Shear Stress and Bending Stress MCQ Quiz in मराठी - Objective Question with Answer for Shear Stress and Bending Stress - मोफत PDF डाउनलोड करा

Explanation:

Stress distribution diagram at the base of a circular chimney:

(a) wind pressure is P acting on the left side of the chimney as seen on the paper:

Let wind pressure is P kN/m² acting on the left side of the chimney as seen on the paper and the height and diameter of the chimney are h and D respectively.

The direct stress is the same base on the chimney because the weight or load of the chimney is uniformly distributed in the base area.

\(Direct\, stress(σ_d)={Weight\, of\, the\, chimney(W)\over Area\, of\, the\, chimney(A)}\)

Bending stress at the base of the chimney due to wind pressure -

\(Bending\, stress(σ_b)={M\over Z}\)

Where M = Moment generated due to wind pressure

Z = Section modulus of chimney

Let the maximum stress at the base of the chimney is σ_max and the minimum stress at the base of the chimney be σ_min.

\(\sigma_{max}=\sigma_{d}+\sigma_{b}\)

\(\sigma_{max}=\sigma_{d}-\sigma_{b}\)

(b) wind pressure is P acting on the right side of the chimney as seen on the paper:

Hence option (3) is not an example of a typical stress distribution diagram at the base of a circular chimney.

Explanation-

Concept-

• M = E x \(\frac{I}{y}\)
• Where M = Moment of resistance, E = Modulus of elasticity, R = Radius of curvature
• I = Distance between center and the extreme fiber
• So from the above expression , It is observed that, Moment of resistance is directly proportional to the moment of inertia.

Given data and Calculation-

Moment of inertia about NA of square section \(I_{1}\)= b⁴/12
y₁ = b/2

Moment of inertia about NA of square section with horizontal diagonal,

\(I_{2} = 2 \times [{{\sqrt2 b\times ({b\over \sqrt2})^3} \over 12}]\)

\(I_{1}= {b^4 \over 12}\)

y₂ = \(b\over{\sqrt2}\)

Ratio of;

\({Z_1 \over Z_2} =\frac {\frac{I_1}{y_1}}{\frac{I_2}{y_2}} = \frac{\frac{b^4 \over 12}{b\over2}}{\frac{b^4 \over 12}{b\over{\sqrt2}}} = \sqrt2\)

Explanation:

Nature of Bending stress:

(i) Cantilever beam: When a cantilever beam is subjected to a hogging bending moment. Then cross-section above the neutral axis (NA) is subjected to tensile stresses and the cross-section below the neutral axis (NA) is subjected to compressive stress.

(ii) Simply supported beam: When simply supported beam is subjected to sagging bending moment, cross-section above NA is subjected to compressive stresses, and cross-section below the neutral axis is subjected to tensile stress. 

Explanation:

Assumptions in the theory of bending:

• The material of the beam is isotropic and homogeneous and follows Hook’s law.
• The stress-induced is proportional to the strain and the stress at any point does not exceed the elastic limit.
• Transverse sections of the beam that were plane before bending remains plane even after bending.
• The beam is initially straight and has a uniform cross-section.
• The modulus of elasticity is the same for the fibers of the beam under tension or compression.
• The beam is subjected to pure bending and therefore bends in an arc of a circle.
• The radius of curvature is large compared to the dimensions of the section.
• There is no resultant pull or push on the cross-section of the beam.
• The loads are applied in the plane of bending.
• The transverse section of the beam is symmetrical about a line passing through the center of gravity in the plane of bending.

Explanation:

Curved beams:

A curved beam is defined as a beam in which the neutral axis in unloaded condition is curved instead of straight.

The distribution of stresses in curved beams is shown in the figure.

There are two factors which distinguish the analysis of straight and curved beams. They are as follows:

• The neutral and centroidal axis of the straight beam is coincident. However, in curved beams, the neutral axis is shifted towards centre of curvature.
• The bending stress is straight beams varies linearly with the distance from the neutral axis. However, in curved beams the stress distribution is hyperbolic.

\(\sigma _{\theta}=-\frac{My}{Ae(R\;-\;y)}\)

where M = bending moment, y = distance of fibre from neutral axis, A = area of cross-section, e = R - R_N, R = radius of centroidal axis and R_N = radius of neutral axis.

Explanation:

For solid circular cross-section
\({{\rm{τ }}_{{\rm{max}}}} = \frac{4}{3}{{\rm{τ }}_{{\rm{avg}}}} \)

where τ_max = maximum shear stress, τ_avg = average shear stress

And shear stress is given by:

τ_avg = \(\frac{F}{\frac{{\rm{\pi }}}{4}{{\rm{d}}^2}}\)

\({{\rm{τ }}_{{\rm{max}}}} = \frac{4}{3}{{\rm{τ }}_{{\rm{avg}}}} = \frac{4}{3} \times \frac{{\rm{F}}}{{\frac{{\rm{\pi }}}{4}{{\rm{d}}^2}}} = \frac{{16{\rm{F}}}}{{3{\rm{\pi }}{{\rm{d}}^2}}}\)

Explanation:

Bending Equation:

\(\frac{σ }{y} = \frac{M}{I} = \frac{E}{R}\)

\(σ = \frac{M}{I}y\)

i.e. σ ∝ y.

The variation of stress in the simple bending of beam is linear. y is the distance from neutral axis. The bending stress varies linearly from zero at the neutral axis to a maximum at the outer surface.

Concept:

Shear stress of beam for rectangular cross-section,

\(\tau = \frac{3}{2} \times \frac{{{V_{\max }}}}{A}\)

Where, V_max = Maximum shear force

Bending stress is given by,

 \(\sigma = \frac{M}{Z}\)

Where, M = maximum Bending moment, Z = Section modulus

Calculation:

From shear force diagram 

Maximum shear force, V_max = P

So, Maximum shear stress,  \(\tau = \frac{3}{2} \times \frac{{{V_{\max }}}}{A}\) 

⇒ \(\tau = \frac{3}{2} \times \frac{{{P}}}{bd}\)        ---(i)

From bending moment diagram,

Maximum bending moment, M = Pa

So, Bensing stress, \(\sigma = \frac{M}{Z}\)

⇒ \(\sigma = \frac{{Pa}}{{\frac{{b{d^2}}}{6}}} = \frac{{6Pa}}{{b{d^2}}}\)       ---(ii)

Ratio of maximum shear stress to maximum bending stress

\(\frac{\tau }{\sigma } = \frac{{\frac{3}{2} \times \frac{P}{{bd}}}}{{\frac{{6Pa}}{{b{d^2}}}}}\)

⇒ \(\frac{\tau }{\sigma } = \frac{d}{{4a}}\)

Explanation:

Nature of Bending stress:

(i) Cantilever beam: When a cantilver beam is subjected to a hogging bending moment. Then cross-section above neutral axis (NA) is subjected to tensile stresses and cross-section below neutral axis (NA) is subjected to compressive stress.

(ii) Simply supported beam: When simply supported beam is subjected to sagging bending moment, cross-section above NA is subjected to compressive stresses and cross-section below neutral axis is subjected to tensile stress. 

From above two concept it is clear that:

(i) The bending stress is zero at neutral axis

(ii) The bending stress in the rectangular beam section varies linearly.

(iii) When a cantilever beam is loaded vertically downwards at its free end, maximum compressive stress will develop at bottom fibre.

Mistake PointsINCORRECT statement asked in the question.

In some applications the shaft is simultaneously subjected to bending moment M and Torque T.

From the simple bending theory equation:

\(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

If σ_b is the maximum bending stresses due to bending moment M on shaft:

\({\sigma _b} = \frac{{32M}}{{\pi {d^3}}}\)

Torsion equation:

\(\frac{T}{J} = \frac{\tau }{r} = \frac{{G\theta }}{L}\)

The maximum shear stress developed on the surface of the shaft due to twisting moment T:

\(\tau = \frac{{16T}}{{\pi {d^3}}}\)

The ratio of the maximum bending stress to maximum shear stress developed in a shaft is

\(\frac{\sigma }{\tau } = \frac{{\left( {\frac{{32M}}{{\pi {D^3}}}} \right)}}{{\left( {\frac{{16T}}{{\pi {D^3}}}} \right)}} = \frac{{2M}}{T}\)

Important Points

Equivalent Bending Moment:

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

Equivalent Torque:

\({T_e} = \sqrt {{M^2} + {T^2}}\)