User: manasahegde234

Reputation:
20
Status:
Trusted
Location:
Last seen:
1 month, 2 weeks ago
Joined:
1 year ago
Email:
m*************@gmail.com

Academic profile

None
None
None
None

Posts by manasahegde234

<prev • 206 results • page 1 of 21 • next >
0
votes
2
answers
224
views
2
answers
Answer: A: Draw the shear stress distribution for various section
... ![enter image description here][1] [1]: https://i.imgur.com/s6hxOD7.jpg ...
written 6 weeks ago by manasahegde23420
0
votes
2
answers
161
views
2
answers
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 6 weeks ago by manasahegde23420
0
votes
2
answers
157
views
2
answers
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 6 weeks ago by manasahegde23420
0
votes
2
answers
146
views
2
answers
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 6 weeks ago by manasahegde23420
0
votes
2
answers
179
views
2
answers
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 6 weeks ago by manasahegde23420
0
votes
2
answers
150
views
2
answers
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 6 weeks ago by manasahegde23420
0
votes
2
answers
113
views
2
answers
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 6 weeks ago by manasahegde23420
0
votes
1
answer
188
views
1
answers
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 6 weeks ago by manasahegde23420 • updated 6 weeks ago by abhishektiwari1712 ♦♦ 0
0
votes
2
answers
101
views
2
answers
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 7 weeks ago by manasahegde23420
0
votes
2
answers
226
views
2
answers
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 7 weeks ago by manasahegde23420

Latest awards to manasahegde234

Popular Question 6 weeks ago, created a question with more than 1,000 views. For Calculate image frequency and rejection ratio.
Popular Question 7 weeks ago, created a question with more than 1,000 views. For Explain TDM and FDM
Popular Question 7 weeks ago, created a question with more than 1,000 views. For Explain TDM and FDM
Popular Question 6 months ago, created a question with more than 1,000 views. For Write short note on vestigial side-band transmission.
Popular Question 6 months ago, created a question with more than 1,000 views. For Derive the expression for frequency modulated wave.
Popular Question 6 months ago, created a question with more than 1,000 views. For Explain TDM and FDM
Centurion 11 months ago, created 100 posts.
Rising Star 11 months ago, created 50 posts within first three months of joining.