## User: Sanket Shingote

Sanket Shingote ♦♦

**180**- Reputation:
**180**- Status:
- Trusted
- Location:
- Dombivli
- Website:
- http://www.kt280.com/
- Twitter:
- sanket1893
- Last seen:
- 5 days, 18 hours ago
- Joined:
- 1 year, 3 months ago
- Email:
- s***************@gmail.com

*"Fighting till the end"* nature makes me perfect team mate.

I am a Textile Technologist from Mumbai's renowned university ICT (Institute of Chemical Technology)

I am currently working with KT280 as an analyst. I have keen interest in branding startups and social media marketing.

I am passionately working on Community Growth on Ques10. Pleased to meet you!

#### Academic profile

#### Posts by Sanket Shingote

<prev
• 162 results •
page 1 of 17 •
next >

0

votes

0

answers

88

views

0

answers

... New-age employers are going all out to keep people engaged and happy in their work spaces. Here are five popular ways that companies are resorting to keep their employees happy.
#1. Professional introduction
Every interviewer asks for an introduction. A candidate should be prepared with the best ...

written 4 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

128

views

1

answers

...
Quadratic element with nodes 3.
$\hspace{1cm}x=\phi, x_1+\phi_2x_2+ \phi_3 x_3 $
where, $\phi,=\frac{1}{2}\xi(\xi-1), \phi_2=\frac{1}{2}\xi (\xi+1),\phi=(1-\xi)(1+\xi)$
![enter image description here][1]
$x=\frac{1}{2}\xi(\xi-1)x_1+\frac{1}{2}\xi(\xi+1)x_2+(1-\xi)( ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

128

views

1

answer

... Coordinates of the nodes of finite element are given by P (4.0) and Q (5.0) Find the expression of x in terms of $\xi$ when :
1) Third node R is taken at (6.0)
2) Third node R is taken at (5.0)
Comment on the result.
**Mumbai University > Mechanical Engineering > Sem 6 > Finite Elemen ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

147

views

1

answers

...
![enter image description here][1]
$L\rightarrow he \rightarrow $ length of element.
$E \rightarrow$Modular of elasticity.
$S\rightarrow $density of material
$A \rightarrow$ cross sectional area .
(i) Natural freoquencies $w_i$ using consistent mass matrix .
The element matrix is given ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

147

views

1

answer

... Consider a uniform cross section bar of length L made up of a material whose Yong's modulus and density are given by E and $\rho$.Estimate the natural frequencies of axial vibration of the bar using both consistent and lumped mass matrices.
![enter image description here][1]
**Mumbai University ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

87

views

1

answers

...
$\hspace{1cm} x_1= 10 mm\hspace{1cm} y_1= 10 mm\\
\hspace{1cm} x_2= 70 mm\hspace{1cm} y_2= 35 mm\\
\hspace{1cm} x_3= 75 mm\hspace{1cm} y_1= 25 mm$
$\beta_1=y_2-y_3 = 10\hspace{2.5cm} r_1= x_3-x_2=5\\
\beta_2=y_3-y_1 = 15\hspace{2.5cm} r_2= x_1-x_3=65\\
\beta_1=y_1-y_2 = 25\hspace{2.5cm} r_3= x_2- ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

87

views

1

answer

... A CST element has nodal coordinates (10,10) , (70, 35) for nodes 1,2 and 3 respectively. The element is 2 mm thick and is of material with properties E=70 GPA. Poisson's ratio os 0.3. After applying the load to the element the nodal deformation were found to be $u_1=0.01mm, \ \ v_1=-0.04mm \ \ u_2= ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

111

views

1

answers

... ![enter image description here][1]
$k_1 =\frac{AE}{L}$
$\begin{bmatrix}
\ 1 & -1 \\
\ -1 & 1 \\
\end{bmatrix}= \frac{200\times200\times10^3}{300}$
$\begin{bmatrix}
\ 1 & -1 \\
\ -1 & 1 \\
\end{bmatrix}=10^3$
$\begin{bmatrix}
\ 166.67 & -166.67 \\
\ -166.67 & 166.67 \\
\end{ ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

111

views

1

answer

... ![enter image description here][1]
**Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis**
**Marks:** 10M
**Year:** Dac 2016
[1]: http://i.imgur.com/QJkZ1Jm.png ...

written 5 months ago by
Sanket Shingote ♦♦

**180**0

votes

1

answer

110

views

1

answers

... ![enter image description here][1]
$\hspace{1cm}E=200 GPa$
$\hspace{1.3cm}= 200\times 10^3 \frac{n}{mw^2}$
| Ele No. | Nodes | L | A | 0 | c | s | cs | $c^2$ | $s^2$ | $\frac{AE}{L}:$ |
|----------|-------|-------|-----|------|--------|-------|------|-----|-----|---- ...

written 5 months ago by
Sanket Shingote ♦♦

**180**#### Latest awards to Sanket Shingote

Centurion
6 months ago,
created 100 posts.

Popular Question
7 months ago,
created a question with more than 1,000 views.
For Get Started With Ques10

Supporter
11 months ago,
voted at least 25 times.

Rising Star
12 months ago,
created 50 posts within first three months of joining.

Autobiographer
13 months ago,
has more than 80 characters in the information field of the user's profile.

Site

- Publications
- Advertise
- RSS
- Stats

Use of this site constitutes acceptance of our User
Agreement
and Privacy
Policy.