## User: Gyanendra pal

Gyanendra pal •

**10**- Reputation:
**10**- Status:
- Trusted
- Location:
- Mumbai
- Last seen:
- 8 months, 2 weeks ago
- Joined:
- 10 months, 3 weeks ago
- Email:
- g*************@gmail.com

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#### Posts by Gyanendra pal

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... ![enter image description here][1]
Maximum B.M
$= \frac{WL^2}{2}=\frac{12\times3^2}{2}=\underline{54\ kNm}$
$M=54\ kNm\\ \theta=35^\circ\\ b=300\ mm\\ d=600\ mm $
Location of N.A
$ I_{XX}=I_{UU}=\frac{bd^3}{12}=\frac{300\times600^3}{12}=5.4\times10^9mm^4\\ I_{YY}=I_{VV}=\frac{db^3}{12}=\frac ...

written 9 months ago by
Gyanendra pal •

**10**• updated 8 months ago by Sanket Shingote ♦♦**250**0

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... **Question** (contd.) plane of loadingis inclined at $35^\circ$ with minor axis in clockwise direction. Also locate the neutral axis position.
**Subject :** Structural Analysis 1
**Topic :** Unsymmetrical Bending
**Difficulty :** Medium ...

written 9 months ago by
Gyanendra pal •

**10**• updated 9 months ago by saxena_archit •**150**0

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... ![enter image description here][1]
Maximum B.M = $\frac{WL^2}{8}=\frac{10\times6^2}{8}={45\ kN}$
$ \underline{M=45\times10^6\ Nmm}\\ \theta=30^\circ$
**Location of N.A**
$I_{XX}=I_{UU}=\frac{bd^3}{12}=\frac{80\times120^3}{12}=11.52\times10^6\ mm^4\\ I_{YY}=I_{VV}=\frac{db^3}{12}=\frac{120\time ...

written 9 months ago by
Gyanendra pal •

**10**• updated 8 months ago by Sanket Shingote ♦♦**250**0

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... If the span of beam is 6m, locate the neutral axis and hence find the stresses at each corners of the beam.
---
**Subject :** Structural Analysis 1
**Topic :** Unsymmetrical Bending
**Difficulty :** HIgh ...

written 9 months ago by
Gyanendra pal •

**10**• updated 8 months ago by Sanket Shingote ♦♦**250**0

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... Here, $\alpha=30^\circ$
$ \therefore \sin\alpha=0.5\ \ \ \ \ \ \ \cos\alpha=0.866$
The momentsof inertia of th section are
$I_{XX}=\frac{100\times150^3}{12}=\underline{28125000\ mm^4}\\ I_{YY}=\frac{150\times100^3}{12}=\underline{12500000\ mm^4} $
![enter image description here][1]
$\tan\beta=\f ...

written 9 months ago by
Gyanendra pal •

**10**• updated 8 months ago by Sanket Shingote ♦♦**250**0

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... **Subject :** Structural Analysis 1
**Topic :** Unsymmetrical Bending
**Difficulty :** High
![enter image description here][1]
[1]: https://i.imgur.com/81BfpmJ.jpg ...

written 9 months ago by
Gyanendra pal •

**10**• updated 9 months ago by saxena_archit •**150**0

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Answer:
A: Explain unsymmetrical bending

...
![enter image description here][1]
1. When the plane of bending does not coincide or parallel to the plane containing the principal centroidal axis of cross section is called as unsymmetrical bending. It is also known as complex or biaxial bending.
2. In unsymmetrical bending, the direction of t ...

written 9 months ago by
Gyanendra pal •

**10**• updated 9 months ago by saxena_archit •**150**0

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... **Subject :** Structural Analysis 1
**Topic :** Unsymmetrical Bending
**Difficulty :** High
...

written 9 months ago by
Gyanendra pal •

**10**• updated 9 months ago by saxena_archit •**150**0

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... ![enter image description here][1]
$ I=I_{XX_1}-I_{XX_2}=\left[ \frac{100\times400^3}{12}\right]-\left[ \frac{(100\times20)(400-2\times20)}{12}\right]\\ I=(533.33\times10^6)-(311.04\times10^6)\\ \underline{I=222.3\times10^6mm^4}$
**Distance of shear centre (e)**
$e=\frac{h^2b^2t}{4I}=\frac{400^ ...

written 9 months ago by
Gyanendra pal •

**10**• updated 8 months ago by Sanket Shingote ♦♦**250**0

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... **Subject :** Structural Analysis 1
**Topic :** Shear Centre
**Difficulty :** High
![enter image description here][1]
[1]: https://i.imgur.com/zY9KOP6.jpg ...

written 9 months ago by
Gyanendra pal •

**10**• updated 9 months ago by saxena_archit •**150**#### Latest awards to Gyanendra pal

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