## User: stanzaa37

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#### Posts by stanzaa37

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Page: Gusseted plate
... Pu=4105$\times10^{3}$N For m20- CFL =20mpa **Step I**. Ap=$\frac{up}{0.6\ast fck}$ =$\frac{4105}{0.6\times 20}$ Ap=342.08$\times10^{3} mm^{2}$ Assume tg=16mm $\neq$tf $\gt$12.7mm Safe Assume ISA. 150$\times$ 115$\times$15 mm ![enter image description here][1] 2) width of plate(sp) = ...
written 11 hours ago by stanzaa370
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... **b) bearign strength of concrete =0.6$\times$fck =0.6$\times$15=9mpa** M15-fck=15 Required-welded gusseted base 1) AP=$\frac{pu}{0.6\times15}=\frac{1700\times10^{3}}{0.6\times 15}$ Ap=183.89$\times 10^{3}mm^{2}$ 2) width of base plate(Bp)=350+2(16) Bp=382mm Assume gusset plate 16mm$\gt$tf 1 ...
written 16 hours ago by stanzaa370
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Page: Batterns Batterns
... 1. Design built up section $\lambda$ e=1.1 $\lambda$act . . .. .IS Pg 51 Claus 7.7.1.4 2. Spacing between channel Ixx=Iyy 3. Spacing between battens $\frac{c}{\gamma yy}\lt 0.7 \ast \lambda e$ & $\frac{c}{\gamma yy}\lt 50$ 4. size of batterns end batterns $\ \ \ \ \ \ \ \ \ \ \ \ ... written 16 hours ago by stanzaa370 0 answers 17 views 0 answers ... **fe 40 and bolt of grade 4.6 case I) Design the column with two channel placed back to back case II) Design the column with two channel place face to face case II) Design the lacing system with side welding connection from channel back to back** Pu=1080$\times 10^{3}$N l=10m=10000 mm leff=kl ... written 1 day ago by stanzaa370 0 answers 10 views 0 answers ... Given Pu = 1800kN = 1800$\times10^{3}$N l=8m=8000mm Left=KL K=0.8......(IS. Pg 45. Clause) Left =0.88$\times8000$=6400mm fe=410fy=250= mpa Required = Built up section **Step I**. Design of column section Pd=Ae$\times$Fcd 1800$\times10^{3}=Ae\times150$Ae=$\frac{1800\times10^{3}}{150 ...
written 2 days ago by stanzaa370
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Page: Built up section
... The limited no of rolled steel section are available in market.The radius of gyration in minor axis is in small giving large slenderness ratio with the result design stress get sufficiently reduce making the section uneconomical Further the rolled steel section do not have sufficient strength to ...
written 2 days ago by stanzaa370
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... **to back in opposite site of gusset plate 2)Angle place same side of gusset** Angle are tack bolted & provided wotj min of two bolts Given P=178KN l=3.05m leff=0 8$\times$3.08 =2464mm consider fed=150mpa Pd=Ae$\times$fcd 178$\times10^{3}=Ae\times150$ back to back connection Ae=1186.67 ...
written 3 days ago by stanzaa370
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... **2) 1 bolt at each end 3) welded at each end** Given ISA=100$\times$100$\times$6 l=3m=3000mm rvv=19.2mm off length =0.854$\ast$length of strut(0.821) Leff=KL=1$\times$3000=3000mm Area=1167mm$^{2}$ 1. Pd=Ae$\times$fcd $\ \ \ \ \$ Ae=1167mm$^{2}$ fcd=$\frac{fy/ymo}{\phi +[\phi^{2}-\lambd ... written 3 days ago by stanzaa370 0 answers 12 views 0 answers ... **2) 1 bolt at each end 3) welded at each end** Given ISA=100$\times$100$\times$6 l=3m=3000mm rvv=19.2mm off length =0.854$\ast$lenght of strut(0.821) Leff=KL=1$\times$3000=3000mm Area=1167mm$^{2}$1. Pd=Ae$\times$fcd$ \ \ \ \ \ $Ae=1167mm$^{2}$fcd=$\frac{fy/ymo}{\phi +[\phi^{2}-\lambd ...
written 3 days ago by stanzaa370
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