Analytical Method of Analysis MCQ Quiz in বাংলা - Objective Question with Answer for Analytical Method of Analysis - বিনামূল্যে ডাউনলোড করুন [PDF]

\({{\rm{\sigma }}_{\rm{n}}} = \frac{{{{\rm{\sigma }}_{{\rm{xx}}}} + {{\rm{\sigma }}_{{\rm{yy}}}}}}{2} + \frac{{{{\rm{\sigma }}_{{\rm{xx}}}} - {{\rm{\sigma }}_{{\rm{yy}}}}}}{2}\:\cos \:2\theta + {{\rm{\tau }}_{{\rm{xy}}}}\:\sin \:2\theta \)

σ_x = 360 MPa, σ_y = -160 MPa, τ_xy = 0

for σ_n = 0

0 = 100 + 260 cos 2θ

θ = 56.30°

σ_xx = 50 MPa, σ_yy = -35 MPa, τ_xy = 0

\(\theta = {\tan ^{ - 1}}\frac{5}{3} = 59.03^\circ\)

Normal stress on an oblique plane is given as:

\({{\rm{\sigma }}_{n1}} = \frac{{{{\rm{\sigma }}_{{\rm{xx}}}} + {{\rm{\sigma }}_{yy}}}}{2} + \frac{{{{\rm{\sigma }}_{{\rm{xx}}}} - {{\rm{\sigma }}_{{\rm{yy}}}}}}{2}\cos 2\theta + {\tau _{xy}}\sin 2\theta \)

σ_n1 = 7.5 - 20 = -12.5 MPa

\({{\rm{\tau }}_{{\rm{xy}}}} = \frac{{{{\rm{\sigma }}_{{\rm{xx}}}} - {{\rm{\sigma }}_{{\rm{yy}}}}}}{2}\sin 2{\rm{\theta }} - {{\rm{\tau }}_{{\rm{xy}}}}\cos 2{\rm{\theta }} = 37.5{\rm{\;MPa}}\)

The maximum principal stress,

\(\begin{array}{l} {\sigma _1} = \frac{{{P_x} + {P_y}}}{2} \pm \sqrt {{{\left( {\frac{{{P_x} - {P_y}}}{2}} \right)}^2} + \tau _{xy}^2} \\ = \frac{{120 + 80}}{2} \pm \sqrt {{{\left( {\frac{{120 - 80}}{2}} \right)}^2} + {{15}^2}} \\ = 100 \pm \sqrt {{{20}^2} + {{15}^2}} \end{array}\)

= 100 ± 25

Tensile stress σ_y = F / A = 105 x 10³ / π x (0.02)²

= 83.55 MN/m²

Now the normal stress on an obliqe plane is given by the relation

σ_q = σ_y sin²q

50 = 83.55 sin²q

q = 50⁰68'

The shear stress on the oblique plane is then given by

τ_q = 1/2 σ_y sin2q

= 1/2 x 83.55 x 10⁶ x sin 101.36

= 40.96 MN/m²

Therefore the required shear stress is 40.96 MN/m²

Explanation:

Graphical Method:

Here,

σ_x = 10 MPa, σ_y = 10 MPa, σ₁ = 15 MPa

As we know:

σ_x + σ_y = σ₁ + σ₂

10 + 10 = 15 + σ₂ ⇒ σ₂ = 5 MPa

Draw a Mohr circle with coordinates (15, 0) and (5, 0)

A(10, τ), B(10, -τ)

The diameter of Mohr circle: σ₁ – σ₂ = 10 MPa

The radius of Mohr circle: \(\frac{{{\sigma _1}~ - {~\sigma _2}}}{2} = 5\;MPa\) 

Centre of Mohr circle: \(\frac{{{\sigma _1} \;+ \;{\sigma _2}}}{2} = 10 \;MPa\)

Coordinates of the centre: O(10, 0)​

So, point A & B will pass through centre of Mohr circle.

Where,

τ = R = 5 MPa

Analytical Method:

Also, \({\sigma _1} = \left( {\frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}}}}{2}} \right) + \sqrt {{{\left( {\frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}} \right)}^2} + \tau _{xy}^2}\) 

\(15 = \left( {\frac{{10 + 10}}{2}} \right) + \sqrt {{{\left( {\frac{{10 - 10}}{2}} \right)}^2} + \tau _{xy}^2}\)

⇒ τ_xy = 5 MPa

T = M/2

\(\begin{array}{l} Principal\;stress = \frac{{16}}{{\pi {D^3}}}\left[ {M + \sqrt {{M^2} + {T^2}} } \right]\\ Max.\;shear\;stress = \frac{{16\sqrt {{M^2} + {T^2}} }}{{\pi {D^3}}} \end{array}\)

 Ratio = \(\frac{{\left[ {M + \sqrt {{M^2} + {T^2}} } \right]}}{{\sqrt {{M^2} + {T^2}} }} = \frac{{1 + \sqrt {\frac{5}{4}} }}{{\sqrt {\frac{5}{4}} }} = \frac{{2 + \sqrt 5 }}{{\sqrt 5 }}\)