## User: manasahegde234

manasahegde234 •

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#### Posts by manasahegde234

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Answer:
A: Assume static conditions.

... ![enter image description here][1]
The areas of three welds are as follows:
$A_1 = (60 t) mm^2 $
$A_2 = (60 t) mm^2 $
$A_3 = (90 t) mm^2$
$A = A_1 + A_2 + A_3 = (210 t mm^2)$
The primary shear stress in the weld is given $\tau=\frac{P}{A}=\frac{50000}{210*t}=\frac{238.1}{t}N/mm^2 (i) $
S ...

written 5 days ago by
manasahegde234 •

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... Step 1: Calculation of permissible tensile stress:
$\sigma_{max}=\frac{S_yt}{fs}=\frac{380}{3.5}=108.57N/mm^2$
Step 2: Calculation of eccentricity(e):
For the cross section XX
$R_N=\frac{\frac{b_i+b_o}{2}h}{\frac{b_ir_o-b_oR_i}{h}*log_e\frac{R_o}{R_i}-(b_i-b_o)}$
$R_N=\frac{\frac{90+30}{2}*120} ...

written 8 days ago by
manasahegde234 •

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... Step 1: Calculation of permissible tensile stress:
$\sigma_{max}=\frac{S_yt}{fs}=\frac{380}{3.5}=108.57N/mm^2$
Step 2: Calculation of eccentricity(e):
For the cross section XX
$R_N=\frac{\frac{b_i+b_o}{2}h}{\frac{b_ir_o-b_oR_i}{h}*log_e\frac{R_o}{R_i}-(b_i-b_o)}$
$R_N=\frac{\frac{90+30}{2}*120} ...

written 8 days ago by
manasahegde234 •

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... Step 1: Calculation of permissible tensile stress:
$\tau_max=\frac{S_ut}{fs}=\frac{200}{3}=66.67N/mm^2$
Step 2: Calculation of eccentricity (e):
b_i=3t
h=3t
R_i=2t
R_o=5t
t_i=t
t=0.75t
$R_N=\frac{t_i(b_i-t)+th}{(b_i-t)log_e(\frac{R_i+t_i}{R_i})+tlog_e\frac{R_o}{R_i}}$
$R_N=\frac{t(3t-0.75 ...

written 22 days ago by
manasahegde234 •

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... Assuming the outer diameter to be 38mm,
$D_o=D+d=Cd+d=d(C+1)$
i.e $d=\frac{D_o}{C+1}=\frac{38}{C+1}$ (a)
We have,
$\tau_max=K\frac{8P_maxD}{\pi d^3}=K\frac{8P_maxC}{\pi d^2}$
Therefore,
$650=K\frac{8*650*c*(c+1)^2}{\pi *38^2}$
$C(C+1)^2K=567.05$ (b)
The problem is solved by trial and error ...

written 22 days ago by
manasahegde234 •

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... ![enter image description here][1]
[1]: https://i.imgur.com/o9vebN9.jpg
Step 1: Permissible shear stress
$\tau=\frac{S_sy}{fs}=\frac{0.5S_yt}{fs}=\frac{0.5*380}{3}=63.33 N/mm^2$
Step 2: Primary and secondary shear forces:
The centre of gravity of three bolts will be at the centre of bolt-2. ...

written 22 days ago by
manasahegde234 •

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... ![enter image description here][1]
[1]: https://i.imgur.com/GdJlsQl.jpg
Step 1. Primary shear force:
$P_1'=P_2'=P_3'=P_4'=\frac{P}{4}=\frac{5*10^3}{4}=1250N$
Step 2 Primary shear force:
$r_1=r_2=r_3=r_4=100mm$
$C=\frac{Pe}{(r_1^2+r_2^2+r_3^2+r_4^2)}=\frac{(5*10^3)(200)}{4(100)^2}=25$
There ...

written 23 days ago by
manasahegde234 •

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... 1. Draw a neat sketch of knuckle joint:
![enter image description here][1]
[1]: https://i.imgur.com/nrajR6C.jpg
2. FInd the diameter of the rod(d):
$d=\sqrt \frac{4P}{\sigma_t \pi}=\sqrt \frac{4(150*10^3)}{\pi *75}=50.462 or 52mm$
3. Using empirical find all dimensions:
Diameter of knuckle ...

written 23 days ago by
manasahegde234 •

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... Step 1: Selection of material:
The rods are subjected to tensile force and strength is the criterion for the selection of rod material. On the basis of strength, the material of the two rods and the cotter is selected as plain carbon steel of Grade 30C8 (Syt=400MPa)
Step 2: Selection of factor of ...

written 23 days ago by
manasahegde234 •

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... The efficiency of a square threaded screw is calculated by,
$\eta=\frac{tan\alpha}{tan(\phi+\alpha} $ (a)
For a self-locking screw,
$\phi \geq \alpha$
Substituting the limiting value for $ \phi=\alpha $ in eqn (a)
$\eta \leq\ \frac{tan\alpha}{tan(\phi+\phi}$
$\eta \leq\ \frac{tan\alpha}{tan2 ...

written 24 days ago by
manasahegde234 •

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created 50 posts within first three months of joining.

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