Mekban Principal Stresses and Strains analitis

  Fakultas Teknik Jurusan Teknik Sipil

Universitas Brawijaya Malang

  Principal Plane

At any point in a strained material, there are three planes, mutually

perpendicular to each other, which carry direct stresses only and no shear

stress. It shows that out of these three direct stresses one will be

maximum the other minimum and the third an intermediate between the

two. These particular planes, which have no shear stress are known as

principal planes Principal Stress

The magnitude of direct stress, across a principal plane, is known as

principal stress Method for Stresses on Oblique Section

• Analytical method
• Graphical method

  Analytical Method for Stresses on an Oblique Section of a Body

Subjected to a Direct Stresses in One Plane

  

Consider rectangular body ABCD of uniform cross-sectional area and unit

thickness subjected to a principal tensile stress. Let, P = tensile force A = cross sectional area Θ = angle which the oblique section makes with normal cross section EF

  The intensity of tensile stress across the section EF: P P P

     cos 

   A A A sec 

  

The magnitude of tensile stress on section EF will be less than p,

because the resisting section has a bigger area. But this stress is

neither normal nor shear stress for section EF. Since the failure of this

body will occur either by tension or by shear , it is therefore essential to

know the normal and tangential stresses across the section EF.

   cos P P  n

  Normal stress across the section EF, Force P P cos

  

  2 p p

     cos cos  cos    n

  Area A A sec  Tangential / shear stress across the section EF, Force P P p sin  p

  

   sin  cos   sin

2  t

  Area A A sec

  2  ₂ Normal stress across the section EF will be maximum when cos

  θ = 1 or θ = 0° Shear stress across the section EF will be maximum when sin 2 θ = 1 or θ

  = 45° and 135 ° Maximum tangential stress: p p p Max p

   sin 2   1   t

  2

  2

  2 Resultant stress:

  2

  2 p  p  p

  R n t

  

A tension member is formed by connecting with glue two wooden scanting

each 7,5 x 15 cm at their end which are cut at an angle of 60°. The member is

subjected to a pull P. Calculate safe value of P if permissible normal and

_{2 2} shear stress in glue are 14 kg/cm and 7 kg/cm respectively.

  Solution :

  14

         

  2   

  2

  2

  2

  2

  2

  7

  2 cm 112 5 ,

  18 30 cos 14 cos

  7 cos sin 67 ,

  16 30 cos 30 sin

  16 17 ,

  17 ,

  Angle of the joint with the normal Θ = 90° - 60 ° = 30 ° P kg cm kg p p p p cm kg p p p p cm kg p cm kg p t n t n 1819 1 , 112 5 ,

  15 5 , 7    Area

      

  

Analytical Method for Stresses on an Oblique Section of a

Body Subjected to a Direct Stresses in Two Mutually

Perpendicular Direction

  

Consider rectangular body ABCD of uniform cross-sectional area and

unit thickness subjected to mutually perpendicular principal tensile

stresses on the face AB, CD, and AD, BC. Also consider an oblique section EF on which we are required to find out the stresses. Let, p = major tensile stress on the face AD and BC ₁ p = minor tensile stress on the face AB and CD ₂ P = tensile force on the section EF (such that P = p x BC) _{1 1 1} P = tensile force on the section EF (such that P = p x GF) _{2 2 2}

Θ = angle which the oblique section makes with normal cross section EG

  Tensile force perpendicular to plane EF cos sin

  P  P   P  n

  1

2 P  p  BC cos  p  GF sin

    n

  1

2 Tensile force tangential to plane EF

  P P sin P cos     t

  1

2 P p BC sin p GF cos

        n

  1

  2

  Normal Stress across the section EF    

  2

  2

  1

  2

  1

  2

  1

  1 p p p p p p p p p p p

  2

  GF GF p BC BC p p

  EF GF p EF BC p p

  EF GF p BC p EF P p n n n n n n n

     

        



    



  2

  1

      

  2 2 cos

     

     

  2 cos

  2

  2 2 cos

  1

  1

  2

  2 sin cos / sin sin

  / cos cos sin cos sin cos

  2

  1

  2

  1

        

  Tangential Stress across the section EF    

  1

  p p p  

  2

       Resultant stress:

     

     

       

  EF GF p BC p EF P p t t t t t t

  EF GF p EF BC p p

  GF GF p BC BC p p

  1 p p p p p p p p

  2

  2

          

  1

  2

  1

  2

  1

  2

  1

  2

  2 cos sin cos sin cos sin / sin cos / cos sin cos sin cos sin

    2 sin

     

2 R t n

  

A point in a strained material is subjected to two mutually

perpendicular tensile stress of 2000 kg/cm2 and 1000 kg/cm2.

Determine the intensities of normal and resultant stress on a plane

inclined at 30° to the axis of the minor stress.

  Example : Solution :

  2 ^{2 1 2 1 2 2 2 1} 1750 60 cos

  2 1000 2000 2 1000 2000

  

2 cos

  2

  2 1000 30 2000 cm kg p p p p p p cm kg p cm kg p n ⁿ

     

     

     

   

  Tangential stress p  p

    _{1 2} p  sin _t 2 

  2 2000  1000

    kg p  sin _t 60  433 ₂ cm

  2 Resultant stress

  2

  2

  2

  2 kg p  p  p  1750  433  1082 ,

  8

  2 R n t cm

  

Analytical Method for Stresses on an Oblique Section of a Body

Subjected to a Direct Stresses in One Plane Accompanied by a

Simple Shear Stress

  

Consider rectangular body ABCD of uniform cross-sectional area and

unit thickness subjected to tensile stress in one plane accompanied by

a shear stress across the face AD, and BC. Also consider an oblique section EF on which we are required to find out the stresses. Let, p = tensile stress on the face AD and BC q = tensile stress across the face AD and BC Θ = angle which the oblique section makes with normal cross section EF

  

From the geometry , we find that the horizontal force acting on AD

  

  AD p P

1 Vertical force acting on AD

     AD q P

  2 Horizontal force acting on GF    GF q P

  3 Normal Force across the section EF    cos sin cos

  3

  2

  n   

1 P P P P

  

Tangential Force across the section EF

   sin cos sin

  3

  2

  t   

1 P P P P

  Normal Stress across the section EF  

  1

      

     

         

  EF P P P EF P p n n n n n n n

  EF P EF P EF P p

  AD AD p p

  GF GF q

AD

AD q

  1 q p p q p p q q p p

  2

  3

  2

           

  3

  2

  2

  / cos cos cos sin cos cos sin cos

  / sin cos / cos sin

  2 cos sin 2 cos

cos sin cos sin cos

  1

     2 sin 2 cos

      

     

     

  Tangential Stress across the section EF  

  3

     

      

      

  EF P P P EF P p t t t t t t t

  EF P EF P EF P p

  AD AD p p

  GF GF q AD AD q

  1 q p p q p p q q p p

  2

  3

  1

  2

  2

   

  

  2

  2

  2

  / cos sin sin cos sin sin cos sin

  / sin sin / cos cos

  1 sin cos cos sin

  2

  1 sin cos 2 sin

  2 cos 2 sin

       

     

  

   

 

        

  The planes of maximum and minimum normal stress maybe found out by equating the tangential stress to zero p q q p q p

  1

  

There are two principal planes, at right angle to each other. Their inclination

with the normal cross section being θ

₁

and

     

     

     

     

  2 2 sin q p p c q p q q p p c q p q

  4

  2 2 sin 4 2 os

  4

  1 4 2 os

  2

  2

  2

  2

2 tan

2 cos 2 sin

  2

  2

  2

  2

  2

  2

  2

      

  1     

  2

  1 2 cos 2 sin

  2

  θ ₂ such that:

  

Values of Principal Stresses maybe found out by substituting the above value

of 2 θ ₁ and 2 θ ₂ in equation:

  2 2 sin 2 2 cos

       

    

   

      

    

     

  q p p p q p p p q p q q p p p p q p p p q p p n ^{n n n n}

  2

  1

  2 2 sin 2 cos

  2

   

  4

  2

  4

  2

  1

  2

  4

  2

  2

  2 ² ₁ ^{2 2 1 2 2 2 2 2 2 1 1 1}

     

  And,  

  2 2 sin 2 cos

       

    

   

      

    

     

  q p p p

q p

p p q p q

q p

p p p q p p p q p p n ^{n n n n}

  2

  1

  2 2 sin 2 2 cos

  2 ² ₂ ^{2 2 2 2 2 2 2 2 2 2 2 2}

  2

  4

  2

  4

  2

  1

  2

  4

  2

  2

  

   Example : A point in a strained material is subjected to a compressive stress of 800 kg/cm2 and a shear stress of 560 kg/cm2. Determine the maximum and minimum intensities of direct stress.

  Solution : kg p   800 ₂ cm kg q  560 ₂

cm

   angle which the principal plane plane  makes with the normal to compressiv e stress 2 q

  2  560 tan 2    1 ,

  4 

  P 800 2 

  54 28 '  

  27 14 '  

  Maximum intensity of direct stress ₂ p p

    ² p    q _{n 1}  

  2

  2  

₂

 800 800    ₂ kg p    560  288 ,

  2 _{n 1}   ₂ cm

  2

  2   Minimum intensity of direct stress

  2 p p

   

  

2

   p q _n  

  2

  2

  2  

  2  800 800

    

  2 kg p    q   1088 ,

  2

  2 cm

  2 _n  

  2

  2  

  

Analytical Method for Stresses on an Oblique Section of a Body

Subjected to a Direct Stresses in Two Mutually Perpendicular

Direction Accompanied by a Simple Shear Stress

  

Consider rectangular body ABCD of uniform cross-sectional area and

unit thickness subjected to tensile stress and shear stress. Consider an oblique section EF on which we are required to find out the stresses. Let, p ₁

  = tensile stress on the face AD and BC p ₂ = tensile stress on the face AB and CD q = shear stress across the face AD and BC

1 Vertical force acting on AD:

2 Horizontal force acting on GF:

3 Vertical force acting on GF:

  Tangential Force across the section EF     cos sin cos sin

  Normal Force across the section EF

  t    

  2

  3

  4

  n

   

  Θ = angle which the oblique section makes with normal cross section EG

From the geometry , we find that the horizontal force acting on AD

  2

  3

  4

  4     sin cos sin cos

  2

     GF p P

     GF q P

     AD q P

  

  

AD q P

1 P P P P P

1 P P P P P

  Normal Stress across the section EF    

  2 ^{1 2 1 2 2 1 1 2 1 2 2 2 1 2 2 2 1 2 1 4 3 2 1 4}

³

^{2 1}

q

p p p p p p p q p p p

p

q p p p q p p p q q p p

      

   

   

   

      

         

       

   

   

  EF P P P P EF P p n ^{n n n n n n n n}

  EF P EF P EF P EF P p

  AD AD q AD AD p p

  GF GF p GF GF q

  / cos sin / cos cos sin cos sin cos sin cos sin cos

      

  / sin sin / sin cos

  2 sin cos sin 2 cos sin cos sin cos sin cos

  1

  2 2 sin 2 cos

  1

  2 2 cos

  2 2 sin 2 2 cos

  2 2 2 cos

  2

  2 sin 2 cos

         

  





 

     

              

      

  Tangential Stress across the section EF  

  2

  1

  4

  3

  2

  1

  4

  

3

  1 q p p p q p p p p q q p p

  1

  GF GF p GF GF q

  AD AD q AD AD p p

  EF P EF P EF P EF P p

  EF P P P P EF P p t t t t t t t

        

       

     

       

  2

  2

     

  1 cos sin cos sin cos sin sin cos cos sin

       

  

     

 

    



     

   

   

  2 cos 2 sin

  2

  / sin cos / sin sin

  2

  / cos cos / cos sin cos sin cos sin cos sin cos sin

  2

  1

  2

  

2

  2

  1

  2

      

  

Maximum and minimum normal stress may be found out by

equating the tangential stress to zero

  1 p  p  q  sin 2  cos 2 

   

  1

  2

  2

  1 p  p sin

2  q cos

  2  

   

  1

  2

  2

  2 q q

   tan 2   

  1 p  p

  

  1

  2 p  p

    

  1

  2

  2

  

There are two principal planes at the right angles to each other. Their

inclination with the normal cross-section being and such that: θ θ _{1 2} q

  

  2 sin 2  

  1

  2

  2 p  p  4 q

  

 

  1

  2  p  p

   

  1

  2 c os 2 

  

  1

  2

  2 p p q

   

  4  

  1

  2 q

  

  2 sin 2  

  2

  2

  2 p  p  q

  4

 

  1

  2  p  p

   

  1

  2 c  os 2 

  2

  2

  2 p p q

   

  4  

  1

  2

  Values of Principal Stresses maybe found out by substituting the above value of 2θ ₁ and 2θ ₂ in equation:

 

   

     

    

   

       

    

   

    

     

     

    

    

   

   

     

     

     

     

   

   

   

   

    

    

         

  4

  2 ² _{2 1 2 1 1} ^{2 2 2 1 2 1 1 2 2 2 1 2 2 2 2 1}

²

^{2 1 2 1 1 2 2 2 1 2 2}

²

^{1 2 1 2 1 2 1 1 2 1 2 1 1}

  2

  2

  4

  2

  1

  2

  4

  2

  2

     

  2

  4

  2

  4

  2

  2 2 sin 2 cos

  2

  2

  

q

p p p p p q p p p p p q p p q q p p p p p p p q p p q q q p p p p p p p p p q p p p p p n ^{n n n n}

  

 

             

     

    

   

       

    

   

    

     

     

    

    

   

   

    

       

     

      

   

   

   

   

    

    

   

  4

  2 ² _{2 1 2 1 2} ^{2 2 2 1 2 1 2 2 2 2 1 2 2 2 2 1}

²

^{2 1 2 1 2 2 2 2 1 2 2}

²

^{1 2 1 2 1 2 1 2 2 1 2 1 2}

  2

  2

  4

  2

  1

  2

  4

  2

  2

     

  2

  4

  2

  4

  2

  2 2 sin 2 cos

  2

  2

  

q

p p p p p q p p p p p q p p q q p p p p p p p q p p q q q p p p p p p p p p q p p p p p n ^{n n n n}

  

 

  

The principal stress pn1 will be maximum whereas the stress pn2

will be minimum. The planes of maximum shear now be found

out. These planes are at right angles of each other and are

inclined at 45° to principal planes. The maximum shear stress will

be given by the relation: p  p n n

  1

  2 p max  t

  2 Example :

A point is subjected to a tensile stress of 60 N/mm2 and a

compressive stress of 40 N/mm2, acting on two mutually

perpendicular planes and a shear stress of 10 N/mm2 on these

planes. Determine the principal stresses as well as maximum shear

stress. Also find out the value of maximum shear stress.

  Solution :

N

p

  

  60

  2

  1 mm N p

  Minor stress  

  40

  2

  2 mm N q

  Shear stress 

  10

  2 mm p

   major principal stress n

  1 p  minor principal stress n

  2

  2 ²

²

₁ ²

²

^{2 1 2 1 1}

       

  2

  40

  60

  2

  2

  2 ^{2 2} ₂ ^{2 2 2 1 2 1 2} mm N p q p p p p p n ⁿ

      

     

     

     

  





     

  40

     

  

  2

  2

  1

  51

  2

  41

  61

  2 max mm N p p p n n t

  



 

  60

  2

  61

    

  10

  2

  40

  60

  2

  40

  60

  2

  2

  mm N p q p p p p p n ⁿ

      

   

  10

    

    

   

     

    

   

    

    

   

  (compressi on)

  41

   Maximum shear stress:

  

A little knowledge that

acts is worth infinitely

more than much knowledge that is idle.