Math, asked by surabdas1234, 1 month ago

. Two rods, made of plain carbon steel 40C8 (Syt = 380 N/mm2 ), are to be connected by means of a cotter joint. The diameter of each rod is 50 mm and the cotter is made from a steel plate of 15 mm thickness. Calculate the dimensions of the socket end making the following assumptions: (i) the yield strength in compression is twice of the tensile yield strength; and (ii) the yield strength in shear is 50% of the tensile yield strength. The factor of safety is 6.​

Answers

Answered by XDPrEm
3

Step-by-step explanation:

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Answered by VaibhavSR
0

Answer:

63.33 and 30mm

Step-by-step explanation:

Given \quad S_{y t}=380 \mathrm{~N} / \mathrm{mm}^{2} \quad(f s)=6 \quad t=15 \mathrm{~mm} d=50 \mathrm{~mm}

Step I Permissible stresses

\sigma_{e}=\frac{S_{y c}}{(f s)}=\frac{2 S_{y t}}{(f s)}=\frac{2(380)}{6}=126.67 \mathrm{~N} / \mathrm{mm}^{2}

\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(380)}{6}=31.67 \mathrm{~N} / \mathrm{mm}^{2}

\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{380}{6}=63.33 \mathrm{~N} / \mathrm{mm}^{2}

Load acting on rods

P=\frac{\pi}{4} d^{2} \sigma_{t} or P=\frac{\pi}{4}(50)^{2}(63.33)

=124348.16 \mathrm{~N}

Inside diameter of socket \left(d_{2}\right)

From Eq. ,

P= {\left[\frac{\pi}{4} d_{2}^{2}-d_{2} t\right] \sigma_{t} }

124348.16=\left[\frac{\pi}{4} d_{2}^{2}-d_{2}(15)\right](63.33)

or d_{2}^{2}-19.1 d_{2}-2500=0

Solving the above quadratic equation,

d_{2}=\frac{19.1 \pm \sqrt{19.1^{2}-4(-2500)}}{2}

\therefore \quad d_{2}=60.45

or 65 \mathrm{~mm}

 Outside diameter of socket

P=\left[\frac{\pi}{4}\left(d_{1}^{2}-d_{2}^{2}\right)-\left(d_{1}-d_{2}\right) t\right] \sigma_{1} 124.348 .16= {\left[\frac{\pi}{4}\left(d_{1}^{2}-65^{2}\right)-\left(d_{1}-65\right)(15)\right](63.33) }  \quad d_{1}^{2}-19.1 d_{1}-5483.59=0

From Eq. .

ord_{1}^{2}-19.1 d_{1}-5483.59=0

Solving the above quadratic equation,,

Dimensions a and

From Eq

From Eq. (4.25 \mathrm{~g}),

c=\frac{P}{2\left(d_{4}-d_{2}\right) \tau} =\frac{124348.16}{2(135-65)(31.67)}

=28.04 \text { or } 30 \mathrm{~mm}

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