## User: Manan Bothra

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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 5 months ago by Manan Bothra0
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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 5 months ago by Manan Bothra0
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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 5 months ago by Manan Bothra0
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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 5 months ago by Manan Bothra0
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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 = $\frac{Total \,\, load}{300} = \frac{(20 ... written 6 months ago by Manan Bothra0 1 answer 215 views 1 answers ...$ 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 6 months ago by Manan Bothra0
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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 6 months ago by Manan Bothra0
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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, $z^2+1=0 \hspace{0.20cm} \& \hspace{0.20cm} z^9 + 9 = 0 \\ \th ... written 6 months ago by Manan Bothra0 2 answers 192 views 2 answers ... Put$ 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)} $Le ... written 6 months ago by Manan Bothra0 2 answers 192 views 2 answers ... Put$ 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 ...