## LESSON 28.

28.1 INRODUCTION

28.2 ELEMENT SUBJECTED TO DIRECT STRESS IN TWO PERPENDICULAR DIRECTIONS

At any point in a material where stress is acting (see fig.-----), it is possible to assume that the point consists of a very small triangular block, such that the stress act across the faces of the block. Consider that direct stresses σx and σy act across the faces LM and MN and that the block has unit depth perpendicular to LMN. Let the stresses τ and σn act on the same plane at an angle Ө to LM.

Fig.--------------

Resolving normal to LN:

σn × LN = σx × LM cosӨ + σy MN sinӨ

σn = σx ×  ${{LM} \over {LN}}$  cosӨ + σy  ${{MN} \over {LN}}$  sinӨ

= σx cos2Ө + σy sin2Ө = ${{{\sigma _x}} \over 2}$ × 2 cos2Ө + ${{{\sigma _x}} \over 2}$ × 2sin2Ө

= ${{{\sigma _x}} \over 2}$ (1- sin2Ө + cos2Ө) +  ${{{\sigma _y}} \over 2}$ (1- cos2Ө + sin2Ө)

= ${{{\sigma _x} + {\sigma _y}} \over 2}$  + σx  $\left[ {{{co{s^2}\theta-si{n^2}\theta } \over 2}} \right]$ σy $\left[ {{{co{s^2}\theta-si{n^2}\theta } \over 2}} \right]$

= ${{{\sigma _x} + {\sigma _y}} \over 2}$  + ${{{\sigma _x} - {\sigma _y}} \over 2}$   cos2Ө-----------------------------------(i)

When Ө = 0, then σn = ${{{\sigma _x} + {\sigma _y}} \over 2}$   + ${{{\sigma _x} - {\sigma _y}} \over 2}$   = σx -----------------------------------(ii)

and, when Ө = П/2, then σn =   -  = σy ----------------------------(iii)

Resolving parallel to LN:

τ × LN = σx × LM sinӨσy × MN cosӨ

τ = σx ×  ${{LM} \over {LN}}$   sinӨσy × ${{MN} \over {LN}}$  cosӨ

= σx cosӨ sinӨσy sinӨ cosӨ  =  (σxσy) sinӨ cosӨ

=  ${{{\sigma _x} - {\sigma _y}} \over 2}$  sin2Ө-----------------------------(iv)

The maximum value of τ occurs when 2Ө = П/2 or Ө = П/4 and then

= ${\tau _{max}}$  =  ${{{\sigma _x} - {\sigma _y}} \over 2}$   ----------------------------------(v)

The resultant stress,

σr =  $\sqrt {{\sigma _n}^2 + {\tau ^2}}$

tanϕ = ${\tau\over {{\sigma _n}}}$ where ϕ is the angle which the resultant stress makes with the normal to the plane is called obliquity.

Example: The principal stresses at a point across two perpendicular planes are 100 MN/m2 (tensile) and 60 MN/m2(tensile).Find the normal, tangential stresses and the resultant stress and its obliquity on a plane at 300 with the major principal plane.

Solution: Given: σx = 100 MN/m2 (tensile);

σy = 60 MN/m2  (tensile); θ = 300

normal stress,  σn${{{\sigma _x} + {\sigma _y}} \over 2}$  +  ${{{\sigma _x} - {\sigma _y}} \over 2}$  cos2Ө

= ${{100 + 60} \over 2}$ + ${{100 - 60} \over 2}$   cos (2×300)

= 80 + 20 cos 600

= 90 MN/m2 (tensile)

Tangential stress, τ   =  ${{{\sigma _x} - {\sigma _y}} \over 2}$ sin2Ө = ${{100 - 60} \over 2}$  sin (2×300)

= 20 sin 600

= 17.32 MN/m2

Hence,             τ   = 17.32 MN/m2

Resultant stress,  σr = $\sqrt {{\sigma _n}^2 + {\tau ^2}}$  =  $\sqrt {{{90}^2} + {{17.32}^2}}$   =  91.65 MN/m2

Hence,                 σr =   91.65 MN/m2

Obliquity ϕ :     tanϕ =  ${\tau\over {{\sigma _n}}}$  =  ${{17.32} \over {90}}$  = 0.1924

Φ = 10.890

Last modified: Wednesday, 11 September 2013, 8:21 AM