qb me5413 structural mechanics jan may.2014

Question Bank April/May 2014 ME5413 Structural Mechanics

SAVEETHA SCHOOL OF ENGINEERING SAVEETHA UNIVERSITY

CHENNAI 602 105 Department of Mechanical Engineering

Academic Year 2013- 2014 / Even Semester

QUESTION BANK

Sub. Code/Name: ME5413-Structural Mechanics Year/Sem: II/IV

UNITISTRESSANDSTRAIN

GeneralConceptsofStressStressAnalysisofAxiallyLoadedBarsStrainStressStrain

RelationsHooksLawThermalStrainandDeformation.

Part A (2 Marks)

1. Define stress.

Ans: Force of resistance per unit area, offered by a body against deformation is known as

stress.

2

2

, - Stress induced in a body in N/m - Force acting on the body in N - Cross sectional area of the body in m

PA

where

PA

=

2. Define strain.

Ans: It is the ratio of change in dimension to the original dimension of the body is called

strain.

, - Strain in a body (no units)

- Change in length of the body in m - Length of the body in m

dLL

where

dLL

=

- Change in length of the body in m

Dept. of Mech. Engg. SSE, Saveetha University Page 1 of 30

strain normal==L

Fig. a. Tensile stress

Fig. b. Compressive stress


3. State Hooks law.

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

to strain produced in the body.

2

- stress induced within elastic limit in N/m - strain produced in the body

4. Define modulus of elasticity.

Ans: The ratio of tensile stress or compressive stress to the corresponding strain is a constant.

This ratio is known as Youngs Modulus or Modulus of Elasticity and is denoted by E.

Tensile Stress Compressive Stress or Tensile Strain Compressive Strain

E =

5. Define shear stress.

Ans: When a body is subjected to two equal and opposite forces acting tangentially across

the resisting section, as a result of which the body tends to shear off the section, then the

stress induced is called shear stress ( ) . Tangential forceResisting area

=

6. Define shear modulus.

Ans: The ratio of shear stress to the corresponding shear strain within elastic limit is known

as shear modulus denoted by C or N

Shear Stress or or Shear Strain

C G N =



7. Define principle of super position.

Ans: When numbers of loads are acting on a body, the resulting strain, according to the

principle of super position will be the algebraic sum of strains caused by individual loads.

8. Define composite bar.

Ans: A bar made up of two or more bars of equal length but of different materials rigidly

fixed with each other is called composite bar.

9. Define thermal stress.

Ans: The stress induced in the body due to change in temperature are known as thermal

stresses.

10. Define modular ratio.

Ans: The ratio of E1/E2 is defined as modular ratio of first material to second material. st

1nd

2

Young's modulus of 1 materialModular ratioYoung's modulus of 2 material

EE

= =

11. Define factor of safety.

Ans: It is the ratio of ultimate stress to the working stress.

u

all

ultimate stress ultimate stressFactor of safety, allowable stress working stress

FS = = =

12. Define shear strain.

Ans: Stress induced in a body when subjected to two equal and opposite forces which are

acting tangentially across the resisting section as a result of which body tends to shear off

across the section is known as shear stress. Corresponding strain is known as shear strain.

13. Define volumetric strain.

Ans: The ratio of change in volume to the original volume of the body (when body is

subjected to a single force or system of forces) is called volumetric strain ( )V .


L


3

3

, - change in the volume of the body in m

- volume of the body in m

VV

VwhereV

V

=

14. Define lateral strain.

Ans: The strain at right angles to the direction of the applied load is known as lateral

strain . ( )laterallateral or

, - change in breadth in m- change in width in m

- breadth of the body in m - width of the body in m

b wb w

wherebw

bw

=

15. Define longitudinal strain.

Ans: When a body is subjected to an axial tensile or compressive load there is an axial

deformation in the length of the body. The ratio of axial deformation to the original length of

the body is known as longitudinal strain ( )longitudinal . longitudinal

,- change in length in m

- length of the body in m

LL

whereL

L

=

16. Define bulk modulus.

Ans: When a body is subjected to the mutually perpendicular like and equal direct stresses,

the ratio of direct stress to the corresponding volumetric strain is found to be constant for a

given material when the deformation is within an elastic limit. This ratio is known as bulk

modulus ( ) . KDirect Stress

Volumetric StrainK =



17. Define Poissons ratio.

Ans: The ratio of lateral strain to the longitudinal strain is a constant for a given material

when the material is stressed within the elastic limit. This ratio is known as Poissons ratio

().

Lateral strainLongitudinal strain

=

18. Find the minimum diameter of a steel wire, which is used to raise a load of 4000 N if the

stress in the rod is not to exceed 95 MN/m2.

Ans: Diameter = 7.32 mm

Part B (16 Marks)

1. (a) A rod 150 cm long and of diameter 2 cm is subjected to an axial pull of 20 kN. If the modulus of elasticity of the material of the rod is 2 x 105 N/mm2. Determine i) the stress

ii) the strain and iii) the elongation of the rod. [6]

(b) Draw the stress-strain curve for mild steel subjected to tension and indicate the salient

points with description. [12]

Ans: i) the stress = 63.662 N/mm2 ii) the strain = 0.000318 and iii) the elongation of the rod

= 0.0477 cm.



2. A tensile test was conducted on a mild steel bar. The following data was obtained from

the test

a. Diameter of the steel bar = 3 cm

b. Gauge length of the bar = 20 cm

c. Load at Elastic Limit = 250 kN

d. Extension of a load of 150 kN = 0.21 mm

e. Maximum Load = 380 kN

f. Total Extension = 60 mm

g. Diameter of the rod at the failure = 2.25 cm

Determine i) the Youngs modulus ii) the stress at elastic limit iii) the percentage elongation,

and iv) the percentage decrease in area. [16]

Ans: i) the Youngs modulus = 202.095 GN/m2 ii) the stress at elastic limit = 353.68 MN/m2

iii) the percentage elongation = 30, and iv) the percentage decrease in area = 43.75.

3. The bar shown in Fig. 1.1 is subjected to a tensile load of 160 kN. If the stress in the middle portion is limited to 150 N/mm2, determine the diameter of the middle portion.

Find also the length of the middle portion if the total elongation of the bar is to be 0.2

mm. Youngs Modulus is given as 2.1 x 105 N/mm2. [16]

6 cm 6 cm

Fig. 1.1

Ans: Length of middle portion = 20.714 cm and diameter of the middle portion = 3.685 cm.

160 kN 160 kN

40 cm a


a



10 kN 50 kN 80 kN 20 kN A B C

4. (a) A brass bar, having cross sectional area of 1000 mm2, is subjected to axial forces as

shown in Fig. 1.2. Find the total elongation of the bar. Take E = 1.05 x 105 N/mm2.

Length of AB is 0.6 m, length of BC is 1 m and length of CD is 1.2 m. [12]

D

Fig. 1.2

(b) The extension in a rectangular steel bar of length 800 mm and of thickness 20 mm,

is found to be 0.21 mm. The bar tapers uniformly in width from 80 mm to 40 mm. If E for the

bar is 2 x 105 N/mm2, determine the axial tensile load on the bar. [4]

Ans: (a) Elongation in length AB = 0.2857 mm (Tensile), elongation in length BC = -0.1904

mm (Compressive), elongation in length BD = 0.2095 mm (Tensile) and total elongation in

length AD = -0.1142 mm (Compressive)

(b) The axial tensile load on the bar = 60.6 kN

5. Three bars made of copper, zinc and aluminum are of equal length and have cross section

500, 750 and 1000 mm2 respectively. They are rigidly connected at their ends. If this

compound member is subjected to a longitudinal pull of 250 kN, estimate the proportional

of the load carried on each rod and the induced stresses. Take E of copper = 1.3 x 105

N/mm2, E of zinc = 1 x 105 N/mm2 and E of aluminum = 0.8 x 105 N/mm2. [16]

Ans:

Material Stress Induced in N/mm2 Load Carried

in kN Copper 147.7 73.85

Zinc 113.625 85.218 Aluminum 90.9 90.9


6. (a) A steel rod of 3 cm diameter and 5 m long is connected to two grips and the rod is

maintained at a temperature of 95C. Determine the stress and pull exerted when the

temperature falls to 30 C, if (i) the ends do not yield (ii) the ends yield by 0.12 cm. Take

E = 2 x 105 MN/m2 and = 12 x 10-6/C. [10] (b) Determine the value of Youngs Modulus and Poissons ratio of a metallic bar of

length 30 cm, and depth and breadth of 4 cm when a bar is subjected to axial compressive

load of 400 kN. The decrease in length is given as 0.075 cm and increase in breadth is 0.003

cm. [6]

Ans: (a) (i) stress = 156 N/mm2 and pull in the rod = 110269.9 N

(ii) stress = 108 N/mm2 and pull in the rod = 76340.7 N

(b) Youngs Modulus = 1 x 105 N/mm2, Poissons ratio = 0.3

7. (a) A rod, which tapers uniformly from 5 cm diameter to 3 cm diameter in a length of 50

cm, is subjected to an axial load of 6000 N. If E = 2 x 105 N/mm2. Find the extension of

the rod. [4]

(b) Derive the expression for Youngs modulus in terms of bulk modulus.

[12]

Ans: (a) Extension of the rod = 0.00127 mm



UNITIIGENERALIZEDHOOKESLAWANDPRESSUREVESSELS

IntroductionConstitutiveRelationship forShearGeneralizedConceptsof Strainand

HooksLawThinWalledPressureVesselsThickWalledCylinders.

Part A (2 Marks)

1. Define resilience.

Ans: The total strain energy stored in a body is commonly known as resilience. Whenever

the straining force is removed from the strained body, the body is capable of doing work.

Hence the resilience is also defined as the capacity of a strained body for doing work on the

removal of the straining force.

2. Define proof resilience.

Ans: The maximum strain energy, stored in a body, is known as proof resilience. The strain

energy stored in the body will be maximum when the body is stressed upto elastic limit.

3. Define modulus of resilience?

Ans: It is defined as the proof resilience of a material per unit volume.

Proof ResilienceModulus of Resilience = Volume of the body

4. Define thin cylinders.

Ans: If the thickness of the wall of the cylindrical vessel is less than 1/20 or 1/50 of its

internal diameter, the cylindrical vessel is known as thin cylinders.

5. Define hoop stress in thin cylinder.

Ans: The stress acting along the circumference of the cylinder is called circumferential stress,

also known as hoop stress.



6. A cylindrical pipe of diameter 1.5 m and thickness 1.5 cm is subjected to an internal fluid

pressure of 1.2 N/mm2. Determine hoop stress.

Ans: Hoop stress = 30 N/mm2.

1

21

2

2,

- hoop stress or circumferential stress in N/m

- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin cylinder in m

pdt

where

ptd

=

7. Define longitudinal stress in thin cylinder.

Ans: The stress acting along the length of the cylinder (i.e., in the longitudinal direction) is

known as longitudinal stress.

8. A cylinder of internal diameter 2.5 m and of thickness 5 cm contains a gas. If the tensile

stress in the material is not to exceed 80 N/mm2, determine the internal pressure of the gas.

Ans: Internal pressure = 3.2 N/mm2.

9. What will be maximum shear stress at any point in a thin cylinder which is subjected to

internal fluid pressure?

Ans: Maximum shear stress ( at any point in a thin cylinder which is subjected to internal fluid pressure is given as

)max



1 2max

1 22

12

2

2 8,

and are tensile stress which are perpendicular to each other.- hoop stress or circumferential stress in N/m- longitudinal stress in N/m

- internal fluid pressure in

pdt

where

p

= =

2N/m - thickness of thin cylinder in m - diameter of thin cylinder in m

td

10. What you meant by efficiency of a joint in thin cylinder?

Ans: In case of a joint (longitudinal and circumferential joint), holes are made in the material

of the shell for the rivets. Due to the holes, the area offering resistance decreases. Due to this,

stress developed in the material of the shell will be more.

1 2

21

22

2

and 2 4,


- longitudinal stress in N/m

- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin

l c

pd pdt t

where

ptd

= =

cylinder in m - efficiency of a longitudinal joint - efficiency of a circumferential joint

l

c

11. What is the expression for circumferential strain in thin cylinder?

Ans: The expression for circumferential strain in thin cylinder is given as

1

12

2

12 2

,- circumferential strain

- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin cylinder in m - Young's modulus of the material in N/m -

pdtE

where

ptdE

=

Poisson's ratio



12. What is the expression for longitudinal strain in thin cylinder?

Ans: The expression for longitudinal strain in thin cylinder is given by

2

22

2

12 2,

- longitudinal strain

- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin cylinder in m - Young's modulus of the material in N/m - Po

pdtE

where

ptdE

=

isson's ratio

13. What is the expression for volumetric strain in thin cylinder?

Ans: The expression for volumetric strain in thin cylinder is given by

2

2

5 22 2,

- volumetric strain

- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin cylinder in m - Young's modulus of the material in N/m - Poi

V

V

pdtE

where

ptdE

=

sson's ratio

14. Define hoop stress in thin sphere.

Ans: Hoop or circumferential stress induced in the material of the shell is given by

1

21

2

4,


- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin cylinder in m

pdt

where

ptd

=

15. A spherical vessel 1.5 m diameter is subjected to an internal pressure of 2 N/mm2. Find

the thickness of the plate required if maximum stress is not to exceed 150 N/mm2 and joint

efficiency is 75%.


Ans: Thickness = 6.67 mm


16. A vessel in the shape of a spherical shell of 1.2 m internal diameter and 12 mm shell

thickness is subjected to pressure of 1.6 N/mm2. Determine the stress induced in the material

of the vessel.

Ans: Stress induced is 40 N/mm2.

17. What is the expression for volumetric strain in thin sphere?

Ans: The expression for volumetric strain in thin sphere is given by

( )

2

2

3 14,

- volumetric strain

- internal fluid pressure in N/m - thickness of thin cylinder in m - diameter of thin cylinder in m - Young's modulus of the material in N/m - Poisson's

V

V

pdtE

where

ptdE

=

ratio

18. Define thick cylinders.

Ans: If the thickness of the wall of the cylindrical vessel is more than 1/20 of its internal

diameter, then it is known as thick cylinders.

19. Write the expression for radial pressure and hoop stress at any radius x for thick cylinder

(Lames equation)?

Ans: The expression for radial pressure at any radius x for thick cylinder is given by

2

2

-

,, - constant obtained from boundary conditions

- internal fluid pressure at any radius in N/m

x

x

bp ax

wherea bp x

=

The expression for hoop stress at any radius x for thick cylinder is given by

2

2

1 0

, - hoop stress at any radius in N/m

, - constant obtained from boundary conditions, which are (i) at , or the pressure of fluid inside the cylinder, and

x

x

x

b ax

wherex

a bx r p p

= +

= =2 (ii) at , 0 or atmosphere pressure.xx r p= =



Part B (16 Marks)

1. (a) Derive an expression for strain energy stored in a body when a load is applied

gradually. [10]

(b) A tensile load of 60 kN is gradually applied to the circular bar of 4 cm diameter and 5

m long. If the value of E = 2 x 105 N/mm2. Determine the strain energy absorbed by the rod.

[6]

Ans: Strain Energy = 35.81 Nm

2. (a) Derive an expression for strain energy stored in a body when a load is applied

suddenly. [10]

(b) Calculate the instantaneous stress produced in a bar 10 cm2 in area and 3 m long by a

sudden application of a tensile load of unknown magnitude. If the extension of the bar due to

suddenly applied load is 1.5 mm. Also determine the suddenly applied load. Take E = 2 x 105

N/mm2 . [6]

Ans: Instantaneous Stress = 100 N/mm2. Load = 50 kN.

3. (a) Derive expression for circumferential stress or hoop stress for a thin cylindrical vessel

subjected to an internal fluid pressure. [8]

(b) A cylindrical pipe of diameter 1.5 m and thickness 1.5 cm is subjected to an internal

fluid pressure of 1.2 N/mm2. Determine: i) longitudinal stress developed in the pipe, ii)

circumferential stress developed in the pipe, and iii) maximum shear stress. [8]

Ans: i) Longitudinal Stress = 30 N/mm2 ii) Circumferential stress = 60 N/mm2 iii) Maximum

shear stress = 15 N/mm2

4. (a) A thin spherical of 1.2 m internal diameter is subjected to an internal pressure of 1.6

N/mm2. If the permissible stress in the plate material is 80 N/mm2 and joint efficiency is

75%, fine the minimum thickness. [4]

(b) A boiler is subjected to an internal steam pressure of 2 N/mm2. The thickness of boiler

plate is 2.6 cm and permissible tensile stress is 120 N/mm2. Find out the maximum diameter,

when efficiency of longitudinal joint is 90% and that of circumferential joint is 40%.

[12]

Ans: (a) Thickness = 8 mm, (b) Maximum diameter = 192 cm



5. A cylindrical thin drum 80 cm diameter and 3 m long has a shell thickness of 1 cm. If the

drum is subjected to an internal pressure of 2.5 N/mm2. Determine i) change in diameter ii)

change in length iii) change in volume. Take E = 2 x 105 N/mm2, Poissons ratio = 0.25.

[16]

Ans: i) change in diameter = 0.035 cm ii) change in length = 0.0357 cm iii) change in

volume = 1507.96 cm3.

6. (a) Derive an expression for change in diameter of thin spherical shell due to an internal

pressure. [8]

(b) A spherical shell of internal diameter 0.9 m and of thickness 10 mm is subjected to an

internal pressure of 1.4 N/mm2. Determine the increase in diameter and increase in volume.

Take E = 2 x 105 N/mm2 and = 1/3. [8] Ans: (b) Increase in diameter = 0.0945 mm, increase in volume = 12028.5 mm3.

7. Find the thickness of metal necessary for a cylindrical shell of internal diameter 160 mm

to withstand an internal pressure of 8 N/mm2, the maximum hoop stress is 35 N/mm2.

[16]

Ans: Thickness of the shell = 20.96 mm

8. A thick spherical shell of 200 mm internal diameter is subjected to an internal fluid

pressure of 7 N/mm2. If the permissible tensile stress in the shell material is 8 N/mm2, find

the thickness of the shell. [16]

Ans: Thickness of the shell = 49.3 mm



UNITIII TORSIONANDCOLUM S N

Introduction Torsion of Circular Elastic Bars Torsion of Inelastic Circular Bars

Columns Example of Instability Criteria for Stability of Equilibrium Column

BucklingTheoryDesignOfColumns.

Part A (2 Marks)

1. Define torsion in shaft.

Ans: A shaft is said to be in torsion when equal and opposite torque are applied at two ends

of the shaft. The torque is equal to the product of the force applied (tangentially to the ends of

the shaft) and radius of the shaft.

Torque = Force Radius

2. Write the relation between torque, polar moment of inertia and shear stress.

Ans: The relation between torque, polar moment of inertia and shear stress is given by

2

2

,- Torque in Nm - maximum shear stress induced at outer surface of the shaft in N/m - shear modulus or modulus of rigidity in N/m - angle of twist in radian - radius of the shaft

T CJ R LwhereT

C

R

= =

4

in m - length of the shaft in m - polar moment of inertia in m

LJ



3. Write down the assumptions made in derivation of shear stress produced in a circular

shaft subjected to torsion.

Ans: 1. The material of the shaft is uniform throughout, 2. Twist along the shaft is uniform, 3.

The shaft is of uniform circular cross section throughout, 4. Cross section of shaft remains

plane after twist, 5. All radii which are straight before twist remain straight after twist.

4. A solid shaft of 150 mm diameter is used to transmit torque. Find the maximum torque

transmitted by the shaft if the maximum shear stress induced to the shaft is 45 N/mm2.

Ans: Torque = 29820.586 Nm

3

2

16,

- torque transmitted by the shaft in Nm - maximum shear stress induced at outer surface of the shaft in N/m

- diameter of the shaft in m

T D

whereT

D

=

5. Define the term torsional rigidity of the shaft.

Ans: Torsional rigidity or stiffness of the shaft is defined as the torque required to produce a

twist of one radian per unit length of the shaft. It is the product of modulus of rigidity (C) and

polar moment of inertia of the shaft (J).

6. Write the expression for maximum torque transmitted by a hollow circular shaft.

Ans: The expression for maximum torque ( )T transmitted by a hollow circular shaft is given by

4 4

2

16,

- maximum shear stress induced at outer surface of the shaft in N/m - outer diameter of the shaft in m - inner diameter of the shaft in m

o i

o

o

i

D DTD

where

DD

=

7. Write down the expression for total strain energy in a solid shaft due to torsion.


Ans: The expression for total strain energy in a solid shaft due to torsion is given by


2

2

2

4,

- total shear strain energy in the shaft in Nm - maximum shear stress induced at outer surface of the shaft in N/m - shear modulus or modulus of rigidity in N/m - volume of the shaf

U VC

whereU

CV

=

3t in m

8. Write down the expression for total strain energy in a hollow shaft due to torsion.

Ans: The expression for total strain energy in a hollow shaft due to torsion is given by

( )2 2 22

2

2

4,

- total shear strain energy in the shaft in Nm - maximum shear stress induced at outer surface of the shaft in N/m - shear modulus or modulus of rigidity in N/m - volume o

o io

U V D DCD

whereU

CV

= +

3f the shaft in m - outer diameter of the shaft in m - inner diameter of the shaft in m

o

i

DD

9. Define a composite shaft.

Ans: The shaft made up of two or more materials, and behaving as a single shaft is known as

composite shaft. The total torque transmitted by a composite shaft is the sum of the torques

transmitted by each individual shaft. But the angle of twist in each shaft will be equal.

1 2

1 2

Total torque transmitted, + Angle of twist,

T T T

== =



10. Define column.

Ans: A vertical member of a structure which is under a compressive load and its length is an

order of magnitude larger than either of its other dimensions such a beam is called a columns.

11. Define a strut.

Ans: Strut is member of a structure which is not vertical or whose one or both ends are

pinned or hinged. Examples of struts are: connecting rods, piston rods etc.

12. List out the stresses by which column fails.

Ans: 1. Direct compressive stress, 2. Buckling Stress, 3. Combined compressive stress and

buckling stress.

13. Define buckling load.

Ans: The load at which the column just buckles is known as buckling load or crippling load.

14. List out the types of End conditions of the column.

Ans: a. Both ends are hinged, b. Both ends are fixed, c. One end is fixed and other end is

hinged, and d. One end is fixed and other is free.

15. A solid round bar 3 m long and 5 cm in diameter is used as a strut with both ends

hinged. Determine the crippling load. Take E = 2 x 105 N/mm2.

Ans: Crippling load = 67.288 kN

16. Define slenderness ratio?

Ans: It is the ratio of effective length of the column (L) to the least radius of gyration (k).

LSlenderness ratio =k



17. Write down the expression of crippling stress in terms of effective length.

Ans: The expression of crippling stress in terms of effective length is given by 2

2

2

2

4

Crippling stress

, - Young's modulus in N/m

I - lease radius of gyration = in mA

- Effective length in m

- cross sectional area in m- least moment of inertia in m

e

e

ELk

whereE

k

LAI

=

Part B (16 Marks)

1. Derive an expression for the shear stress produced in a solid circular shaft which is

subjected to torsion. [16]

2. A shaft ABC of 500 mm length and 40 mm external diameter is bored, for a part of its

length AB, to a 20 mm diameter and for the remaining length BC to a 30 mm diameter bore.

If the shear stress is not to exceed 80 N/mm2, find the maximum power, the shaft can transmit

at a speed of 200 rpm. If the angle of twist in the length of 20 mm diameter bore is equal to

that in the 30 mm diameter bore, find the length of the shaft that has been bored to 20 mm

and 30 mm diameter. [16]

Ans: Power transmitted = 14.39 kW, L1 = 289 mm, L2 = 211 mm

3. (a) In a hollow circular shaft of outer and inner diameter of 20 cm and 10 cm respectively,

the shear stress is not to exceed 40 N/mm2.find the maximum torque which the shaft can

safely transmit. [4]

(b) A solid shaft of diameter 80 mm is subjected to a twisting moment of 8 MN mm and a

bending moment of 5 MN mm at a point. Determine: i) principal stresses and ii) position of

the plane on which they act. [12]


Ans: (a) Torque = 58904.86 Nm (b) major principal stress = 143.57 N/mm2, minor principal

stress = 44.1 N/mm2 and position of plane = 118.59


4. A hollow shaft, having an inside diameter 60% of its outer diameter, is to replace a solid

shaft transmitting the same power at the same speed. Calculate the percentage saving in

material, if the material to be used in also the same. [16]

Ans: Percentage saving in material= 29.88 %.

5. A solid circular shaft of 10 cm diameter of 4 m is transmitting 112.5 kW power at 150

r.p.m. Determine (a) the maximum shear stress induced in the shaft, (b) strain energy stored

in the shaft. (Take C = 8 104 N/mm2) [16]

Ans: Max. Shear Stress = 36.5 N/mm2, Strain Energy = 130793 Nmm.

6. A solid round bar 3 m long and 5 cm in diameter is used as a strut. Determine the

crippling load when (i) both ends are hinged, (ii) Both ends are fixed and (iii) one end is fixed

and other end is hinged. Take E = 2105N/mm2 [16]

Ans: i) crippling load both ends are hinged = 67. 288 kN

ii) crippling load both ends are fixed = 269.152 kN

iii) crippling load one end is fixed and other end is hinged = 13.458 N

7. Using Eulers formula, calculate the critical stresses for a series struts having

slenderness ratio 40, 120, 200 under the following conditions: (i) Both ends hinged, (ii) Both

ends fixed. [16]

Ans: i) critical stresses when both ends hinged a) for slender ratio 40, 1264.54 N/mm2, b)

for slender ratio 120, 140.5 N/mm2, c) for slender ratio 200, 50.58 N/mm2.

ii) critical stresses when both ends fixed a) for slender ratio 40, 5058.16 N/mm2, b) for

slender ratio 120, 562.02 N/mm2, c) for slender ratio 200, 202.32 N/mm2.



UNITIVBEAMS

CalculationofReactionSFD,BMDforCantileverBeams,SimplySupportedBeamsand

OverhangingBeamsBeamDeflectionbyDirectIntegrationandMomentAreaMethod.

Part A (2 Marks)

1. Define shear force.

Ans: The algebraic sum of vertical forces at any section of the beam to the right or left of the

section is known as shear force.

2. Define bending moment.

Ans: The algebraic sum of moments of all the forces acting to the right or left of the section

is known as bending moment.

3. Define shear force diagram.

Ans: The diagram which shows the variation of the shear force along the length of a beam is

called shear force diagram (SFD).

4. Define bending moment diagram.

Ans: The diagram which shows the variation of the bending moment along the length of the

beam is called bending moment diagram (BMD).

5. Define beam.

Ans: Beam is a structural member which is supported along the length and subjected to

external loads acting transversely i.e., perpendicular to the centre line of the beam.

6. What is meant by transverse loading on beams?

Ans: If a load is acting on the beam which is perpendicular to the centre line of it then it is

called transverse loading.

7. List the type of beams.

Ans: a. Simply supported beam, b. Overhanging beam, c. Cantilever beam, d. Continuous

beam, e. Beam fixed at one end and simply supported at the other end and f. Fixed beam.



8. List the various types of loads.

Ans: a. Point Load, b. Uniformly distributed Load, and c. Uniformly varying Load

9. Define a cantilever beam?

Ans: Beam which is free at one end and fixed at other end is called cantilever beam.

10. Define overhanging beam.

Ans: If the end portion of the beam is extended beyond the support, such beam is known as

overhanging beam.

11. What is simply supported beam?

Ans: A beam supported or resting freely on the supports at its both ends is known as simply

supported beam.

12. What is meant by point or concentrated load?

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

13. What is uniformly distributed load?

Ans: If a load which is spread over a beam in such a manner that rate of loading w is

uniform throughout the length, then it is called uniformaly distributed load (UDL).

14. Define shear stress distribution.


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

distribution.


15. Define the term point of contra flexure.

Ans: The point where bending moment is zero after changing its sign is known as point of

contra flexure or point of inflexion.

16. Write down the expression for slope at the supports of a simply supported beam carrying

a point load at the centre.

Ans: 2

16A BWL

EI = =

Where,

A , B - Slope at the supports (A & B) of simply supported beam in radian W - Point load at the centre in N

L - Length of beam in mm

E - Youngs modulus in N/mm2

I - Moment of inertia in mm4

17. Write down the expression for deflection at the centre of simply supported beam carrying

point load at the centre.

Ans: 48cWLy

EI=

Where,

yc - Deflection at the centre of simply supported beam in mm

W - Point load at the centre in N

L - Length of beam in mm

E - Youngs modulus in N/mm2

I - Moment of inertia in mm4

18. Define Mohr's theorem.

Ans: The change of slope between any point is equal to the net area of the bending moment

diagram between these point divided by EI.


Point of contra flexure


19. Name the areas in which moment area method is conveniently used.

Ans: a. Problems on cantilevers, b. simply supported beam carrying symmetrical loading and

c. Beams fixed at both ends.

Part B (16 Marks)

1. A cantilever beam of length 2 m carries the point loads as shown in Fig. 4.1. Draw the

shear force and bending moment diagram. [16]

300 N 500 N 800 N

A B C D

0.8 m0.7 m0.5 m

Fig. 4.1

2. A cantilever of length 5 m is loaded as shown in Fig. 4.2. Draw the shear force and

bending moment diagram for the cantilever. [16]

1.0 m 1.5 m 2.0 m

3 kN 1 kN/m

2.5 kN A E D C B

0.5 m

Fig. 4.2

3. A simply supported beam of length 10 m carries the uniformly distributed load and two

point loads as shown in Fig.4.3. Draw S.F and B.M diagram for the beam. Also calculate the

maximum bending moment. [16]

50 kN 40 kN

A C D B

10 kN/m

4 m 4 m 2 m

Fig. 4.3

4. State the assumptions made in the theory of simple bending and derive the simple

bending equation. [16]


RA RB


5. Derive an expression for deflection and slope of a beam subjected to uniform bending

moment. [16]

6. (a) Draw the shear force and bending moment diagram for a simply supported beam

carrying a uniformly distributed load. [8]

(b) Calculate slope and deflection of simply supported beam carrying uniformly

distributed load by Mohrs Theorem [8]

7. (a) Draw the shear force and bending moment diagram for a cantilever of length L

carrying a point load at the free end [8]

(b) A beam 6 m long, simply supported at its ends is carrying a point load of 50 kN at its

centre. The moment of inertia of the beam is given as 78 x 106 mm4. If E = 2.1 x 105 N/mm2,

calculate i) Deflection at the centre of the beam ii) Slope at the supports. [8]

Ans: (b) i) Deflection at the centre of the beam = 13.736 mm ii) slope at the supports =

3.935

8. A cantilever of length 2 m carries a point load of 20 kN at the free end and another load

20 kN at its centre. If E = 105 N/mm2 and I = 108 mm4 for the cantilever, then determine by

moment area method, the slope and deflection of the cantilever at the free end. [16]

Ans: slope at the free end = 0.005 radians and deflection at the free end = 7 mm



UNITVSTRESS STRAINTRANSFORMATION&YIELDANDFRACTURECRITERIA

Transformation of Stress Principal Stress in Two Dimension Problems Maximum

ShearStressesinTwoDimensionalProblemsMohrsCircle.

Part A (2 Marks)

1. Define Principal plane.

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

2. Define Principal stress.

Ans: The normal stresses acting on a principal plane are known as principal stress.

3. Name the methods by which stresses on oblique section can be determined?

Ans: a. Analytical Method and b. Graphical Method.

4. What is Mohrs circle?

Ans: It is method of finding normal, tangential and resultant stresses on an oblique plane.

5. Write down the conditions under which Mohrs circle will be drawn.

Ans: (a). Body subjected to two mutually perpendicular principal tensile stresses of unequal

intensities. (b). A body subjected to two mutually perpendicular stress which are unlike and

unequal, and (c). A body subjected to two mutually perpendicular stress accompanied by

simple shear stress.

6. What is the radius of the Mohrs circle?

Ans: Radius of the Mohrs circle is equal to the maximum shear stress

7. What are the planes along which the greatest shear stresses occur?

Ans: Greatest shear stress occurs at the planes which is inclined at 45 to its normal.

8. Define obliquity.

Ans: The angle made by the resultant stress with the normal of the oblique plane, is known as

obliquity.

9. A rectangular bar of cross section area 10000 mm2 is subjected to an axial load of 20 kN.

Determine the normal stress on section which is inclined at an angle of 300.

Ans: Normal Stress, 2cosn = = 1.5 N/mm2, where direct stress, PA = = 2 N/mm2.

10. A rectangular bar of cross-sectional area 10000 mm2 is subjected to an axial load of 20

kN. Determine the shear stresses on a section which is inclined at an angle of 30 with normal cross-section of the bar.



Ans: Shear stress = 0.866 N/mm2

11. Find the diameter of a circular bar which is subjected to an axial pull of 160 kN, if the

maximum allowable shear stress on any section is 65 N/mm2.

Ans: Diameter = 39.58 mm

12. Give the expressions for normal and tangential stresses on an inclined plane when it is

subjected to an axial pull.

Ans: Normal stress, 2cosn = and Tangential stress, sin 22t =

13. What is the angle between planes of maximum and minimum normal stresses?

Ans: The planes of maximum and minimum normal stresses are at an angle of 90 to each

other.

14. At a point in a strained material the principal stresses are 100 N/mm2 (tensile) and 60

N/mm2 (compressive). Determine the normal stress on a plane inclined at 50 to the axis of

major principal stress.

Ans: Normal Stress = 125 N/mm2

15. At a point within a body subjected to two mutually perpendicular directions, the stresses

are 80 N/mm2 and 40 N/mm2 tensile. Each of the above stresses is accompanied by a shear

stress of 60 N/mm2. Determine the normal stress on an oblique plane inclined at an angle of

45 with the axis of minor tensile stress.

Ans: Normal Stress = 120 N/mm2

16. What is the condition for maximum shear stress, when a member is subjected to two

direct stresses in two mutually perpendicular directions accompanied by a simple shear

stress?

Ans: Condition for maximum shear stress is given by


2 1

22 1

2

tan 22

,, two direct stresses mutually perpendicular direction in N/mm an oblique plane inclined at angle with the axis of minor stress in radian simple shear stress in N/mm

where

=


Part B (16 Marks)

1. Derive an expression for normal stress and tangential stress for a member subjected to

direct stress in one plane. [16]

2. a) A rectangular bar of cross- sectional area of 11000 mm2 is subjected to a tensile load P

as shown in Fig. 5.1. The permissible normal and shear stresses on the oblique plane BC are

given as 7 N/mm2 and 3.5 N/mm2 respectively. Determine the safe value of P. [8]

C

PP 60 B

Fig. 5.1

b) The tensile stresses at a point across two mutually perpendicular planes are 120 N/mm2

and 60 N/mm2. Determine the normal, tangential and resultant stresses on the plane inclined

at 300 to the axis of the minor principal stress. [8]

Ans: (a) Safe value of axial pull Stress = 8.083 N/mm2, and P = 88.91 kN

(b) Normal Stress = 105 N/mm2, Tangential Stress = 25.98 N/mm2, Resultant Stress =

108.16 N/mm2.

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 a plane at right angles to the former. Each of the

accompanied by a shear stress of 63 N/mm2. And that associated with the former tensile stress

tends to rotate the block anticlockwise. Find 1) direction and magnitude of each of the

principal stresses. 2) magnitude of the greatest shear stress. [16]

Ans: Major Principal Stress = 148.936 N/mm2, Minor principal stress = 8.064 N/mm2,

direction of principal stress = 3143 or 121 43 and Magnitude of the greatest shear stress

= 70.436 N/mm2


4. The principal stresses at a point in a bar are 200 N/mm2 (tensile) and 100 N/mm2

(Compressive). Determine the resultant stress in magnitude and direction on a plane inclined

at 600 to the axis of major principal stress. Also determine the maximum intensity of shear

stress in the material at the point. Solve using Mohrs circle method. [16]


Ans: Normal Stress = 125 N/mm2, Tangential Stress = 129.9 N/mm2, Resultant Stress =

180.27 N/mm2, Maximum shear stress =150 N/mm2 and inclination of the resultant stress is

46.6

5. An elemental cube is subjected to tensile stresses of 30 N/mm2 and 10 N/mm2 on two

mutually perpendicular planes and a shear stress of 10 N/mm2 on these planes. Draw the

Mohrs circle of stresses. [16]

Ans: Major Principal Stress = 34.2 N/mm2, Minor Principal Stress = 5.86 N/mm2, Maximum

Shear Stress = 14.1 N/mm2 and location of principal planes are 22.5 or 112.5

6. At a certain point in a strained material, the intensities of stresses on two planes at right

angles to each other are 20 N/mm2 and 10 N/mm2 both tensile. They are accompanied by a

shear stress of magnitude 10 N/mm2. Solve using Mohrs Circle. [16]

Ans: Major Principal Stress = 26.2 N/mm2, Minor Principal Stress = 3.82 N/mm2 and

location of principal planes are 31.85 or 121.85


qb me5413 structural mechanics jan may.2014

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