Hooke’s Law

Hooke’s Law:

It states that when the material is loaded within the elastic limit, the stress is directly proportional to the strain .i.e, the ratio of stress to the corresponding strain is a constant within the elastic limit and the constant is known as Young’s Modulus or Modulus of Elasticity.

Therefore

                                                                Stress α strain

                                                                Stress = E (strain)

                                                                E= stress/strain

                                                                E=σ/e

Where E= Young’s Modulus  σ = stress of the material , e = strain in the material

PROBLEMS ON STRESS, STRAIN AND YOUNG’S MODULUS

  

1)    An elastic rod of 25mm in diameter ,200mm long extends by 0.25mm under a tensile load of 40 KN. Find the intensity of stress, strain and elastic modulus of the material.

Step1: Data

Length of the rod=200mm

Diameter of the rod=25mm

Extension =0.25mm

Load=40KN=40x10³N

Step 2: Calculation of area of the material

A= πd²/4

   = π (25)²/4

   = 490.87mm²

Step 3: Calculation of intensity of stress

σ = load applied/ area of material

   = 40x10³/490.87

   = 81.48N/mm²

Step 4: Calculation of strain

e = extension of rod /original length

   = 0.25/200

   = 1.25x10^-3

2)  A cast iron column has the external diameter of 300mm and 20mm thick. Find the safe compressive load on the column with factor of safety of 5, if the crushing strength of the material is 550N/mm².

Step1: Data

External diameter = 300mm

Thickness = 20mm

FOS = 5

Crushing strength = 550N/mm²

Step2: Calculation of internal diameter

Internal diameter = External diameter + 2(thickness)

                                300+2(20)

                                340 mm

Step3: Calculation of area of the material

A= π (D-d) ²/4

   = π (340-300)²/4

  = 1256.64 mm²

Step 4: Calculation of safe stress

FOS = Ultimate stress/safe stress

 5       = 550/ safe stress

Safe stress = 550/ 5

Safe stress = 110 N/mm² 

Step 5: Calculation of Safe Load

Safe stress = Safe load /Area of cross section

110            = Safe load/1256.64

Safe load   = 138.230KN

    

3)  A 30 mm diameter steel rod when subjected to an axial force was subjected to a strain of 0.6x10^-3. Find the tensile force caused by the above strain. Take E=200KN/mm²

Step1: Data

Diameter =30 mm

Strain      = 0.6x10^-3

Young’s modulus E=200KN/mm² = 2X10⁵ N/mm²

Step2: Calculation of area of the material

A= πd²/4

   = π (30)²/4

   = 706.86 mm²

Step 3: Calculation of intensity of stress

E= σ/e

2X10⁵= σ/0.6x10^-3

σ = 120 N/mm²

Step 4: Calculation of Load

Stress = Load /Area of cross section

120    = Load/706.86

Load = 84.82KN

4) Find the maximum and minimum stresses produced in thestepped bar as shown in the fig  due to an axially applied compressive load of 12KN.

Step1: Data

Compressive load=12 X 10³N.

Minimum diameter = 12mm

Maximum diameter = 25mm

Step2: Calculation of area of the material

Minimum area= A= πd²/4

                              = π (12)²/4

                              = 113.09mm²

Maximum area =A = πd²/4

                               = π (25)²/4

                               = 490.87mm²

Step 3: Calculation of intensity of stress

Maximum stress: = Load / Minimum Area of cross section

                             = 12X1000/113.09

                             =106.11 N/mm²

Minimum stress: = Load / Maximum Area of cross section

                             = 12X1000/490.87

                          =24.45 N/mm²

Step1: Data

Diameter =3 mm

Load=495 N

Extension=0.07mm

Gauge length=100mm

Step2: Calculation of area of the material

A= πd²/4

   = π (3)²/4

   = 7.068mm²

Step 3: Calculation of intensity of stress

Stress = Load /Area of cross section

           = 495/7.068

Stress = 70.03 N/mm²

Step 4: Calculation of strain

Strain = Extension/gauge length

           = 0.07/100

           = 0.7X10^-3

Step5: Calculation of Young’s modulus

    E = σ/e

       = 70.03/0.7X10^-3

       = 1.00X10⁴ N/mm²

6) A wooden tie is 75mm wide,150mm deep and 1.5m long.It is subjected to a axial pull of 45000N.The stretch of the member is found to be 0.6380 mm. Find the Young’s modulus for the material.

Step1: Data

Width (b) =75mm

Depth (d) =150mm

Length(L) =1.50m=1.5X1000

Load    = 45000N

Stretch = 0.6380mm

E =?

Step2: Calculation of area of the material

A= bXd

   =75X150

   =11.25X10³ mm²

Step 3: Calculation of intensity of stress

Stress = Load /Area of cross section

           = 45000/11.25X10³

Stress = 4 N/mm²

Step 4: Calculation of strain

Strain = Stretch / length

            = 0.6380/100

= 0.7X10^-3

 Step5: Calculation of Young’s modulus

E = σ/e

    = 4/0.7X10^-3

    = 5.71X10³N/mm²

7)    A load of 4000N has to be raised at the end of the steel wire. If the unit stress in the wire must not exceed 80N/mm² what is the minimum diameter required? What will be the extension of 3.5 m length of wire? Take E=2x10⁵N/mm²

Step1: Data:

Length(L) =3.50m=3.5X1000mm

Load    = 4000N

Stress   = 80N/mm²

 E=2x10⁵N/mm²

Minimum diameter=??

Extension=??

Step2: Calculation of area of the material

Stress = Load /Area of cross section

80 = 4000/Area of cross section

Area of cross section = 50mm²

Step3: Calculation of Minimum diameter of material

A= πd²/4

50 = π(d)²/4

 d =   7.978mm

Step 4: Calculation of strain

E = σ/e

2x10⁵ = 80/e

e   = 0.4X10^-3

Step5: Calculation of Extension of material

Strain = Extension/gauge length

 0.4X10^-3 = Extension/3.5X1000

Extension= 1.4mm

8) A wooden tie is 60 mm wide ,120 mm deep and 1.5 m long.It is subjected to a axial pull of 30KN. The stretch of the member is found to be 0.625 mm. Find the Young’s modulus of the tie material.

Step1: Data

Width (b) =60 mm

Depth (d) =120 mm

Length(L) =1.50m=1.5X1000

Load    = 30000N

Stretch = 0.6250 mm

E =? 

Step2: Calculation of area of the material

A= bXd

   =60X120

  =7.2X10³ mm²

Step 3: Calculation of intensity of stress

Stress = Load /Area of cross section

           = 30000/7.2X10³

Stress = 4.166 N/mm²

Step 4: Calculation of strain

Strain = Stretch / length

          = 0.625/1.5X1000

         = 0.416X10^-3

Step5: Calculation of Young’s modulus

E = σ/e

   = 4.166/0.416X10^-3

   = 1.00X10⁴ N/mm²

9) A 20mm diameter brass rod was subjected to the tensile load of 40KN. The extension of the rod was found to be 254 divisions in the 200mm extensometer. If each division is equal to 0.001mm, find the elastic modulus of the brass.

Step 1: Data

Diameter =20 mm

Tensile load = 40KN=40X1000

Gauge length =200 mm

No of divisions = 254

least count = 0.001mm

Extension = 254X0.001 = 0.254 mm

Step2: Calculation of area of the material

A= πd²/4

   = π (20)²/4

   = 314.16 mm²

Step 3: Calculation of intensity of stress

Stress = Load /Area of cross section

           = 40000/314.16

Stress = 127.32 N/mm²

Step 4: Calculation of strain

Strain = Stretch / length

            = 0.254/200

            = 1.27X10^-3

Step5:Calculation of Young’s modulus

E = σ/e

   = 127.32/ 1.27X10^-3

   = 1.00X10⁵ N/mm²

10) A hollow steel column has an external diameter of 250mm and an internal diameter of 200mm. Find the safe axial compressive load for the column if the safe compressive stress is 120N/mm².

Step 1: Data

External diameter = 250mm

Internal diameter = 200mm. 

Safe Compressive stress = 120N/mm² 

Step2: Calculation of area of the material

A= π(D-d)²/4

= π (250-200)²/4

= 1963.49 mm²

Step 3:Calculation of intensity of stress

Safe Stress = Safe Load /Area of cross section

120           = Safe Load/1963.49

Safe Load = 235.61 KN

11) A hollow steel column of external diameter 250mm has to support an axial load of 2000KN.If the ultimate stress for the steel column is 480N/mm².Find the internal diameter of the column allowing the factor of safety of 4.

Step 1: Data

External diameter = 250mm

Ultimate stress = 480N/mm²

Axial load = 2000KN

Factor of safety = 4.

Internal diameter = ? 

Step 2: Calculation of working Stress

FOS=Ultimate stress /working stress

4 = 480/working stress

working stress= 120N/mm².

Step 3: Calculation of area of the material

Stress = Load /Area of cross section

 120    = 2000X1000/Area of cross section

Area of cross section = 16.66X10³mm²

Step 4 : Calculation of internal diameter of  material

A= π(D-d)²/4

16.66X10³= π (250-d)²/4

d= 104.35 mm

12) A solid alloy bar of 40mm in diameter is used as a tie. If the permissible tensile stress in the material is 320N/m², determine the capacity of the bar. If the hollow steel bar with internal diameter of 20mm is used instead of solid alloy bar determine its external diameter. For steel hollow bar the permissible stress is 150N/mm².

Step 1: case 1: solid bar

Diameter = 40mm

stress = 320N/mm²

Step 2: Calculation of area of material

A= πd²/4

   = π (40)²/4

   =1256.63mm²

Step 3: Calculation of intensity of load

Stress = Load /Area of cross section

 320    =Load/1256.63

Load = 402.12KN

Step 4: Case 2: Hallow bar

internal diameter: 20mm

stress = 150N/mm²

Step 5: Calculation of area of material

Stress = Load /Area of cross section

150= 402.12X1000/Area of cross section

Area of cross section=2680.8mm²

A= π(D-d)²/4

2680.8= π (D-20)²/4

D         = 38.42mm

  13)    Fig shows a steel wire and a copper wire of each of length 1m and carrying an axial load of 5KN.

a)      For the same diameter of the wires find the ratio of extension of the copper wire to the extension of the steel wire.

b)      For the same extension of the wires ,find the ratio of the diameter of the copper wire to the diameter of the steel wire .TakeE_s=2x10⁵N/mm² and E_c=1.2x10⁵N/mm²

Step 1: For the same diameter of the wires

we know that E = σ/e

Given E_s=2x10⁵N/mm² and E _c=1.2x10⁵N/mm²

      Therefore, Es/E c= (Extension of steel)/ (Extension of copper)

(Extension of steel)/ (Extension of copper) = (2x10⁵) /(1.2x10⁵)

                                                                                                        = 1.67

Step 2 ^: For the same extension of the wires 

Es/E c=Ac/As

Es/E c=(dc/ds)²

dc/ds = 1.29

ds/dc=0.774

     14)      The following data refer to a mild steel specimen tested in laboratory:

a)      Diameter of the specimen=25mm

b)      Length of the specimen=300mm

c)      Extension under a load of 15KN=0.045mm

d)     Load at yield point=127.65KN

e)      Maximum load=208.6KN

f)       Length of the specimen after failure= 375mm

g)      Neck diameter=17.75mm

Determine young’s modulus, yield point, ultimate stress, percentage elongation, percentage reduction in area, safe stress with FOS of 2.

step 1: Calculation of area of material

A= πd²/4

   = π (25)²/4

   =490.87mm²

Step 2: Calculation of intensity of stress

Stress = Load /Area of cross section

           =15X1000/490.87

Stress = 30.55N/mm²

Step 3: Calculation of strain

Strain = Stretch / length

            = 0.045/300

            = 0.15X10^-3

Step4: Calculation of Young’s modulus

E = σ/e

     = 30.55/0.15X10^-3

     = 2.03X10⁵ N/mm²

Step5: Yield point stress

Stress = Load /Area of cross section

           =127.65X1000/490.87

Stress = 260.04N/mm²

Step6: Ultimate stress

Stress = Load /Area of cross section

           =208.6X1000/490.87

Stress = 424.96N/mm²

Step 7: percentage elongation

percentage elongation = {(length at failure-initial length)/length at failure)}X100

                                     = {(375-200)/375}X 100

                                     = 46.67%

Step 8: percentage reduction in area

final area=A= πd²/4

                      = π (17.75)²/4

                      =247.44mm²

percentage reduction in area= {(initial area-final area)/initial area} X100

                                              = {490.87-247.44)/490.87}X100

                                              = 49.59%

14) A bar of diameter 20 mm and length 100 mm extends by 0.2 mm. If E of the material is 2x10⁵N/mm², what amount of load applied to the rod. If an extension of 20% greater is required for the same load applied above, how much the diameter of the bar need to be reduced.

 Step 1: Data:

      case 1: Diameter = 20 mm

                Length = 100 mm

                Extension = 0.2 mm  

                E= 2x10⁵N/mm²

                      Load =??

             

    Step 2: Calculation of area of material:

A= πd²/4

   = π (20)²/4

   = 314.16 mm²

     Step 3: Calculation of strain

Strain = Extension / length

            = 0.2/100

            = 2X10^-3

Step4: Calculation of intensity of stress

E = σ/e

2x10^5 = σ / 2X10^-3

σ =400N/mm²

Step5: Calculation of intensity of load

Stress = Load /Area of cross section

400   =Load/314.16

Load = 125.66KN

Step6: Case 2

Extension is 20% more: 

Total extension =1.2X0.2 = 0.24mm

Load = 125.66KN

Diameter = d =??

Length =  100 mm

      Step 7: Calculation of strain

Strain = Extension / length

         = 0.24/100

 Strain = 2.4X10^-3

Step8: Calculation of intensity of stress

        E = σ/e

2x10⁵ = σ / 2.4X10^-3

σ = 480N/mm²

Step9: Calculation of diameter

Stress = Load /Area of cross section

480    =125.66X1000/A

          A       =261.79 mm²

       261.79 = πd²/4

            = π(d)²/4

     d     = 18.25 mm

Step 10 : percentage reduction in diameter

(initial diameter-final diameter)/initial diameterX100

{(20-18.25)/20} X 100

= 8.75%

    15)      A short piece of steel pipe is to carry a compressive load of 1000KN with a factor of safety of 2 against yielding. If the thickness of the pipe is to 1/5^th of the outside diameter, Find the minimum required outside diameter Take yield stress of the steel as 280MPa.

     

Step 1: Data:

Load = 1000KN

Factor of safety = 2 

Outside diameter = D

Thickness = 1/5^th D

 yield stress = 280MPa = 280N/mm²

Step 2 : Calculation of Working stress

FOS= yield stress /working stress

2 = 280/working stress

Working stress= 140N/mm².

Step 3: Calculation of internal diameter

       d = D-2t

       d = D-2(1/5) D

       d = 0.4D

    Step4: Calculation of intensity of stress

       Stress = Load /Area of cross section

        140    =1000X1000/A

          A       =7142.85 mm²

        

     Step 5: Calculation of diameters

 A= π(D-d)²/4

7142.85= π (D-0.4D)²/4

D         = 119.2 mm

d          = 71.52 mm

 16)  A hollow steel tube is used to carry a compressive load of 150KN. The yield stress for steel is 250N/mm² and factor of safety is 2 determine the thickness of the steel specimen if the external diameter is 100mm.

    Step 1: Data:

     Load = 150KN

     yield stress = 250N/mm²

      factor of safety = 2

     external diameter = 100 mm.

      

      Step 2: Calculation of Working stress

FOS= yield stress /working stress

2 = 250/working stress

      Working stress= 125 N/mm².  

        

      Step3: Calculation of intensity of stress

       Stress = Load /Area of cross section

        125    =150X1000/A

             A       =1200 mm²

       

        Step 5: Calculation of diameters

 A= π(D-d)²/4

1200= π (100-d)²/4

d      = 60.91 mm

Step 6 : Calculation of thickness

t = {(D-d)}/2

t = {(100- 60.91)}/2

t = 19.54 mm

     17)      The tensile test was conducted on a mild steel bar. The following details are

a)      Diameter of the steel bar=16mm

b)      Load at proportionality limit= 72KN

c)      Load at failure=80KN

d)     Diameter of the rod at failure=12mm

e)      Gauge length of the bar =80mm

f)       Extension at the load of 60KN=0.115mm

g)      Final length of the bar=104mm

Determine young’s modulus, proportionality limit stress, true breaking stress, percentage elongation of the material.

step 1: Calculation of area of material

A= πd²/4

   = π (16)²/4

   =201.06 mm²

Step 2: Calculation of intensity of stress

Stress = Load /Area of cross section

           =60 X1000/201.06

Stress = 298.41N/mm²

Step 3: Calculation of strain

Strain = Stretch / length

          = 0.115/80

           = 1.43X10^-3

Step4: Calculation of Young’s modulus

E = σ/e

     = 298.41/1.43X10^-3

     = 2.07X10⁵ N/mm²

Step5: proportionality limit stress

Stress = Load /Area of cross section

           =72X1000/201.06

Stress = 358.10N/mm²

Step6: True breaking stress

Stress = Load /Area of cross section

           =80X1000/201.06

Stress = 397.89N/mm²

Step 7: percentage elongation

percentage elongation = {(length at failure-initial length)/length at failure)}X100

                                     = {(104-80)/104}X 100

                                     =23.07%

Step 8: percentage reduction in area

final area=A= πd2/4

                        = π (12)2/4

                        =113.09mm2

percentage reduction in area= {(initial area-final area)/initial area} X100

                                              = {201.06-113.09)/201.06}X100

                                               = 43.75%