User: stanzaa37

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stanzaa370
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Posts by stanzaa37

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Page: Horizontal chain passes over a pully carried by a bracket the bracket is attach to stanction by 6 no of 20 mm bolts as shown in fig find load w which can be carried from the free end chain safely
... 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 ...
page ddss(76) written 12 hours ago by stanzaa370
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Page: bolted connection supporting eccentric load P as shown if the load on any bolt is not to exceed 29kN calculate maximum value of P
... ![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}{\ ...
page ddss(76) written 13 hours ago by stanzaa370
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Page: Determine safe load p that can be carried by than as shown bolts are 20mm $\phi$ 4.6 grade.Thickness of bracket plate is 10mm
... ![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 $\frac{e}{3do}=\frac ...
page ddss(76) written 13 hours ago by stanzaa370
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Page: Bracket connection
... Type I - when bolt group are subjected to shear moment in their shear plain the load that is ecentric with respective to centroid of the bold group can be replace with a force acting through the center of bolt group and moment with the magnitude M=P.e p=load in kN e= eccentric distance mm Step ...
page ddss(76) written 14 hours ago by stanzaa370
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Page: determine capacity of ISHB 400 82.2 kg/m with 1 cover plate of 400$\times$20 mm on each flange is used as column of effective height 6.5m design suitable slab base take m20 concrete and sbc of soil
... **180kN/m$^{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}}$=180Kpa **Step I** Requirement 1.design column 2 design base 3.design gusset base Pd ...
page ddss(76) written 1 day ago by stanzaa370
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Page: Gussted base
... When load on column is large or when column is subjected to moment along with axial load gusseted base is provided It consist of base plate gusset plate connecting angle provide on either side of column and web clear angle **Step I** Cal required area of base plate a) Ap=$\frac{Pu}{0.6\ast fc ...
page ddss(76) written 1 day ago by stanzaa370
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Page: Design slab base for built up column consisting of 2ISLC 350 back to back separated distance 220mm & carrying factored load of 1800 kN used m20 & fe 410 take bearing capacity of soil 350kN/m$^{2}$
... 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$\times10^{3}mm ...
page ddss(76) written 1 day ago by stanzaa370
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Page: Slab Base
... The basic function of this is to distribute the concentrated load from the column over a longer area initially the column load is distributed over a base plate & then to supporting concrete & finally to the sail below such that the bearing capacity of soil is not exceeded . The size of the ...
page ddss(76) written 1 day ago by stanzaa370
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Page: design built up column of eff length 5.9m It is sub to factored axial comp load of 2000 kN provide 2 channel back to back with battens by site welding used steel grade fe 410
... Given L=5.9m = 5900 mm Pu=2000$\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 ...
page ddss written 2 days ago by stanzaa370
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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 $ ...
page ddss(76) written 2 days ago by stanzaa370

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