A 25 tooth spur gear having a form factor of 0-402- module of 2 mm face width of 45 mm and pressure angle of 25- to operate at 900 rpm- with - yt-172 MPa and velocity factor- kf-1-5- the allowable bending load is

Given data: y = 0.402; σ _yt=172 MPa d = m N = 2 ×25 = 50 mm V=π dN=π× 0.05× 15=2.356m/s=463.77 pm we know 1p=m F_b=σ _ytbymc_v =11.5(172× 45× 0.402× 2) ...

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Standard VIII

Physics

Normal Force

A 25 tooth sp...

Question

A 25 tooth spur gear having a form factor of 0.402, module of 2 mm face width of 45 mm and pressure angle of $25^{\circ}$ to operate at 900 rpm. with $σ_{y t} = 172 M P a$ and velocity factor, $k_{f} = 1.5,$ the allowable bending load is

9.334 kN

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4.149 kN

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2.149 kN

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6.222 kN

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Solution

The correct option is B 4.149 kN
Given data:
y = 0.402;

$σ_{y t} = 172 M P a$

d = m N = 2 $\times$ 25 = 50 mm

$V = π d N = π \times 0.05 \times 15 = 2.356 m / s = 463.77 p m$

we know $\frac{1}{p} = m$

$F_{b} = \frac{σ_{y t} b y m}{c_{v}}$

= $\frac{1}{1.5} (172 \times 45 \times 0.402 \times 2)$

= 4.149 kN

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A 25 tooth spur gear having a form factor of 0.402, module of 2 mm face width of 45 mm and pressure angle of 25∘ to operate at 900 rpm. with σyt=172MPa and velocity factor, kf=1.5, the allowable bending load is

The correct option is B 4.149 kNGiven data: y = 0.402; σyt=172MPa d = m N = 2 ×25 = 50 mm V=πdN=π×0.05×15=2.356m/s=463.77pm we know 1p=m Fb=σytbymcv =11.5(172×45×0.402×2) = 4.149 kN

A 25 tooth spur gear having a form factor of 0.402, module of 2 mm face width of 45 mm and pressure angle of $25^{\circ}$ to operate at 900 rpm. with $σ_{y t} = 172 M P a$ and velocity factor, $k_{f} = 1.5,$ the allowable bending load is