User: Manan Bothra

gravatar for Manan Bothra
Reputation:
0
Status:
Trusted
Location:
Last seen:
5 months, 3 weeks ago
Joined:
2 years, 6 months ago
Email:
b***********@gmail.com

Academic profile

None
None
None
None

Posts by Manan Bothra

<prev • 208 results • page 1 of 21 • next >
0
votes
1
answer
219
views
1
answers
Answer: A: Design battened system. Draw neat sketches to show details. Use 4.6 grade 20mm d
... 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
0
votes
1
answer
879
views
1
answers
Answer: A: Also design curtailment of flange plates, end bearing stiffener and connection b
... 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
0
votes
1
answer
214
views
1
answers
Answer: A: Also design suitable welded connection between web and flange plates
... 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
0
votes
1
answer
255
views
1
answers
Answer: A: Provide suitable curtailment of flange plates if necessary. Also design the weld
... 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
0
votes
1
answer
316
views
1
answers
Answer: A: Assuming the depth of plate girder restricted to 1500 mm and NO intermediate sti
... 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
0
votes
1
answer
215
views
1
answers
Answer: A: Evaluate $ \int_{- \infty}^{\infty} \frac{x^2+x+2}{x^4+10x^2+9} dx $ using
... $ 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
0
votes
1
answer
185
views
1
answers
Answer: A: Evaluate $ \int_{- \infty}^{\infty} \frac{dx}{(x^2+1)^2} $
... $ 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
0
votes
1
answer
163
views
1
answers
Answer: A: Evaluate $ \int_{0}^{\infty} \frac{1}{(x^2+1)(x^2+9)} dx $
... $ 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
0
votes
2
answers
192
views
2
answers
Answer: A: Evaluate $ \int_{0}^{2 \pi} \frac{d \theta}{3 + 2cos \theta} $
... 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
0
votes
2
answers
192
views
2
answers
Answer: A: Evaluate $ \int_{0}^{2 \pi} \frac{d \theta}{3 + 2cos \theta} $
... 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

Latest awards to Manan Bothra

Centurion 10 months ago, created 100 posts.