NOORUL ISLAM COLLEGE OF ENGINEERING

KUMARA COIL

AE45 AIRCRAFT STRUCTURES I

IV SEMESTER, AERONAUTICAL ENGINEERING

2009-2010

J.SAJI SOUNDARA RAJ

HOD LECTURER/AERO

2MARKS QUESTIONS WITH ANSWERSUnit1

1. Differentiate Truss and frame.

Truss FrameTruss is defined as number of members riveted together to carry the horizontal, vertical and inclined loads in equilibrium.

Frame is defined as number of members welded together to carry the horizontal, vertical loads in equilibrium.

2. Differentiate the perfect and imperfect frames?

Sl.No. Perfect frame Imperfect frame1.

2.

3.

Perfect frame have sufficient frame or enough members to carry the load.

It satisfies the formula n= 2j-3.

Eg. Triangular frame.n=3, j=3n=2j-33=2x3-3,3=3

Imperfect frame have less or more members to carry the load than the required numbers.

It does not satisfy the formula n= 2j-3.

Eg. Square framen=4, j=4n=2j-34=2x4-3,4≠5

3. Differentiate the deficient frame and redundant frame?

Sl.No. Deficient frame Redundant frame1. If the number of members are less

than the required of members.n < 2j-3

If the number of members are more than the required number of members.n > 2j-3

4. Define plane truss and space truss. Give some examples.A plane truss is a two dimension truss structure composed of number of bars

hinged together to form a rigid frame work, all the members are lie in one plane. Eg.: Roof truss in industries.

A space truss is a three dimension truss structure composed of number of bars hinged together to form a rigid frame work, all the members are lie in different plane. Eg.: Transmission line towers, crane parts.

5. What are the methods used to analyze the plane & space frames? Analytical method

1. Method of joints2. Method of sections (method of moments)3. Tension co-efficient method.

Graphical method.

6. What are the hints to be followed while analyzing a simply supported truss using method of joints?

The support reactions are determined first. The analysis is to be started from the free end where there is a maximum of

two unknown forces, using the condition of equilibrium Σ Fx = 0 and Σ Fy =0. All the members are assumed to be tensile. Consider tensile forces as positive and compressive as negative.

7. Give relation between the number of members and joints in a truss and explain its uses.n=2j-3, where n= number of members, j= number of joints. This relation is used

to find the type of the frames. Perfect frame is only solved by method of joints.

8. What are the hints to be followed while analyzing a cantilever truss using method of joints?

There is no need to find the support reactions. The analysis is to be started from the free end where there is a maximum of two

unknown forces, using the condition of equilibrium Σ Fx = 0 and Σ Fy =0. All the members are assumed to be tensile. Consider tensile forces as positive and compressive as negative. The force convention is, upward force assigns positive sign and downward force

assigns negative sign.

9. What are assumptions made in the analysis of a truss?1. In a frame or truss all the joints will be pin jointed.2. All the loads will be acting at the joints only.3. The self-weight of the members of the truss is neglected. Only the live load is

considered. 4. The frame is a perfect one.

10. What are the types of framed structures? Explain each type?(1)Efficient or perfect structure: A structure is said to be perfect. It satisfies the equations n=2j-3 n=no of member

j=no of joints

(2) Deficient or imperfect structure: A structure is said to be deficient, when n<2j-3

(3) Redundant frame: If a structure is said to be redundant, when n > 2j-3

11. What is cantilever truss? What is simply supported truss?If anyone of the member of the truss is fixed and the other end is free, it is called a

cantilever truss. There is no reaction force at the fixed end.

12. What are the conditions of equilibrium used in the method of joints? Why?The conditions of equilibrium used in the method of joints are, Σ Fx = 0,

Σ Fy =0. One of the assumptions is all the joints are pin jointed, there is no moment. The equilibrium condition Σ Mx =0 is not used.

13. Explain with examples the statically determinate structures.If the structure can be analyzed and the reactions at the support can be determined

by using the equations of static equilibrium such as Σ Fx = 0, Σ Fy =0 and Σ Mx =0, then it is called as a statically determinate structure. Example: Simply supported beam, pin jointed truss nor frame.

14. Differentiate the statically determinate structures and statically determinate structures.

Sl.No. Statically determinate structures Statically indeterminate structures1. Conditions of equilibrium are

sufficient to analyze the structureConditions 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.

15. Where are truss type structures found in an aircraft?Fuselage, Bi-plane, Tri-plane and wing etc.,

16. Define neutral axis? Neutral axis is defined as the line of intersection of the neutral layer with the transverse section. Then the stress will be compressive at any point above the neutral axis and tensile below the neutral axis.

17. What are the types of supports? Explain each support?

Roller support Hinged support Fixed support Sliding support

Roller Support:It has only one support reaction.

Hinged support:It has two support reactions.

Fixed support: It has two support reactions.

Sliding support:It has two support reactions.

18. What are types of load? Explain each load? Point load

One point is acted upon by a concentrated load. Uniformly distributed load

Load is distributed uniformly Uniformly varying load

Load is varying throughout the beam

19. What is the procedure followed for method of joints? Find the support reactions using moment method Each and every joints is treated separately as a free body in

equilibrium Tensile forces is considered as +ve Compressive force is considered as –ve

20. What is structure? Structure is composed of several bars or rods joined together in a particular

fashion, then each bar is known as member or structure.

21. What are the assumptions made in frame for calculation? (1) All members are pin joint (2) The frame is load only at the joints. (3) The frame is the perfect frame

(4) The self weight of the members is neglected.

22. What are the different types of beams? Simply supported beam Cantilever beam Overhanging beam Fixed beam Continuous beam Propped cantilever beam

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Unit II1. Define: Continuous beam.

A continuous beam is one, which is supported on more than two supports. For usual loading on the beam hogging (-ve) moments causing convexity upwards at the supports at the supports and sagging (+ve) moments causing concavity upwards occur at mid span.

2. 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 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.

3. Give the procedure for analyzing the continuous beams with fixed ends using three moment equations.

The three moment equations, for the fixed end of the beam, can be modified by imagining a span of length lo and moment of inertia, beyond the support and applying the theorem of three moments as usual.

4. Define: Moment distribution method.(Hardy cross method).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.

5. Define Constant strength of beam.If the flexural rigidity (EI) is constant over the uniform section, it is called constant

strength beam.

6. Define composite beam.A structural member composed of two or more dissimilar materials jointed together to

act as a unit. The resulting system is stronger than the sum of its parts. The composite action can better utilize the properties of each constituent material.

Ex. : Steel- concrete composite beam, steel-wood beam.

7. Write the two conditions for the analysis of composite beam.(i) Strain (Stress /E) in all the material are same (e1=e2)

[(e1 =P1/ A1E1); (e2 =P2/ A2E2)]

(ii) The total load = P1 + P2 + P3 (P1 = Stress x Area)

8.. Write down the most general form of the 3- moment equation and all the notations: (MAl1/ E1I1) + 2MB (l1/E1I1 + l2/E2I2) + (MCl2/E2I2) + 6(a1x1/E1I1l 1 + a2x2/E1I1l1) = 6[(d1l1)/(d2l2)].where,MA = Hogging bending moment at AMB = Hogging bending moment at BMC = Hogging bending moment at C

l1 = Length of span between supports A,Bl2 = Length of span between supports A,Bx1 = C.G of bending moment diagram from support Ax2 = C.G of bending moment diagram from support Ca1 = Area of bending moment diagram between supports A, Ba2 = Area of bending moment diagram between supports B, Cd1 = Sinking at support A with compare to sinking at support Bd1 = Sinking at support C with compare to sinking at support B

9. Define stiffness factor in moment distribution method. It is the moment required to rotate the end through one unit angle without translate the far end.

10. Define the ‘carry over’ factor used in the moment distribution method. The ratio of moment produced at a joint to moment applied at the other joint without displacing is called carry over factor

11. A solid cylinder 100cm long & 5cm in diameter is subjected to a tensile force of 80KN. One part of this cylinder of length L1 is made of steel (E = 210Gpa) and the other part of length L2 is made of aluminium (E=70Gpa). Determine the length L1 & L2 so that the two parts elongate to an equal amount.

Solution: Stress = P/A Strain = dl/l E = (P/A) / (dl/l) dl/l = (P/A) / E dl = (Pl/AE)Elongation is same for two parts. (P1l1)/(A1E1) = (P2l2)/(A2E2) l1/210 = l2/70 l1 = 3l2

l1+l2 = 100 4l2 = 100 l2 = 25cm l1 =3x25 l1 =75cm.Ans: l1 =75cm. l2 = 25cm.

12. Define and give S.I. units for

(a) Flexural rigidity(b) Stiffness

(a) Flexural Rigidity: It is defined as the product of modulus of elasticity and the second moment of inertia. Its unit is N-mm2

(b) Stiffness: In a mechanical vibrating system, the stiffness is the restoring force per unit displacement. For example measure of the strength of a spring. Its unit is N/mm.

13. A 20cm long tube 15cm internal diameter & 1cm thick is surrounded by a brass tube of same length and thickness. The tubes carry an axial load of 150KN. Estimate the load carried by each. Es=210GPa, Eb=100GPa.

Solution: es = eb

ps/Es = pb/Eb

[Ps/(0.172 –0.15

2 )] / (210x109) = [Pb /(0.192- 0.172

)] /(100x109) 74404x10-10xPs = 1.388x10-9xPb

Ps =1.86549 Pb

Ps + Pb =150000 1.86549 Pb + P =150000 Pb =52347.06 N Ps =97652.94 N.Ans: Ps =97652.94 N Pb =52347.06 N.

14. Explain with examples the statically indeterminate structures.If the forces on the members of a structure cannot be determined b using

conditions of equilibrium (Σ Fx = 0, Σ Fy =0 and Σ Mx =0), it is called statically indeterminate structures.Example: Fixed beam, continuous beam.

15. Define: Distribution factor. When several members meet at a joint and a moment is applied at the joint to

produce rotation 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 jointIf there is 3 members, Distribution factors = k1 / (k1 + k2 + k3), k2 / (k1 + k2 + k3), k3 / (k1 + k2 + k3)

16. Give the values of (6A1x1/l1), (6A2x2/l2) for different type of loading.Type of loading (6A1x1 / l1) (6A2x2 / l2)UDL for entire span wl3 / 4 wl3 / 4Central point loading (3/8) Wl2 (3/8) Wl2

Uneven point loading (wa / l) / (l2 –a2) (wa / l) / (l2 –a2)

17. Write down the Clapeyron’s three moment equations for the continuous beam with sinking of the supports.

MAl1+ 2MBl2 +MCl2= (6a1x1/ l1+ 6a2x2/ l2) -6EI (d1/l1+d2/l2)

Where,MA = Hogging bending moment at AMB = Hogging bending moment at BMC = Hogging bending moment at Cl1 = Length of span between supports A,Bl2 = Length of span between supports B,Cx1 = C.G of bending moment diagram from support Ax2 = C.G of bending moment diagram from support Ca1 = Area of bending moment diagram between supports A, Ba2 = Area of bending moment diagram between supports B, Cd1 = Sinking at support A with compare to sinking at support Bd1 = Sinking at support C with compare to sinking at support B

18. Write down the Clapeyron’s three moment equations for the fixed beam.

MA + 2MB = (6ax) / l2

MA = Hogging bending moment at AMB = Hogging bending moment at Bl = Length of span between supports A,Bx = C.G of bending moment diagram from support Aa = Area of bending moment diagram between supports A, B

19. Write down the Clapeyron’s three moment equation for the continuous beam carrying UDL on both the spans.

MAl1+ 2MBl2 +MCl2= (6a1x1/ l1+ 6a2x2/ l2) = (w1 l13)/4 + (w2 l2

3)/4where,

MA = Hogging bending moment at AMB = Hogging bending moment at BMC = Hogging bending moment at Cl1 = Length of span between supports A,Bl2 = Length of span between supports B,C

20. Find the distribution factor for the given beam.

Joint Member Relative stiffness Sum of relative stiffness Distribution factor

A AB 4EI/L 4EI/L (4EI/L)/ (4EI/L)=1B BA

BC3EI/L4EI/L

3EI/L+4EI/L=7EI/L (3EI/L)/ (7EI/L)=3/7(4EI/L)/ (7EI/L)=4/7

C CBCD

4EI/L4EI/L

4EI/L+4EI/L=8EI/L (4EI/L)/ (8EI/L)=4/8(4EI/L)/ (8EI/L)=4/8

D DC 4EI/L 4EI/L 4EI/L+4EI/L=1

--------------------------------Unit III

1. Define strain energy.When an elastic body is under the action of external forces the body deforms and

work is done by these forces. If a strained, perfectly elastic body is allowed to recover slowly to its unstrained state. It is capable of giving back all the work done by these

external forces. This work done in straining such a body may be regarded as energy stored in a body may be regarded as energy stored in a body and is called strain energy or resilience.

2. Define proof resilience.The maximum energy stored in the body within the elastic limit is called proof

resilience.

3. Write the formula to calculate the strain energy due to axial loads (tension)

Limit 0 to L

where , P= Applied tensile load L= Length of the member A= Area of member E= Young’s modulus

3. A beam of Aluminium has I=500cm4, L=2m and fixed as a cantilever. Find deflection at the tip for a load of 1000N at the tip. (E=70GPa). δ = 5wl4/384EI = (5x500x24)/(384x70x109x500x0.014) = 2.976x10-4m Ans. = 0.2976 mm

4. If the above beam is fixed as a simply supported beam and subjected to a distributive load of 500N/m .Find the maximum deflection.

Deflection = 5/384 ( wl4/EI) = 5/384 [( 500x24) / (70x109 x 500x10-8)]

Ans: = 0 .2976mm

5. Write down strain energy expression for (a) Tension (b) Bending. (a) U= 1/2 ∫ (σ2/AE)dx

(b) U= 1/2 ∫ (M2/EI)dx

7. State Castigliano’s theorem.If ‘U’ is the total strain energy in any structure due to the

application of external loads W1,W2,W2,……….Wn at points A1,A2,A3,……….An in the direction of Ax1,Ax2,Ax3,……….Axn and due to the couples M1,M2,M3……….Mm at points B1,B2,B3,……….Bm respectively , then the deflection at the points A1,A2,A3,……….An in the direction of Ax1,Ax2,Ax3,……….Axn are ∂U/∂W1, ∂U/∂W2,

∂U/∂W3…………….∂U/∂Wn and the angular positions of the couples are ∂U/∂M1, ∂U/∂M2,

∂U/∂m3…………….∂U/∂Mm at the respective points of application.

8. State Maxwell’s reciprocal theorem.Ans: It states that “work done by the first system of loads due to the deflection

by a second system of loads equals work done by the second system of loads due to the deflection caused by a first system of loads”.

9. A solid cube of steel (E=210Gpa) is subjected to a tension 200MPa, find the strain energy per unit volume. U = σ2 / 2E = (200x106)2 / 2x210x109

Ans. = 1.96x104 N/m2

10. A solid cube of steel (G=80GPa) is subjected to a shear of 56MPa. Find the strain energy per unit volume. U = τ2 / 2G = (56x106)2 / 2x 80x109

Ans. = 95.238x 103N/m2

11. Write down strain energy expressions for various loading cases. U= 1/2 ∫ (σ2/AE)dx for tension U= 1/2 ∫ (M2/EI)dx for bending U= 1/2 ∫ (T2/2GJ)dx for torsion

12 .A Cantilever beam of length ’L’ subjected to a tip load P, find the deflection at the tip using Castigliano’s theorem . M= - (P+W)x∂M/∂W = -xδ = 1/EI ∫ M ∂M/∂W dx Limit 0 toL =1/EI ∫ -(P+w)x.(-x) dx Limit 0 toL Put W=0 =1/EI ∫ Px2 dx = 1/EI .L3/3 .P δ = PL3/3EI

13. Explain unit load method with an example.The external load is removed and the unit load is applied at the point where the

deflection or rotation is to found.For example, to find the force of each member in a redundant frame.

14. Find the slope at the support of a simply supported beam of length L and subjected to a uniformly distributed load by unit load method.θ = (wL3) / (24EI)where,θ = Slope or anglew = Uniformly distributed load/unit runL= Length of the beamE= Young’s modulus of the material of the beamI= area moment of inertia

15. Give an example of how energy methods can be used to analyze indeterminate structures?

By using energy methods we can analyze indeterminate structures. In a redundant frame, by unit load method we can solve.

16. 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.

17. Define: Modulus of resilience.The proof resilience of a body per unit volume, i.e. the maximum energy stored in

the body within the elastic limit per unit volume.

18. Define trussed beam.A beam strengthened by providing ties and struts is known as trussed beams.

19. Compare the unit load method and Castigliano’s first theorem.In the unit load method, one has to analyze the frame twice to find the load and

deflection, while in the latter method, only one analysis is needed.

20. Write the general formula for beam using unit load method.

δ = ∫(mM /EI )dx Limit 0 to l

For slope θ = ∫(mcM /EI )dx Limit 0 to lwhere,M= moment due to unit loadmc = moment due to unit coupleEI= Flexural rigidity

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UNIT IV

1. What are the types of column failure?1. Crushing failure

The column will reach a stage, when it will be subjected to the ultimate crushing stress, beyond this the column will fail by crushing. The load corresponding to the crushing stress is called crushing load. This type of failure occurs in short column.

2. Buckling failureThis kind of failure is due to lateral deflection of the column. The load at which

the column just buckles is called buckling load or crippling load. This type of failure occurs in long column.

2. What are the factors affect the strength of the column? 1. Slenderness ratio

Strength of the column depends upon the slenderness ratio, it is increased the compressive strength of the column decrease as the tendency to buckle is increased.2. End conditions

Strength of the column depends upon the end conditions also.

3. What are the assumptions followed in Euler’s equation?1. The material of the column is homogeneous, isotropic and elastic.2. The section of the column is uniform through out.3. The column is initially straight and load axially.4. The effect of the direct axial stress is neglected.5. The column fails by buckling only.

4. What are the limitations of the Euler’s formula?1. It is not valid for mild steel columns. The slenderness ratio of mild steel

column is less than 80.2. It does not take direct stress. But in excess of load it can withstand under direct compression only.

5. Write the Euler’s formula for different end conditions.1. Both ends fixed.

PE= π2EI/(0.5L)2

2. Both ends hinged PE= π2EI/(L)2

3. One end fixed other end hinged PE= π2EI/(0.7L)2

4. One end fixed other end free. PE= π2EI/(2L)2

6. Define equivalent length of the column.The distance between adjacent points of inflection is called equivalent length of

the column. A point of inflection is found at every column end, that is free to rotate and every point where there is a change of the axis. i.e. there is no moment in the inflection points.

ORThe equivalent length of the given column with given end conditions, is the length

of an equivalent column of the same material and cross section with hinged ends, and having the value of the crippling load equal to that of the given column.

7. What are the uses of south well plot? (column curve)

The relation between the buckling load and slenderness ratio of a column is known as south well plot.

The south well plots clearly shows the decrease of buckling load when increase of slenderness ratio.

It gives the exact value of slenderness ratio of column subjected to a particular amount of buckling load.

8. What is beam column? Give examples.Column having transverse load in addition to the axial compressive load are

termed as beam column.Ex. Engine shaft, Wing of an air craft.

9. Write the expression for the maximum deflection developed in a beam column carrying central point load with axial load hinged at both ends.δ max = δo / [1-(P/PE)]where,

δo = QL3 / 48 EI Q= Central point load P= axial load

PE= Euler’s load

10. Write the expressions for the maximum bending moment and max. stress developed in a beam column carrying central point load hinged at both ends. M max = Mo [1- 0.18 (P/ PE)] / [1-( P/ PE)] σmax =P/A +(Mmax/!) where,

Mo =QL/4Q= central point loadP=axial loadPE= Euler’s loadZ= section modulus

11. Write the expressions for the maximum deflection developed in a beam column carrying uniformly distributed load with axial load, hinged at both ends.δ max = δo / [1-( P/PE)]

where,δo = 5wL4/384 EI

w=uniformly distributed load /unit run P=Axial Load PE= Euler’s load

12. Write the expressions for the maximum bending moment and the max. stress developed in a beam column carrying uniformly distributed load with axial load, hinged at both ends.M max = Mo [1+ 0.03 (P/ PE)] / [1-(P/ PE)] σmax =P/A +(Mmax/Z)

where, Mo = wL2 /8w= uniformly distributed load/unit runP=axial loadPE= Euler’s loadZ= section modulus

13. Differentiate between long & short columns.Long columns: The length of the column is greater than 30 times its diameter. Slenderness ratio is greater than 120. They are subjected to buckling stress only. Short columns: The column has length less than 8 times its diameter. Slenderness ratio is less than 32. Buckling is generally negligible.

14. Give 2 examples of aircraft structural parts that act as beam column.1. Fuselage2. Wing.

15. Explain (a) Slenderness ratio (b) Radius of Gyration.

(a) Slenderness ratio: It is the ratio between the unsupported length of the column to the minimum radius of gyration of the cross sectional ends of the column. (b) Radius of Gyration: The radius of gyration is the distance from the specified axis to the center of mass.

16. State the importance and application of Rankine’s formula. The Rankine Hypothesis is designed to link the acceptable theoretical expression available for both very short and very long struts to obtain an expression applicable to struts / columns of all dimensions.PR = ( Aσc ) / [1+(alc

2/k2)]wherea = σc /(π2E), which is Rankine’s constant.

17. What is slenderness ratio? What is its relevance in columns? It is the ratio between the unsupported length of the column to the minimum radius of gyration of the cross sectional ends of the column. The strength of the column depends upon the slenderness ratio. If the slenderness ratio is increased the compressive strength of a column decreases as the tendency to buckle is increased.

18. What is an ideal column?

Ideal column is a column in which axial load acts on it.A member of structure or bar which carries an axial compressive load is called the

strut. If the strut is vertical i.e. inclined at 90◦ to the horizontal is known as column, pillar or stanchion.

19. The load carrying capacity of a column (increases / decreases / remains unchanged) because of eccentricity.Ans.

The load carrying capacity of a column decreases because of eccentricity.

20. Write Euler’s formula for maximum stress for a initially bent column?σmax = P/A +( Mmax/Z)

= P/A + Pa/ [1-( P / PE)]ZwhereP = Axial loadPE = Euler’s loade= eccentricityZ= section modulusEI = Section modulus

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Unit V

1. What are the types of failures?1. Brittle failureFailure of a material represents direct separation of particles from each other,

accompanied by considerable deformation.2. Ductile failureSlipping of particles accompanied by considerable plastic deformation.

2. List out different theories of failure.1. Maximum principal stress theory (Rankine’s theory)2. Maximum principal strain theory (St.Venant’s theory)3. Maximum shear stress theory (Tresca’s theory or Guest’s theory)4. Maximum shear strain theory (Von-Mises-Hencky theory or Distortion energy theory)5. Maximum strain energy theory(Beltrami theory or Haigh’s theory)

3. Define Maximum principal stress theory.According to this theory, the failure of the material is assumed to take place when

the value of the maximum principal stress (σ 1) reaches a value to that of the elastic limit stress (fs) of the material σ 1 = fy.

4. Define maximum principal strain theory (St. Venant’s theory)According to this theory, the failure of the material is assumed to take place when

the value of the maximum principal strain (e1) reaches a value to that of the elastic limit strain (fy /E)of the material,

e1 = fy /EIn 3D, e1= 1/E[σ1 –(1/m)( σ2 + σ3)] = fy /E → [σ1–(1/m)( σ2 + σ3)] = fyIn 2D, σ3 =0 → e1= 1/E[σ1 –(1/m)( σ2)] = fy /E → [σ1–(1/m)( σ2)] = fy

5. Explain Octahedral shear stress theory. toct = (N/3)(( s1-s2)2+(s2-s3)2+(s3-s1)2)1/2 =(Ö2/3) sy If it is two dimensional, s3 = 0 (2s1

2-2s1s2+2s22)1/2 = (Ö2/N) sy

Ö2 (s12-s1s2+s2

2)1/2 = (Ö2/N) sy (s1

2-s1s2+s22)1/2 = sy / N .

6. State two failure theories applicable to brittle material.1. Maximum principal stress theory.2. Octahedral shear stress theory.

7. What is the failure criterion according to maximum shear stress theory? This theory implies that failure will occur when the max. shear stress tmax in the complex system reaches the value of the shear stress in simple tension at the elastic limit

i.e., tmax = (s1-s2)/2 = (set/2) in simple tension (or) s1-s2 = set .

8. Define Maximum shear strain theory ( Von-Mises – Hencky theory or Distortion energy theory)

According to this theory, the failure of the material is assumed to take place when the maximum shear strain exceeds the shear strain determined from the simple tensile test.

In 3D, shear strain energy due to distortion U= (1/12G)[( s1-s2)2 +(s2-s3)2+(s3-s1)2]Shear strain energy due to simple tension, U= fy2 / 6G

(1/12G)[( s1-s2)2 +(s2-s3)2+(s3-s1)2] = fy2 / 6G [( s1-s2)2 +(s2-s3)2+(s3-s1)2] = 2 fy2 In 2D [( s1-s2)2 +(s2-0)2+(0-s1)2] = 2 fy2

9. Define Maximum strain energy theory (Beltrami theory)According to this theory, the failure of the material is assumed to take place when

the maximum strain energy determined from the simple tensile test.In 3 D, strain energy due to deformation U= (1-2E) [s1

2 +s22+s3

2 –(1/m)( s1 s2+ s2 s3 + s3 s1)]Strain energy due to simple tension, U= fy2/2E (1-2E) [s1

2 +s22+s3

2 –(2/m)( s1 s2+ s2 s3 + s3 s1)] = fy2/2E [s1

2 +s22+s3

2 –(2/m)( s1 s2+ s2 s3 + s3 s1)] = fy2

In 2D, [s12 +s2

2 –(2/m)( s1 s2)] = fy2

10. What are the theories used for ductile failures.1. Maximum principal strain theory (St. Venant’s theory)

2. Maximum shear stress theory (Tresca’s theory)3. Maximum shear strain theory (Von-Mises-Hencky theory or Distortion energy

theory)

11. What are the limitations of Maximum principal stress theory ( Rankine’s theory)1. This theory disregards the effect of other principal stresses and effect of

shearing stresses on other planes through the element.2. Material in tension test piece slips along 45◦ to the axis of the test piece, where

normal stress is neither maximum nor minimum, but the shear stress is maximum.3. Failure is not a brittle, but it is a cleavage failure.

12. Write the limitations of maximum shear stress theory. (Tresca’s theory)This theory does not give the accurate results for the state of stress of pure shear

in which the maximum amount of shear is developed. (in torsion test)

13. Write the limitations of Maximum shear strain theory (Von-Mises-Hencky thory or Distortion energy theory)

It cannot be applied for the materials under hydrostatic pressure.

14. Write the limitations of Maximum strain energy theory (Beltrami theory or Haigh’s theory)

This theory does not apply to brittle materials for which elastic limit in tension and in compression are quite different.

15. Write the failure theories and its relationship between tension and shear. 1. Maximum principal stress theory (Rankine’s theory) ζ y = fy2. Maximum principal strain theory (St.Venant’s theory) ζ y = 0.8 fy3. Maximum shear stress theory (Tresca’s theory or Guest’s theory) ζ y = 0.5 fy

4. Maximum shear strain theory (Von-Mises-Hencky theory or Distortion energy theory) ζ y = 0.577 fy5. Maximum strain energy theory(Beltrami theory or Haigh’s theory) ζ y = 0.817 fy

16. Write the volumetric strain per unit volume. fy2 /2E

17. Define octahedral stresses.A plane, which is equally inclined to the three axes of reference, is called

octahedral plane. The normal and shearing stress acting on this plane are called octahedral stresses. toct = 1/3 [( s1-s2)2+(s2-s3)2+(s3-s1)2]1/2

18. Define Plasticity ellipse.The graphical surface of a Maximum shear strain theory (Von-Mises-Hencky

theory or Distortion energy theory) is a straight circular cylinder. The equation is 2D is

s12 - s1 s2 + s2

2 = fy2 which is called Plasticity ellipse.

------------------------

1) Determine the forces in the members of the truss indicated in the fig.

Solution: Taking moment about ‘A’ R2 X 12 = 2 X 3 + 1 X 6 + 9 X 3 R2 = 6 + 6 + 27 12 = 39 12 = 3.25 KN

Upward forces = Downward forces

R2 + R1 = 2 + 1 + 3 3.25 + R1 = 6 R1 = 6 – 3.25 R1 = 2.75 KNJoint A : Equating vertical forces :

R1 + PAB sin 60 = 0 2.75 + PAB sin 60 =0 PAB = - 2.75 Sin 60

PAB = - 3.175 KNEquating Horizontal Forces : PAB cos 60 + PAE = 0

2 3.175 cos 60 + PAE = 0 PAE = 1.588 KN

Joint D:Equating Vertical forces

RD + FCD sin 60 = 0 3.25 + FCD sin 60 = 0FCD = - 3.75 KN

FCD = 3.75 KN (compression)

Equating horizontal forces

- FCD Cos 60 - FED = 0 - 3.75 X Cos 60 – FED = 0 FED = - 1.875kN

Joint E :F5 Sin 60 + F3 Sin 60 - 1 = 0

F5 = 1 – F3 Sin 60 Sin 60

= 5.05 KNJoint B :F1 Sin 60 – F3 Sin 60 = 2

- F3 = 2 + 1.59 Sin 60 Sin 60 = -3.899 KN

F4 + F3 Cos 60 – F1 Cos 60 = 0 F4 = 1.59 Cos 60 + 3.899 Cos 60 F4 = 2.74 KN

RESULT :

F1 = 1.59 KN F5 = 0.2887 KN F2 = -3.175 KN F6 = 1.875 KN F3 = 0.866 KN F7 = - 3.75 KN F4 = - 2.02 KN

2) Find the support reactions and force in each member of a truss .

Solution: As the truss is supported on roller at ‘B’ there will be support reaction vertical at ‘B’. The support reaction at ‘A’ will be the vertical and horizontal component at ‘A’.

Moment at A : Ra X 0 + RbX 4 = ( 12 X 1.5 ) + (18 X 2 ) 4 X Rb = 54 Rb = 13.5 KN

Upward Force = Down ward force Rav + Rb = 18 Rav +13.5 =18 Rav = 4.5 KN

Right = left

RAH = 12 KN

At Joint A:

θ = tan-1 1.5

2θ = 36.86˚

Equating Vertical Forces: FAC Sin 36.86 + Rav = 0 FAC Sin 36.86 + 4.5 = 0

FAC = - 7.5 KN

Equating Horizontal forces :

FAC Cos 36.86 + FAD - RAH = 0 (-7.5 X Cos 36.86 ) + FAD -12 = 0 FAD = 18 KN

At Joint B:

Equating Vertical Forces: FBC Sin 36.86 + 13.5 = 0 FBC = - 22.5 KN

Equating Horzontal forces :

-FBC Cos 36.86 - FBD = 0FBD = - ( - 22.5 ) X Cos 36.86

FBD = 18 KN

At Joint D:

Equating Vertical Forces: FDC -18=0

FDC = 18 KNRESULT :

FAD = FDA = 18 KNFAC = FCA = - 7.5 KNFBC = FCB = -22.5 KNFBD = FDB = 18 KNFDC = FCD = 18 KN

3. Fig shows a truss with a span of 5m and carrying a load of 5KN at its apes. Find the force in all the members of the truss.

Soln:

R1 + R2 = 5 KN

Sin 30 = AB 5

AB = 5 Sin 30 = 2.5m

In ∆ ABD , Cos 60 = BD 2.5 BD = 1.25m

Take Moment About B,

5 X 1.25 - R2 X5 = 0 R2 = 5 X 1.25 5 = 1.25 KN

R1 = 5 – 1.25 = 3.75 KN Consider Joint B : F2 + F1 Cos 60 = 0 R1 + F1 Sin 60 = 0 F1 Sin 60 = - R1 F1 = -3.75 Sin 60

= - 4.330 KN

F2 = - F1 Cos 60 = 4.330 X Cos 60 = 2.165 KNConsider Joint C :F3 Cos 30 - F2 = 0

R2 + F3 Sin 30 = 0

R2 = - R2 Sin 30 = -2.5 KN

RESULT :

Force AB = - 4.33 KN (compressive)

Force BC = 2.165 KN (Tensile)

Force AC = -2.5 KN (compressive)

4) Fig. shows a warren girder consisting of seven members each of 4m length supported at its ends and loaded as shown. Determine the stresses in the members by method of joints?

Solution :

R1 = R2 = 5 KN

Joint A :

Resolving the forces vertically,

PAB Sin 60 + 5 = 0 PAB = - 5 . Sin 60 = - 5.7736 KN

Resolving the forces horizontally,

PAB Cos 60 + PAE = 0 PAE = - PAB Cos 60 = -(-5.7736) X 0.5 =2.8868 KN

Joint B:

Resolving the forces vertically,

PAB Sin 60 + PBESin 60 + 5 = 0 -5.7736 sin 60 + PBESin 60 + 5 = 0 PBESin 60 = 5.7736 sin 60 -5 = 0 PBE = 0

Resolving the forces horizontally,

PBC + PBE cos 60 – PAB cos 60 = 0 PBC + 0 –(-5.7736) cos 60 = 0 PBC = -5.7736 X 0.5 = -2.8868 KN

RESULT:

Member Force(KN) Member Force(KN)

ABBCCDDE

-5.7736-2.8868-5.77362.8868

AEBECE

2.886800

5) Fig shows a truss of 8m span and loaded as shown. Find out the forces in all the members of the truss using method of joints.

Solution :

Taking moment about A, we get

RE X 8 = 2 X 2 + 1 X 2 = 6 RE = 6 = 0.75 KN 8Also, RAV + RE = 2 RAV = 2- RE = 2 – 0.75 = 1.25 KN

Horizontal reaction at A, RAH = 1 KN

Joint A:

Resolving the forces vertically, PAB sin 45 + 1.25 = 0 PAB = - 1.25 Sin 45 = - 1.77 KN

Resolving the forces horizontally, PAF + PAB cos 45 = 1 PAF – (- 1.77) cos 45 = 1 PAF – 1+ 1.77 cos 45 =2.25 KNJoint E Resolving the forces vertically, PED sin 45 + 0.75 = 0 PED = -0.75 Sin 45 = - 1.06 KN

Resolving the forces horizontally,-PED cos 45 – PEF = 0

PEF = -PED cos 45 = -(-1.06 X 0.707) = 0.75 KNJoint B : Resolving the forces vertically, -2 + PBC cos 45 – PBF sin 45 – PBA sin 45 = 0 -2 + PBC cos 45 – PBF sin 45 –(-1.77) sin 45 = 0 PBC– PBF = 0.75 0.707 = 1.06 KN Resolving the forces horizontally, PBC sin 45 + PBF cos 45 – PBA cos 45 = 0 PBC + PBF = - 1.77 PBC = 1.06 – 1.77 2 = - 0.36 KN PBF = - 1.42 KNJoint D: Resolving the forces vertically, PDC sin 45 – PDF sin 45 – PDE sin 45 = 0 PDC – PDF – PDE = 0 PDC – PDF = - 1.06Resolving the forces horizontally,

-PDC cos 45 – PDF cos 45 + 1 + PDE cos 45 = 0-PDC X 0.707 + PDF X0.707 = 1- 1.06 X 0.707 PDC + PDF = 0.25 0.707

= 0.35 KNSolving 1 and 2 , we get PDC = - 0.35 KN PDF = 0.71 KNJoint C : Resolving the forces vertically,

5 PCF - PCD cos 45 – PCB cos 45 = 0 PCF = 0.35 cos 45 + 0.36 cos 45 = 0.5 KNRESULT:

Member Force(KN)

ABBCCDDEEFFABFFDCF

-1.77-0.36-0.35-1.060.752.25-1.420.710.5

Unit2

1. Fig. shows a continuous bean with load. Assume E1 is constant throughout the beam. Drawn SFD & BMD.

Considering the span ABM= Wab/lBM at A MA =0Mx = 2 x 1-8 x1.8/3.6

= 1.8 KN-mMB =0for span BCMB =0 MC=0My= 4x 1.8 x0.6/2.4

= 1.8 KN.m

Apply the theorem of three moments in AB&BC We know that MAl1 +2MB (l1+l2) +Mcl2 +6 [ a1 x1/l1+a2x2/l2]

a1 = ½ bh = ½ x 3.6 x 1.8=3.24 m2a2 = ½ bh = ½ x 2.4 x 1.8

=2.16m2X1= 3.6/2 = 1.8 m.

Centroid from B= a+l/3

Centroid from C= b+l/3

Centroid from B = 0.6+2.4/3= 1m

Centroid from C=2.4+1.8/3= 1.4m

Substitute known values --------------- (1)Consider the beam of span AB &BCA and C are end supports.

MA = MC =0 (support freely)0+2MB (3.6 +2.4)+0 +6 [3.24 x 1.8/3.6+ 2.16 x1.4/2.4] 12 MB = -6 (1.62 +1.26)MB = - 17.28/12 MB = - 1.44kN-mTo find support reactions. Taking moment about B in span ABRAx 3.6 - 2 x 1.8 -MB =0RA x 3.6 - 3.6 +1.44 =0

RA = 3.6 - 1.44/3.6 = 2.16/3.6 =0.6 kNTaking moment about B in span BCRC x 2.4 -4 x 0.6 –MB =0RC x2.4 -2.4 +1.44 =0Rc = 2.4 -1.44/204 = 0.96/2.4 =0.04kN

RA +RB+RC = 2+4 = 0.6 + RB +0.4 = 6RB = 6-0.6-0.4= 5 kN.

Shear forceF at CR =0Fat CL =0.4 KNFat YR =0.4 kN

F at YL =0.4 -4 = - 3.6 kNF at BR = -3.6 kN. F at BR = - 3.6 kN. F at BL = - 3.6 +5 = +1.4 kN. F at XR = +1.4 kN. F at XL = 1.4 -2= -0.6 kN. F at AR = - 0.6 kN. F at AL = - 0.6 +0.6 =0

2.For continuous beam ABCD Loaded page 570 as find (i) memorials at the supports; ii) Reaching at the supports; Drew the B.M and S.F diagrams also.

Solution For span AB MA =0MB=0Max. B.M = Wab/l = 80 x 2 x 3/5

= 96 kN. When a continuous beam has overhangs on one side on n both sides then

the moment for the support that the beam overhangs is determined by treating the overhanging portion as a cantilves. The nest of the analysis is done by applying the theorem of three moments as per usual procedure.

3. Analyze the continuous beam as shown in fig. Draw BMD.

For span AB

MA= 0

MB = - Wa

= - 2000 x 1

= -2000 Nm

Span DEME =0 MD = - Wa= - 1000 x2 = - 2000 Nm

Span BC

Max. BM =

=

= 4000 Nm Span CDMc = M0 = 0

MC = RD x (D x1 ) – triangular load D x x1 x Dx /3 = 4000 x 6 x 3/4 - 6000 x 3 /3 = 18000 – 6000 = 12000 Nm. We know 3 moments equation

MA + 2MB ( ) +MC + 6 =0

span BC & CD a1 = 2 /3 x base x height = 2 /3 x 4x 4000 = 10666.67m2

a1 = 10666.61 x 2 = 21333.34 m3

a2 =2 [(RD ) – ]dx

= 2 (6000 x - 2000/9x3)dx

= 2(6000x2 /2 - 2000/7x4/4)

= 2 ( 3000 x 32 - 2000/9 x 34/4

= 45000 m2

a2 = 45000 x 3 = 135000 m3

-2000 x 4 +2 Mc x10 + (-2000 x 6)+6 (21333.34 /4 + 45000 x 3/6 ) - 8000 +20 Mc – 12000 + 6 (5333.33 + 22500) = 166999.98

= 20 MC = 8000 +12000 = 18

Mc = = - 7349.99 kN.m

4.A continuous beans ABC of constant EI, spam AB cause a load of 1000N and a central clockwise manner of 3000Nm is spam BC, spam AB is 10m, span BC is 15m. Draw SFD & BMD.

SolutionFor Span ABRA=0RB=0

wlR max = _______

4

1000x10= ___________ =25000 Nm

4

Bending moment for span BCRc x 15 – 30000=0

Rc=30000/15

= 2000 NMoment about CRB x 15 – 30000=0

RB=30000/15 = 2000 N(Mc)y Right = 2000 x 7.5 = 15000Nm(Mc)y = 2000 x 7.5 – 30000(Mc)yl = -15000 NmSpan AB & BCA&C are and supports MA = Mc =0MA li +2 MB (l1+l2) + Mcl2+6[a1 x1 /l1+ a2x2/l2]=0

A1 = ½ x 10 x 25000= 125000

X1= 10/2 = 5mapply in(1)0+ 2mb (25) +6 [ 125000 x 5/10]

+[ ½ x 7.5 x 15000 x 2/3 x 7.5+ ½ x 7.5 x (-15000) x (7.5 + 1/3 x 7.5)/15]=050MB + 6 [ 62500 – 18750] = 050MB - 262500MB = - 5250 NmTaking moment about B (Left)RA x 10 – 10000 x 5 + 5250 =0RA = 4475NTaking Moment about B (Right)Rc x 15 – 30000 + 5250 =0Rc = 1650 NRA +RB + Rc = 10000

4475RB = 10000- ___________

1650= 3875 N

Shear forceFat CR =0Fat CC = 1648.5 NFat YR = 1648.5NFat BR = 1648.5 NFat Bc = 1648.5 + 3878.925 = 5527.25NF at XR = 5527.25 NF at X L = 5527.25 – 10000

F at AR = -4472.25 NF at AL= 4472.75+4472.75 =0

5. Draw BMD for propped cantilever beam.

For span ABMA =0MB =0Mmax = Wl / 4

= 10000 x 8 /4 = 20000N

For span BCMB =0Mc=0

For span CDMC =0MD =0

Mcentre =

=5000 x 8x8/8

= 40000 Non

For span DEMD = -5000 x 4 x2= - 40000 Nm

MF =0We know the equation(When at the same level)

Considering span AoA & AB

a1 = 0x1 = 0

a2 = ½ x 8 x 20000

= 80000 NmX2 = 4m

MAo x 0 + 2MA

8MA + 4MB = -12000 -------------------------------- (1)

Span AB and BC

a1 = ½ x 8 x 20000= 80000

= 4ma2 =0

= 4

4MA + 24 MB + 8 Mc =-120000

Span BC & CDIn this span MD = – 40000 Nm

a1 =0

a2 =

=

= 4m

8MB + 24 Mc – 160000 + 32000 =08MB + 24 MC = -160000 ---------------------------------------------------(iii)

From (i), (ii) and (iii)MC=-6551.72 NmMA= -14827 Nm

MC=-344 Nm

UNIT-3ENERGY METHODS

1. Find the vertical deflection at point C of the frame shown in figure. Solution:

Element From Limit Mxx ∂m/∂p

BC C (O,L) PX X

AB B (O,L) PL L

To find vertical deflection,

= 0L∫ ((px/EI) (x) (dx)) + ((pl/EI) (L) (dx))

= (p/EI) 0L∫ ((x^2dx) + (L^2))

= (p/EI) 0L∫ ((L^3/3) + (L^2) (L))

= (P/EI) ((L^3+3L^3)/3)

δ = ((Pl^3)/EI) ((1+3)/3)

= (4/3) ((Pl^3)/EI) δ = (4PL^3)/3EI

2. A cantilever beam of length L is subjected to a tip load P, find the deflection at the tip using Castigliano's theorem.

Solution:

M = - (P+W) X

(∂M/∂W) = -X

δ = (1/EI) 0L∫ (M) (∂M/∂W) dx

= (1/EI) 0L∫ (-(P+W) X) (-X) dx

Put W = 0,

δ = (1/EI) 0L∫ (PX^2) dx

δ = (P/EI) (X^3/3)

δ = (PL^3)/ (3EI)

3. Using unit load method finds the deflection at the free end of the cantilever beam.

Solution:

Mx = m = -PX

M = -x

δ = (1/EI) (-PX) (-X) dx

= (P/EI) (X^2) dx

= (P/EI) [(X^3)/ (3)]

δ = (PL^3)/ (3EI)

4. Using unit load method find the deflection at free end of a cantilever beam carrying udl load. Take E = 200000 Mpa, I = 8250000000 mm^4.

Solution: Mx = (WX^2/2)

m = -X

δ = (1/EI) ∫ (mM) dx

= (1/EI) ∫ (-X) (Wx^2/2) dx

= (1/EI) ∫ (WX^3/2) dx

= (W/2EI) [(X^4)/4]

= (WL^4)/ (8EI)

Substitute the value,

δ = (25 x 12^4 x 1000)/ (8 x 2 x 10^18 x 825)

= 0.0392

= 39.2 mm.

5. Using unit load method finds the slope at point ‘c’ of the beam shown in figure. Take E=200000N/mm^2; I=75000000mm^4.

Solution:

θ = 1/EI ∫Mc. M dx

Portion Origin Limits M

BC B 0 to 3 -40x

AC C 0 to 3 -40(3+x)

θ = θBC + θAC

Portion Origin Limits Mc

BC B 0 to 3 0

AC C 0 to 3 -1

= (1/EI) -40x (0) dx + (1/EI) -40(3+x) (-1) dx

= 0 + (1/150*10^2) 0

3∫ (120+40x) dx = (120*3) (1/150*10^2) + (1/150*10^2) * 40 (X2/2)0

3

= 360(1/150*10^2) + (1/150*10^2) * 180

= 540/ (1/150*10^2) θ = 0.036 radians

Unit4

1. Solid round bar and 2.5m long is used as a shut one and of the strut is fixed, while

its other and is hinged. Find the safe compressive load for this strut, using Euler’s

formula. Assume E = 200EN/m2 and F.O.S. = 3.

End Condition : Owed and formed other end hinged.

c = 2

= 414.1

5.2

= 1.7680m.

Euler’s crippling load is given by the relation.

P Euler = 2

2

c

EL

= 2

2

c

EL

= 2

492

768.1

)06.0(64

10200 xxx

= 401.735KN

Safe compressive load = SOF

PEuler

..

= 3

735.401

= 133.912 KN.

2. I section joints400mmx200mmx20mm and 6m long strut with both ends fixed. What

is Euler’s crippling load for the column? Take Young’s modulus for the joint as

200Gpa.

Solution

Element Area mm2 x(mm) y(mm) h for Ixx h for Iyy

1. 200 x 20 =

4000

100 10 190 0

2. 20 x 360 =

7200

100 200 0 0

3. 200 x 20 =

4000

100 390 -190 0

400072004000

10040002002001004000

xxx

x

= 100 mm

321

332211

aaa

yayayay

400072004000

39040002007200104000

xxx

y

= 200mm.

Ixx1 = 12

3bd

= 12

20200 3x

= 133333.33 mm4.

Ixx2 = 12

3bd

= 12

36020 3x

= 77760000 mm4.

Ixx3 = 12

3bd

= 12

20200 3x

= 133333.33 mm4.

Iyy1 = 12

3bd

= 12

20020 3x

= 13333333.33 mm4.

Iyy2 = 12

20360 3x

= 240000 mm4.

Iyy3 = 12

20020 3x

= 13333333.33 mm4.

Ixx = ∑( Ixx + Ah2)

= [(133333.33 + 4000 x 1902) + 77760000 + 7200 x 02 + 133333.33 +

4000 x (-190)2]

= 366.8 x 106 mm4.

Iyy = ∑( Iyy + Ah2)

= [(13333333.33 + 0 + 240000 + 13333333.33

= 26.9 x 106 mm4.

Since Iyy is less than Ixx, therefore the joint will tend to tuckle in yy direction.

Thus we shall take the value of I as Iyy = 26.9 x 106mm4. Moreover as the column is

fixed at its both ends, therefore equivalent length of the column.

c = 2

= 2

6000

= 3000 mm.

Crippling load for the column

P Euler = 2

2

c

EL

= 2

32

3000

1026410200 xxxx

= 5.89 x 106 N

= 5899.83 KN

Crippling stress or buckling stress

= Area

oadCripplingl

= 4000720014000

1089.5 6

x

= 387.5 N

3. I section 3cm x 2mm top and bottom flange and 5cm x 2mm middle web is used as a

column of length 60cm with one end fixed and the other end free. If E = 70 GPa (i)

Calculate the loads the column can carry (ii) Device the formula used.

Solution

Element Area mm2 x(mm) y(mm) h for Ixx h for Iyy

1. 2 x 30 = 60 15 1 26 0

2. 50x 3 = 150 15 27 0 0

3. 2 x 30 = 60 15 53 -26 0

321

332211

aaa

xaxaxax

6010060

1560151001560

xxx

x

= 15 mm

321

332211

aaa

yayayay

6010060

536027100160

xxx

y

= 27 mm.

Ixx1 = 12

3bd

= 12

230 3x

= 20 mm4.

Ixx2 = 12

3bd

= 12

502 3x

= 20833.33 mm4.

Ixx3 = 12

3bd

= 12

230 3x

= 20 mm4.

Iyy1 = 12

3bd

= 12

302 3x

= 4500 mm4.

Iyy2 = 12

250 3x

= 33.33 mm4.

Iyy3 = 12

302 3x

= 4500 mm4.

Ixx = ∑( Ixx1 + A1h12)

= [(20 + 60 x 262) + (20833.33+ 100 x 02) + (20 +60 x(-26)2]

= 101993.33 mm4.

Iyy = ∑( Iyy1 + A1h12)

= [(4500 + 60 x 0) + 33.33 + 4500

= 9033.33 mm4.

I min = 9033.33 mm4

One end fixed other end free

Le = 2l

= 2 x 600

= 1200 mm

Crippling load = 2

2

c

EL

= 2

32

1200

33.90331070 xxx

= 4333.94 N

4, A column of length 100cm with both ends pinned has I section. The top, bottom

flange and the web have dimensions 10cm x 2cm each. The material is Aluminum with E

= 70 GPa find the buckling stress. Derive the formula used.

3210

521052105210

xxxxxx

x

= 5 cm

3210

1321072101210

xxxxxx

y

= 7 cm.

Ixx1 = 12

3bd

= 12

210 3x

= 6.67 cm4.

Ixx2 = 12

3bd

= 12

102 3x

= 166.67 cm4.

Ixx3 = 12

3bd

= 6.67 cm4.

Ixx = (6.67 + 10 x 2 x 62) + (166.67+ 2 x 10 x 02) + (6.67 + 2)

= 726.67 + 726.67

= 1453.34 cm4.

Iyy = 333.33 + 6.67

= 340 cm4.

P = 2

2

c

EI

= 24

92

100

340

10

1070x

xx

= 2.348 x 106 N

= 2348 KN.

Buckling stress = 3210

2348

xx

= 39.13 KN/cm2

= 01.001.0

13.39

x

= 391300 KN/m2.

5. Calculate the safe load an a hollow cast iron column of 150mm external diameter and

100mm internal diameter. Column length = 10m with me end fixed and the other end

hinged. Use Euler’s formula with a F.O.S and E = 95 GN/m2

2

= 2

10

= 7.07 m

PEuler = 2

2

EL

= 374 KN

Safe load = SOF

PEuler

..

= 74.8 KN.

6. A T – section 150mm x 120mm x 20mm is used as a strut of 4m long with hinged at

its both ends. Calculate the crippling load and buckling stress, if young’s modulus for

the material be 200 GPa.

Solution

Element Area mm2 x(m) y(m) h for Ixx h for Iyy

1. 150x20=3000 75 110 -24 0

2. 20x100=2000 75 50 36 0

21

2211

aa

xaxax

20003000

72000753000

xx

x

= 75 mm

21

2211

aa

yayay

20003000

5020001103000

xx

y

= 86 mm.

Ixx1 = 12

3bd

= 12

20150 3x

= 100000 mm4.

Ixx2 = 12

3bd

= 12

10020 3x

= 1.66 x 106 mm4.

Iyy1 = 12

3bd

= 12

1502 3x

= 5.625 x 106 mm4.

Iyy2 = 66.67 x 103 mm4.

Ixx = ∑( Ixx1 + A1h12)

= [(100000 + 3000 x (-24)2) + (1.66 x 106+ 2000 x 362)

= 6.08 x 106 mm4.

Iyy = 5.625 x 106 + 66.67 x 103

= 5.691 x 106 mm4.

Both end hinged Le =

Crippling load P= 2

2

c

EI

= 2

632

4000

10691.510200 xxxx

= 702 x 103 N

= 702 KN.

Unit 5

1) In a material at a point the major principal stress is 180 MN/m2 and the minor

principal stress is compressive. If the tensile yield pt of the steel is 225 MN/m2. Find the

value of Minor principle stress at which yielding will commence according to each of the

following criteria of failure.

1) Max shear stress

2) Max Total strain energy

3) Max shear strain energy

Take Poisson’s ratio = 0.26

Solution:

T1 = 180.mN/m2

Tt = 225mN/m2

1/m = 0.26

(i)Maximum shear stress

180 - = 225

Compressive)

(ii). Maximum total strain energy theory

= 1802+ (- 2)2

=

2252 =

=

= 96.08 MN/m2

(iii). Maximum shear strain

3) 2

101250 2 = (180-(-

101250 = (180+ +

68850 = (180+

68850 = 1802+360

360 - 36450 =0

= 72.24a MN/m2

2). A circular shaft of tensile yield strength 300Mpa is subjected to a combined state of

loading defined by a bending moment M= 15 KNm. & Torque T= 15 KNm. Calculate the

diameter, d. which the load must have in order to achieve a factor of safety N= 2. Apply

the following theories.

1) Max strain theory

2) Max shear stress theory

3) Max Distortion energy theory

Given:

M = 15 KNm

T = 1.5 KNm

I =

N = 2

(1) M= I

=

M =

b =

T=

=

=

=

=

T1, T2 =

T1 =

Tt = =

= 150 x 103KN/M2

According to maximum shear stress theory

Tt =

150 x 103 = -

150 x 103 =

150 x 103 =

d3 =

d = 0.1129m

d = 112.900

(2) Maximum distortion theory (Shear strain energy theory)

Now according to shear strain energy theory

2 t2 =

=

=

= 2

2 t2 = 2 (

2 t2 =

= x (

= +

-

=

= 4(

=

= (

150x103 =

d3 =

d = .11045m

d= 110.45mm

(3) Maximum Strain Theory

= -

=

=

=-0.3 (15+

d= 0.1089 m d= 108.9 mm

3). A circular shaft of tensile yield strength 300 MPa is subjected to combined state of loading defined by a bending moment M=50 KNm and torque t=20KN-m. Calculate the diameter d, which the bar must have in order to achieve a factor of safety N=3. Apply the following theories (i) Max shear stress theory. (ii) Maximum distortion theory (iii). Octahedral shear stress theory

Solution:

M = 50 KNm

T = 20 KNm

I =

N = 3

(i) M=I

=

M =

b =

T=

=

=

=

=

T1, T2 =

T1 =

Tt = =

= 150 x 103 KN/M2

According to max shear stress theory

Tt =

150 x 103 = -

150 x 103 =

d= 0.156m

(ii) Maximum shear strain theory or distortion energy theory:

2 t2 =

=

=

= 2

2 t2 = 2 (

2 t2 =

= x (

= +

-

=

= 4(

σt2 =

= (

d= 0.155m

Maximum Strain Theory

= -

=

d = 0.156m

4) Find the slope at the support of a simply supported beam of length L and subjected to a

uniformly distributed load by unit load method.

RA =

RB =

M = RA x – wx

= x-

RA + RB = 0

Moment about A

RB L = 1

RB =

RA + RB = 0

RA + =0

RA + -

M0 = RA x + 1

= x +1

=

=

=

=

=

=

=

=

=

5). In figure Load P is supported by a vertical bar DB of length L and Cross section A

and by two equally inclined bars of length L and cross section area A1 Determine the

forces in the bars and also the ratio Which will make the forces in all the bars

numerically equal.

Equating horizontal forces

FAB Cos 45 + FBC Cos45 = o

FAB = FBC

Equating Vertical forces

- FAB Sin45 +FBD-p- FBC Sin 45 = 0

- FAB + FBD – P – FBC =0

Substitute FBC = FAB

=

` =

Total strain energy stored in the frame

WT = WAB + WBC + W BD

=

=

WT =

We Know

LAB = LBC = LBD

EAB = EBC = EBD

FAB can be determined by the condition of Minimum strain energy

, The Condition

FBD=P+ FAB

4AFABL + 2A1p L+4A1FABL=0

4FABL (A + 2A1) = - 2A1 P L

FAB =

FAB =

FAB = P+ FBA

= P +

= P -

= P -

=

=

FAB = -

FAB = -

Defining the forces are numerically equal , the

FAB = FBD

= -