User: manasahegde234

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Posts by manasahegde234

<prev • 203 results • page 1 of 21 • next >
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Answer: A: Draw the shear stress distribution for various section
... ![enter image description here][1] [1]: https://i.imgur.com/s6hxOD7.jpg ...
written 3 months ago by manasahegde23420
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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 manasahegde23420
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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 manasahegde23420
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Answer: A: Draw the shear stress distribution diagram for the following fig.
... 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 manasahegde23420
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Answer: A: Derive an expression for strain energy in a member subjected to an axial load
... 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 manasahegde23420
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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 manasahegde23420
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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 manasahegde23420
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Answer: A: Problem on Torsion for finding internal and external diameter of shaft.
... 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 manasahegde23420 • updated 3 months ago by Abhishek Tiwari ♦♦ 30
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Answer: A: Problem on strain energy.for gradually applied load.
... 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 manasahegde23420
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Answer: A: Problem on Shear Stress T-Section
... ![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 manasahegde23420

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