KTU B.Tech Solved ModelQuestions -CE202 STRUCTURAL ANALYSIS- I

**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 look aesthetic and will cause psychological upsetting of the occupants.

b. Exessive deflection may cause cracking in the materials attached to the

structure. For

example, if the deflection of a floor beam is excessive, the floor finishes and partition walls supported on the beam may get cracked and unserviceable.

**2****.What is meant by cambering technique in structures?**

Cambering is a technique applied on site, in which a slight upward curve is

made in the structure/ beam during construction, so that it will straighten out and attain the straight shape during loading. This will considerably reduce the

downward deflection that may occur at later stages.

**3****.Name any four methods used for computation of deflections in structures.**

1. Double integration method 2. Macaulay’s method

3. Conjugate beam method 4. Moment area method

5. Method of elastic weights 6. Virtual work method- Dummy unit load method

7. Strain energy method 8. Williot Mohr diagram method

**4****. State the difference between strain energy method and unit load method in the determination of deflection of structures.**

In strain energy method, an imaginary load P is applied at the point where the

deflection is desired to be determined. P is equated to zero in the final step and the deflection is obtained.

In unit load method, an unit load (instead of P) is applied at the point where the deflection is desired.

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

` 1. The external & internal forces are in equilibrium.

2. Supports are rigid and no movement is possible.

3. The materials is strained well with in the elastic limit.

**6****.Distinguish between pin jointed and rigidly jointed structure.**

Pin jointed structure Rigidly jointed structure

1. The joints permit change of angle between connected member.

The members connected at a rigid joint will maintain the angle between them even under deformation due to loads.

2. The joints are incapable of transferring any moment to the connected

members and vice-versa. Members can transmit both forces and moments between themselves through the joint.

3. The pins transmit forces between connected member by developing shear. Provision of rigid joints normally increases the redundancy of the structures.

**7****.What is meant by thermal stresses**?

Thermal stresses are stresses developed in a structure/member due to change in temperature. Normally, determine structures do not develop thermal stresses.

They can absorb changes in lengths and consequent displacements without developing stresses.

**8****. What is meant by lack of fit in a truss?**

One or more members in a pin jointed statically indeterminate frame may be a

little shorter or longer than what is required. Such members will have to be forced in place during the assembling. These are called members having Lack of fit. Internal forces can develop in a redundant frame (without external loads) due to lack of fit.

**9****. Write down the two methods of determining displacements in pin jointed plane frames by the**

unit load concept.

The methods of using unit loads to compute displacements are, i) dummy unit load method.

ii) using the principle of virtual work.

**1****0****. 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. The changes in lengths of members may result in displacement of joints. But these would not result in internal stresses or changes in external reactions.

**1****1****. Distinguish between ‘deck type’ and ‘through type’ trusses.**

A deck type is truss is one in which the road is at the top chord level of the

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

A through type truss is one in which the road is at the bottom chord level of the trusses.

When we travel on the road way, we would see the web members of the trusses

on our left and right. That gives us the impression that we are going` through’ the

bridge.

**1****2****. Define static indeterminacy of a structure.**

If the conditions of statics i.e., ΣH=0, ΣV=0 and ΣM=0 alone are not sufficient

to find

either external reactions or internal forces in a structure, the structure is called a statically indeterminate structure.

**1****3****. Differentiate the statically determinate structures and statically indeterminate structures**?

statically determinate structures statically indeterminate structures

1. Conditions of equilibrium are sufficient to analyze the structure Conditions of equilibrium are insufficient to analyze the structure

2. Bending moment and shear force is independent of material and cross sectional area. Bending moment and shear force is dependent of material and independent of cross sectional area.

3. No stresses are caused due to temperature change and lack of fit. Stresses are caused due to temperature change and lack of fit.

**1****4****. Define : Trussed Beam.**

A beam strengthened by providing ties and struts is known as Trussed Beams.

**15****. Define: Unit load method.**

The external load is removed and the unit load is applied at the point, where the

deflection or rotation is to found.

**1****6****. Give the procedure for unit load method.**

1. Find the forces P1, P2, ……. in all the members due to external loads.

2. Remove the external loads and apply the unit vertical point load at the joint if the

vertical deflection is required and find the stress.

3. Apply the equation for vertical and horizontal deflection.

** INFLUENCE LINES**

1. **Where do you get rolling loads in practice**?

Shifting of load positions is common enough in buildings. But they are more pronounced in bridges and in gantry girders over which vehicles keep rolling.

**2****. Name the type of rolling loads for which the absolute maximum bending moment occurs at the midspan of a beam.**

(i) Single concentrated load

(ii) udl longer than the span (iii) udl shorter than the span (iv) Also

when the resultant of several concentrated loads crossing a span, coincides with a concentrated load then also the maximum bending moment occurs at the centre

of the span.

**3****. What is meant by absolute maximum bending moment in a beam**? When a given load system moves from one end to the other end of a girder,

depending upon the position of the load, there will be a maximum bending moment for every section. The maximum of these bending moments will usually occur near or at the midspan. The maximum of maximum bending moments is called the absolute maximum bending moment.

**4****. What is the absolute maximum bending moment due to a moving udl longer than the span of a simply supported beam?**

When a simply supported beam is subjected to a moving udl longer than the

span, the absolute maximum bending moment occurs when the whole span is loaded.

Mmax max = wl2/8

5**. State the location of maximum shear force in a simple beam with any kind of loading.**

In a simple beam with any kind of load, the maximum positive shear force

occurs at the left hand support and maximum negative shear force occurs at right hand support.

**6****.What is meant by maximum shear force diagram?**

Due to a given system of rolling loads the maximum shear force for every

section of the girder can be worked out by placing the loads in appropriate

positions. When these are plotted for all the sections of the girder, the diagram that we obtain is the maximum shear force diagram. This diagram yields the ‘design shear’ for each cross section.

**7****. What is meant by influence lines?**

An influence line is a graph showing, for any given frame or truss, the variation

of any force or displacement quantity (such as shear force, bending moment, tension, deflection) for all positions of a moving unit load as it crosses the structure from one end to the other.

**8****. What are the uses of influence line diagrams?**

(i) Influence lines are very useful in the quick determination of reactions, shear

force, bending moment or similar functions at a given section under any given system of moving loads and

(ii) Influence lines are useful in determining the load position to cause maximum value of a given function in a structure on which load positions can vary.

**9****. What do you understand by the term reversal of stresses?**

In certain long trusses the web members can develop either tension or

compression depending upon the position of live loads. This tendancy to change the nature of stresses is called reversal of stresses.

**1****0****. State Muller-Breslau principle.**

Muller-Breslau principle states that, if we want to sketch the influence line for

any force quantity (like thrust, shear, reaction, support moment or bending moment) in a structure,

(i) We remove from the structure the resistant to that force quantity and

(ii) We apply on the remaining structure a unit displacement corresponding to that force quantity.

The resulting displacements in the structure are the influence line ordinates

sought.

**1****1****. What is the necessity of model analysis?**

(i) When the mathematical analysis of problem is virtually impossible.

(ii) Mathematical analysis though possible is so complicatedand time consuming that the model analysis offers a short cut.

(iii) The importance of the problem is such that verification of mathematical

analysis by an actual test is essential.

**1****2****. Define similitude.**

Similitude means similarity between two objects namely the model and the

prototype with regard to their physical characteristics:

• Geometric similitude is similarity of form

• Kinematic similitude is similarity of motion

• Dynamic and/or mechanical similitude is similarity of masses and/or forces.

**1****3****. What is the principle of dimensional similarity?**

Dimensional similarity means geometric similarity of form. This means that all

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

**1****4****. Name any four model making materials.**

Perspex, plexiglass, acrylic, plywood, sheet araldite and bakelite are some of

the model making materials. Micro-concrete, mortar and plaster of paris can also be used for models.

**1****5****. What is ‘dummy length’ in models tested with Begg’s deformeter.**

Dummy length is the additional length (of about 10 to 12mm) left at the extremities of the model to enable any desired connection to be made with the gauges.

**1****6****. What are the three types of connections possible with the model used**

**wi****t****h Begg’s deformeter**.

(i) Hinged connection (ii) Fixed connection (iii) Floating connection

**1****7****. What is the use of a micrometer microscope in model analysis with**

**Begg’s deformeter.**

Micrometer microscope is an instrument used to measure the displacements of any point in the x and y directions of a model during tests with Begg’s deformeter.

**1****.What is an arch? Explain.**

** ARCHES**

An arch is defined as a curved girder, having convexity upwards and supported at its ends.

The supports must effectively arrest displacements in the vertical and horizontal directions. Only then there will be arch action.

**2****.State Eddy’s theorem**.

Eddy’s theorem states that “ The bending moment at any section of an arch is

proportional to the

vertical intercept between the linear arch (or theoretical arch) and the centre line

of the actual arch.”

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

For a three hinged parabolic arch, the degree of static indeterminancy is zero. It

is statically determinate. 5.Explain with the aid of a sketch, the normal thrust and radial shear in an arch rib. H A B H R N

Let us take a section X of an arch. (fig (a) ). Let θ be the inclination of the tangent at X. If H is the horizontal thrust and V the vertical shear at X, from the free body of the RHS of the arch, it is clear that V and H will have normal and radial components given by,

N= H cosθ + V sinθ

R= V cosθ – H sinθ

**4****.Which of the two arches, viz. circular and parabolic is preferable to carry a uniformly distributed load? Why?**

Parabolic arches are preferably to carry distributed loads. Because, both, the shape of the arch and the shape of the bending moment diagram are parabolic. Hence the intercept between the theoretical arch and actual arch is zero everywhere. Hence, the bending moment at every section of the arch will be zero. The arch will be under pure compression which will be economical.

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

An arch is essentially a compression member which can also take bending

moments and shears.

Bending moments and shears will be absent if the arch is parabolic and the loading uniformly distributed.

A cable can take only tension. A suspension bridge will therefore have a cable and a stiffening girder.

The girder will take the bending moment and shears in the bridge and the cable, only tension.

Because 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 have only cable tension and no bending moment on the girder.

**6****.Under what conditions will the bending moment in an arch be zero throughout.**

The bending moment in an arch throughout the span will be zero, if

(i) the arch is parabolic and (ii) the arch carries uniformly distributed load throughout the span.

**7****.Draw the ILD for bending moment at a section X at a distance x from the left end of a three hinged**

parabolic arch of span ’l’ and rise ‘h’.

Mx = µ x – Hy

µ x Hy

(+) (-)

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

**8****. Indicate the positions of a moving point load for maximum negative and positive bending moments in a three hinged arch.**

Considering a three hinged parabolic arch of span ‘l’ and subjected to a moving

point load W, the position of the point load for

a. Maximum negative bending moment is 0.25l from end supports. b. Maximum positive bending moment is 0.211l from end supports.

**9****. Draw the influence line for radial shear at a section of a three hinged arch.**

Radial shear is given by Fx = H sinθ – V cosθ,

where θ is the inclination of tangent at X. l sinθ

l – x cosθ 4r

**1****0****. Distinguish between two hinged and three hinged arches.**

Two hinged arches Three hinged arches

1. Statically indeterminate to first degree Statically determinate

2. Might develop temperature stresses Increase in temperature causes increasein central rise. No stresses.

3. Structurally more efficient Easy to analyse. But in costruction, the central

hinge may involve additional expenditure.

4. Will develop stresses due to sinking of supports Since this is determinate, no stresses due to support sinking.

**1****1****. Explain the effect of yielding of support in the case of an arch.**

Yielding of supports has no effect in the case of a 3 hinged arch which is

determinate. These displacements must be taken into account when we analyse 2 hinged or fixed arches under

∂U = ∆H instead of zero ∂H

∂U = ∆VA instead of zero ∂VA

Here U is the strain energy of the arch and ∂H and ∆VA are the displacements

due to yielding of supports.

**1****2****. Write the formula to calculate the change in rise in three hinged arch if there is a rise in temperature.**

Change in rise = l2 + 4r2 α T 4r

where l = span length of the arch r = central rise of the arch

α = coefficient of thermal expansion

T = change in temperature

**1****3****. In a parabolic arch with two hinges how will you calculate the slope of the arch at any point.**

Slope of parabolic arch = θ = tan-1 4r (l – 2x) l2

where θ = Slope at any point x (or) inclination of tangent at x.

l = span length of the arch r = central rise of the arch

**1****4****. How will you calculate the horizontal thrust in a two hinged parabolic arch if there is a rise in temperature.**

Horizontal thrust = l α TEI l y2dx 0

where l = span length of the arch

y = rise of the arch at any point x

α = coefficient of thermal expansion

T = change in temperature

E = Young’s Modulus of the material of the arch

I = Moment of inertia

**1****5****. What are the types of arches according to the support conditions.**

i. Three hinged arch

ii. Two hinged arch

iii. Single hinged arch

iv. Fixed arch (or) hingeless arch

**1****6****. What are the types of arches according to their shapes.**

i. Curved arch

ii. Parabolic arch iii. Elliptical arch iv. Polygonal arch

** **

**SLOPE-DEFLECTION METHOD**

**1****.What are the assumptions made in slope-deflection method?**

(i) Between each pair of the supports the beam section is constant.

(ii) The joint in structure may rotate or deflect as a whole, but the angles between the members meeting at that joint remain the same.

**2****. How many slope deflection equations are available for a two span continuous beam?**

There will be 4 nos. of slope-deflection equations, two for each span.

**3****. What is the moment at a hinged end of a simple beam?**

Moment at the hinged ends of a simple beam is zero.

**4****. What are the quantities in terms of which the unknown moments are expressed in slope-deflection method?**

In slope-deflection method, unknown moments are expressed in terms of

(i) slopes (θ) and (ii) deflections (∆)

**5****. How do you account for sway in slope deflection method for portal frames?**

Because of sway, there will br rotations in the vertical members of a frame.

This causes moments in the vertical members. To account for this, besides the equilibrium, one more equation namely shear equation connecting the joint-

moments is used.

**6****. Who introduced slope-deflection method of analysis?**

Slope-deflection method was introduced by Prof. George A.Maney in 1915.

**7****. Write down the general slope-deflection equations and state what each term represents?**

A B

General slope-deflection equations:

MAB = M’AB + 2EI/L( 2θA + θB + 3∆ / l )

MBA = M’BA + 2EI/L( 2θB + θA + 3∆ / l)

where, M’AB , M’BA = Fixed end moment at A and B respectively due to the

given loading.

θA , θB = Slopes at A and B respectively

∆ = Sinking of support A with respect to B

**8****. Mention any three reasons due to which sway may occur in portal frames.**

Sway in portal frames may occur due to

(i) unsymmetry in geometry of the frame

(ii) unsymmetry in loading or

(iii) Settlement of one end of a frame.

**9****. How many slope-deflection equations are available for each span?**

Two numbers of slope-deflection equations are available for each span,

describing the moment at each end of the span.

**1****0****. State the limitations of slope deflection method.**

(i) It is not easy to account for varying member sections

(ii) It becomes very cumbersome when the unknown displacements are large in number.

**1****1****. Why is slope-deflection method called a ‘displacement method’?**

In slope-deflection method, displacements (like slopes and displacements) are

treated as unknowns and hence the method is a ‘displacement method’.

**1****2****. Define degrees of freedom.**

In a structure, the number of independent joint displacements that the structure

can undrgo are known

as degrees of freedom.

**1****3****. A rigid frame is having totally 10 joints including support joints. Out of slope-deflection and moment distribution methods, which method would you prefer for analysis? Why?**

Moment distribution method is preferable.

If we use slope-deflection method, there would be 10 (or more) unknown displacements and an equal number of equilibrium equations. In addition, there

would be 2 unknown support momentsper span and the same number of slope- deflection equations. Solving them is difficult.

**1****4****. What is the basis on which the sway equation is formed for a structure?**

Sway is dealt with in slope-deflection method by considering the horizontal

equilibrium of the whole frame taking into account the shears at the base level of columns and external horizontal forces.

The shear condition is MAB + MBA – Ph + MCD + MDC + P = 0

** **

**M****OMENT DISTRIBUTION METHOD**

**1****.What is the difference between absolute and relative stiffness? **Absolute stiffness is represented in terms of E, I and l, such as 4EI / l. Relative stiffness is represented in terms of I and l, omitting the constant E.

Relative stiffness is the ratio of stiffness to two or more members at a joint.

**2****. Define: Continuous beam.**

A Continuous beam is one, which is supported on more than two supports. For

usualloading on the beam hogging ( – ive ) moments causing convexity upwards at

the supports and sagging ( + ve ) moments causing concavity upwards occur at mid span.

**3****.What are the advantages of Continuous beam over simply supported beam?**

1. The maximum bending moment in case of continuous beam is much less

than in case of simply supported beam of same span carrying same loads.

2. In case of continuous beam, the average bending moment is lesser and hence lighter materials of construction can be used to resist the bending moment.

**4****. In a member AB, if a moment of –10 KNm is applied at A, what is the moment carried over to B?**

Carry over moment = Half of the applied moment

∴Carry over moment to B = -10/5 = -5 KNm

**5****. Define: Moment distribution method.( Hardy Cross mrthod).**

It is widely used for the analysis of indeterminate structures. In this method, all

the members of the structure are first assumed to be fixed in position and fixed end moments due to external loads are obtained.

**6****. Define: Stiffness factor.**

It is the moment required to rotate the end while acting on it through a unit

rotation, without translation of the far end being (i) Simply supported is given by k = 3 EI / L (ii) Fixed is given by k = 4 EI / L

where, E = Young’s modulus of the beam material.

I = Moment of inertia of the beam

L = Beam’s span length.

**7****. Define: Distribution factor.**

When several members meet at a joint and a moment is applied at the joint to

producerotation without translation of the members, the moment is distributed among all the members meeting at that joint proportionate to their stiffness.

Distribution factor = Relative stiffness / Sum of relative stiffness at the joint

If there is 3 members, Distribution factors = k1 , k2 , k3 k1 + k2 + k3 k1 + k2 + k3 k1 + k2 + k3

**8****. Define: Carry over moment and Carry over factor.**

Carry over moment: It is defined as the moment induced at the fixed end of the

beam by the action of a moment applied at the other end, which is hinged. Carry over moment is the same nature of the applied moment.

Carry over factor ( C.O) : A moment applied at the hinged end B “ carries over” to the fixed end A, a moment equal to half the amount of applied moment and of the same rotational sense.

C.O =0.5

**9****. Define Flexural Rigidity of Beams.**

The product of young’s modulus (E) and moment of inertia (I) is called

Flexural Rigidity (EI) of Beams. The unit is N mm 2

**1****0****. Define: Constant strength beam.**

If the flexural Rigidity (EI) is constant over the uniform section, it is called

Constant strength beam.

**1****1****. What is the sum of distribution factors at a joint?**

Sum of distribution factors at a joint = 1.

**1****2****. What are the situations where in sway will occur in portal frames?**

a. Eccentric or unsymmetric loading

b. Unsymmetrical geometry

c. Different end conditions of the columns d. Non-uniform section of the members

e. Unsymmetrical settlement of supports f. A combination of the above

**1****3****. What are symmetric and antisymmetric quantities in structural behaviour?**

When a symmetrical structure is loaded with symmetrical loading, the bending

moment and deflected shape will be symmetrical about the same axis. Bending moment and deflection are symmetrical quantities.