## User: stanzaa37

stanzaa37 •

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#### Posts by stanzaa37

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... the bolts are 20mm $\phi$ m and 4.6 grade
![enter image description here][1]
**Step I**
list bolt value=$\frac{fub\times Anb\times hn}{\sqrt{3}ymb}$
=$\frac{400\times245\times 1}{\sqrt{3}\times 1.25}$
Vdsb=45.26kN
**Step II**
dirct shera force per bolt=$\frac{w}{6}$
in horizontal direction
d ...

written 12 hours ago by
stanzaa37 •

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... ![enter image description here][1]
**Step I**
List bolt value
Shera capacity of bolt=$\frac{fub\times Anb\times hn}{\sqrt{3}\times ymb}$
=400
bolt value=29 kN
**Step II**
Direct shera force=$\frac{f}{4(no.of \ bolt)}$=0.25p
**Step III**
force in bolt due to tortion
F2=$\frac{\sum(P.e)rn}{\ ...

written 13 hours ago by
stanzaa37 •

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... ![enter image description here][1]
**Step I**
list bolt value
shear capacity of bolt=$\frac{fub\times Anb\times nn}{\sqrt{3}Ymb}$
=$\frac{400\times 245\times 1}{\sqrt{3} \times 1.25}$
Vdsb=45.26 kN
bearing capacity of bolt=2$\times Kb\times d\times t\times $fu
find kb
$\frac{e}{3do}=\frac ...

written 13 hours ago by
stanzaa37 •

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Page:
Bracket connection

... Type I - when bolt group are subjected to shear moment in their shear plain the load that is ecentric with respective to centroid of the bold group can be replace with a force acting through the center of bolt group and moment with the magnitude
M=P.e
p=load in kN
e= eccentric distance mm
Step ...

written 14 hours ago by
stanzaa37 •

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... **180kN/m$^{2}$ Design gusseted base of above column**
Given ISHB 400 at 82.2 kg/m
cover plate on each flange =400$\times$20
effective length of column=6500mm
SBC of soil=150$\frac{kN}{m^{2}}$=180Kpa
**Step I**
Requirement
1.design column
2 design base
3.design gusset base
Pd ...

written 1 day ago by
stanzaa37 •

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Gussted base

... When load on column is large or when column is subjected to moment along with axial load gusseted base is provided
It consist of base plate gusset plate connecting angle provide on either side of column and web clear angle
**Step I** Cal required area of base plate
a) Ap=$\frac{Pu}{0.6\ast fc ...

written 1 day ago by
stanzaa37 •

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... Given Pu=1800 kN
=1800$\times 10^{3}$N
bearing strenght of concrete =0.6$\times$20
=18mpa
SBC=350 KN/m$^{2}$
Required-Design slab base
**Step 1** Required area of base plate
a) Ap=$\frac{Pu}{0.6\times 20}=\frac{1800\times 10^{3}}{0.6\times 20}=150\times10^{3}mm^{2}$
AP=150$\times10^{3}mm ...

written 1 day ago by
stanzaa37 •

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Slab Base

... The basic function of this is to distribute the concentrated load from the column over a longer area initially the column load is distributed over a base plate & then to supporting concrete & finally to the sail below such that the bearing capacity of soil is not exceeded .
The size of the ...

written 1 day ago by
stanzaa37 •

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... Given
L=5.9m = 5900 mm
Pu=2000$\times 10^{3}$N
required two channel back to back with battern & site welding $\gamma$ mw=1.5
1. Pu=Aea$\times $ fcd
2000$\times 10^{3}$=Area$\times$ 150
Area=13333.33mm$^{2}$
Area of each section=$\frac{13333.33}{2}=6666.67mm^{2}$
provide fcd=190mpa
Pu=A ...

written 2 days ago by
stanzaa37 •

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Page:
welded connection

... Step
1. Design built up column
2. Spacing
3. spacing of battens
4. Size of batten
end batten $ \ \\ \ \ \\ \ \\ \ \ $ Intermedient batten
1. depth $ \ \ \ \ \ \ \ \ \ \ \\ \ $ depth d$\frac{3}{3}d^{'}$
2. Length $ \ \ \ \ \ \ \ \ \ $ length
3. thickness $ \ \ \ \ $ thickness
$ ...

written 2 days ago by
stanzaa37 •

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