ME2254-STRENGTH OF MATERIALS Questions Bank 2014

Anna University, Chennai

QUESTION BANK

ME2254-STRENGTH OF MATERIALS DEPARTMENT OF MECHANICAL ENGINEERING

SRINIVASAN ENGINEERING COLLEGE- PERAMBALUR

UNIT- I

STRESS STRAIN DEFORMATION OF SOLIDS

1. A rod of 150 cm long and diameter 2.0cm is subjected to an axial

pull of 20 KN. If the modulus of elasticity of the material of the rod is 2x

105 N/mm2 Determine 1. Stress 2. Strain 3. the elongation of the rod

(16)

2. The extension in a rectangular steel bar of length 400mm and thickness

10mm is found to 0.21mm .The bar tapers uniformly in width from

100mm to 50mm. If E for the bar is 2x 105 N/mm2 ,Determine the axial

load on the bar (16)

3. Find the young’s modulus of a rod of diameter 30mm and of length

300mm which is subjected to a tensile load of 60 KN and the extension of the rod is equal to 0.4 mm

(16)

4. The extension in a rectangular steel bar of length 400mm and thickness

3mm is found be 0.21mm .The bar tapers uniformly in width from

20mm to 60mm E for the bar is 2x 105 N/mm2 Determine the axial load on the bar. (16)

5. The ultimate stress for a hollow steel column which carries an axial load of 2Mn is 500 N/mm2 .If the external diameter of the column is

250mm, determine the internal diameter .Take the factor of safety as 4.0

(16)

6. A steel rod of 20mm diameter passes through a brass tube of 30 mm inside diameter and 40mm outside diameter . Two nuts one on each side are tightened until a stress of 10 n/mm2 is induced in the rod. F1.2xind the stresses in the rod and the tube if the assembly is heated through 600C. Take

Es = 2x105N/mm2, αs =1.2x10-5/0C

Eb = 0.8x105N/mm2, αb =1.2x10-5/0C

7. A solid steel rod of 5 m length and 10 mm diameter is subjected to an axial load of 5 kN . Find the stress induced in the rod if the load is applied (a) gradually,(b) suddenly and (c) impact after falling from a height of 150 mm. Also find the strain energy stored in the rod under given conditions. Take E= 200 kN/mm2.

8. A rod of diameter 10 mm and length 1.5 mm hangs vertically from the ceiling of a roof. A collar is attached at its lower end on which a load of 250 N falls from a height of 200 mm. Find the strain energy absorbed and the instantaneous deflection of the rod. Take E = 200 kN/ mm2.

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UNIT II

BEAMS – LOADS AND STRESSES

1. Three planks of each 50 x200 mm timber are built up to a symmetrical I section for a beam. The maximum shear force over the beam is 4KN. Propose an alternate rectangular section of the same material so that the maximum shear stress developed is same in both sections. Assume then width of the section to be 2/3 of the depth. (16)

2. A beam of uniform section 10 m long carries a udl of KN/m for the entire length and a concentrated load of 10 KN at right end. The beam is freely supported at the left end. Find the position of the second support so that the maximum bending moment in the beam is as minimum as possible. Also compute the maximum bending moment (16)

3. A beam of size 150 mm wide, 250 mm deep carries a uniformly distributed load of w kN/m over entire span of 4 m. A concentrated load 1 kN is acting at a distance of 1.2 m from the left support. If the bending stress at a section 1.8 m from the left support is not to exceed 3.25

N/mm2 find the load w. (16)

4. A cantilever of 2m length carries a point load of 20 KN at 0.8 m from the fixed end and another point of 5 KN at the free end. In addition, a u.d.l. of 15 KN/m is spread over the entire length of the cantilever. Draw the S.F.D, and B.M.D. (16)

5. A Simply supported beam of effective span 6 m carries three point

loads of 30 KN, 25 KN and 40 KN at 1m, 3m and 4.5m respectively from the left support. Draw the SFD and BMD. Indicating values at salient points.

(16

)

6. A Simply supported beam of length 6 metres carries a udl of

20KN/m throughout its length and a point of 30 KN at 2 metres from the right support. Draw the shear force and bending moment diagram.Also find the position and magnitude of maximum Bending moment.

(16)

7. A Simply supported beam 6 metre span carries udl of 20 KN/m for left half of span and two point loads of 25 KN end 35 KN at 4 m and 5 m from left support. Find maximum SF and BM and their location drawing SF and BM diagrams.

(

16)

8. A horizontal beam of 10 m long is carrying a usdl load of 1 kN/m. The beam is supported on two supports 6 m apart. Find thee position of the supports so that the bending moment on the beam is as small as possible. Also draw the shear force and bending moment diagram.

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UNIT- III TORSION

1. Determine the diameter of a solid shaft which will transmit 300 KN at 250 rpm. The maximum shear stress should not exceed 30 N/mm2 and twist should not be more than 10 in a shaft length 2m.

Take modulus of rigidity = 1x 105N/mm2. (16)

2. The stiffness of the closed coil helical spring at mean diameter 20 cm is made of 3 cm diameter rod and has 16 turns. A weight of 3 KN is dropped on this spring. Find the height by which the weight should be dropped before striking the spring so that the spring may be compressed

by 18 cm. Take C= 8x104 N/mm2. (16)

3. It is required to design a closed coiled helical spring which shall deflect 1mm under an axial load of 100 N at a shear stress of 90 Mpa. The spring is to be made of round wire having shear modulus of 0.8 x

105 Mpa. The mean diameter of the coil is 10 times that of the coil wire.

Find the diameter and length of the wire. (16)

4. A steel shaft ABCD having a total length of 2400 mm is contributed by three different sections as follows. The portion AB is hollow having outside and inside diameters 80 mm and 50 mm respectively, BC is solid and 80 mm diameter. CD is also solid and 70 mm diameter. If the angle of twist is same for each section, determine the length of each portion and the total angle of twist. Maximum permissible shear stress is 50 Mpa and shear modulus

0.82 x 105 MPa (16)

5. The stiffness of close coiled helical spring is 1.5 N/mm of compression under a maximum load of 60 N. The maximum shear stress in the wire of the spring is 125 N/mm2. The solid length of the spring (when the coils are touching) is 50 mm. Find the diameter of coil, diameter of wire and number of coils.

C = 4.5 (16)

6. A solid shaft of length 3.5 m and diameter 25 mm rotates at a frequency of

40 Hz. Find the maximum power to be transmitted by the shaft, assuming the maximum shear stress in the shaft is limited to 40 Mpa and angle of twist does not exceed 60. Take G = 80 Gpa

7. Find the maximum torque that can be applied safely to a shaft of diameter

300 mm. The permissible angle of twist is 1.50 in a length of 7.5 m and a maximum shear is limited to 42 Mpa. Take G = 84.4 Gpa.

8. A solid steel shaft has to transmit 100 kW at 160 r.p.m. Taking allowable shear stress as 70 Mpa, find the suitable diameter of the shaft. The maximum torque transmitted in each revolution exceeds the mean by 20%.

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UNIT-IV BEAMDEFLECTION

1. A simply supported beam of 10 m span carries a uniformly distributed load of 1 kN/m over the entire span. Using Castigliano’s theorem, find the slope at the ends. EI = 30,000 kN/m2.

(16)

2. A 2m long cantilever made of steel tube of section 150 mm external diameter and10mm thick is loaded. If E=200 GN/m2 calculate (1) The value of W so that the maximum bending stress is 150 MN/m (2) The maximum deflection for the loading (16)

3. A beam of length of 10 m is simply supported at its ends and carries two point loads of100 kN and 60 kN at a distance of 2 m and 5 m respectively from the left support. Calculate the deflections under each load. Find also the maximum deflection. Take I = 18 X 108 mm4 and E =

2 X 105. (16)

4. i) A column of solid circular section, 12 cm diameter, 3.6 m long is hinged at both ends. Rankine’s constant is 1 / 1600 and σc = 54 KN/cm2. Find the buckling load. ii) If another column of the same length, end conditions and rankine constant but of 12 cm X 12 cm square cross- section, and different material, has the same buckling load, find the value of σc of its material. (16)

5. A beam of length of 6 m is simply supported at its ends. It carries a uniformly distributed load of 10 KN/m as shown in figure. Determine the deflection of the beam at its mid-point and also the position and the maximum deflection. Take EI=4.5 X 108 N/mm2. (16)

6. An overhanging beam ABC is loaded as shown is figure. Determine the deflection of the beam at point C. Take I = 5 X 108 mm4 and E = 2 X

105 N/mm2. (16)

7. A cantilever of length 2 m carries a uniformly distributed load of 2.5

KN/m run for a length of 1.25 m from the fixed end and a point load of 1

KN at the free end. Find the deflection at the free end if the section is rectangular 12 cm wide and 24 cm deep and E=1 X 104 N/mm2

(16)

8. A cantilever of length 2m carries a uniformly distributed load 2 KN/m over a length of 1m from the free end, and a point load of 1 KN at the free end. Find the slope and deflection at the free end if E = 2.1 X 105 N/mm2 and I = 6.667 X 107 mm4 . (16)

9. Determine the section of a hollow C.I. cylindrical column 5 m long with ends firmly built in. The column has to carry an axial compressive load of 588.6 KN. The internal diameter of the column is 0.75 times the

external diameter. Use Rankine’s constants. a = 1 / 1600, σc = 57.58

KN/cm2 and F.O.S = 6. (16)

(10). A cantilever beam of length L carries a udl of intensity w per unit length over its entire span and a point load P at its free end. Find the maximum deflection of the beam using castingliano’s theorem.

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UNIT-V

ANALYSIS OF STRESSES IN TWO DIMENSIONS

1. A Thin cylindrical shell 3 m long has 1m internal diameter and 15 mm metal thickness. Calculate the circumferential and longitudinal stresses induced and also the change in the dimensions of the shell, if it is subjected to an internal pressure of1.5 N/mm2 Take E = 2x105

N/mm2 and poison’s ratio =0.3. Also calculate change in volume. (16)

2. A closed cylindrical vessel made of steel plates 4 mm thick with plane ends, carries fluid under pressure of 3 N/mm2 The diameter of the cylinder is 25cm and length is 75 cm. Calculate the longitudinal and hoop stresses in the cylinder wall and determine the change in diameter, length and Volume of the cylinder. Take E =2.1x105 N/mm2 and 1/m =

0.286.

(16)

3. A rectangular block of material is subjected to a tensile stress of 110

N/mm2 on one plane and a tensile stress of 47 N/mm2 on the plane at right angle to the former plane and a tensile stress of 47 N/mm2 on the

plane at right angle to the former. Each of the above stress is

accompanied by a shear stress of 63 N/mm2 Find (i) The direction and

magnitude of each of the principal stress (ii) Magnitude of greatest shear stress (16)

4. At a point in a strained material, the principal stresses are100 N/mm2 (T) and 40 N/mm2 (C). Determine the resultant stress in magnitude and

direction in a plane inclined at 600 to the axis of major principal stress. What is the maximum intensity of shear stress in the material at the point? (16)

5. A rectangular block of material is subjected to a tensile stress of 210

N/mm2 on one plane and a tensile stress of 28 N/mm2 on the plane at right angle to the former plane and a tensile stress of 28 N/mm2 on the

plane at right angle to the former. Each of the above stress is

accompanied by a shear stress of 53 N/mm2 Find (i) The direction and

magnitude of each of the principal stress (ii) Magnitude of greatest shear stress (16)

6 .A closed cylindrical vessel made of steel plates 5 mm thick with plane ends, carries fluid under pressure of 6 N/mm2 The diameter of the cylinder is 35cm and length is 85 cm. Calculate the longitudinal and hoop stresses in the cylinder wall and determine the change in diameter, length and Volume of the cylinder. Take E =2.1x105 N/mm2 and 1/m =

0.286.

(16)

7. At a point in a strained material, the principal stresses are 200 N/mm2 (T) and 60 N/mm2 (C) Determine the direction and magnitude in a plane inclined at 600 to the axis of major principal stress. What is the maximum intensity of shear stress in the material at the point

(16)

8. At a point in a strained material, the principal stresses are 100 N/mm2 (T) and 40 N/mm2 (C) Determine the direction and magnitude in a plane inclined at 600 to the axis of major principal stress. What is the maximum intensity of shear stress in the material at the point

(16)

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ME2254-STRENGTH OF MATERIALS Two Marks Questions With Answers 2014

Anna University, Chennai

2 MARK QUESTIONS WITH ANSWERS

ME2254-STRENGTH OF MATERIALS DEPARTMENT OF MECHANICAL ENGINEERING SRINIVASAN ENGINEERING COLLEGE-PERAMBALUR

UNIT- I

STRESS STRAIN DEFORMATION OF SOLIDS

1. Define stress.

When an external force acts on a body, it undergoes deformation. At the

same

time the body resists deformation. The magnitude of the resisting force

is numerically equal to the applied force. This internal resisting force per unit area is called stress. Stress σ= Force/Area, P/A Unit N/mm2

2. Define strain

When a body is subjected to an external force, there is some change of

dimension in the body. Numerically the strain is equal to the ratio of change in length to the original length of the body

Strain = Change in length/Original length

e = ∂L/L

3. State Hooke’s law.

It states that when a material is loaded, within its elastic limit, the stress

is directly proportional to the strain.

E=σ/e

E= Stress / Strain, unit is N/mm2

Where,

E - Young‟s modulus

σ - Stress e - Strain

4. Define shear stress and shear strain.

The two equal and opposite force act tangentially on any cross sectional

plane of the body tending to slide one part of the body over the other part. The stress induced is called shear stress and the corresponding

strain is known as shear strain.

5. Define Poisson’s ratio.

When a body is stressed, within its elastic limit, the ratio of lateral strain

to the longitudinal strain is constant for a given material. Poisson‟ ratio (μ or 1/m) = Lateral strain /Longitudinal strain

6. State the relationship between Young’s Modulus and Modulus of

Rigidity.

E = 2G (1+1/m)

Where, E – Young’s Modulus, K - Bulk Modulus, 1/m – Poisson’s ratio

7. Define strain energy

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.

8. Give the relationship between Bulk Modulus and Young’s

Modulus.

E = 3K (1-2ν)

Where, E – Young’s Modulus, K - Bulk Modulus, ν – Poisson’s ratio

9. What is compound bar?

A composite bar composed of two or more different materials joined

together

such that system is elongated or compressed in a single unit.

10. Define- elastic limit

Some external force is acting on the body, the body tends to

deformation. If the

force is released from the body its regain to the original position. This is called elastic limit

11. Define – Young’s modulus

The ratio of stress and strain is constant with in the elastic limit.

E = Stress/Strain

12. Define Bulk-modulus

The ratio of direct stress to volumetric strain.

K = Direct stress/Volumetric strain

13. Define- lateral strain

The strain right angle to the direction of the applied load is called lateral

strain.

13. Define- longitudinal strain

When a body is subjected to axial load P, The length of the body is

increased. The axial deformation of the length of the body is called longitudinal strain.

14. What is principle of super position?

The resultant deformation of the body is equal to the algebraic sum of

the deformation of the individual section. Such principle is called as principle of super position

15. Define- Rigidity modulus

The shear stress is directly proportional to shear strain.

N = Shear stress/Shear strain

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UNIT II

BEAMS - LOADS AND STRESSES

1. Define beam?

BEAM is a structural member which is supported along the length and

subjected To external loads acting transversely (i.e) perpendicular to the center line of the beam.

2. What is mean by transverse loading on beam?

If a load is acting on the beam which perpendicular to the central line of

it then it is called transverse loading.

3. What is Cantilever beam?

A beam one end free and the other end is fixed is called cantilever beam.

4. What is simply supported beam?

A beam supported or resting free on the support at its both ends.

5. What is mean by over hanging beam?

If one or both of the end portions are extended beyond the support then

it is called over hanging beam.

6. What is mean by concentrated loads?

A load which is acting at a point is called point load.

7. What is uniformly distributed load.

If a load which is spread over a beam in such a manner that rate of

loading „w‟ is uniform through out the length then it is called as udl.

8. Define point of contra flexure? In which beam it occurs?

Point at which BM changes to zero is point of contra flexure. It occurs in

overhanging beam.

9. What is mean by positive or sagging BM?

BM is said to positive if moment on left side of beam is clockwise or

right side of the beam is counter clockwise.

10. What is mean by negative or hogging BM?

BM is said to negative if moment on left side of beam is

counterclockwise or right side of the beam is clockwise.

11. Define shear force and bending moment?

SF at any cross section is defined as algebraic sum of all the forces

acting either side of beam. BM at any cross section is defined as algebraic sum of the moments of all the forces which are placed either

side from that point.

12. When will bending moment is maximum?

BM will be maximum when shear force change its sign.

13. What is maximum bending moment in a simply supported beam

of span ‘L’ subjected to UDL of ‘w’ over entire span?

Max BM =wL2 /8

14. In a simply supported beam how will you locate point of maximum bending moment?

The bending moment is max. When SF is zero. Write SF equation at that

point and equating to zero we can find out the distances „x‟ from one end .then find maximum bending moment at that point by taking all

moment on right or left hand side of beam.

15. What is shear force?

The algebraic sum of the vertical forces at any section of the beam to the

left or right of the section is called shear force.

16. What is shear force and bending moment diagram?

It shows the variation of the shear force and bending moment along the

length of the beam.

17. What are the types of beams?

1. Cantilever beam

2. Simply supported beam

3. Fixed beam

4. Continuous beam

5. over hanging beam

18. What are the types of loads?

1. Concentrated load or point load

2. Uniform distributed load

3. Uniform varying load

19. In which point the bending moment is maximum?

When the shear force change of sign or the shear force is zero

20. Write the assumption in the theory of simple bending?

1. The material of the beam is homogeneous and isotropic.

2. The beam material is stressed within the elastic limit and thus obey

Hooke’s law.

3. The transverse section which was plane before bending remains plains after bending also.

4. Each layer of the beam is free to expand or contract independently

about the layer, above or below.

5. The value of E is the same in both compression and tension.

21. Write the theory of simple bending equation?

M/ I = F/Y = E/R

M - Maximum bending moment

I - Moment of inertia

F - Maximum stress induced

Y - Distance from the neutral axis

E – Young’s modulus R – Radius of curvature

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UNIT – III TORSION

1. Define Torsion

When a pair of forces of equal magnitude but opposite directions acting

on body, it tends to twist the body. It is known as twisting moment or torsion moment or simply as torque.

Torque is equal to the product of the force applied and the distance

between the point of application of the force and the axis of the shaft.

2. What are the assumptions made in Torsion equation

The material of the shaft is homogeneous, perfectly elastic and obeys

Hooke’s Law.

Twist is uniform along the length of the shaft

The stress does not exceed the limit of proportionality

The shaft circular in section remains circular after loading

Strain and deformations are small.

3. Define polar modulus

It is the ratio between polar moment of inertia and radius of the shaft. =

J/R polar moment of inertia = J Radius R

4. Write the polar modulus for solid shaft and circular shaft.

polar moment of inertia = J Radius R

J = D4 / 32

5. Why hollow circular shafts are preferred when compared to solid circular shafts?

The torque transmitted by the hollow shaft is greater than the solid shaft.

For same material, length and given torque, the weight of the hollow shaft will be less compared to solid shaft.

6. Write torsion equation

T/J=Cθ/L=q/R

T-Torque, θ-angle of twist in radians

J- Polar moment of inertia

C-Modulus of rigidity

L- Length

q- Shear stress R- Radius

7. Write down the expression for power transmitted by a shaft

P=2NT/60

N-speed in rpm

T-torque

8. Write down the expression for torque transmitted by hollow shaft

T= (/16) (D4-d4)/d4

T-torque

q- Shear stress

D-outer diameter d- Inner diameter

9. Write down the equation for maximum shear stress of a solid circular section

in diameter ‘D’ when subjected to torque ‘T’ in a solid shaft.

T=/16 D3 Where T-torque, q-Shear stress and D-diameter

10. Define tensional rigidity

Product of rigidity modulus and polar moment of inertia is called

torsional rigidity

11. What is composite shaft?

Some times a shaft is made up of composite section i.e. one type of shaft

is sleeved over other types of shaft. At the time of sleeving, the two shafts are joined together, that the composite shaft behaves like a single shaft.

12. What is a spring?

A spring is an elastic member, which deflects, or distorts under the

action of load and regains its original shape after the load is removed.

13. State any two functions of springs.

1. To measure forces in spring balance, meters and engine indicators.

2. To store energy.

14. What are the various types of springs?

i. Helical springs

ii. Spiral springs iii. Leaf springs

iv. Disc spring or Belleville springs

15. Classify the helical springs.

1. Close – coiled or tension helical spring.

2. Open –coiled or compression helical spring.

16. What is spring index (C)?

The ratio of mean or pitch diameter to the diameter of wire for the spring

is called the spring index.

17. What is solid length?

The length of a spring under the maximum compression is called its

solid length. It is the product of total number of coils and the diameter of wire.

Ls = nt x d Where, nt = total number of coils.

18. Define spring rate (stiffness).

The spring stiffness or spring constant is defined as the load required per

unit deflection of the spring. K= W/y

Where W –load and y – Deflection

19. Define pitch.

Pitch of the spring is defined as the axial distance between the adjacent

coils in uncompressed state. Mathematically

Pitch=free length/ n-1

20. Define helical springs.

The helical springs are made up of a wire coiled in the form of a helix

and are primarily intended for compressive or tensile load.

21. What are the differences between closed coil & open coil helical

springs?

The spring wires are coiled very closely, each turn is nearly at right

angles to the axis of helix The wires are coiled such that there is a gap between the two consecutive turns. Helix angle is less than 10o Helix angle is large (>10o)

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UNIT-IV BEAMDEFLECTION

1. What are the methods for finding out the slope and deflection at a section?

The important methods used for finding out the slope and deflection at a

section

in a loaded beam are

1. Double integration method

2. Moment area method

3. Macaulay‟s method

The first two methods are suitable for a single load, where as the last one is suitable for several loads.

2. Why moment area method is more useful, when compared with double integration?

Moment area method is more useful, as compared with double

integration method because many problems which do not have a simple mathematical solution can be simplified by the ending moment area

method.

3. Explain the Theorem for conjugate beam method?

Theorem I :“The slope at any section of a loaded beam, relative to the

original

axis of the beam is equal to the shear in the conjugate beam at the

corresponding section”

Theorem II: “The deflection at any given section of a loaded beam,

relative to the

original position is equal to the Bending moment at the corresponding

section of the conjugate beam”

4. Define method of Singularity functions?

In Macaulay’s method a single equation is formed for all loading on a beam, the equation is constructed in such away that the constant of

Integration apply to all portions of the beam. This method is also called

method of singularity functions.

5. What are the points to be worth for conjugate beam method?

1. This method can be directly used for simply supported Beam

2. In this method for cantilevers and fixed beams, artificial constraints need to be supplied to the conjugate beam so that it is supported in a manner

consistent with the constraints of the real beam.

6. What are the different sections in which the shear stress distribution is to be obtained?

Rectangular section

Circular section

I- section

T- section

Miscellaneous section

7. What do you mean by shear stress in beams?

The stress produced in a beam, which is subjected to shear forces is

known as stresses.

8. What is the formula to find a shear stress at a fiber in a section of a beam?

The shear stress at a fiber in a section of a beam is given by

F = shear force acting at a section

A = Area of the section above the fiber

Y = Distance of C G of the Area A from Neutral axis

I = Moment of Inertia of whole section about N A

b = Actual width at the fiber

9. What is the shear stress distribution rectangular section?

The shear stress distribution rectangular section is parabolic and is given by

q = F/2I [d2 /4 – y2]

d = Depth of the beam

y = Distance of the fiber from NA

10. What is the shear stress distribution Circular section?

q = F/3I [R2-y2]

11. State the main assumptions while deriving the general formula for shear sresses.

The material is homogeneous, isotropic and elastic

The modulus of elasticity in tension and compression are same. The shear stress is constant along the beam width

The presence of shear stress does not affect the distribution of bending stress.

12. Define: Shear stress distribution

The variation of shear stress along the depth of the beam is called shear

stress distribution

13. What is the ratio of maximum shear stress to the average shear stress for the

rectangular section?

Qmax is 1.5 times the Qavg.

14. What is the ratio of maximum shear stress to the average shear stress in the case of solid circular section?

Qmax is 4/3 times the Qave.

15. What is the shear stress distribution value of Flange portion of the I-section?

q= f/2I * (D2/4 - y)

D-depth

`y- Distance from neutral axis

16. What is the value of maximum of minimum shear stress in a rectangular cross

section?

`Qmax=3/2 * F/ (bd)

17. What is the shear stress distribution for I-section?

The shear stress distribution I-section is parabolic, but at the junction of

web and flange, the shear stress changes abruptly. It changes from F/8I [D2 –d2] to B/b x F/8I [D2 –d2] where D = over all depth of the section

d = Depth of the web

b = Thickness of web

B = Over all width of the section.

18. How will you obtained shear stress distribution for unsymmetrical section?

The shear stress distribution for Unsymmetrical sections is obtained after

calculating the position of N A.

19 Where the shear stress is max for Triangular section?

In the case of triangular section, the shear stress is not max at N A. The

shear stress is max at a height of h/2

20. Where shear stress distribution diagram draw for composite section?

The shear stress distribution diagram for a composite section, should be

drawn by calculating the shear stress at important points.

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UNIT V

ANALYSIS OF STRESSES IN TWO DIMENSION

1. State principle plane.

The planes, which have no shear stress, are known as principal planes.

These planes carry only normal stresses.

2. Define principle stresses and principle plane.

Principle stress: The magnitude of normal stress, acting on a principal plane is known as principal stresses.

Principle plane: The planes which have no shear stress are known as

principal planes.

3. What is the radius of Mohr’s circle?

Radius of Mohr‟s circle is equal to the maximum shear stress.

4. What is the use of Mohr’s circle?

To find out the normal, resultant stresses and principle stress and their

planes.

5. List the methods to find the stresses in oblique plane?

1. Analytical method

2. Graphical method

6. Define thin cylinder?

If the thickness of the wall of the cylinder vessel is less than 1/15 to 1/20

of its

internal diameter, the cylinder vessel is known as thin cylinder.

7. What are types of stress in a thin cylindrical vessel subjected to internal pressure?

These stresses are tensile and are know as Circumferential stress (or

hoop stress ) and Longitudinal stress.

8. What is mean by Circumferential stress (or hoop stress) and

Longitudinal stress?

The stress acting along the circumference of the cylinder is called

circumferential stress (or hoop stress) whereas the stress acting along the length of the cylinder is known as longitudinal stress.

17. What are the formula for finding circumferential stress and longitudinal stress?

Circumferential stress (f1) is given by as f1 = p x d / 2t x η1 and the

longitudinal stress (f2) is given by f2 = p x d / 2t x ηc

.

18. What are maximum shear stresses at any point in a cylinder? Maximum shear stresses at any point in a cylinder, subjected to internal fluid pressure is given by f1 –f2 / 2 = pd / 8t

19. What are the formula for finding circumferential strain and longitudinal strain?

The circumferential strain (e1) and longitudinal strain (e2) are given by

e1 = pd / 2tE (1- 1/2m), e2 = pd / 2tE (1/2 – 1/m).

20. What are the formula for finding change in diameter, change in length and change volume of a cylindrical shell subjected to internal fluid pressure p?

d = pd2 /2tE (1 – 1/2m),

L = pdL /2tE (1/2 – 1/m),

V = pd /2tE (5/2 – 2/m) x volume,

21. What are the formula for finding principal stresses of a thin cylindrical shell Subjected to internal fluid pressure p and a torque?

Major Principal Stress = f1 + f2 / 2+ {(f1 - f2 /2)2 + fs2}

Minor Principal Stress = f1 + f2 / 2 + {(f1 - f2 /2)2 + fs2}

Maximum shear stress = ½ [Major Principal Stress - Minor Principal

Stress]

Where f1 = Circumferential stress, f2 =Longitudinal stress,

fs =shear stress due to torque.