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SBM COLLEGE OF ENGINEERING AND TECHNOLOGY, DINDIGULDEPARTMENT OF MECHANICAL ENGINEERING

Sub : Strength of materials Year : IISem : IV Acc.Year : 2011-2012Faculty Name : Nandagopal.K Subject Code : 10122ME405

QUESTION BANK UNIT I

( STRESS, STRAIN AND DEFORMATION OF SOLIDS)

TWO MARK QUESTIONS

1. Define rigid materials & deformable materials. (Nov 2009)

2. Define strength, stiffness & stability (or) Material properties. (Nov 2003)

3. Define stress, strain & express the mathematical relations between them. (June 2007)

4. State Hookes law. (Nov 2005, June 2009, Apr 2010)

5. Define Modulus of rigidity (or) shear modulus. (Apr 2004, Nov 2006, June 2007)6. Define bulk modulus. (Jun 2006, Nov 2006, Nov 2007)

7. Define poissons ratio & Modular ratio. (Apr 2003, Jun 2006, Nov 2007, June 2009)

8. Define strain energy (or) resilience due to axial load. (Jun 2006)

9. Define proof resilience. (Apr 2010)

10. Define modulus of resilience. ( Apr 2010)

11. Define thermal stress & thermal strain. (June 2009, Nov 2009)

12. What are the governing limitations in compound bar? (Nov 2003)

13. What do you understand by a composite bar? (Apr 2003)

14. Define strain energy. (Nov 2005)

15. Define elasticity & plasticity

16. What are the types of stress?

17. Define simple stress & compound stress.

18. What are the types of strain?

19. Define tensile stress & tensile strain.

20. Define compressive stress & compressive strain.

21. Define shear stress & shear strain. 22. Define factor of safety & load factor.

23. Define yield strength.

24. Define volumetric strain.

25. Relation between E&G, E&K.

26. What are the types of elastic constants & the relation between them?

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SIXTEEN MARK QUESTIONS1. Find the Value of P and change in length of each component and the total change in length of the

bar shown in fig, take E = 200 kN/mm 2. (Apr 2003)(June 2009)

2. A bar 30 mm X 30mm X 250mm long was subjected to a pull of 90 kN in the direction of its

length. The extension of the bar was found to be 0.125 mm, while the decrease in each lateral

dimension was found to be 0.00375 mm. Find the youngs modulus, Poissons ratio, rigidity

modulus and bulk modulus for the material of the bar. (Apr 2003) (Apr 2010)

3. A steel plate 300 mm long, 60 mm wide and 30 mm deep is acted upon by the forces shown in

fig. Determine the change in volume from first principle. E= 200 kN/mm2

and poissons ratio =0.3. (Nov 2003)

4. The composite bar shown in fig, is rigidly fixed at the ends. An axial pull of P=15 kN is applied

at B 20C. find the stresses in each material at 80C. s = 11 x 10-6 per C , a = 24 x 10

-6 per C,

Es= 210 kN/mm2, E a = 70 kN/mm

2. (Nov 2003)

5. A steel rod of 20 mm passes centrally through a copper tube of 50 mm external diameter and 40

mm internal diameter. The tube is closed at each end by rigid plates. If the temperature of the

assembly is raised by 50C, calculate the stress developed in copper and steel. Take E s = 200

kN/mm 2, E c = 100 kN/mm2, s = 12 x 10

-6 per C, c = 18 x 10-6 per C. (Nov 2005)(Jun

2008)(June 2009), (Nov 2009).

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6. Find the stresses in each section of the bar and the total extension of the bar, Take E = 2.1 x 10 5

N/mm 2. (Nov 2006)

7. A mild steel rod of 20 mm diameter and 300 mm long is enclosed centrally inside a hollow upper

tube of external diameter 30 mm and internal diameter of 25 mm. The ends of the tube and rods

are brazed together and the composite bar is subjected to an axial pull of 40 kN. If E for steel and

copper is 200 GN/m 2 and 100 GN/m 2 respectively, find the stresses developed in the rod and

tube. Also find the extension of the rod. (June 2007).

8. An axial pull of 35000 N is acting on a bar consisting of three lengths as shown in fig. If the

youngs modulus = 2.1 x 10 5 N/mm 2, Determine stress in each section and Total extension of the

bar. (Nov 2007).

9. A reinforced concrete column 500 mm X 500 mm in section is reinforced with 4 steel bars of 20

mm diameter, one in each corner. The column is carrying a load of 750 kN. Determine thestresses in concrete and steel bars. Take E s = 210 GPa and E c = 14 GPa. Also, calculate load

carried by steel and concrete. (Nov 2007)(Nov 2009).

10. A compound bar is constructed from three bars each 50 mm wide and 12 mm thick fastened

together to form a bar 50 mm wide and 36 mm thick. The middle bar is of aluminum for which E

= 70 GPa and the outer bars are of brass for which E = 100 GPa. If the bars are initially fastened

at 18C and the temperature of the whole assembly is then raised to 50C, determine the stresses

set up in brass and aluminum. Take coefficient of linear expansion as 18 x 10 -6 for brass and 22 x

10-6

for aluminum. What will be the changes in these stresses if an external compressive load of 15 kN is then applied on the bar? (Apr 2008)

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11. A 25 mm diameter bar is subjected to an axial tensile load of 100 kN. Under the action of this

load a 200 mm gauge length is found to extend 0.19 mm. Determine the modulus of elasticity of

the material. If in order to reduce weight whilst keeping the external diameter constant, the bar is

bored axially to produce a hollow cylinder of uniform thickness, what is the maximum of bore

possible given that the maximum allowable stress is 240 MPa? The load can be assumed to

remain constant at 100 kN. What will be the change in the outer diameter of the bar under the

above load? Taking poissons ratio = 0.3, also calculate the bulk modulus and the shear modul;us

of the material. (Apr 2008)

12. A steel bar 4 m long is acted upon by forces shown in fig. Determine the total elongation of the

bar Take E = 205 kN/mm 2. (June 2009)

13. A compound tube consists of a steel tube of 140 mm internal diameter and 5 mm thickness and an

outer brass tube of 150 mm internal diameter and 5 mm thick. The two tubes are of same length.

Compound tube carries an axial load of 600 kN. Find the stresses carried by each tube and

amount of shortening. Length of the tube is 120 mm. E s= 2 x 105 N/mm 2, E b = 1 x 10

5 N/mm 2.

(June 2009).

14. A steel rod of 30 mm diameter is subjected to a pull of 75 kN. The measured extension on gaugelength of 50 mm is 0.10 mm and change in diameter is 0.004 mm. Calculate Youngs modulus,

Poissons ratio and Bulk modulus. (Nov 2009).

15. The load P is applied on the bars as shown in Fig. Find the safe load P if the stresses in brass and

steel are not to exceed 60 N/mm 2 and 120 N/mm 2 respectively. E for steel = 200 kN/mm 2, E for

brass = 100 kN/mm 2. The copper rods are 40 mm x 40 mm in section and the steel rod is 50 mm x

50mm in section. Length of steel rod is 250 mm and copper rod is 150 mm. (Nov 2009).

16. Derive the expression between Elastic constants (E, G and K) for elastic and isotropic materials.

(Apr 2010)

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TWO MARK ANSWERS

1. Define rigid materials & deformable materials. (Nov 2009)

Rigid materials: A body in which the relative positions of any two arbitrary points is invariantunder any conditions.

Deformable materials: A body which may deform under certain conditions. The conditionswhich involved in the deformation may be due to application of force, temperature, torque etc..

2. Define strength, stiffness & stability (or) Material properties. (Nov 2003)Strength: Ability of a material to resist the externally applied force without breaking (or)

yielding is known as strength of materials.

Stiffness: Is the property of a material to resist elastic deformation (or) deflection. Modulus of elasticity is measured as stiffness.

Stability: Ability of a material to withstand high load without major deformation.

3. Define stress, strain & express the mathematical relations between them. (June 2007)Stress: When an external forces act on a body, it under goes deformation. The internal

resistance offered by the material against the deformation is called the stress. ( Resistance forceoffered by a body against the deformation )

Stress () = Load (P) / Area (A)

Strain: strain is the ratio of the change in dimension to the original dimension.Strain (e) = change in length () / original length ()

4. State Hookes law. (Nov 2005, June 2009, Apr 2010)When the material is loaded within its elastic limit, the stress is directly proportional to

the strain.

E = Stress (f) (or) conventional stress / Strain (e) (or) conventional strain

5. Define Modulus of rigidity (or) shear modulus. (Apr 2004, Nov 2006, June 2007)It is the ratio between shear stress to the corresponding shear strain within its elastic limit.

Modulus of rigidity (G) = Shear stress ( s) / shear strain (e s)

6. Define bulk modulus. (Jun 2006, Nov 2006, Nov 2007)Bulk modulus (K) = Direct stress (Shear stress (or) Normal stress) / volumetric strain.

7. Define poissons ratio & Modular ratio. (Apr 2003, Jun 2006, Nov 2007, June 2009)Poissons ratio: It is the ratio between lateral strain to the corresponding longitudinal

strain within its elastic limit. Poissons rati o (1/m) = Lateral strain (e b (or) e d )/ Longitudinal strain(e l)

Modular ratio : The ratio between the youngs modulus of the two materials. Modular ratio = Youngs modulus of steel (E s)/ Youngs modulus of copper (E c)

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8. Define strain energy (or) resilience due to axial load. (Jun 2006)Whenever a body is strained, some amount of energy is absorbed in the body. The energy

which is absorbed in the body due to straining effect is known as strain energy.Total strain energy stored in the body is known as resilience.

9. Define proof resilience. (Apr 2010)Maximum strain energy which can be stored in the bar without permanent deformation is

called its proof resilience.Proof resilience = ( 2max / 2E) x Volume.

10. Define modulus of resilience. ( Apr 2010)Maximum strain energy which can be stored in the bar per unit volume is called itsmodulus of resilience.

Modulus of resilience = ( 2max / 2E)

11. Define thermal stress & thermal strain. (June 2009, Nov 2009)Thermal stress: Stress in the body due to change in temperature are known as thermal stress

(or) temperature stress. t = T E

Thermal stress depends on a) Co-efficient of linear expansion, b) Change of temperature, c)Youngs modulus.

Thermal strain: Stress in the body due to change in temperature are known as thermal strain(or) temperature strain.

e = Tl = T L T- change in temprature (t 2-t 1)

12. What are the governing limitations in compound bar? (Nov 2003)Elongation (or) contraction of individual material of a composite member is equal. So,

the strain induced in those materials are also equal. e 1 = e 2 (or) P 1l1 / A 1E1 = P 2l2 / A 2E2.

Sum of loads carried by individual materials of the composite member is equal to thetotal load applied on the member.

13. What do you understand by a composite bar? (Apr 2003)A composite bar may be defined as a bar made up of two (or) more different material

joined together in such a manner that the system extends (or) contracts as a whole unit equally,when subjected to axial pull (or) push.

14. Define superposition.When a number of loads are acting on a body, the resulting strain according the principle

of superposition, will be the algebraic sum caused by individual loads.

Total deformation of the body will be equal to the algebraic sum of deformation of theindividual sections.

15. Define creep.If the member subjected to a constant stress, the elongation or deflections continue to

increase with time. This phenomenon is referred to as creep.

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16. Gauge length.If the member subjected to a constant stress, the elongation or deflections continue to

increase with time. This phenomenon is referred to as creep.

17. Define strain energy. (Nov 2005)Whenever a body is strained, some amount of energy is observed in the body. The energy

which is observed in the body due to straining effect is known as strain energy.

18. What is strain energy stored in a body when the load is applied gradually.The strain energy stored, U = (P 2 / 2E) x Volume.

19. Classification of materialRigid materials (or) rigid body Deformable materials (or) Deformable body

Elastic material ( Elastic body) Plastic material (Plastic body)

20. Define elasticity & plasticityElasticity: The property of a material to regain its original shape after deformation when the

applied force is removed is known as elasticity (or) elastic body.Plasticity: The property of a material which enables the formation of permanent deformation

without rupture due to applied force is known as plasticity (or) plastic body.

21. What is mean by elastic limit of the material?For every material the property of assuming (or) regaining its previous shape & size is

exhibited on the removal of the loading, when the intensity of stress is within a elastic limit iscalled as elastic limit.

(Or)There is always a limiting value of load up to which the strain totally disappears on the

removal of load, the stress corresponding to this load is called elastic limit.

22. What is mean by linear elastic material & non-linear elastic material?Linear elastic material: A material is elastic usually implies the stress is directly proportional

to strain, such materials are linear elastic material.

Non-linear elastic material: A material which is elastic but responding in a non linear manner (when unloaded, returning back along the loading path to its initial stress-free state of deformation) is called as non linear elastic material.

23. Define isotropic material & An-isotropic material.Isotropic material: Materials having the same elastic properties in all directions are called

isotropic materials.

An -isotropic material: Materials having different physical properties in all directions arecalled an-isotropic materials.

24. Define homogenous material & give example.Same elastic properties at all point in the body. (Ex: Stainless steel column, wood)

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25. Define load & types of loads.A load may be defined as the combined effect of external forces acting on a body.

Tensile load Compressive load Twisting load Bending load

26. Define Hoop-stress.The tensile stress which is induced circumferentially along the circular ring is called is

called hoop stress.f Hoop = ([(D-d ) /d ] x E )N/m

2 D-larger diameter of ringd- smaller diameter of ringE-Youngs modulus

27. What are the types of stress?Tensile stress Compressive stress Shear stress (or) tangential stress Bending stress Torsional stress

28. Define simple stress & compound stress.Simple stress: When a body is subjected to an external force in one direction only, the stress

in the body is called simple stress.

Compound stress: When a body is subjected to an external force in more than one direction,the stress in the body is called compound stress.

29. What are the types of strain?Tensile strain

Compressive strain Shear strain (or) tangential strain

30. Define tensile stress & tensile strain.When a member is subjected to equal & opposite axial pulls, the length of the member is

increased. The stress induced at any cross section of the member is called tensile stress and thecorresponding strain is known as tensile strain.

Tensile stress = Tensile load / Area Tensile strain = Increase in length / Original length (Ductile Large tensile strain steel,

Brittle relatively small strain C.I, Concrete)

31. Define compressive stress & compressive strain.

When a member is subjected to equal & opposite axial pushes, the body length tends toshorten by increasing its diameter. The stress induced at any cross section of the member is calledcompressive stress and the corresponding strain is known as compressive strain.

Compressive stress = Compressive load / Area Compressive strain = Decrease in length / Original length

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32. Define shear stress & shear strain.When a body is subjected to two equal & opposite forces acting tangentially across the

resisting section, the body tends to shear off across the cross section. The stress induced is calledshear stress and the corresponding strain is called shear strain.

33. Define factor of safety & load factor.Factor of safety = Ultimate stress / Working stressLoad factor = Ultimate load / Working load

34. What are factors of safety & load factor.Reliability of the material

Nature of loading Effect of corrosion & wear

35. Define yield strength.In ductile materials, the force which is applied on the material at the yield point, when the

material is in semi plastic stage.

36. Define yield stress & ultimate stress.Yield stress = Safe load / AreaUltimate stress = Maximum load / Area

37. Distinguish between yield stress &ultimate stress.

Sl.No. Ultimate stress (maximum) Yield stress (safe) 1.

2.

Ultimate stress = Maximum load / Area

Ultimate stress is greater than safe stress

Yield stress = Safe load / Area

yield stress is less than ultimatestress

38. Define volumetric strain.Volumetric strain (e v)= change in volume(V) / original volume (V)

39. What is the volumetric strain for rectangular bar?Volumetric strain (e v)= (l)/l + (b)/b +(d)/d = e l+e b+ e d

40. What is the volumetric strain for cylindrical rod?Volumetric strain (e v)= (l)/l + 2*(d)/d = e l+2*e d

41. What is the volumetric strain for sphere?Volumetric strain (e v)= 3*(d)/d = 3* e d

42. What are the types of elastic constants & the relation between them?E-Youngs modulus, G-Modulus of rigidity, K-Bulk

modulusE = [9KG / (9K+G)]

43. Relation between E&G, E&K.E = 2G(1+1/m) or 2G(1+) E-Youngs modulus , G -Modulus of rigidity, K-Bulk

modulusE = 3K(1-2/m) or 3K(1- 2) 1/m - Poissons ratio

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UNIT 1: FORMULAS

1. Stress, = P/A;

2. Strain, e = l/l;

3. Youngs modulus (or) Modulus of elasticity (E) = stress / strain = /e = Pl/ l*A;

(/e = Linear stress / Linear strain (or) Tensile stress / Tensile strain (or) Compressive stress /

Compressive strain)

4. Analysis of bar with varying cross-section

Change in length l = (l 1 l2 l3) = (P 1l1/A1*E 1) (P 2l2/A2*E 2) (P 3l3/A3*E 3);

(+ Tension, - - Compression)

5. Change in length (Rectangular, circular bar, Hollow bar, tapered bar, tapered rectangular

section)

l = Pl / AE;

(Rectangular, A=l*b; circular, A= / 4D 2; hollow bar, A = (/4 *D 2-d2); Tapered bar, A = (/4

*D 1*D 2); Tapered rectangular section, A = (1/t *(a-b)log e (a/b)) # If thickness is given

means, d=D-2t;

6. Volumetric strain, e v = V/V;

7. Volumetric strain for rectangular bar (e v) = (l/l) + (d/d) + (b/b) = (e l) + (e d) + (e b);

8. Volumetric strain for cylindrical rod (e v) = (l/l) + 2x(d/d) = (e l) + 2*(e d);

9. Volumetric strain for sphere (e v) = 3x(d/d) = 3x(e d);

10. Poissons ratio (1/m) = Lateral strain (or) secondary strain / Linear strain (or) Primary

strain;

11. Lateral strain (or) secondary strain = b/b (Or) d/d;

12. Linear strain (or) Primary strain = l/l;

13. Thermal stress, f, = TE

14. Thermal strain, e = T

15. Percent age contraction of area = [(A-A 1) / A] *100; A1 Area of fracture

16. Relation between E, G & K. E = [9KG / (9K+G)]

17. Relation between E&G, E&K.E = 2G(1+1/m) or 2G(1+) E-Youngs modulus , G -Modulus of rigidity, K-Bulk modulusE = 3K(1-2/m) or 3K(1- 2) 1/m - Poissons ratio

18. Define Hoop-stress.Hoop = ([(D-d ) /d ] x E )N/m

2 D-larger diameter of ringd- Smaller diameter of ringE-Youngs modulus

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19. Define proof resilience.Proof resilience = ( 2max / 2E) x Volume.

20. Define modulus of resilience.Modulus of resilience = ( 2max / 2E)

21. Composite bare1=e 2 (or) l 1 = l2 (P 1l1/A 1E 1 = P 2l2/A2E 2)P=P 1+P 2

22. Thermal stress & strain when supports not yield:Thermal stress, f t = T E, where T - change in temperatures (t 2-t1)Thermal strain, e = T Change in length due to thermal stress, l = T L

23. Thermal stress & strain when supports yield:Thermal stress, f t = ( T (L -))/L*E, where T - change in temperatures (t 2-t1)Thermal strain, e = ( T(L -)/L)

24. Thermal stress in composite bar(Free expansion of steel + Expansion due to tensile stress in steel = Free expansion of copper

- Contraction due to tensile stress in copper)( s . T.L) + (f s / E s).L = ( b . T.L) - ( b / E b).L