## User: manasahegde234

manasahegde234 •

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#### Posts by manasahegde234

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... ![enter image description here][1]
[1]: https://i.imgur.com/s6hxOD7.jpg ...

written 3 months ago by
manasahegde234 •

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Answer:
A: Problem on bending stress

... ![enter image description here][1]
1. Calculation of $\bar y$ and $I_{NA}$:
$\bar y_{top}=\bar y_t=\frac{120*20*\frac{20}{2}+200*20*(20+\frac{200}{2})+20*120*(20+200+\frac{20}{2})}{120*20+200*20+120*20}$
$y_t=120mm$
$y_t= y_{bottom}=120mm$ (symmetry about x-axis}
$I_{N-A}=I_{big rectangle}-I_{s ...

written 3 months ago by
manasahegde234 •

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Answer:
A: Problem on shear stress.

... ![enter image description here][1]
Let the safe point load at mid span i.e \frac{2}{2}=1m be W (N)
Now, moment of inertia of section about neutral axis
$I=\frac{100*(50+50+50)^3}{12}=2.8125*10^7 mm^4$
Maximum shear force F=\frac{W}{2}
Now, shear stress at the glue joint=$q_{1-1}=\frac{F a bar ...

written 3 months ago by
manasahegde234 •

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... Note
1.For commonly used section like Rectangle. Circle, T-Section, Channel Section, I-Section, Shear Stress is Maximum at Neutral Axis
2.If the plane or section eg: 1-1, at which Shear Stress is required, is above N.A., we have to consider the area above the plan for calculating $A\bar y$
$q_{av ...

written 3 months ago by
manasahegde234 •

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... Application of external loads to a member causes deformation of the member but the member has a natural tendency to oppose the deformation and in doing so it develops internal stresses. These internal stresses have the capacity to do work and as such, the member has energy stored in it. Thus work do ...

written 3 months ago by
manasahegde234 •

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Answer:
A: Problem on bending stress

... 1. Moment of inertia:
$(I_{NA})_I=\frac{bd^3}{12}=\frac{10*10^3}{12}=833.33mm^4$
When depth is mode twice the width,
![enter image description here][1]
$(I_{NA})_II=\frac{bd^3}{12}=\frac{10*20^3}{12}=6666.67$
Section of modulus=Z=$\frac{I_{NA}}{y_{max}}$
$Z\propto I_{NA}$
When $I_{NA}$ increa ...

written 3 months ago by
manasahegde234 •

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Answer:
A: Problem on Shear Stress

... ![enter image description here][1]
Let F be the shear force in N
Average shear stress $q_{avg}=\frac{total shear force}{total area}$
$q_{avg}=\frac{F}{b*d}$
Now, shear stress at (1)-(1) section is $q_{1-1}$,
$q_{1-1}=\frac{F}{Ib}*A*\bar y$
$=\frac{F*(b*(\frac{h}{2}-y))*[y+\frac{1}{2}(\frac{h}{ ...

written 3 months ago by
manasahegde234 •

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... Note:
If Power formula used for shaft then Torque will be Mean Torque i.e Tavg.
In Torsional Equation, Torque is Maximum Torque i.e Tmax.
For Hollow shaft in torsional equation radius will be half of the externl diameter.
Data: A hollow shaft of diameter ratio $\frac{3}{8}$
i.e $di=\frac{3}{8} ...

written 3 months ago by
manasahegde234 •

**20**• updated 3 months ago by Abhishek Tiwari ♦♦**30**0

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... Data: L=2000mm, b=500mm, d=40mm
P=600KN=$600*10^3 N$
E=200GPa=$200*10^3 N/mm^2$
$p=\frac{P}{A}=\frac{600*10^3}{50*40}=300N/mm^2$
Strain energy =$\frac{p^2}{2E}*AL$
$u=\frac{300^2}{2*200*10^3}*(50*40)*2000$
$u=900*10^3 Nmm$ ...

written 3 months ago by
manasahegde234 •

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... ![enter image description here][1]
Shear force $F=60*10^3 N$
**1. Calculation of moment of inertia and** $\bar y_{base}$
$\bar y_{base}$ = $\frac{20*100*\frac{100}{2}+20*120*(100+\frac{20}{2})}{20*100+20*120}$
= 82.72mm
$A =20*100+20*120=4400mm$
$I_{base}=\frac{20*100^3}{3}+\frac{120*20^3}{12} ...

written 3 months ago by
manasahegde234 •

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