Engineering Mechanics 3(2+1)

LESSON 28.

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 σ_x and σ_y act across the faces LM and MN and that the block has unit depth perpendicular to LMN. Let the stresses τ and σ_n act on the same plane at an angle Ө to LM.

 

 

 

 

 

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

 

Resolving normal to LN:

σ_n × LN = σ_x × LM cosӨ + σ_y MN sinӨ

σ_n = σ_x ×  \[{{LM} \over {LN}}\]  cosӨ + σ_{y  \[{{MN} \over {LN}}\]}  sinӨ

     = σ_x cos²Ө + σ_y 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)

     = \[{{{\sigma _x} + {\sigma _y}} \over 2}\]  + σ_{x  \[\left[ {{{co{s^2}\theta-si{n^2}\theta } \over 2}} \right]\]}  – σ_{y \[\left[ {{{co{s^2}\theta-si{n^2}\theta } \over 2}} \right]\]}

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

When Ө = 0, then σ_n = \[{{{\sigma _x} + {\sigma _y}} \over 2}\]   + \[{{{\sigma _x} - {\sigma _y}} \over 2}\]   = σ_x -----------------------------------(ii)

and, when Ө = П/2, then σ_n =   -  = σ_y ----------------------------(iii)

Resolving parallel to LN:

τ × LN = σ_x × LM sinӨ – σ_y × MN cosӨ

τ = σ_x ×  _{\[{{LM} \over {LN}}\]}   sinӨ – σ_y × _{\[{{MN} \over {LN}}\]}  cosӨ

  = σ_x cosӨ sinӨ – σ_y sinӨ cosӨ  =  (σ_x – σ_y) 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,

σ_r =  \[\sqrt {{\sigma _n}^2 + {\tau ^2}}\]

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: σ_x = 100 MN/m² (tensile);

                             σ_y = 60 MN/m²  (tensile); θ = 30⁰

normal stress,  σ_n = \[{{{\sigma _x} + {\sigma _y}} \over 2}\]  +  \[{{{\sigma _x} - {\sigma _y}} \over 2}\]  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,  σ_r = \[\sqrt {{\sigma _n}^2 + {\tau ^2}}\]  =  \[\sqrt {{{90}^2} + {{17.32}^2}}\]   =  91.65 MN/m²

Hence,                 σ_{r =}   91.65 MN/m²

 

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

                    Φ = 10.89⁰