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## User: Gyanendra pal

Reputation:
10
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Trusted
Location:
Mumbai
Last seen:
5 months, 3 weeks ago
Joined:
7 months, 3 weeks ago
Email:
g*************@gmail.com

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#### Posts by Gyanendra pal

<prev • 125 results • page 1 of 13 • next >
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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 6 months ago by Gyanendra pal10 • updated 5 months ago by Sanket Shingote ♦♦ 220 1 answer 185 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 6 months ago by Gyanendra pal10 • updated 6 months ago by saxena_archit150 1 answer 330 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 6 months ago by Gyanendra pal10 • updated 5 months ago by Sanket Shingote ♦♦ 220
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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 6 months ago by Gyanendra pal10 • updated 5 months ago by Sanket Shingote ♦♦ 220
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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] $\tan\beta=\f ... written 6 months ago by Gyanendra pal10 • updated 5 months ago by Sanket Shingote ♦♦ 220 1 answer 139 views 1 answer ... **Subject :** Structural Analysis 1 **Topic :** Unsymmetrical Bending **Difficulty :** High ![enter image description here][1] [1]: https://i.imgur.com/81BfpmJ.jpg ... written 6 months ago by Gyanendra pal10 • updated 6 months ago by saxena_archit150 1 answer 241 views 1 answers ... ![enter image description here][1] 1. When the plane of bending does not coincide or parallel to the plane containing the principal centroidal axis of cross section is called as unsymmetrical bending. It is also known as complex or biaxial bending. 2. In unsymmetrical bending, the direction of t ... written 6 months ago by Gyanendra pal10 • updated 6 months ago by saxena_archit150 1 answer 241 views 1 answer ... **Subject :** Structural Analysis 1 **Topic :** Unsymmetrical Bending **Difficulty :** High ... written 6 months ago by Gyanendra pal10 • updated 6 months ago by saxena_archit150 1 answer 139 views 1 answers ... ![enter image description here][1]$ 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 6 months ago by Gyanendra pal10 • updated 5 months ago by Sanket Shingote ♦♦ 220
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