## User: Manan Bothra

Manan Bothra •

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- Joined:
- 2 years, 8 months ago
- Email:
- b***********@gmail.com

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#### Posts by Manan Bothra

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... Given
Fractured load = 1600 kN, Effective length = 5200 mm, 4.6 grade 20 mm diameter bolts.
To know: Design batten system, using two channels placing back to back.
For double channels slenderness ratio to be assumed between 40-80.
Therefore, assume allowable compression stress = 190 N/mm$^2$
...

written 8 months ago by
Manan Bothra •

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... Given,
L = 20m, wd = 100 kN/m (factored including self weight)
To know: Design of welded plate girder
**1) Load calculation**
Total factored load on girder = 100 kN/m
Maximum bending moment = $\frac{100 \times 20^2}{8}$ = 5000 kN.m
Maximum shear force = $ \frac{100 \times 20}{2} $ = 1000 kN
* ...

written 8 months ago by
Manan Bothra •

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... Given,
L = 25m, w = 60 kN/m including self-weight
To know, Design of welded plate girder.
**1) Load calculation**
Total factored load on girder = 60 kN/m
Maximum bending moment = $\frac{60 \times 25^2}{8}$ = 4687.5 kN.m
Maximum shear force = $ \frac{60 \times 25}{2} $ = 750 kN
**2) Design of ...

written 8 months ago by
Manan Bothra •

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... Given,
L = 25m
w = 100 kN/m (Superimposed)
**1) Load calculation**
Total factored load on girder = 1.5x100 = 150 kN/m
Maximum bending moment = $\frac{150 \times 25^2}{8}$ = 11,718.75 kN.m
Maximum shear force = $ \frac{150 \times 25}{2} $ = 1875 kN
**2) Design of web**
Section classification
...

written 8 months ago by
Manan Bothra •

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... Given:
L = 12 m, DL = 20 kN/m (excluding self wt), LL = 20 kN/m, 2 point load of 600 kN at 4m from each support, D = 1500mm restricted.
To know: Design of welded plate girder with end bearing stiffeners.
1) Load calculations:
Assume self wt of girder/m = $ \frac{Total \,\, load}{300} = \frac{(20 ...

written 8 months ago by
Manan Bothra •

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...
$ f(x) = \frac{x^2+x+2}{x^4+10x^2+9} \\
f(z) = \frac{z^2+z+2}{z^4+10z^2+9}
$
For poles $ z^4 + 10z^2 + 9 = 0 $
$
z^4 + 9z^2 + z^2 + 9 = 0 \\
z^2(z^9+9) + 1(z^2+9) = 0 \\
(z^2+1)(z^2+9) = 0 \\
\therefore z^2 = -1 \hspace{0.20cm} \& \hspace{0.20cm} z^2 = -9 \\
z = \pm i \hspace{0.20cm} \& ...

written 8 months ago by
Manan Bothra •

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...
$
f(x) = \frac{1}{(x^2+1)^2} \\
f(z) = \frac{1}{(z^2+1)^2}
$
For poles $ z^2 = -1 \implies z = \pm i $
$
f(z) = \frac{1}{[(z-i)(z+i)]^2} = \frac{1}{(z-i)^2(z+i)^2}
$
z = i and z = -i are the two poles of which z = i is the only pole which lies inside and if of order 2.
Residue of f(z) at z = ...

written 8 months ago by
Manan Bothra •

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...
$
f(x) = \frac{1}{(x^2+1)(x^2+9)} \\
f(z) = \frac{1}{(z^2+1)(z^+9)}
$
Consider,
$
\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+9)} \\
= \int_c \frac{1}{(z^2+1)(z^+9)}
$
Consider, $ f(z) = \frac{1}{(z^2+1)(z^+9)} $
For poles, $ z^2+1=0 \hspace{0.20cm} \& \hspace{0.20cm} z^9 + 9 = 0 \\
\th ...

written 8 months ago by
Manan Bothra •

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... Put $ z = e^{i \theta} \\
cos\theta = \frac{z^2 + 1}{2-z} \\
d\theta = \frac{dz}{iz} $
The equation can be written as,
$
\int_c \frac{1}{3 + 2(z^2+1/2z)} \frac{dz}{iz} \\
= \int_c \frac{z}{3z+z^2+1} \frac{dz}{iz} \\
= \int_c \frac{dz}{i(z^2+3z+1)} \\
= \frac{1}{i}\int_c \frac{dz}{(z^2+3z+1)}
$
Le ...

written 8 months ago by
Manan Bothra •

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... Put $ z = e^{i \theta} \\
cos\theta = \frac{z^2 + 1}{2-z} \\
d\theta = \frac{dz}{iz} $
The equation can be written as,
$
\int_c \frac{1}{3 + 2(z^2+1/2z)} \frac{dz}{iz} \\
= \int_c \frac{z}{3z+z^2+1} \frac{dz}{iz} \\
= \int_c \frac{dz}{i(z^2+3z+1)} \\
= \frac{1}{i}\int_c \frac{dz}{(z^2+3z+1)}
$
Le ...

written 8 months ago by
Manan Bothra •

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