Effect on Shaft MCQ Quiz in मराठी - Objective Question with Answer for Effect on Shaft - मोफत PDF डाउनलोड करा

Explanation:

Following are the assumptions made in the theory of Simple Bending:

• The material of the beam is homogenous and isotropic.
• The beam is initially straight, and all the longitudinal fibres bend in circular arcs with a common centre of curvature.
• Members have symmetric cross-sections and are subjected to bending in the plane of symmetry.
• The beam is subjected to pure bending and the effect of shear is neglected.
• Plane sections through a beam, taken normal to the axis of the beam remain plane after the beam is subjected to bending.
• The radius of curvature is large as compared to the dimensions of the beam.

Explanation:

Concept:

The power transmission by circular shaft:

• The main purpose of a shaft is to transmit power from the source to driver machines in factories and workshops.
• Power is the time rate of doing work. 
• Consider a shaft of diameter d, subjected to a torque T, while the shaft is rotating at N rpm or ω rad/sec, subjected to end couples which cause turning effect.
• Power transmitted by circular shaft is:   \(P\,=\,\frac{2\pi NT}{60}\)

• Normally, the torque is expressed in units of Nm and power in units of Watt (W) where 1 W is equal to 1 J of work per Second.

Solution:

Given, diameter of solid circular shaft

d = 200 mm , N = 120 rpm , torque on the shaft (T) = 80000 Nm.

∴  Power transmitted by circular shaft is:

 \(P\,=\,\frac{2\pi NT}{60}=\,\frac{2\times\pi\times 120\times 80000}{60} \,=\, 320000\pi \,\,Watt=\,320\pi \) kW

Explanation:

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

Torque per radian twist is known as torsional stiffness (k)

\(k=\frac{T}{\theta}=\frac{GI_p}{L}\)

The parameter GI_p is called torsional rigidity of the shaft. 

Torsional rigidity is also defined as torque per unit angular twist per unit length

\(GI_p=\frac{T}{\theta/L}\)

Concept:

Power in a rotating shaft can be calculated from the equation-

P = ω × T

where \(\omega=\frac{2\pi N}{60}\)

The relation between shear stress and Torque acting in a shaft can be found by the Torsional equation.

\(\frac{{\rm{T}}}{{\rm{J}}} = \frac{{{{\rm{τ }}_{{\rm{}}}}}}{{\rm{r}}} = \frac{{{\rm{Gθ }}}}{{\rm{L}}}\)

Calculation:

Given:

d = 2 m, P = 120 W and N = 180 rpm.

We know that

P = ω × T

\(P=\frac{2\pi NT}{60}\)

\(120=\frac{2\pi\ \times\;180\;\times \;T}{60}\)

\(\therefore T=\frac{20}{\pi}\;Nm\)

Using torsional equation

\(\frac{{\rm{T}}}{{\rm{J}}} = \frac{{{{\rm{τ }}_{{\rm{}}}}}}{{\rm{r}}} \)

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

\(\therefore\tau=\frac{16\;\times\;\frac{20}{\pi}\;}{\pi\ \times\;2^3}\Rightarrow\frac{40}{\pi^2}\;N/m^2\)

Concept:

\({τ _{max}} = \frac{{16T}}{{\pi {d^3}}}\)  

\(T = {τ _{max}} × \frac{{\pi {d^3}}}{{16}}\)

∴ T ∝ d³ 

P = ωT

∴ P ∝ ω,

∴ P ∝ T i.e P ∝ d³

where, τ_max = Max shear stress developed, T = Torque, d = Diameter of shaft, P = Power transmitted, ω  = Angular velocity 

Calculation:

Given:

P₁ = 90 kW, d₂ = 2d₁ , ω₂ = 0.5ω₁

Power P ∝ ωd³

\(\frac{{P_1}}{{P_2}}=\frac{{ω_1~\times~d_1^3 }}{{ω_2~\times~d_2^3}}=\frac{{90}}{{P_2}}=\frac{{ω~\times~d^3}}{{0.5ω~\times~8d^3}}\)

∴ P₂ = 360 kW.

Concept:

Bending stress can be calculated by

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

where σ_b = Bending stress, E = Young's modulus of elasticity, I = Moment of inertia of the cross-section, R = Radius of curvature, y = Distance of outer fiber from the neutral axis

Calculation:

Given:

diameter of rod (d) = 10 mm, Curvature diameter (D) = 2000 mm, E = 2 × 10⁵ N/mm²

Radius of curvature, \(R=\frac{D}{2}\) = 1000 mm

\(y = \frac{d}{2}\) = 5 mm

\(σ_b = \frac{y~\times~E }{R}\)

\(σ_b = \frac{5~\times~2~\times~10^5 }{1000}\)

σ_b = 1000 N/mm²

Concept:

Torsion Equation of a shaft is given by,

\(\frac{T}{J} = \frac{τ }{r} = \frac{{Gθ }}{L}\)

Where, T = Torque, J = Polar moment of inertia, fs or τ = Shear stress, r = Radius of shaft, G = Shear modulus, θ = Angle of twist and L = Length of shaft

Polar moment of inertia for a circular shaft of diameter d is \(\frac{{\pi {d^4}}}{{32}}\)

Polar modulus is defined as the ratio of the polar moment of inertia to the radius of the shaft. It is also called as torsional section modulus. It is denoted by Zp.

Polar section modulus of shaft is given by,

\({Z_p} = \frac{J}{r} = \frac{{\frac{{\pi {D^4}}}{{32}}}}{{\frac{D}{2}}} = \frac{{\pi {D^3}}}{{16}}\)

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

\(τ_{max} = \frac{{T}}{{Z_p}} = \frac{{16T}}{{\pi {D^3}}}\)

∴ τmax ∝ 1/D3

Concept:

The torsional equation for the shaft is given by,

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

Torque per radian twist over the length is known as torsional stiffness (k)

\(k=\frac{T}{\theta}=\frac{GJ}{L}\)

Here,

The parameter GJ is called the torsional rigidity of the shaft. 

Torsional rigidity is also defined as torque per unit angular twist over the length of the shaft

\(GJ=\frac{T}{\theta/L}\)

Explanation:

Twisting moment impart an angular displacement of one end cross-section with respect to the other end and it will setup shear stresses on any cross section of the bar perpendicular to its axis.

Torsion Equation of a shaft is given by,

\(\frac{T}{J} = \frac{τ }{r} = \frac{{Gθ }}{L}\)

Polar section modulus of shaft is given by,

\({Z_p} = \frac{J}{r} = \frac{{\frac{{\pi {D^4}}}{{32}}}}{{\frac{D}{2}}} = \frac{{\pi {D^3}}}{{16}}\)

Where, T = Torque, J = Polar moment of inertia, τ = Shear stress, r = Radius of shaft, G = Shear modulus, θ = Angle of twist and L = Length of shaft

Shear stress distribution across the section of the circular shaft is shown below.

Using Torsinal formula

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

Forgiven T and J; τ ∝ r i.e.

The shear stress distribution is linear.

Shear stress at centre, τcentre = 0

Concept:

Power (P)

P = T × W

Torsion equation

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

Calculation:

Given:

P = 40 kW, N = 500 rpm, D = 40 mm

As, P = T × W

\(40 \times {10^3} = {\rm{T}} \times \frac{{2{\rm{\pi }} \times 500}}{{60}} ⇒ {\rm{T}} = 763.9{\rm{\;Nm}} = 763.9 \times {10^3}{\rm{Nmm}}\)

Now,

\(\frac{T}{J} = \frac{\tau }{R} ⇒ \tau = \frac{{TR}}{J} = \frac{{763.9 \times {{10}^3} \times 20 \times 32}}{{\pi \times {{40}^4}}}\)