## User: Gyanendra pal

Reputation:
10
Status:
Trusted
Location:
Mumbai
Last seen:
11 months, 4 weeks ago
Joined:
1 year, 1 month ago
Email:
g*************@gmail.com

None
None
None
None

#### Posts by Gyanendra pal

<prev • 125 results • page 1 of 13 • next >
1
308
views
1
... ![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 12 months ago by Gyanendra pal10 • updated 11 months ago by Sanket Shingote ♦♦ 250 1 answer 308 views 1 answer ... **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 12 months ago by Gyanendra pal10 • updated 12 months ago by saxena_archit150 1 answer 479 views 1 answers ... ![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 12 months ago by Gyanendra pal10 • updated 11 months ago by Sanket Shingote ♦♦ 250
1
479
views
1
... 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 12 months ago by Gyanendra pal10 • updated 11 months ago by Sanket Shingote ♦♦ 250
1
260
views
1
... 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] $\tan\beta=\f ... written 12 months ago by Gyanendra pal10 • updated 11 months ago by Sanket Shingote ♦♦ 250 1 answer 260 views 1 answer ... **Subject :** Structural Analysis 1 **Topic :** Unsymmetrical Bending **Difficulty :** High ![enter image description here][1] [1]: https://i.imgur.com/81BfpmJ.jpg ... written 12 months ago by Gyanendra pal10 • updated 12 months ago by saxena_archit150 1 answer 489 views 1 answers ... ![enter image description here][1] 1. When the plane of bending does not coincide or parallel to the plane containing the principal centroidal axis of cross section is called as unsymmetrical bending. It is also known as complex or biaxial bending. 2. In unsymmetrical bending, the direction of t ... written 12 months ago by Gyanendra pal10 • updated 12 months ago by saxena_archit150 1 answer 489 views 1 answer ... **Subject :** Structural Analysis 1 **Topic :** Unsymmetrical Bending **Difficulty :** High ... written 12 months ago by Gyanendra pal10 • updated 12 months ago by saxena_archit150 1 answer 239 views 1 answers ... ![enter image description here][1]$ 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 13 months ago by Gyanendra pal10 • updated 11 months ago by Sanket Shingote ♦♦ 250
1