as sbowoin Fig 1512shyLet E1 == Modulus of elasticity of part 1

11 = Moment of inertia of the part I M 1 = Moment of resistance for part 1

O == StresS in part 1 Z == Modwusof section for part 1

E] 12 M2bull al ~ = C6~nding values for part 2 arid R Radius of the bend up beam

We know rhal lhe moment of resiSllJnCe for beam J Fig 1512

Similarly M2 = 02 X Zz Total moment of reistance of the composite section

M = MI +M2 =(01 X ZI+ (tl1 x~) (i)

We also know that at any distance from the neutral axis the strain in both the materialgt will be the

same tll = 01 or El IEl

where m = EF ie Modulus ratio J2

From the above two relations we can find out the total moment of resistance of a composite beam or stres~s in the two materials But if the $eCrlQns of both the materials are not symmetrical then one area of the components is converted into an ~uiaIent area of fIle other

13

155 Beams of Composite sectiOn (FI1tched Beams) A composite section maybedefmedaS a seetioumade up of two or more differem materials

joined together in sucb a manner that tbcentybehave like a single piece ano each material bends to the same radius of curvature Such beams ate used when a beam ofone material if used alone requires quite a large cross--sectional area which doesnoistiitthe space available A material is then reinforced with some other mareriat of higher su-ength inorderlo reducemiddot the cross-sectional area of the beam and to suit the space available (as is done in the ca~ of reinfoned cement concrete beams)

]n such cases the total moment of te$istance will beequalto the sum of themoments of individual sections

Coasidef a beam ofa compositpoundsection made up of two differentmateriaJs as sbowoin Fig 1512shyLet E1 == Modulus of elasticity of part 1

11 = Moment of inertia of the part I M 1 = Moment of resistance for part 1

O == StresS in part 1 Z == Modwusof section for part 1

E] 12 M2bull al ~ = C6~nding values for part 2 arid R Radius of the bend up beam

We know rhal lhe moment of resiSllJnCe for beam J Fig 1512

Similarly M2 = 02 X Zz Total moment of reistance of the composite section

M = MI +M2 =(01 X ZI+ (tl1 x~) (i)

We also know that at any distance from the neutral axis the strain in both the materialgt will be the

same tll = 01 or El IEl

where m = EF ie Modulus ratio J2

From the above two relations we can find out the total moment of resistance of a composite beam or stres~s in the two materials But if the $eCrlQns of both the materials are not symmetrical then one area of the components is converted into an ~uiaIent area of fIle other

13

ExA~e 164 AnJ~Jlaquo~wiIhreclangularends liasthelol1owing~ Flanglaquo=15q ~20m1fl lb 300 nun 10 mm

FilUl rile maximum sheiinnli~d~opedin the ~amfora-shear force l1j50 iN SoIlJTlOH Given F1~WidIh(ll) lSOmm Flange thic~s 1--150--4

20 mm ~ Depth of web (d) 300 nunWtdth of web =10 rom Overall depth of the section (0) 340 mm and shearing force (f) T 0 kN 50 x 10 N I

We know that moment of inertia of the I-section about ilS centre of gravity and parallel to l-X axis 340

IIOmm

Fig1610

= 50XI0 middotmiddotmiddotflreg [(340)2 _ (300)2) + IOX(3002] Nmm2 (1163xlOb)x H) _ 8 8

= t 68 Nfmrn2= 18 MFa Am

3

14

Q25

-r-~=2 Stiffness of a Spring -~~ j

The load required to produce a unit deflection in a spring is called spring stiffness or niffness of a spring

~~ Ty~ of Springs We have already discussed that~ spring isU$edfor absorbing energy due to resilience Thus in

general me springs are of the folloWing tWo ljpeS depending upon the type of resiijenee 1 Bending spring and ~ Torsion spnng

284 Bending Springs A spring wbiehis stibjecred to bendingoolyand the resilience is also due to it is known asmiddotmiddot

bending sprillg Laminated springs or leaf~ate also called bending springs

285 Torsion Spri~~~ cgtc ~~ A spring which issub~Ji) ng moment only and the resilience is also duetO

it is known as a torsioll$p~g also called torsion springs Some springs are bull subjected to bending aswcltaa1orsf~lt

28~6 Forms of $poop ~t f _ - ~ lt

Though there are many fotnlSltrf~ringswhich are made by the manufacturers yet the following types of springs are CQmmoJibr~inamp~g practice

t Carriage springs or teafc~~ ~ Helical springs

plical Type Leaf Springs f two types (l ) semi-etlipticaltypeamp (ie simply

(2) quaner-ellipticaf(e~~ caniilever) types

~

i

Carriagg Springs or Leaf Springs

The carriage springs are widely used inmilway wagons coaches and road vehicles these days These are used to absorb shocks which give an unpleasant feeling to me passengers The en gy absorbed by a laminated spring during a shock is released immediately without doing any useful ork

A laminated spring in iLgt simplest form consists of a number of parallel strips of a meta1 having different lengths but same width and placed one over the other in laminations as shown in Fig 281

15

All the plates are initially bent to the same radius and are free 10 slide one over the other When the spring i$ loaded to the designed load all tbeplates become flat and the central deflectiltm disappears The purpose of this type of arrartgementof plates is to make the spring ofuniform strength throughout This is achieved by tapering the ends of the laminations The semi-elliptical type spring rests on the axis of the vehicle and its top plate is pinned at lheends to the chassis ofthe vehicle

Now consider a carnage spring pinned al its both ends and carrying an upward load at its centre as shown in Fig 281

Let =Span ofthe spring

t -= Thiclmess of plates

b = Width of the plates n= Number of piates

W = wad acting on the spring

(J l Maximum bending stress developed in the plates

() = OriginiIJ deflection of tbe top spring and

R Radius of the spring

W Ii 2 T

IW

Fig 281 Carriage spring A linle consideration ill shoYthat the load will be acting On the spring on the lowermostplate

and it wiII be shared equally on me two ends of the top plate as shown in Fig 281 We know that the bending moment at the centre of the span due to lhis load

WIM= (i)

4shyand momnt resisred by one plate

0 == J M =21 y l I y j

3 ox~

__1_2 amiddot bt bi t )1=- and v=shy 12 2

2 Total moment resisled by n plates

6 (

M = nab (ii6

For detailgt please refer to Art

J6

Since1he ~bei1ding~t due ~lqadisequaLtothetotalmiddotresistingmomentmiddottberefmemiddot equating (i) and (it) middotmiddotd

~ s~ lt1 1or

From the geo~ ofthe~1~we~that~celamiddot~ j

(5 = (ibullbull~~

We allro know that in the case of a bending beam _ E or R=Ey =i ( l~ y - R 0 20 y= ~

Substituting this value ofRio~ (iii) ~ l1ql2

amp --- E t 4-Et middotampx~

20

Now substitufulgtnelalue ofOtn1he aWve equation

a _ 3Wl x 1 = 3wi~ - 2nbl 4Et 8Enbf

17

All the plates are initially bent to the same radius and are free 10 slide one over the other When the spring i$ loaded to the designed load all tbeplates become flat and the central deflectiltm disappears The purpose of this type of arrartgementof plates is to make the spring ofuniform strength throughout This is achieved by tapering the ends of the laminations The semi-elliptical type spring rests on the axis of the vehicle and its top plate is pinned at lheends to the chassis ofthe vehicle

Now consider a carnage spring pinned al its both ends and carrying an upward load at its centre as shown in Fig 281

Let =Span ofthe spring

t -= Thiclmess of plates

b = Width of the plates n= Number of piates

W = wad acting on the spring

(J l Maximum bending stress developed in the plates

() = OriginiIJ deflection of tbe top spring and

R Radius of the spring

W Ii 2 T

IW

Fig 281 Carriage spring A linle consideration ill shoYthat the load will be acting On the spring on the lowermostplate

and it wiII be shared equally on me two ends of the top plate as shown in Fig 281 We know that the bending moment at the centre of the span due to lhis load

WIM= (i)

4shyand momnt resisred by one plate

0 == J M =21 y l I y j

3 ox~

__1_2 amiddot bt bi t )1=- and v=shy 12 2

2 Total moment resisled by n plates

6 (

M = nab (ii6

For detailgt please refer to Art

J6

Since1he ~bei1ding~t due ~lqadisequaLtothetotalmiddotresistingmomentmiddottberefmemiddot equating (i) and (it) middotmiddotd

~ s~ lt1 1or

From the geo~ ofthe~1~we~that~celamiddot~ j

(5 = (ibullbull~~

We allro know that in the case of a bending beam _ E or R=Ey =i ( l~ y - R 0 20 y= ~

Substituting this value ofRio~ (iii) ~ l1ql2

amp --- E t 4-Et middotampx~

20

Now substitufulgtnelalue ofOtn1he aWve equation

a _ 3Wl x 1 = 3wi~ - 2nbl 4Et 8Enbf

17

Since1he ~bei1ding~t due ~lqadisequaLtothetotalmiddotresistingmomentmiddottberefmemiddot equating (i) and (it) middotmiddotd

~ s~ lt1 1or

From the geo~ ofthe~1~we~that~celamiddot~ j

(5 = (ibullbull~~

We allro know that in the case of a bending beam _ E or R=Ey =i ( l~ y - R 0 20 y= ~

Substituting this value ofRio~ (iii) ~ l1ql2

amp --- E t 4-Et middotampx~

20

Now substitufulgtnelalue ofOtn1he aWve equation

a _ 3Wl x 1 = 3wi~ - 2nbl 4Et 8Enbf

17