V ch H ig h Te ch D rRR D rSR Engineering College Ava di 2

VTHT AVADI III SEM

Mechanical Engineering Department 1 Strength of Materials

Vel Tech High Tech Dr.RR Dr.SR Engineering College

Avadi – 600 062

A Course Material

on

Strength of Materials

Mechanical Engineering Department

I


VTHT AVADI III SEM


VTHT AVADI III SEM


CONTENT

S.No Particulars Page

1 Unit – I 5

2 Unit – II 42

3 Unit – III 117

4 Unit – IV 155

5 Unit – V 214

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UNIT-1

PART-A

1. A rod of diameter 30mm and length 400mm was found to elongate 0.35mm when it was subjected to a load of 65KN. compute the modulus of elasticity of the material of this rod. (EVEN 2011)

Sol: Elongation ∂L = PL/AE E=PL/A∂L

= 65x10³x400 π/4 x(30)²x0.35

= 105.092 N/mm² 2. What is strain energy and write its unit in S.I. System? (EVEN 2011)

When an elastic material is deformed due to application of external force, internal resistance is developed in the material of the body. Due to deformation, some work is done by the internal resistance developed in the body, which is stored in the form of energy. This energy is known as strain energy. It is expressed in N-m.

3. State Hooke‟s law. (EVEN 2010, EVEN 2013)

It states that when a material is loaded, within its elastic limit, the stress is directly proportional to the strain.

Stress α strain

α e

E = / e Unit is N/mm²

Where, E – Young‘s Modulus, – Stress,

e - Strain.

4. Define Bulk modulus. (EVEN 2010)

When a body is stressed, within its elastic limit, the ratio of direct stress to the corresponding volumetric strain is constant. This ratio is known as Bulk Modulus.

Bulk Modulus = Direct stress/ Volumetric strain

5. Define Poison‟s Ratio. (EVEN 2009)

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‘s ratio, or 1/m = Lateral strain/Longitudinal strain.

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6. What is Thermal stress? (EVEN 2009) When a material is free to expand or contract due to change in temperature, no stress and

strain will be developed in the material. But when the material is rigidly fixed at both the ends, the change in length is prevented. Due to change in temperature, stress will be developed in the material. Such stress is known as thermal stress.

7. The strain induced in an MS bar of rectangular section having width equal to twice the

depth is 2.5x10^-5. The bar is subjected to a tensile load of 4KN. Find the section dimensions of the bar Take E=0.2x10^6 N/mm². (ODD 2008)

Given: e= 2.5x10^-5, P=4x10^3 N, E=0.2x10^6 N/mm²

To find: Breadth b, Depth d

Sol: Stress, = E x e = 0.2x10^6 x 2.5x10^-5

A = P/ = 4x10^3/5

= 800 mm²

Area, A = b x d [b = 2d]

= 2d x d

800 = 2d²

d = 20mm; b=2d=2 x 20 = 40 mm.

Result: 1. Breadth b = 40mm 2. Depth d= 20mm

8. Define Proof Resilience and Modulus of Resilience. (ODD 2008, EVEN 2013)

The maximum strain energy that can be stored in a material within its elastic limit is known as proof resilience.

It is the Proof resilience of the material per unit volume.

Modulus of resilience = Proof resilience

Volume of the body

9. The Young‟s modulus and the shear modulus of material are 120Gpa and 45Gpa respectively. What is its Bulk modulus? (EVEN 2008)

Given:

Young‘s Modulus E = 120 G Pa = 120x10^9 Pa =120x10^9 N/m² = 120x10³ N/mm²

= 5 N/mm²

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Shear Modulus, G = 45Gpa = 45 x 10³ N/mm²

Solution: Young‘s Modulus, E = 9KG

3K + G

120x10³ = 9 x K x 45 x 10³

3 K + 45 x 10³

120x10³ [3 K + 45 x 10³] = 9 x K x 45 x 10³

3 K + 45 x 10³ = 3.375 K

45 x 10³ = 0.375 K

10. Calculate the instantaneous stress produced in a bar of cross sectional area 1000 mm²

and 3m Long by the sudden application of a tensile load of unknown magnitude, if the instantaneous Extension is 1.5 mm. Also find the corresponding load. Take E = 200 G pa. (EVEN 2008)

Given: A = 1000 mm²; L = 3 m = 3000 mm; ∂L = 1.5 mm; E = 200GPa = 200 x 10³ N/mm²

Solution: ∂L = x L / E

1.5 = x 3000 / 200 x 10³

= 2P / A P = x A / 2

= 100 x 1000

2

11. Give the relation between modulus of elasticity and modulus of rigidity. (EVEN 2007)

E = 2G [1 + 1/m]

Where, E – Young‘s Modulus, N/mm², G – Modulus of rigidity, N/mm²,

Bulk Modulus, K = 120 x 10³ N/mm²

= 100 N/mm²

P = 50 x 10³ N

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1/m – Poisson‘s ratio. 12. Write the concept used for finding stresses in compound bars. (EVEN 2007)

(i) Elongation or contraction in each bar is equal. So, the strains induced in those bars are also equal.

Change in length of bar (1) = Change in length of bar (2)

P1L1 = P2L2

A1E1 A2E2

(ii) The sum of loads carried by individual materials of a composite member is equal to the

Total load applied on the member.

Total load, P = Load carried by bar (1) + Load carried by bar (2)

13. Estimate the load carried by a bar if the axial stress is 10 N/mm² and the diameter of bar is 10 mm. (ODD 2006)

Given: Stress, = 10 N/mm²; Diameter D = 10 mmSolution: Stress = Load / Area

= Load / π/4 x (D) ²

10 = Load / π/4 x (10) ²

14. What is the strain energy stored when a bar of 6 mm diameter 1 m length is subjected

to an axial Load of 4 KN, E = 200 KN/mm². (ODD 2006)

Given: Diameter D = 6 mm; Area, A = π/4 x (D) ² = π/4 x (6) ² = 28.274 mm²;

Load, P = 4 KN = 4 x 10³ N; Length, L = 1 m = 1000 mm;

Young‘s Modulus E = 200 KN/mm² = 200 x 10³ N/mm² Solution:

Volume of the bar, V = A x L

= 28.274 x 1000

Stress, = Load / Area = P/A = 4 x 10³ / 28.274

P = P1 + P2

Load, P = 785.39 N

V = 28274 mm³

= 141.472 N/mm²

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Strain energy stored, U = ²/2E x V = (141.472)²/2x200x10³ x 28274

15. A circular rod 2 m long and 15 mm diameter is subjected to an axial tensile load of 30 KN. Find the elongation of the rod if the modulus of elasticity of the material of the rod is

120 KN/mm².(EVEN 2006)

Given:

L = 2 m = 2000 mm; D = 15 mm; Load P = 30 KN = 30 x 10³ N;

Modulus of Elasticity E = 120 KN/mm² = 120 x 10³ N/mm²

To find: (a) Stress (b) Strain (c) Elongation.

Solution:

Stress = Load / Area = P/A = 30 x 10³/ π/4 x (D) ² = 30 x 10³/ π/4 x (15) ²

Young‘s modulus, E = stress/strain = /e; e = /E = 169.85/120 x 10³ = 0.00141

Strain e = Change in length / Original length

e = ∂L/L; ∂L = e x L = 0.001415 x 2000 = 2.83 mm

16. Define Strain energy and write its unit. (EVEN 2006)

When an elastic material is deformed due to application of external force, internal resistance is developed in the material of the body. Due to deformation, some work is done by the internal resistance developed in the body, which is stored in the form of energy. This energy is known as strain energy. It is expressed in N-m.

17. Define Factor of safety. (ODD 98)

It is defined as the ratio of ultimate tensile stress to the permissible stress (working Stress).

Factor of Safety = Ultimate stress / Permissible stress

18. Define Modulus of Rigidity. (EVEN 2003)

U = 1414.7 N-mm

Stress = 169.85 N/mm²

Strain e = 0.001415

∂L = 2.83

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When a body is stressed, within its elastic limit, the ratio of shearing stress to the corresponding shearing strain is constant. This ratio is known as Modulus of rigidity.

Modulus of rigidity Or Shear modulus, G = Shearing stress / Shearing strain.

19. Define Bulk modulus. (ODD2000)

When a body is subjected to a uniform direct stress in all the three mutually Perpendicular directions, the ratio of the direct stress to the corresponding volumetric strain is found to be a constant is called as the bulk modulus of the material and is denoted by K.

20. What do you understand by a compound bar? (ODD 2004)

A composite member is composed of two or more different materials which are joined together so that the system is elongated or compressed as a single unit.

21. What are the types of elastic constants? (EVEN 2000)

There are three types of elastic constant.

1. Modulus of Elasticity or Young‘s Modulus, E

2. Bulk Modulus, K

3. Shear Modulus or Modulus of Rigidity, G.

22. Define strain energy density. (EVEN 2004, 2003)

Strain energy density is defined as the maximum strain energy that can be stored in a material within the elastic limit per unit volume. It is also known modulus of resilience.

23. What is stability? (ODD 2003)

The stability may be defined as an ability of a material to withstand high load without major deformation.

24. Give the relation for change in length of a bar hanging freely under its own weight.

(EVEN 2005)

Change in length, ∂L = PL/AE

Where, P – Axial load.

L – Length of the bar.

E – Young‘s Modulus of the bar.

A – Area of the bar.

25. A brass rod 2 m long is fixed at both its ends. If the thermal stress is not to exceed 76.5

N/mm², calculate the temperature through which the rod should be heated. Take the values of α and E as 17 x 10^-6/K and 90Gpa respectively. (EVEN 2005)

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Solution: Thermal stress, = α T E

76.5 = 17 x 10^-6 x T x 90 x 10³

26. Determine the Poisson‟s ratio and bulk modulus of a material for which young‟s modulus is 1.2 x 10^5 N/mm² and modulus of rigidity is 4.8 x 10^4 N/mm². (ODD 2004)

Solution: Young‘s modulus, E = 2G (1+1/m) 1.2 x 10^5 = 2 x 4.8 x 10^4 (1+1/m)

27. Give the relationship between Bulk modulus and Young‟s modulus. (EVEN‟96, ODD 97)

E = 3K (1+2/m)

Where, E – Young‘s modulus, K – Bulk modulus,

1/m – Poisson‘s ratio. 28. Define shear stress and shear strain (ODD2010)

The two equal and opposite forces act tangentially on any cross sectional plane of a body tending to slide one part of the body over the other part. The stress induced in that section is called shear stress and the corresponding strain is known as shear strain.

29. State the principle of superposition. (ODD2010)

The total deformation is equal to the algebraic sum of the deformation of the individual sections. This principle of finding out the resultant deformation is known as principle of superposition

The change in length of such member is given by,

∂L = P1L1 + P2L2 + P3L3 + …………

AE

30. State Volumetric strain. (EVEN 2002)

Volumetric strain is defined as the ratio of change in volume to the original volume of the body.

Volumetric strain ev = Change in volume

Original volume

T = 50 K

Poisson‟s ratio, 1/m = 0.25

ev = ∂v/v

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31. Define tensile stress and tensile strain.

The stress induced in a body, when subjected to two equal and opposite pulls, as a result of which there is an increase in length, is known as tensile stress. The ratio of increase in length to the original length is known as tensile strain.

32. Define compressive stress and compressive strain.

The stress induced in a body, when subjected to two equal and opposite pushes, as a result of which there is a decrease in length, is known as compressive stress. The ratio of increase in length to the original length is known as compressive strain.

33. Define stress and strain. (ODD 97)

Stress:

The force of resistance per unit area, offered by a body against deformation is known as stress.

Strain:

The ratio of change in dimension to the original dimension when subjected to an external load is termed as strain and is denoted by e. It has no unit.

34. Define modulus of rigidity (EVEN2003)

The ratio of shear stress to the corresponding shear strain when the stress is within the elastic limit is known as modulus of rigidity or shear modulus and is denoted by C or G or N.

35. Give example for gradually applied load and suddenly applied load.

Example for gradually applied load:

When we lower a body with the help of a crane, the body first touches the platform on which it is to be placed. On further releasing the chain, the platform goes on loading till it is fully loaded by the body. This is the case of gradually applied load.

Example for suddenly applied load:

When we lower a body with the help of a crane, the body is first of all, just above the platform on which it is to be placed. If the chain breaks at once at this moment the whole load of the body begins to act on the platform. This is the case of suddenly applied load.

36. What is resilience?

The strain energy stored by the body within elastic limit, when loaded externally is called resilience.

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37. Distinguish between suddenly applied and impact load.

When the load is applied all of a sudden and not step wise is called is suddenly applied load. The load which falls from a height or strike and body with certain momentum is called falling or impact load.

38. Define proof resilience?

The maximum strain energy stored in a body up to elastic limit is known as proof resilience.

39. Define modulus of elasticity. (EVEN 98)

The ratio of tensile stress or compressive stress to the corresponding strain is known

as modulus of elasticity or young‗s modulus and is denoted by E. 40. State principal plane. (EVEN98)

The planes which have no shear stress are known as principal planes. These planes carry only normal stresses.

41. Define Elasticity. (EVEN 2012)

Elasticity is the tendency of solid materials to return to their original shape after being deformed. Solid objects will deform when forces are applied on them. If the material is elastic, the object will return to its initial shape and size when these forces are removed.

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UNIT-I 16 MARK QUESTIONS

PART-B

1. A circular rod of diameter 20 mm and 500mm long is subjected to a tensile force of 45 KN. the modulus of elasticity may be taken as 200 KN/mm2. Find stress, strain and

elongation of the bar due to the applied load. [EVEN 2011]

Given data:

Load p =45 KN = 45 x 103 N

Young‘s modulus E = 200 KN /mm2

Length of the rod, l = 500 mm

Diameter of the rod = 20 mm

Data asked:

Stress, strain, elongation CIRCULAR ROD

Solution:

Cross sectional area; A = πd2/4 =π x 202/4 = 314.159 mm2

Stress =load / area =45x103/314.5 = 143. 24 N/mm2

Strain, e = stress / young‘s modulus =143.24/200x103 = 0.0007162

Elongation δl=Pl/AE = (45x103 x 500) / (314.159 x 200x103) = 0.358 mm

Result:

Stress = 143.24 N/mm2

Strain = 0.0007162

Elongation δl = 0.358 mm

2. A hollow steel rod of tube is to be

used to carry an axial compressive load

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of 140 KN. The yield stress for steel is 250 N/mm2. A factor of safety of 1.75 is to be used in

the design. The following three classes of the tubes of external diameter 101.6 mm are

available. Which section do you recommended from 3.65 mm, 4.05mm and 4.85 mm?

Given data:

Yield stress = 250 N/mm2

Factor of safety = 1.75

External diameter of the tube = 101.6 mm

Load P = 140 KN = 140x10 N HOLLOW STEEL ROD

Data asked:

Which section do you recommended from the following given thickness 3.65 mm, 4.05mm,

4.85mm

Solution:

Working stress or permissible stress = yield stress/FOS = 250/1.75 = 142.85 N/mm2

Cross sectional area A = load / working stress = 140x10 /142.85 = 980 mm2

Also area of the hollow tube = π/4(D2-d2) = 980

π/4 (101.6 – d2) = 980 or d2 = 9074.78

d = 95.26 mm, thickness t = (D – d)/2 = (101.6 – 95.2)/2 =3.169 mm

Hence light section, t = 3.65 mm is enough.

Result:

Thickness t = 3.65 mm is enough

3. The safe stress for a hollow steel column which carries an axial load of 2.1x10 KN is 125

MN/m2. If the external diameter of the column is 30 cm. determine the internal diameter.

[ODD-2007]

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Given data:

Axial load P = 2.1 x 10 KN

Stress = 125 MN/m2 = 125 N/mm2

External diameter of the column=30 cm = 300mm

Data asked:

Internal diameter of the column d

Solution:

Area A = load/stress =2.1x106/125 =16800 mm2 HOLLOW STEEL COLUMN

But area of the hollow A = (π/4) (D2 – d2) = 16800

= (π/4) (300 2- d2) = 16800

d2 = 68598.73;

d=262mm

Result:

Internal diameter of the column d = 262 mm

4. Define the elastic constants and write the relationship between them Modulus of

elasticity, modulus of rigidity and bulk modulus are the three elastic constants. [ODD-2007]

Modulus of elasticity: it is defined as the ratio of stress to strain within the elastic limit and is

usually denoted by letter „E ‗

Modulus of elasticity E = stress/strain = /e, N/mm2

Modulus of rigidity: it is defined as the ratio of shearing stress to shearing strain within the

elastic limit and is denoted by the letter „G‟ or „N‟

Modulus of rigidity G = shear stress/ shear strain = / Ф

Bulk modulus:

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It is defined as the ratio of identical pressure ‗P‘ acting in three mutually perpendicular directions to corresponding volumetric strain is denoted by letter ‗K‘

Bulk modulus K = pressure ‗p‘ acting in three mutually perpendicular direction/volumetric

strain=p/ev

Relationship between the three elastic constants:

Wkt the relationship between elastic constant and rigidity modulus E = 2G (1+1/m) ------- (1)

The relationship between elastic constant and bulk modulus K = 3K (1- 2/m) --------------- (2)

From the equation (1) we get 1/m = (E/2G)-1 substituting it in equation (2)

We get, E = 3K (1- 2((E/2G)-1))

= 3K (1-(E/G) +2)

= 3K (3-(E/G)) = 9K-(3KE/G)

E (1+3K/G) = 9K , E ((E+3K)/G) = 9K

E = 9GK/ (G+3K)

5. A bar of 20 mm diameter is tested in tension. It is observed that when a load of 37.7 KN

is applied, the extension measured over a gauge length of 200 mm is 0.12 mm and

contraction in diameter is 0.0036 mm. find the poison‟s ratio and elastic constants E, G and K. (EVEN 2002)

Given data:

Load P = 37.7 KN = 37.7x103 N

Length l = 200mm

Extension δl =0.12mm

And contraction in diameter δd = 0.0036 mm

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Data asked:

Poison‘s ratio and the value of three elastic constants BAR

Solution:

Area A = (π/4) d2 = (π/4) x202= 314.5 mm2

Linear strain, e = δl/l = 0.12/200 = 0.0006

Lateral strain el = δd/d = 0.0036/20 =0.0018

Wkt, δl = Pl/AE or we can change it as E = Pl/A δl = (37.77x103x200) / (314.5x0.12)

=200004.71 N/mm2

E = 2G [1+ (1/m)] or G = E/2[1+ (1/m)] = 200004.71/2(1+0.3)

Rigidity modulus G = 76924.89 N/mm2

Also we know that E = 3K (1 – (2/m)) or K = E /(1 – (2/m))

K = 200004.71/3(1- 2x0.3) = 166670.59 N/mm2

Result:

Young‘s modulus E = 200004.71 N/mm2

Rigidity modulus G = 76924.89 N/mm2

Bulk modulus K = 166674.59 N/mm2

6.A circular rod of 100 mm diameter and 500mm long is subjected to a tensile force of 1000

KN .Determine the modulus of rigidity , bulk modulus and change in volume if poison‟s ratio = 0.3 and young‟s modulus E = 2x105 N/mm2 (EVEN 2005)

Given data:

Diameter of rod d = 100 mm

Length of the rod l = 500 mm

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Tensile force P = 1000 KN = 1000 x 103 N

Poisson‘s ratio 1/m = 0.3

Young‘s modulus E = 2x105 N/mm2

Data asked:

Modulus of rigidity,

bulk modulus and

change in volume

Solution: CIRCULAR ROD

Wkt E = 2G [1+ (1/m)] = 3K (1 – (2/m))

Rigidity modulus G = E/2[1+ (1/m)] = 2x10 5/2(1+0.3) = 0.7692x10 5 N/mm2

Bulk Modulus K = 2x105/3(1- 2x0.3) = 1.6667x105 N/mm2

Longitudinal stress = P/A =1000x103/ (π/4) x 1002 =127.32 N/mm2

Linear strain ex = stress/young‘s modulus =127.32/2x10 5 = 63.66x10 -5

Lateral strain ey = - (1/m) ex and ez = - (1/m) ex

wkt volumetric strain ev = ex + ey + ez = ex(1 – (2/m))

Volumetric strain ev = 63.66x10-5(1 – 2x0.3) = 25.46x10-5

Also wkt volumetric strain ev = δV/V

Therefore change in volume, δV = ev x V =25.66x10-5 x (π/4) x d2x l

= 25.66x10-5 x (π/4) x1002 x500 = 1000 mm3

Result:

Modulus of rigidity G = 0.7692x105 N/mm2

Bulk modulus K = 1.6667x1055 N/mm2 and

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Change in volume δV = 1000 mm3

7. An axial pull of 35000N is applied on a bar consists of three lengths as shown in fig . the

young‟s modulus E = 2.1x`105 N/mm2.determine (1) stresses in each section (2) total

extension of the bar [ODD-2007 & 2012]

Given data:

Axial pull P = 3500 N

Young‘s modulus E = 2.1x105 N/mm2

Data asked:

Stresses in each section

and total extensions of the bar

Solution:

Tensile stresses section 1

= load/area at section 1

= P/A = 35000/ (π/4) x202 STEPPED BAR

= 111.46 N/mm2

Tensile stresses in section 2


= P/A = 35000/ (π/4) x302

= 49.53 N/mm2

Tensile stress in section 3


= 35000/ (π/4) x502

= 17.83 N/mm2

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Total extension of the bar δl

= (P/E) [l1/A1+l2/A2+l3/A3]

= (35000/2.1x105) [(200/314) + (250/706.5) + (220/1962.5)]

δl = 0.1838 mm

Result:

Tensile stress in section1 = 111.46 N/mm2



Total extension δl = 0.1838 mm

8. The load „p‟is applied on the bars as shown in fig. find the safe load „p‟ of the stresses in the brass and steel are not to exceed 60 N/mm2 and 120 N/mm2 respectively. E –for the

steel = 200N/mm2 and E- for brass = 100 KN/mm2.the copper rods are 40mm x 40 mm in

section and the steel rod is 50mm x 50mm in section length of the steel is 250mm and

copper rod is 150mm [ODD-2009]

Given data:

Stress in brass c = 60 N/mm2

Stress in steel s = 120 N/mm2

E –for steel Es = 200KN/mm2 =200x103 N/mm2

E –for brass Ec = 100KN/mm2 = 100x103 N/mm2

Cross section of the copper rod = 40mmx40mm

Cross section of the steel rod = 50mmx50mm

Length of the copper rod lc = 150mm

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Length of the steel rod ls = 250mm COMPOSITE BAR

Asked:

Find the safe load ‗P‘

Solution:

Area of the copper component Ac = 2(40x40) = 3200mm2

Area of the steel component As = 50x50 = 2500 mm2

Decrease in length of copper = decrease in length of steel

δls =δlc , eclc = esls or (es /ec) = (lc/ls)= 150/250=0.6

( s/ c) = (es/ec)(Es/Ec) = 0.6x(200x103/100x103) = 1.2

s=1.2 c

Where Pc-reaches 60 N/mm2 Ps- will reach 1.2x60 N/mm2 which is less than the permissible

value.

Where P = c Ac + s As = (60x3200) + (72x2500) = 372000 N = 372 KN

Result:

The load is applied on the bar ‗P‘ = 372 KN

9. A steel rod of 25mm diameter is placed inside a copper tube of 30mm internal diameter

and 5mm thickness and the ends are rigidly fixed. The assembly is subjected to a

compressive load of 250KN. Determine the stresses induced in the steel rod and copper

tube. Take the modulus of elasticity of the steel and the copper as 200 GPa and 80 GPa

respectively. [ODD-2006]

Given data:

Diameter of the steel rod = 20mm

Inside diameter of the copper tube d = 30mm

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Outside diameter of the copper tube D = 40mm

(ie calculated from the given thickness t = 5 mm)

Asked:

Stresses induced in the steel rod and copper tube STEEL ROD

Solution:

Area of the steel rod As = (π/4) d2 = (π/4) x252 = 490.6mm2

Area of the copper tube Ac = (π/4) (D2 – d2) = (π/4) (402 – 302) = 549.5 mm2

From the static equilibrium condition

Ps + Pc = P ----------- (1)

Where Ps –load on steel and Pc—load on copper

Wkt, δls = δlc or (Ps l/AsEs) = (Pc l/AcEc)

Both sides length are same since we can eliminate both side ‗l ‘

(Ps/AsEs) = (Pc//AcEc) or Ps = Pc [(AsEs)/ (AcEc)]

Ps = Pc [(490.6x200x103)/ (549.5x80x103)] = 2.23Pc

Ps = 2.23Pc substitute this on equation (1) we get

2.23Pc + Pc = 250x103

3.23Pc = 250x103 or Pc = 77.39KN

The remaining value from ‗P‘ is Ps i.e., Ps = P – Pc = 250x103 – 77.29 = 172.61KN

Stress in steel rod s = load on steel /area of steel rod = Ps /As = 172.61x103/490.6= 351 N/mm2

Stress in copper tube c = load on copper tube/area of copper tube= Ps/As

=77.39x103/549.5=140.8 N/mm2

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Result:

The stress on the steel rod c= 351 N/mm2

The stress on the copper rod c = 140.8 N/mm2

10. Find the total strain energy stored in a steel bar of diameter 50 mm and length 300mm

when it is subjected to an axial load of 150 KN. Take the modulus of elasticity of steel as

200x10 3MPa [ODD-2006]

Given data:

Diameter of the steel rod d = 50mm

Length of the steel rod l = 300mm

Axial load on the steel rod P = 150 KN= 150x103 N STEEL BAR

Modulus of elasticity of steel E = 200x103MPa =200x103 N/mm2

Asked:

Total strain energy stored

Solution:

Total strain energy stored = ( 2/2E) xAl

Stress in the bar =load /Area= 150x103/ (π/4) d2= 150x103/(π/4)x502 =76.43 N/mm2

(76.43)2

Total strain energy stored = x (π/4) x502x300 =8598 N mm

2x200x103

Result:

Total strain energy stored = 8598 N mm

11. Two vertical rods one of steel rod and other of copper rod are each rigidly fixed at the

top and 600mm apart. Diameters and length of the rods are 25mm and 5 m respectively. A

cross bar fixed to the radio at the lower end carries a load of 7 KN such that the cross bar

remains horizontal even after loading. Find the steps in each rod and the portion of the

VTHT AVADI III SEM


load on the cross bar. Assume the modulus of elasticity for steel and copper as 200 KN/mm2

and 100 KN/mm2 respectively. [EVEN-2011]

Given data:

Diameter of the rods dc = ds = 25 mm

Length of the rods ls = lc = 5m =5000mm

Load P= 7KN =7x103 N

Es =200x103 and Ec = 100x103 N/mm2

Data asked:

The stresses in each rod and the position of the

load on the cross bar

Solution:

Wkt, change in length of steel = change in

length of copper ie, δls = δlc COMPOSITE BAR

(Ps ls /AsEs) = (Pc lc /AcEc) here ls = lc also As = Ac

Since the cross bar remains horizontal the extension of the steel and copper rods are equal. Also

these rods have the same original length.

i.e., (Ps /Es) = (Pc /Ec) or Ps = (Es/Ec) Ps= (200x103/100x10 3) Pc

Ps = 2Pc also wkt , Ps + Pc = P

Therefore 2Pc + Pc = P or 3Pc = 7000N ie, Pc = 2333.33N

Also Ps = 7000 – 2333.33 Ps = 4666.66N

Stress in the copper rod c = load on copper rod/area of copper rod = 2333.33/490.5=4.75

N/mm2

Stress in the steel rod s =load on steel rod/area of steel rod=4666.6/490.6=9.51 N/mm2

Now taking the moment about the copper rod and equating the same, we get

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7000 x = Ps x 600

x = (4666.6x600)/7000 = 399.99mm

Result:

Stress in copper rod c = 4.75 N/mm2

Stress in steel rod s = 9.51 N/mm2

The position of the load on the cross bar ‗x‘ = 399.99 mm

12. A bar of copper section 8 mm x 8 mm is subjected to an axial pull of 7000N. The lateral

dimensions of the bar are found to be changed to 7.9985mm x7.9985mm. If the modulus of

rigidity of the material is 0.8x105 N/mm2. Determine the Poisson‟s ratio and modulus of elasticity [EVEN-2012]

Given data:

SQUARE BAR

After pulling the dimension changed from 8mmx8mm to 7.9985mmx7.9985mm

The bar is square in cross section

Therefore the contraction in lateral side = 8mm - 7.9985mm = 1.5x10 -3 mm

Asked:

Poison‘s ratio and modulus of elasticity

Solution:

Wkt stress = load /area of cross section= 7000/ (8x8) =109.37 N/mm2

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The value of the rigidity modulus are given since we can use the relation

E = 2N (1+ (1/m)) =2 x 0.8x105(1+ (1/m)) or E =1.6x105+(1.6x105x(1/m))----(1)

Also wkt, 1/m = el /e=el / ( /E) = E(el / )

But el = δx/x = 1.5x10-3/8 = 1.875x10-4

1/m =E x 1.875x10-4/109.375 or .E = (1/m)x109.37/1.875x10-4 = 583333.3(1/m) --- (2)

Equating the equations (1) and (2) we get

583333.3(1/m) = 1.6x105+1.6x105(1/m)

583333.3(1/m) – 1.6x105(1/m) = 1.6x105

1/m = 0.3779

Substituting this value in equation (2) we get

E = 583333.3x0. 3779

E = 220472.4 N/mm2

Result:

The poison‘s ratio 1/m = 0.3779

The modulus of elasticity E = 220472.4 N/mm2

13. A steel tube of 30 mm external diameter and 20 mm internal diameter encloses a copper

rod of 15mm diameter to which it is rigidly joined at each end. If at a temperature of 10 0C

there is no longitudinal stress. Calculate the stress in the rod and tube when the

temperature is raised to 2000C. Take E- for steel and copper as 2.1x105 N/mm2 and 1x105

N/mm2 respectively. The value of coefficient of linear expansion for steel and copper is

given as 11x10-6 per 0C and 18x10-6 per 0C respectively. [EVEN-2012]

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Given data:

Diameter of the copper rod dc = 15mm

Internal diameter of the steel tube d = 20mm

External diameter of the steel tube D = 30mm

E - for steel tube

Es = 2.1x105 N/mm2 and

E- for copper

Ec = 1x105 N/mm2

Coefficient of linear expansion for steel

αs = 11 x10-6 per0C

Coefficient of linear expansion for copper

αc = 18 x10-6per0C STEEL TUBE

Asked:

The stresses in the rod and tube

Solution:

Area of the copper rod Ac = (π /4) x152 = 56.25π mm2

Area of the steel tube As = (π/4) x (D2 – d2) = (π/4) x (302 – 202) = 125π mm2

As the value of ‗α‘ for copper is more than that of steel, hence the copper rod would more expand more than the steel tube if it were free. Since the two are joined together the copper will

be prevented from expanding its full amount and will be put in compression the steel being put in

tension

For equilibrium of the system, compressive load on copper= tensile load on steel

i.e, c = s (As/Ac) = s x (125π/56.25π) = 2.22 s

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Wkt, the copper rod and steel tube will actually expand by the same amount

Now actual expansion of the steel = free expansion of steel + expansion due to tensile stress

Actual expansion of steel = αsTl + ( s /Es) l

Actual expansion of the copper= free expansion of copper – contraction due to compressive

stress

Actual expansion of the copper = αcTl – ( s/Es) l

But, actual expansion of steel = actual expansion of copper

αsTl + ( s /Es)l = αTl – ( c/Ec)l or αsT + ( s /Es) = αcT – ( c/Ec)

(11x10-6x190) + ( s /2.1x105) = (18x10-6x190) – (2.22 s /1x105)

( s /2.1x105) + (2.22 s /1x105) = (18x10-6x190) – (11x10-6x190)

(0.476 s +2.22 s) 10-5 = 5x10-6x190

2.696 s = (7x10-6x190)/10-5 or s = 49.33 N/mm2

Therefore, c = 2.22 s = 2.22x49.35 =109.5 N/mm2

Result:

The stress in copper rod c = 109.5 N/mm2

The stress in steel tube s = 49.35N/mm2

14. A bar ABCD of steel is 600mm long and the two ends AB and CD are respectively

30mm and 40mm in diameter and each is 150mm in length the middle portion BC being

25mm in diameter. Determine the final length of the bar, when subjected to an axial

compressive load of 120 KN, where young‟s modulus E = 2.1x 105N/mm2 [EVEN-2007]

Given data:

Axial compressive load,

P = 120KN

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Modulus of elasticity

E = 2.1x105 N/mm2

Asked:

Determine the final length of the bar

Solution:

The total contraction by the axial compressive load STEPPED BAR

= (P/E) [(l1/A1) + (l2/A2) + (l3/A3)]

Contraction

δl = (120x103/2.1x105) [(150/ (π/4)302)) + (150/ (π/4)402) + (300/ (π/4)252)]

= 0.538968 mm

The final length of the bar

= original length – contraction

= 600 – 0.58968 = 599.46 mm

Result:

The final length of the bar l = 599.46 mm

15.A load of 2MN is applied on a short concrete column 50mm x 50mm. the column is

reinforced with fair steel are of 10mm diameter one in each corner. Find the stresses in the

concrete and the steel bars. Take E- for steel as 2.1x105 N/mm2 and for concrete as

1.4x104N/mm2 (ODD 2007)

Given data:

Total load applied P = 2MN= 2x106N/mm2

Diameter of the steel bar = 10mm

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Size of the concrete column = 500mmx500mm

E-for steel Es =2.1x105N/ mm2

E-for concrete Econ =1.4x105N/mm2 SHORT CONCRETE COLUMN

Asked:

Find the stresses in the concrete and steel bar

Solution:

Area of the concrete column Ac = 500x500 = 250000mm2

Area of the steel bar As = (π/4)102=314.15mm2

Area of the concrete = area of column – area of steel bar = 250000 – 314.15

= 249685.8mm2

Strain in steel = strain in concrete

( s/Es) = ( c/Ec)

Or s = (Es/Ec) c = (2.1x105/1.4x104) c = 15 c

Ps + Pcon = P or sAs + con Acon = P

15 cx314.15 + conx249685.8 = 2x106

254398 c = 2x106

Therefore c = (2x106/254398) = 7.86 N/mm2

Also s = 15 c=15x7.86 = 117.8 N/mm2

Result:

The stress in concrete con =7.86 N/mm2

The stress in steel s = 117.8 N/mm2

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16.A short bar of length 100mm tapers uniformly from a diameter 40mm to a diameter

30mm and carries an axial compressive load of 200KN. Find the change in length of the bar

[ODD-2009]

Given data:

Taper pin one end diameter = 40mm

And another end diameter = 30mm

Length l = 100mm

Assume young‘s modulus E = 2x105 N/mm2

Asked:

Change in length of the bar

Solution: TAPER BAR

The change in length of the circular taper rod δl = 4Pl/πEd1d2

i.e. δl = (4x200x103x100)/(πx2x105x40x30)=0.10615mm

Result:

The change in length of the circular taper rod δl = 0.10615mm

17.calculate the value of the stress and strain in portion AC and CB of the steel bar shown

in fig. A close fit exits at both of rigid support at room temperature and the temperature is

raised by 750C. Take E-200GPa and α = 12x10-6/0C for steel area of cross sections of AC is

400mm2 and of BC is 800mm2. [AMIE ODD 2002]

Given data:

Rise in temperature t = 750C

Young‘s modulus E = 200GPa

= 200x109/106

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= 200x103 N/mm2

Area of the section AAC = 400 mm2

Area of the section ABC = 800 mm2

Asked:

The value of the stresses in portion AC and BC STEPPED BAR

Solution:

Elongation in portion AC

δAC = lACαT =300x12x10-6x75=270x10-3mm

Strain in portion AC

= δlAC/lAC=270x10-3/300 = 900x10-6

Compressive stress in AC

= E x strain in portion AC

=200x103x900x10-6=180 N/mm2

Strain in portion

CB=δlCB/lCB=300x12x10-6x75/300=900x10-6

Compressive stress in portion CB =200x103x900x10-6 = 180 N/mm2

Result:

The stress in portion AC AC = 180 N/mm2

The stress in portion BC BC = 180 N/mm2

18.calculte the modulus of rigidity and bulk modulus of the cylindrical bar of diameter 25

mm and length 1.6 m. if longitudinal strain in a bar during tensile stress is four times the

lateral strain. Determine the change in volume when the bar is subjected to a hydrostatic

pressure of 100 N/mm2. Take E = 1x105N/mm2 [ODD-98]

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Given data:

Diameter d=25 mm

Length l = 1.6m =1600mm

Direct stress or hydrostatic pressure =100 N/mm2

Asked:

The modulus of rigidity, bulk modulus and volumetric strain

Solution: CYLINDRICAL BAR

Longitudinal strain or linear strain = 4x lateral strain

Or lateral strain /linear strain =1/m = 1/4= 0.25

wkt, young‘s modulus

E = 2G[1+(1/m)]

Or rigidity modulus

G =E/2(1+ (1/m))

= 1x105/2(1+0.25)

= 0.4x105N/mm2

Also wkt, young‘s modulus E= 3K (1 – (2/m))

Or bulk modulus K = E/3(1 – 2x0.25) = 1x105/3(1 – 2x0.25) = 0.66x105N/mm2

Bulk modulus K = direct stress/volumetric strain= /δV

Or volumetric strain δV/V=direct stress/bulk modulus = /K

Therefore, change in volume δV = ( /K) V= (100/0.66x105)(π/4)x252x1600=1178 mm3

Result:

The modulus of rigidity G = 0.4x105 N/mm2

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Bulk modulus K = 0.66x105 N/mm2

Change in volume δV = 1178 mm3

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

in fig. determine the change in volume. Take E = 200 KN/mm2 and Poisson‟s ratio = 0.3 (EVEN 2003)

Given data:

Length l = 300mm

Width b = 60mm

Thickness t= 30mm

Load in x-direction

= 50KN

=50x103N [+ve since tensile load]

Load in y-direction

= -80KN STEEL PLATE

= -80x103N [-ve since compressive load]

Load in z-direction

= 75KN

= 75x103 [+ve since tensile load]

Young‘s modulus

E = 2x105N/mm2

Poisson‘s ratio (1/m) = 0.3

Data asked:

Change in volume δV

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Solution:

Stress in x-direction x = load in x-direction /cross sectional area in x-direction

= (50x103)/ (60x30) = 27.77N/mm2

Stress in y-direction y = load in y-direction /cross sectional area in y-direction

= (-80x103)/ (300x60) = -8.88N/mm2

Stress in z-direction z = load in z-direction /cross sectional area in z-direction

= (75x103)/ (300x60) = 4.16N/mm2

Wkt direct strain δV/V = 1/E ( x + y + z) [1-2/m]

Or change in volume δV = 1/E ( x) + y + z) [1-2/m] V

= (1/2x105) (27.77 – 8.88 + 4.16)x[1 – 2x0.3] 300x60x30

= 24.89 mm3

Result:

The change in volume δV = 24.89mm3

20. reinforced concrete column 500 in section is reinforced with 4 steel bars of 25 diameter one

in each corner,the colummn is carrying a load of 1000kN.Find the stresses in the concrete and ste

A

^ 3

el bars

Take E= 210 N/mm^2 and 14 10^3N/mm.(MAY-13)

s

s c

Given:

area of the column=500 500

=0.25 10^6mm^2

diameter of the steel d =25mm,for steel bar

E =210 10^3N/mm^2,E =14 10^3

N/mm^2

c s

To Find:

1.stresses in concrete and steel bars, ,

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s

s

s

s

Solution:

1: area of steel bar

A = ( ) ̂ 2 2

A = (25) ̂ 2 2

A =490.87mm^2

for four steel bars A =4 490.87

=1963.48mm^2

d

2: area of concrete,

Ac=area of column -area of steel

Ac=0.25 10^6-1963.48

Ac=2.48 10^5mm^2

3:load(P):

P=1000KN

total load=load on steel+load on concrete

P=Ps+Pc 1

by using formula

Ls=Lc

hence L is neglected,

PsLs PcLc

AsEs AcEc

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(1963.48 210 10 ̂ 3) (2.48 10 ̂ 5 14 10 ̂ 3)

Ps=0.118Pc 2

sub 2in 1

P=0.118Pc+Pc

1000 10^3=1.118Pc

Pc=8.94 10^5 N/mm^2

Ps=0.118Pc

Ps=0.118 8.94 10^5

Ps=1.05 10^5 N/m

Ps Pc

m^2

s

s

s

c

c

stress on steel bar

=

1.05 10 ̂ 5 =

1963.48

=53.47 N/mm^2

stress on concrete =

8.94 10 ̂ 5 =

2.48 10 ̂ 5

Ps

As

Pc

Ac

c =3.60 N/mm^2

c

s

Result:

1.) =3.60 N/mm^2

2.) =53.47 N/mm^2

21. a solid circular bar of diameter 20 mm when subjected to an axial tensile load of 40KN,

the reduction in diameter of the rod was observed as 6.4 x 103 mm. The bulk modulus of the

material of the bar is 67Gpa. Determine the following:

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(i)Young‟s Modulus (ii) Poisson‟s ratio (iii) Modulus of rigidity (iv) Change in length per metre (v) Change in volume of the bar per metre length.(EVEN-2013)

Given:

D = 20mm

P = 40 KN

δd = 6.4 x 103 mm

K = 67Gpa = 67 x 103 N/mm2

To Find:

(i)Young‘s Modulus (ii) Poisson‘s ratio (iii) Modulus of rigidity (iv) Change in length per metre (v) Change in volume of the bar per metre length.

Solution:

l

t

l

34

t

t

4

l

4l

l

lateralstrain Poisson 's ratio

linear strain

e1 (1)

m e

d 6.4 10 e 3.2 10

d 20

sub (e ) valuein equation (1)

1 3.2 10

m e

e 3.2 10 m

Tensilestress PE (

Tensilestrain or linear strain e Ae

(i)Poisson'sratio(μ)

3

2 2 4l

5

P )

A

40 10 127.3E

20 e 20 3.2 m4 4

mE 3.9 10 (2)

2E 3K(1 )

m

π π

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5 3

5

3

m 2 E 3K( )

m

mE 3k(m 2)

3.9 10 3 67 10 (m 2)

3.9 10 (m 2)

3 67 10

5

3

3.9 10 m 2

3 67 10

m 1.94 2

m 3.94

1 0.25

m

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5

5

53 6

2 2

sub m valuein equation (2)

mE 3.9 10

0.25E 3.9 10

3.9 10 N NE 1560 10 1.5 10 mm mm0.25

(ii)Young'sModulus E

6

63 2 5 2

3

3

1E 2G(1 )

m

1.5 10 2G(1 0.25)

1.5 10 G 600x10 N mm 6 10 N mm

2(1 0.25)

P

AE l

l

Pl 40 10 1000l 0.0816mm

AE 1560 10 4

2

(iii)Modulusof Rigidity

(iv)Change in lengthper metre

π(20)

3

4

2 2 3

4

3

(V) Changein volume v

v l d 2

v l d

v 0.0816 6.4 10 2

v 1000 20

v 5.584 10

v

v d 1000 (20) 1000 314159.26 mm4 4

v 314159.26 5.584 10

v 175.33mm

π π

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Result:

(i)Poisson'sratio(μ)= 0.25

(ii)Young'sModulus E =6

2N1.5 10

mm

(iii)Modulusof Rigidity =5 26 10 N mm

(iv)Changeinlengthpermetre 0.0816mm

(V)Changein volume v =175.33mm3

UNIT – II BEAM DEFLECTION

PART-A:

1. What are the different types of loads?

1. Point load 2.Uniformly distributed load (udl) 3.Uniformly varying load (uvl).

2. What are the different types of beams? (EVEN 2009)

1. Cantilever beam 2.Simply supported beam 3. Fixed Beam 4.Over hanging beam

5.Contionuous beam.

3. Define Shear force and bending moment.(EVEN 2013)

Shear is total load acting at a point (upward load – downward load). Bending moment at a point

is product of load and distance (anticlockwise moment – clockwise moment).

4. What are the assumptions in the theory of simple bending? (EVEN 2010)

1. Beam is straight.

2. Each layer expand and contract independently.

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3. Load acts normal to the axis of the beam.

4. Beam material is homogeneous.

5. Young‘s modulus is same for tension and compression

5. Define point of contra flexure. (EVEN 2010,2013)

It is a point where the bending moment changes its sign from +ve to –ve or –ve to +ve. At that

point bending moment is Zero.

6. Define flexural rigidity.

It is the product of moment of inertia and young‘s modulus (EI).

7. Define section modulus. (EVEN 2011)

It is the ratio of moment of inertia over the neutral axis (Z=I/y). It is denoted by ―Z‖. It is also known as strength of the section.

8. Where will be the maximum bending moment in simply supported beam?

Bending moment is maximum where shear fore is zero.

9. Where will be the maximum bending stress in the beam?

The bending stress is the maximum at the ends of the section. And Zero at neutral axis.

10. Where will be the maximum Shear stress in a beam?

The maximum shear stress is the maximum at neutral axis and zero at the ends.

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11. What is the value of bending moment corresponding to a point having a zero shear

force? (EVEN 2010)

The value of bending moment is maximum when the shear force changes its sign or zero. In a

beam, that point is considered as maximum bending moment.

12. Define the term point of contra flexure? (EVEN 2010& 2013)

The point where the BM changes its sign or zero is called the point of contra flexure.

13. What are SF and BM diagrams?

SF diagram shows the variation of forces along the length of the beam. BM diagram shows the

variation of bending moment along the length of the beam.

14. Write the relation between SF and BM?

The rate of charge of BM is equal to the SF at the section,

DM/dx = -F

15. In SSB, how do you locate the point of maximum bending moment?

The BM is maximum when the SF changes its sign or zero. Write the SF equation at that point

and equating it to zero we can find out the distance ―x‖ from one end. Then find the maximum BM at that point by taking moment of all forces left or right from that point.

16. Define beam.

Beam is the structural member which is supported along the length and subjected to external

loads acting transversely i.e., perpendicular to the centre line of the lateral dimensions.

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17. Define shear force and bending moment at a section?(ODD 2012)

Shear force: SF at any cross section is the algebraic sum of all the forces acting either sides of a

beam.

Bending moment: BM at a cross section is the algebraic sum of the moment of all the forces

which are placed either side from that point.

18. State the theory of simple bending?

If a beam is bend only due to application of constant bending moment and not due to shear, and

then it is called simple bending.

19. Is bending stress a direct stress or shear stress?

Direct stress

20. What are the assumptions made in the theory of bending? (EVEN 2007)

i) The material is perfectly homogeneous and isotropic. It obeys hooks law.

ii) The value of young‘s modulus is the sane in tension as well as in compression.

iii) The radius of curvature of the beam is very large compared to the cross section

dimensions of the beam.

iv) The resultant force on a transverse section of the beam is zero.

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21. Write down the bending moment equation. (EVEN 2009)

M bE

The bending equation is = = =

I y R

M – Bending moment, I – Moment inertia of the section, b– Bending stress at that section

y- Distance from the neutral axis, E- young‘s modulus of the material, R-radius of curvature of

the beam

22. A rectangular beam 150 mm wide and 200 mm deep is subjected to a shear force of 40

KN. Determine the average shear stress and maximum shear stress. (EVEN 2008)

F 40 x 103

Average shear stress qave = = = 1.33 N/mm2

Bd150 x 200

Maximum shear stress q max= 1.5 qave

= 1.5 x 1.33

= 2 N/mm2

23. A rectangular beam 150 mm wide and 250 mm deep is subjected to a Maximum shear

force of 30 KN. What is the shear stress to distance of 25 mm above the neutral axis?

(EVEN 2008)

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y=25 mm

[ ] = 1.152 N/mm2

24. The section modulus with respect to xx- axis of rectangleof width b and depth d is

And in case of circle the section modulus is (ODD 2007)

bd2 πd3

6 , 32

25. Write down relations for maximum shear force and bending moment in case of a

Cantilever beam subjected to uniformly distributed load running over entire span. (EVEN

2007)

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Maximum shear force = w l

w l2

Maximum bending moment = 2

26.A cantilever beam of 3 m long carries a load of 20 KN at its free end. Calculate the shear

force and bending moment at a section 2 m from the free end. (EVEN 2006)

X 20 KN

Shear force is 20 KN

Bending moment is 20 x 2 = 40 KN-m

X

2m

3 m

27. Draw the shear force diagram for a cantilever beam of span 4 m and carrying a point

Load of 50 KN at Point (ODD 2006)

50 KN

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2 m 2 m

50

50 kn

28. Mention and sketch any two types of supports for the beams. (EVEN 2011)

1. Hinged or pinned support

2. Roller support

3. Fixed support

29. A simply supported beam is subjected to u.d.l of w per unit length throughout its length write the value maximum bending moment.

Maximum bending moment=WL2/8

30.Define clear span and effective span.

The horizontal distance between the supporting walls is called the clear span of the beam. The horizontal distance between the lines of action of end –reaction is called effective span.

31.What is moment of resistance?

SFD

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The couple produced in a flexural member due to internal forces is called as moment of

resistance.

32.State the theory of simple bending?

If a length of a beam is subjected to a constant bending moment and no share force (i.e. zero

shear force) then the stresses will be set up in that length of the beam due to B.M. only and that length of the beam is said to be in pure bending or simple bending. The stresses set up in that length of beam are known as bending stress.

33.Define neutral axis of a cross section. (ODD 2012)

The line of intersection of the neutral surface on a cross-section is called the neutral axis of a

cross-section. There is no stress at the axis.

34. What will be the shape of bending moment and shear force diagrams for different types of load.

35. What is meant by shear flow? (EVEN-2013)

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

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UNIT-II 16 MARKS QUESTION:

1. A cantilever beam of length 3 m carries the point loads as shown in fig. draw the shear

force and at bending moment diagram for the

cantilever beam.[EVEN 2006)

GIVEN DATA:

LB =1.5m; WB =300N

LC=2.5m; WC =400N

LD =3m; WD =500N

TO DRAW:

SFD and BMD

SHEAR FORCE:

SF at D = 500N

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SF at C = 500+400 =900N

SFat B =500+400+300 =1200N

SF at A =1200N

BENDING MOMENT:

BM at D=0

BM at C = -500x0.5=250Nm

BM at B=-500x1.5-400x1=-1150Nm

BM at A=-500x3-400x2.5-300x1.5=-2950Nm

RESULT:

SFD and BMD are shown in fig respectively.

VTHT AVADI III SEM


2. Draw the bending moment and shear force diagram for a cantilever beam to a point load

of 10 KN at mid span. Take span equal to 4m.

GIVEN DATA:

LB =4m;

LC=2 m;

WC =10KN

TO DRAW:

SFD and BMD

SHEAR FORCE:

SF at B = 0

SF at C =10KN

SF at A =10KN

VTHT AVADI III SEM


BENDING MOMENT:

BM at B =0

BM at C =0

BM at A =-10x2=-20KNm

RESULT:


VTHT AVADI III SEM


3.A cantilever beam of length 2m carries a point load of 1KN at its free end , and another

load of 2KN at a distance of 1m from the free end .draw the SF and BM diagram for the

cantilever .[ODD 2003]

GIVEN DATA:

LC =2m; WC =1KN

LB =1m; WB =2KN

TO DRAW:

SFD and BMD

SHEAR FORCE:

SF at C =+1KN

SF at B =+3KN

SF at A =+3KN

VTHT AVADI III SEM


BENDING MOMENT:

Taking moment from right side of the beam section considered.

BM at C

=-WCx0 =0

BM at B

=-WCx(LC-LB)xWBx0

=-1x1-2x0

=-1KN/m

BM at A

=-WCxLC-WBxLB

=-1x2-2x1

=-4KN/m

RESULT:


4.A cantilever of length 2m carries a UDL of3KN/m.Draw the SF and BM diagrams.

VTHT AVADI III SEM


[APRIL 1995]

GIVEN DATA:

L =2m;

w =3KN/m

TO DRAW:

SFD and BMD

SHEAR FORCE:

For UDL

SF =3KN/mxlength

SF at B =3x0 =0

SF at A =3KN/mx2m =6KN

VTHT AVADI III SEM


BENDING MOMENT:

For UDL

BM =Forcexdistance

BM at B

=-3x02/2

=0

BM atA

=-3x22/2

=-6KN/m

RESULT:


VTHT AVADI III SEM


5.A cantilever of length 4m carries a UDL of 3KN/m run over the whole length and two

point loadsof 4KN and 2.5KN are place 1m and 2m respectively from the fixed end. Draw

the shear force and BM diagrams. [ODD 2001]

GIVEN DATA:

As shown in fig.

TO DRAW:

SFD and BMD

SHEAR FORCE:

SF at D =0

SF at C

=2.5KN+3KN/mx2m

=8.5KN

SF at B

=+4KN+2.5KN+3KN/mx3m

=+15.5KN

SF at A

VTHT AVADI III SEM


= 4KN+2.5KN+3KN/mx4

=+18.5KN

BENDING MOMENT:

BM at D =0

BM at C

=-2.5x0-3x2x1

=-6KN-m

Similarly,

BM at B

=-4x0-2.5x1-3x3x1.5

=-16KN/m

BM at A

=-4x1-2.5x2-3x4x2

=-33KN/m

RESULT:


VTHT AVADI III SEM


6. A Cantilever beam of length 5m is loaded as shown in fig. Draw the SF and BM

diagrams.

GIVEN DATA:

As Shown in fig.

TO DRAW:

SFD and BMD

SHEAR FORCE:

SF at D

=3KN

SF at C

=3+2

=+5KN

SF at B

=3KN+2KN+3KN/mx1.5m

=+9.5KN

SF at A =+11.5KN

VTHT AVADI III SEM


BENDING MOMENT:

BM at D =0

BM at C

=-3KNx2m

=-6KN/m

BM at B

=-3x3.5m-2x1.5m-3KN/mx1.5mx0.75

=-10.5-3-3.375

=-16.875KN/m

BM at A

=-3x5-2x3-3x1.5x(0.75+1.5)-2x1.5

=-15-6-10.125-3

=-34.125KN/m.

RESULT:


VTHT AVADI III SEM


7.a beam freely supported over an effective span of 5 m carries point loads 3KN, 4.5KN

and 7KN at 1, 2.5 and 3.5m respectively from the left hand support. Construct the SF and

BM diagrams.

GIVEN DATA:

As shown in fig

TO DRAW:

SFD and BMD

SOLUTION:

To find RA and RE,

Taking moment about A,

REx5 =3x1+4.5x2.5+7x3.5

=38.75/5

=7.75KN

We know that,

RA+RE =3+4.5+7 =14.5KN

7.75+RA =14.5

RA =+6.75KN

SHEAR FORCE:

VTHT AVADI III SEM


SF at E =-RE =-7.75KN

SF at D =-7.75+7 =-0.75KN

SF at C =-7.75+7+4.5 =3.75KN

SF at B =-7.75+7+4.5+3 =6.75KN

SF at A =6.75KN =RA

BENDING MOMENT:

BM at E =0

BM at D =+REx1.5m =7.75x1.5 =11.625KN/m

BM at C =7.75x2.5-7 =12.375KN/m

BM at B =7.75x4x-17.5-6.75 =+6.75KN/m

BM at A =0

RESULT:


VTHT AVADI III SEM


8. A beam of 8m span simply supported at its end carries loads of 2KN and 5KN at a

distance of 3m and 6m respectively from right support. In addition, the beam carries a

UDL of 4KN/m for its entire length. Draw the SF and BM diagrams. [EVEN 95]

GIVEN DATA:

As shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:


RDx8m

=2KNx5m+5KNx2m+4KN/mx8mx8/2

RD =148/8 =18.5KN

RA+RD =2KN+5KN+4KN/mx8m

18.5+RA =39

RA =20.5KN

SHEAR FORCE:

SF at D =-RD =-18.5KN

SF at C =-RD+4KN/mx3m+2KN =-4.5KN

VTHT AVADI III SEM


SF at B =-RD+4KN/mx6m

SF at A =+RA =20.5KN

BENDING MOMENT:

BM at D =0

BM at C =RDx3m-4KN/mx3mx1.5

=18.5x3-18 =37.5KN/m

BM at B =RDx6m-2KNx3m-4KN/mx6mx3

=18.5x6-6-72 =33KN/m

BM at A =0

MAXIMUM BENDING MOMENT

RA-5KN-4KN/mxXm =0

20.5-5-4X =0

X =3.875m

Maximum BM

=RDx4.125-2x1.125-4KN/mx4.125x4.125/2

=40.03KN/m

RESULT:


VTHT AVADI III SEM


9. A simply a supported beam of length 5m carries a uniformly varying load of 800N/m run

at one end of zero at the other end. Draw the SF and BM diagrams for the beam. Also

calculated the position and magnitude of maximum bending moment. [ODD 2000]

GIVEN DATA:

As shown in fig.

TO DRAW:

SFD and BMD.

Find maximum BM and its position.

SOLUTION:


RBx5 =0.5x5x800x0.7x5 =1333.33N

RA+RB =0.5x5x800 =2000N

RA =2000-RB =2000-1333.33 =666.67N

SHEAR FORCE:

SF at B =- RB =-1333.33N

SF at A = RA =666.67N

VTHT AVADI III SEM


BENDING MOMENT:

BM at A =0

BM at B =0

MAXIMUM BENDING MOMENT

SFX =-1333.33+0.5x800xx xx/5 =0

X =4.08m from end B.

The maximum BM,

Mmax =1333.33x-0.5x(800xx /5)xxx0.7xx

Substituting the x value in the above equation

Mmax

=1333.33x4.08-0.5x(800x4.08/5)x4.08x0.7x4.08

Mmax =1817.73N/m

RESULT:

SFD and BMD are shown in fig respectively

10. Analyse the beam as shown in fig. and draw the SFD and BMD. [ODD 97]

VTHT AVADI III SEM


GIVEN DATA:

As shown fig.

TO FIND:

I.Maximum BM

ii. Draw SFD and BMD

SOLUTION:


REx7 =3x2x(1+5)+5x4+0.5x3x4x0.3x3

RE=62/7

=8.857KN

RA+RE =0.5x3x4+5+3x2

RA =17-8.857

=8.143KN

SHEAR FORCE:

SF at E =-RE =-8.857KN

SF at D =-8.857+3X2=-2.857KN

VTHT AVADI III SEM


SF at C =-2.857+5 =+2.143KN

SF at B =2.143KN

SF at A =RA =8.143KN

BENDING MOMENT:

BM at E =0

BM at D =REX2-3X2X1

=8.857X2-6 =11.714KN/m

BM at C = 8.857X3-12 =14.57KN/m

BM at B =8.857X4-18-5 =12.428 KN/m

BM at A =0

RESULT:


VTHT AVADI III SEM


11.A beam 12m long is supported at two points 2mfrom each end , so that there are two

equal overhanging portions. It carries concentrated loads of 4KN, 3KN and 5KN at 1m, 8m

and 12m respectively from the left end. Draw the SF and BM diagrams. What are the

values of maximum BM and maximum SF? [EVEN 2012)

GIVEN DATA:

As shown in fig.

TO DRAW:

i. SFD and BMD

ii. Maximum BM & SF

SOLUTION:

Taking moment about C,

RE=8 =5x10m+3x6mx-4x1m =64

RE =8KN

RE=RC =4+3+5 =12

RC =4KN

SHEAR FORCE:

SF at F =5KN

SF at E =5-8 =-3KN

VTHT AVADI III SEM


SF at D =-3+3 =0

SF at C =0-4 =-4KN

SF at B =-4+4 =0

SF at A =0

BENDING MOMENT:

BM at F =0

BM at E=-5X2 =-10KN/m

BM at D =-5X4+8X2 =-4KN/m

BM at C =-5X10+8X8-3X6 =-4KN/m

BM at B =-5X11+8X9-3X7+4X1 =0

BM at A =0

RESULT:


12. A horizontal beam AD, 10m long carries a

UDL of 12KN/m length run together with a

concentrated load 30KN at the left end A . The

beam is supported at a point B which is 2.5m

VTHT AVADI III SEM


from A and C which is in the right hand half of the beam and 3m from D. Plot the SF and

BM diagrams.[EVEN 2011]

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

Taking moment about B,

RCx4.5m =12KN/mx7.5x7.5/2-30KN =225KN/m

RC =50KN

RA+RC =12KN/mx10+30KN =150KN

RA =100KN

SHEAR FORCE:

SF at D =0

SF at C =12KN/mx3m-50KN =-14KN

SF at B =-14KN+12KN/mx4.5m-100KN

=-60KN

SF at A =-30KN

VTHT AVADI III SEM


BENDING MO MENT:

BM at D =0

BM at C =-12KN/Mx3x1.5 =-54KN/m

BM at B =-12KN/Mx7.5x3.75+50x4.5

=-112.5KN/m

BM at A =0

Maximum BM

12KN/mX x-RC =0

X =50/12 =4.166m

Maximum BM

=-12x4.166x4.166/2+50x (4.166-3)

=-45.83KN/m

RESULT:


13.A beam AB of length 7m is simply supported at two supports which are 5m apart with

an overhang of 2m on the right side of the beam as shown in fig. The beam carries a UVL

of 6KN/m over the entire length of SSB and a concentrated load of 4KN at the right end of

the beam. Draw SFD and BMD. Locate maximum BM. [ODD 98]

VTHT AVADI III SEM


GIVEN DATA:

As shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:


RBx5 =0.5x5x6x0.3x5+4x7 =53

RB =10.6KN

RA+RB =0.5x5x6+4 =19KN

RA =19-10.6 =8.4KN

SHEAR FORCE:

SF at C =4KN

SF at B =4-10.6 =-6.6KN

SF at A =RA =8.4KN

BENDING MOMENT:

BM at C =0

BM at B =-4x2 =-8KN/m

VTHT AVADI III SEM


BM at A =0

Maximum BM

SFX =4-RB(x-2)+0.5x(x)x1.2(x-2) =0

4-10.6x +21.2-0.6(x-2)2 =0

x2+13.66x-38 =0

By using quadratic formula,

x =2.37m or -16.02m

Maximum BM,

Mmax =-4xx+RB(x-2)-0.5x(x-2)x1.2x(x-2)x0.3x(x-2)

=-9.48+3.922-0.02

Mmax =-5.658KN/m

RESULT:


14.A beam of length 10 m is simply supported at its ends carries two concentrated load of 5

KN each at a distance 3 m and 7 m from the left support and also uniformly distributed

load of 1 KN/m between the point loads draw SFD and BMD and calculate the maximum

bending moment.(EVEN 2011)

GIVEN DATA:

AS shown in fig.

VTHT AVADI III SEM


TO DRAW:

SFD and BMD

SOLUTION:

Taking moment about A

RD=5*7+1*4*(4/2+3) +5*3

RD*10=70

RD=7

W.K.T

RA+RD=5+1*4+5

RA=7 KN

SHEAR FORCE:

SFat D= -RD = -7KN

SF at C (without point load) = - RD = -7

SF at C (with point load) = - RD = -7 +5 = -2 KN

SF at B (without point load) = RA= 7

SF at B (without point load) = RA -5 = 7- 5 = 2KN

SF at A =RA =7 KN

BENDING MOMENT:

VTHT AVADI III SEM


BMD at D=0

BM at C= RD*3 = 7*3 = 21 KN-m

BM at B = RA*3 =7*3 =21 KN-m

BM at A =0

The maximum bending moment is situating at a distance of x from the point A, where the shear

forces change its sign.

X =3+2 =5m

MAX BENDING MOMENT:

Max binding moment= RD * 5 *2 – (1*2)*2/2

= 7* 5- 5*2 -2 =23 KN-m

RESULT:


VTHT AVADI III SEM


VTHT AVADI III SEM


15. Draw the S.F and B.M diagram for beam shown in fig. (ODD 2006).

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

TAKING MOMENT ABOUT “A”

RB * 8 = 2000*10+4000*5 -1600*3

VTHT AVADI III SEM


RB * 8 = 35200

RB =4400 N

RA +RB =1600+4000+2000

RA+ 4400 =7600

RA = 3200 N

SHEAR FORCE:

SF at E =2000 N

SF AT B =2000 -4400 =-2400 N

SF AT D = -2400+4000 =1600 N

SF AT A = 1600-3200 = -1600 N

SF AT C = 1600 – 1600 =0

BENDING MOMENT:

BM AT E =0

BM AT C =0

BM AT B = -2000 *2= -4000 N-m

BM AT D = -2000 *5 + 4400 *3 =3200 N-m.

BM AT A = (-200 *10) + (4400*8)-(4000*5)= -4800 N-m.

RESULT:


VTHT AVADI III SEM


VTHT AVADI III SEM


VTHT AVADI III SEM


16. For the SSB loaded as shown in fig. draw the shear force diagram and bending moment

diagram obtain maximum bending moment.. (EVEN 2010).

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

TAKING MOMENT ABOUT “B”

RB *7 = 20 *5.5 +(15 * 3) (3/2+1)

RB = 31.8 KN

RA + RB =15 *3 +20

RA +31.8 = 65

RA = 33.2 KN

SHEAR FORCE:

SF AT B = -31.8 KN

SF AT E =-31.8 +20 = -11.8

SF AT D = -11.8

VTHT AVADI III SEM


SF AT C = -11.8 + 15* 3 =33.2 KN

SF AT A =33.2 KN

BENDING MOMENT:

BM AT A =0

BM AT B =0

BM AT C =33.2 *1 =33.2 KN

BM AT D =33.2 *4 – (15 * 32/2) = 65.3 KN- m

BM AT E = 33.2 * 5.5 – (15* 3)*3 =47.7 KN-m

TO FIND MAXIMUM BENDING MOMENT:

Maximum bending moment occur at a point where

shear force is zero. Form the SFD it can be seen that

SF takes zero value in the region is

SFX = 33.2 – (15 *(x-1)) =0

33.2 -15X +15 =0

X =3.21

BM AT 3.21 m

MAX = 33.2 *3.21 – 15(3.21- 1)2/2

= 69.94 KN-m

Maximum bending moment = 69.94 kN-m

VTHT AVADI III SEM


RESULT:


VTHT AVADI III SEM


VTHT AVADI III SEM


17. For the beam shown in fig draw the SFD and BMD.(EVEN 2008).

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

TAKING MOMENT ABOUT A:

RB *5 = (4 *5 * 5/2)+ (2*6)+(7*4)+(5*1)

RB =19 KN

RA +RB = 5+4+7+2+4*5

RA =15 KN

VTHT AVADI III SEM


SHEAR FORCE:

SF AT C = 2 KN

SF AT B = 2 – 19 = -17KN

SF AT D = -17 *4 = -13 KN

SF AT E = -6 +(4*3) = 6 KN =6+ 5 =11 KN

SF AT A = 15 KN

BENDING MOMENT:

BM AT C = 0

BM AT A =0

BM AT B = -2*1 =-2 KN-m

BM AT D = -2 +19 = 17+(-4*1) =13 KN-m

RESULT:


VTHT AVADI III SEM


VTHT AVADI III SEM


18. Draw the shear force and bending moment diagram of the beam loaded as shown in fig.

(ODD 2008)

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

RA =5 KN

VTHT AVADI III SEM


RB =70 KN

SHEAR FORCE:

FC =25 KN

FB=35 KN

FB =-35KN

FA =5 KN

BENDING MOMENT:

MC =0

MB= -60 KN-m

MA =0

RESULT:


VTHT AVADI III SEM


VTHT AVADI III SEM


19. A simply supported beam of 4 m span is

carrying load as shown in fig. the SFD and BMD

draw.(EVEN 2010)

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

VTHT AVADI III SEM


TAKING MOMENT ABOUT “A”

RB* 4 = 4 * 1.5 + 2 *1 *(1/2 +1.5)

RB =2.5 KN

RA +RB =4+2 *1

RA =6-2.5 =3.5 KN

SHEAR FORCE:

SF AT B = -2.5 KN

SF AT C =-2.5 KN

SF AT D = -2.5 +(2*1) = -0.5 KN (WITHOUT POINT LOAD)

= -0.5 +4 = 3.5 KN

SF AT A = 3.5 KN

BENDING MOMENT:

BM AT A =0

BM AT D = 0

BM AT C = RB *1.5 =3.75 KN-m

BM AT D = RB *2.5 -2 * 1 * ½

= 6.25 -1 = 5.25 KN- m

RESULT:


VTHT AVADI III SEM


20. A cantilever 3.6 m long carrier load

of 30, 70, 40 and 60 KN at distance of a

0.6, 1.5, and 2.4m respectively from the

free end. Draw the SF and BM diagrams

for the cantilever. (EVEN 2012).

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SHEAR FORCE:

SFat E =30 KN

VTHT AVADI III SEM


SFat D = 30+70 =100 KN

SFat C = 100 +40 =140 KN

SF at B = 140+60 =200 KN

Join all the values by straight horizontal liner.

BENDING MOMENT:

BM AT E =0

BM AT D

= -30 *0.6

= -18 KN-m

BM AT C

= -30*1.5 -70 *0.9

= -108 KN-m

BM AT B

= -30 * 2.4 -70 *1.8 -40 *0.9

= -234 KN-m

BM AT A

= -30 *6 -70 *5.4 -40 *4.5 -60 *3.6

VTHT AVADI III SEM


= -954 KN-m

Join the values by straight inclined lines as.

RESULT:


21. Draw the shear force and bending moment diagram

for the beam show in fig.(ODD 2011)

GIVEN DATA:

AS shown in fig.

TO DRAW:

SFD and BMD

SOLUTION:

TAKING MOMENT ABOUT A

RC *8 = ½ * 2.5 *5 *[(1/3 *2.5) +8] + 10 * 8 *8/2 + 20*6

RC *8 =495.2

VTHT AVADI III SEM


RC =61.9 KN

RA + RC =1/2 *2.5 *5 +10 *8 +20 = 106.25 KN

RA =106.25 -61.9 = 44.35 KN

SHEAR FORCE:

SF AT D =0

SF AT C (without reaction Rc)

= ½ *2.5 *5 = 6.25 KN

SF AT C (with reaction)

= 6.25 – 61.9 = -55.65 KN

SF AT B (without point load)

= -55.65 +10 *2 = -35.65 KN

SF AT B (with point load)

= -35.65 +20 = -15.65 KN

SF AT A = RA = 44.35 KN

Join the values CD by parabolic curve and

all other valves by inclined straight lines.

BENDING MOMENT:

BM AT D = 0

BM AT C

VTHT AVADI III SEM


= -1/2 *2.5 *5 *1/3 *2.5

= -5.208 KN-m

BM AT B

= -1/2 *2.5 *5 *[(1/3*2.5)+2] +61*2 * (-10*2*2/2)

= 36.09KN-m

Join all the valves AB and C by parabolic curves and CD by cubic curves.

RESULT:


22. The cross section of the beam is shown. This beam is of cantilever type and carries a

UDL of 16 kN/m. if the span of the beam is 2.5 m; determine the maximum tensile and

compressive stresses in the beam. (EVEN 2007, 2008,

2009).

GIVEN:

UDL = 165 KN/m

L =2.5 m

TO FIND:

The maximum tensile stresses =?

VTHT AVADI III SEM


The maximum compressive stresses =?

SOLUTION:

AREA OF SECTION (1), a1 =50*10 =500 mm2

AREA OF SECTION (2), a2 =15 *35 = 525 mm2

Y1 for section (1) form bottom most layer

Y1 = 35 + 10/2 = 40 mm

Y2 for section (2) form bottom most layer

Y2 =35 /2 = 17.5 mm

For unsymmetrical section, the centre of gravity of the section is placed y mm form the bottom

face. The y may be calculated by6 using the following formula.

Y =

=

=28.47 mm

Moment of inertia of rectangle (1) about an axis through its C.G and parallel to x-x axis .

Ig1 =bd3/12

= 50* 103/12

= 4166.667 mm4

VTHT AVADI III SEM


From parallel axis theorem moment of inertia of the rectangle (1) from x- x axis.

I1 =Ig1 +a1 h12

Where,

L1 = distance between C.G of the section (1) form reference line and the y.

= y1 – y

= 40-28.47

=11.53 mm

I1= 4166.667 +500 *11.53 2

= 70637.11 mm4

Similarly for section (2)

I2 =IG2 +a2h22

I2 = 15 *352/12 + 525 (28.47 -17.5)2

= 116772.72 mm4

Now moment of inertia of whole section about x axis

IXX = I1 +I2

= 70637.11+116772.72

=187409.83 mm4

M= 16 * 2.5 /2 =50 KN-m

W.K.T,

VTHT AVADI III SEM


M/I=

The maximum compressive bending stress is on the top most layer of the beam.

The distance from y to top layer is

= 45 -28.47

=16.53 mm

b = M/I *y

= 50 * 16.53/187409.83

= 4.4 *103 KN/m2

The maximum tensile stress is on the bottom most layer of the beam.

b = M/I *y

=50*28.47/187409.83

= 7.59*103KN/m2

RESULT:

b= 4.4 *103 KN/m2

b= 7.59*103KN/m2

23. a timber beam of rectangular section is to support a load of 20 KN uniformly

distributed over a span of 36 m, when the beam is simply supported. If the depth of the

section is to be twice the breadth and the stress in the timber is not exceed 7 n /mm3 modify

the cross section of the beam if it carries a concentrated 30 KN placed at the mid span with

the same ration of breadth to depth.(EVEN 2006) (ODD 2006)

VTHT AVADI III SEM


GIVEN:

Load, w= 20 KN =20000 N

Length l =3.6m

b =7 N/mm2

WP= 30 KN =30000 N

D= 2 b

TO FIND:

Case (1) uniformly distributed load, 20KN

(1) B, (2) D

Case(2). Point load 30 KN

(1). B (2) D

Solution :

case1.

Carried a uniformly distributed load

Maximum bending moment (BM)=wl2/8

=20000*(3.6)2/8

BM=3.24 *1010 N-mm

Moment of inertia,

I=bd3/12

=b(2b) 3/12

I=8 b4/12

VTHT AVADI III SEM


Z=1/y= 8b3/12

Moment of resistance,

m= b*z

=7* 8b3/12

M= 56*b3/12

Equating the moment of resistance to maximum bending movement

3.24*1010 =56*b3/12

B=1907.711 mm

D= 2b =2* 1907.11 =3815.42 mm

Case2.

Beam carries a point load at the center

Maximum bending movement BM= wl/4

= 30000*3.6/4 =27000 n-m

= 27000*103 N-mm

Equating the movement of resistance to max bending movement

56 *b3/12 = 27000000

B=179.52 mm

D=2b= 359.04 mm

RESULT:

(i) B=1907.711 mm D=3815.42 mm

(ii) B=179.52 mm

D=2b= 359.04 mm

VTHT AVADI III SEM


24. A beam of rectangular cross section 50 mm wide and 150 deep is used. As cantilever 6m

long and subjected to a uniformly distributed load of 2 KN/m over the entire length.

Determine the bending stress at 50mm from the top fibre, at the mid span of the beam.

Also, calculate the maximum bending stress. (ODD 2008)

GIVEN:

Cross section 50 mm wide and 150 deep

Long= 6m

Uniformly distributed load of 2 KN/m

TO FIND:

Maximum bending stress, =?

SOLUTION:

B=50mm

D=150mm

L=600mm

W=2000N/mm

I=14.06*106mm4

Maximum bending moment

VTHT AVADI III SEM


Mmax =wl2/2

=2*(3000)2/2

=9*106N-mm

Maximum bending stress, = b/YMAX

b= (36*106)* 75/14.06*106

=192.02 N/mm2

Bending moment at mid span M/I= mid/ymid

=9*106 *(25)/14.06*106

=16N/mm2

RESULT:

Maximum bending stress=192.02 n/mm2

25. A simply supported timber beam of span 6m carries a UDL of 12 KN/m over the entire

span and a point load of a KNat 2.5 m from the left support. If the bending stress in the

timber is not to exceed 8 N/mm2design a suitable section for the beam. The depth of beam

equals twice the breadth. (ODD 2001)

GIVEN:

L=6 m

UDL, w1 =12 KN/m

Bending stress =8 N/mm2

D=2b

TO FIND:

VTHT AVADI III SEM


B=D=?

SOLUTION:

RC *6 = 12 * 6 *6/2 +9 *2.5

RC =39.75 KN

RA + RC =12 *6+9 =81 KN

RA =81 – 39.75 =41.25 KN

S.F CALCULATION:

SF AT C = -RC = -39.75 KN

SF AT B (without point load)

= -RC+12 *3.5= 2.25 KN

SF AT A = RA= 41.25 KN

From SFD, the SF changes its sign at a distance of x –m form c.

Therefore, the maximum BM liesat that point.

Maximum BM calculation

The SF equation at that point is,

SFX= -RC +12 *x=0

X=39.75/12 =3.3125 m form end c.

Therefore maximum BM,

MMAX =RC *x -12*x*x/2

=39.75 * 3.3125 -12 *(3.3125)2/2

VTHT AVADI III SEM


MMAX= 65.83*106 N-mm

Y=D/2=B

I=bd3/12

=8b4/12

By using the relation

M/I= b/Y

65.83*106/8/12*B4=8/b

B3 =12.3*106

B=231.12 mm

D=2b=462.24mm

RESULT:

B=231.12 mm

D=2b=462.24mm

VTHT AVADI III SEM


26. Calculate the maximum bending stress and shear stress in a cantilever beam of span 6m

which carries a uniformly distributed load of 5 KN/m over a distance of 4m form the free

end. The cross section of the beam is a rectangle of breadth 100mm and depth 150mm.

(EVEN 95)

GIVEN:

L=6m

W=5KN/n

B=100mm

D=150mm

TO FIND:

Maximum bending stress and shear stress

SOLUTION:

For the given cantilever beam, the maximum BM is acting at fixed, end.

Therefore the maximum BM,

Mmax =5 *4 * 4/2 =40 KN-m

=40*106 N-mm

Y=d/2 =150/2 =75 mm

I=bd3/12

=100*(150)3/12

=28.125*106 mm4

By using the relation

VTHT AVADI III SEM


M/I= b/Y

40*106/28.125*106 = b/75

b= 106.67 N/mm2

Maximum shear stress,

Qmax=3/2 * F/bd

F=5*4=20 KN at A

Qmax= 3 * 20000/2* (100*150)

=2 N /mm2

RESULT:

Qmax=2 N /mm2

27. A simply supported beam of span 6 m is carrying a uniformly distributed load of 2

KN/m over the entire span. Calculate the magnitude of shear force and bending moment at

every section, 2 m from the left support. Also draw shear force and bending moment

diagrams. (EVEN 2013)

Given:

Span l = 6 m

Load w = 2 KN/m

To Find:Shear force and bending moment at every section, 2m from left support.

VTHT AVADI III SEM


Solution:

Taking moment about ‗A‘

RB × 6 = 2 × 2 × RB× 6 =12

RB =

= 6 KN

RA +RB= 2 × 6

RA +RB= 12

RA = 12 - RB = 12 – 6 = 6 KN

Shear Force Calculation:

Shear at RB= -6 KN

Shear at RD = -6 + 2 × 2 = - 2 KN

Shear at RC = -6 + 2 × 4 = 2 KN

Shear Forces at RA = 6 KN

Bending Moment Calculation:

B.M at B = 0

B.M at D = RB× 2 – 2 × 2 × = 12 – 4 = 8 KN-m

B.M at C = RB× 4 – 2 × 4 × = 24 -16 = 8 KN-m

B.M at A = 0

Maximum Bending Moment

VTHT AVADI III SEM


B.M (max)=

(OR)

B.M (max)at

=

=

= 18 – 9 = 9 KN- m

RESULT:


28. State the assumption made in the theory of simple bending equation and drives the

simple bending equation (EVEN-2013)

There exists a define relationship among applied moment, bending stresses and bending

deformation (radius of curvature). This relationship can be derived in two steps:

(i) Relationship between bending stresses and radius of curvature.

(ii) Relationship between applied bending moment and radius of curvature.

(i) Relationship between bending stresses and radius of curvature: Consider an elemental

lengthABof the beam as shown in Fig .Let EFbe the neutral layer and CDthe bottom

most layer. If GHis a layer at distance y from neutral layer EF, initially AB = EF = GH =

CD.

VTHT AVADI III SEM


Let after bending A, B, C, D, E, F, G and H take positions A′, B′, C′, D′, E′, F′, G′ and H′

respectively as shown in Fig. 10.2(b). Let R be the radius of curvature and be the angle subtendedby C′A′ and D′B′ at centre of radius of curvature. Then,

EF = E′F′, since EF is neutral axis

= R ... (i)

Strain in GH =Final length – Initial length

Initial length

G 'H ' GH

GH

But GH EF(The initial length)

R

G 'H ' (R y)

(R y)Strain in layer GH R

R

y

R

Since strain in GH is due to tensile forces, strain in GH = f/E ...(iii)

VTHT AVADI III SEM


Wheref is tensile stress and E is modulus of elasticity.

From eqns. (ii) and (iii), we get

f y

E R

f E

y R

(ii) Relationship between bending moment and radius of curvature:Consider an elemental

areaδaat distance y from neutral axis as shown in Fig.

From eqn., stress on this element is

VTHT AVADI III SEM


2

2

2

E f y

R

Forceonthiselement

E y a

R

Moment of resis tan ceof this elemenl forceabout neutralaxis

E y ay

R

E y a

R

Total moment resisted by thesec tion M 'is given by

E M ' y a

R

E y a

R

From thedefination of inertia about centroidalaxis

2

, we

know

I y a

EM ' I

R

From equilibrium condition,M where M is applied moment

E M I

R

M E

I R

M f E

I y R

VTHT AVADI III SEM


UNIT – III TORSION

PART-A

2. Write down the simple torsion formula with the meaning of each symbol for circular cross section.(EVEN 2011, 2010, 2009)

T Cθ τ= =

J L RWhere,

T – Torque N-mm

J – Polar moment of inertia mm4

– Shear stress N/mm2

R – Radius mm

C – Modulus of rigidity N/mm2

- Angle of twist radian

l- Length mm

2.Define stiffness of spring and mention its unit in SI system.(EVEN 2011, 2010, ODD 2006)

The stiffness of the spring is defined as the load required producing unit deflection.

C d4

Stiffness, K = N/mm

64 R3 n

3. Compare closed and open coiled helical springs. (EVEN 2009, 2013)

Closed coiled helical springs Open coiled helical springs

1. Adjacent coils are very close to each other. 1. Large gap between adjacent coils

2. Only tensile load can carry. 2. Tensile and compressive loads can carry.

VTHT AVADI III SEM


3. Helix angle is negligible 3. Helix angle considerable.

4. What is the maximum shear stress produced in a bolt of diameter 20mm when it is tightened by a spanner which exerts a force of 50 N with a radius of action of 150 mm?

(ODD 2008)

Solution: Torque, T = 50 x 150 = 7500 N-mm

16 T 16 x 7500

Shear stress, = = = 4.775 N/mm2

π D3 π x (20)3

5. A close coiled helical spring of 10 mm in diameter having 10 complete turns, with mean diameter 120 mm is subjected to an axial load of 200 N. Determine the maximum shear

stress and stiffness of the spring. Take G = 9 x 104 N/mm2. (ODD 2008)

Given: d = 10 mm; n = 10; D = 120 mm → R = 60 mm; W = 200 N

8 W D 8 x 200 x 120

Solution: Shear stress, = = = 61.12 N/mm2

π d3 π x (10)3

6. Define polar modulus of a section. What is the polar modulus value for a hollow circular

section of 100 mm external diameter and 40 mm internal diameter? (EVEN 2008)

It is the ration between polar moment of inertia and the maximum radius of a circular section.

π π

Zp= (D4 – d4) = x (1004 - 404) = 9561300 mm3

32 32

VTHT AVADI III SEM


7. A close coiled helical spring is to carry an axial load of 500 N. Its mean coil diameter is to be 10 times its wire diameter. Calculate this diameter if the maximum shear stress in the

material is to be 80MPa.(EVEN 2008)

Given: W = 500 N; D = 10d; = 80MPa = 80 N/mm2

Solution: 8 W D 8 x 200 x 10d

Solution: Shear stress, = =

π d3 π d3

12.732

80 = d2

d = 12.62 mm

D = 10d = 10 x 12.62 = 126.2 mm

8. Express the strength of a solid shaft.(ODD 2007)

Strength of a solid shaft is given by

π

Torque, T= x x D3

16

9.Give the expression for finding deflection of a closed coiled helical spring.(ODD 2007)

64 W R3 n

Deflection, δ =

C d4

10.What do you mean by torsional rigidity of a shaft? Also, give the expression for finding

power transmitted by a shaft.(EVEN 2008)

The product of modulus of rigidity (C) and polar moment of inertia (J) of a shaft is known as torsional rigidity or stiffness of a shaft.

2πNT

VTHT AVADI III SEM


Power, P =

60000 T → Torque in N-m., N → Speed in rpm, P → Power in KW

11.How will you find maximum shear stress induced in the wire of a close-coiled helical

spring carrying an axial load?(EVEN 2007)

16 W R

Maximum shear stress, = π d3

12.Find the minimum diameter of shaft required to transmit a torque of 29820 Nm if the

maximum shear stress is not to exceed 45 N/mm2(ODD 2007)

Given: T = 29820 N-m; = 45 N/mm²

Solution: π

Torque, T= x x D3

16

π

29820 x 106 = x 45 x D3

16

D = 150 mm

13.Find the torque which a shaft of 50 mm diameter can transmit safely, if the allowable shear stress is 75 N/mm2(EVEN 2006)

Given: Shear Stress, = 75 N/mm²; Diameter D = 50 mmSolution: π

Torque, T= x x D3

16

: π

Torque, T= x 75 x (50)3

16

VTHT AVADI III SEM


T = 1839843.75 N-m

14. Differentiate open coiled helical spring from the close coiled helical spring and state the type of stress induced in each spring due to an axial load.(EVEN 2006, 2013)

Closed coiled helical springs Open coiled helical springs

1. Adjacent coils are very close to each other. 1. Large gap between adjacent coils

2. Only tensile load can carry. 2. Tensile and compressive loads can carry.

3. Helix angle is negligible 3. Helix angle considerable.

4. The pitch between two adjacent turns is small. 4. The pitch between two adjacent turns is high.

15.What are the assumptions made in torsion equation? (EVEN 2004)

1. The material of the shaft is homogenous, perfectly elastic and obeys Hooke‘s law.

2. Twist is uniform along the length of the shaft.

3. The stress does not exceed the limit of proportionality

4. The shaft circular in section remains circular after loading.

5. Strain and deformations are small.

16. Why hollow circular shafts are preferred when compared to solid circular

shaft?(EVEN 99)

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

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

17. What is the power transmitted by circular shaft subjected to an torque of 700 KN-m at

110 rpm(ODD 2003)

2πNT 2 x π x 110 x 700

Power, P = = = 8063.42 KW

60 600

18. Calculate the maximum torque that a shaft of 125 mm diameter can transmit, if the

maximum angle of twist is 1º in a length of 1.5 m. Take C = 70 x 103 N/mm².(EVEN 2005)

Given data:

Diameter, D = 125 mm; Angle of twist, θ = 1º x π/180 = 0.017 rad; Length, l = 1.5 m = 1500 mm

Modulus of rigidity, C = 70 x 103 N/mm².

VTHT AVADI III SEM


To Find: Maximum torque, Tmax

Solution: Torsional equation

T Cθ τ= =

J L R

π/32 [D4]

T = x 70 x 103 x 0.017

15000

Tmax= 19.01 x 106N-mm

19. A helical spring is made of 4 mm steel wire with a mean radius of 25 mm and number of turns of coil 15. What will be deflection of the spring under a load of 6 N. Take C = 80 x

103 N/mm².(EVEN2010)

Given: d = 4 mm; R = 25 mm; n = 15; W = 6 N; C = 80 x 103 N/mm².

Solution: Axial deformation,

64 W R3 n 64 x 6 x 253 x 15

Deflection, δ = = = 4.39 mm.

C d4 80 x 103 x 44

20. Give shear stress and deflection relation for close coiled helical spring.(EVEN 2004)

64 W R3 n

Deflection, δ =

C d4

16 W R

Shear stress, = π d3

4 π R2 n

δ = C d x

VTHT AVADI III SEM


21. What are the uses of closed coiled helical spring?(EVEN 2000)

Railway wagons, cycle seating, pistols, brakes etc.

22. What is meant by spring constant or spring index? (EVEN 2010)

Spring constant is the ratio of mean diameter of the spring to the diameter of the wire.

23. What is meant by spring? (ODD 2011)

Spring is a device which is used to absorb energy by taking very large change in its form without permanent deformation and then release the same when it is required.

24. Classify the spring.(EVEN 2011)

(i) Torsion spring (ii) Bending spring.

25. Define Torsion.

It is the angle of twist due to the load.

26. Define torsional rigidity. (EVEN 2012)

It is a product of modulus of rigidity and polar moment of inertia (GIp).

27. How does the shear stress vary across a solid shaft?

The stress is zero at the centre (neutral axis) and maximum at the perimeter.

28. For same weight, which shaft will carry more torque, a solid one or a hollow one? Why?

Hollow shaft will carry more torque because polar inertia will be more for hollow shaft to solid shaft for the weight and length.

29. What are the types of spring? (EVEN 2012)

Springs are following types.

1. Semi elliptical Leaf spring.

2. Quarter elliptical Leaf spring.

3. Closed coil helical spring

4. Open coil helical spring

30. What are leaf springs?

Several plates are fastened together one over the other to form a layer of plates, such arrangements are known as leaf springs.

VTHT AVADI III SEM


31. What is the difference between closed coil and open coil spring? (EVEN 2013)

Closed Coil Open coil

Angle of helix is less than 10o Angle of helix greater than 20o

It is used for tensile load It is used for both tensile and compressive load

Eg: Brake, accelerator Eg: Shock absorber, ballpoint pen

32. Define Wahl‟s factor.

The effect of direct shear and change in coil curvature a stress factor is defined, which is known as Wahl's factor‘s = Wahl‘s factor, if we take into account the Wahl's factor then the formula for

the shear stress becomes

33. What are the conditions to design a circular shaft?

1. The stress should be within the limit of the torque.

2. Angle of twist should be within the torque.

34. Define torsional energy or torsional resilience.

It is the strain energy stored due to angular twist. It is the product of Average torque and twist.

35.Define polar modulus.

Polar modulus is defined as the ratio of the polar moment of inertia to the radius of

the shaft. It is also called torsional section modulus and is denoted by Zp.

36.Define torsional rigidity.(EVEN 2013)

Let a twisting moment T produce a twist of ∅ radian in a length l then

The Product of modulus of rigidity G &polar moment inertia J is called torsional rigidity or stiffness of shaft.

Where G—modulus of rigidity of the material

VTHT AVADI III SEM


37. Define stiffness of a spring? In what unit it is measured?

Stiffness of a spring is defined as load per unit deflection. It is denoted by K and unit is N/mm.

38.Write the equation for strain energy stored in a shaft due to torsion.

39.What is a spring? State various types of spring.

Springs are elastic members which distort under load and regain their original shape when load isremoved.

Types of springs:

1. Helical springs

a. Closed-coiled spring b. open-coiled helical spring

2. Leaf spring

a. full-elliptic b.semi elliptic ,c. cantilever

VTHT AVADI III SEM


3. Torsion spring

40. Compute the torsional rigidity of a 100mm diameter, 4m length shaft

C=80KN/mm2.(EVEN 2013)

Given:

Diameter (d) = 100mm

Length (l) = 4m=4000mm

Modulus of rigidity (C) = 80KN/mm2 = 80x103 N/mm2

To Find:

Torsional rigidity (CJ)

Solution:

Torsional rigidity = modulus of rigidity(C) x polar moment of inertia (J)

Polar moment of inertia: J = d4 mm4 =

× (100)4 = 9817477.042 mm4

Torsional rigidity = C×J = 80×103 × 9817477.042= 7.85×1011 N/mm2.

Result: Torsional rigidity (CJ) = 7.85×1011 N/mm2.

VTHT AVADI III SEM


16 MARK QUESTIONS

PART-B

1. A solid shaft has totransmit 75 KW at 200 r.p.m. taking allowable shear stress as 70

N/mm2, find suitable diameter for the shaft, if the maximum torque transmitted at each

revolution exceeds the mean by 30%.[ODD-2008]

Given data:

Power transmitted WKW P 310 7575

R.P.M of the shaft N=200 rpm

Shear stress 2/ 70 mmN

mean meanmean

meanmean

TTT

TT Ttorque Maximum

.3 1.3 0

30 00

max

Data asked:

Suitable diameter of the shaft

VTHT AVADI III SEM


Solution:

NmmT

N

P TTorqueor

NT Pd transmittePower

mean

mean

mean

3580980 2002

60000 1075

2

60000

60000

2

3

NmmT

TTT

TT Ttorque Maximum

mean meanmean

mean mean

4655274 3580980.3 1

.3 1.3 0

30

max

0 0

max

3max

1

3max

1

3

16

16

4655274 16 69.57 70

70

Wkt Maximum torque transmitted by the solid shaft T d

Tor Diameter of the shaft d

d mm mm

Result:

The diameter of the solid shaft d = 70 mm

2. A hollow steel shaft of outside diameter 75 mm is transmitted a power of 300 KW at 2000

rpm. Find the thickness of the shaft if the maximum shear stress is not to exceed 40 N/mm2.

[ODD-2009]

Given data:

Out diameter of the shaft do = 75 mm

Power transmitted P = 300 KW

Speed of the shaft N= 2000 rpm

VTHT AVADI III SEM


Maximum shear stress = 40 N/mm2

Data asked:

Thickness of the shaft.

Solution:

NmmT

N

P TTorqueor


mean

mean

mean

143312120002

60000 10300

2

60000

60000

2

3

VTHT AVADI III SEM


3 4max

4

3

3

4

1

4

116

161

1433121 16 0.53198

40 75

1 0.53198

0.46802 0.8271

0.8271

0.8271

0.8271 75 6

o

i

o

o

i

o

i o

i

Wkt Maximum torque transmitted by the solid shaft T d k

dwhere k

d

Tor k

d

or

k or

k

die

d

d d

d

2

75 62 13 6.5

2 2 2o i

mm

d dwkt thickness t mm

Result:

Thickness of the hollow shaft t =6.5 mm

3. A shaft is required to transmit 75 KW power at 100 r.p.m and the maximum twisting

moment is 30% greater than the mean. Find the diameter of the steel shaft if the maximum

stress is 70 N/mm2. Also determine the angle of twist in a length of 3m of the shaft. Assume

the modulus of rigidity for the steel as 90 KN/mm2. [EVEN-2011]

Given data:

232

2

3

/10 90/90mod

3000 3

/ 70max

100

10 7575

mm Nmm KNG rigidityof ulus

mm mshafttheof length

mm Nstress imum

rpm Nspeed

WKW Pd transmittepower

VTHT AVADI III SEM


Data asked:

The angle of twist

Solution:

NmmT

N

P TTorqueor


mean

mean

mean

1790490 1002

60000 1075

2

60000

60000

2

3

NmmT

TTT

TT Ttorque Maximum

mean meanmean

mean mean

2327627 1790490.3 1

.3 1.3 0

30

max

0 0

max

mm mmd

T dshafttheof Diameteror

dT shaftsolidtheby dtransmitte torqueMaximum Wkt

5165. 5070

162327627

16

16

3

1

3

1

max

3max

G rtwist ofAngle

or G

r wktAlso

0

3

5 180

.0915 0

.0915 01090

3000

5. 25

70

radiantwist ofAngle

Result:

VTHT AVADI III SEM


05 twistof angleThe

4. A solid circular shaft transmits 75KW power at 200 rpm. Calculate the shaft diameter, if

the twist in the shaft is not to exceed 10 in 2 m length of shaft and shear stress is limited to

50 N/mm2. Take modulus of rigidity C= 1x105N/mm2[EVEN-2012]

Given data:

radiantwist ofAngle

mm NC rigidityof ulus


mm Nstress shearimum

rpm Nspeed


.01744 0180

1 1

/10 1mod

2000 2

/ 50max

200

10 7575

0

25

2

3

Data asked: The shaft diameter d

Solution:

3

2

60000

60000

2

75 10 60000 3580980

2 200

mean

mean

mean

NTPower transmitted P

Por Torque T

N

T Nmm

2

3max

1

3max

1

3

1 max 50 /

16

16

2327627 16 62

50

Diameter of the when imun shear stress N mm


Tor Diameter of the shaft d

d mm

VTHT AVADI III SEM


C

T Jinertia ofmoment PolarOr

dJ inertiaof momentPolar Jwhere C

J

T relationthe gU

than moreexceed beshouldtwistwhen shaftthe ofDiameter

4

0

32sin

12

mm mmd Diameter

C

T dor

C

T die

21 73 . 7232

01744. 010 1

2000 2327627

32

32

4

1

5

4

1

4

The suitable diameter of the shaft must be bigger value among the above two diameter values

mmshaft d of the Diameter 73

Result:

The diameter of the shaft d= 73 mm

5. What must be the length of a 5 mm diameter aluminium wire so that it can be twisted

through one complete revolution without exceeding a shearing stress of 42 MN/m2? Take

C=27GNm2 [EVEN-2012]

Given data: 6

2 2

6

92 3 2

6

10max 42 / 42 42 /

10

10mod 27 / 27 27 10 /

10

imum shear stress MN m N mm

ulus of rigidity C GN m N mm

0360 2

5

Angle of twist for one complete revolution radian

Diameter of the shaft d mm

Data asked:

VTHT AVADI III SEM


shaftthe ofLength

Solution:

T

J CLength or

C

J

T relationthe wkt

444

3

3

.32 61 532 32

.3 1030 542 16

16

mmdJ inertiamoment Polar

NmmT

dT torquewhere

m mm

T

J CLength

.1 9 10 . 10092.3 1030

.32 61 21027 3

Result:

mshafttheof Length 1 .10

6. A hollow shaft with diameter ratio 3/8 is required to transmit 500 KW at 100 rpm, the

maximum torque being 20% greater than the mean. The maximum shear stress is not to

exceed 60 N/mm2 and the twist in length of 3 m is not to exceed 1.40. Calculate the

minimum diameters required for the shaft. C=84 KN/mm2. [ODD-2009]

Given data:

VTHT AVADI III SEM


.375 0 8

3

.2 1

.02442 0180

4 . 1 4 . 1

/10 84mod

3000 3

/ 60max

100

10 500500

max

0

23

2

3

o

i

mean

d

d Kratio Diameter

TT torqueMaximum

radiantwist ofAngle




rpm Nspeed


Data asked:

The minimum diameters required

Solution:

NmmT

N

P TTorqueor


mean

mean

mean

47770700 1002

60000 10500

2

60000

60000

2

3

NmmT

TT dtransmitte torqueMaximum mean

57324840 47770700.2 1

.2 1

max

max

VTHT AVADI III SEM


2

3 4max

1

3

max

4

1

3

0 4

1 max 60 /

116

16

1

57295778.4 16 17

60 1 0.375

o

o


Wkt Maximum torque transmitted by the solid shaft T d k

Tor Outside diameter of the shaft d

k

d

0.58 171

0.375 171 63.9 64i o

mm mm

Inside diameter of the shaft d k d mm mm

C


kdJ inertiaof momentPolar Jwhere C

J

T relationthe gU


o

44

0

1 32

sin

.4 12

mm mmd diameterOutside

kC

T dor

C

T kdie

o

o

o

8 172 . 171.375 01

32

02442. 010 84

30004. 57295778

1

32

1 32

4

1

43

4

1

4

44

mm mmd kd shaftthe ofdiamter Inside oi 65 5. 64375 172 . 0


Result:

The outside diameter of the shaft do = 172 mm

The inside diameter of the shaft di = 65 mm

VTHT AVADI III SEM


7. Determine the diameter of the solid shaft which transmits 90KW at 160 rpm. Also

determine the length of the shaft if the twist must not exceed 10 over the entire length. The

maximum shear stress is limited to 60 N/mm2. Take the value of modulus of rigidity=8x104

N/mm2 [EVEN-2011]

Given data:

radiantwist ofAngle



rpm Nspeed


o 01745 .0180

1 1

/10 8mod

/ 60max

160

10 9090

24

2

3

Data asked:

Diameter of the shaft and length of the shaft

Solution:

(i)Diameter of the shaft d

NmmT

N

P TTorqueor


mean

mean

mean

5371480 1602

60000 1090

2

60000

60000

2

3

mmd

T dshafttheof Diameteror

dT shaftsolidtheby dtransmitte torqueMaximum Wkt

.8 76 60

16 5371480

16

16

3

1

3

1

max

3max

(ii)Length of the shaft:

VTHT AVADI III SEM


mm

R CLength or

C

Rrelation theg U

.6 893 60

.4 38 01745. 010 8

sin

4

Result:

The diameter of the shaft d= 76.8 mm

Length of the shaft l =893.6 mm

8. A closed coiled helical spring is to carry a load of 500 N. its mean coil diameter is to be 10

times that of the wire diameter. Calculate this diameter if the maximum shear stress in the

material of the spring is to be 80 MN/m2. [EVEN-2012]

Given data:

22 / 80/80

,10

500

mm Nm MNstress shear

diameter wired whered Ddiameter coilmean

NW springon Load

Data asked:

The diameters of mean coil and wire

Solution:

VTHT AVADI III SEM


.15 159 80

5 50016

80

5 50016

16

16sin

2

3

3

3

d or

dd

WR dor

dWR relationthe gU

Or wire diameter d= 12.6mm

Mean coil diameter D = 10d = 10x12.6=126mm

Result:

Wire diameter d = 12.6 mm

Mean coil diameter D = 126 mm

9. Thestiffness of a close coiled helical spring is 15N/mm of compression under a maximum

load of 60 N; the maximum shearing stress produced in the wire of the spring is 125

N/mm2. The solid length of the spring (when the coils are touching) is given as 5 cm find

(i) Diameter of wire

(ii) mean diameter of the coils and

(iii) Number of coils required, take modulus of rigidity is 4.5x104N/mm2.

[EVEN-2011}

Given data:

Stiffness of the helical spring = 15 N/mm

Maximum load W= 60 N

Shearing stress = 125 N/mm2

The solid length of the spring = 5cm=50 mm

Data asked:

VTHT AVADI III SEM


Diameter of wire, mean diameter of the coils and Number of coils required.

Solution:

The solid length of the spring L = 5 cm=50 mm (when the coils are touching)

i.e., number of coils 150

d dn

Wkt,

2.409 0 60

1125

16

1

16

16

33

3

3

ddR radiuscoil mean

W dR or

dWR

Also wkt, 364 3

4

nR

Cd sStiffness

mmd wireof Diameter

dor

d

d d

dd

d

get weon andequation Subtitute

.42 06 3 . 137

137 5. 150 0684. 064

10 5. 4

50 0684. 064

5 10 . 4.5 1

50 409. 064

10 5. 4.5 1

, 321

4

1

44

9

4 4

3 3

4 4

. 15.61 14 42. 3

50 50

.72 3242. 3.409 0 2

.409 0 223

3

sayd

ncoils ofNumber

mmD

dR Dcoil theof diametermean

Result:

VTHT AVADI III SEM


The diameter of the wire d= 3.42 mm

Mean diameter of the coil D= 32.72 mm

The number of coils n = 15

10. A helical spring in which the mean diameter of the coils is 12 times the wire

diameter, is to be designed to absorb300 J energy with an extension of 150 mm. the

maximum shear stress is not to exceed 140 N/mm2. Determine the mean diameter of the

spring, diameter of the wire which forms the spring and the number of turns. Assume

the modulus of rigidity of the material of the spring as 80 KN/mm2.[ODD-2009]

Given data:

Mean diameter of the coil D = 12 d, where d-wire diameter

Energy absorbed E = 300 J

Extension = 150mm

Shear stress = 140 N/mm2

Modulus of rigidity G = 80 KN/mm2

Data asked:

Mean diameter of the spring, diameter of the wire and number of turns

Solution:

Wkt,

3

3 3

1 300*10

2

300 10 2 300 10 2 4000

150

Energy absorbed W

or Load W N

VTHT AVADI III SEM


Result:

3

3

3

2

3 4

4 3

3

sin 16

16

16 4000 12

140

16 4000 12 1747

140

41.7

12 41.7 500.4

16

16

150 80 10 41.

U g the relation WR d

WR or d

dd

or d

diameter of the wire d m

mean diameter of coil D mm

WR n C d or n

Cd WR

Number of coils n

4

3

7 36.35 37

16 4000 250.2

Result:

41.7

12 41.7 500.4

diameter of the wire d m

mean diameter of coil D mm

11. A hollow circular shaft 20mm thick transmits 300 KW at 200 rpm. Determine the inner

diameter of the shaft if the shear strain is not to exceed 8.6x10-4. Take C=80GN/m2, trial

and error method can be used. [ODD-2010]

Given data:

234

24

4

3

8 / . 6810 806 10 . 8

modmax

/10 8mod

6 10 . 8

200

10 300300

mm N

ulus shearstrain shearstress shearimum


strain Shear

rpm Nspeed


Solution:

VTHT AVADI III SEM


NmmT

N

P TTorqueor


mean

mean

mean

143312102002

60000 10300

2

60000

60000

2

3

Shear stress is given, hollow shaft

4

343 1

16 )1 (

16 o

ioo

d

ddkdT Torque

4

3

4

6 3

6 6

6

40 1

16

4014.32 10 68.8 1

16

, min 100

11.7 10 14.32 10

min 108

14.3 10

o

o

o

o

o

o

o

o

dTorque T d

d

dd

d

By trial and error method assu g d mm

Torque T Actual value

but assu g d mm

Torque T Actu

, 108

40 108 40 68

68

o

i o

i

al torque value so that the outside diameter d mm

Also the inside diameter d d mm

The inside diameter d mm

Result:

12. A solid shaft is subjected to a torque of 100 N-m.Find the necessary shaft diameter if

the allowable shear stress is 100 N/mm2 and the allowable twist is 30 per 10 diameter length

of the shaft. Take C = 1x105 N/mm2[ODD-2008]

Given data:

VTHT AVADI III SEM


d shaftthe ofDiameter

asked Data

dLength

twist ofAngle



NNm Td transmitteTorque

:

10

05233 .0180

3 3

/ 100max

/10 1mod

10 100100

0

2

25

3

Solution:

2

3max

1

3max

13 3

1 max 60 /

16

16

100 10 16 17.38 18

100



Tor Outside diameter of the shaft d

d mm mm

C


dJ inertiaof momentPolar Jwhere C

J

T relationthe gU


4

0

32sin

32

mm mmd

d Diameter

C

T dor

C

T die

48 13 . 1232

05233. 010 1

10 10100

32

32

3

1

5

3

4

1

4

VTHT AVADI III SEM



mmshaft d of the Diameter 18

Result:

The diameter of the shaft d= 18 mm

13. 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 is also same. [EVEN-2010]

Data given:

For hollow shaft

Inside diameter di = 0.6 do [Outside diameter]

Let the solid shaft diameter be d

Both solid and hollow shafts transmit the same power at same speed

Data asked:

Percentage saving in material

Solution:

Polar modulus of the solid shaft 16

3d

Zsolid

VTHT AVADI III SEM


Polar modulus of the hollow shaft

16 8704. 0

.6 0116

1 16

3

43

43

o

o

ohollow

d

d

kd

Z

Since both the shafts should have the same polar modulus,

hollow ZsolidZ

.1489 1 8704. 0

1

16 16.8704 0

3

33

d

d

or dd

o

o

.047 1 d

do

1001

100

solid

hollow

solid

hollow solid

A

A

A

A Amaterial insavings Percetage

VTHT AVADI III SEM


o

o

kd

d

d

kd

d

d d

material insavings Percetage

o

o

io

.84 29 100.6 01.047 1 1

100 11

100 1

1

100

4

4 1

22

2

2

2

22

2

22

Result:

The percentage savings in material= 29.84%

14. Derive the derivation of shear stress produced in a circular shaft subjected to

torsion. [EVEN-2012]

Consider a shaft of length l, radius R fixed at one end and subjected to a torque T at the

other end is shown in fig.

CIRCULAR SHAFT

VTHT AVADI III SEM


Let `0‘ be the centre of the circular section, `B‘ a point on surface and AB be the line on the shaft parallel to the axis of the shaft.

When the shaft is subjected to torque ` T‘, B is moved to B‘.

twistangle of also called and is DOD'

Now distortion at the outer surface due to torque T=DD‘ from fig.

, Shear stress at the outer surface

Wkt Modulus of rigidity C Rshear strain at the outer surface R

1'

tan sin[tan'

'

DDsurfaceouter atstrain Shear

ce smallvery isCD

DD

DD

shaftthe ofLength

surfacetheat Distortion

length unitper Distortionsurface outerat strainShear

R

Cor

VTHT AVADI III SEM


15. The expression for the maximum torque transmitted by the circular solid shaft.

The maximum torque transmitted by the circular solid shaft, is obtained from the

maximum shear stress induced at the outer surface of the solid shaft. Consider a shaft

subjected to a torque T as shown in (EVEN 2012)

Let - be the maximum shear stress induced at the outer surface

R-be the radius of the shaft

q-be the shear stress at a radius ‗r‘ from the centre.

CIRCULAR SOLID SHAFT.

VTHT AVADI III SEM


Consider an elementary circular ring of thickness ‗dr‘ at a distance ‗r‘ from the centre as shown in fig

dA = 2πr dr

r

q

RHere

dA q

ring ofArea ringon actingstress shearring elementaryon forceTurning

R

rqr radiusthe atstress Sear

,

dr rR

rdr R

r

22

2

Turning moment due to the turning force on ring

dT=Turning force on ring x Distance of ring from the axis

dr rR

r drrR

dT

3

2

2

2

3

3

34

0

4

0

3

0

3

0

1622

24 2

4 22

2

dd

T

RR

R

r

R drr

R

dr rR

dTT momentturning totalThe

RR

RR

VTHT AVADI III SEM


3

16 dT Torque or moment turning total The

316.)

5 A hollow shaft with diameter ratio is required to transmit 450kw at 120rpm.The shearing

stress in the shaft must not exceed 60 N/mm^2 and twist in length of 2.5m is not to exceed 1 .

Calculate m

inimum external diameter of the shaft.take c=80 kN/mm^2.(MAY-13)

Given:

3 = 0.6 = d= 0.6D; c=80 10^3 N/mm^2;

5

mm^2

d

D

P=450kw = 450 10^3 w

N=120 rpm

L=2.5m 2.5 1000 = 2500mm

=1 = 1 = 0.01744 180

TO FIND:

a.) minimum external diameter of shaft

2a.) P=

60

2 450 10^3=

60

450 10 ̂ 3 60 = T

2 120

T=35.82 10^3 Nm

T=35.82 10^6 Nmm

NT

T

b.) Torque,

^ 4 ^ 4 T= ]

16

^ 4 (0.6 ) ̂ 4 35.82 10^6= 60 [ ]

16

^ 4 0.1296 ^ 4 35.82 10^6=11.775 ]

D d

D

D d

D

D d

D

VTHT AVADI III SEM


c.) Angle of twist( ):

. ,T=35.82 10^6 Nmm

j= [ ^ 4 ^ 4] 32

c=80 10^3 N/mm^2;; =0.01744;

T c

j l

D d

l=2500mm

35.82 10 ̂ 6 80 10 ̂ 3 0.01744

2500[ ^ 4 ^ 4] 32

[ ^ 4 ^ 4] 64.18 10 ̂ 6 32

D^4 d^4= 654.06 10^6

D^4[1-0.1296]=654.06 10^6

[0.8704]D^4=654.06 10^6

D d

D d

D^4=751.44 10^6

D=165.56 mm

d=99.34 mm

^ 435.82 10^6=11.775 [1-0.1296]

35.82 10^6=11.775 D^3[0.8704]

D^3=3497473.462

D=151.75 mm - ext diameter

d=0.6 151.7

D

D

5

d=91.05 mm -int diameter

Result: External diameter of shaft D=165.56 mm

VTHT AVADI III SEM


17.Derive a relation for deflection of a closed coiled helical spring subjected to an axial

downward load w.(May-13)

In closed coil helical spring an axial pull or thrust produces only torsion.The type of stresses

that are produced in this type of spring are

a.) Direct shear stress

b.) Torsional shear stress

c.) Bending shear stress

In a closed coil helical spring,direct shear and bending stress are negligible,when subjected to axial pull w.

Direct shear stress and torsional shear stress are neglected when subjected to a couple at one end.

Consider a spring of mean coil radius 'R'

wire diamter,d

number of coils ,n

total length of wire,l

axial deflection of spring at the free end under load w;

angle of rotation of the free axial couple M;

consider a closed coil helical spring subjected to an axial load of W at one end,

The twisting moment exerted on the wire,

T= axial force mean radius of spring

T=W R;

From torsion equation,

.

Angle of twist for the whole length of spring;

=

w.k.t

T=W.R,l=2 Rn;j= d^4 32

T c

j l

Tl

cj

VTHT AVADI III SEM


2=

. ^ 4

32

64 ^ 2. = radians

^ 4

thus the free end will twist htrough an angle ,consequently the free end will have an axial moment R .

If is axial movement or deflection,

WR Rn

c d

WR n

cd

64 ^ 2. =R =R

^ 4

64 ^ 3. = ;;(or)

^ 4

8 ^ 3. =

^ 4

WR n

cd

WR n

cd

WR n

cd

he stiffness K of the spring is al load reqiured to produce unit deflection

K=

K= 64 ^ 3.

( ) ^ 4

therefore, relation for a

W

W

WR n

cd

closely coiled helical spring subjected to axial load W,

8 ^ 3. =

^ 4

WR n

cd

VTHT AVADI III SEM


UNIT-4

PART-A

1. A cantilever beam of spring 2 m is carrying a point load of 20 KN at its free end.

Calculate the slope at the free end. Assume EI=12×103 KN-m2. (EVEN 2006)

Slope at the free end,

øB=WL2/2EI

=20× (2)2/2×12×103

ØB=0.0033 rad

2. Calculate the maximum deflection of a simply supported beam carrying a point load of

100 KN at midspan. Span =6m, EI=20000 KN/m2. (ODD 2006)

Maximum deflection,

Ymax=WL3/48 EI

Ymax=100× (6)3/48×20000

=0.0225m

3. What is the maximum deflection in a simply supported beam subjected to uniformly

distributed load over the entire span? (EVEN 2007)

The maximum deflection in beam,

Yc=5WL4/384 EI

VTHT AVADI III SEM


4.In a simply supported beam of 3m span carrying uniformly distributed load throughout

the length, the slope at the supports is 10, what is the maximum deflection in the

beam?(EVEN 2008)

Slope, ØA=WL3/24EI= 10

ØA=WL3/24EI= π/180

Maximum deflection, y max = 5 WL4/384 EI

= WL4 5/25 EI 16

= π/180 * 5×3/16

ymax= 0.0164m

5.State the expression for slope and deflection at the free of a cantilever beam of length „L‟ subjected to a uniformly distributed load of „w‟ per unit length.(ODD 2008)

Slope at B,

ØB=WL3/6EI

Maximum deflection, y B =WL3/8EI

CANTILEVER BEAM

6. What is the relation between slope, deflection and radius of curvature of a beam? (EVEN

2010)

1/R =d2y/dx2

R= radius of curvature

Slope,= dy/dx

Y = Deflection.

7. Where the slope and deflection will be maximum for the cantilever with point load at its

free end?

VTHT AVADI III SEM


Both slope and deflection will be maximum at the free end.

8. What is the slope at the support for a SSB of constant EI and span L carrying central

concentrated load?

Slope at the support,

øa=øb=WL2/16EI

9. State the two theorems in the moment area method.(EVEN 2010)

Mohr‟s theorem 1: The change of slope between any two points is equal to the net area of the BM diagram

between these points divided by EI.

Mohr‟s theorem 2: The total deflection between any two points is equal to the moment of the area of the

BM diagram between these two points about the last point divided by EI.

10. Name any method which employs BMD for the calculation of slope and deflection.

Moment area method

11. Calculate the effective of a long column, whose actual length is 4 m, when,

a) Both ends are fixed.

b) One end is fixed while the other end is free.(MAY /JUNE 2006)

a) When both ends are fixed,

Effective length, L=4/2= 2m

b) When one end is fixed and the other end is free.

Effective length, L=2×4= 8m.

12.Find the critical load of an Euler‟s column having 4m length, 50mm×100mm cross section and hinged at both the ends E=200KN/mm2.(ODD 2006)

Young‘s modulus, E=200KN/mm2

P=π2EI/L2

P=128.51 KN.

13. What is crippling load? Give the effective length of columns when both ends hinged and

when both ends fixed? (EVEN 2007)

The load at which the column just buckles is known as buckling load or critical load or

crippling load.

VTHT AVADI III SEM


Column with both ends are hinged:

Effective length, L=l(actual length)

Column with both ends are fixed:

Effective length, L=l/2

14. State any two assumptions made in Euler‟s column theory. (EVEN 2008)

1. The cross section of the column is uniform throughout its length,

2. The length of the column is very long as compared to its cross sectional dimensions.

15. State Rankine‟s formula foe crippling load.

Crippling load,

p= cA/ (1+a(L/K) 2

16. State the limitations of Euler‟s formula.(EVEN 2012)

Crippling stress= π2E/ (L/k) 2

For example, column with both ends hinged, L=1.

Crippling stress=π2E/ (L/k) 2

L/k- slenderness ratio.

If the slenderness ratio is small, the crippling stress will be high. We know that for the column

material, crippling stress cannot be more than the crushing stress. It is thus obvious, that the

Euler‘s formula gives a valve of crippling stress greater than the crushing stress when the slenderness ratio is less than a certain limit.

17. Define column.

A structural member which is subjected to axial compressive load is known as column (or) strut.

In column, the member of structure is vertical and both of its ends are rigidly fixed.

VTHT AVADI III SEM


Example:A vertical pillar between roof and floor.

18. What are the assumptions of double integration method?(EVEN 2009)

1. The equation is based on the bending moment.

2. The effect of shear force is very small and thus neglected.

3. Beams are uniform

4. Inertia is uniform.

5. Material is homogenous.

19.Define Slope Theorem & Deflection Theorem.

Slope Theorem:It is the ratio of area of bending moment diagram over the flexural rigidity is

called Mohr first theorem, to find the slope (A/EI).

Deflection Theorem:It is the product of slope and centroidal distance from a point, to find the

deflection (Ax/EI).

20. What is a conjugate beam?

It is hypothetical beam; the load is derived from the bending moment diagram of the actual beam

to find the slope and deflection. It is useful for varying I section along the span. The reaction at

the support from the BM diagram load will give the slope.

21. What are the assumptions of Euler? (EVEN 2010)

1. Column is straight.

2. Load is axial.

3. Self weight is neglected.

4. Column material is homogeneous.

5. Column fails due to buckling.

VTHT AVADI III SEM


6. EI(flexural rigidity) is uniform

22.What are the different types of ends and their equivalent length in the column?(EVEN

2011)

1. Both ends hinged. (Le=L)

2. Both ends fixed. (Le=L/2)

3. One end hinged other end fixed. (Le=1.414L (root 2*L))

4. One end fixed other end free. (Le=2L)

23. Define: Column and strut.

A column is a long vertical slender bar or vertical member, subjected to an axial Compressive load and fixed rigidly at both ends. A strut is a slender bar or a member in any position other than vertical, subjected to a Compressive load and fixed rigidly or hinged or pin jointed at one or both the ends.

24. 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 failure: This 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 or critical load. This type of failure occurs in long column.

25. What is slenderness ratio (buckling factor)? What is its relevance in column?(EVEN 2013)

It is the ratio of effective length of column to the least radius of gyration of the cross sectional ends of the column.

Slenderness ratio = l eff / r

l eff = effective length of column

r = least radius of gyration Slenderness ratio is used to differentiate the type of column. 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.

26. What are the factors affect the strength column?

1. Slenderness ratio

VTHT AVADI III SEM


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.

27. Differentiate short and long column (EVEN 2011)

Short column Long column

1. It is subjected to direct compressive stresses it is subjected to buckling stress only only. 2. Failure occurs purely due to crushing only. Failure occurs purely due to bucking only.

3. Slenderness ratio is less than 80 Slenderness ratio is more than 120.

4. Its length to least lateral dimension is less It‘s length to least lateral dimension is moreThan 30. (L / D › 30) than 8. (L / D ‹ 8)

28.What is meant by Double-Integration method? (EVEN 2013) Double-integration method is a method of finding deflection and slope of a bent beam. In this method the differential equation of curvature of bent beam,

Is integrated once to get slope and twice to get deflection. Here the constants of integration C1

and C2 are evaluated from known boundary conditions.

29.Write the maximum value of deflection for a cantilever beam of length of length L, constant EI and carrying concentrated load W at the end.

Maximum deflection at the end of a cantilever due to the load =WL3/3El

30.Write the maximum value of deflection for a simply supported beam of a length L, constant EI and carrying a central concentrated load W.(EVEN 2012) Maximum deflection at a mid span of simply supported beam due to a central load

WL3/48𝐸𝐼31. Write the value of fixed end moment for a fixed beam of span L and constant EI

carrying central concentrated load W. Fixed end moment due to central concentrated load W =𝑊𝐿/8

VTHT AVADI III SEM


32.What are the different methods used for finding deflection and slope of beams? (EVEN

2011)

(i) Double integration method (ii) Mecaulay‘s method (iii) Strain energy method (iv) Moment area method (v) Unit load method

33.Write the differential equation of deflection of a bent beam.

34.What is meant by elastic curve?

The deflected shape of a beam under load is called elastic curve of the beam, within elastic limit.

35.When Mecaulay‟smethod is preferred?

This method is preferred for determining the deflections of a beam subjected to several

concentrated loads or a discontinuous load.

36.What are the boundary conditions for a simply supported end?

The boundary conditions for a simply supported end beam are: (i) Deflection at the support is zero. (ii) Slop exists at all points except at the point where deflection is maximum. (iii) Bending moment is zero at the support

37.What are the boundary conditions for a fixed end?

Both deflection and slope at the fixed support are zero.

38.Define the term slope.

VTHT AVADI III SEM


Slope at any point on the bent beam is the angle through which the tangent at that point makes with the horizontal.

39.What is meant by deflection of beams?

When a flexural member is subjected to transverse loads, the longitudinal axis of the beam deviates from its original position because of the bending of the beam. This deviation at any cross section is called as deflection.

40. What are the values of slop and deflection for a cantilever beam of length „l‟ subjected to load „W‟ at free end?

41.How the differential equation is written for the beams of varying cross section?

If a beam is of varying cross-section and varies uniformly according to some law, the expression

EId2y/dx2= Mxcan be arranged in the formd2y/dx2=Mx/Elxin which Mxand Lx are functions of x.

42.When do you prefer Moment Area Method?

Even though the moment area method can be used for problems on slopes and deflections, it is convenient to use this method for the following types of problems (with varying cross-section) (i) Cantilever beams (ii) Simply supported beams carrying symmetrical loading (iii) Beams fixed at both ends.

43. What is the value of maximum deflection for a fixed beam of span „l‟, carryingconcentrated load W at midspan?

Maximum deflection under the load = 𝑊𝑙3/192𝐸𝑙44. What is the value of maximum deflection for a fixed beam of span „l‟, carryinguniformly distributed load W per meter run?

Maximum deflection at mid span =𝑊𝑙4/384𝐸𝑙

VTHT AVADI III SEM


45.What is the slope at the support for a simply supported beam of constant EI and span L carrying central concentrated load?

Slope at the support due to central concentrated load, w = 𝑊𝑙2/16𝐸𝑙46.Write the support moment for a fixed beam of constant EI and span L carrying uniformly distributed load W per unit length over the entire length

Support moment due to u.d.l= 𝑊𝑙2/12

47.A cantilever beam of constant EI and span L carries a u.d.l of W unit length throughout

Its length, what is the slope at the free end?

Slope at the free end =𝑊𝑙2/6𝐸𝑙48.Write the deflection at the free end of a cantilever beam of constant EI and span L

carrying u.d.l of W/meter length.

Maximum deflection at the free end of a cantilever due to u.d.l of W/m =𝑊𝑙3/8𝐸𝑙49.What is meant by determinate beams?

The beams whose external reacts can be determined with the help of equations of static equilibrium alone are called determinate beams.

50. What is meant by indeterminate beams?

The beams whose support reactions cannot be obtained with the help of static equations of Equilibrium alone is called indeterminate beams.

51. Give examples for determinate and indeterminate beams Determinate beams: cantilever and simply supported beams Indeterminate beams: fixed end beams, continuous beams and propped cantilever beams.

52.What are the units of slope and deflection?

Slope = radians

Deflections = mm

53.What are the values of slope and deflection for a simply supported beam of length l

VTHT AVADI III SEM


subjected to moment at both the ends.

SIMPLY SUPPORTED

54.There are two beams one simply supported and other fixed beam carry concentrated load W at the mid span. Their spans are equal. Compare deflections.

55. Write the crippling load and effective length for column for different end condition.

56.Define crippling load?

The load at which the column just buckles is called ―buckling load‖ this is also knownas critical load or crippling load.

57.What is effective length of column?(ODD 2012)

The effective length of a given with given end condition is the length of an equivalent column of the same material and section with hinged ends having the value of the crippling load

VTHT AVADI III SEM


to that of the given column.

58.Write radius of gyration for a solid circular cross section at diameter d?

59.What is a beam column?

It is a member which is subjected with axial thrust and lateral loads.

60. Write the equation for Euler‟s critical stress.

61.Define core or kern of the section.

Core: The middle portion of a section

Kern: it is an area within which the line of action of the force p must cut the cross section if the

stress is not to become tensile. Rectangular section kern is b/6 and circular section kern is d/4

16 MARKS:

PART-B

1. A beam is simply supported as its ends over a span of 10 m and carries two concentrated

loads of 100 KN at a distance of 2 m and 5 m respectively from the left support. Calculate

a) slope at the left support: b) slope and deflection under the 100 KN load. Assume

EI=36×104 KN-m2. (EVEN 2006)

Given:

EI= 36×104KN-m2

To find:

a) Slope at the left support.

D C A B

X

X

60

kN

100

kN

2

m 5

m

10

m

x

R R

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b) Slope and deflection under 100 KN load.SIMPLY SUPPORTEDBEAM

Solution:

Macaulay‟s method:

Let RA and RB be the reaction at the left and right supports,

Taking moments about A,

RB× 10 = (60×5) + (100×2)

Wkt,

RA + RB = 100KN + 60KN

RA + 50KN = 160KN

RA=110KN

Taking bending moment at x:

EI d2y/dx2 = RA× x -100(x-2) -60(x-5)

EI d2y/dx2 = 110× x -100(x-2) -60(x-5)

Integrating we get,

EI dy/dx = (110 × x2)/2 + c1-100(x-2)2/2 -60(x-5)2/2

EI dy/dx = (55 × x2) + c1-50(x-2)2 -30(x-5)2

Integrating again, we get

EIy = 55 (x3/3) + c1x + c2 -50(x-2)3/3 -30(x-5)3/3

At x= 0, deflection, y= 0

Substituting x= 0, y= 0 values in eqn 2 up to first dotted line.

C2=0

At x= 0, deflection, y= 0

Substituting x= 10, y= 0 values in eqn 2

1

2

VTHT AVADI III SEM


0= 55/3(10)3 + c1(10) + c2 – 50(10-2)3/3 – 30(10-5)3/3

= 18.33 ×103 + c1(10) + 0-8533.33-1250=C1 = -855

Sub c1,c2 and EI values in eqn (1),

Slope equation

36× 104dy/dx = 55x2 + (-855) -50(x-2)2 -30(x-5)2

a) Slope at the left support:

Slope at the left support, i.e., at A, x= 0

Substituting x= 0 is eqn 3 up to first dotted line

36 × 104dy/dx = 55 (10)2 – 855

dy/dx = -855/ (36 × 104)

slope at A1øA = dy/dx = -2.375 × 10-3 rad

b) Slope at 100 KN load:

Sub x= 2m in eqn 3 up to second dotted line

36 × 104(dy/dx) = 55(2)2 – 855 – 50 (2.2)2

36 × 104 (dy/dx) = -635

dy/dx = -1.76 × 10-3 rad

slope at 100 KN, øc = dy/dx = -1.76 × 10-3 rad.

c) Deflection at 100 KN:

Wkt, deflection equation,

EIy = 55(x3/3) + c1x + c2 -50(x-2)3/3 -30(x-5)3/3

Substituting EI, c1 and c2 values

36× 104 y = 55(x3/3) + (-855x) + 0 -50(x-2)3/3 -30(x-5)3/3

Deflection at 100KN, sub x=2 in eqn 3 up to second dotted line.

36 × 104 y = 55(2)3/3 – 855(2) – 50(2-2)3/3

36 × 104 y = -1563.33 KN-mm

Y = -1563.33/ (36 × 104)

Y= -4.34 mm

Deflection at 100KN, yc = -4.34 mm

Result:

Deflection at 100KN, yc = -4.34 mm

3

3

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2. A horizontal beam AB is simply supported at A and B 6 m apart. The beam is subjected

to a clockwise couple of 300 KN-m at a distance of 4 m from the left end A. if E=2×108

mm4. Determine, using Macaulay‟s method. (ODD 2006)

a) The deflection at a point where couple is acting.

b) The maximum deflection

Given:

E= 2 × 105 N/mm2

I = 2 × 108 N/mm2

To find:

Deflection at c=?

Maximum deflection =?

Solution:SIMPLY SUPPORTEDBEAM


RB× 6 = 300

RB = 50KN

But

RA + RB = 0

RA = -RB = -50 KN

BM eqn between AC,

Mx = EI ( d2y/dx2)

= 50 x – 120

Integrating eqn (1) with respect to x

EI (dy/dx) = 50(x2/2) + c1 -120x

Integrating eqn (2), w.r.t. ‘x‘ EIy = 25(x3/3) + c1x + c2 -120 (x2/2)

EIy = 25(x3/3) + c1x + c2 -60 (x2)

Applying the following boundary conditions,

a) When x= 0 , y= 0 and

b) When x= 6m , y= 0

Applying the first boundary condition in eqn (3)

1

2

3

A B C

4 m

6 m

2 m

300 KN-m

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0 = c2

Now equation (3) reduces to

EIy = 25(x3/3) + c1x – 60 x2

Applying the secondary boundary condition eqn 4,

0 = 25 × (6)3/3 + c1 (6) -60(6)2

6 c1 = 60 (6)2 – (25/3) × (6)3

C1 = 60

The equation (4) becomes,

EI y = 25(x3/3) + 60x – 60x2

The deflection at c when x= 2m

EI y = 25(2000)2/3 + 60(2000) – 60(2000)2

Deflection, y = 6.643 × 1010 / (2 × 10 5 × 2 × 108)

= 0.00166 mm

The slope becomes zero where the maximum deflection occurs

Therefore, dy/dx = 0

Applying this condition in eqn (2)

0 = 25x2 + 60 -120x (c1 = 0)

25x2-120x + 60 = 0

X = 120 +(√120 – 4× 25 × 60)/(2 × 25)

= 0.567 m (or) 4.233 m

At x= 0.567 m, then the deflection y= 0.000037 mm <yc

At x= 4.233 m, the deflection y= 0.016 mm >yc

Therefore, the maximum deflection occurs at a distance of 4.233 m from B,

Ymax = 0.016 mm

Result:

Ymax = 0.016 mm

4

5

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3.A beam AB of length 8 m is simply supported at its ends and carries two point loads of 50

KN and 40 KN at a distance of 2 m and 5 m respectively from left support A. determine,

deflection under each load, maximum deflection occurs. Take E=2×105 N/mm2 and

I=85×106mm4. (EVEN 2007)

Given:

E = 2 × 105 N/mm2

I = 85 × 106 N/mm4

To find:

Ymax=?

Solution:

Reactions at supports,

Taking moment about ASIMPLY SUPPORTEDBEAM

RB× 8 = (50 × 2) + (40 × 5)

RB = 37.5 KN

RA + RB = 50 + 40

RA = 52.5 KN

Macaulay‘s method: Mx = EI (d2y/dx2) = 52.5x -50(x-2) -40(x-5)

Integrating,

EI (dy/dx) = 52.5(x2)/2 + c1 -50(x-2)2/2 -40(x-5)2 /2

Again integrating,

EI y = 52.5(x3)/6 + c1 x + c2 -50(x-2)3 /6 -40(x-5)3/6

Applying the boundary conditions,

At x= 0, y= 0

C2 = 0

1

2

40 50

A C D B 2 m

5 m

8 m

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At x= 8, y= 0

0= 52.5(8)3/6 + c1(8) + 0 – 50(8-2)3/6 – 40(8-5)3/6

0 = 4480 + 8 c1 – 1800 – 180

0 = 2500 + 8c1

C1 = -312.5

Sub c1 value in slope equation

dy/dx = 1/(EI) [52.5 x3/6 -312.5 – 50 (x-2)3/6 - 40 (x- 5)3]

sub c1 ,c2 value in deflection equation

y= 1/(EI) [52.5 x3/6 -312.5x – 50 (x-2)3/6 - 40 (x- 5)3/6]

deflection,

at c, x = 2m

sub x= 2 in eqn (4)

yc = -2.205mm

at D , x = 6

yD= 4.008mm

maximum deflection:

put dy/dx = 0 in eqn (3)

x = 1.32m from A

sub x= 1.32m in eqn (4)

sub x= 1.32 m in eqn (4)

ymax = -3.7mm

Result:

ymax = -3.7mm

4. A beam is loaded as shown in fig. determine the deflection under the load points. Take

E=200 Gpa and I=160×106 mm4. (ODD 2007)

Given:

E=200 Gpa

I=160×106 mm4

ToFInd:

Maximum deflection, ymax = ?

Solution:

Taking moment about ASIMPLY SUPPORTEDBEAM

3

4

15 kN

3m 6m 4m

10 kN

A C D

B

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RB× 13 = (10 × 3) + (15 × 9)

RB = (10 × 3) + (15 × 9)/ 13

RB = 12.67 KN

RA + RB = (10 + 15)

RA = 25 – 12.69

RA = 12.30 KN.

Macaulay‘s method

Mx = EI (d2y /dx2) = 12.30x | -10(x-3) | -15(x-9)

Integrating,

EI dy/dx = 12.3x2/2 + c1 | -10(x-3)2/2 | -15(x-9)2/2

Again integrating

EIy = 12.3 x3/6 + c1x + c2 | -10(x-3)3/6 |-15(x-9)3/6

Applying boundary conditions

At x= 0, y= 0

At x= 13, y= 0

0 = 12.3 (13)3/6 + c1× 13 + 0 – 10(13.3)3/6 – 15(13-9)3/6

C1 = -20

Sub c1 value in slope equation,

dy/dx = (1/EI) [12.3x2/2 – 205 – 10(x-1)2/2 – 15(x-9)2/2]

sub c1, c2 values in deflection equation

y = (1/EI) [12.3x3/6 -205x- 10(x-1)3/6 -15 (x-9)3/6]

deflection,

at c, x= 3

sub x= 3 in eqn (4)

yc = (1/(2 × 105 × 160 × 106))

[12.3 × 33/6 - 205 × 3 – 10 (3-1)3/6 – 15 (3-9)3 /6]

= (1/(2 × 105 × 160 ×10 6)) [53.35 – 615 – 13.33 + 540]

Yc = (1/(2 × 105 × 160 × 106)) [34.98]

Yc = =1.09mm

At D, x= 10 in equation,

YD = (1/ (2 × 105× 160 × 106/6) [12.3 × 10)3/6 - 2.5 × 10 – 10 ×(9)3/6 – 12(10 – 9)3/6]

YD = -3.8 mm

Maximum deflection, put dy/dx = 0 in eqn (3)

3

4

2

1

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0 = (1/EI) [12.3x3/2 – 205 – 10(x-1)2/2 – 15(x-9)2/2]

X= 2.12m from A Sub x= 2.12m in eqn (4)

Ymax = 2.49m

Maximum deflection, ymax = 2.49m

Result:

Maximum deflection, ymax = 2.49m

5. In the beam shown in fig. determine the slope at the left end C and the at 1 m from left

end. Take EI=0.65 MNm2. (EVEN 2008)

Given:

EI=0.65 MNm2

To Find:

Determine the slope = ?

Solution:OVERHANGING BEAM

MB) = (3 ×20) + (1.2 × 20) + (30 × 1.2 × 1.8)

RA × 2.4 = 0

RA = 62KN

RB = 14KN

Add UDL of 30KN/m above and below DB.

VTHT AVADI III SEM


OVERHANGING BEAM

EI (d2y/dx2) = -20x |+62(x- 0.6) – (30/2) (x-0.6)2 |-20(x-1.8) + 30 (x-1.8)2/2

EI dy/dx = -20x2/2 + c1 |+62(x-0.6)2/2 -15(x-0.60)3/3 | -20 (x- 1.8)2/2 + 15(x-1.8)3/3

EIy = -10x3/3 +c1x +c2 | +31(x- 0.6)3/3 - 5(x-0.6)4/4 |-10 (x- 1.8)3/3 + 5(x-1.8)4/4

At x= 0.6m, y= 0 and at x= 3m, y=0

Sub c1= -3.72, c2= 2.952

EI (dy/dx) at c(x-0) = -3.72

(dy/dx)C = (-3.72/(0.65 × 103)) = -0.00572

EIy at 1m = -10/3 × 13 -3.72 (1) + 2.952 + 31/3 + 0.43– (5/4) × 0.44

= -3.472

Y at 1m = (-3. 472/(0.65 × 103)) = -0.00534m

Result:Y at 1m= -0.00534m

6. A simply supported beam is loaded as shown in fig. is 200mm wide and 400 mm deep.

Find the slopes at the supports. Deflections under and location and magnitude of the

maximum deflection. Take E=2×104 N/mm2. (ODD 2008)

Given:

E=2×104 N/mm2 20 kN

1 m 1 m 2 m

10 kN

A C D

B

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To Find:

Ymax=?

Solution:

Taking moment about A,SIMPLY SUPPORTEDBEAM

RB × 4 = (10 × 1) + (20 × 2)

RB = 50/4

RB = 12.5 KN

RA + RB = 10 + 20

RA = 30 – 12.5

RA = 17.5 KN

Moment of inertia,

I = bd3/12

= 200 × (400)3/12

I = 1.07 × 109 mm4

Macaulay‘s method:

Mx = EI (d2y/dx2) = 17.5x | -10(x-1) | -20(x-2)

Integrating,

EI (dy/dx) = 17.5 x2/2 + c1 | -10 (x-2)2/2 | -20(x-2)2/2

EI (dy/dx) = 17.5 x2/2 + c1 | -5(x-2)2|-10(x-2)2

Again integrating,

EIy = 17.5x3/6 + c1x + c2 |-5(x-1)3/3 |-10(x-2)3/3

Applying the boundary conditions

2

1

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At x= 0, y=0 ,c2=0

At x=4,y=0,c1= -25.75

Sub c1 value in sloping equation

(1) = dy/dx = (1/EI) [17.5x2/2 -28.75 – 5(x-1)2 -10(x-2)2

Sub c1 ,c2 value in deflection equation,

(2) =y= (1/EI) [17.5x3/6 -28.75x – 5(x-1)3/3 -10(x-2)3/3]

Slop at supports,

At A, x= 0

Sub x= 0 in equation (3)

dy/dx=øA= - 0.0013 rad

At B, x= 4m

Sub x= 4 in equation (3),

dy/dx =øB = 0.0012 rad

deflection,

At c, x= 1m

Sub x=1 in eqn (4)

Yc = -1.05 × 10-6m

Yc = -1.05mm

At D, x= 3m

Sub x= 3 in equation (4)

YD = -1.13 × 10-6m

YD = -1.13mm

Maximum deflection,

3

4

VTHT AVADI III SEM


Put dy/dx = 0 in eqn (3)

X = 1.95 mm from A

Sub x= 1.95 m from eqn (4),

Ymax = -1.68mm

Result:

Ymax = -1.68mm

7.A simply supported beam AB of span 4 m, carrying a load of 100 KN at its mid span C

has half cross sectional moment of inertias 24×106mm4 over the left half of the span and

48×106mm4 over the right half. Find the slope at two supports and the deflection under the

load. Take E=200Gpa. (EVEN 2010)

Given:

Span AB(L) = 4m

Load (w) = 100KN

Moment of inertia of (IAC) = 24 × 106mm4

=24 × 10-6mm4SIMPLY SUPPORTEDBEAM

Moment of inertia of (IBC) = 48 × 106mm4

= 48 × 10-6 mm4

Modulus of elasticity (E) = 200Gpa

= 200 × 106 KN/m2

100 kN

I 2I

2 2

B A

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SIMPLY SUPPORTEDBEAM

To Find:

Yc = ?

Solution:

Slope at the two supports,

Wkt, the bending moment will be zero at A and B and increase by a straight line low to

(wl/4) = (100 × 4)/4 = 100 KN –m

Now draw the conjugate beam. Taking moments about A,

RA × A = (1/EI) (0.5 × 100 × 2 × (4/3)) + (1/2EI) (0.5 × 100 × 2 × (8/3))

= (400/3EI) + (400/3EI) = (800/3EI)

RA = (800/3EI × 4) = (200/3EI)

RA = (1/EI) (0.5 × 100 × 2) + (1/EI) (0.5 × 100 × 2) - (200/3EI)

= (100/EI) + (50/EI) – (200/3EI)

RA= (250/3EI)

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Wkt, shear force at A,

FA = +RA = + (250/3EI)

Slope at A,

IA = (250/3EI) = (250/ (3 × (200 × 106) × (24 × 10-6))

= 0.017 rad

Similarly, shear force at B,

FB = -RB = (200/3EI)

Slope at B,

IB = - FB = - (-200/3EI)

= (200/EI)

= (200/ ((3 × (200 × 106) × (24× 10-6)

IB = 0.014 rad

Deflection under the load,

Wkt, bending moment at c

Conjugate beam,

Mc = (250/3EI) × 2 – (1/EI) × (0.2 × 1000 × 2 × (2/3))

= (500/3EI) - (200/3EI)

= (300/ 3EI)

= (100/EI)

Deflection at c,

Yc = Mc= (100/ ((200 × 106) × (24 × 10-3)

Yc = 0.0208 m (or) 20.8mm.

VTHT AVADI III SEM


Result:

Yc = 0.0208 m

8) A hollow cylindrical column of 150 mm external diameter and 20 mm thick, 6 m long is

hinged at one end fixed at the other end. Find the ratio of Euler and Rankine‟s critical load. Take E=8×104 N/mm2. Fc=550 N/mm2 and Rankine‟s constant as 1/1600.(ODD 2008).

Given:

External diameter, D=150mm

Wall thickness, t = 20 mm

Internal diameter, d = D -2t

=150 - 40

= 110 mm

Length, L= 6m = 6000mmHOLLOW CYLINDRICAL COLUMN

Young‘s modulus, E = 80 KN/mm2 = 80 × 103 N/mm2

Rankine‘s constant, a =1/1600

Crushing stress , c = 567 N/mm2

To find:

a) Euler‘s formula

b) Rankine‘s formula

Solution:

Case (i):

Crippling load by using Euler‘s formula area of the column,

A =π/4 (D2 – d2)

=π/4 (1502 – 1102)

A = 8168.14 m2

l l d

Φ D

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Moment of inertia,

I =π/64 (D4 – d4)

=π/64 (1504 – 1104)

I = 17.66 × 106mm4

Least radius of gyration,

K = √I/A

= √17.66 × 106/8168.14

K =46.49 mm

Wkt, for both ends are hinged,

Effective length,

L = l

Crippling load,

pE = π2 EI/L2

= π2 × 80 ×103 ×17.66 × 106/ (6000)2

PE = 387.33 × 103 N

Case (ii):

Crippling load by using Rankine‘s formula

Crippling load, PR = c× A/ (1 + a (1/K) 2

= 567 × 8168.14/ (1 + (1/1600) (6000/46.49)2

PR = 405.89 × 103 N

Crippling load by Euler‘s formula/crippling load by Rankine‘s formula = (PE/PR)

= 387. 33 × 103/(405.89 × 103)= 0.954

VTHT AVADI III SEM


Result:

(i) PE = 387 * 103 N

(ii) PR = 405.89 * 103N

(iii) PE/PR = 0.954

9. For the cantilever beam shown. Find the deflection and slope at the free end. EI=10000

KN/m2. (ODD 2009)

Given:

EI=10000 KN/m2

To find:

Slope =?

Deflection =?

Solution:

A1 = (0.5 × 1 × (2/EI)) = (1/EI)

X1 = (2/3) × 1 = (2/3) m

A2 = (1/EI) × 1 = (1/EI)

X2 = 1.5m

A3 = (0.5 × 1 × (2/EI)) = (1/EI)

X3 = 1 + (2/3) × 1 = (5/3)m

Slope at c = A1 + A2 + A3CANTILEVER BEAM

=(1/EI) + (1/EI) + (1/EI)

= (3/EI)

Deflection at c

= A1 X1 + A2 X2 + A3 X3

2 kN 2 kN

2 EI EI

1 m C

B A 1 m

2 kN 2 kN

2 EI EI

1 m C

B A 1 m

M/EI

2/EI 3/EI

1/EI 2

3

1

VTHT AVADI III SEM


= (1/EI) × (2/3) + (1/EI) × 1.5 + (1/EI) × (5/3)=(11.50/3EI)

Slope = (3/10000) = 3 × 10-4 radians

Deflection = (11.5/30000) = 3.83 × 10-4 m

Result:

Slope = (3/10000) = 3 × 10-4 radians

Deflection = (11.5/30000) = 3.83 × 10-4 m

10. The external and internal diameters of a hollow cast iron column are 50 mm and 40 mm

respectively. If the length of this column is 3 m and both of its ends are fixed, determine the

crippling load using Euler formula taking E=100 GPA. Also determine the Rankin load for

the column assuming fc= 550 Mpa and α=1/1600. (EVEN 2009)

Given:

D= 50mm

d= 40mm

l= 3m = 3000mm

E = 100 Gpa = 100 × 103 N/mm2

α = a = 1/1600

To find:

crippling load

(i) Rankine‘s formula

(ii) Euler‘s formula

Solution:

Crippling load when both ends are fixed,

P = 4π2 EI/l2

I = moment of inertia

= (π/64) (D4 – d4)

= (π/64) (504 – 404)

VTHT AVADI III SEM


I = 181.13 × 103mm4

P = (4 × 3.14 × 3.14 ×100 ×103 × 181.13 × 103)/(3000)2

P = 79.45 × 103 N

Wkt, crippling load by using Rankine‘s formula,

P = c A/(1 + a(L/K)2)

A = π/4 (D2 – d2)

= π/4 (502 – 402)

A = 706.858 mm4


K = √I/A

= (181.13 × 103/706.85)1/2

K = 16 mm

P = (550 × 706.85)/(1 + (1/1600) ((3000/2)/16)2 for (L =l/2)

P = 59.87 × 103N

Result:

P = 79.45 × 103 N Euler‘sLoad

P = 59.87 × 103N Rankine‘sLoad

11. A) Write the expressions for Euler‟s critical load of long columns for different end conditions.

1. Column with both ends are hinged

Critical load, P = π2EI/ (L) 2

2. Column with one end is fixed and the other end is free

Critical load, P =π2EI/ (4L) 2

VTHT AVADI III SEM


3. Both ends of the column are fixed

Critical load, p = 4π2EI/ (L) 2

4. Column with one end is fixed and the other end is hinged

Critical load, P = 2π2EI/ (L) 2

B) A hollow cylindrical column of 150 mm external diameter and 15 mm thick, 3 m long is

hinged at one end fixed at the other end. Find the ratio of Euler and Rankine‟s critical load. Take E=8×104 N/mm2. Fc=550 N/mm2 and Rankine‟s constant as 1/1600.(ODD 2008).

Given:

External diameter, D =150mm



=150 - 15×2

= 120 mm

Length, L= 3m = 3000mm

Young‘s modulus,E = 80 KN/mm2 = 80 × 103 N/mm2


Crushing stress, c = 550 N/mm2

To find:

The ratio of Euler‘s&Rankine‘s critical load =?

Solution:

Case (i):


A =π/4 (D2 – d2)

=π/4 (1502 – 1202)

A = 6.36 ×103mm2

Moment of inertia,

VTHT AVADI III SEM


I =π/64 (D4 – d4)

=π/64 (1504 – 1204)

I = 14.67 × 106mm4


K = √I/A

= √14.67 × 106/6.36 × 103

K =48.02 mm


Effective length,

L = l/√2

= 3000/ (√2)

L = 2121.32mm

Crippling load, Euler formula,

pE = 2π2 EI/L2

=2× π2 × 80 ×103 ×14.67 × 106/ (3000)2

PE = 2.57× 106 N

Case (ii):


Crippling load,

PR = c× A/ (1 + a (L/K) 2

= 550 × 6.36 × 103/ (1 + (1/1600) (2121.332/48.02)2

PR = 1.57 × 106N

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Crippling load by Euler‘s formula/crippling load by Rankine‘s formula = (PE/PR)

= 2.57× 106/ (1.57 × 106)

= 1.63

Result:

Crippling load by Euler‘s formula/crippling load by Rankine‘s formula = (PE/PR)= 1.63

12. Find Euler‟s crippling load for a hollow cast iron column of 20 mm external diameter, 25 m thick and 6 m long hinged at both ends. Compare the load with crushing load

calculated from Rankine‟s formula. Fc=550 N/mm2. Rankine‟s constant=1/1600, E=1.2×105

N/mm2. (EVEN 2009).

Given:




=200 – (25+25)= 150 mm

Length,= 6m = 6000mm

Young‘s modulus, E = 1.2 × 105 N/mm2



To find:

Crippling load

a) Euler‘s formula (PE)

b) Rankine‘s formula (PR)

Solution:

Case (i):


A =π/4 (D2 – d2)

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=π/4 (2002 – 1502)

A = 13.74 ×103mm2

Moment of inertia,

I =π/64 (D4 – d4)

=π/64 (2004 – 1504)

I = 53.68 × 106mm4


K = √I/A

= √53.68 × 106/13.74 × 103

K =62.51 mm



PE = π2 EI/L2

= π2 × 1.2×105 ×53.68 × 106/(6000)2

PE = 1.76× 106 N

Case (ii):


Crippling load,

PR = c× A/ (1 + a (L/K) 2

= 550 × 13.74 × 103/(1 + (1/1600)(6000/62.51)2

PR = 1.118 × 106 N

Result:

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PE = 1.76× 106 N

PR = 1.118 × 106 N

13. A 1.2m long column has a circular cross- section of 45 mm diameter. One of the ends of

the column is fixed in direction and position and the other end is free. Taking factor of

safety as 3, calculate the safe load using. (MAY/ JUNE 2012).

a) Rankin‟s formula, take yield stress=560 N/mm2 and Rankin‟s constant a=1/1600.

b) Euler‟s formula, elastic modulus= 1.2× 105N/mm2.

Given:


Length, l= 1.2m = 1200mm

Young‘s modulus, E = 1.2 × 105 N/mm2



FOS = 3

To find:

Calculate the safe load using:

a) Euler‘s formula

b) Rankine‘s formula

Solution:

Case (i):


A =π/4 (D2)

Euler‘s formula young‘s modulus for cast iron 1.2 x 105 N/mm2

=π/4 (452)

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A = 1590.43mm2

Moment of inertia,

I =π/64 (D4)

=π/64 (454)

I = 20.12 × 104mm4


K = √I/A

= √20.12 × 104/1590.43

K =11.25 mm

Wkt, for one end fixed and other end is free,

Effective length,

L = 2l

= 2×1200

L = 2400mm


pE = π2 EI/4L2

= π2 × 1.2 ×105 ×20.12 × 104/4 × (1200)2

PE = 0.41× 105 N

Safe load= crippling load/ factor load

= 0.41 × 105/3

Safe load = 13.7 × 103 N

Case (i):

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Crippling load,

PR = c× A/ (1 + a (L/K) 2

= 560 × 1590.43/(1 + (1/1600)(2400/11.25)2

PR = 0.30 × 1066N

Wkt,

Safe load = crippling load/ factor of safety

Safe load = (0.30 × 105)/5

Safe load = 10.06× 103 N

(ii) case 2 :

Crippling load by using Euler‘s formula

Crippling load,

P = π2 EI/4l2

= π2 x 1.2x105 x 20.12 x104 / 4 x (1200)2

Safe Load = crippling load / factor of safety

= 0.41 x 105 /3

Safe Load = 13.7 x 103 N

Result:

Safe load by using Rankine‘s formula: = 10.06× 103 N

Safe load load by using Euler‘s formula = 13.7 x 103 N

14. An I section joint 400mm×200mm×20mm and 6m long is used as a strut with both ends

fixed. What is Euler‟s crippling load for the column? Take E=200 GPA. (ODD 2007)

VTHT AVADI III SEM


Given:

Outer depth, d2 = 400mm

Thickness = 20mm

Inner width, d1 = 400-20-20

= 360mm

Outer width, b2 = 200mm

Inner width, b1 = 200 – 20

= 180mm

Length, L =6m = 6000mm

Young‘s modulus, E= 200Gpa = 2 x105 N/mm2

To find:

Crippling load=?I SECTION

Solution:

Moment of inertia about x-x axis

Ixx = (1/12) [b2d23 - b1 d1

3]

= (1/12) [200 ×(400)3 – 180 × (360)3]

Ixx =0.366 × 109 mm4

Moment of inertia about y-y axis

Iyy = (1/12) × 360 × 203 + [(200)3 × 20 × 2]

Iyy = 320 × 106mm4

Here ,Iyy<Ixx

So, least value of the moment of inertia

200 mm

20 mm

400

mm

x x

y

y

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Iyy = I = 320 × 106mm4

I = 320 × 106mm4

Wkt,

Critical load,

P = 4π2 EI/ (L)2 (both ends are fixed)

= 4 × π2 × 2 ×105 × 320 × 106/(6000)2

P= 70.18 × 106N

P = 70.18 MN

Result:

Critical load,P = 70.18 MN

15.A steel cantilever 6 m long carries two point loads, 15KN at the free end and 25 KN at a

distance of 2.5m from the free end.

Find:

(i) Slope at the free end

(ii) Deflection at the free end.

Take I = 1.3 × 103 mm4 and E = 2 × 105 N/mm2(EVEN 2011)

Given data:

W1= 15 KN

W2 = 25 KN

a= 2.5m

To find:

(i) Slope at the free end c.(Øc)

(ii) Deflection at the free end c. (yc)

Solution:

VTHT AVADI III SEM


Slope at the free end due to the load 15KN alone = W1L2/ (2EI)

Slope at the free end due to the load 25KN alone = W2 (L – a) 2/ (2EI)

Total slope at the free end,

Øc = W1L2/(2EI) + W2 (L – a)2/(2EI)

= (15000 × (6000)2) + 25000 × (6000 – 2500)2/(2 ×2 ×105 ×1.3 × 108)

Øc = 0.01627 rad.

Double integration method:

Deflection at the free end due to the load 15KN alone = W1L3/(3EI)

Deflection at the free end due to the load 25KN alone

= W1L3/ (3EI) + W2(L – a) 3/ (3EI) + W2 (L – a) 2/ (2EI) a

= [(15000 × 60003/3×2×105×1.3×108) + (25000 × (6000 – 2500)3/ (3×2×105 ×1.3×108) × 2500]

= 41.54 + 13.74 + 14.723

Yc = 70.003 mm or 70mm

Result:

Øc = 0.01627 rad.

Yc = 70.003 mm or 70mm

16. A cantilever of length 4m carries a UDL of 8KN/m length over the centre length. If the

section is rectangular of 150mm × 260mm. find the deflection and slope at the free end.

Take E = 2.1 × 105 N/mm2.(EVEN 2011)

Given data:

Length, L = 4m = 4000mm

UDL,

W = 8 KN/m

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Cross section = 150mm × 260mm

E = 2.1 × 105 N/mm2

To find:

(i) Slope at the free end

(ii) Deflection at the free end

Solution:

For the given cross section,

Moment of inertia,

I =bd3/12 = 150 × 2603/12

I = 2.197 × 108 mm4

Double integration method:

Maximum slope will be at free end of the beam.


ØB = WL3/ (6EI)

= 8 × 40003/ (6 × 2.1 × 105× 2.197 ×108)

= 0.00185 rad.

Deflection at the free end will be the maximum deflection,

Deflection at the free end,

YB = WL4/ (8EL)

= 8 × 40004/ (8 × 2.1×105 ×2.197 × 108)

= 5.548mm

Macaulay‟s method:

For the cantilever beam with UDL,

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Bending moment, Mx = -Wx2/2

Integrating the above equation,

Slope equation,

EI dy/dx = - Wx3/6 + c1

Integrating again, we get,

Deflection equation,

EI y = -WX2/24 + c1X + c2

Applying the following be for cantilever

(i) When x= 4m, slope dy/dx= 0

(ii) When x= 4m, deflection y= 0.

Applying the first B.C to the slope equation,

0= - 8 × 40002/6 + c1

C1 = 8.53×1010

Applying, the second B.C to the deflection equation,

0 = -8 × 40004/24 + 8.53×1010× 4000 +c2

C2 = -2.56×1014

Sub, the value of x, c1 and c2 in (1) we get slope and deflection.

At the free end, x= 0

Slope ØB= dy/dx = c1/EI = (8.53 × 1010)/ (2.1×105×2.197×108)

ØB = 0.00185 rad.

Deflection, y = c2 / (EI) = (- 2.56×1014)/ (2.1×105 × 2.197×108)

Y= -5.548mm

Moment area method:

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Draw the BMD for the given beam BM at B= 0

BM at A= WL2/2 = (8×40002)/2

MA = 6.4 × 107 N/mm

Area of BMD = (1/3) ×L× (WL2/2)

= (1/3) × 4000× (8×40002)/2

=8.53×1010

According to moment area method,


ØB = area of BMD EI

= 8.53×1010/ (2.1×105 × 2.197×108)

ØB = 0.00185 rad.

X = (3/4) × L = (3/4) × 4000 = 3000mm

Deflection, yc = (AX/EI) = (8.53×1010 × 3000)/ (2.1× 105×2.197×108)

Yc = 5.548mm

Conjugate beam method:

Draw the BMD and then draw the conjugate beam,

i.e, M/(EI) diagram as shown in fig.

Total load on the conjugate beam = area of the (M/EI) diagram

= (1/3) × 4000 × (-1387×10-6)

P = -0.00185

Slope of B = shear force at B for conjugate beam

ØB = -P = 0.00185rad

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Deflection at B,

YB = BM at B for conjugate beam

= -P × (3/4) ×4000 = 0.00185 × (3/4) × 4000

YB = 5.548mm

Result:

ØB = 0.00185rad

YB = 5.548mm

17. A cantilever beam 50mm wide and 80 mm deep is 2m long. It carries a UDL over the

entire length along with a point load of 5KN at its free end. Find the slope at the free end

when the deflection is 7.5mm at the free end when the deflection is 7.5mm at the free end.

Take E = 2× 105 N/mm2(EVEN 2012)

Given data:

Cross section = 50mm × 80mm

Length, L= 2m

Point load, W = 5KN = 5000N

Deflection, yB = 7.5mm

To find:

Slope at the free end.

Solution:

Moment of inertia,

I = bd3/12

I = 50 × 803/12

= 2.13 × 106mm4

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Deflection at the free end due to UDL

= WL4/ (8EI)

Deflection at the free end due to point load,

= WL3/ (3EI)

Total deflection yB = WL4/ (8EI) + WL3/ (3EI)

7.5 = WL4/ (8EI) + (WL3/3EI)

7.5 = WL4/ (8EI) + (5000 × 20003)/ (3× 2×105×2.13×106)

WL4/ (8EI) = -190.39 or WL3/ (EI) = -0.0952

Slope at the free end due to UDL, = WL3/ (6EI)

Total slope, ØB = WL3/ (6EI) + WL2/ (2EI)

= (1/6) × (-0.0952) + (5000× 20002)/ (2×2×105×2.13×106)

= 0.0076 rad.

Result:

Total slope, ØB= 0.0076 rad.

18. A horizontal beam is freely supported at its ends 8 mm apart and carries a UDL of 15

KN/m over the entire span. Find the maximum deflection. Take E = 2 × 105N/mm2and I = 2

× 109mm4.(EVEN 2009).

Given data:

L = 8m = 8000mm

W = 15KN/mm = 15 N/mm

E= 2×105 N/mm2SIMPLY SUPPORTED BEAM

I = 2× 109 N/mm4

To find: Maximum deflection, ymax =?

VTHT AVADI III SEM


Solution:

Macaulay‘s method BM equation,

EI d2y/dx2 = 60X – 15(X2/2)

Integrating the above equation twice,

Slope equation, EI dy/dx = 60(x2/2) – 15(x3/6) +c1

=30 x2 – 15(x3/6) + c1.


EI.y = 30(x3/3) – 15(x4/24) +c1x +c2

= 10x3 – 15(x4/24) + c1x + c2

Applying the following boundary conditions

(i) When x= 0, y=0

(ii) When x= 800mm ,y= 0

Applying first B.C to the deflection equation,

0= 10(0)3 – 15(0)4/24 + c1 (0) +c2

C2 = 0

For the second,

0 = 10 (8000)3 – 15(8000)4/24 + c1 (8000)

C1 = 3.1936 × 1011

Maximum deflection occurs at x= 4000mm

Yc = (1/EI) [10x3 – 15(x4/24) +c1x ]

= (1/2×105×2×109) [10(4000)3 – 15(4000)4/24 + 3.1936 ×1011 ×4000]

Ymax= 2.79mm

Result:

VTHT AVADI III SEM


Maximum deflection, ymax = 2.79mm

19. A SSB of hollow circular section of inter diameter of 220 mm and internal diameter of

150mm has a span of 6m. It is subjected to central point load of 50 KN and a UDL of

5KN/m run. Determine the maximum deflection E= 200 KN/mm2

Given data:

D = 200mm

d= 150mm

L= 6m = 6000mm

E= 200 KN/mm2 = 200×103 N/mm2

To find:

Maximum deflection, ymax=?

Solution:

Moment of inertia I = (π/64) (D4 - d4)

= (π/64) (2004 – 1504)

= 53689327.58 mm4

Total deflection, ymax= deflection to point load + deflection due to UDL

= WL3/48EI + 5WL4/384EI

= (1/200×103×53689327.58) [50000 × 60003/48 + 5×5×60004/384]

Ymax= 28.81mm

Result:

Maximum deflection Ymax= 28.81mm

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20. A steel pipe 50mm internal diameter and wall thickness 2mm is simply supported on a

span of 6mm. if the deflection is limited to 1/480 of the span. Calculate the maximum UDL

that can carry. E =2×105 N/mm2.

Given data:

d= 50mm

t= 2mm

Ymax =1/480 ×L

L= 6m = 6000mm

E= 2×105 N/mm2

To find:

Maximum UDL =?

Solution:

External diameter

D= d +2r

= 50 + (2×2) = 54mm

Moment of inertia

I = (π/64) (D4 – d4)

= (π/64) (544 – 504)

= 110596.63 mm4

Deflection ymax= (1/480) × 6000

= 12.5mm

But, ymax = (5WL4/384EI)

Ymax= 5 × W × (6000)4/ (384 × 2×105 × 110596.63)

VTHT AVADI III SEM


12.5 = 5 × W × (6000)4/ (384×2×105×110596.63)

W =16 .38 N/m

Result:

Maximum UDL W =16 .38 N/m

21. A SSB of span 6m carries UDL 5 KN/m over a length of 3m extending from left end.

Calculate deflection at midspan. E = 2× 105 N/mm, I = 6.2 × 106 mm4.

Given data:

L= 6m = 6000mm

W= 5 KN/m = 5N/mm

E = 2× 105 N/mm2

I = 6.2 × 106 mm4

To find:

Deflection at mid span. Yc=?SIMPLY SUPPORTED BEAM

Solution:


RB × 6000 = 5× 3000× (3000/2)

RB = 3750N.

RA + RB = 5×3000

RA = 11250N

BM equation,

EI d2y/dx2 = 3750x – 5(x-3000)2/2


6 m

A B C

RA = 11250 RB = 3750 N

5

VTHT AVADI III SEM


EI dy/dx = 3750x2 /2 + c1 -5(x- 3000)3/6

EI y = 3750 x3/6 +c1 x + c2 – 5(x- 3000)4/24

Applying the following boundary conditions

(i) When x= 0, y= 0

(ii) When x= 6000m, y= 0

Applying first B.C

0 = 3750(0)3/6 + c1 (0) + c2 -5(0-3000)4/24

C2 = 1.6875 × 10 13

For second B.C

0 = [3750 × (6000)3/6 + c1(6000) + 1.6875 × 1013 – (5(6000) – 3000)4/24 ]

C1 = -2.25 × 1010

Deflection at mid span x= 3000 mm

Ymax = (1/2×105×6.2×106) [3750(3000)3/6

– 2.25×1010×3000+1.6875×1013– 5(3000- 3000)4/24

Ymax= -27.22mm

Result:

Ymax= -27.22mm

22.A 3m long cantilever of uniform rectangular cross section 150mm wide and 300mm deep

is loaded with a point load with a point load of 3KN at the free end and a UDL of 2KN/m

over the entire length. Find the maximum deflection E= 210 KN/mm2. Use Macaulay‟s method.

Given data:

W = 3KN

W= 2KN/m =2/1000 KN/m

3 kN 2kN/m

A

C

x

x

VTHT AVADI III SEM


=2N/mm

L=3m = 3000mm=xCANTILEVER BEAM

E = 210 KN/mm2 = 210 × 103 N/mm2

b= 150mm

d= 300mm

To find:

Maximum deflectionYmax=?

Solution:

Moment of inertia for rectangular cross section

I = bd3/12 = 15 × 3003/12

I = 3375 × 105 mm4

Bending moment at xx,

Mx = -Wx –wx2/2


Slope equation,

EI dy/dx = -Wx2/2 – wx4/6 + c1


EI y = -Wx3/6 -wx4/24 + c1x + c2

At x= 3m, y and dy/dx= 0

Applying boundary conditions, from (2)

0= -3000(3000)2/2 - 2(3000)3/6 + c1

C1 = 2.25 × 1011 N/mm2

VTHT AVADI III SEM


0= -3000(3000)3/6 – 2(3000)4/24 + 2.25 × 1011× 3000 +c2

C2 = -6.5475 × 1014 N/mm2

The deflection equation becomes,

EI y = -Wx3/6 – wx4/24 + 2.25 × 1011x – 6.5475 × 1014

But maximum deflection occurs at the free end, when x= 0,

Ymax= (1/EI) [- 6.5475 × 1014]

= (1/210 × 103× 3375 × 1010) [-6.5475 × 1014]

Ymax= 9.24 mm

Result:

Maximum deflectionYmax=9.24 mm

23. Derive the Euler‟s formula for a long column with one end fixed and the other is free. Write the assumptions also. (EVEN-2013)

Assumptions made in the Euler‘s column theory.

1. The cross section of the column is uniform thought its length.

2. The material of the column is perfectly elastic, homogenous and obeys hooks‘ law.

3. The lengths of the column is very long as compared to its cross sectional dimensions.

4. The column is initially straight and the compressive load is applied axially.

5. The failure of column occurs due to buckling alone.

6. The self weight of column is negligible.

VTHT AVADI III SEM


Consider a column RS of length l, and uniform cross sectional area A, carrying a critical load P

and R.

The column is fixed at one end S and the other end is free.

Due to application of critical load P, the column will deflect as shown in fig.

SR is the original position of the column and SR‘ is the deflected positions due to critical load .

Consider any section at a distance at a distance x from the fixed end S.

Let y is a deflection at this section and q is a deflection at free end R.

The moment due to crippling load at is section is given by

Moment = Load * Distance

Moment = P * (q-y)

VTHT AVADI III SEM


2

2

2

2

2

2

2

2

2

2

d y Moment EI

dx

d yP q -y) ΕΙ

dx

d yEI Py pq

dx

DividingbyEI,

d y Py pq

dx EI EI

d y P pq y

dx EI EI

Thesolutionoftheabovedifferentialequationis

P Py A cos(x ) Bsin(x ) q 1

EI EI

Case(i):

At end S, x = 0, y=0 and slope dy/dx=0 substituting these values in equ 1

0= A cos 0°+Bsin 0°+q

0=A+q

A=-q

At S, x=0 and dy/dx=0

So, differentiating equation 1 with respect to x,

VTHT AVADI III SEM


P P1 y A cos(x ) Bsin(x ) q

EI EI

dy P P P PA( 1)sin(x ) Bcos(x ) 0

dx EI EI EI EI

P P0 Asin(0 Bcos(0 )

EI EI

P0 0 B

EI

P 0

EI

B 0

Substituting A=-q and B=0 values in equation (1)

VTHT AVADI III SEM


Py q cos(x ) q 2

EI

Py q cos(x ) q

EI

case(ii) :

At R '(freeend), x l and y q

subthesevaluesin equ. 2

Pq q cos(l ) q

EI

Pq q a cos(l )

EI

P0 q cos(l )

EI

Pq cos(l ) 0

EI

Pq 0or cos(l ) 0

EI

Butq 0

Pcos(l ) 0

EI

Pcos(l ) cos o

EI 2

3 5r cos or cos

2 2

P 3 5l or or

EI 2 2 2

VTHT AVADI III SEM


2

2

2

2

Now taking least significant value,

Pl

EI 2

P

EI 2l

P

EI 4l

EI P

4l

24. A horizontal beam of length „l‟ and flexural rigidity EI carries a point load W at its midspan.The beam is rigidly fixed at is left end and partially fixed at its right end in

such a way that the fixing moment at the rigidly fixed left end is Wl/b. If the supports

are at the same level, determine the fixing moment and slope at the right end.(EVEN-

2013)

VTHT AVADI III SEM


B C C

2 2

B

2

C

2

C B

3

C

2 2

B

3 2

3

Deflection at freeend,

y y (a)

Deflection at freeend,

W(L a) W(L a) y a

3EI 2EI

where,

W(L a)

2EI

W(L a)

2EI

W(L a)y

3EI

L When a

2

W(L a) W(L a) y a

3EI 2EI

W L W L L(L ) (L )

3EI 2 2EI 2 2

W L W ( ) (

3EI 2 2EI

2

3 3

3

B

L)

2

WL WL

24EI 16EI

5WL y

48EI

VTHT AVADI III SEM


UNIT-5

PART-A

1. What are assumptions involved in the analysis of thin cylindrical shells? (EVEN 2011)

1. The material of the cylinder is Homogeneous, isotropic and obeys Hook‘s law. 2. The hoop stress distribution in thin cylinder is uniform over the cross section

from inner to outer surface since the thickness of the cylinder is thin.

3. Weight of fluid and material of the cylinder is not taken into account.

2. Define principle planes. (EVEN 2011, 2010)

The planes which have no shear stress are known as principle planes.

3. List out the modes of failure in thin cylindrical shell due to an internal pressure .

(EVEN 2010)

When these stresses exceed the permissible limit, the cylinder is likely to fail in the

followings two ways

(i) It may split up into two semi circular halves along the cylinder axis.

(ii) It may split up into two cylinders

4. What is Mohr‟s circle method? (EVEN 2009)

Another method which is frequently used to find out the normal, tangential and resultant

stresses in oblique plane is Mohr‘s circle method. It is also a graphical method.

5. What is principle stress? (EVEN 2009)

The magnitude of normal stress, acting on a principle planes are known as principle

stresses.

6. Define principle planes and principal stresses. (ODD 2008, 2007, 2006, 2011, 2013)

The planes which have no shear stress are known as principle planes.

VTHT AVADI III SEM


The magnitude of normal stress, acting on a principle planes are known as principle

stresses.

7. A storage tank of internal diameter 280 mm is subjected to an internal pressure of 2.5

Mpa. Find the thickness of the tank, if the loop and longitudinal stresses are 75 Mpa and

45 Mpa respectively. (ODD 2008)

Given: Hoop stress, c = 75 MPa

Longitudinal stress, d = 45 Mpa

Diameter, d = 280 mm

Pressure, P = 2.5 MPa

Solution: Hoop stress is greater than longitudinal stress. So we can use

P d

c = 2 t

t = p d / 2 c = 2.5 x 280

2 x 75

8. A spherical shell of 1 m internal diameter undergoes a diametrical strain of 10-4 due

to internal pressure. What is the corresponding change in its internal volume? (EVEN

2008)

Change in volume, dV = ev x V

= 3 x e x V

= 3 x 1 x 10-4 x π / 6 (1000)3

t = 4.66 mm

dV = 157.079 mm3

VTHT AVADI III SEM


9. A Thin cylinder closed at both ends is subjected to an internal pressure of 2 MPa. It is

internal diameter is 1 m and the wall thickness is 10 mm. What is the maximum shear

stress in the cylinder material? (EVEN 2008)

Given: P = 2 MPa = 2 x 106 Pa = 2 x 106 N/m2 = 2 N/mm2 ; d = 1 m = 1000 mm ; t = 10 mm

P d

Solution: Hoop stress, c =

2 t

= 2 x 1000

2 x 10

Longitudinal stress, d = P d

4 t

= 2 x 1000

4 x 10

max – min 100 - 50

Maximum shear stress, max = =

2 2

c = 100 N/mm2

d = 50 N/mm2

max= 25 N/mm2

VTHT AVADI III SEM


10. A boiler of 800 mm diameter is made up of 10 mm thick plates. If the boiler is

subjected to an internal pressure of 2.5 MPa, determine circumferential and longitudinal

stress. (ODD 2007)

Given : d = 800 mm; t = 10 mm; p = 2.5 MPa = 2.5 N/mm2

P d

Solution: Circumferential stress, c = = 2.5 x 800

2 t 2 x 10

11. 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 the longitudinal stress developed in the pipe.

(EVEN 2007)

Given: diameter, d = 1.5 m = 1500 mm; thickness t = 1.5 cm = 15 mm; p = 1.2 N/mm2

Solution: Longitudinal stress, d = P d

4 t

= 1.2 x 1500

4 x 15

c = 100 N/mm2

d = 30 N/mm2

VTHT AVADI III SEM


12. How will you find major principal stress and minor principal stress? Also mention how

to locate the direction of principal planes. (EVEN 2007)

(i)Position of principal planes,

2q

Tan 2θ = 1 – 2

(ii) Major Principal stress

1 + 2 1

n1 = + ( 1 – 2) 2 + 4 q2

2 2

(iii) Minor principal stress

1 + 2 1

n2 = - ( 1 – 2) 2 + 4 q2

2 2

13. Find the thickness of the pipe due to an internal pressure of 10 N/mm2. If the

permissible stress is 120 N/mm2 . The diameter of pipe is 750 mm. (ODD 2006)

Given: Pressure P = 10 N/mm2 ; Stress, = 120 N/mm2 ; Diameter, d = 750 mm

P d

Solution: Circumferential stress, c =

VTHT AVADI III SEM


2 t

10 x 750

120 =

2 x t

14. The principal stress at a point is 100 N/mm2 (tensile) and 50 N/mm2 (compressive)

Respectively. Calculate the maximum shear stress at this point.

Solution: 1 – 2 150 – (-50)

Maximum shear stress, ( t ) max = =

2 2

15. A spherical shell of 1 m diameter is subjected to an internal pressure 0.5 N/mm2.

Find the thickness if the allowable stress in the material of the shell is 75 N/mm2 . (EVEN

2006)

Given: Pressure, P = 0.5 N/mm2; Diameter, d = 1 m =1000 mm; Max. Stress, = 75

N/mm2

P d

Solution: Maximum stress, d =

4 t

t =31.25 mm

( t )max = 75 N/mm2

VTHT AVADI III SEM


0.5 x 1000

75 =

4 x t

16. In a thin cylinder will the radial stress vary over the thickness of wall? (EVEN 95)

No, In thin cylinder radial stress developed in its wall is assumed to be constant since the

wall thickness is very small as compared to the diameter of cylinder.

17. What do you understand by the term wire winding of thin cylinder? (ODD 99)

In order to increase the tensile strength of a thin cylinder to withstand high internal;

pressure without excessive increase in wall thickness, they are sometimes pre-stressed by

winding with a steel wire under tension.

18. What is the effect of riveting a thin cylindrical shell? (EVEN 99)

Riveting reduces the area offering the resistance. Due to this the circumferential and

longitudinal stresses are more. It reduces the pressure carrying capacity of the shell.

19. Differentiate between thin cylinder and thick cylinder. (EVEN 98, 2000)

Thick Cylinder Thin Cylinder

1.Ratio of wall thickness to the diameter

of cylinder is less than 1/20

1. Ratio of wall thickness to the diameter

of cylinder is more than 1/20.

t =1.67 mm

VTHT AVADI III SEM


2. Circumferential stress is assumed to

be constant throughout the wall

thickness.

2. Circumferential stress various from

inner to outer wall thickness.

20. Distinguish between cylindrical shell and spherical shell. (EVEN ‟99)

Spherical shell Cylindrical shell

1. Circumferential stress is twice the

longitudinal stress.

1. Only hoop stress presents.

2. It withstands low pressure than

spherical shell for the same diameter

2. It withstands more pressure than

cylindrical shell for the same diameter.

21. In a thin cylindrical shell if hoop strain is 0.2 x 10-3 and longitudinal strain is 0.05 x 10-3,

find out volumetric strain. (ODD 99)

Volumetric strain, ev = 2 ec + ea

= 2 (0.2 x 10-3) + 0.05 x 10-3

ev = 0.45 x 10-3

22. Write the equation for the change in diameter and length of a thin cylinder shell, when

subjected to an internal pressure. (ODD 99)

P d2 1

Change in diameter, ∂d = (1 - )

2 t E 2 m

P d l

Change in length, ∂l = (1/2 – 1/m)

VTHT AVADI III SEM


2 t E

23. Write down the volumetric strain in a thin spherical shell subjected to internal pressure

„P‟ (EVEN 96, 97)

3 p d

Volumetric stain, ev = 3 x ec = (1 – 1/m)

4 t E

24. Write the expression for hoop stress and longitudinal stress in thin cylinder due to

Pressure p (ODD‟96, EVEN‟98)

, c = 𝑷𝒅/2𝒕 d =𝑷𝒅/𝟒𝒕

25. Define Thin Cylinder or Sphere.(EVEN -2012)

A Cylinder whose thickness is 20 times less than its diameter is known as thin cylinder.

26. What are the different types of stresses in cylinder?

1. Circumferential stress of Hoop stress.

2. Longitudinal Stress

27. Define Hoop Stress. (EVEN 2012, 2013)

Hoop stress is the stress induced by fluid or gas inside the cylinder perpendicular to the

length of the pipe. The thickness of the cylinder is decided based on hoop stress value because

hoop stress is two times more than the longitudinal stress.

VTHT AVADI III SEM


28. Define longitudinal stress.

Longitudinal stress is the stress induced by the fluid or gas along the length of the

Cylinder.

29. Define Oblique.

It is the angle between the resultant stress and normal stress.

30. What are the planes along which the greatest shear stresses occur? (EVEN 2001)

Greatest shear stress occurs at the planes which inclined at 45º to its normal.

31. A bar of cross sectional area 6000 mm2 is subjected to a tensile load of 50 KN applied

at each end. Determine the normal stress on a plane inclined at 30º to the direction of

loading.

Given: Area, A = 600 mm2; Load, P = 50 KN; θ = 30º;

Load 50 x 103

Solution: Stress, = = = 83.33 N/mm2

Area 600

Normal stress, n = cos2 θ = 83.33 x cos2 30º = 62.5 N/mm2

VTHT AVADI III SEM


32. Write the expressions for a normal stress on an inclined plane in a block which is

subjected to two mutually perpendicular normal stresses and shear stresses? (EVEN 2000)

1+ 2 1- 2

Normal stress, n = + ( ) cos 2θ + q sin 2θ

2 2

33. At a point in a strained material is subjected to a compressive stress of 100 N/mm2 and

shear stress of 60 N/mm2. Determine graphically or otherwise the principal stresses. (ODD

2001)

Given: = -100 N/mm2 ; q = 60 N/mm2

Major Principal Stress

1

n1 = + 2 + 4 q2

2 2

-100 1

n1 = + (-100) 2 + 4 x 602

2 2

n1 = 28.1 N/mm2

Minor principal stress

1

VTHT AVADI III SEM


n2= - 2 + 4 q2

2 2

-100 1

n2 = - (-100) 2 + 4 x 602

2 2

n2 = -128.1 N/mm2

34. What is the radius of Mohr‟s circle? (ODD ‟98, EVEN ‟96)

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

35. Give the expression for stresses on an inclined plane when it is subjected to a axial pull.

(ODD‟99)

Normal stress, n = cos2 θ

Shear stress, t = /2 sin2θ

Resultant stress, res = n 2 + t

2

36. Write the maximum value of shear stress in thin cylinder?

Where

P-internal fluid pressure

d-diameter

VTHT AVADI III SEM


t-thickness of the cylinder.

37. For what purpose are the cylindrical and spherical shells used?

The cylindrical and spherical shells are used generally as containers for storage of liquids

and gases under pressure

38. When is the longitudinal stress in a thin cylinder is zero?

In case of cylinders with open ends, e.g. in a pipe with water flowing through it under

pressure, longitudinal stress is zero

39. What are assumptions made in the analysis of thin cylinders?

Radial stress is negligible.

Hoop stress is constant along the thickness of the shell.

Material obeys Hooke‘s law.

Material is homogeneous and isotropic

40. What is the operating pressure in a thin cylinder and thick cylinder?

For thin cylinder the operating pressure is up to 30MN/m2

For thick cylinder the operating pressure is up to 250MN/m2 or more.

41. Give two methods to compute principal stresses?

1. Analytical method 2. Graphical method

VTHT AVADI III SEM


16 MARKS QUESTION

PART-B

1.A hollow cylindrical drum 750 mm in diameter and 2.5 m long has a shell thickness of 10

mm. if the drum is subjected to an internal pressure of 2.6 N/mm2.Determine 1. Change in

diameter 2. Change in length 3. Change in volume. Take E= 2.1x105 N/mm2 and poisson‟s ratio = 0.3 [EVEN-2011]

Solution:

mm d

m tE

pd ddiameter inChange

wkt

.296 0

.3 0 2

1 1

1 10 . 2102

6 750 . 2

2

1 1

2

5

2

2

mm

m tE

pdl llength inChangth

.232 0

.3 0 2

1

10.1 2 102

2500 750.6 2

1

2

1

2

5

:

750

10

2.5 2500

1' 0.3

Diameterof cylinder d mm

Thickness of cylinder t mm

Length of cylinder l m mm

poison s ratio m

Given datá

VTHT AVADI III SEM


3

5

9

392

2

.9 974296

.3 0 22

5

10 1. 2102

104 10 . 16 750 . 2

10 104. 12500 7504

4

2

2

5

2

mmv

v volumein Change

mm

ld vVolume but

m tE

pdv vvolume inChange

Result:

Change in diameter δd = 0.296mm

Change in length δl = 0.232 mm

Change in volume δv =974296.9 mm3

2. A closed cylindrical drum 600 mm in diameter and 2 m long has a shell thickness of 12

mm. If it carries a fluid under a pressure of 3 N/mm 2, calculate the longitudinal and hoop

stress in the drum wall and also determine the change in diameter, change in length and

change in volume of the drum. Take E = 200Gpa and 1/m = 0.3 [ODD-2008]

VTHT AVADI III SEM


.3 01

'

/10200200mod'

/ 3Pr

2000 2

12

600

:

23

2

m ratios poison

mm NGPa Eulus sYoung

mm Np fluidof essure

mm ml drumthe ofLength

mmz tdrum ofThickness

mm ddrum Diameterof

datá Given

To find:

Longitudinal and hoop stress, change in diameter, change in length and change in volume

Solution:

2

2

/ 75122

600 3

2

5 / . 3712 4

600 3

4

mm Nt

Pd stressHoop

mm Nt

Pd stressal Longitudin

c

l

mm d

m tE


wkt

.19 0

.3 0 2

1 1

102122

600 3

2

1 1

2

5

2

2

VTHT AVADI III SEM


mm

m tE


.15 0

.3 0 2

1

102 122

2000 6003

1

2

1

2

5

3

5

8

382

2

.25 402909

.3 0 22

5

102122

655 10 . 5600 3

10 655. 52000 6004

4

2

2

5

2

mmv

v volumein Change

mm

ldv Volumebut

m tE


Result:

Longitudinal stress l = 37.5 N/mm2

Hoop stress c = 75 N/mm2



Change in volume δv =402909.25 mm3

3. A cylinder has an internal diameter of 230 mm, wall thickness 5 mm and is 1 m long. It is

found to change in internal volume by 12x10-6 m3 when filled with a liquid at a pressure P.

Taking E = 200GPa and poisson‟s ratio = 0.25. Determine the stresses in the cylinder, thechange in its length and internal diameter. [EVEN-2008]

VTHT AVADI III SEM


.25 01

'

/10200200mod'

10 1210 1012 1012

1000 1

5

230

:

23

33963 6

m ratios poison

mm NGPa Eulus sYoung

mmmv volumeInternal

mm mldrum ofLength

mm tdrum ofThickness

mm ddrum Diameterof

datá Given

To find:

Longitudinal and hoop stress, change in diameter and change in length.

Solution:

mmd

m tE


wkt

.0289 0

.25 0 2

1 1

1025 2

25 230 . 1

2

1 1

2

5

2

2

mm

m tE


.0359 0

.25 0 2

1

102 52

1000230.25 1

1

2

1

2

5

Result:

2

2

75 / . 285 2

25 230 . 1

2

375 / . 145 4

25 230 . 1

4

mm Nt

Pd stressHoop

mm Nt


c

l

VTHT AVADI III SEM


Longitudinal stress l = 14.375 N/mm2

Hoop stress c = 28.75 N/mm2



4. A cylindrical shell of 1 m long 150 mm internal diameter having thickness of as 7 mm is

filled with fluid at atmosphere pressure. If additional 25 cc of fluid is pumped into the

cylinder, find the pressure exerted by the fluid on the cylinder shell and the resulting hoop

stress. Assume E = 2x105 N/mm2 and poisson‟s ratio=0.25 (EVEN-12)

.27 01

'

/102mod '

102525

1000 1

7

150

:

25

33

m ratios poison

mm NE uluss Young

mmccv fluidadditional ofvolume

mm mldrum ofLength

mm tshell ofThickness

mm dshell ofDiameter

datá Given

Solution:

2

5

73

372

2

/.47 13

.27 0 22

5

1027 2

10 767. 1150 1025

767 10 . 11000 1504

4

2

2

5

2

mm Nppressure Internal

p

mm

ld vVolume but

m tE


236 / . 1447 2

47 150 . 13

2 mm N

t

Pd stressHoop c

VTHT AVADI III SEM


Result:

Pressure on the cylinder shell p= 13.47 N/mm2

Hoop stress c = 144.36 N/mm2

5. A thin cylindrical shell of 3 m long, 1.2 m in diameter is subjected to an internal pressure

of 1.67 N/mm2. If the thickness of the shell is 13 mm, find the circumferential and

longitudinal stresses. Also find the maximum shear stress and change in dimensions of the

shell. Assume E=2x10 5 N/mm2 and 1/m=0.28 [EVEN-2007]

.28 01

'

/102mod '

67 / . 1int

3000 3

13

12002. 1

:

25

2

m ratios poison

mm NE uluss Young

mm Np pressureernal

mm mldrum ofLength

mm tshell ofThickness

mm md shellof Diameter

datá Given

2

2

/08. 77132

67 1200 . 1

2

54 / . 3813 4

67 1200 . 1

4

:

mm Nt

Pd stressHoop

mm Nt


Solution

c

l

VTHT AVADI III SEM


5

55

4

4

55

2max

478 10 . 8.28 0102

.08 77

102

.54 38

.398 01200314 10 . 3

10 34. 3.28 0102

.54 38

102

.08 77

/ 27. 192

.54 38 08. 77

2

mE Ee strainal Longitudin

mmded

d

dewkt

mE Ee strainntial Circumfere

mm Nstress shearMaximum

cl

l

c

c

lc

c

lc

454

5

476 10 . 7478 10 . 834 10 . 322,

.254 0 3000478 10 . 8

lcv

l

l

e ee strainVolumetric

mmle l

l

le strainntial Circumfere

Also wkt

3

94

392

2

.3 2536506

10 393. 3476 10 . 7

393 10 . 33000 12004

4

mmv

v volumein Change

mm

ld vVolume but

v ev volumein Change v

Result:

Change in volume = 2536506 mm3

6. An element has a tensile stress of 600 N/mm2 acting on two mutually perpendicular

planes and shear stress of 100N/mm2 on these planes. Find the principle stress and maximum shear stress. [ODD-11]

2

22

21

/ 100

/ 6000

/ 600

:

mm Nq

mm N

mm N

Given data

VTHT AVADI III SEM


stress shearimum andplanes incipalstresses incipal

find To

maxPr ,Pr

:

Solution:

222

22

2121

1

/ 700100 4600 6002

1

2

600 600

42

1

2,

mm N

qstress principalMajor n

222

22

2121

2

/ 500100 4600 6002

1

2

600 600

42

1

2,

mm N

qstress principalMinor n

22

21max 4

2

1 qstress shearMaximum

max stressshear Maximum 222

/ 100100 4600 6002

1 mm N

Result:

max stressshear Maximum 100N/mm2

22

21

/ 500,

/ 700,

mm Nstress principalMinor

mm Nstress principalMajor

n

n

7. The stresses that a point in a strained material is px = 200 N/mm2 and py = -150 N/mm2

and q = 80 N/mm2. Find the principal plane and principal stresses. Using Graphical method and verify with analytical method. (ODD‟2012)

2

22

21

/ 80

/ 150

/ 200

:

mm Nq

mm N

mm N

Given data

VTHT AVADI III SEM



find To

maxPr ,Pr

:

Solution:

Graphical method: 1. Draw a horizontal line and set Off OA and OB equal to p1 and p2 on opposite side to the

scale, since both the stresses are opposite each other. 2. Bisect BA at C. 3. Draw perpendicular line AS from A which is equal to shear stress. 80 N/mm2 to the same

scale Or draw BR.

22

21

/ 165,

/ 215,

mm NOQ stressprincipal Minor

mm NOH stressprincipal Major

n

n

plane incipalPr 000 28 .10225. 12.5 242 orSAC

Analytical method:

VTHT AVADI III SEM


222

22

2121

1

4 / . 21780 41502002

1

2

150200

42

1

2,

mm N


222

22

2121

2

47 / . 16780 41502002

1

2

150200

42

1

2,

mm N


Result:

22

21

4 / . 167,

4 / . 217,



n

n

plane incipalPr 00 28 .10228. 12 or

8. 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 above

stresses is accompanies by a shear stress of 63 N/mm2. Determine principal stresses, principal planes and the maximum shear stress. (ODD‟2010)

2

22

21

/ 63

/ 47

/ 110

:

mm Nq

mm N

mm N

Given data


find To

maxPr ,Pr

:

Solution:

00

21

.8 10228. 12.56 24 2

150200

80 222 tanPr

oror

qplane incipal

VTHT AVADI III SEM


222

22

2121

1

9 / . 14863 447 1102

1

2

47 110

42

1

2,

mm N


222

22

2121

2

/06. 863 447 1102

1

2

47 110

42

1

2,

mm N


22

21max 4

2

1 qstress shearMaximum


63 447 1102

1

=70.43 2/ mm N

Result:

22

21

/06. 8,

9 / . 148,



n

n

plane incipalPr 00 71 .71 121 . 31 or

max stressshear Maximum 70.43N/mm2

9. A rectangular block of material is subjected to a tensile stress of 90 N/mm2 along x-axis

and a compressive stress of 45 N/mm2 on a plane at right angles to it, together with shear stresses of 80 N/mm2 on the same plane. Calculate (ODD‟09) (i) The direction of principal planes

(ii) The magnitude of principal stresses (iii) The magnitude of greatest shear stress.

00

21

.71 71 121 . 31.43 63 2

2 47110

63 222 tanPr

oror

qplane incipal

VTHT AVADI III SEM


2

22

21

/ 80

/ 45

/ 900

:

mm Nq

mm N

mm N

Given data


find To

maxPr ,Pr

:

Solution:

222

22

2121

1

17 / . 12780 445 902

1

2

45 90

42

1

2,

mm N


222

22

2121

2

17 / . 8280 445 902

1

2

45 90

42

1

2,

mm N



80 445 902

1

max stressshear Maximum 104.67N/mm

Result:


max stressshear Maximum 104.67N/mm2

10. A certain point in a strained material, the stresses on two planes at right angles to each

other are 80MN/m2 and 60 N/m2 both tensile. They are accomplished by a shear stress of 20

00

21

.9 9 114 . 24.84 49 2

.185 1 4590

80 222 tanPr

oror

qplane incipal

22

21

17 / . 82,

9 / . 127,



n

n

VTHT AVADI III SEM


MN/m2. Find graphically or otherwise the location of principal planes and evaluate the principal stresses. (EVEN‟2009)

22

222

221

/ 20/20

/ 60/60

/80 /80

:

mm Nmm MNq

mm Nmm MN

mm MNmm MN

data Given


find To

maxPr ,Pr

:

Solution:

222

22

2121

1

36 / . 9220 460 802

1

2

60 90

42

1

2,

mm N


222

22

2121

2

637 / . 4720 460 902

1

2

45 90

42

1

2,

mm N


Result:

22

21

63 / . 47,

36 / . 92,



n

n


11.) A thin clyinder is 3.5m long ,90cm in diameter and thickness of metal is 12mm and

it is subjected to an internal pressure 2.8 N/mm^2.Calculate the change in dimensions

of the clyinders,maximum intensity of shear induced.E=200Gpa,poisson's ratio is 0.3 .(May-13)

00

21

.7 7 121 . 31.43 63 2

2 6090

20 222 tanPr

oror

qplane incipal

VTHT AVADI III SEM


GIVEN:

l=3.5m=3.5 1000=3500mm;

d=90cm=900mm;

t=12mm;

P=2.8N/mm^2;E=200Gpa=200 10^3 N/mm^2(or)=2 10^5 N/mm^2;

1 0.3

m

TO FIND:

a.) calculate change in dimensions( v, L, d)

b.) maximum intensity of the shear stress

SOLUTION:

A.) Change in diameter( d)

^ 2 1 d= (1 );

2 2

2.8 900 ̂ 2 1 d= [1 (0.3)];

2 12 200 10 ̂ 3 2

d=0.401625mm.

pd

tE m

B.)change in length( l):

1 1l= ( )

2 2

2.8 900 3500 1 l= ( 0.3)

2 12 200 10 ̂ 3 2

l=0.3675mm

pdl

tE m

VTHT AVADI III SEM


C.) change in volume:

5 2v= [ ]

2 2

2.8 900 2.225 10 ̂ 9 v= [2.5 2(0.3)]

2 12 200 10 ̂ 3

v= ^ 2. 4

v=2.225 10^9 mm^3

v=1168125 1.9

v=2.219 10^6 mm^3

pdv

tE m

d l

aLongitudinal stress = 4

2.8 900

4 12

52.5 N/mm^2a

pd

t

c

c

c

Hoop stress = 2

2.8 900 =

2 12

=105 N/mm^2

pd

t

VTHT AVADI III SEM


a

c

RESULT:

v=2.219 10^6 mm^3

l=0.3675 mm

d=0.401625 mm

=5.25 N/mm^2

=105 N/mm^2

12.) The normal stress at a point on two perpendicular planes are 140 mpa(tensile),

100 mpa(compresssive).Determine the shear stress on these planes if the maximum

principle stress is limited to 150 mpa(tensile).Determine:

a.)Minimum princple stress

b.)Maximum shear stress

c.)Normal,shear,resultant shear stress on a planes which is inclined at 30 anticlockwise

to Xplane.(May-13)

b

1

GIVEN:

=140 mpa=140 N/mm^2(tensile)

100 mpa=100 N/mm^2(comprressive)

=30 ;q=150 mpa=150 N/mm^2

TO FIND:

a.)Minimum principle stress

b.)Maximum shear stress and its plane

c.)normal,shear,resultant stresses and planes inclined at 30 anticlockwise to x.

VTHT AVADI III SEM


1 2 1 2n

n

SOLUTION:

a.)NORMAL STRESS:

= .cos 2 qsin22 2

40 140 = (0.5) 129.9

2 2

114.9N/mm^2n

2

2t max 1 2)

max

1 2 2n 1 2

2

b.)MAXIMUM SHEAR STRESS:

1) = ( 4 ^ 2

2

( ) 192.093 / ^ 2

C.)MINIMUM PRINCIPLE STRESS:

1= ( ) 4 ^ 2

2 2

172.09 / ^ 2n

q

t N mm

q

N mm

1 2t

.)TANGENTIAL STRESS:

= sin 2 cos 22

100 140 sin(60) 150cos(60)

2

=178.92 N/mm^2

t

t

d

q

2 2

2 2114.9 178.92

212.6 / ^ 2

n t

res

res N mm

res

E.)RESULTANT STRESS:

2 21 1 2

1

1 ( ) 4 ^ 2

2 2

212.03 / ^ 2

n

n

q

N mm

VTHT AVADI III SEM


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