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7/13/2019 Two Marks - Print

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UNIT-I

DEFLECTION OF DETERMINATE STRUCTURES

1. Why is it necessary to compute deflections in structures?

Computation of deflection of structures is necessary for the following reasons:

a. If the deflection of a structure is more than the permissible, the structure will not lookaesthetic and will cause psychological upsetting of the occupants.

b. Eessi!e deflection may cause cracking in the materials attached to the structure. "or

eample, if the deflection of a floor beam is ecessi!e, the floor finishes and partitionwalls supported on the beam may get cracked and unser!iceable.

#. What is meant by cambering techni$ue in structures?

Cambering is a techni$ue applied on site, in which a slight upward cur!e is made in thestructure%beam during construction, so that it will straighten out and attain the straight shape

during loading. &his will considerably reduce the downward deflection that may occur at later

stages.

'. (ame any four methods used for computation of deflections in structures.

1. )ouble integration method #. *acaulay+s method'. Conugate beam method -. *oment area method

. *ethod of elastic weights /. 0irtual work method )ummy unit load method

2. 3train energy method 4. Williot *ohr diagram method

-. 3tate the difference between strain energy method and unit load method in the determination of

deflection of structures.

In strain energy method, an imaginary load 5 is applied at the point where the deflection isdesired to be determined. 5 is e$uated to 6ero in the final step and the deflection is obtained.

In unit load method, an unit load 7instead of 58 is applied at the point where the deflection is

desired.

. What are the assumptions made in the unit load method?

9 1. &he eternal internal forces are in e$uilibrium.

#. 3upports are rigid and no mo!ement is possible.'. &he materials is strained well with in the elastic limit.

/. ;i!e the e$uation that is used for the determination of deflection at a gi!en point in beamsand frames.

)eflection at a point is gi!en by,lI =

due to the applied loadsm = moment at a section > due to a unit load applied at that point I and in

the direction


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ofthe

)esired displacementEI = fleural rigidity


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2. Write down the e$uations for moments due to the eternal load for beam shown in "ig.

># >'>1 1 BA >#

>'1


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1#. What is the effect of temperature on the members of a statically determinate plane truss?

In determinate structures temperature changes do not create any internal stresses. &he

changes in lengths of members may result in displacement of oints. Aut these would not result ininternal stresses or changes in eternal reactions.

1'. )istinguish between Gdeck types+ and Gthrough type+ trusses.@ deck type is truss is one in which the road is at the top chord le!el of the trusses. We

would not see the trusses when we ride on the road way.

@ through type truss is one in which the road is at the bottom chord le!el of the trusses.When we tra!el on the road way, we would see the web members of the trusses on our left and

right. &hat gi!es us the impression that we are going9 through+ the bridge.

1-. )efine static indeterminacy of a structure.

If the conditions of statics i.e., H=


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>

@ns: >

A CE

-m > >> > >

@ 1

)

@pply unit force in the hori6ontal direction at ). m !alues are tabulated as below:

5ortion m imits

)C 1 < to -m

CE 1- < to 'm

EA 1- ' to /m

A@ 1 < to -m

12. )ifferentiate the statically determinate structures and statically indeterminate structures?

3l.(o statically determinate structures statically indeterminate structures

1. Conditions of e$uilibrium are sufficient

to analy6e the structure

Conditions of e$uilibrium are insufficient to

analy6e the structure

#. Aending moment and shear force is

independent of material and cross

sectional area.

Aending moment and shear force is dependent

of material and independent of cross sectional

area.'. (o stresses are caused due to

temperature change and lack of fit.

3tresses are caused due to temperature change

and lack of fit.

14. )efine : &russed Aeam.@ beam strengthened by pro!iding ties and struts is known as &russed Aeams.

1D. )efine: Fnit load method.&he eternal load is remo!ed and the unit load is applied at the point, where the

deflection or rotation is to found.

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UNIT-II

INFLUENCE LINES

1. Where do you get rolling loads in practice?

3hifting of load positions is common enough in buildings. Aut they are more pronounced in bridgesand in gantry girders o!er which !ehicles keep rolling.

#. (ame the type of rolling loads for which the absolute maimum bending moment occurs at the midspan of

beam.7i8 3ingle concentrated load 7ii8 udl longer than the span 7iii8 udl shorter than the span 7i!8 @l

when the resultant of se!eral concentrated loads crossing a span, coincides with a concentrated load then al

the maimum bending moment occurs at the centre of the span.

'. What is meant by absolute maimum bending moment in a beam?

When a gi!en load system mo!es from one end to the other end of a girder, depending upon tposition of the load, there will be a maimum bending moment for e!ery section. &he maimum of the

bending moments will usually occur near or at the midspan. &he maimum of maimum bending moments

called the absolute maimum bending moment.

-. Where do you ha!e the absolute maimum bending moment in a simply supported beam when a series ofwheel loads cross it?

When a series of wheel loads crosses a simply supported beam, the absolute maimum bendi

moment will occur near midspan under the load Wcr , nearest to midspan 7or the hea!iest load8. If Wcrplaced to one side of midspan C, the resultant of the load system B shall be on the other side of CJ and W

and B shall be e$uidistant from C. (ow the absolute maimum bending moment will occur under W cr .

Wcr and B coincide, the absolute maimum bending moment will occur at midspan.

. What is the absolute maimum bending moment due to a mo!ing udl longer than the span of a simplysupported beam?

When a simply supported beam is subected to a mo!ing udl longer than the span, the absolute

maimum bending moment occurs when the whole span is loaded.*ma ma = w l

2

4

/. 3tate the location of maimum shear force in a simple beam with any kind of loading.In a simple beam with any kind of load, the maimum positi!e shear force occurs at the left hand

support and maimum negati!e shear force occurs at right hand support.

2. What is meant by maimum shear force diagram?

)ue to a gi!en system of rolling loads the maimum shear force for e!ery section of the girder can

worked out by placing the loads in appropriate positions. When these are plotted for all the sections of t

girder, the diagram that we obtain is the maimum shear force diagram. &his diagram yields the Gdesishear+ for each cross section.

4. What is meant by influence lines?@n influence line is a graph showing, for any gi!en frame or truss, the !ariation of any force

displacement $uantity 7such as shear force, bending moment, tension, deflection8 for all positions of

mo!ing unit load as it crosses the structure from one end to the other.

D. What are the uses of influence line diagrams?

7i8 Influence lines are !ery useful in the $uick determination of reactions, shear force, bendingmoment or similar functions at a gi!en section under any gi!en system of mo!ing loads and


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7ii8 Influence lines are useful in determining the load position to cause maimum !alue of a gi!en

function in a structure on which load positions can !ary.

1 in a simply supported beam @A of span Gl+ m

1

@ > A 7l8

7l8l K

%l

11. )raw the I) for bending moment at any section > of a simply supported beam and mark the ordinates.

1

@ > A

7l8

7 l 8

l

1#. What do you understand by the term re!ersal of stresses?

In certain long trusses the web members can de!elop either tension or compression depending uponthe position of li!e loads. &his tendency to change the nature of stresses is called re!ersal of stresses.

1'. 3tate *ullerAreslau principle.

*ullerAreslau principle states that, if we want to sketch the influence line for any force $uantity7like thrust, shear, reaction, support moment or bending moment8 in a structure,

7i8 We remo!e from the structure the resistant to that force $uantity and

7ii8 We apply on the remaining structure a unit displacement corresponding to that force $uantity.&he resulting displacements in the structure are the influence line ordinates sought.

1-. 3tate *awellAetti+s theorem.@ A C

l l

B@ BA BC

1

In a linearly elastic structure in static e$uilibrium acted upon by either of two systems of eternal

forces, the !irtual work done by the first system of forces in undergoing the displacements caused by the

second system of forces is e$ual to the !irtual work done by the second system of forces in undergoing the

displacements caused by the first system of forces.


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*awell Aetti+s theorem helps us to draw influence lines for structures.

1. What is the necessity of model analysis?

7i8 When the mathematical analysis of problem is !irtually impossible.

7ii8 *athematical analysis though possible is so complicated and time consuming that the moanalysis offers a short cut.

7iii8 &he importance of the problem is such that !erification of mathematical analysis by an actu

test is essential.

1/. )efine similitude.

3imilitude means similarity between two obects namely the model and the prototype with regard to

their physical characteristics:

;eometric similitude is similarity of form

inematic similitude is similarity of motion

)ynamic and%or mechanical similitude is similarity of masses and%or forces.

12. 3tate the principle on which indirect model analysis is based.

&he indirect model analysis is based on the *uller Areslau principle.

*uller Areslau principle has lead to a simple method of using models of structures to get the

influence lines for force $uantities like bending moments, support moments, reactions, internal shears,thrusts, etc.

&o get the influence line for any force $uantity, 7i8 remo!e the resistant due to the force, 7ii8 apply aunit displacement in the direction of the 7iii8 plot the resulting displacement diagram. &his diagram is the

influence line for the force.

14. What is the principle of dimensional similarity?)imensional similarity means geometric similarity of form. &his means that all homologous

dimensions of prototype and model must be in some constant ratio.

1D. What is Aegg+s deformeter?

Aegg+s deformeter is a de!ice to carry out indirect model analysis on structures. It has the facility toapply displacement corresponding to moment, shear or thrust at any desired point in the model. In addition,pro!ides facility to measure accurately the conse$uent displacements all o!er the model.

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UNIT-III

ARCHES

1. What is an arch? Eplain.@n arch is defined as a cur!ed girder, ha!ing con!eity upwards and supported at its ends.

&he supports must effecti!ely arrest displacements in the !ertical and hori6ontal directions. nly then there

will be arch action.

#. What is a linear arch?If an arch is to take loads, say W1, W#, and W' 7fig8 and a 0ector diagram and funicular polygon are

plotted as shown, the funicular polygon is known as the linear arch or theoretical arch.

p

W#

W1 W#W'

$ L B

W1 ) W'

5 L B 3 t 5 C E 3

r

s

H

0ector)iagram

@ &

A 3pace

)iagram

&he polar distance Got+ represents the hori6ontal thrust. &he links @C, C), )E, and EA will be und

compression and there will be no bending moment. If an arch of this shape @C)EA is pro!ided, the

will be no bending moment."or a gi!en set of !ertical loads W1, W#..etc., we can ha!e any number of linear arches

depending on where we choose G+ or how much hori6ontal thrust 7ot8 we choose to introduce.

'. 3tate Eddy+s theorem.Eddy+s theorem states that M&he bending moment at any section of an arch is proportional to

the !ertical intercept between the linear arch 7or theoretical arch8 and the centre line of the actual

arch.N

A* = rdinate #' scale factor

>

W#W1

W'o# @ctual arch


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o' &heoretical

arch o1

>

-. What is the degree of static indeterminacy of a three hinged parabolic arch?

"or a three hinged parabolic arch, the degree of static indeterminancy is 6ero. It isstatically determinate.


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. Eplain with the aid of a sketch, the normal thrust and radial shear in an arch rib.

H @ A H

B

(

et us take a section > of an arch. 7fig 7a8 8. et be the inclination of the tangent at >. If H is thori6ontal thrust and 0 the !ertical shear at >, from the free body of the BH3 of the arch, it is clear thatand H will ha!e normal and radial components gi!en by,

(= H cos K 0 sin

B= 0 cos H sin

/. Which of the two arches, !i6. circular and parabolic is preferable to carry a uniformly distributed load? Wh

5arabolic arches are preferably to carry distributed loads. Aecause, both, the shape of the arch and tshape of the bending moment diagram are parabolic. Hence the intercept between the theoretical arch a

actual arch is 6ero e!erywhere. Hence, the bending moment at e!ery section of the arch will be 6ero. &arch will be under pure compression which will be economical.

2. What is the difference between the basic action of an arch and a suspension cable?

@n arch is essentially a compression member which can also take bending moments and shears.Aending moments and shears will be absent if the arch is parabolic and the loading uniformly distributed.

@ cable can take only tension. @ suspension bridge will therefore ha!e a cable and a stiffening girde

&he girder will take the bending moment and shears in the bridge and the cable, only tension.Aecause of the thrusts in the cables and arches, the bending moments are considerably reduced.

If the load on the girder is uniform, the bridge will ha!e only cable tension and no bending momenton the girder.

4. Fnder what conditions will the bending moment in an arch be 6ero throughout.

&he bending moment in an arch throughout the span will be 6ero, if

7i8 the arch is parabolic and 7ii8 the arch carries uniformly distributed load throughout the span.

D. )raw the I) for bending moment at a section > at a distance from the left end of a three hinged

parabolic arch of span +l+ and rise Gh+.* = O Hy

Hy7K8 78

x(l-x)/ l x(l-x)/ l

1


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Considering a three hinged parabolic arch of span Gl+ and subected to a mo!ing point load W, the

position of the point load fora. *aimum negati!e bending moment is


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1. Eplain the effect of yielding of support in the case of an arch.Pielding of supports has no effect in the case of a ' hinged arch which is determinate. &hese

displacements must be taken into account when we analyse # hinged or fied arches underU = H instead of 6ero

H

F = Q0@ instead of 6ero

0@Here F is the strain energy of the arch and H and 0@ are the displacements due to yielding of supports.

1/. Write the formula to calculate the change in rise in three hinged arch if

Change in rise l#

K -r#

&

-r

where l = span length of the arch

r = central rise of the arch

= coefficient of thermal epansion& = change in temperature

12. In a parabolic arch with two hinges how will you calculate the slope of the arch at any point.

3lope of parabolic arch = = tan1

-r 7lO #x8

l#

where = 3lope at any pointx 7or8 inclination of tangent atx.l = span length of the arch

r = central rise of the arch

14. How will you calculate the hori6ontal thrust in a two hinged parabolic arch if there is a rise in temperature

Hori6ontal thrust = l &E Il

y

#

dx

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