asked me that how i caculated the weight of 280' length in Bbs software.Answer: Dear Dia^2 * 2.67 / 2.204 e.g.3/4^2 = 0.75^2 = 0.5625 * 2.67= 1.5018/2.204= 0.68143kg/Rft.280'-0" * 0.681=190.80Kg

Kg/ft and kg/meeter k lye?

d^2/162.162kg/m (mm unit wala)

0.00618 D^2 . In MKS .Kg /m .And2.675 d^2 . . In Fpslbs/ft

Dear length meter mai ho ya feet mai. Ye matter nhi krta. Steel dia mm mai hai ya inch mai, ye matter krta hai.

How to calculate stirrups for beams?

Dear stirrups dia and C/C hamesha mention hota hai e.g. #3@6"C/C. Is ka matlab #3 mean 3 soot ta 10mm ka steel 6" center to center dale gai. Ferz kere agar ap beam ya column ki stirrups calculate ker rehe to clear span 10'-0" hai to 21 stirrups dale gi. 10'/0.5=20 jo b answer ai os mai 1 plus ker do.

If dia is given in inches then which formula will apply to calculate steel weight ?

Dia^2 * 2.67 / 2.204 e.g.3/8^2 = 0.375^2 = 0.140625 * 2.67= 0.3754688/2.204= 0.170kg/Rft.

And if #4 of steel bar which formula will apply ? Dia^2/24 right ?

Dear osi formula se b answer aa jai ga. Jo btaya hai.

Dear jo formula ap ne likha os se answer galat ata hai. Answer 0.303kg/Rft ana chahiye for #4

Ahm and c/l ka formula yeh ha.2(a+b)+24db CL: Cut lengthDia^2 /52.8 for Rft for expl (4)^2/52.8=0.303 (i dnt know about this formula but ur answer is also right.)

Mirza sab (span /center 2 center plus 1 ) ye 1 plus kyun krty hain ?

0 pr b to aik bar dale gi na. Say span 10' ka hai aur spacing 6"c/c hai to ap 10' ko inch mai kr lo. To pehla bat 0" pr mean start pt 2nd bar 6" pr and 3rd bar 12" pr. Aise ginte jao. 21 bar ain gai. Mean 20+1

L-2C devide by centre to centre distance +1 = number of stirrups

ho sakhta ha stirrup ki length kase maloom kare gy beam ki width 18" ha concrete cover 1.5"

agar beam 18" x 24" hai to 15" x 21" ka ring hoga aur cut length (1.25*2)+(1.75*2)+(0.17*2) = 6.33/rft

0.17*2 yeh kya ha

Dear hook hoti stirrup ki 2" ki. Wese to is ki length ka general note mai diya hota. Normally 6d hota. Agar na hoto ap 2" rekh skte.

Beam 10*1.5*2Ratio 1:2:4

Wet Volume10*1.5*2 = 30 CFT

Dry Volume30*1.54 = 46.2 CFT

Sum of Ratio1+2+4 = 7

Quantity of Cement17*46.2 = 6.6 CFT6.61.25 = 5.28 Bags

Quantity of Sand27*46.2 = 13.2 CFT

Quantity of Crush47*46.2 = 26.4

Yehi ho ga na

0.17*4 ho ga phir kyun 4 hook hn gi na

Hum ko sir ny hook ka formula 4d bataya tha

Ji sehi hai

4 hook nhi 2 hoti hai aur 4d b ho skta ye designer ki merzi. Normally 6d hota hai.

Beam column main 1 main bar hoti hai dosri anchor bar main aur anchor ka normally dia kya hota ha

Mirza sab one cubic feet concrete only not steel Ok is me sand cement aggregate ki ratio kea ho gi

ye to element of structure k hisb se hogi na. Agr lean concrete to 1:4:8 agar foundation to ye 3000 psi ya 1:2:4 ho skti aur isi terha columns 1:1.5:3 ho skti.

column mai sb hi main bars hoti aur bar dia depend krta load pr. Bar dia kiya hoga ye apni job nahi qs ki job k jo b bar hai jis b shape mai hai os ki length nikalna.

Isi post prHamza Hassanne cement sand crush nikala wo dekh lo ap.

Ji wo 1:2:4 ratio ki working hai.

Quantity surveyor ka kya kya kaam hota hai ak project may ? Preparation of IPCsBar bending schedulesSub contractors billEscalationCliamsRate analysisTenderingPreparation of BOQETC..

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Now i know... for finding... steel weight... there is two method...Dia square / 24 /2.204 for kg/foot...example.... 4 no steel...4x4/24/2.204, =0.303 kg/ftnow mm k ly...dia square / 162.162example... 12mm steel12x12/162.162 = 0.88 kg/mthank you admins nd who help me... thanks alot.

QUANTITIES REQUIRED:Plastering (CM 1:3) = 1.50 bags / 10 m 2Plastering (CM 1:5) = 1.05 bags / 10 m 2Ceiling Plastering (CM 1:3)= 48 kg / 10 m 2Brick work (CM 1:5) = 86 Kg / 10 m 3Brick work (CM 1:6) 9" thick = 80.64 Kg / 10 m 3Brick work (CM 1:3) 4" thick = 15.46 Kg / 10 m 3Lime for white washing = 10 Kg/100 m 2Painting = 10 ltr/ 100 m 2Distemper 1 st coat = 6.5 Kg / 100 m 2Distemper 2 nd coat = 5.0 kg / 100 m 2Snowcem 1 st coat = 30 Kg / 100 m 2Snowcem 2 nd coat = 20 Kg / 100 m 2Paint ready mixed one coat = 10 ltr / 100 m 2Weathering Course = 7.68 Kg / m 2Flooring = 8.10 kg / m 2Pressed tiles for weathering course (CM 1:3) = 7.68 Kg / 10 m 2Granolithic floor finish = 8.10 Kg / 10 m 2

Hooghoudt Drain-Spacing EquationHooghoudt's equation can be written as:.Q L2 = 8 Kb d (Di - Dd) (Dd - Dw) + 4 Ka (Dd - Dw)2where:Q = steady state drainage discharge rate (m/day)Ka = hydraulic conductivity of the soil above drain level (m/day)Kb = hydraulic conductivity of the soil below drain level (m/day)Di = depth of the impermeable layer below drain level (m)Dd = depth of the drains (m)Dw = steady state depth of the watertable midway between the drains (m)L = spacing between the drains (m)d = equivalent depth, a function of L, (Di-Dd), and rr = drain radius (m)Steady (equilibrium) state conditionIn steady state, the level of the water table remains constant and the discharge rate (Q) equals the rate of groundwater recharge (R), i.e. the amount of water entering the groundwater through the watertable per unit of time. By considering a long-term (e.g. seasonal) average depth of the water table (Dw) in combination with the long-term average recharge rate (R), the net storage of water in that period of time is negligibly small and the steady state condition is satisfied: one obtains a dynamic equilibrium.Derivation of the equationFor the derivation of the equation Hooghoudt used the law of Darcy, the summation of circular potential functions and, for the determination of the influence of the impermeable layer, de method of mirror images and superposition.Hooghoudt published tables for the determination of the equivalent depth (d), because the function (F) in d = F (L,Di-Dd,r) consists of long series of terms.Crop yield and seasonal average depth of the water tableDetermining:the discharge rate (Q) from the recharge rate (R) in a water balance as detailed in the article: hydrology (agriculture)the permissible long term average depth of the water table (Dw) on the basis of agricultural drainage criteriathe soil's hydraulic conductivity (Ka and Kb) by measurementsthe depth of the bottom of the aquifer (Di)the design drain spacing (L) can be found from the equation in dependence of the drain depth (Dd) and drain radius (r).Drainage criteriaOne would not want the water table to be too shallow to avoid crop yield depression nor too deep to avoid drought conditions. This is a subject of drainage research. The figure shows that a seasonal average depth of the water table shallower than 70 cm causes a yield depressionThe figure was made with the SegReg program for segmented regression.Equivalent depthIn 1991 a closed-form expression was developed for the equivalent depth (d) that can replace the Hooghoudt tables:[6]d = L / 8 { ln(L/r) + F(x) }where:x = 2 (Di Dd) / LF(x) = 4e2nx / n (1 e2nx), with n = 1, 3, 5, . . .Extended useTheoretically, Hooghoudt's equation can also be used for sloping land.[7] The theory on drainage of sloping land is corroborated by the results of sand tank experiments.[8] In addition, the entrance resistance encountered by the water upon entering the drains can be accounted for.Definitions of drainage of sloping land and entrance resistanceGeometry of drains in sloping land used in the theoryCorroboration with drainage experiments in a sloping sand tankDefinition of entrance resistanceAmplificationAmplification of parameters compared to HooghoudtThe drainage formula can be amplified [9] to account for (see figure on the right):the additional energy associated with the incoming percolation water (recharge), see groundwater energy balancemultiple soil layersanisotropric hydraulic conductivity, the vertical conductivity (Kv) being different from the horizontal (Kh)drains of different dimensions with any width (W)EnDrain: shape of the watertable with drainageComputer programThe amplified drainage equation uses an hydraulic equivalent of Joule's law in electricity.It is in the form of a differential equation that cannot be solved analytically (i.e. in a closed form) but the solution requires a numerical method for which a computer program is indispensable.The availability of a computer program also helps in quickly assessing various alternatives and performing a sensitivity analysis.The blue figure shows an example of results of a computer aided calculation with the amplified drainage equation using the EnDrain program.[10] It shows that incorporation of the incoming energy associated with the recharge leads to a somewhat deeper water table

Attached with the topography and SI report for the project.

Project brief:

We are appointed as a design consultant to undertake the design of a double storey detached house on the proposed site (attached topographical map). The detached will have overall 3000 sq ft and an enclosed car park area for 3 vehicles. The build up area will consist of 3 bedrooms (with bathroom), one game room, one maid bedroom, one main living area, one dry kitchen, one wet kitchen, one common bathroom and one common toilet.

For the next submission,-we will need to propose and review 2 overall conceptual plans for the house. Perform conceptual design using either masonry, timber, or steel construction (a combination thereof). Then recommend the choice and outlining reasons and justification.

- For the selected scheme, propose and carry out both geotechnical and structural detailed design, such as: load analysis, overall stability checks, structural design, connection design, earthwork, foundation (pad or pile), retaining structure, access road to the building, drainage structure, proposed utility alignment (sewer, electricity, water). engineering calculations, drawings should be included

We are also required to design a septic tank and drainage for discharging sewer & runoff from the property into the nearby drainage system.

- A construction program and perform risk analysis where appropriate, bills of quantity, cost estimation, specifications, etc.

-Perform an environmental impact assessment where necessary

- Undertake a feasibility study of two distinct and viable arrangements for the detached house which includes work programme, conceptual options with estimated cost.

-Overall and detailed design of prefer scheme

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